LIBRARY OF
WELLESLEY COLLEGE
PRESENTED BY
TVof Horaford
HIGHER ALGEBRA
A SEQUEL TO
ELEMENTARY ALGEBRA EOR SCHOOLS.
s.
HIGHER ALGEBRA
A SEQUEL TO
ELEMENTARY ALGEBRA FOR SCHOOLS
BY
H. S. HALL, M.A.,
FORMERLY SCHOLAR OF CHRIST'S COLLEGE, CAMBRIDGE,
MASTER OF THE MILITARY AND ENGINEERING SIDE, CLIFTON COLLEGE ;
AND
S. K KNIGHT, B.A.,
FORMERLY SCHOLAR OF TRINITY COLLEGE, CAMBRIDGE,
LATE ASSISTANTMASTER AT MARLBOROUGH COLLEGE.
FOURTH EDITION.
Honfcon:
MACMILLAN AND CO.
AND NEW YOKE.
1891
[The Right of Translation is reserved.}
2 TO ■
4 2 2 12/
First Printed 1887.
Second Edition with corrections 1888.
Third Edition revised and enlarged 1889.
Reprinted 1890. Fourth Edition 1891.
}
PREFACE.
The present work is intended as a sequel to our Elementary
Algebra for Schools. The first few chapters are devoted to
a fuller discussion of Ratio, Proportion, Variation, and the
Progressions, which in the former work were treated in an
elementary manner ; and we have here introduced theorems
and examples which are unsuitable for a first course of
reading.
From this point the work covers ground for the most
part new to the student, and enters upon subjects of special
importance : these we have endeavoured to treat minutely
and thoroughly, discussing both bookwork and examples
witli that fulness which we have always found necessary in
our experience as teachers.
It has been our aim to discuss all the essential parts
as completely as possible within the limits of a single
volume, but in a few of the later chapters it has been im
possible to find room for more than an introductory sketch ;
in all such cases our object has been to map out a suitable
first course of reading, referring the student to special treatises
for fuller information.
In the chapter on Permutations and Combinations we
are much indebted to the Rev. W. A. Whitworth for per
mission to make use of some of the proofs given in his
Choice and Chance. For many years we have used these
proofs in our own teaching, and we are convinced that this
vi PREFACE.
part of Algebra is made far more intelligible to the beginner
by a system of common sense reasoning from first principles
than by the proofs usually found in algebraical textbooks.
The discussion of Convergency and Divergency of Series
always presents great difficulty to the student on his first
reading. The inherent difficulties of the subject are no
doubt considerable, and these are increased by the place it
has ordinarily occupied, and by the somewhat inadequate
treatment it has hitherto received. Accordingly we have
placed this section somewhat later than is usual; much
thought has been bestowed on its general arrangement, and
on the selection of suitable examples to illustrate the text ;
and we have endeavoured to make it more interesting and
intelligible by previously introducing a short chapter on
Limiting Values and Vanishing Fractions.
In the chapter on Summation of Series we have laid
much stress on the " Method of Differences" and its wide and
important applications. The basis of this method is a well
known formula in the Calculus of Finite Differences, which in
the absence of a purely algebraical proof can hardly be con
sidered admissible in a treatise on Algebra. The proof of the
Finite Difference formula which we have given in Arts. 395,
396, we believe to be new and original, and the development
of the Difference Method from this formula has enabled us to
introduce many interesting types of series which have hitherto
been relegated to a much later stage in the student's reading.
We have received able and material assistance in the
chapter on Probability from the Rev. T. C. Simmons of
Christ's College, Brecon, and our warmest thanks are due
to him, both for his aid in criticising and improving the
text, and for placing at our disposal several interesting and
original problems.
It is hardly possible to read any modern treatise on
Analytical Conies or Solid Geometry without some know
PKEFACE. yii
ledge of Determinants and their applications. We have
therefore given a brief elementary discussion of Determi
nants in Chapter xxxm., in the hope that it may provide
the student with a useful introductory course, and prepare
him for a more complete study of the subject.
The last chapter contains all the most useful propositions
in the Theory of Equations suitable for a first reading. The
Theory of Equations follows so naturally on the study of
Algebra that no apology is needed for here introducing pro
positions which usually find place in a separate treatise. In
fact, a considerable part of Chapter xxxv. may be read
with advantage at a much earlier stage, and may conveniently
be studied before some of the harder sections of previous
chapters.
It will be found that each chapter is as nearly as possible
complete in itself, so that the order of their succession can
be varied at the discretion of the teacher ; but it is recom
* mended that all sections marked with an asterisk should be
reserved for a second reading.
In enumerating the sources from which we have derived
assistance in the preparation of this work, there is one book
to which it is difficult to say how far we are indebted.
Todhunter's Algebra for Schools and Colleges has been the
recognised English textbook for so long that it is hardly
possible that any one writing a textbook on Algebra at the
present day should not be largely influenced by it. At the
same time, though for many years Todhunter's Algebra has
been in constant use among our pupils, we have rarely
adopted the order and arrangement there laid down; in
many chapters we have found it expedient to make frequent
use of alternative proofs; and we have always largely sup
plemented the text by manuscript notes. These notes,
which now appear scattered throughout the present work,
have been collected at different times during the last twenty
H. H. A. b
Viii PREFACE.
years, so that it is impossible to make definite acknowledge
ment in every case where assistance has been obtained from
other writers. But speaking generally, our acknowledge
ments are chiefly due to the treatises of Schlomilch, Serret,
and Laurent; and among English writers, besides Todhunter's
Algebra, we have occasionally consulted the works of De
Morgan, Colenso, Gross, and Chrystal.
To the Rev. J. Wolsienholme, D.Sc, Professor of Mathe
matics at the Royal Indian Engineering College, our thanks
are due for his kindness in allowing us to select questions
from his unique collection of problems ; and the consequent
gain to our later chapters we gratefully acknowledge.
It remains for us to express our thanks to our colleagues
and friends who have so largely assisted us in reading and
correcting the proof sheets ; in particular we are indebted to
the Rev. H. C Watson of Clifton College for his kindness in
revising the whole work, and for many valuable suggestions;
in every part of it.
**' 1887 " H. S. HALL, j
S. R. KNIGHT. <
PREFACE TO THE THIRD EDITION.
In this edition the text and examples are substantially
the same as in previous editions, but a few articles havej
been recast, and all the examples have been verified again.
We have also added a collection of three hundred Miscel
laneous Examples which will be found useful for advanced
students. These examples have been selected mainly but
not exclusively from Scholarship or Senate House papers ;
much care has been taken to illustrate every part of the
subject, and to fairly represent the principal University and
Civil Service Examinations.
March, 1889.
CONTENTS.
CHAPTER I. ratio.
Commensurable and incommensurable quantities .
Ratio of greater and less inequality
i
a _c _e _ /pa n + qc n +re n + ...\ n
b~d~f~"'~\pb n + qd n + rf n +...J '
a 1 + a 2 + a 3 +... + a n
b l + b 2 + b. i + ... + b n
Cross multiplication
Eliminant of three linear equations
Examples I
lies between greatest and least of fractions
flh
V
PAGE
2
3
a
n
8
9
10
CHAPTER II. proportion.
Definitions and Propositions 13
Comparison between algebraical and geometrical definitions 1G
Case of incommensurable quantities 17
Examples II. 19
CHAPTER III. VARIATION.
;
If Ace B, then A = mB 21
Inverse variation 22
Joint variation 23
li Ace B when G is constant, and A & C when B is constant, then
A=mBG 23
Illustrations. Examples on joint variation . . . . . .21
Examples III 20
b1
CONTENTS.
CHAPTER IV. ARITHMETICAL PROGRESSION.
Sum of n terms of an arithmetical series
Fundamental formulae
Insertion of arithmetic means
Examples IV. a
Discussion of roots of dn~ + (2a d)n2s =
Examples IV. b
PAGE
28
29
31
31
33
35
f' CHAPTER V. GEOMETRICAL PROGRESSION.
Insertion of geometric means ....
Sum of n terms of a geometrical series .
Sum of an infinite geometrical series
Examples V. a.
Proof of rule for the reduction of a recurring decimal
Sum of n terms of an arithmeticogeometric series
Examples V. b
38
39
40
41
43
44
45
CHAPTER VI. HARMONICAL PROGRESSION. THEOREMS CONNECTED
WITH THE PROGRESSIONS.
Reciprocals of quantities in H. P. are in A. P.
Harmonic mean ......
Formulae connecting A. M., G.M., H.M.
Hints for solution of questions in Progressions
Sum of squares of the natural numbers
Sum of cubes of the natural numbers .
2 notation
Examples VI. a. .....
Number of shot in pyramid on a square base
Pyramid on a triangular base
Pyramid on a rectangular base
Incomplete pyramid .....
Examples VI. b
5J
CHAPTER VII. scales of notation.
Explanation of systems of notation
Examples VII. a.
Expression of an integral number in a proposed scale .
Expression of a radix fraction in a proposed scale .
57
59
59
01
CONTENTS.
XI
The difference between a number and tho sum of its digits is divisible
by r  1
Proof of rule for " casting out the nines "
Test of divisibility by r + 1
Examples VII. b
CHAPTER VIII.
nationalising the denominator of
sjb + jc + s/d
Rationalising factor of fJa±Z/b
Square root of a + Jb + *Jc + Jd
Cube root of a + *Jb
Examples VIII. a.
Imaginary quantities
J ax J b=  sjab .
If a + ib = 0, then a = Q, b = .
If a + ib = c + id, then a = c, b = d
Modulus of product is equal to product of moduli
Square root of a + ib
Powers of i .
Cube roots of unity ; 1 + w f or =
Powers of u .
Examples VIII. b.
SURDS AND IMAGINARY QUANTITIES.
a
PAGE
62
C3
64
65
07
68
69
70
72
74
75
75
75
77
77
79
79
80
81
CHAPTER IX. THE THEORY OF QUADRATIC EQUATIONS.
A quadratic equation cannot have more than two roots ... 83
Conditions for real, equal, imaginary roots 84
b c
Sum of roots = — , product of roots =  85
a a
Formation of equations when the roots are given ..... 86
Conditions that the roots of a quadratic should be (1) equal in magni
tude and opposite in sign, (2) reciprocals 88
Examples IX. a 88
For real values of x the expression ax 2 + bx + c has in general the same
sign as a ; exceptions 90
Examples IX. b 92
Definitions of function, variable, rativnnl integral function ... 93
Condition that ax 2 + 2hxy+ by 2 + 2gx + 2fy + c may be resolved into two
linear factors 9 i
Condition that ax 2 + bx + c = and a'x + b'x + c' = may have a common
root 96
Examples IX. c. 96
Xll
CONTENTS.
CHAPTER X. MISCELLANEOUS EQUATIONS.
Equations involving one unknown quantity .
Reciprocal equations
Examples X. a
Equations involving two unknown quantities
Homogeneous equations ....
Examples X. b
Equations involving several unknown quantities
Examples X. c.
Indeterminate equations ; easy numerical examples
Examples X. d
page
97
100
101
103
104
106
107
109
111
113
CHAPTER XL permutations and combinations.
Preliminary proposition • • .115
Number of permutations of n things r at a time 115
Number of combinations of n things r at a time 117
The number of combinations of n things r at a time is equal to the
number of combinations of n things ?irata time . . .119
Number of ways in which m + n +p + ... things can be divided into
classes containing m, n, p, ... things severally .... 120
Examples XI. a 122
Signification of the terms 'like' and 'unlike' ..... 124
Number of arrangements of n things taken all at a time, when p things
are alike of one kind, q things are alike of a second kind, &c. . 125
Number of permutations of n things r at a time, when each may be
repeated 126
The total number of combinations of n things 127
To find for what value of r the expression n G r is greatest . . . 127
Ab initio proof of the formula for the number of combinations of n
things r at a time 128
Total number of selections of p + q+r+ ... things, whereof p are alike
of one kind, q alike of a second kind, &c 129
Examples XI. b 131
CHAPTER XII. mathematical induction.
Illustrations of the method of proof 133
Product of n binomial factors of the form x + a 134
Examples XII 135
CONTENTS. Xlii
CHAPTER XIII. BINOMIAL THEOREM. POSITIVE INTEGRAL INDEX.
PAGE
Expansion of (x + a) 11 , when n is a positive integer .... 137
General term of the expansion 139
The expansion may be made to depend upon the case in which the first
term is unity 140
Second proof of the binomial theorem 141
Examples XLII. a 142
The coefficients of terms equidistant from the beginning and end
are equal 143
Determination of the greatest term 143
Sum of the coefficients 146
Sum of coefficients of odd terms is equal to sum of coefficients of even
terms 146
Expansion of multinomials 146
Examples XIII. b. 147
CHAPTER XIV. BINOMIAL THEOREM. ANY INDEX.
Euler's proof of the binomial theorem for any index
General term of the expansion of (1 + x)' 1 .....
Examples XIV. a
Expansion of (lrx) n is only arithmetically intelligible when x<l
The expression (.rf?/)' 1 can always be expanded by the binomial
theorem
General term of the expansion of (1  .r) _n ....
Particular cases of the expansions of (1  x)~ n
Approximations obtained by the binomial theorem
Examples XIV. b.
Numerically greatest term in the expansion of (l + x) n .
Number of homogeneous products of r dimensions formed out
letters
Number of terms in the expansion of a multinomial
Number of combinations of n things r at a time, repetitions being allowed 166
Examples XIV. c 107
of n
150
153
155
155
157
157
158
159
161
162
164
105
CHAPTER XV. MULTINOMIAL THEOREM.
General term in the expansion of (a + bx + ex 2 + dx 3 + ...) p , when ^ is a
positive integer 170
General term in the expansion of (a + bx + cx + <lv :i + ...) n , when //
is a rational quantity 171
Examples XV 173
XIV
CONTENTS.
CHAPTER XVI. LOGARITHMS.
PAGE
Definition. N=a)og a N 175
Elementary propositions 176
Examples XVI. a 178
Common Logarithms • • .179
Determination of the characteristic by inspection ..... 180
Advantages of logarithms to base 10 181
Advantages of always keeping the mantissa positive .... 182
Given the logarithms of all numbers to base a, to find the logarithms
to base b . . . 183
log a &xlog 6 a = l 183
Examples XVI. b 185
CHAPTER XVII. EXPONENTIAL AND LOGARITHMIC SERIES.
Expansion of a x . Series for e
( l\ n
e is the limit of ( 1 +  ) , when n is infinite
V V
Expansion of log,, (1 + x)
Construction of Tables of Logarithms .
Rapidly converging series for log,, (n + 1)  log e n
The quantity e is incommensurable
Examples XVII
187
188
191
192
194
195
195
CHAPTER XVIII. INTEREST AND ANNUITIES.
Interest and Amount of a given sum at simple interest . . . .198
Present Value and Discount of a given sum at simple interest . . 198
Interest and Amount of a given sum at compound interest . . . 199
Nominal and true annual rates of interest 200
Case of compound interest payable every moment .... 200
Present Value and Discount of a given sum at compound interest . . 201
Examples XVIII. a 202
Annuities. Definitions 202
Amount of unpaid annuity, simple interest 203
Amount of unpaid annuity, compound interest 203
Present value of an annuity, compound interest 204
Number of years' purchase 204
Present value of a deferred annuity, compound interest . . . .205
Fine for the renewal of a lease 206
Examples XVIII. b 206
CONTENTS. XV
CHAPTER XIX. INEQUALITIES.
PAGE
Elementary Propositions 208
Arithmetic mean of two positive quantities is greater than the geometric
mean 209
The sum of two quantities being given, their product is greatest when
they are equal : product being given, the sum is least when they are
equal 210
The arithmetic mean of a number of positive quantities is greater than
the geometric mean 211
Given sum of a, &, c, ...; to find the greatest value of a m b n c p 212
Easy cases of maxima and minima 212
Examples XIX. a 213
The arithmetic mean of the ?/i th powers of a number of positive
quantities is greater than m th power of their arithmetic mean,
except when m lies between and 1
If a and b are positive integers, and a>b, ( 1 +  ) > ( 1 + ^ )
? 1> * > » >0 'Vrr > vrrf
Examples XIX. b
b
y
'a + b\ a+b
214
216
217
217
218
CHAPTER XX. LIMITING VALUES AND VANISHING FRACTIONS.
Definition of Limit 220
Limit of a + a x x + a 2 x" + a 3 x 3 + ... is a when x is zero .... 222
By taking x small enough, any term of the series a + a r r + a^x + ...
may be made as large as we please compared with the sum of all
that follow it; and by taking x large enough, any term may be
made as large as we please compared with the sum of all that
precede it . 222
Method of determining the limits of vanishing fractions . . . 221
Discussion of some peculiarities in the solution of simultaneous
equations 226
Peculiarities in the solution of quadratic equations .... 227
Examples XX 228
CHAPTER XXI. CONVERGENCE AND DIVERGENCY OF SERIES.
Case of terms alternately positive and negative ..... 230
u
Series is convergent if Lim ~ n is less than 1 232
«u i
XVI
CONTENTS.
Comparison of 2rt n with an auxiliary series 2v n .
The auxiliary series ^p + 2P + 3~p +
Application to Binomial, Exponential, Logarithmic Series
Limits of
log 71
n
and nx n when n is infinite
Product of an infinite number of factors
Examples XXI. a. .
u v
wseries is convergent when v series is convergent, if
u
Series is convergent if Lim ]n I — —  1 )
< \u n +i J
Series is convergent if Lim ( n log — — ) > ]
Series 20 (n) compared with series 2a ,l {n)
The auxiliary series 2
jii
v
nl
n (log n) p
Series is convergent if Lim \n ( — ~  1 J  l log n
Product of two infinite series
Examples XXI. b. .
PAGE
. 234
. 235
. 237
. 238
. 238
. 241
. 243
. 244
. 245
. 247
. 248
248
249
252
CHAPTER XXII. UNDETERMINED COEFFICIENTS.
If the equation f(x)=0 has more than n roots, it is an identity . . 254
Proof of principle of undetermined coefficients for finite series . . 254
Examples XXII. a 256
Proof of principle of undetermined coefficients for infinite series . . 257
Examples XX1T. b 2C0
CHAPTER XXIII. PARTIAL FRACTIONS.
Decomposition into partial fractions
Use of partial fractions in expansions
Examples XXIII
261
265
265
CHAPTER XXIV. recurring series.
Scale of relation 267
Sum of a recurring series 269
Generating function 269
Examples XXIV 272
CONTENTS.
XV ii
CHAPTER XXV. continued fractions.
PAGE
Conversion of a fraction into a continued fraction .... 273
Convergents are alternately less and greater than the continued fraction 275
Law of formation of the successive convergents 275
Pn&wlPnl4n=(~ 1 ) n 27G
Examples XXV. a. 277
The convergents gradually approximate to the continued fraction . . 278
Lhnits of the error in taking any convergent for the continued fraction 279
Each convergent is nearer to the continued fraction than a fraction
with smaller denominator . 280
Pp' P p'
— ,:> or <x~, according as> or < — . 281
qq q q
Examples XXV. b 281
CHAPTER XXVI. indeterminate equations of the first
DEGREE.
Solution of axbi/ = c
Given one solution, to find the general solution
Solution of ax + by = c
Given one solution, to find the general solution
Number of solutions of ax + by = c .
Solution of ax + by + cz = d, a'x + b'y + c'z = d' .
Examples XXVI. ... .
284
286
286
287
287
289
290
CHAPTER XXVII. recurring continued fractions.
Numerical example 292
A periodic continued fraction is equal to a quadratic surd . . . 293
Examples XXVII. a 21)4
Conversion of a quadratic surd into a continued fraction . . . 295
The quotients recur 296
The period ends with a partial quotient 2a x 297
The partial quotients equidistant from first and last are equal . . 298
The penultimate convergents of the periods 299
Examples XXVII. b. . . 301
CHAPTER XXVIII. indeterminate equations of the second
DEGREE.
Solution of ax 2 + 2hxy + by* + 2gx + 2fy + c =
The equation # 2  Ny 2 =l can always be solved
303
304
xviii CONTENTS.
PAGE
Solution of x 2  Ny 2 = 1
. 305
General solution of x 2  Ny 2 = 1
. 306
Solution of x 2  n 2 y 2 = a
. 308
Diophantine Problems
. 309
Examples XXVIII
. 311
CHAPTER XXIX. summation of series.
Summary of previous methods • • .312
u n the product of n factors in A. P 314
u n the reciprocal of the product of n factors in A. P 316
Method of Subtraction 318
Expression of u n as sum of factorials 318
Polygonal and Figurate Numbers 319
Pascal's Triangle 320
Examples XXIX. a 321
Method of Differences 322
Method succeeds when u n is a rational integral function of n . . 326
If a n is a rational integral function of n, the series 2a n x' 1 is a recurring
series 327
Further cases of recurring series 329
Examples XXIX. b 332
Miscellaneous methods of summation ....... 331
Sumof series l r + 2 r + S r +...+n r 336
Bernoulli's Numbers 337
Examples XXIX. c 338
CHAPTER XXX. theory of numbers.
Statement of principles
Number of primes is infinite
No rational algebraical formula can represent primes only .
A number can be resolved into prime factors in only one way
Number of divisors of a given integer
Number of ways an integer can be resolved into two factors .
Sum of the divisors of a given integer
Highest power of a prime contained in In .
Product of r consecutive integers is divisible by [r
Fermat's Theorem NPi l=M(p) where p is prime and N prime
Examples XXX. a
Definition of congruent ....
to 2?
341
342
342
342
343
343
344
345
345
347
348
350
CONTENTS.
XIX
If a is prime to b, then a, 2a, 3a, ... (6 1) a when divided by 6 leave
different remainders ......
(p(abcd...)=<p(a)(p(b)<p(c) <p(d)
♦PO»'(ii)(iJ)(il)
Wilson's Theorem : 1 + \p  1 = M (p) where p is a prime
A property peculiar to prime numbers ....
Wilson's Theorem (second proof)
Proofs by induction .......
Examples XXX. b.
PAGE
350
352
352
354
354
355
35G
357
CHAPTER XXXI. the general theory of continued
FRACTIONS.
Law of formation of successive convergents .
— —  —  ... has a definite value if Lim ■■■ '" n4 ' 1 >0 .
a,+ a. 2 +
The convergents to
\ h
y n+l
a l ~ a 2~
. . . are positive proper fractions in ascend
ing order of magnitude, if a n <kl + b n
General value of convergent when a n and b n are constant
Cases where general value of convergent can be found .
is incommensurable, if — <1
cl
a x + a 2 +
Examples XXXI. a
Series expressed as continued fractions .
'Conversion of one continued fraction into another
Examples XXXI. b
CHAPTER XXXII. probability.
Definitions and illustrations. Simple Events .
(Examples XXXII. a
/Compound Events
I Probability that two independent events will both happen is pp' .
[ The formula holds also for dependent events .
Chance of an event which can haj^pen in mutually exclusive ways
Examples XXXII. b
Chance of an event happening exactly r times in n trials
Expectation and probable value .......
"Problem of points" . .......
359
362
363
364
365
366
367
369
371
372
373
376
377
378
379
381
383
385
386
388
XX
CONTENTS.
Examples XXXII. c.
Inverse probability
Statement of Bernoulli's Theorem .
P P
Proof of formula Q r = ^rjjn
Concurrent testimony ....
Traditionary testimony ....
Examples XXXII. d
Local Probability. Geometrical methods
Miscellaneous examples
Examples XXXII. e
PAOE
389
391
392
396
899
401
402
405
CHAPTER XXXIII. dktkrminants.
Eliminant of two homogeneous linear equations ..... 409
Eliminant of three homogeneous linear equations . . . .410
Determinant is not altered by interchanging rows and columns . . 410
Development of determinant of third order 411
Sign of a determinant is altered by interchanging fcw< adjacent rows or
columns . . . . . . 412
If two rows or columns are identical, the determinant vanishes . . 112
A factor common to any row or column may be placed outside . . 412
Cases where constituents are made up of a number of terms . . . 413
Keduction of determinants by simplification of rows or columns . .111
Product of two determinants 417
Examples XXXIII. a 419
Application to solution of simultaneous equations 422
Determinant of fourth order ... 423
Determinant of any order . . . 42jl
Notation Sia^^ ... . ... 425
Examples XXXIII. b. . . . \r,
CHAPTER XXXIV. miscellaneous theorems and examples.
Keview of the fundamental laws of Algebra
f(x) when divided by x  a leaves remainder
Quotient of / (x) when divided by x  a
Method of Detached Coefficients .
Horner's Method of Synthetic Division .
Symmetrical and Alternating Functions
Examples of identities worked out
List of useful formula? .
/»
429
432
433
434
434
435
437
438
CONTENTS.
XXI
Examples XXXIV. a
Identities proved by properties of cube roots of unity
Linear factors of a 3 + 6 3 + c 3  Sabc
Value of a n + b n + c n when a + b + c = Q
Examples XXXIV. b. .
Elimination
Elimination by symmetrical functions
Euler's method of elimination
Sylvester's Dialytic Method .
Bezout's method ....
Miscellaneous examples of elimination
Examples XXXIV. c. . . .
PAGE
438
440
441
442
442
444
444
445
446
446
447
449
CHAPTER XXXV. theory of equations.
Every equation of the n th degree has n roots and no more . . . 452
Kelations between the roots and the coefficients 452
These relations are not sufficient for the solution 454
Cases of solution under given conditions 454
Easy cases of symmetrical functions of the roots 455
Examples XXXV. a. 456
Imaginary and surd roots occur in pairs 457
Formation and solution of equations with surd roots .... 458
Descartes' Kule of Signs 459
Examples XXXV. b 460
Value of /(.r + //). Derived Functions 462
Calculation of f(x+h) by Horner's process 463
/ ( x) changes its value gradually 464
If f(a) and/ (b) are of contrary signs, f(x) = has a root between
a and 6 464
An equation of an odd degree has one real root 465
An equation of an even degree with its last term negative has two real
roots 465
If / (x) = has r roots equal to a, f (x) = has r  1 roots equal to a . 466
Determination of equal roots 467
/'(*)_ I 1 , 1 , 468
J (X) xa xb xc
Sum of an assigned power of the roots . 468
Examples XXXV. c 470
Transformation of equations 471
Equation with roots of sign opposite to those of f(x) = . . . 471
Equation with roots multiples of those of f{x) =0 . • • • 472
XX11
CONTENTS.
Equation with roots reciprocals of those of / (x) =
Discussion of reciprocal equations ....
Equation with roots squares of those of f(x) = .
Equation with roots exceeding by h those of f (x) =
Bemoval of an assigned term
Equation with roots given functions of those of f{x).
Examples XXXV. d
Cubic equations. Cardan's Solution
Discussion of the solution
Solution by Trigonometry in the irreducible case .
Biquadratic Equations. Ferrari's Solution .
Descartes' Solution
Undetermined multipliers
Discriminating cubic ; roots all real
x y
Solution of three simultaneous equations
+
+
a+\ b + \ c + \
PAGE
. 472
. 473
. 475
. 475
. 476
. 477
. 478
. 480
. 481
. 482
. 483
. 484
. 486
. 486
=1, &c. . 487
Examples XXXV. e.
Miscellaneous Examples
Answers
488
490
525
HIGHER ALGEBRA.
CHAPTER I.
RATIO.
1. Definition. Ratio is the relation which one quantity
bears to another of the same kind, the comparison being made by
considering what multiple, part, or parts, one quantity is of the
other. •
The ratio of A to B is usually written A : B. The quantities
A and B are called the terms of the ratio. The first term is
called the antecedent, the second term the consequent.
2. To find what multiple or part A is of B, we divide A
by B ; hence the ratio A : B may be measured by the fraction
^ , and we shall usually find it convenient to adopt this notation.
In order to compare two quantities they must be expressed in
terms of the same unit. Thus the ratio of £2 to 15s. is measured
.... ■ 2x20 8
by the traction — ^ — or  .
Note. A ratio expresses the number of times that one quantity con
tains another, and therefore every ratio is an abstract quantity.
3. Since by the laws of fractions,
a ma
b = mJ'
it follows that the ratio a : b is equal to the ratio ma : mb ;
that is, the value of a ratio remains unaltered if the antecedent
and the consequent are multiplied or divided by the same quantity.
H. H. A. 1
2 HIGHER ALGEBRA.
4. Two or more ratios may be compared by reducing their
equivalent fractions to a common denominator. Thus suppose
_ _ x xt a a V i x bx ,
a : o and x : y are two ratios. JNow  = ~ and — = =— : hence
J h by y by 3
the ratio a : b is greater than, equal to, or less than the ratio
x : y according as ay is greater than, equal to, or less than bx.
5. The ratio of two fractions can be expressed as a ratio
Ch C
of two integers. Thus the ratio — : — is measured by the
° b a
a
fraction — , or =— : and is therefore equivalent to the ratio
c be
d
ad : be.
6. If either, or both, of the terms of a ratio be a surd
quantity, then no two integers can be found which will exactly
measure their ratio. Thus the ratio J'2 : 1 cannot be exactly
expressed by any two integers.
7. Definition. If the ratio of any two quantities can be
expressed exactly by the ratio of two integers, the quantities
are said to be commensurable ; otherwise, they are said to be
incommensurable.
Although we cannot find two integers which will exactly
measure the ratio of two incommensurable quantities, we can
always find two integers whose ratio differs from that required
by as small a quantity as we please.
J5 2236068... ™ A1I ,
Thus V = . = 559017...
4 4
, , . J5 559017 , 559018
and therefore — > mm(> and < jooOOOO ;
so that the difference between the ratios 559017 : 1000000 and
J 5 : 4 is less than 000001. By carrying the decimals further, a
closer approximation may be arrived at.
8. Definition. Ratios are compounded by multiplying to
gether the fractions which denote them ; or by multiplying to
gether the antecedents for a new antecedent, and the consequents
for a new consequent.
Example. Find the ratio compounded of the three ratios
2a : Sb, Q>ab : 5c 2 , c : a
KATIO.
m . . , ,. 2a Gab c
The required ratio = x „ x 
6b be 1 a
_4a
~ DC '
9. Definition. When the ratio a : b is compounded with
itself the resulting ratio is a 2 : b 2 , and is called the duplicate ratio
of a : b. Similarly a 3 : b 3 is called the triplicate ratio of a : b.
Also a 2 : b 2 " is called the subduplicate ratio of a : b.
Examples. (1) The duplicate ratio of 2a : 3b is 4a 2 : 96.
(2) The subduplicate ratio of 49 : 25 is 7 : 5.
(3) The triplicate ratio of 2x : 1 is 8a; 3 : 1.
10. Definition. A ratio is said to he a ratio of greater
inequality, of less inequality, or of equality, according as the
antecedent is greater than, less than, or equal to the consequent.
11. A ratio of greater inequality is diminished, and a ratio of
less inequality is increased, by adding the same quantity to both
its terms.
a , ,, .. , , , a + x
Let T be the ratio, and let = be the new ratio formed by
6 b + x J
erms.
a a + x ax — bx
adding x to both its terms.
Now
b b + x b(b+x)
x(a — b)
~b(b + x) }
and a  b is positive or negative according as a is greater or
less than b.
H. /, y a a + x
ence it a > b, T > ^ ;
o b + x
d. « j a a ~t~ x
it a <b, 7 < ;
b b + x
which proves the proposition.
Similarly it can be proved that a ratio of greater inequality
is increased, and a ratio of less inequality is diminished, by taking
the same quantity from both its terms.
12. When two or more ratios are equal many useful pro
positions may be proved by introducing a single symbol to
denote each of the equal ratios.
1—2
4 HIGHER ALGEBRA.
The proof of the following important theorem will illustrate
the method of procedure.
// a c e
b d f '
1
, . . . /pa n + qc n + re 11 + . . .\ n
each of these ratios = ( — r j s ) ,
J \pb n + qd n + rt n + . . . /
where p, q, r, n are any quantities ivhatever.
ace j
Ijet 7  = —, — ~>— • • • — & ')
b d J
then a — bk, c = dk, e =fk, ...;
whence pa n =pb n ^ qc" = qd"k n , re n = rf"k",... ;
pa" + gc" + re n + ... _pb"k" + qd"k n + rf"k n +
''' pb H + qd"+rf+... ' fb n + qcl n +r/ i +...
= k";
i
'pa" + qc" + re" + . . .\ n _ , a c
e
= k = ^ = ,=
2)b" + qd" + ?'/" + .../ b d
By giving different values to p, q, r, n many particular cases
of this general proposition may be deduced ; or they may be
proved independently by using the same method. For instance,
a _c e
b~d'f'"
each of these ratios
b+d +f+
a result of such frequent utility that the following verbal equi
valent should be noticed : When a series of fractions are equal,
each of them is equal to the sum of all the numerators divided by the
sum of all the denominators.
(I C €>
Example 1. If  =  = , shew that
b d J
a z b + 2c 2 e  Sae 2 / _ ace
~b 4 + 2^/36/ 3 ~bdf
Let «££X;.
Let 6 _ rf _^._A,,
then a = bk, c = dk, e =fk ;
RATIO. 5
a a 6+2c»g3qgy _ W +2d?fk*  3bf 3 k 3
* '* k 4 + 2tlf  Bbf 3 "" fc 4 + 2r//  36/8
... a c e
ace
= bdf'
Example 2. If  = f =  , prove that
a b c
* 2 + a 2 y 2 + & 2 3 2 + c 2 _ ( .c + y +2 ) 2 + (a + & + c) 2
#+a y + b z + c a; + ?/ + 2 + a + &+c
x it z
Let  = r =  = A; , so that x = «£, ?/ = 6/c, 2; = ch ;
a y c
„ s a + a 3 aW+a* (k* + l)a
then = —  = L_ _ — ' ;
a: + a ah + a Jc+1
x*+a* y a +y • g a +e»_ (ife a +l)o (& 2 + l)& (fc 2 + l)c
ar + a ?/ + & z + c ' /c + 1 £ + 1 & + 1
Jfc 2 + l)(a + 6 + c)
fc+1
Jfc 8 (a+6+c) 8 +(a+ 6+c) a
&(a + & + c) + a + 6 + c
_ (lea + kb + he) % + (a + b + c) 2
(ka + kb + kc)+a + b + c
_ (x+y+z)*+( a+ b + cf
x+y+z+a+b+c
13. If an equation is homogeneous with respect to certain
quantities, we may for these quantities substitute in the equation
any others proportional to them. For instance, the equation
lx 3 y + mxifz + ny 2 z 2 —
is homogeneous in x, y, z. Let a, j3, y be three quantities pro
portional to x, y, % respectively.
x 11 z
Put h = — = 75 =  , so that x  ak, y = /3k, z = yk ;
a £ y
then Ia 3 f3k 4 + ma(3 2 yk* + n^y'k 4 = 0,
that is, 7a 3 /? + ma/3 2 y + nj3 2 y 2 = ;
an equation of the same form as the original one, but with
a, /?, y in the places of x, y, z respectively.
6 HIGHER ALGEBRA.
14. The following theorem is important.
. . . . . i Q
If y* , y~ , .— ,.... r n be unequal fractions, of which the de
1 2 3 n
nominators are all of the same sign, then the fraction
a, + a 8 + a 3 + ... + a n
b l +b 2 +b 3 + •'• +b n
lies in magnitude between the greatest and least of them.
Suppose that all the denominators are positive. Let ■=* be the
least fraction, and denote it by k ; then
a
b
— /c i .'.a — ko «
' r r >
a
y 1 > k : .. a> kb
b l
a
i '
b k; .. a 2 > kb 2 ;
2 >
a
and so on;
.*. by addition,
a
,+«2 + « 3 + + a n > ( b l +b , + K + +K) k '>
a l + a 2 + a 3 + + a u . a r
b.+b 9 + b.+ +b ' b
1 2 3 n r
Similarly we may prove that
a l + a 2 + a 3 + + a n a t
6 . +*.+*■ + +K < V
where ^ is tlie greatest of the given fractions.
In like manner the theorem may be proved when all the
denominators are negative.
15. The ready application of the general principle involved
in Art. 12 is of such great value in all branches of mathematics,
that the student should be able to use it with some freedom in
any particular case that may arise, without necessarily introducing
an auxiliary symbol.
Example 1. If  — X — = V  = z ,
b + ca c + ab a + bc
prove that x + y + z = *&+*)+? (*+*)+ * (*+V)
a + b + c 2(ax + by + cz)
RATIO.
t i e ,i • e i sum °f numerators
Each of the given fi actions = — = __
sum of denominators
_ x + y + z
' a + b + c " ( ''
Again, if we multiply both numerator and denominator of the three
given fractions by y + z, z + x, x + y respectively,
each fractions \ {l j + z)  = ?(« + *> __.  '(* + * )
(y + z)(b + ca) (z + x) (c + ab) (x + y) (a+be)
sum of numerators
sum of denominators
= x (y +z) + y (z + x ) +z {x + y)
2ax + 2by + 2cz
.'. from (1) and (2),
x + y + z _x (y + z)+y (z + x)+z (x + y)
a + b + c~ 2 (ax + by + cz)
Example 2. If
(2).
prove that
l(mb + ncla) m(nc + lamb) n (la + mb  nc) '
I m n
x(by + czax) y (cz + axby) z(ax + by cz)
We have
x y z
I m n
mb + nc — la nc + lamb la + mb — nc
v z
+
m n
= '"2/a"
= two similar expressions ;
ny + mz _lz + nx _ mx + ly
a b c
Multiply the first of these fractions above and below by .r, the second by
y, and the third by z ; then
nxy + mxz _ Jyz + nxy _ mxz + lyz
ax by cz
= _2lyz
by + cz ax
= two similar expressions ;
I m n
x (by + cz ax) y (cz + axby) z (ax + bycz)'
8 HIGHER ALGEBRA.
16. If we have two equations containing three unknown
quantities in the first degree, such as
a l x + b l y + c l z=Q (1),
a 2 x + b 2 y + c 2 z = (2),
we cannot solve these completely ; but by writing them in the
form
X II
we can, by regarding  and  as the unknowns, solve in the
z z
ordinary way and obtain
x b l c 2  b 2 c i y __ c x a 2  c 2 a l >
% " afi 2  a 2 b l ' z " afi 2 a 2 b l '
or, more symmetrically,
x y
,(3).
b x c 2  b 2 c x c l a 2  c 2 a, afi 2  a_p x '
It thus appears that when we have two equations of the type
represented by (1) and (2) we may always by the above formula
write down the ratios x : y : z in terms of the coefficients of the
equations by the following rule :
Write down the coefficients of x, y, z in order, beginning with
those of y; and repeat these as in the diagram.
Multiply the coefficients across in the way indicated by the
arrows, remembering that in forming the products any one
obtained by descending is positive, and any one obtained by
ascending is negative. The three results
h i c z h fv c x a 2 c 2 a n a A a 2 b >
are proportional to x, y, z respectively.
This is called the Rule of Cross Multiplication,
RATIO. 9
Example 1. Find the ratios of x : y : z from the equations
7x=4y + Qz t 3z = 12x + Uy.
By transposition we have 7x  Ay  82 = 0,
12x + llySz = 0.
"Write down the coeilicients, thus
4 8 7 4
11 3 12 11,
whence we obtain the products
(4)x(3)llx(8), (8)xl2(3)x7, 7 x 11  12 x (4),
or 100, 75, 125;
x y z
•'* 100 ~ ^75~"125'
x , ,. x y z
that is,  = * = ? .
4 3 5
Example 2. Eliminate x, y, z from the equations
a 1 a; + ^ 1 ?/ + c 1 2 = (1),
a^ + ^y + c^^O (2),
Ogaj+fegy+c^^O (3).
From (2) and (3), by cross multiplication,
*__ _ y „ j* .
k> C 3 " Vs C 2«i ~ C 3«2 «2 6 3 ~ ll ih '
denoting each of these ratios by k, by multiplying up, substituting in (1),
and dividing out by A, we obtain
Oj (Va  6 3 c a) + & i (^'"3  c 3 a a) + ( 'i (« A  " A) = °
This relation is called the eliminant of the given equations.
Example 3. Solve the equations
ax + by + cz = (1),
x+ y+ z = (2),
hex + cay + abz = (b  c) (ca) (ab) (3).
From (1) and (2), by cross multiplication,
x y z
 = — ^— = T — k, suppose :
bc ca ab
.. x = k (b c), y — k (c  a), z — k(a b).
Substituting in (3),
k {bc(bc) + ca (c  a) + ab (a  b)} ={b c) (c  a) {a  b),
k{{bc)(c a) {a  &) \ = (be) [e  a) {a  b) ;
.. fcsslj
^ln'nce x = c b, y — ar, z = b  a.
10 HIGHER ALGEBRA.
17. If in Art. 16 we put z = 1, equations (1) and (2) become
a x x + b x y + c t = 0,
v + h 2 y + c 2 = ° >
and (3) becomes
x y
b x c 2  b 2 c l c x a a  c 2 a i aj> %  a 2 b ] '
a l b 2 a b l * afi 2 a 2 b l
Hence any two simultaneous equations involving two un
knowns in the first degree may be solved by the rule of cross
multiplication.
Example. Solve 5x3y 1 = 0, x + 2y = 12.
By transposition, 5x  3y  1 = 0,
x + 2y 12 = 0;
x y 1
*'• 36 + 2 =  1 + 60 ~ 10 + 3 ;
38 59
whence x = is' y = lS'
EXAMPLES. I.
1. Find the ratio compounded of
(1) the ratio 2a : 36, and the duplicate ratio of 9b 2 : ab.
(2) the subduplicate ratio of 64 : 9, and the ratio 27 : 56.
2a /6a?
(3) the duplicate ratio of j :  M  , and the ratio Sax : 2by.
2. If #+7 : 2 (# + 14) in the duplicate ratio of 5 : 8, find x.
3. Find two numbers in the ratio of 7 : 12 so that the greater
exceeds the less by 275.
4. What number must be added to each term of the ratio 5 : 37
to make it equal to 1 : 3 \
5. If x : y=3 : 4, find the ratio of 7x4y : 3x+y.
6. If 15 (2a 2  y 2 ) = *7xy, find the ratio of x : y.
RATIO. 11
7 If ?=£ = «
2rt 4 & 2 + 3a 2 e 2 5eV " l
prove that _^__^__ = _
8. If v = =  7 , prove that j is equal t<»
6ca a
y
9. If
a _ y
q + rp r+pq p + qr
shew that (q  r) x + (r  p) y + (p  q) z = 0.
10. If — — == —  =  , find the ratios of x : y : z.
xz z y '
ii # if y+ z = z+ ' v ==  r+ ^
pb + qc pc + qa pa + qb'
Khew tliat 2 (*+?+*) _ (6+o).r+( C +«)y+(« ± i) i
a+o + c 6c + <?aa6
12. If i'=^ = 2 ,
a o c
3
.tfS + a 3 y 3 + 6 3 z 3 + c 3 _(.y + ?/ + *) 3 + (q + &+c)
shew tliat — „ „ t t   ;,, + ., ., — , \« ■ /_ . j. , _\s •
.rfa 2 y* + b 2 2 2 + c (>c+y + 5) i + (a + o + c)
2y + 2g.v _ 2g + 2.cy _ 2A+2yg
1<J. II — i — J
BheW that 26 + 2ca = 2c + 2a" 6 = 2a + 26c "
14. If (a 2 +6 2 + c 2 ) (.i 2 +y 2 + ^ 2 ) = («.v+^ + ^) 2 ,
shew that x : a=y : b = z : c.
15. If I (my + rut  Ix) = m (nz + Ix  my) = n (Ix + my  nz\
y+zx z+xy x+yz
prove  — j = = —   — •
1 I m n
16. Shew that the eliminant of
ax + cy + bz = Q, cx + by + az = 0, bx + </y + c; = 0,
is a 3 + & 3 + c 3 3«6c = 0.
17. Eliminate x, y, z from the equations
ctx + hy + (/z = 0, hx + by\fz = 0, gjc+fy+C2=0.
12 HIGHER ALGEBRA.
18. If x = cy + bz, y = az+cx } z=bx + ay,
X II z 2
shew that j—* = y i •> = n 9 •
1  a 1 \b l \—c L
19. Given that a(y + z)=x, b(z + x)=y, c(x+y)=z,
prove that bc + ca + ab + 2abc = l.
Solve the following equations :
20. 3x4y + 7z = 0, 21. x+y= z,
2xy2z = 0, 3x2y+17z = 0,
to?f+£=l8. x* + 3f + 2z s = l67.
22. tyz + 3sa?=4an/, 23. 3x 2  2y 2 + oz 2 = 0,
2tys  Sac = 4ry, 7a*  3y 2 I5z 2 = 0,
ff+2y+32=19. 5.04^ + 73 = 6.
24. If .* +^L^ *„<>,
I m n
Ja+Jb + Jb+Jo + </c+V« '
shew that — — = ?== = =
(ab)(c\/ab) (b  c) (a  V be) (c  a) (b  \J ac)
Solve the equations :
25. ax + by + cz = 0,
bcx + cay + abz = 0,
xyz + abc (a 3 x + b 3 y + &z) = 0.
26. a.£+&y + C2=a 2 # + & 2 y + 6' 2 2==0,
x + y + z + (bc)(ca) (ab) = 0.
27. If a(y+x)=x, b(z + x)=y, c(x+y)=z,
X 2 ?/ 2 s 2
prove that —  — 7— = , — * =
1 a {I be) b(lca) c(lab)
28. If ax + ky+gz = 0, kx + by + fz^0, gx+fy + cz = 0,
prove that
x 2 y 2 z 2
^ bcf 2 cag 2 abh 2
(2) (be f 2 ) {ea  g 2 ) (ab  h 2 ) = (fg  eh) (gk  af) (A/ bg).
CHAPTER II.
PROPORTION,
18. Definition. When two ratios are equal, the four
quantities composing them are said to be proportionals. Thus
ft c
if  =  , then a, b, c, d are proportionals. This is expressed by
saying that a is to b as c is to d, and the proportion is written
a : b : : c : d ;
or a : b — c : d.
The terms a and d are called the extremes, b and c the means.
19. If four quantities are in proportion, the product of the
extremes is equal to the product of the means.
Let a, b, c, d be the proportionals.
Then by definition — =. — •
J b d
whence ad = be.
Hence if any three terms of a proportion are given, the
fourth may be found. Thus if a, c, d are given, then b = — .
Conversely, if there are any four quantities, a, b, c, d, such
that ad = be, then a, b, c, d are proportionals ; a and d being the
extremes, b and c the means ; or vice versa.
20. Definition. Quantities are said to be in continued
proportion when the first is to the second, as the second is
to the third, as the third to the fourth ; and so on. Thus
a, b, c, d, are in continued proportion when
a b c
bed
14 HIGHER ALGEBRA.
If three quantities a, b, c are in continued proportion, then
a : b — b : c \
.. ac = b 2 . [Art. 18.]
In this case b is said to be a mean proportional between a and
c \ and c is said to be a third proportional to a and b.
21. If three quantities are proportionals the first is to the
third in the duplicate ratio of the first to tJie second.
Let the three quantities be a. b, c: then T = .
1 be
a a b
Now = r x 
cue
a a a 2
= b X 6 = F ,;
that is, a : c = a 2 : b 2 .
It will be seen that this proposition is the same as the definition
of duplicate ratio given in Euclid, Book v.
22. If a : b  c : d and e 'f=g : h, then will ae : bf= eg : dh.
„ a c , e g
.b or F =  and ■>=!:
b d j h
ae eg
•'* bf = dh y
or ae : bf= eg : dh.
Cor. If a : b = c : d,
and b : x = d : v/,
then a : x = c : y
This is the theorem known as ex cequali in Geometry.
23. If four quantities a, b, c, d form a proportion, many
other proportions may be deduced by the properties of fractions.
The results of these operations are very useful, and some of
them are often quoted by the annexed names borrowed from
Geometry.
PROPORTION. 15
(1) If a : b = c : d, then b : a = d : c. [Invertendo.]
For  =  ; therefore 1 f = = 1 r _ ;
b d' b d'
that is  =  :
a c
or b : a = d : c.
(2) If . a : b = c : d, then a : c = b : d. [Alternando.]
For acZ  be ; therefore — j = — ;
•! , . a b
that is,  = , :
c a
or a : c = b : d.
(3) If « : 6 = c : d, tlien a + b : b = c + d : d. [Componeudo.']
lor 7 = , : therefore s + 1 = , + 1 :
o d o d
a + b c + d
that is 7 — = — =— :
o d
or a + b : 6 = c + d : d.
(4) If a : 6 = c : d, then ab : b = c d : d. [Divideudo.]
For = = , : therefore  — 1 =  7  1 :
b d b d
. , " . a — bcd
that is, —7— = — =— 3
or a  b : b  c  d : d.
(5) If « : 6  c : df, then a +6 : a — b=c+d:c— d.
For by (3) r = j ;
1 1 / * \ a — bc—d
and by (4) j ^j
. , . . . « + & c + d
.'. by division, = = •
J ab cd'
or a + b : ab = c + d : cd.
This proposition is usually quoted as Componeiuh a) id JJivi
dendo.
Several other proportions may be proved in a similar way.
16 HIGHER ALGEBRA.
24. The results of the preceding article are the algebraical
equivalents of some of the propositions in the fifth book of Euclid,
and the student is advised to make himself familiar with them
in their verbal form. For example, dividendo may be quoted as
follows :
When there are four proportionals, the excess of the first above
the second is to the second, as the excess of the third above the
fourth is to the fourth.
25. We shall now compare the algebraical definition of pro
portion with that given in Euclid.
Euclid's definition is as follows :
Four quantities are said to be proportionals when if any equi
multiples whatever be taken of the first and third, and also any
equimultiples wJiatever of the second and fourth, the multiple of
the third is greater than, equal to, or less than the multiple of the
fourth, according as the multiple of the first is greater than, equal
to, or less than the multiple of the second.
In algebraical symbols the definition may be thus stated :
Four quantities a, b, c, d are in proportion when p>c = qd
according as p>a = qb, p and q being any positive integers tcJudever.
I. To deduce the geometrical definition of proportion from
the algebraical definition.
a c u
Since z   , by multiplying both sides by  , we obtain
pa 2 )C
qb qd '
hence, from the properties of fractions,
pc = qd according as pa = qb,
which proves the proposition.
II. To deduce the algebraical definition of proportion from
the geometrical definition.
Given that pc = qd according as pa = qb, to prove
a c
b = ~d'
PROPORTION. 17
If j is not equal to  , one of them must be the greater.
Suppose g > ^ ; then it will be possible to find some fraction 2
which lies between them, q and <p being positive integers.
Hence  > 
b p
P
0).
(2>
and  < ?
From (1) pa>qb;
from (2) 2)c<qd\
and these contradict the hypothesis.
Therefore y and  are not unequal; that is  = • which proves
the proposition.
26. It should be noticed that the geometrical definition of pro
portion deals with concrete magnitudes, such as lines or areas,
represented geometrically but not referred to any common unit
of measurement. So that Euclid's definition is applicable to in
commensurable as well as to commensurable quantities ; whereas
the algebraical definition, strictly speaking, applies only to com
mensurable quantities, since it tacitly assumes that a is the same
determinate multiple, part, or parts, of b that c is of d. But the
proofs which have been given for commensurable quantities will
still be true for incommensurables, since the ratio of two incom
mensurables can always be made to differ from the ratio of two
integers by less than any assignable quantity. This lias been
shewn in Art. 7 ; it may also be proved more generally as in the
next article.
27. Suppose that a and b are incommensurable; divide b
into m equal parts each equal to /?, so that b = m/3, where m is a
positive integer. Also suppose f3 is contained in a more than n
times and less than n+ 1 times;
i, a nB . (n+1) B
then  > ^ and < * /^ ,
o mp imp
that is, =■ lies between — and ;
o m m
so that j differs from — by a quantity less than — . And since we
H. H. A. 2
18 HIGHER ALGEBRA.
can choose B (our unit of measurement) as small as we please, m can
1
be made as great as we please. Hence — can be made as small
as we please, and two integers n and m can be found whose ratio
will express that of a and b to any required degree of accuracy.
28. The propositions proved in Art. 23 are often useful in
solving problems. In particular, the solution of certain equa
tions is greatly facilitated by a skilful use of the operations com
ponendo and dividendo.
Example 1.
If (2ma + 6mb + Snc + 9wtZ) (2ma  Gmb  Snc + 9nd)
= (2ma  6mb + Snc  9/uZ) (2mm + Gmi  Snc  dnd),
prove that a,b, c, d are proportionals.
2ma + Gmb + Snc + 9nd _ 2ma + Qmb  Snc  9nd
2ma  bmb + Snc  \)nd 2 ma  6mb  Snc + 9nd '
.*. componendo and dividendo,
2 (2ma + Snc) _ 2 {2ma  Sue)
2~{Gmb + 9nd) ~ 2 (Smb  9m/) '
2ma + Snc Gmb + ( .)nd
Alternando, n  =— =  — : —  — ,.
2maSnc bmbvna
Again, componendo and dividendo,
Ama _ \2mb
One ~ lQnd ;
a b
whence  = , ,
c a
or a : b — c : d.
Example 2. Solve the equation
Jx+l + Jx^l _ 4a; 1
Jx + l Jx1 2
We have, componendo and dividendo,
Jx+l _ 4a; + 1
.r + l_ 16a; 2 + 8a;+l
*'• x  1 " 16a; 2 24a; + 9 *
Again, componendo and dividendo,
2x _ 32a; 2  16a; + 10
2 ~ ~ 32a;  8
16a; 2 8a; + 5
" X ~ 16a; 4 '
whence 16a; 2  4a; = 16a; 2  8a; + 5 ;
5
• •
x = .
PROPORTION. 19
EXAMPLES. II.
1. Find the fourth proportional to 3, 5, 27.
2. Find the mean proportional between
(1) 6 and 24, (2) 36'0a 4 and 250a 2 6 2 .
X II x
3. Find the third proportional to '  f  and  .
y x y
If a : b = c : d, prove that
4. a 2 c + ac 2 : b 2 d + bd 2 = (a + c) 3 : (b + df.
5. pa 2 + <?6 2 : £>a 2 — qb 2 =pc 2 + qd 2 : pc 2 — qd 2 .
6. ac : bd=*Ja 2 + c 2 : *Jb 2 + d 2 .
7. \/a 2 ~+"c 2 : \/^+d^=j S /ac + < ^ : ^Jbd+j.
If a, 6, c, o? are in continued proportion, prove that
8. a : 6 + ^=03 : <?d+d\
9. 2a + 3(i : 3a4d=2a 3 + 3b 3 : 3a 3 46 3 .
10. (a 2 + b 2 + c 2 ) (b 2 + c 2 + d 2 ) = (aft + &c 4 c^) 2 .
11. If b is a mean proportional between a and c, prove that
• a 2_fr2 + c 2
a 2 6 2 + c 2
12. If a : 6=c : d, and e : /=# : h t prove that
ae + bf : aebf=cg + dh : cgdh.
Solve the equations :
2afi3afi+a;+l 3^^ + 50713
13.
14.
15.
20733072071 307 3 07 2 507+13*
Zx*+x 2  2o7  3 _ 5o7 4 + 2o7 2 7o7 + 3
3ot*  x 2 + 2o; + 3 ~~ 5o? 4  2o7 2 + 7o  3 '
(m\n)x — (a b) (m + n)x + a + c
(mn)x — (a + 6) (wi — n)x + ac'
16. If a, &, c, o? are proportionals, prove that
7 . (a — b)(a — c)
+ d=b + c + K ^ .
a
a
17. If a, b, c, d, e are in continued proportion, prove that
(ab + be + cd + e&) 2 = (a 2 + 6 2 + c 2 + rf 2 ) (6 2 + c 2 + d 2 + e 2 ).
2—2
20 HIGHER ALGEBRA.
18. If the work done by x — 1 men in x + 1 days is to the work done
by x + 2 men in x  1 days in the ratio of 9 : 10, find x.
19. Find four proportionals such that the sum of the extremes is
21, the sum of the means 19, and the sum of the squares of all four
numbers is 442.
20. Two casks A and B were filled with two kinds of sherry, mixed
in the cask A in the ratio of 2 : 7, and in the cask B in the ratio of
1 : 5. What quantity must be taken from each to form a mixture
which shall consist of 2 gallons of one kind and 9 gallons of the other \
21. Nine gallons are drawn from a cask full of wine; it is then
filled with water, then nine gallons of the mixture are drawn, and the
cask is again filled with water. If the quantity of wine now in the cask
be to the quantity of water in it as 16 to 9, how much does the cask
hold?
22. If four positive quantities are in continued proportion, shew
that the difference between the first and last is at least three times as
great as the difference between the other two.
23. In England the population increased 15*9 per cent, between
1871 and 1881; if the town population increased 18 per cent, and the
country population 4 per cent., compare the town and country popula
tions in 1871.
24. In a certain country the consumption of tea is five times the
consumption of coffee. If a per cent, more tea and b per cent, more
coffee were consumed, the aggregate amount consumed would be 1c per
cent, more ; but if b per cent, more tea and a per cent, more coffee
were consumed, the aggregate amount consumed would be 3c per cent,
more : compare a and b.
25. Brass is an alloy of copper and zinc ; bronze is an alloy
containing 80 per cent, of copper, 4 of zinc, and 16 of tin. A fused
mass of brass and bronze is found to contain 74 per cent, of copper, 16
of zinc, and 10 of tin : find the ratio of copper to zinc in the composition
of brass.
26. A crew can row a certain course up stream in 84 minutes;
they can row the same course down stream in 9 minutes less than
they could row it in still water : how long would they take to row down
with the stream ?
CHAPTER III.
VARIATION.
29. Definition. One quantity A is said to vary directly
as another B, when the two quantities depend upon each other in
such a manner that if B is changed, A is changed in the same
ratio.
Note. The word directly is often omitted, and A is said to vary
as B.
For instance : if a train moving at a uniform rate travels
40 miles in 60 minutes, it will travel 20 miles in 30 minutes,
80 miles in 120 minutes, and so on; the distance in each case
being increased or diminished in the same ratio as the time.
This is expressed by saying that when the velocity is uniform
the distance is ptroportional to the time, or the distance varies as
the time.
30. The symbol oc is used to denote variation \ so that
A on B is read "A varies as B."
31. If A. varies as B, tlien A is equal to B multiplied by some
constant quantity.
For suppose that a, a lt a„, a 3 ..., b, b x , b 2 , b 3 ... are corresponding
values of A and B.
mi i i /» •,• a b a b a b ,
Inen, by deimition, — = = : — = — ; — = r ; and so on,
«, V ^ K »3 K
/. si = ■=* = y^= ,.j each being equal to T .
6, 6 2 6 3 b
TT any value of A .
Hence = r . _ =— = is always the same :
the corresponding value ot B
that is, — — 7u, where m is constant.
.'. A=mB.
22 HIGHER ALGEBRA.
If any pair of corresponding values of A and B are known,
the constant m can be determined. For instance, if A = 3 when
^=12,
we have 3 =m x 12;
and A = \B.
32. Definition. One quantity A is said to vary inversely
as another Z?, when A varies directly as the reciprocal of B.
Thus if A varies inversely as B, A = ^ , where m is constant.
The following is an illustration of inverse variation : If 6 men
do a certain work in 8 hours, 12 men would do the same work in
4 hours, 2 men in 24 hours ; and so on. Thus it appears that
when the number of men is increased, the time is proportionately
decreased; and viceversa.
Example 1. The cube root of x varies inversely as the square of y ; if
x=8 when y = 3, find x when y = l^.
By supposition £/x= — , where m is constant.
if
Tit
Putting x = 8, y = 3, we have 2 =n>
.*. ?;t = 18,
v 18
and *jx = — , ;
r
hence, by putting y = ^, we obtain a; = 512.
Example 2. The square of the time of a planet's revolution varies as
the cube of its distance from the Sun; find the time of Venus' revolution,
assuming the distances of the Earth and Venus from the Sun to be 91J and
66 millions of miles respectively.
Let P be the periodic time measured in days, D the distance in millions
of miles ; we have P 2 a D 3 ,
or P*=kD 3 ,
where k is some constant.
For the Earth, 365 x 365 = k x 91± x 91 x 91£,
4x4x4
whence k =
365
. p2 _ 4 x 4 x 4
" r ~ 365 " '
VARIATION. 23
For Venus, pa^i^ili x 66 x 66 x 6G ;
3 b.)
whence P = 4x66
:■:
/264
V 365
= 264 x a/*7233, approximately,
= 264 x 85
= 2244.
Hence the time of revolution is nearly 224£ days.
33. Definition. One quantity is said to vary jointly as a
number of others, when it varies directly as their product.
Thus A varies jointly as B and C, when A = mBC. For in
stance, the interest on a sum of money varies jointly as the
principal, the time, and the rate per cent.
3 i. Definition. A is said to vary directly as B and in
versely as C, when A varies as ^ .
35. 7/*A varies as B when C is constant, and A varies as C
when B is constant, then tvill A vary as BC ivhen both B and C
vary.
The variation of A depends partly on that of B and partly on
that of C. Suppose these latter variations to take place sepa
rately, each in its turn producing its own effect on A ; also let
a, b, c be certain simultaneous values of A, B, C.
1 . Let C be constant while B changes to b ; then A must
undergo a partial change and will assume some intermediate value
a\ where
"= (1)
2. Let B be constant, that is, let it retain its value b, while C
changes to c ; then A must complete its change and pass from its
intermediate value a' to its final value a, where
From (1) and (2) — x  = — x  :
x ' a a b c
that is, A = = . BC,
be
or A varies as BC.
24 HIGHER ALGEBRA.
36. The following are illustrations of the theorem proved in
the last article.
The amount of work done by a given number of men varies
directly as the number of days they work, and the amount of
work done in a given time varies directly as the number of men ;
therefore when the number of days and the number of men are
both variable, the amount of work will vary as the product of
the number of men and the number of days.
Again, in Geometry the area of a triangle varies directly as
its base when the height is constant, and directly as the height
when the base is constant ; and when both the height and base
are variable, the area varies as the product of the numbers
representing the height and the base.
Example. The volume of a right circular cone varies as the square of the
radius of the base when the height is constant, and as the height when the
base is constant. If the radius of the base is 7 feet and the height 15 feet,
the volume is 770 cubic feet ; find the height of a cone whose volume is 132
cubic feet and which stands on a base whose radius is 3 feet.
Let h and r denote respectively the height ani radius of the base
measured in feet ; also let V be the volume in cubic feet.
Then V=mr 2 h, where m is constant.
By supposition, 770 — m x 7 2 x 15 ;
22
whence m = — ;
.*. by substituting V= 132, r = S, we get
22
132= xOxft;
— X
whence 7i= 14 ;
and therefore the height is 14 feet.
37. The proposition of Art. 35 can easily be extended to the
case in which the variation of A depends upon that of more than
two variables. Further, the variations may be either direct or
inverse. The principle is interesting because of its frequent oc
currence in Physical Science. For example, in the theory of
gases it is found by experiment that the pressure (p) of a gas
varies as the "absolute temperature" (t) when its volume (v) is
constant, and that the pressure varies inversely as the volume
when the temperature is constant ; that is
2? oc t, when v is constant ;
VARIATION. 25
and p cc  , when t is constant.
v
From these results we should expect that, when both t and v are
variable, we should have the formula
p cc  , or pv = kt, where k is constant ;
and by actual experiment this is found to be the case.
Example. The duration of a railway journey varies directly as the
distance and inversely as the velocity; the velocity varies directly as the
square root of the quantity of coal used per mile, and inversely as the
number of carriages in the train. In a journey of 25 miles in half an hour
with 18 carriages 10 cwt. of coal is required; how much coal will be
consumed in a journey of 21 miles in 28 minutes with 16 carriages?
Let t be the time expressed in hours,
d the distance in miles,
v the velocity in miles per hour,
q the quantity of coal in cwt.,
c the number of carriages.
We have t oc  ,
v
and v oc *!l ,
c
whence t oc — ,
or t — — 7 , where k is constant.
Substituting the values given, we have
1 _ k x 18 x 25
2 ~ jm ;
that is, k =
25x36"
v/lO . cd
Hence t = ^ — ^— T .
2o x 36 Jq
Substituting now the values of t, c, d given in the second part of the
question, we have
28 710x16x21 .
60" 25x36^2 '
a •• / n/10x 16x21 ^ ,
that is, s/q=— 15x28 =5^10,
whence q — = = 6.
Hence the quantity of coal is 6cwt.
26 HIGHER ALGEBRA.
EXAMPLES. III.
1. If x varies as y, and #=8 when y = 15, find x when y = 10.
2. If P varies inversely as Q, and P=7 when # = 3, find P when
3. If the square of x varies as the cube of y, and x—3 when y = 4,
find the value of y when #=y .
3 10
4. A varies as B and C jointly; if A = 2 when # =  and C=— ,
find when A = 54 and i? = 3.
5. If .4 varies as C, and i? varies as C, then J. ±Z? and \/ AB will
each vary as C.
C
6. If J. varies as BC, then Z> varies inversely as 7 .
2
7. P varies directly as Q and inversely as R\ also P = ~ when
o
^ =  and R =— : find (^ when P=a/48 and jR=\/<5.
8. If a' varies as y, prove that x 2 +y 2 varies as x 2 y\
9. If y varies as the sum of two quantities, of which one varies
directly as x and the other inversely as x ; and if y = 6 when x=4, and
y = 31 when x = 3 ; find the equation between x and y.
10. If 3/ is equal to the sum of two quantities one of which varies
as x directly, and the other as x 2 inversely; and if y = 19 when x=2, or
3 ; find y in terms of x.
11. If A varies directly as the square root of B and inversely as
the cube of C, and if 4 = 3 when .£=256 and C=2, find B when A = 24
and C=g .
12. Given that x + y varies as z +  , and that x — y varies as z — ,
z z
find the relation between x and z, provided that z =2 when x =3 and
y = \.
13. If J. varies as B and C jointly, while B varies as Z> 2 , and C
varies inversely as A, shew that A varies as D.
14. If y varies as the sum of three quantities of which the first is
constant, the second varies as .r, and the third as x 2 ; and if y = when
x=l, y = l when x=2, and y = 4 when x = 3; find y when x=7.
15. When a body falls from rest its distance from the starting
point varies as the square of the time it has been falling : if a body falls
through 402^ feet in 5 seconds, how far does it fall in 10 seconds ?
Also how far does it fall in the 10 th second?
VARIATION. 27
16. Given that the volume of a sphere varies as the cul>c of its
radius, and that when the radius is 3& feet the volume is 179rj cubic
feet, find the volume when the radius is 1 foot 9 inches.
17. The weight of a circular disc varies as the square of the radius
when the thickness remains the same; it also varies as the thickness
when the radius remains the same. Two discs have their thicknesses
in the ratio of 9 : 8 ; find the ratio of their radii if the weight of the
first is twice that of the second.
18. At a certain regatta the number of races on each day varied
jointly as the number of days from the beginning and end of the regatta
up to and including the day in question. On three successive days
there were respectively 6, 5 and 3 races. Which days were these, and
how long did the regatta last?
19. The price of a diamond varies as the square of its weight.
Three rings of equal weight, each composed of a diamond set in gold,
have values «£«., £b, £c> the diamonds in them weighing 3, 4, 5 carats
respectively. Shew that the value of a diamond of one carat is
the cost of workmanship being the same for each ring.
20. Two persons are awarded pensions in proportion to the square
root of the number of years they have served. One has served 9 years
longer than the other and receives a pensio?i greater by ,£50. If the
length of service of the first had exceeded that of the second by 4 years
their pensions would have been in the proportion of 9 : 8. How long
had they served and what were their respective pensions ?
21. The attraction of a planet on its satellites varies directly as
the mass (M)of the planet, and inversely as the square of the distance
(D) ; also the square of a satellite's time of revolution varies directly
as the distance and inversely as the force of attraction. If m v d v t v
and m 2 , d 2 , £ 2 , are simultaneous values of J/, D, T respectively, prove
that
Hence find the time of revolution of that moon of Jupiter whose
distance is to the distance of our Moon as 35 : 31, having given
that the mass of Jupiter is 343 times that of the Earth, and that the
Moon's period is 27*32 days.
22. The consumption of coal by a locomotive varies as the square
of the velocity; when the speed is 10 miles an hour the consumption of
coal per hour is 2 tons : if the price of coal be 10s. per ton, and the other
expenses of the engine be lis. 3c/. an hour, find the least cost of a journey
of 100 miles.
CHAPTER IV.
ARITHMETICAL PROGRESSION.
38. Definition. Quantities are said to be in Arithmetical
Progression when they increase or decrease by a common dif
ference.
Thus each of the following series forms an Arithmetical
Progression :
3, 7, 11, 15,
8, 2, 4, 10,
a, a + d, a + 2d, a + 3d,
The common difference is found by subtracting any term of
the series from that which follows it. In the first of the above
examples the common difference is 4 ; in the second it is — 6 ; in
the third it is d.
39. If we examine the series
a, a + d, a + 2d, a + 3d, . . .
we notice that in any term the coefficient of d is always less by one
than the number of the term hi the seiies.
Thus the 3 rd term is a + 2d;
6 th term is a + 5d ;
20 th term is a + I9d',
and, generally, the p th term is a + ( p — \)d.
If n be the number of terms, and if I denote the last, or
?i th term, we have I = a + (n — 1) d.
40. To find the sum of a number of terms in Arithmetical
Progression.
Let a denote the first term, d the common difference, and n
the number of terms. Also let I denote the last term, and s
ARITHMETICAL PROGRESSION. 29
the required sum ; then
8 = a+(a + d) + (a + 2d) + ... + (I  2d) + (ld) + l;
and, by writing the series in the reverse order,
s = I + (I  d) + (I  2d) + ... + (a + 2d)+ (a + d) + a.
Adding together these two series,
2s = (a + l) + (a + l) + (a + l)+ ... to n terms
= n (a + I),
•'• s = ^(a + l) (1);
a
and l~a + (nl)d (2),
.. s = ^{2a + (nl)d\ (3).
41. In tlie last article we have three useful formula; (1),
(2), (3) ; in each of these any one of the letters may denote
the unknown quantity when the three others are known. For
instance, in (1) if we substitute given values for s, n, I, we obtain
an equation for finding a ; and similarly in the other formulae.
But it is necessary to guard against a too mechanical use of these
general formulae, and it will often be found better to solve simple
questions by a mental rather than by an actual reference to the
requisite formula.
Example 1. Find the sura of the series 5^, GJ, 8, to 17 terms.
Here the common difference is 1^; hence from (3),
the sum = \ 2 x — + 16
= y (11+20)
17x31
~2
= 263£.
Example 2. The first term of a series is 5, the last 45, and the sum
400 : find the number of terms, and the common difference.
If n be the number of terms, then from (1)
400 = " (5 + 4r>);
whence n = 10.
*li
30 HIGHER ALGEBRA.
If d be the common difference
45= the 16 th term = 5 + 15d;
whence d = 2f .
42. If any two terms of an Arithmetical Progression be
given, the series can be completely determined; for the data
furnish two simultaneous equations, the solution of which will
give the first term and the common difference.
Example. The 54 th and 4 th terms of an A. P. are  61 and 64 ; find the
23 rd term.
If a be the first term, and d the common difference,
 61 = the 54 th term = a + 53d ;
and 64 = the 4 th term = a + 3d ;
5
whence we obtain d= jr, a = Hh. ;
and the 23 rd term = a + 22d = 16£.
43. Definition. When three quantities are in Arithmetical
Progression the middle one is said to be the arithmetic mean of
the other two.
Thus a is the arithmetic mean between a — d and a + d.
44. To find the arithmetic mean betiveen two given quantities.
Let a and b be the two quantities ; A the arithmetic mean.
Then since a, A, b are in A. P. we must have
b  A = A — a,
each being equal to the common difference ;
a + b
whence A —
2
45. Between two given quantities it is always possible to
insert any number of terms such that the whole series thus
formed shall be in A. P. ; and by an extension of the definition in
Art. 43, the terms thus inserted are called the arithmetic means.
Example. Insert 20 arithmetic means between 4 and 67.
Including the extremes, the number of terms will be 22 ; so that we have
to find a series of 22 terms in A.P., of which 4 is the first and 67 the last.
Let d be the common difference ;
then 67 = the 22 nd term = 4 + 21d ;
whence d = S, and the series is 4, 7, 10, 61, 64, 67 ;
and the required means are 7, 10, 13, 58, 71, 64.
ARITHMETICAL PROGRESSION. 31
46. To insert a given number of arithmetic means betiveen
two given quantities.
Let a and b be the given quantities, n the number of means.
Including the extremes the number of terms will be u + 2 ;
so that we have to find a series of n + 2 terms in A. P., of which
a is the first, and b is the last.
Let d be the common difference ;
then b = the (n + 2) th term
whence d = r :
71+ 1 '
and the required means are
b — a 2 (b — a) nib — a)
a +  , a H * — =' , a + — * — _ ' .
n+l n+l n+l
Example 1. The sum of three numbers in A.P. is 27, and the sum of
their squares is 293 ; find them.
Let a be the middle number, d the common difference ; then the three
numbers are a  d, a, a + d.
Hence ad + a + a + d = 27 ;
whence a = 9, and the three numbers are 9  d, 9, $ + d.
.. (9rf) 2 + 81 + (9 + d) 2 = 293;
whence d=±5;
and the numbers are 4, 9, 14.
Example 2. Find the sum of the first p terms of the series whose
w"' term is 3n  1.
By putting n=l, and n=p respectively, we obtain
first term = 2, last term =3p — 1 ;
.. sum=(2 + 3i>l)=(3p + l).
EXAMPLES. IV. a.
1. Sum 2, 3, 4J,... to 20 terms.
2. Sum 49, 44, 39,... to 17 terms.
3 2 7
3. Sum, , — ,... to 19 terms.
4 o I —
32 HIGHER ALGEBRA.
7
4. Sum 3, , If,... to n terms.
o
5. Sum 3'75, 35, 325,... to 16 terms.
6. Sum Tl, 7, 6J,... to 24 terms.
7. Sum 13, 31, 75,... to 10 terms.
6 12
8. Sum . , 3 x /3, 75 »... to 50 terms.
3 4
9. Sum j= , tt , V 5 ;.. to 25 terms.
10. Sum a  36, 2a  56, 3a  76, . . . to 40 terms.
11. Sum 2a  6, 4a  36, 6a  56,. . . to n terms.
n tt + 6 3a6 , ,, ,
12. Sum £ , a, — ^ — ,... to 21 terms.
13. Insert 19 arithmetic means between  and — 9.
14. Insert 17 arithmetic means between 3^ and — 41§.
15. Insert 18 arithmetic means between  36.17 and S.v.
16. Insert as arithmetic means between x 2 and 1.
17. Find the sum of the first n odd numbers.
18. In an A. P. the first term is 2, the last term 29, the sum 155;
find the difference.
19. The sum of 15 terms of an A. P. is 600, and the common differ
ence is 5 ; find the first term.
20. The third term of an A. P. is 18, and the seventh term is 30 ;
find the sum of 17 terms.
21. The sum of three numbers in A. P. is 27, and their product is
504 ; find them.
22. The sum of three numbers in A. P. is 12, and the sum of their
cubes is 408 ; find them.
23. Find the sum of 15 terms of the series whose n th term is 4?i4 1.
24. Find the sum of 35 terms of the series whose p ih term is ^ + 2.
25. Find the sum of p terms of the series whose n th term is  + b.
26. Find the sum of n terms of the series
2 a 2  1 3 6a 2 5
, 4a , , . . .
a a a
ARITHMETICAL PROGRESSION. 33
47. In an Arithmetical Progression when s, a, d are given,
to determine the values of n we have the quadratic equation
s = ^ <2a + (n l)d\ ;
when both roots are positive and integral there is no difficulty
in interpreting the result corresponding to each. In some cases
a suitable interpretation can be given for a negative value of n.
Example. How many terms of the series 9, 6, 3,... must be
taken that the sum may be G6 ?
Here ? {18 + (»l) 3}=66;
that is, nlnU = Q,
or (nll)(n+4)=0;
.'. ?i=ll or  4.
If we take 11 terms of the series, we have
 9,  6,  3, 0, 3, 6, 9, 12, 15, 18, 21 ;
the sum of which is 66.
If we begin at the last of these terms and count backwards four terms, the
sum is also 66; and thus, although the negative solution does not directly
answer the question proposed, we are enabled to give it an intelligible meaning,
and we see that it answers a question closely connected with that to which
the positive solution applies.
48. We can justify this interpretation in the general case in
the following way.
The equation to determine n is
dn 2 + (2ad)n2s = (1).
Since in the case under discussion the roots of this equation have
opposite signs, let us denote them by n and  n . The last
term of the series corresponding to n l is
a + (n l  1 ) d ;
if we beirin at this term and count backwards, the common
difference must be denoted by  d, and the sum of yi., terms is
{2 (« + »,!</) + (», !)(</)}
and we shall shew that this is equal to 6.
H. H. A. 3
34 HIGHER ALGEBRA.
For the expression =  ? \ 2a + (2n { — n 2 — l)dl
= ^ 1 2an 2 + 2n x n 2 d  n 2 (n 2 + 1) d
1
= ^ I 2n x n 2 d  (da*  2a  d .n 2 )\
= l(4s2s) = s,
since — n 2 satisfies dn 2 + (2a — d) n— 2s = 0, and — n } n 2 is the
product of the roots of this equation.
49. When the value of n is fractional there is no exact num
ber of terms which corresponds to such a solution.
Example. How many terms of the series 26, 21, 16, ...must be taken to
amount to 71 ?
n
Here ~ {52 + (nl)(5)} = 74;
that is, 5)i 2  57u + 148 = 0,
or (n4)(5n37) = 0;
.*. ?i = 4 or 1%.
Thus the number of terms is 4. It will be found that the sum of 7 terms
is greater, while the sum of 8 terms is less than 74.
50. We add some Miscellaneous Examples.
Example 1. The sums of n terms of two arithmetic series are in the
ratio of 7?t + l : 4« + 27; rind the ratio of their 11 th terms.
Let the first term and common difference of the two series be a v d x and
a„, d 2 respectively.
" We have M^* £+1
2a 2 + {nl)d 2 4?i + 27
Now we have to find the value of — — tttt', hence, by putting n—21, we
a 2 + l0d 2 ' x
obtain
2^ + 20^ _ 1 48 _ 4 _
2a 2 + 20d 2 ~ 111 "~ 3 '
thus the required ratio is 4 : 3.
Example 2. If S u S 2 , S&...S,, are the sums of n terms of arithmetic
series whose first terms are 1, 2, 3, 4,... and whose common differences are
1, 3, 5, 7,... ; find the value of
#L + <Sf 2 +£ 3 +. .. + £„.
ARITHMETICAL PROGRESSION.
We have S^ {2 + (n  1)} = n AH+D ,
.S>^{2 i >+(»l)(2 i ;l)}= ?  i {(2pl)n+l};
w
•. the required sum=  {(n + l) + (3n + l) + (2/> 1 . n + 1)}
m
?l ~
=  { (n + 3n + 5n + . . .2p  1 . ;/) + p)
= 1 ±{n(l + 3 + 5+...2 1 >l)+p}
= r 2 (»l> 2 +P)
EXAMPLES. IV. b.
1. Given a= 2, c?=4 and .5 = 100, find n.
2. How many terms of the series 12, 16, 20,... must be taken to
make 208 ?
3. In an A. P. the third term is four times the first term, and the
sixth term is 1 7 ; find the series.
4. The 2 n * 1 , 31 st , and last terms of an A. P. are 7j, 5 and 6j
respectively ; find the first term and the number of terms.
5. The 4 th , 42 nd , and last terms of an A. P. are 0,  95 and  1 25
respectively ; find the first term and the number of terms.
6. A man arranges to pay off a debt of £3600 by 40 annual
instalments which form an arithmetic series. When 30 of the instal
ments are paid he dies leaving a third of the debt unpaid: find the
value of the first instalment.
7. Between two numbers whose sum is 2£ an even number of
arithmetic means is inserted; the sum of these means exceeds their
number by unity : how many means are there 2
8. The sum of n terms of the series 2, 5, 8,... is !>">0 : find n.
3—2
36 HIGHER ALGEBRA.
9. Sum the series  r , _ , , .— , ... to n terms.
10. If the sum of 7 terms is 49, and the sum of 17 terms is 289,
find the sum of n terms.
11. If the p th , q th , r ih terms of an A. P. are a, b, c respectively, shew
that (qi')a + (rp)b+(pq)c = 0.
12. The sum of p terms of an A. P. is q, and the sum of q terms is
p ; find the sum ofp + q terms.
13. The sum of four integers in A. P. is 24, and their product is
945 ; find them.
14. Divide 20 into four parts which are in A. P., and such that the
product of the first and fourth is to the product of the second and third
in the ratio of 2 to 3.
15. The p th term of an A. P. is q, and the q th term is p ; find the
m tb term.
16. How many terms of the series 9, 12, 15,... must be taken to
make 306?
17. If the sum of n terms of an A. P. is 2n + 3n 2 , find the ? tth term.
18. If the sum of m terms of an A. P. is to the sum of n terms as
'in 2 to ?i 2 , shew that the m th term is to the n th term as 2m — 1 is to 2n — 1.
19. Prove that the sum of an odd number of terms in A. P. is equal
to the middle term multiplied by the number of terms.
20. If 5 = n (pn  3) for all values of n t find the p th term.
21. The number of terms in an A. P. is even ; the sum of the odd
terms is 24, of the even terms 30, and the last term exceeds the first by
10 1 : find the number of terms.
22. There are two sets of numbers each consisting of 3 terms in A. P.
and the sum of each set is 15. The common difference of the first set
is greater by 1 than the common difference of the second set, and the
product of the first set is to the product of the second set as 7 to 8 : find
the numbers.
23. Find the relation between x and y in order that the ? ,th mean
between x and 2y may be the same as the ? th mean between 2x and y,
n means being inserted in each case.
24. If the sum of an A. P. is the same for p as for q terms, shew
that its sum for p + q terms is zero.
CHAPTER V.
GEOMETRICAL PROGRESSION.
51. Definition. Quantities are said to be in Geometrical
Progression when they increase or decrease by a constant factor.
Thus each of the following series forms a Geometrical Pro
gression :
3, G, 12, 24,
1  1 I I
3' 9' 27'
a, ar, ar 2 , ar 3 ,
The constant factor is also called the common ratio, and it is
found by dividing any term by that which immediately iwecedes
it. In the first of the above examples the common ratio is 2 ; in
the second it is —  ; in the third it is r.
o
52. If we examine the series
a, ar, ar 2 , ar 3 , ai A ,
we notice that in any term the index of r is always less by one
tlian the number of the term in the series.
Thus the 3 rd term is ar 2 ;
the 6 th term is ar s ;
the 20 th term is ar 19 ;
and, generally, the p ih term is a?^ 1 .
If n be the number of terms, and if I denote the last, or n ,h
term, we have l = ar"~\
53. Definition. When three quantities are in Geometrical
Progression the middle one is called the geometric mean between
the other two.
38 HIGHER ALGEBRA.
To find the geometric mean between two given quantities.
Let a and b be the two quantities ; G the geometric mean.
Then since a, G, b are in G. P.,
b _G
G~ a'
each being equal to the common ratio ;
.. G 2 = ab;
whence G = Jab.
54. To insert a given number of geometric means between
two given quantities.
Let a and b be the given quantities, n the number of means.
In all there will be n + 2 terms ; so that we have to find a
series of n + 2 terms in G. P., of which a is the first and b the last.
Let r be the common ratio ;
then b = the (n + 2) th term
«r" +1 ;
" ~a'
i
■••"©■" <»
Hence the required means are of, a? 2 ,... ar n , where r has the
value found in (1).
Example. Insert 4 geometric means between 100 and 5.
We have to find 6 terms in G. P. of which 160 is the first, and 5 the
sixth.
Let r be tbe common ratio ;
tben 5 = the sixth term
= 160?' 5 ;
. 1
* ' ~32'
whence r= o'
and the means are 80, 40, 20, 10.
GEOMETRICAL PROGRESSION. o!)
55. To find the sum of a number of terms in Geometrical
Progression.
Let a be the first term, r the common ratio, n the number of
terms, and s the sum required. Then
8 = a + car + ar 2 + + ar n ~ 2 + ar"~ l ;
multiplying every term by r, we have
rs = ar + ar 2 + + ar"~ 2 + ar"" 1 + ar*,
Hence by subtraction,
rs — s = ar n — a ;
.. (rl)s = a(r"l);
,..5fe^a (i).
r  1
Changing the signs in numerator and denominator,
.?S=*3 (2).
1 r
Note. It will be found convenient to remember both forms given above
for s. using (2) in all cases except when? 1 isj^ositive and greater than 1.
Since ar'^ 1 ^ 1, the formula (1) may be written
rla
S= 7T :
a form which is sometimes useful.
2 3
Example. Sum the series  , 1, , to 7 terms.
3
The common ratio =   ; hence by formula (2)
the sum = —
(23
II
2187]
128 I
2 2315 2
~ 3 X 128 * 5
403
40 HIGHER ALGEBRA.
n 111
56. Consider the series 1, r, ^ 2 , ~ 3 ,
The sum to n terms =
2
>
H 1
.2 2
27
2 «i •
From this result it appears that however many terms be
taken the sum of the above series is always less than 2. Also we
see that, by making n sufficiently large, we can make the fraction
njr^i as sma U a s we please. Thus by taking a sufficient number
of terms the sum can be made to differ by as little as we please
from 2.
In the next article a more general case is discussed.
57. From Art. 55 we have s = \
1 r
a ar"
1 — t 1 — r '
Suppose r is a proper fraction; then the greater the value of
ar' 1
n the smaller is the value of ?•", and consequently of ; and
therefore by making n sufficiently large, we can make the sum of
n terms of the series differ from ^ by as small a quantity as
we please.
This result is usually stated thus : the sum of an infinite
number of terms of a decreasing Geometrical Progression is ^ :
1 — r
or more briefly, the sum to infinity is
a
1r'
Example 1. Find three numbers in G. P. whose sum is 19, and whose
product is 216.
Denote the numbers by , a, ar; then  x a x ar = 216 ; hence a = 6, and
r r
the numbers are  , 6, 6r.
r
GEOMETRICAL PROGRESSION. 41
6
 + 6 + 6r=19;
r
.. 613r + 6r 2 = 0;
3 2
whence r =  or  .
Thus the cumbers are 4, 6, 9.
Example 2. The sum of an infinite number of terms in G. P. is 15, and
the sum of their squares is 45 ; find the series.
Let a denote the first term, r the common ratio ; then the sum of the
(l ' ci^
terms is ; and the sum of their squares is z „ .
1  r 1 r*
Hence ,—=15 (1),
1  r
a 2
1 _ 7 2 = 45 ( 2 )
Dividing (2) by (1) ~ = 9 (3),
l + r
and from (1) and (3) z = 5;
2
whence r=x , and therefore a = 5.
an, +1 ... 10 20
Thus the series is o, — , — ,
o y
EXAMPLES. V. a.
112
1. Sum ,,,... to 7 terms.
A O 9
2. Sum 2, 2^, 3i,... to 6 terms.
3. Sum ^t, l£, 3,... to 8 terms.
4. Sum 2, 4, 8,... to 10 terms.
5. Sum 16'2, 54, 18,... to 7 terms.
6. Sum 1, 5, 25,... to p terms.
7. Sum 3, 4, — ,... to 2n terms.
o
8. Sum 1, N /3, 3,... to 12 terms.
1 8
9. Sum j , 2, jr ,... to 7 terms.
v /2 ' s '2
42 HIGHER ALGEBRA.
11 3
10. Sum ~, 3, j,. to ^ terms.
4
11. Insert 3 geometric means between 2^ and  .
13. Insert 6 geometric means between 14 and  — .
12. Insert 5 geometric means between 3f and 40.
I
64
Sum the following series to infinity :
14. , 1, ?,... 15. 45, 015, 0005,...
16. 1665, 111, 74,... 17. 3" 1 , 3~ 2 , 3',...
18. 3, v /3, 1,... 19. 7, N /42, 6,...
20. The sum of the first 6 terms of a G. P. is 9 times the sum of
the first 3 terms ; find the common ratio.
21. The fifth term of a G. P. is 81, and the second term is 24; find
the series.
22. The sum of a G. P. whose common ratio is 3 is 728, and the
last term is 486 ; find the first term.
23. In a G. P. the first term is 7, the last term 448, and the sum
889 ; find the common ratio.
24. The sum of three numbers in G. P. is 38, and their product is
1728; find them.
25. The continued product of three numbers in G. P. is 216, and
the sum of the product of them in pairs is 156 ; find the numbers.
26. If S p denote the sum of the series l+r p + r 2p +... ad inf., and
s p the sum of the series 1 — r p + r 2p  ... ad inf., prove that
/Op + Sp == ^*ij'2p'
27. If the p th , q th , r th terms of a G. P. be a, b, c respectively, prove
that a« r 6 r *c*«=l.
28. The sum of an infinite number of terms of a G. P. is 4, and the
sum of their cubes is 192 ; find the series.
58. Recurring decimals furnish a good illustration of infinite
Geometrical Progressions.
Example. Find the value of "423.
•423 =4232323
4 23 23
~ io + iooo + iooooo +
~io + IP + 10 5+ ;
• •
that,*, «3 10 , 103
GEOMETRICAL PROGRESSION. 43
4 23
23 / 1 1 \
+ io 3 V + io 2 + io* + )
_4 23 1
_ 10 + 10 :: ' _ _1_
10"
4_ 23 100
"io" 1 "!^ 3 " 99
4 _23
~~ 10 + 990
_419
" 990 '
which agrees with the value found by the usual arithmetical rule.
59. The general rule for reducing any recurring decimal to
a vulgar fraction may be proved by the method employed in the
last example ; but it is easier to proceed as follows.
Tojind the value of a recurring decimal.
Let P denote the figures which do not recur, and suppose
them j> in number; let Q denote the recurring period consisting of
q figures ; let D denote the value of the recurring decimal ; then
;
>
D = 'PQQQ
.. 10>xD = P'QQQ
and 10T+' *D = PQQQQ
therefore, by subtraction, (10 p+y  lC) D = PQP;
that is, 10" (10'  1) D = PQ  P ;
. D _ PQP
' ' (10'' 1)10'''
Now 10" 1 is a number consisting of q nines; therefore the
denominator consists of q nines followed by p ciphers. Hence
we have the following rule for reducing a recurring decimal to a
vulgar fraction :
For the numerator subtract the integral number consisting of
the nonrecurring fgures from the integral number consisting of
the nonrecurring and recurring figures ; for the denominator take
a number consisting of as many nines as there are recurring jig n n 8
followed by as many ciphers as there are nonrecurring figures.
44 HIGHER ALGEBRA.
60. To find the sum ofn terms of the series
a, (a + d) r, (a + 2d) r 2 , (a + 3d) r 3 ,
in which each term is the product of corresponding terms in an
arithmetic and geometric series.
Denote the sum by S ; then
S=a+(a + d)r+(a + 2d)r 2 + ... + (a + n~^ld)r"' ;
.. rS= ar + (a + d)r 2 + . . . +(a+ n2d)r n ~ l + (a + n ld)r n .
By subtraction,
S(l  r) = a + (dr + dr 2 + . . . + dr"" 1 )  (a + nlct) r n
dr(\r n ~ l ) ,  N „
= a + — ^ =  (a + n  Id) r ;
1 — r v ' '
a dr(lr" 1 ) _ ( a + n~^\d ) r"
•'• lr + (1r) 2 T^r '
Cor. Write S in the form
a
dr dr" (a + n~ld)r\
lr + (lry~ (lr)* T^r ;
then if r<l, we can make r" as small as we please by taking n
sufficiently great. In this case, assuming that all the terms which
involve r n can be made so small that they may be neglected, we
obtain z — + 7^ r „ for the sum to infinity. We shall refer
1r (1 r) J
to this point again in Chap. XXI.
In summing to infinity series of this class it is usually best to
proceed as in the following example.
Example 1. If x <1, sum the series
l + 2ar + 3x 2 + 4x 3 + to infinity.
Let S = l + 2a; + 3a: s + 4a s + ;
.. xS= x + 2x* + 3x*+ ;
S(lx) = l + x + x 2 + x*+
• •
1
~lx ]
• a I
GE0METK1CAL PROGRESSION. 45
Example 2. Sum the series 1 +  + , + — . + . . . to n terms.
o o 0"*
T » i ^ 7 10 Sn2
Let S== i +i+ _ + _ + + _..
1 1 4 7 3n  5 3n2
•"• 5* 5 + 52 + 53+ + 57^+ 5 ,— J
4 , /3 3 3 3 \ 3n2
1 3 A 1 1 1 \ 3n
, 3 * "
= 1 +
_2
f 5 7 " 1 " +5«»j " "~5«~
'■>i
D
N
' 3 / 1\ 3«2
= 1 + I (1 " 5 — J  5.
7 12w + 7 _
~ 4 ~ 4 . 5* ;
35 12/t+7
•'* 6 ~ 16 16 . 5" 1 '
EXAMPLES. V. b.
1. Sum 1 4 2a 4 3a 2 + 4a 3 4 . . . to n terms.
3 7 15 31
2. Sum 1 +  4 77. + ^, + c^rr. + . . • to infinity.
4 lb 64 zoo
3. Sum 1 + 3.r + 5d' 2 + 7o? + 9.z 4 + ... to infinity.
, 2 3 4
4. Sum 1 +  4  2 + 3 + . . . to n terms.
3 5 7
5. Sum 1 + + 7 + Q + ... to infinity.
2 4 o
6. Sum l + 3^ + 6 l f 2 410ji >3 4... to infinity.
7. Prove that the (n + l) th term of a G. P., of which the first term
is a and the third term b, is equal to the (2»+l) th term of a G. P. of
which the first term is a and the fifth term 6.
8. The sum of 2n terms of a G. P. whose first term is a and com
mon ratio r is equal to the sum of n of a G. P. whose first term is b and
common ratio r 1 . Prove that b is equal to the sum of the first two
terms of the first series.
46 HIGHER ALGEBRA.
9. Find the sum of the infinite series
l + (l + b)r + (l + b + b 2 )r 2 + {l + b + b 2 + b 3 )r 3 +...,
r and b being proper fractions.
10. The sum of three numbers in G. P. is 70 ; if the two extremes
be multiplied each by 4, and the mean by 5, the products are in A. P. ;
find the numbers.
11. The first two terms of an infinite G. P. are together equal to 5,
and every term is 3 times the sum of all the terms that follow it; find
the series.
Sum the following series :
12. .r+a, ,v 2 + 2<x, .r 3 + 3a. .. to u terms.
13. x (x + if) + x 2 (x 2 + y 2 ) + a? (a* 3 + if) + . . . to n terms.
14. « + o j 3«   , ha + — +... to 2p terms.
2 3 2 3 2 3
15. 3 + ^2 + 33 + 34 + 35 + p +  to mfinit y
16.
454545 . „ .,
7 ~ 72 + 73  74 + 75  76 +  to llifinit y
17. If a, b, c, d be in G. P., prove that
(b  cf + (c  a) 2 + (d  b) 2 = {a d) 2 .
18. If the arithmetic mean between a and b is twice as great as the
geometric mean, shew that a : 6 = 2 + ^/3 : 2^3.
19. Find the sum of n terms of the series the r th term of which is
(2rfl)2'\
20. Find the sum of 2n terms of a series of which every even term
is a times the term before it, and every odd term c times the term
before it, the first term being unity.
21. If S n denote the sum of n terms of a G. P. whose first term is
a, and common ratio r, find the sum of S lf S 3 , /8' 5 ,.../8 r 2B _ 1 .
22. If S v JS 2 , S 3 ,...S P are the sums of infinite geometric series,
whose first terms are 1, 2, 3,..,j2, and whose common ratios are
2' 3' 4 ' * ' ' ^Ti respectively,
prove that &\ + S 2 + S 3 + . . . + S p =f (p + 3).
23. If r < 1 and positive, and m is a positive integer, shew that
(2»i + l)r wl (lr)<lr 2wi + 1 .
Hence shew that nr n is indefinitely small when n is indefinitely great.
CHAPTER VI.
HARMON ICAL PROGRESSION. THEOREMS CONNECTED WITH
THE PROGRESSIONS.
61. DEFINITION. Three quantities a, b, c are said to be in
Harmonical Progression when  = 7 — .
c o — c
Any number of quantities are said to be in Harmonical
Progression when every three consecutive terms are in Har
monical Progression.
62. The reciprocals of quantities in Harmonical Progression
are in A rithmetical Progression.
By definition, if «, b, c are in Harmonical Progression,
a a — b
~c^~b^~c' }
.'. a(b — c) = c (a — b),
dividing every term by abc,
1111
c b b a'
which proves the proposition.
63. Harmonical properties are chiefly interesting because
of their importance in Geometry and in the Theory of Sound :
in Algebra the proposition just proved is the only one of any
importance. There is no general formula for the sum of any
number of quantities in Harmonical Progression. Questions in
H. P. are generally solved by inverting the terms, and making use
of the properties of the corresponding A. P.
48 HIGHER ALGEBRA.
64. To find the harmonic 7tiean between two given quantities.
Let a, b be the two quantities, H their harmonic mean;
then  , ~ , T are in A. P. ;
a 11 b
1 I I I
''11 a~b IV
2 11
H~ a + &'
,, 2ab
a + b
Example. Insert 40 harmonic means between 7 and ^ .
Here 6 is the 42 na term of an A. P. whose first term is  ; let d be the
common difference ; then
6 = ^ + 41d ; whence d = .
2 3 41
Thus the arithmetic means are  ,  ,  ; and therefore the har
7
monic means are 3£, 2\,...~.
*
65. If A, G> II be the arithmetic, geometric, and harmonic
means between a and b, we have proved
a + b
A = ~Y~ (!)•
G = Jab (2).
H=^ (3).
a+b v '
_, „ . Tr a + b 2ab 7 ~ 2
Therefore All = —  — . T = ab = G :
2 a+b
that is, G is the geometric mean between A and //.
From these results we see that
. ~ a + b ,z a + b 2 Jab
A  G=~Jab = g^_
HARMONICA! PROGRESSION. 49
which is positive if a and b are positive; therefore the arithmetic
mean of any two positive quantities is greater than their geometric
mean.
Also from the equation G*A1I, we see that G is inter
mediate in value between A and 11; and it lias been proved that
A > G, therefore G > II ; that is, the arithmetic, geometric, and
harmonic means between any tioo positive quantities are in descending
order of magnitude.
66. Miscellaneous questions in the Progressions afford scope
for skill and ingenuity, the solution being often neatly effected
by some special artifice. The student will find the following
hints useful.
1. If the same quantity be added to, or subtracted from, all
the terms of an A P., the resulting terms will form an A. P. with
the same common difference as before. [Art. 38.]
2. If all the terms of an A.P. be multiplied or divided by
the same quantity, the resulting terms will form an A. P., but
with a new common difference. [Art. 38.]
3. If all the terms of a G.P. be multiplied or divided by the
same quantity, the resulting terms will form a G.P. with the
same common ratio as before. [Art. 51.]
4. If a, b, c, d... are in G.P., they are also in continued pro
portion^ since, by definition,
a b c 1
bed r '
Conversely, a series of quantities in continued proportion may
be represented by x, aw, xr' 2 ,
Example 1. If a 2 , b 2 , c 2 are in A. P., shew that b + c, c + a, a + b are
in H. P.
By adding ab + ac + bc to each term, we see that
a* + ab + ac + bc, b 2 + ba + bc + ac, c' 2 + ca + cb + ab are in A.P. ;
that is {a + b) (a + c), {b + c)(b + a), (c + a) (c + b) are in A. P.
.., dividing each term by (a + b)(b + c) (c + a),
. , . are m A. P. :
b + c c + a a + b
that is, b + c, c + a, a + b are in H. P
H. H. A. 4
50 HIGHER ALGEBRA.
Example 2. If I the last term, d tlie common difference, and s the sum
of n terms of an A. P. be connected by the equation Sds={d + 2l) 2 , prove that
d = 2a.
Since the given relation is true for any number of terms, put n= 1 ; then
a = l = s.
Hence by substitution, 8ad = {d + 2a) 2 ,
or (d2ay = 0;
.. d — 2a.
Example 3. If the p th , q th , r th , s th terms of an A. P. are in G. P., shew that
p  q, q  r, r  s are in G. P.
"With the usual notation we have
a + (pl)d_ a + (gl )d_a+ (rl)d
^V(q^lJd^T¥^l)~daT(^l)d LAlt  bb  (4)J '
.*. each of these ratios
{a + (pl)d}{a+(ql)d\ _ {a+ (q 1) d]  \a + (r 1) d}
~ {a + (q  1) d\  {a+ (r 1) d) " {a+(r  1) d]  {a + {s  1) d\
= pq^qr
qr r — 8 '
Hence p  q, q  r, r  s are in G.P.
67. The numbers 1, 2, 3, are often referred to as the
natural numbers ; the n th term of the series is n, and the sum of
the first n terms is  (n +1).
68. To find the sum of the squares of the first n natural
numbers.
Let the sum be denoted by S ; then
£=l 2 + 2 2 + 3' + +n 2 .
We have n 3  (n  l) a = 3n 2  3n+ 1 ;
and by changing n into n—l,
( n _ \y _ (n _ 2) 3 = 3(™  l) 2  3(w  1) + 1 ;
similarly (w  2) 3  (71  3) 3 = 3(w  2) 2  3(n  2) + 1 ;
3 3 2 3 =3.3 2 3.3+l;
2 3 l 3 =3.2 2 3.2 + l;
1 3 0 3 =3.1 2 3.1 + 1.
THE NATURAL NUMBERS. 51
Hence, by addition,
^ 3 = 3(l 2 + 2 3 + 3 2 + ...+»')3(l + 2 + 3 + ...+n) + *
3n(n + l)
— 6b — + it.
a
. •. 3aS = n  n + ■ ± —
a
= n(n + 1) (n — 1 4 ;;);
. „ n(n+ l)(2n + l)
•• S= 6 ■
C9. To fiml the sum of the cubes of the frst n natural
numbers.
Let the sum be denoted by S ', then
£=l 3 +2 3 + 3 3 + +n\
We have n*  (n  l) 4 = 4?i 3  Gn 2 + 4n  1 ;
(n  1 ) 4  (?*  2) 4 = 4 (n  1 ) 3  6 (n  1) 8 + 4 (n  1)  1 ;
( w _ 2) 4  ( w  3)« = 4(ra 2) 3  6 (n 2) 2 + 4 (n  2)  1 ;
3 4 2 4 = 4.3 3 6.3 2 + 4.3l;
2 4 l 4 = 4.2 3 G.2 2 + 4.2l;
1 4 0 4 = 4.1 3 6.1 2 + 4.11.
Hence, by addition,
w 4 = 4#6(l 2 + 2 2 +...+?i 2 ) + 4(l +2 + .. + 7i)n;
.\ 4S = n 4 + n + 6(l 2 + 2 2 +...+7t 2 )±(\+2 + ...+n)
= n* + 7i + 7i (n+ 1) (2n+ 1) 2n(7i + 1)
= 7i (71 + 1) (?r 7i+\ + 2n+\2)
= 7i (n + 1 ) (?r + n) ;
, g _ w'(n + l) a _f W (n+l) )'
' • * " 4 ~ ( 2 J *
Tims ^Ae s?<m o/* Me cz^es of the f7'st n natural 7iumhers is
equal to the squa7'e of the siwi of these 7iumbers.
The formulae of this and the two preceding articles may be
applied to find the sum of the squares, and the sum of the cubes
of the terms of the series
«, a + d, a + 2d,
4—2
52 HIGHER ALGEBRA.
70. In referring to the results we have just proved it will
be convenient to introduce a notation which the student will fre
quently meet with in Higher Mathematics. We shall denote the
series
1 + 2 + 3 + . . . + n by 2,n ;
1* + 2* + 3* + ... +n a by %n* ;
l 3 + 2 3 + 3 3 +... +« 8 by 2n 3 ;
where 2 placed before a term signifies the sum of all terms of
which that term is the general type.
Example 1. Sum the series
1 . 2 + 2 . 3 + 3 . 4 + . . .to n terms.
The w th term=ra(n+l)=n 2 +«; and by writing down each term in a
similar form we shall have two columns, one consisting of the first n natural
numbers, and the other of their squares.
.•. the sum = 2m 2 + 2?i
_w(m+1) (2m + 1) n(n + l)
T
6 ' 2
n(n+l) j2n+l )
n(n + l)(n+2)
3
Example 2. Sum to n terms the series whose M th term is 2' 1_1 + 8m 3  6m 2 .
Let the sum be denoted by S ; then
S = 2 2» 1 + 82>i 3 62n 2
_ 2" 1 8m 2 (m + 1) 2 _ 6m ( m+1)(2m + 1)
" 2  1 + ~~ 4~ 6
= 2»  1 +« (m + 1) {2m (m + 1)  (2m + 1) }
= 2' l l + n(n + l)(2n 2 l).
EXAMPLES. VI. a.
1. Find the fourth term in each of the following series :
(1) 2, 2J, 3i,...
(2) 2, 21, 3,...
(3) 2, 2f, 3i,...
2. Insert two harmonic means between 5 and 11.
2 2
3. Insert four harmonic means between  and — .
o 1«3
EXAMPLES OX THE riiOGltESSlOXS. 53
4. If 12 and 9 : l are the geometric and harmonic means, respect
ively, between two numbers, find them.
5. If the harmonic mean between two quantities is to their geo
metric means as 12 to 13, prove that the quantities are in the ratio
of 4 to 9.
6. If a, b, c be in H. P., shew that
a : a — b = a + c : a — c.
7. If the iii lh term of a H. P. be equal to n, and the u lh term be
equal to m, prove that the (m + n) th term is equal to
m n
m + n
8. If the p th , <7 th , r th terms of a H. P. be a, b, c respectively, prove
that (<j  r) be + (r — p) ca + (pq) ub = 0.
9. If b is the harmonic mean between a and c, prove that
1 111
j + , =  +  .
o — a b — c a c
Find the sum of n terms of the series whose n th term is
10. 3n*n. 11. n s +^n. 12. »(»+2).
13. » 2 (2»+3). 14. 3" 2". 15. 3 (4' l + 2; i 2 )4/i :J .
16. If the (m+iy\ (?^+l) th , and (r+ l) th terms of an A. P. are in
( i. P., and on, n y r are in H. P., shew that the ratio of the common
2
difference to the first term in the A. P. is — .
n
17. If I, m, n are three numbers in G. P., prove that the first term
of an A. P. whose £ th , m th , and ?i th terms are in H. P. is to the common
difference as m\\ to 1.
18. If the sum of n terms of a series be a + bu + cri 2 , find the n th
term and the nature of the series.
19. Find the sum of n terms of the series whose n th term is
4?i(?i 2 +l)(6>i 2 fl).
20. If between any two quantities there be inserted two arithmetic
means A u A ; two geometric means G ly G 2 ; and two harmonic means
H 1 , 7/ 2 ; shew that 6^0',, : II 1 H. 2 = A l + A 2 : I^ + IL,.
21. If p be the first of n arithmetic means between two numbers,
and q the first of n harmonic means between the same two numbers,
prove that the value of q cannot lie between p and f — J p.
22. Find the sum of the cubes of the terms of an A. P., and shew
that it is exactly divisible by the sum of the term.
54) HIGHER ALGEBRA.
Piles of Shot and Shells.
71. To find the number of shot arranged in a complete
pyramid on a square base.
Suppose that each side of the base contains n shot ; then the
number of shot in the lowest layer is n 2 \ in the next it is (n—l) 2 ;
in the next (n2) 2 ; and so on, up to a single shot at the
top.
.. S^n 2 + (nl) 2 + (n2) 2 +... + l
= n(n+l)(2n + l)
6
72. To find the number of shot arranged in a complete
pyramid the base of which is an equilateral triangle.
Suppose that each side of the base contains n shot ; then the
number of shot in the lowest layer is
n + (n  1) + (n  2) + + 1 j
xi • n(n + 1) 1 , 2
that is, — V« or  [n + n) .
— —
In this result write n — 1, » — 2, for n, and we thus obtain
the number of shot in the 2nd, 3rd, layers.
.. S=i($n* + 2,n)
M >(n + l)(» + 2) [Art7a]
73. To find the number of shot arranged in a complete
pyramid the base of which is a rectangle.
Let m and n be the number of shot in the long and short side
respectively of the base.
The top layer consists of a single row of m — (n — l), or
m — n+1 shot ;
in the next layer the number is 2 (in — n + 2);
in the next layer the number is 3 (in — n + 3) ;
and so on ;
in the lowest layer the number is n (m — n + n).
PILES OF SHOT AND SHELLS. 55
.. S= (m 01 + 1) + *2(mn + 2) + 3(ww + 3) + ... +n(rnn + n)
= (m  n) (1 + 2 + 3 + ... + n) + (l 2 + 2 2 + 3 s + ... + n 2 )
(wi  n) n (n + 1 ) w (n + 1) (2n +1)
2 + 6
= n ( n + 1 ){3(mn) + 2n + l}
_n(n + l) (3m n+ 1)
= 6 '
74. To find the number of shot arranged in an incomplete
2>yramid the base of which is a rectangle.
Let a and b denote the number of shot in the two sides of the
top layer, n the number of layers.
In the top layer the number of shot is ab ;
in the next layer the number is (a + 1) (6 + 1) ;
in the next layer the number is (a + 2) (b + 2) \
and so on ;
in the lowest layer the number is (« + n  1) (b + n — 1)
or ab + (a + b)(nl) + ()il) 2 .
.. S = abn + (a + 6) % (n 1) + % (n Vf
= abn + ("l)w(a + 6) + ( nl)n(2 .nl + 1 )
2 6
=  {6ab + 3 (a + b) (n  1) + (w  1) (2m  1)}.
75. In numerical examples it is generally easier to use the
following method.
Example. Find the number of shot in an incomplete square pile of 16
courses, having 12 shot in each side of the top.
If we place on the given pile a square pile having 11 shot in each side of
the base, we obtain a complete square pile of 27 courses;
and number of shot in the complete pile = ^ = <)'.)30 ; [Art. 7 1 .]
11 x 12 x 23
also number of shot in the added pile= „ = 506;
.*. number of shot in the incomplete pile =6424.
56 HIGHER ALGEBRA.
EXAMPLES. VI. b.
Find the number of shot in
1. A square pile, having 15 shot in each side of the base.
2. A triangular pile, having 18 shot in each side of the base.
3. A rectangular pile, the length and the breadth of the base con
taining 50 and 28 shot respectively.
4. An incomplete triangular pile, a side of the base having 25 shot,
and a side of the top 14.
5. An incomplete square pile of 27 courses, having 40 shot in each
side of the base.
6. The number of shot in a complete rectangular pile is 24395 ; if
there are 34 shot in the breadth of the base, how many are there in its
length ?
7. The number of shot in the top layer of a square pile is 169,
and in the lowest layer is 1089 ; how many shot does the pile contain ?
8. Find the number of shot in a complete rectangular pile of
15 courses, having 20 shot in the longer side of its base.
9. Find the number of shot in an incomplete rectangular pile,
the number of shot in the sides of its upper course being 11 and 18,
and the number in the shorter side of its lowest course being 30.
10. What is the number of shot required to complete a rectangular
pile having 15 and 6 shot in the longer and shorter side, respectively, of
its upper course?
11. The number of shot in a triangular pile is greater by 150 than
half the number of shot in a square pile, the number of layers in each
being the same; find the number of shot in the lowest layer of the tri
angular pile.
12. Find the number of shot in an incomplete square pile of 16
courses when the number of shot in the upper course is 1005 less than
in the lowest course.
13. Shew that the number of shot in a square pile is onefourth the
number of shot in a triangular pile of double the number of courses.
14. If the number of shot in a triangular pile is to the number of
shot in a square pile of double the number of courses as 13 to 175 ; find
the number of shot in each pile.
15. The value of a triangular pile of 16 lb. shot is ,£51 ; if the
value of iron be 10s. 6d. per cwt., find the number of shot in the
lowest layer.
16. If from a complete square pile of n courses a triangular pile of
the same number of courses be formed ; shew that the remaining shot
will be just sufficient to form another triangular pile, and find the
number of shot in its side.
\
CHAPTER VII.
SCALES OF NOTATION.
76. The ordinary numbers with which we are acquainted in
Arithmetic are expressed by means of multiples of powers of 10;
for instance
252 x 10 + 5;
4705 = 4 x 10 3 + 7 x 10 2 + x 10 + 5.
This method of representing numbers is called the common or
denary scale of notation, and ten is said to be the radix of the
scale. The symbols employed in this system of notation are the
nine digits and zero.
In like manner any number other than ten may be taken as
the radix of a scale of notation ; thus if 7 is the radix, a number
expressed by 2453 represents 2x7 3 + 4x7" + 5x7 + 3; and in
this scale no. digit higher than 6 can occur.
Again in a scale whose radix is denoted by r the above
number 2453 stands for 2r 3 + 4? ,2 + hr + 3. More generally, if in
the scale whose radix is r we denote the digits, beginning with
that in the units' place, by a tt , a,, a 2 ,...aj then the number so
formed will be represented by
a r n + a ,r n ~ 1 + a y~~ + . . . + a/ 2 + a,r + a,
where the coefficients a , a ,,...«,. are integers, all less than r, of
which any one or more after the first may be zero.
Hence in this scale the digits are r in number, their values
ranging from to r — 1 .
77. The names Binary, Ternary, Quaternary, Quinary, Senary,
Septenary, Octenary, Nonary, Denary, Undenarv, and Duodenary
are used to denote the scales corresponding to the values fae»,
three,... twelve of the radix.
58 HIGHER ALGEBRA.
In the undenary, duodenary, . . . scales we shall require symbols
to represent the digits which are greater than nine. It is unusual
to consider any scale higher than that with radix twelve ; when
necessary we shall employ the symbols t, e, T as digits to denote
' ten ', ' eleven ' and ' twelve '.
It is especially worthy of notice that in every scale 10 is the
symbol not for ' ten ', but for the radix itself.
78. The ordinary operations of Arithmetic may be performed
in any scale ; but, bearing in mind that the successive powers of
the radix are no longer powers of ten, in determining the carrying
figures we must not divide by ten, but by the radix of the scale
in question.
Example 1. In the scale of eight subtract 371532 from 530225, and
multiply the difference by 27.
530225 136473
371532 27
136473 1226235
275166
4200115
Explanation. After the first figure of the subtraction, since we cannot
take 3 from 2 we add 8 ; thus we have to take 3 from ten, which leaves 7 ; then
6 from ten, which leaves 4 ; then 2 from eight which leaves 6 ; and so on.
Again, in multiplying by 7, we have
3x7 = twenty one = 2x8 + 5;
we therefore put down 5 and carry 2.
Next 7x7 + 2 = fifty one = 6x8 + 3;
put down 3 and carry 6 ; and so on, until the multiplication is completed.
In the addition,
3 + 6 = nine = lx8 + l;
we therefore put down 1 and carry 1.
Similarly 2 + 6 + l = nine=l x 8 + 1;
and 6 + l + l = eight = lx8 + 0;
and so on.
Example 2. Divide 15et20 by 9 in the scale of twelve.
9)15<?£20
lee96...G.
Explanation. Since 15 = 1 x T + 5 = seventeen = 1 x9 + 8,
we put down 1 and carry 8.
Also 8 x T + e = one hundred and seven = e x 9 + 8 ;
we therefore put down e and carry 8; and so on.
SCALES OF NOTATION. 59
Example 3. Find the .square root of 442641 in the scale of seven.
134
442641(646
34
1026
G02
1416112441
12441
EXAMPLES. Vila.
1. Add together 23241, 4032, 300421 in the scale of five.
2. Find the sum of the nonary numbers 303478, 150732, 2G4305.
3. Subtract 1732765 from 3673124 in the scale of eight.
4. From 3^756 take 2e46t2 in the duodenary scale.
5. Divide the difference between 1131315 and 235143 by 4 in the
scale of six.
6. Multiply 6431 by 35 in the scale of seven.
7. Find the product of the nonary numbers 4685, 3483.
8. Divide 102432 by 36 in the scale of seven.
9. In the ternary scale subtract 121012 from 11022201, and divide
the result by 1201.
10. Find the square root of 300114 in the quinary scale.
11. Find the square of tttt in the scale of eleven.
12. Find the G. C. M. of 2541 and 3102 in the scale of seven.
13. Divide 14332216 by 6541 in the septenary scale.
14. Subtract 20404020 from 103050301 and find the square root of
the result in the octenary scale.
15. Find the square root of ce^OOl in the scale of twelve.
16. The following numbers are in the scale of six, find by the ordi
nary rules, without transforming to the denary scale :
(1) the G. C. M. of 31141 and 3102 ;
(2) the L. C. M. of 23, 24, 30, 32, 40, 41, 43, 50.
79. To express a given integral number in any proposed scale.
Let iV be the given number, and r the radix of the proposed
scale.
Let a u , a n a 2 ,...a t be the required digits by which iV is to be
expressed, beginning with that in the units' place; then
N= a r" + a ,r" _1 + ... + ar~ + a.r + a lt .
n n — 1 2 10
We have now to find the values of « , a,, "_...",,
60
HIGHER ALGEBRA.
Divide N by r, then the remainder is a , and the quotient is
a r" l + a
n 71—1
r n 2 + ... +a r + a l .
If this quotient is divided by r, the remainder is a i ;
if the next quotient a 2 ;
and so on, until there is no further quotient.
Thus all the required digits a , a x , a g1 ...a n are determined by
successive divisions by the radix of the proposed scale.
Example 1. Express the denary number 5213 in the scale of seven.
7)5213
7)7447 5
7)106. 2
7 )15. 1
2 1
Thus 5213 = 2x7 4 +lx7 3 + lx7 + 2x7 + 5;
and the number required is 21125.
Example 2. Transform 21125 from scale seven to scale eleven.
e)21125
e)1244T t
~e)Gl
3. t
.. the required number is 3t0t.
Explanation. In the first line of work
21 = 2x7+l=fifteen = lx<? + 4;
therefore on dividing by e we put down 1 and carry 4.
Next 4x7 + 1 = twenty nine = 2 x e + 7 ;
therefore we put down 2 and carry 7 ; and so on.
Example 3. Reduce 7215 from scale twelve to scale ten by working in
scale ten, and verify the result by working in the scale twelve.
r 7215 f)7215 1
JL2 «)874. 1
.4
.2
In scale
of ten

80
12
1033
12
t)t^.
t)10.
1.
In scale
of twelve
J
1 12401
Thus the result is 12401 in each case.
Explanation. 7215 in scale twelve means 7 x 12 3 + 2 x 12 2 + 1 x 12 + 5 in
scale ten. The calculation is most readily effected by writing this expression
in the form [{(7 x 12 + 2) } x 12 + 1] x 12 + 5 ; thus we multiply 7 by 12, and
add 2 to the product; then we multiply 86 by 12 and add 1 to the product;
then 1033 by 12 and add 5 to the product.
SCALES OF NOTATION. f>l
80. Hitherto we have only discussed whole numbers; but
fractions may also be expressed in any scale of notation ; thus
2 5
•25 in scale ten denotes — + — , ;
10 10*'
2 5
•25 in scale six denotes = + ^ ;
G
2 5
•25 in scale r denotes — h — .
r r
Fractions thus expressed in a form analogous to that of
ordinary decimal fractions are called radixfractions, and the point
is called the radixpoint. The general type of such fractions in
scale r is
~ 3 ~ 3 ~ >
r r r
where 6 , b 2 , 6 a , ... are integers, all less than r, of which any one
or more may be zero.
81. To express a given radix fraction in any proposed scale.
Let F be the given fraction, and r the radix of the proposed
scale.
Let b , b , b 3 ,... be the required digits beginning from the
left ; then
F JX + b A+ b ^ 3+
r r r
We have now to find the values of 6 p b 2 , 6 3 ,
Multiply both sides of the equation by r ; then
rF=b+ 2 + h l+ ;
Hence b l is equal to the integral part of rF ; and, if we denote
the fractional part by F x , we have
Hi. + J +
Multiply again by r\ then, as before, b is the integral part
of rF x ; and similarly by successive multiplications by r, each of
the digits may be found, and the fraction expressed in the pro
posed scale.
62 HIGHER ALGEBRA.
If in the successive multiplications by r any one of the
products is an integer the process terminates at this stage, and
the given fraction can be expressed by a finite number of digits.
But & if none of the products is an integer the process will never
terminate, and in this case the digits recur, forming a radix
fraction analogous to a recurring decimal.
13
Example 1. Express ^ as a radix fraction in scale six.
13 ft 13x3 7.
16 x6= 8 = 4 + 8'
7 a 7x3 Kj. 1
1 a lx3 I.. 1
^x6 = 3.
4 5 13
.. the required fraction = g + ^ + p + Qi
= •4513.
Example 2. Transform 1606424 from scale eight to scale five.
We must treat the integral and the fractional parts separately,
5 )16064 '24
5 )2644 ... 5
5)440... 4 1*44
5)71. ..3 J>_
5)13... 2 264
2...1 £_
404
5_
024
After this the digits in the fractional part recur; hence the required
number is 2123401240.
82. In any scale of notation of which the radix is r, the sum
of the digits of any whole number divided by r  1 will leave the
same remainder as the whole number divided by r — 1.
Let N denote the number, a , a lt a 2 , a n the digits begin
ning with that in the units' place, and S the sum of the digits;
then
N = a Q + a x r + a 2 r 2 + + a„_/ , ~ 1 + ar n ;
S=a + a x +a 2 + + a n _ l + a n
r .tfS=a 1 (rl) + a 2 (r°l)+ + •„_, (i*  1) + «, (f  1).
SCALES OF NOTATION. 03
Now every term on the right hand side is divisible by r — 1 •
iVS
. * . = — an integer ;
r 1 6 y
that is, =/ +
r  1 r  1 '
when; / is some integer j which proves tlie proposition.
Hence a number in scale r will be divisible by ?• — 1 when the
sum of its digits is divisible by r — 1.
83. By taking ?=10 we learn from the above proposition
that a number divided by 9 will leave the same remainder as the
sum of its digits divided by 9. The rule known as " casting out
the nines " for testing the accuracy of multiplication is founded
on this property.
The rule may be thus explained :
Let two numbers be represented by da + b and 9c f d, and
their product by P; then
P^Slac + %c + 9ad + bd.
Hence — has the same remainder as ^ ; and therefore the
s?nn of the digits of /*, when divided by 9, gives the same
remainder as the sum of the digits of bd, when divided by 9. If
on trial this should not be the case, the multiplication must have
been incorrectly performed. In practice b and d are readily
found from the sums of the digits of the two numbers to be
multiplied together.
Example. Can the product of 31256 and 8127 be 263395312 ?
The sums of the digits of the multiplicand, multiplier, and product are 17,
21, and 31 respectively; again, the sums of the digits of these three numbers
are 8, 3, and 7, whence M = 8x3 = 24, which has 6 for the sum of the
digits; thus we have two different remainders, 6 and 7, and the multiplication
is incorrect.
84. If N denote any number in the scale of 'r, and D denote
the difference, supposed positive, between the sums of the digit* in the
odd and the even places; then N — D or N + D is a multiple, of
r+ 1.
64 HIGHER ALGEBRA.
Let a , «!, a , a n denote the digits beginning with that
in the units' place; then
JV= a + a,r + a r 2 + ar* + + a ,r" 1 + a r".
.. A r a + a A a 2 + a 3  ...=<*, (r+1) + « 2 (r 2  1) + a 3 (r 3 + 1) + ...;
and the last term on the right will be a w (r"+l) or a n (r n — 1)
according as n is odd or even. Thus every term on the right is
divisible by r + I ; hence
^ ! % i ' = an mteoer.
r + 1 °
NOW a 0~ a } +fl 2~ CC 3 + —^D)
.'. =— is an integer;
which proves the proposition.
Cor. If the sum of the digits in the even places is equal to
the sum of the digits in the odd places, D = 0, and N is divisible
by r + 1.
Example 1. Prove that 4 •41 is a square number in any scale of notation
whose radix is greater than 4.
Let r be the radix ; then
441 = 4 +  + i=(2 + Y;
r r z \ rj
thus the given number is the square of 2*1.
Example 2. In what scale is the denary number 24375 represented by
213?
Let r be the scale ; then
13 7
2 +  + 2=24375=2^;
r r 16
whence " 7r 2  16r48 = ;
that is, (7r+12)(/4) = 0.
Hence the radix is 4.
Sometimes it is best to use the following method.
Example 3. In what scale will the nonary number 25607 be expressed
by 101215 ?
The required scale must be less than 9, since the new number appears
the greater ; also it must be greater than 5 ; therefore the required scale
must be 6, 7, or 8; and by trial we find that it is 7.
SCALES OF NOTATION. 65
Example 4. By working in the duodenary scale, find the height of a
rectangular solid whose volume is 364 cub. ft. 1048 cub. in., and the area of
whose base is 46 sq. ft. 8 sq. in.
The volume is 364^i? cub. ft., which expressed in the scale of twelve is
264734 cub. ft.
The area is 46^ 4 sq. ft., which expressed in the scale of twelve is 3<08.
We have therefore to divide 264*734 by StOS in the scale of twelve.
3*08)264734(7e
22*48
36274
36274
Thus the height is 7ft. lliu.
EXAMPLES. VII. b.
1. Express 4954 in the scale of seven.
2. Express 624 in the scale of five.
3. Express 206 in the binary scale.
4. Express 1458 in the scale of three.
5. Express 5381 in powers of nine.
6. Transform 212231 from soale four to scale five.
7. Express the duodenary number 398e in powers of 10.
8. Transform 6£12 from scale twelve to scale eleven.
9. Transform 213014 from the senary to the nonary scale.
10. Transform 23861 from scale nine to scale eight.
11. Transform 400803 from the nonary to the quinary scale.
12. Express the septenary number 20665152 in powers of 12.
13. Transform ttteee from scale twelve to the common scale.
3
14. Express — as a radix fraction in the septenary scale.
15. Transform 17 "15625 from scale ten to scale twelve.
16. Transform 200 "2 11 from the ternary to the nonary scale.
17. Transform 71*03 from the duodenary to the octenary scale.
1552
18. Express the septenary fraction — — as a denary vulgar fraction
in its lowest terms.
19. Find the value of *4 and of '42 in the scale of seven.
20. In what scale is the denary number 182 denoted by 222?
25
21. In what scale is the denary fraction — denoted by 0302?
H. H. A. 5
66 HIGHER ALGEBRA.
22. Find the radix of the scale in which 554 represents the square
of 24.
23. In what scale is 511197 denoted by 1746335 ?
24. Find the radix of the scale in which the numbers denoted by
479, 698, 907 are in arithmetical progression.
25. In what scale are the radixfractions *16, "20, '28 in geometric
progression ?
26. The number 212542 is in the scale of six; in what scale will it
be denoted by 17486?
27. Shew that 148'84 is a perfect square in every scale in which the
radix is greater than eight.
28. Shew that 1234321 is a perfect square in any scale whose radix
is greater than 4 ; and that the square root is always expressed by the
same four digits.
29. Prove that 1331 is a perfect cube in any scale whose radix is
greater than three.
30. Find which of the weights 1, 2, 4, 8, 16,... lbs. must be used to
weigh one ton.
31. Find which of the weights 1, 3, 9, 27, 81,... lbs. must be used
to weigh ten thousand lbs., not more than one of each kind being used
but in either scale that is necessary.
32. Shew that 1367631 is a perfect cube in every scale in which the
radix is greater than seven.
33. Prove that in the ordinary scale a number will be divisible by
8 if the number formed by its last three digits is divisible by eight.
34. Prove that the square of rrrr in the scale of s is rm^OOOl, where
q, r, s are any three consecutive integers.
35. If any number N be taken in the scale ?*, and a new number N'
be formed by altering the order of its digits in any way, shew that the
difference between N and N' is divisible by r — 1.
36. If a number has an even number of digits, shew that it is
divisible by r+1 if the digits equidistant from each end are the same.
37. If in the ordinary scale S t be the sum of the digits of a number
JV, and 3# 2 be the sum of the digits of the number 3iV, prove that the
difference between aS^ and >S' 2 is a multiple of 3.
38. Shew that in the ordinary scale any number formed by
writing down three digits and then repeating them in the same order
is a multiple of 7, 11, and 13.
39. In a scale whose radix is odd, shew that the sum of the
digits of any number will be odd if the number be odd, and even if
the number be even.
40. If n be odd, and a number in the denary scale be formed
by writing down n digits and then repeating them in the same order,
shew that it will be divisible by the number formed by the n digits,
and also by 9090... 9091 containing n \ digits.
CHAPTER VIII.
SURDS AND IMAGINARY QUANT1T1KS.
85. In the Elementary Algebra, Art. 272, it is proved that
the denominator of any expression of the form rr r can be
Jb + Jc
rationalised by multiplying the numerator and the denominator
by Jb — Jc, the surd conjugate to the denominator.
Similarly, in the case of a fraction of the form
Jb + Jc + Jd '
where the denominator involves three quadratic surds, we may by
two operations render that denominator rational.
For, first multiply both numerator and denominator by
Jb + Jc — Jd; the denominator becomes (Jb + Jc) 2 — (Jd) 2 or
b + c  d + 2 J be. Then multiply both numerator and denominator
by (b + c  d) — 2 J be; the denominator becomes (b + c  d) 2 — Abe,
which is a rational quantity.
Example. Simplify
12
3+^/52^/2
The expression  12 ( 3 + y/5 + V*)
me expression _ (3+ ^ 5)S _ (V2 )»
^ 12(3 + ^/5 + 2^/2 )
6 + 6^5
2 (3 + v /5+ 2^ /2)^51)
U/5+l)U/5l)
2+V5+V10V2
= 1+^5+^/10^2.
5—2
68 HIGHER ALGEBRA.
86. To find the factor which will rationalise any given bino
mial surd.
Case I. Suppose the given surd is £]a  $b.
Let ZJa = x, ?Jb = y, and let n be the l.c.m. of p and q ; then
x n and y n are both rational.
Now x n — y n is divisible by x  y for all values of n, and
ar _ y" = (xy) {x n ~' +x n ~ 2 y + x n Sf + + y"" 1 ).
Thus the rationalising factor is
.X'"" 1 + x n ~ 2 y + x n ~y + + y"~ l ;
and the rational product is x n — y'\
Case II. Suppose the given surd is p Ja + fjb.
Let x, y, n have the same meanings as before; then
(1) If n is even, x n — y" is divisible by x + y, and
x n  y n = (x + y) (x n ~ l  x n ~ 2 y + + xf~*  y"" 1 ).
Thus the rationalising factor is
ur l ary + + ay"' 3 r 1 ;
and the rational product is x" — y n .
(2) If n is odd, x" + y" is divisible by x + y, and
x n + y n = (x + y) (x n ~ x  x n Sj +  xy n ~ 2 + y n ~').
Thus the rationalising factor is
x n ~ 1 x n ~ 2 y+ xy n  2 + y" 1 ;
and the rational product is x" + y n .
Example 1. Find the factor which will rationalise ^/S + ^/5.
i i
Let x = 3 2 , y = 5 5 ; then x b and y 6 are both rational, and
x e  y 6 — (x + y) (x 5  x*y + xhj 2  xhj 3 + xy i  y 5 ) ;
thus, substituting for x and y, the required factor is
541 3223 14 5
32  32 . 53+ 32 . 53  32 . 53+ 32 . 53  53,
5 13 2 14 5
or 329. 5 5 + 32~. 5315 + 32. 535 5 ;
6 6
and the rational product is 32  5 s " = 3 3  5 2 = 2.
SURDS AND IMAGINARY QUANTITIES. 69
Example 2. Express (&+&) * \5 5 9 s )
as an equivalent fraction with a rational denominator.
i i i
To rationalise the denominator, which is equal to 5" 3*, put 5 2 = x,
3 4 = y ; then since x 4 y A = [x  y) (x* + xhj + xy 2 + if')
3 2 1 12 3
the required factor is 5 + 5 2 . 3* + 5 ] . 3* + 3~ 4 ;
4 4
and the rational denominator is 5 2  3* = 5 2  3 = 22.
/ i i\ / 3 2 1 12 3\
+Vl • V 5 5 + 3 V \& + 5 2 . 3^ + 5 . 3~ 4 + 3 V
.•. the expression = ' — i — ' — '
4 3 1 2 2 13 4
_ 5 2 + 2 . 5 2 .3 j + 2. 5 ,J .3 4 "+2.5 5 .3 j + 3 t
22
3 1 113
_l l + o 2 . 3 j + 5.3 2 +5 2 .3 j
11
87. We have shewn in the Elementary Algebra, Art. 277,
how to find the square root of a binomial quadratic surd. We
may sometimes extract the square root of an expression contain
ing more than two quadratic surds, such as a + Jb + Jc 4 Jd.
Assume Ja + Jb + Jc + Jd = Jx + Jy + Jz j
.'. a + Jb + Jc + Jd = x + y + z + 2 Jxy + 2 Jxz + 2 Jyz.
If then 2 Jxy = Jb, 2 Jxz = Jc, 2 Jyz = Jd,
and if, at the same time, the values of x, y, z thus found satisfy
x + y + z = a, we shall have obtained the required root.
Example. Find the square root of 21  4^/5 + 8^/3  4^/15.
Assume V 21  V 5 + V 3  V 15 = sl x + Jy  sl z \
.'. 21  4^5 + 8^/3  4J15 = x + y + z + 2 Jxy 2 Jxz  2 Jyz.
Put 2jxy = 8JS, 2jxl = 4J15, 2jyz = ±Jo\
by multiplication, xyz = 240 ; that is Jxyz=4 s /lo ;
whence it follows that ,Jx = 2j3, Jy = 2, „Jz = s /5.
And since these values satisfy the equation x + y + z = 21, the required
root is 2^3 + 2^/5.
70 HIGHER ALGEBRA.
88. If J a, + Jh = x + Jy, then ivill J a, — Jh = x — Jy,
For, by cubing, we obtain
ci + Jb=x 3 + 3x 2 ^/y + 3xy + y Jy.
Equating rational and irrational parts, we have
a = x 3 + 3xy, Jb = 3x 2 Jy + y Jy \
.'. a Jb = x 3  3x 2 Jy + 3xy y Jy;
that is, J a  Jb = x— Jy.
Similarly, by the help of the Binomial Theorem, Chap. XIII. ,
it may be proved that if
Ja + Jb = x + Jy, then J a  Jb = x Jy,
where n is any positive integer.
89. By the following method the cube root of an expression
of the form a ± Jb may sometimes be found.
Suppose Ja + Jb = x + Jy ;
then Ja Jb = x Jy.
.. Jtf^b=x 2 y (1).
Again, as in the last article,
a = x 3 + 3xy (2).
The values of x and y have to be determined from (1) and (2).
In (1) suppose that J a 2 — b=c; then by substituting for y in
(2) we obtain
a = x 3 + 3x (x 2 — c) ;
that is, kx 3 — 3cx — a.
If from this equation the value of x can be determined by
trial, the value of y is obtained from y = x 2 — c.
Note. We do not here assume sjx + s ly for the cube root, as in the
extraction of the square root; for with this assumption, on cubing we should
have
a + Jb = xjx + Sxjy + Syjx + yjy y
and since every term on the right hand side is irrational we cannot equate
rational and irrational parts.
SUKDS AND IMAGINARY QUANTITIES. 71
Example. Find the cube root of 72  32 x /5.
Assume sf 72 '62^5 = x  ^/y ;
then ^72 + 32 s /5 = x + s /y.
By multiplication , ^5184  1024 x 5 = a; 2  y ;
that is, ± = x'y (1).
Again 72  32^/5 = .c 3  fkcPJy + Sxy  y^'y ;
whence 72 = x 3 + 3.t// (2).
From (1) and (2) , 72 = x :i + Sx (x  4) ;
that is, ar } 3x = 18.
By trial, we find that x = S; hence y = o, and the cube root is 3^/5.
90. When the binomial whose cube root we are seekhi"
consists of two quadratic surds, we proceed as follows.
Example. Find the cube root of 9 N /3 + ll v /2.
By proceeding as in the last article, we find that
.. the required cube root =*J3 ( 1 + * / J
= v /3+v/2.
91. We add a few harder examples in surds.
4
'Example 1. Express with rational denominator
N V9^3 + r
4
The expression = — ^
3 fj  3 3 + 1
(J + l)
l3 3 + l) (3 3 3 ri + l)
iM±i]3 3 + l
3 + 1 ~ d +1 "
72 HIGHER ALGEBRA.
Example 2. Find the square root of
l(xl) + f j2x <i 7xi.
The expression = \ {3x  3 + 2 J(2x + l)(x4) }
a
= ±{(2x + l) + {x±) + 2jc2x + l){x£)};
hence, by inspection, the square root is
Example 3. Given ^5 = 223607, find the value of
(5
J2 + J7 3J5'
Multiplying numerator and denominator by >J2,
^62^/5
the expression
2 + ^146^/5
n/51
2 + 3^/5
EXAMPLES. Villa.
Express as equivalent fractions with rational denominator
i 1 2 ^
L  1 + V2V3' A J2+J3J5'
3. 1  = . 4 ^^
sfa + s/b + s/a + b' *Jal\/2a + *Ja + l
^10 + ^5^/3 fi (j3 + x /5)(j5 + ^/2)
Find a factor which will rationalise :
i i
7. #3 a/2. 8. ^/5 + ^/2. 9. 06+6*.
10. N 3 /3l. 11. 2 + 4/7. 12. 4/5^3.
SUKDS AND IMAGINARY QUANTITIES. 73
Express with rational denominator:
16 * /3 17 v 8 + ^ 4 i« W
Find the square root of
19. 162 N /202 v /28 + 2 N /l3.>. 20. 24+4^154^212^35.
21. G + ,/12^24,/8. 22. 5 x /10 N /15 + N /G.
23. a+36+4+4^/a4^62V3oS
24. 21+3 N /8  6 N /3  (5 Jl  v '24  N /56 + 2 N /21.
Find the eube root of
25. 10+6 JZ. 26. 38 + 17^5. 27. 9970^/2.
28. 38^14100^2. 29. 54^3 + 41^5. 30. 135^387^6.
Find the square root of
31. a + x + \J%ax + x 2 . 32. 2a  \/3a 2  2ab  b' 2 .
i i
33. l + « 2 + (l+a 2 + a 4 ) 2 . 34. l+(l« 2 )" 2 .
35. If a = — —j , 6 = — i— , find the value of 7« 2 + 1 1 ab  lb' 2 .
36. If x = jl'jl , y = S /z7^ » find the value of 3<t " 2 ~ 5j y + 3 ^
Find the value of
V2615J3 / 6 + 2V3
5V2V38T573' V3319 N /3
1 1 2
39. (28  10 N /3) 2  (7 + 4 v /.3) " 2 . 40. (26 + 15 s fzf  (26 + 15 N /3)
41. Given s /b = 223607, find the value of
lOx/2 N /10 + N /18
N /18  a/3+V5 N /8 + V3  V5 '
42. Divide x*+ 1 + 3# #2 by *  1 + */2.
43. Find the cube root of 9a6 2 + (b 2 + 24a 2 ) ^6* 3a 8 .
44. Evaluate V^' 2  1 , when 2.r; = Ja+ i ,
xs/x*\ \'"
■_>
3
74 HIGHER ALGEBRA.
Imaginary Quantities.
92. Although from the rule of signs it is evident that a
negative quantity cannot have a real square rootlet imaginary
quantities represented by symbols of the form J a, J 1 are of
frequent occurrence in mathematical investigations, and their
use leads to valuable results. We therefore proceed to explain
in what sense such roots are to be regarded.
When the quantity under the radical sign is negative, we can no
longer consider the symbol J as indicating a possible arithmetical
operation ; but just as J a may be defined as a symbol which obeys
the relation J a x Ja = a, so we shall define J— a to be such that
J— a x J— a =  a, and we shall accept the meaning to which this
assumption leads us.
It will be found that this definition will enable us to bring
imaginary quantities under the dominion of ordinary algebraical
rules, and that through their use results may be obtained which
can be relied on with as much certainty as others which depend
solely on the use of real quantities.
93. By definition, J I x J I =  1.
.. Ja.Jl x Ja. Jl^a^l);
that is, ( J a . J 1) 2 =  a.
Thus the product J a . J— 1 may be regarded as equivalent to
the imaginary quantity J— a.
94. It will generally be found convenient to indicate the
imaginary character of an expression by the presence of the
symbol J 1 ; thus
JZjtf = Jja 2 x (  1) = a J7 JT.
95. We shall always consider that, in the absence of any
statement to the contrary, of the signs which may be prefixed
before a radical the positive sign is to be taken. But in the use
of imaginary quantities there is one point of importance which
deserves notice.
SURDS AND IMAUiNAltY QUANTITIES. 75
Since ( a) x ( b) — ab,
by taking the square root, we have
J a x J b = ± Jab.
Thus in funning the product of J a and J— b it would appear
that either of the signs + or — might be placed before Jab.
This is not the case, for
J a x J b = J a . J I x \/b . Jl
=  Jab.
96. It is usual to apply the term c imaginary ' to all expres
sions which are not wholly real. Thus a+bj—l may be taken
as the general type of all imaginary expressions. Here a and b
are real quantities, but not necessarily rational.
97. In dealing with imaginary quantities we apply the laws
of combination which have been proved in the case of other surd
quantities.
Example 1. a + b J  1 ± (c + d J  1) = a ± c + (b ± d) J  1.
Example 2. The product of a 4 b J  f and c + djl
= (a + bj^l)(e + dj^l)
= ac  bd 4 (be + ad) */  1.
98. If a + b J — 1 = 0, 2/tew a = 0, em(£ b = 0.
For, if a + b J^\ = 0,
then bJ—\=a;
.'. —6" = a";
.'. a 2 + b 2 = 0.
Now « 2 and b 2 are both positive, therefore their sum cannot
be zero unless each of them is separately zero ; that is, a — 0,
and 6 = 0.
99. 7f& + bJ^T = c + d l J~ 1, then a = c, andh <I
For, by transposition, a — c + (b — d) J 1 = ;
tlierefore, by the last article, a — c = 0, and 6 — ^ = 0;
that is a = c, and 6 </.
76 HIGHER ALGEBRA.
Thus in order that two imaginary expressions may be equal it
is necessary and sufficient that the real parts should be equal, and
the imaginary parts should be equal.
100. Definition. When two imaginary expressions differ
only in the sign of the imaginary part they are said to be
conjugate.
Thus a — b J — 1 is conjugate to a + b J — 1.
Similarly ^2 + 3^1 is conjugate to J '2  3 J 1.
101. The sum and the product of two conjugate imayinary
expressions are both real.
For ( a + b J \ +ab J\ = 2a.
Again (a + b J 1) (a  b J 1) = a 2  ( b 2 )
= a 2 + b 2 .
102. Definition. The positive value of the square root of
a 2 + b 2 is called the modulus of each of the conjugate expressions
a + b J — 1 and a — b J — 1.
103. The modulus of the product of two imaginary expres
sions is equal to the product of their moduli.
Let the two expressions be denoted by a+bj— 1 and c+dJ—\.
Then their product = ac — bd + (ad + bc) J — 1, which is an
imaginary expression whose modulus
= J(ac  bd) 2 + (ad + be) 2
= Ja s c* + b*<f + a*d* + b s (?
= J (a 2 + b 2 ) (c 2 + dr)
= J a 2 + b 2 x Jc 2 + d 2 ;
which proves the proposition.
104. If the denominator of a fraction is of the form a + bj— 1,
it may be rationalised by multiplying the numerator and the
denominator by the conjugate expression a — b J — 1.
SURDS AND IMAGINARY QUANTITIES. 77
For instance
c + dj  1 (c + dJ\)(abJ \ )
+ b J=l ~(a + b J~i)(abJT)
ac + bd + (ad — be) J — 1
2 72
a + b
a
ac + bd ad — be
i
a + b~ a' + b
• _• v
Thus by reference to Art. 97, we see that the sum, difference,
product, and quotient of two imaginary expressions is in each case
an imaginary expression of the same form.
105. To find the square root of a + h J — 1.
Assume J a + b V— 1 =x + y s/ — 1,
where x and y are real quantities.
By squaring, a + bj—\=x 2 y 2 + 2xy J — 1 ;
therefore, by equating real and imaginary parts,
x 2 y 2 = a (1),
^!/ = b (2);
.. (x 2 + y 2 ) 2 = (x 2  ff + (2xyY
2 , 2.2
= a + b ;
.• . x s + if = J a 2 + 6" (3).
From (1) and (3), we obtain
. Ja 2 + b 2 + a . Ja 2 + b 2 a
x * — 2 .»"=*— i ;
Thus the required root is obtained.
Since x and y are real quantities, x 2 + y is positive, and therefore in (3)
the positive sign must be prefixed before the quantity *Ja 2 + b.
Also from (2) we see that the product xy must have the same sign as b ;
hence x and y must have like signs if b is positive, and unlike signs if b is
negative.
78 HIGHER ALGEBRA.
Example 1. Find the square root of  7  24 J  1.
Assume J 7 21*/ l = x + y J 1 ;
then 724 N / T l = ^ 2 2/ 2 + 2^ N / T l;
••• * 2 ?/ 2 =7 (1),
and 2a;?/ = 24.
= 49 + 576
= 625;
.\ar + 2/ 2 = 25 (2).
From (1) and (2), x = 9 and ?y 2 = 16 ;
.. x= ±3, ?/== ±4.
Since the product xy is negative, we must take
x  3, y =  4 j or # =  3, ?/ = 4.
Thus the roots are 3  4 „/  1 and 3 + 4 *J  1 ;
that is, 7  7  247"TT= ± (3  4 J ~i).
Example 2. To find the value of ^/  64a 4 .
It remains to find the value of \/ ±*J  1.
Assume
J + Jl = x + yJl;
then
+ J~l=x*y + 2xyJ^l;
.. .t 2  ?y 2 = and 2.17/ = 1 ;
whence
1
%/2
V = ^ ; or.x=i, !/= L;
.•.v/ + v/l = ±^(l + Vl).
Similarly J  J^l= ± * (1  ^ ZI)
.•.^±^^[=±^(1=1=^31);
and finally ^64a 4 = ± 2a (1 ± ^/^T).
SURDS AND IMAGINARY QUANTITIES. 70
106. The symbol J — 1 is often represented by the letter i; but
until the student has had a little practice in the use of imaginary
quantities he will find it easier to retain the symbol J — 1. It is
useful to notice the successive powers of J — 1 or i ; thus
(71)^1, i*=ij
and since each power is obtained by multiplying the one before it
by J — 1, or ?', we see that the results must now recur.
107. We shall now investigate the properties of certain imagi
nary quantities which are of very frequent occurrence.
Suppose x  ^1 ; then x 3 = 1, or x 3 — 1 = ;
that is, (x  1 ) (x 2 + x+ 1)^0.
. \ either x — 1 = 0, or x 2 + x + 1 = ;
W3
whence 35=1, or x = .
It may be shewn by actual involution that each of these
values when cubed is equal to unity. Thus unity has three cube
roots,
l+JZTs 1733
2 "' 2 ~'
two of which are imaginary expressions.
Let us denote these by a and ft ; then since they are the roots
of the equation
x 2 + x + l =0.
their product is equal to unity ;
that is, aft= 1 ;
. \ aft = a 2 ;
that is, . ft = a 2 , since a 8 = 1.
Similarly we may shew that a = ft 2 .
108. Since each of the imaginary roots is thr, square of thr
other, it is usual to denote the three cube roots of unity by 1, <d, «>~.
80 HIGHER ALGEBRA.
Also a) satisfies the equation x 2 + x + 1 = ;
. •. 1 + to + w 2 = ;
that is, the sum of the three cube roots of unity is zero.
Again, to . o> 2 — w = 1 ;
therefore (1) the product of the two imaginary roots is unity ;
(2) every integral power of w 3 is unity.
109. It is useful to notice that the successive positive
integral powers of a> are 1, to, and w 2 ; for, if n he a multiple of 3,
it must be of the form 3m ; and to" = w 3m — 1.
If n be not a multiple of 3, it must be of the form 3m + 1 or
3m + 2.
rt> n 1 n 3m + 1 3m
It n = dm + 1 , W — o> =to .to = to.
KO . O " 3m +2 3m 2 2
w = om +, w=w =w .w=to.
110. We now see that every quantity has three cube roots,
two of which are imaginary. For the cube roots of a 3 are those
of a 7 " x 1, and therefore are a, ao>, aw 2 . Similarly the cube roots
of 9 are ^9, o> ^9, a> 2 ^9, where ^ 7 9 is the cube root found by the
ordinary arithmetical rule. In future, unless otherwise stated,
the symbol %ja will always be taken to denote the arithmetical
cube root of a.
(9 4. 3 / _ 1 \2
Example 1. Reduce * , . to the form A + Bj  1.
2 + N /l
The expression
^ 49 + 12^1
2 + v / := ~l
(5 + 12 N /Jl)(2V^l)
(2+ Jl)(2JT)
10 + 12 + 29 J~l
4 + 1
2 29 /— r
= 5 + W 1;
which is of the required form.
Example 2. Resolve x 3 + y 3 into three factors of the first degree.
Since x 3 + if = (x + y) (x 2 xy + y 2 )
.'. x* + y 9 = (x + y) (x + toy) (x + ury) ;
for w + w 2 = 1, and w 3 = l.
SURDS AND IMAGINARY QUANTITIES. 81
Example 3. Shew that
(a + wb + arc) (a + w 2 6 + toe)  a 2 + b 2 + c 2  60  ca  ab.
In the product of a + wb + arc and a + orb + wc,
the coefficients of Zr and c 2 are or, or 1 ;
the coefficient of be = w 2 + o> 4 = or + o> =  1 ;
the coefficients of ca and a& = o> 2 + o> =  1 ;
.*. (a + cob + arc) {a + urb + wc) = a 2 + b 2 + c 2  be  ea  ab.
Example 4. Shew that
(l + ««>)'(lM+U?)*=0.
Since 1 + w + o> 2 = 0, we have
(1 + u w 2 ) 3  (1  w + w a )3=(  2m 2 ) 3  (  2o>) :J
= 8o> 6 + 8a/ {
= 8 + 8
= 0.
EXAMPLES. VIII. b.
1. Multiply 2 \/~~3 + 3 V 3 ^ by 4 *J^3 5 a/^2.
2. Multiply 3 V ^7  5 V^ by 3 V^+ 5 V^.
3. Multiply e^ 1 +e'^~ 1 by e^ _1 e V*.
A AT IX' I l+V 3 ^ . lV^
4. Multiply a; 5 by # = .
Express with rational denominator :
_ 1 3 a/~2~ + 2 *J~h
0. , — • o. , , .
3V2 3V22V5
3 + 2 V~l 32 V^l a+rV^l ax»fi
25\/ r l 2 + 5V :r l' aWisTi a+a?V^l"
g (.f + Vl)a (W l)^ ia (a + VljaCftVl)^
.vV1 .r + V1 (a + \/l) 2 (« Vl) a#
11. Find the value of (  \/  l) 4n + 3 , when w is a positive integer.
12. Find the square of Jd + 40 V"T+ V9  40 V ?.
H. H. A. (j
82 HIGHER ALGEBRA.
Find the square root of
13. S + ISV^L 14. llCOV 17 !". 15. 47 + 8V3.
16. SV^l. 17. a 2 l+2a^^l.
18. ±ab2(a 2 b 2 )*/^T.
Express in the form A + iB
iy * 23r ZU> 2V3i\/2' zu 1T
09 a + O 8 9<* (^ + ^) 2 («^) 2
3 — i a — ib a + io
If 1, co, g> 2 are the three cube roots of unity, prove
24. (l+co 2 ) 4 = co. 25. (lco + co 2 )(l+coor) = 4.
26. (1  co) (1  co ) (1  CO 4 ) (1  co 5 )  9.
27. (2 + 5co + 2co 2 ) 6 = (2 + 2co + 5a> 2 ) 6 = 729.
28. (lco + co 2 )(lco 2 + co 4 )(lco 4 + co 8 )... to 2>i factors = 2 2 ».
29. Prove that
A 3 +yZ + £> _ 2#g Z = (x+y+z) (# + i/a> + Za> 2 ) (x +y<o 2 + Za).
30. If x=a+b t y — aw + Z>co 2 , s=«co 2 + 6co,
shew that
(1) xyz=a 3 +b 3 .
(2) ^ 2 + ?/ 2 + 5 2 = 6a6.
(3) a 3 +y 3 +s 3 =3(a 3 +& 3 ) k
31. If ax + cy + bz = X, ex + by + az = I", Zu + ay + gs = if,
shew that ( a 2 + 6 2 + c 2  be  ca  ab) (x 2 +y 2 + z 2 yz zx  xy)
= X 2 +Y 2 + Z 2  YZ XZ XY.
CHAPTER IX.
THE THEORY OF QUADRATIC EQUATIONS.
111. After suitable reduction every quadratic equation m;iy
be written in the form
(Lif J rbx + C — (1),
and the solution of the equation is
 b ± Jtf^iac
x = \ (2).
2a v '
We shall now prove some important propositions connected
with the roots and coefficients of all equations of which (1) is
the type.
112. A quadratic equation cannot have more than tiro roots.
For, if possible, let the equation ax 2 + bx + c = have three
different roots a, f3, y. Then since each of these values must
satisfy the equation, we have
aa 2 t ba + c~0 (1),
afi 2 + bp + c = (2),
ay 2 + by + c = (3).
From (1) and (2), by subtraction,
a(a 2 p 2 ) + b(aP) = 0;
divide out by a — fi which, by hypothesis, is not zero; then
a (a + ft) + b = 0.
Similarly from (2) and (3)
a {fi + y) + b = ;
.•. by subtraction a (a  y) — ;
which is impossible, since, by hypothesis, a is not zero, and u is
not equal to y. Hence there cannot be three different roots.
G— 2
84 HIGHER ALGEBRA.
113. In Art. Ill let the two roots in (2) be denoted by a and
/3, so that
b + Jb 2  Aac —b Jb 2 — Aac _
a =
, P =
2a ' ' 2a '
then we have the following results :
(1) If b 2  Aac (the quantity under the radical) is positive,
a and (3 are real and unequal.
(2) If b 2 — Aac is zero, a and ft are real and equal, each
reducing in this case to — 77 .
2a
(3) If b 2  Aac is negative, a and ft are imaginary and unequal.
(4) If b 2 — Aac is a perfect square, a and /3 are rational and
unequal.
By applying these tests the nature of the roots of any
quadratic may be determined without solving the equation.
Example 1. Shew that the equation 2x 2 6a; + 7 = cannot be satisfied
by any real values of x.
Here a = 2, b =  6, c — 7 ; so that
&2_ 4ac= (_6)24.2.7=20.
Therefore the roots are imaginary.
Example 2. If the equation a? 2 + 2 (k + 2) x + 91c = has equal roots, find l\
The condition for equal roots gives
(fc + 2) 2 = 9£,
fc2_5ft + 4=0,
(fc4)(fcl)=0j
.. k = A, or 1.
Example 3. Shew that the roots of the equation
x 2  2p3 +p 2 q 2 + 2qr r 2 =
are rational.
The roots will be rational provided ( 2p) 2  4 (p 2  q 2 + 2qrr 2 ) is a
perfect square. But this expression reduces to 4 (q 2 2qr + r 2 ), or 4:(qr) 2 .
Hence the roots are rational.
  , D .  b + Jb 2  Aac b Jb 2  Aac
114. Since a= ^ ' ?= 2a '
we have by addition
 b + Jb 2  Aac b Jb 2  Aac
* + ^ 2a
__M_b
2a a
0);
THE THEORY OF QUADRATIC EQUATIONS. 85
and by multiplication we have
n = ( b + J¥~r^ ) (_. 5 _ j,/r^ c)
_ 4ac c
~4a 2 = « : (2).
By writing the equation in the form
2 ° c
a a '
these results may also be expressed as follows.
unity, * qUadratiC e( l uation «**" ** coefficient of the first term is
its 4**5d? the roots is equal t0 the coefficient of ■ with
(ii) the product of the roots is equal to the third term.
*£%* £»"& ££&£**"■ not contain the 8nta0 ™
115. Since __ =a+ £ and £ *
the equation ar+  a; +  = may be written
x 2 (a + f])x + ap = (1).
Hence any quadratic may also be expressed in the form
x 2  (sum of roots) x + product of roots = (2).
Again, from (1) we have
(xa)(xp) = Q (3).
We may now easily form an equation with given roots.
Example 1. Form the equation whose roots are 3 and  2.
The equation is (*  3) (*+ 2)=0,
or ««* 6=0.
When the roots are irrational it is easier to use the following
metnou. ~>
86 HIGHER ALGEBRA.
Example 2. Form the equation whose roots are 2 + ^3 and 2  ^3.
We have sum of roots = 4,
product of roots = 1 ;
. • . the equation is x  Ax + 1 = 0,
by using formula (2) of the present article.
116. By a method analogous to that used in Example 1 of
the last article we can form an equation with three or more given
roots.
7
Example 1. Form the equation whose roots are 2,  3, and ^ .
o
The required equation must be satisfied by each of the following sup
positions :
7
#2=0, # + 3 = 0, £ = ();
therefore the equation must be
(*2)(*+3)(*)=0j
that is, [x  2) (x +3) (5a; 7) =0,
or 5a; 3 2a; 2 37a; + 42 = 0.
Example 2. Form the equation whose roots are 0, ±«, j .
The equation has to be satisfied by
c
x = 0, x = a. x=a, x=} ;
b
therefore it is
x (x J a) (x  a) ( x   j = ;
that is, x (x 2  a 2 ) (bx  c) = 0,
or bx 4  ex 3  a~bx + acx = 0.
117. The results of Art. 114 are most important, and they
are generally sufficient to solve problems connected with the
roots of quadratics. In such questions the roots should never be
considered singly, but use should be made of the relations ob
tained by writing down the sum of the roots, and their product,
in terms of the coefficients of the equation.
Example 1. If a and /3 are the roots of xpx + q = 0, find the value of
(l)a 2 + /3 2 , (2)a s + /3 3 .
We have a + (2=p,
a
a(3 = q.
.. a 2 + /3 2 =(a + /3) 2 2a/3
=p*2q.
THE THEORY OF QUADRATIC EQUATIONS. 87
Again, a ;{ + ft ■ = (a + /3) (a 2 + p" 2  a/3)
=i>{(a + /3) a 3a/3]
=*(?» 89).
Example 2. If a, p" are the roots of the equation /.r'' mx + 7i = 0, find the
equation whose roots are — ,  .
p a
"We have sum of roots = +  = ^ ,
p a. ap
product of roots =  . =1 :
p a
.. by Art. 115 the required equation is
or apx 2  (a 2 + p~ 2 ) £ + 0/3 = 0.
O O 7
As in the last example o 2 +j8 B = =— , and a/3 =   .
L t
., ,. n ?ji 2 2wZ ?i _
.*. the equation is x = — x + y = 0,
V v it
or n /x 2  ( m 2  2nZ) x + nl = 0,
Example 3. When .r = ^ , find the value of 2x 3 + 2x 2 7x+l'2 ;
and shew that it will be unaltered if £ be substituted for x.
a
Form the quadratic equation whose roots are ^ ;
the sum of the roots = 3 ;
17
the product of the roots = — ;
hence the equation is 2.r 2  6.r + 17 = ;
.*. 2x 2 6x + 17 is a quadratic expression which vanishes for either of the
, 3*5^/"!
values ^ •
m
Now 2a* + 2.t 2  Ix + 72 = x (2.r 2  C>.c + 17) + 4 (2.r 2  Cx + 17) + 4
=xx0+4x0+4
= 4;
which is the numerical value of the expression in each of the supposed cases.
88 HIGHER ALGEBRA.
118. To find Vie condition that the roots of the equation
ax 2 + bx + c = should be (1) equal in magnitude and opposite
in sign, (2) reciprocals.
The roots will be equal in magnitude and opposite in sign if
their sum is zero ; hence the required condition is
_  = 0, or b = 0.
a
Again, the roots will he reciprocals when their product is
unity ; hence we must have
c i
— = 1. or c = a.
a
The first of these results is of frequent occurrence in Analyti
cal Geometry, and the second is a particular case of a more
general condition applicable to equations of any degree.
Example. Find the condition that the roots of ax 2 + bx + c = may be (1)
both positive, (2) opposite in sign, but the greater of them negative.
b c
We have a + B= — , a8= .
a a
(1) If the roots are both positive, a/J is positive, and therefore c and a
have like signs.
Also, since a + fi is positive, — is negative; therefore b and a have unlike
signs.
Hence the required condition is that the signs of ft and c should be like,
and opposite to the sign of b.
(2) If the roots are of opposite signs, a/3 is negative, and therefore c and
a have unlike signs.
Also since a +/3 has the sign of the greater root it is negative, and there
fore  is positive; therefore b and a have like signs.
(X
Hence the required condition is that the signs of a and b should be like,
and opposite to the sign of c.
EXAMPLES. IX. a.
Form the equations whose roots are
. 4 3 m n pq p + q
o / n m p+q p—y
4. 7±2 N /5. 5. ±2«/3~5. 6. p±2s/Tq.
THE THEORY OF QUADRATIC EQUATIONS. 89
7. 3±5l 8. a±ib. 9. ±i(a b).
10. 3, , i. 11. , 0, . 12. 2±«/3, 4.
13. Prove that tlio roots of the following equations are real :
( 1 ) x 2  2ax +a a  6 a  c 2  0,
(2) (a  b + c) r +4 (a  6) .v + (a  b  c) = 0.
14. If the equation x 2  15 m(2x8) = lias equal roots, find the
values of m.
15. For what values of m will the equation
x 2  2x (1 + 3//0 + 7 (3 + 2m) =
have equal roots ?
16. For what value of m will the equation
x* — bx m  1
axc m + 1
have roots equal in magnitude but opposite in sign ?
17. Prove that the roots of the following equations are rational:
(1) (a + cb)x 2 + 2cx + (b + ca) = 0,
(2) abc 2 x 2 + 3a 2 cx + b 2 ex  6a 2 ab + 2b' 1 = 0.
If a, /3 are the roots of the equation ax 2 + bx + c = 0, find the values of
18. »,+£. 19. aW + aV. 20. (f) 2 .
Find the value of
21. a 3 + s 2  X + 22 when .r = 1 + 2/.
22. x 3  Zx 2  8x +15 when x = 3 + ? .
23. .t 3  «.r 2 + 2a 2 .r + 4« 3 when = 1  J  3.
a
24. If o and /3 are the roots of x*+px+q=O t form the equation
whose roots are (a ft) 2 and (a + /3) 2 .
25. Prove that the roots of (x — a) (.»; b) = h 2 are always red.
26. I f .', , x % are the roots of ctx*+bx + c = 0, find tho value i >f
(1) (ax l + b) 2 + («x i + b) 2 ,
(2) (ax^byt+iaxt+b)*.
90 HIGHER ALGEBRA.
27. Find the condition that one root of ax 2 + bx\c = shall be
n times the other.
28. If a, (3 are the roots of ax 2 + bx + c = 0, form the equation whose
roots are a 2 + /3 2 and o~ 2 +/3 2 ,
29. Form the equation whose roots are the squares of the sum and
of the difference of the roots of
2x* + 2 (m + n) x + m 2 + n 2 =0.
30. Discuss the signs of the roots of the equation
px 2 + qx + r = 0.
119. The following example illustrates a useful application
of the results proved in Art. 113.
x + 2x — 11
Example. If x is a real quantity, prove that the expression ■■ .
can have all numerical values except such as lie between 2 and 6.
Let the given expression be represented by y, so that
a 2 + 2:r ll_
2(s3) ~ y;
then multiplying up and transposing, we have
rr 2 +2.r(l?/) + 6f/ll = 0.
This is a quadratic equation, and in order that x may have real values
4(1 i/) 2 4(Gy — 11) must be positive; or dividing by 4 and simplifying,
?/ 2  8*/ + 12 must be positive ; that is, (y  6) (y  2) must be positive. Hence
the factors of this product must be both positive, or both negative. In the
former case y is greater than 6; in the latter y is less than 2. Therefore
y cannot lie between 2 and 6, but may have any other value.
In this example it will be noticed that the quadratic expression
y 2 — 8y + 12 is positive so long as y does not lie between the roots
of the corresponding quadratic equation y 2 — Sy + 12 = 0.
This is a particular case of the general proposition investigated
in the next article.
120. For all real values of x tlie expression ax 2 + bx+c has
the same sign as a, except when the roots of the equation ax 2 +bx + c =0
are real and unequal, and x has a value lying between them.
Case I. Suppose that the roots of the equation
ax 2 + bx + c =
are real ; denote them by a and ft, and let a be the greater.
THK THEORY OF QUADRATIC EQUATIONS. 91
Then ((.r 2 4 bx + c — a ( x* +  X + )
V a aj
= a {x 2  (tt + ft) X + aft
= a (x  a) (x  ft).
Now if x is greater than a, the factors x — a, x — ft are both
positive ; and if x is less than ft, the factors x — a, x — ft are both
negative; therefore in each case the expression (x — a)(x — ft) is
positive, and ax 2 + bx + c lias the same sign as a. But if x has a
value lying between a and ft, the expression (./•  a) (x  ft) is
negative, and the sign of ax" + bx + c is opposite to that of a.
Case II. If a and ft are equal, then
ax 2 + bx + c = a(x— a) 2 ,
and (x  a) 2 is positive fur all real values of x ; hence ax 2 + bx + c
has the same sign as a.
Case III. Suppose that the equation ax 2 + bx + c = Q lias
imaginary roots ; then
ax 2 + bx + c — alx* x + 
a a\
{•
(/ b\ iaeb')
But b 2 — 4«c is negative since the roots are imaginary ; hence
iacb* . .
is positive, and the expression
4a 2
( x + &)
2 Aac — b 2
^ +
\a 2
is positive for all real values of x ; therefore ax 2 + bx + c has the
same sign as a. This establishes the proposition.
121. From the preceding article it follows that the expression
ax 2 + bx + c will always have the same sign whatever real value x
may have, provided that b 2  Aac is negative or zero; and if this
condition is satisfied the expression is positive or negative accord
ing as a is positive or negative.
Conversely, in order that the expression ax 2 + bx + c may be
always positive, b 2 — Aac must be negative or zero, and a must be
positive; and in order that ax 2 + bx + c may be always negative
I 1  Aac must be negative or zero, and a must be negative.
92 HIGHER ALGEBRA.
Example. Find the limits between which a must lie in order that
ax 2  Ix + 5
5x 2  Ix + a
may be capable of all values, x being any real quantity.
_ ax 1 lx + 5
Put i o rr =v;
then (a5?/)a: 2 7.r(l?/) + (5a?/):=0.
In order that the values of x found from this quadratic may be real, the
expression
49 (1  y)' 2  4 (a  5y) (5  ay) must be positive,
that is, (49  20a) y 2 + 2 (2a 2 + l)y + (49  20a) must be positive ;
hence (2a 2 + 1) 2  (49  20a) 2 must be negative or zero, and 49  20a must be
positive.
Now (2a 2 + 1) 2  (49  20a) 2 is negative or zero, according as
2 (a 2  10a + 25) x 2 (a 2 + 10a  24) is negative or zero ;
that is, according as 4 (a  o) 2 (a + 12) (a  2) is negative or zero.
This expression is negative as long as a lies between 2 and  12, and for
such values 49  20a is positive; the expression is zero when a = 5,  12, or 2,
but 49 20a is negative when a = 5. Hence the limiting values are 2 and
 12, and a may have any intermediate value.
EXAMPLES. IX. b.
1. Determine the limits between which n must lie in order that
the equation
2ax (ax + nc) + (n 2  2) c 2 =
may have real roots.
x . 1
2. If x be real, prove that 5 — ' must lie between 1 and  r^ .
x x l — bx + 9 11
^•2 .77 4 1
3. Shew that = — ' lies between 3 and  for all real values of x.
x + x+\ 3
sb 3 + 34a?— 71
4. If x be real, prove that • — ^ — ■= = can have no value between
x x 1 + 2# — 7
5 and 9.
5. Find the equation whose roots are
s]a ± sja  b
6. If a, /3 are roots of the equation x 2 — px+q=0, find the value of
(1) atitfpifl + ptfPaia),
(2) ( a p)* + (Pp)\
THE THEORY OF QUADRATIC EQUATIONS. !>:{
7. If the roots of ht?+ nx+n=0 be in the ratio of p : q t prove that
8. If x be real, the expression n . r admits of all values
2 (x  n)
except such as lie between 2)i and 2m.
9. Tf the roots of the equation ax 2 + 2hx + c=() be a and (3, and
those of the equation Ax' y  + 2Ux+C=0 be affi and fi + d, prove that
b*ae_ B*AC
~~a 2 ~ A 2 '
10. Shew that the expression — — —5 will be capable of all
p + 3x  4x* L
values when x is real, provided that p has any value between 1 and 7.
.#42
11. Find the greatest value of n * n  for real values of x.
& 2x 2 + 3x + 6
12. Shew that if x is real, the expression
(x 2 bc)(2xbc)~ i
has no real values between b and c.
13. If the roots of ax 2 + 2bx + c = be possible and different, then
the roots of
(a + c) (ax 2 + 2bx + c) = 2 (ac  b 2 ) (.r 2 + 1)
will be impossible, and vice versa.
14. Shew that the expression  fl  {) ,! will be capable of all
(ox — a) (ex — a)
values when x is real, if a 2  b 2 and c 2 — d 2 have the same sign.
*122. We shall conclude this chapter with some miscellaneous
theorems and examples. It will be convenient here to introduce
a phraseology and notation which the student will frequently
meet with in his mathematical reading.
Definition. Any expression which involves x, and whose
value is dependent on that of x, is called a function of X.
Functions of x are usually denoted by symbols of the form f(x),
F(x),<f>(x).
Thus the equation y =f(x) may be considered as equivalent
to a statement that any change made in the value of as will pro
duce a consequent change in ;//, and vice versd. The quantities x
and y are called variables, and are further distinguished as the
independent variable and the dependent variable.
94 HIGHER ALGEBRA.
An independent variable is a quantity which may have any
value we choose to assign to it, and the corresponding dependent
variable has its value determined as soon as the value of the inde
pendent variable is known.
%
123. An expression of the form
pjs" + p x x n ] + pjf 2 + . . . + p n _ t x + p n
where n is a positive integer, and the coefficients p lt , p lt p a ,...p n do
not involve x, is called a rational and integral algebraical function
of x. In the present chapter we shall confine our attention to
functions of this kind.
*124. A function is said to be linear when it contains no
higher power of the variable than the first ; thus ax + b is a linear
function of x. A function is said to be quadratic when it
contains no higher power of the variable than the second ; thus
ax 2 + bx + c is a quadratic function of x. Functions of the third,
fourth,... degrees are those in which the highest power of the
variable is respectively the third, fourth, Thus in the last
article the expression is a function of x of the n th degree.
*125. The symbol fix, y) is used to denote a function of two
variables x and y ; thus ax + by + c, and ax 2 + bxy + cy 2 + dx + ey +f
are respectively linear and quadratic functions of x, y.
The equations fix) = 0, fix, y) — are said to be linear, quad
ratic, ... according as the functions f(x), f(x, y) are linear, quad
ratic,....
*126. We have proved in Art. 120 that the expression
ax 2 + bx + c admits of being put in the form a (x — a) (x — fi),
where a and j3 are the roots of the equation ax 2 + bx + c — 0.
Thus a quadratic expression ax 2 + bx\ c is capable of being
resolved into two rational factors of the first degree, whenever
the equation ax 2 + bx + c = has rational roots ; that is, when
b 2  iac is a perfect square.
*127. To find the condition that a quadratic function ofx,y
may be resolved into two linear factors.
Denote the function hy f(x, y) where
f{x, y) = ax z +'2hxy + by 2 + 2gx+ 2fy + c.
THE THEORY OF QUADRATIC EQUATIONS. 95
Write this in descending powers of x, and equate it to zero;
thus
ax* + 2x (hy + y) + by 2 + 2fy + c  .
Solving this quadratic in x we have
_  (h + (j) ± J{hy + y)*  a (by 2 + 2fy + c)
x — ,
a
< >v ax + hy + g = ± Jy 2 (h*  ab) + 2y (hy — a/) + (g 2  ac).
Now in order that J\.r, y) may be the product of two linear
factors of the form px + qy + r, the quantity under the radical
must be a perfect square ; hence
(kg  a/) 2 = (h  ab) {<f  ac).
Transposing and dividing by a, we obtain
abc + 2fyh — af 2 — by 2 — ch 2 = ;
which is the condition required.
This proposition is of great importance in Analytical Geometry.
*128. To find the condition that the equations
ax 2 + bx + c — 0, ax 2 f b'x + c 
may have a common root.
Suppose these equations are both satisfied by x a ; then
aa. 2 + ba + c = 0,
a'a 2 +b'a + c' = 0;
.*. by cross multiplication
a" a 1
be — b'c ca — c'a ab' — ab '
To eliminate a, square the second of these equal ratios and
equate it to the product of the other two ; thus
a a
1
(ca — c'a) 2 (be — b'c) ' (ab' — ab) '
.'. (ca — ca) 2 = (be — b'c) (ab' — ab),
which is the condition required.
It is easy to prove that this is the condition that the two
quadratic functions ax 2 + bxy + cy 2 and a'x 2 + b'xy + c'y' may have
a common linear factor.
06 HIGHER ALGEBRA.
^EXAMPLES. IX. c.
1. For what values of m will the expression
y 2 + 2xy + 2x + my  3
be capable of resolution into two rational factors ?
2. Find the values of m which will make 2.v 2 + mxy + 3y 2  5y  2
equivalent to the product of two linear factors.
3. Shew that the expression
always admits of two real linear factors.
4. If the equations
x 2 + px + q = 0, x 2 + p'x + q' =
have a common root, shew that it must be either
p'l'p'q nr 9q
qq pp
5. Find the condition that the expressions
Lv 2 + mxy + ny 2 , l'x 2 + m'xy f n'y' 1
may have a common linear factor.
6. If the expression
%a? + 2Pxy + 2y 2 + 2ax  4y + 1
can be resolved into linear factors, prove that P must be one of the
roots of the equation P' 2 + 4aP + 2d 1 + 6 = 0.
7. Find the condition that the expressions
ax 2 + 2hxy + by 2 , a'x 2 + 2k'xy + b'y 2
may be respectively divisible by factors of the form y mx, my + x.
8. Shew that in the equation
x 2  Zxy + 2y 2  2x  3y  35 = 0,
for every real value of x there is a real value of y, and for every real
value of y there is a real value of x.
9. If x and y are two real quantities connected by the equation
9x 2 + 2xy +y 2  92.r  20y + 244 = 0,
then will x lie between 3 and 6, and y between 1 and 10.
10. If (ax 2 + bx + c)y\a'x 2 + b'x + e' = 0, find the condition that x
may be a rational function of y.
CHAPTER X.
MISCELLANEOUS EQUATIONS.
129. In this chapter we propose to consider some mis
cellaneous equations ; it will be seen that many of the^l Z
solved by the ordinary rules for quadratic equJtions, but others
require some special artifice for their solution
_ 3_ 3
Example I. Solve 8x 2n 8x~^=63.
Multiply by .r 2n and transpose; thus
 i.
8x n  63x 2 '*8 = 0;
— L
(a?"8)(8x^+l) = 0;
 1
«2n = 8, or;
8'
2n
■=(*)* «(p)*;
..*=«■», or A.
Example 2. Solve 2 /+ 3 / = ^ 6
V « V « « I
.•.%+! = * + «■■■
2<% 2 6a 2 ?/& 2 </ + 3a& = 0;
(2ay~&)(ty3a)=0;
6 3a
* & 2 9a 2
6a
4a ' " l &a *
H. H. A.
98 HIGHER ALGEBRA.
Examples. Solve (*5)(a: 7)(« + 6) (* + 4) = 504.
We have (x 2  x  20) (x 2  x  42) = 504 ;
which, being arranged as a quadratic in x 2  x, gives
(a 2  x) 2  62 (x 2  x) + 336 =
.. (x 2 a:6)(x 2 x56) =
.. X * X Q = f or a 2 a; 56 =
whence x = S, 2, 8, 7.
130. Any equation which can be thrown into the form
ax 2 + bx + c + p J ax 2 + bx + c — q
may be solved as follows. Putting y = J ax 2 + bx + c, we obtain
Let a and ft be the roots of this equation, so that
J ax 2 + bx + c = a, J ax 2 +bx + c = ft ;
from these equations we shall obtain four values of x.
"When no sign is prefixed to a radical it is usually understood
that it is to be taken as positive; hence, if a and ft are both
positive, all the four values of x satisfy the original equation.
If however a or ft is negative, the roots found from the resulting
quadratic will satisfy the equation
ax 2 + bx + c — p J ax 2 + bx + c = q,
but not the original equation.
Example. Solve x 2  ox + 2 Jx 2  5z + 3 = 12.
Add 3 to each side ; then
rc 2 5a; + 3 + 2 N /^5a; + 3 = 15.
Putting Jx 2 5x+3 = y, we obtain y 2 + 2y  15 = ; whence y = 3 or  5.
Thus *Jx 2  5x + 3 = + 3, or Jx 2 6x + S = 5.
Squaring, and solving the resulting quadratics, we obtain from the first
ic=6 or 1; and from the second x^
satisfies the given equation, but the second pair satisfies the equation
ing quadratics, we obtain from the first
: = — ^ . The first pair of values
x 2  5x 2 Jx 2 5x + 3 = 12.
MISCELLANEOUS EQUATIONS. 99
131. Before clearing an equation of radicals it is advisable
to examine whether any common factor can be removed by
division.
Example. Solve *J x' 2  lax + 10a' 2  Jx + ax 6a — x 2a.
We have
*J(x2a)(x5a)  J{x2a) (x+Sa) = x2a.
The factor *J x  2a can now be removed from every term ;
.'. sjx 5a Jx + 3a = „Jx  2a ;
x  5a + x + 3a  2 *J(x  5a) (x + 3a) = x  2a ;
x = 2 Jx' 2  2ax  15a' 2 ;
3ar8aa;60a 2 = 0;
{x 6a) (3a; + 10a) =0;
c 10a
x — ba, or — — .
Alsoxby equating to zero the factor Jx  2a, we obtain x = 2a.
On trial it will be found that x = 6a does not satisfy the equation : thus
the roots are — — and 2a.
D
The student may compare a similar question discussed in the Elementary
Algebra, Art. 281.
132. The following artifice is sometimes useful.
Example. Solve J'3x  4.x + 34 + JSx' 2  4x  11 = 9 (1).
We have identically
(3x4a; + 34)(3a; 2 4xll) = 45 (2).
Divide each member of (2) by the corresponding member of (1); thus
J'dx  4.r + 34  JSx 2  4x  11 = 5 (3).
Now (2) is an identical equation true for all values of x, whereas (1) is an
equation which is true only for certain values of x ; hence also equation (3)
is only true for these values of x.
From (1) and (3) by addition
v /3x 2 4a; + 34 = 7;
whence as = 3, or .
7—7
100 HIGHER ALGEBRA.
133. The solution of an equation of the form
ax 4 ± bx 3 ± ex 2 ± bx + a = 0,
in which the coefficients of terms equidistant from the beginning
and end are equal, can be made to depend on the solution of a
quadratic. Equations of this type are known as reciprocal equa
tions, and are so named because they are not altered when x is
changed into its reciprocal  .
For a more complete discussion of reciprocal equations the
student is referred to Arts. 568 — 570.
Example. Solve 12a; 4  56x 3 + 89a; 2  56.x + 12 = 0.
Dividing by x 2 and rearranging,
12/W 2 ) 56^+^+89 = 0.
Put x + =z: then a; 2 + — = z 2 2;
x x
.. 12 (z 2  2) 56^ + 89 = 0;
whence we obtain z =  , or = .
2 u
1 5 13
a; 2 6
13 2
By solving these equations we find that x = 2,  ,  ,  .
134. The following equation though not reciprocal may be
solved in a similar manner.
Example. Solve 6a; 4  25a; 3 + 12a; 2 + 25a; + 6 = 0.
We have 6 (^ 2 +^i)  25 fx  \ + 12 = 0;
whence 6(a; — j 25 (a; — 1+24 = 0;
.. 2 (^^3 = 0, or 3 fx ]80;
whence we obtain a; = 2,   , 3,   .
135. When one root of a quadratic equation is obvious by
inspection, the other root may often be readily obtained by
making use of the properties of the roots of quadratic equations
proved in Art. 114.
MISCELLANEOUS EQUATIONS. 101
Example. Solve ( 1  a) {x + a)  2a ( 1  ar) = 0.
This is a quadratic, one of whose roots is clearly a.
Also, since the equation may be written
2ax + (1  a 2 ) x  a (1 + a 2 ) = 0,
the product of the roots is   ; and therefore the other root is — — .
EXAMPLES. X. a.
Solve the following equations :
1. a 2 2x~ 1 = 8. 2. 9 + a 4 = 10a'.
1 3 1 _J
3. 2jx + 2x § =5 4. 6a?*~7**8a7 *.
2 1 JL 1
5. 3"+6=5#». 6. 3.f 2n .r ri 2=0.
7  »>/;+Vj"* a \/S + \/?'
■2*.
i
9. 6 x /a=5a 2 13. 10. 1+8.^ + 9^ = 0.
11. 32*+ 9« 10. 3*. 12. 5 (5* + 5*) = 26.
13. 2 2 * + 8 + 1 = 32.2'. 14. 2 2 * + 3 57 = 65(2*l).
15. ,/*■+£* 16. ^.#=5A
17. (x  7) (a  3) {x + 5) (.v + 1 ) = 1 680.
18. (x + 9) (x  3) (x  7) {x + 5) = 385.
19. x (2x + 1 ) (.v  2) (2a  3) = 63.
20. (2a7)(a 2 9)(2a + 5) = 91.
21. A' 2 + 2 >/a 2 + 6a = 24  6x.
22. 3a 2  4a + s /'3x i 4x6 = l8.
23. 3a 2 7 + 3 (N /3s a 16a? + 21 = 16a;.
24. 8 + 9 J(&v 1) (x 2) = 3.c 2  7a.
25 . ^ 2 +s /,,,_ 5 , +3= ^:.
102 HIGHER ALGEBRA.
26. 7.^hs±i_c» W .Y.
x \*j x j
27. J4x 2 7xlb  >Jx 2 3x= Jx 2  9.
28. >/2^ 2 9^ + 4 + 3 J%vl = J2x 2 + 2\xl\.
29. V 2 ^ 2 + 5^7 + V3(a; 2 7a; + 6)  J7x 2 6xl = 0.
30. s /a 2 + 2ax3x 2  Ja 2 + ax6x 2 = ^/Sa 2 + 3ao;  9a; 2 .
31. J2x 2 + bx  2  V2# 2 + 5a;  9 = 1.
32. x/3^ 2  2x + 9 + x/3.r 2  2o; 4 = 13.
33. V2^ 2 7a;+l  ./2a; 2 9a; + 4 = 1.
34. >/ 3 ^ 2  7^  30  */2o; 2  7x  5 = a?  5 .
35. o^ + a?4o; 2 + a;+l = 0.
8
9
36. x* + ^x 2 + l = 3x 3 +3x. 37. 3*+l3(« s +#)=2tf*.
38. 10(o7 t +l)63a?(a; 2 l) + 52a; 2 = 0.
x+J\2ax _*Ja+\ a + 2a; + J 'a 2  4x 2 _ bx
x  <Jl2ax sJctV a+2x J a 2  4a; 2 ~ a '
. 1 a; + sjx 2  1 a;  ^/.r 2  1 ,— — 
41. , \ = 8x jx 2  3x + 2.
x  sjx 2  1 x + a/^ 2  1
42. >/^+#I = . 43. £".+ ./».
Jtfx 2 a; 2 l V #
44. 2* 2 : 2 2 * = 8 : 1. 45. a 2 *(a 2 + I) = (a?* + a*)a.
46 ^/a?5 = V3a?7 18 (7a;  3) _ 250 V^+T
3a;7 xb ' ' 2a;+l " 3n /7^3
2 2 1
48. (a + a;) 3 " + 4 (a  *)■ = 5 (a 2  a; 2 )" 3 .
49. >/a; 2 + aa;l  Jx 2 + bxl = J a  K /b.
50. ^B+»^El. 8 a
#  V^' 2  1 # + \/a' 2  1
51. .v 4  2.v3 + a; = 380. 52. 27^ + 2U + 8 =
MISCELLANEOUS EQUATIONS. 108
136. We shall now discuss some simultaneous equations of
two unknown quantities.
Example 1. Solve x+2+y+S+ J(x + 2) (y + 3) = 39.
(z + 2) 2 + (y + 3) 2 + (:r + 2)( 2 / + 3)=741.
Put x + 2 = m, and y + 3 = v ; then
u+v + Juv = Sd (1),
w 2 + v 2 + wv = 741 (2),
hence, from (1) and (2), we obtain by division,
u + v  Juv = 19 (3) .
From (1) and (3), u+t?=29;
and Juv = 10,
or wv = 100;
whence w = 25, or 4; v = 4, or 25 ;
thus x = 23, or 2; y=l, or 22.
Example 2. Solve .r 4 + y*= 82 (1),
ary=2 (2).
Put # = w + t>, and y = u v;
then from (2) we obtain v = l.
Substituting in (1), (w + l) 4 + (u 1) 4 = 82;
.. 2(m 4 + 6m 2 + 1) = 82;
u 4 + 6u 2 40 = 0;
whence w 2 = 4, or — 10 ;
and u= ±2, or ± >/~ 10*
Thus x=s, l, i± V^iO;
ysal, 3, li^10.
JEa;ampZe3. Solve f^  — ^ = 2A (1),
e Sxyx + y 10
7x + 5y = 29 (2).
From (1), 15 (2a; 2 + Sxy + y*  3z 2 + Axy  y) = 38 (3.x 2 + 2xy  y*) ;
.. 129o; 2 29xy38?/ 2 = 0;
.. {Sx2y)(iBx + 19y)=0.
Hence Sx = 2y (3),
or 43# = 19y (1).
104 HIGHER ALGEBRA.
From (3),
x y __7x + 5y
2 = 3 = 29
Again, from (4),
= 1, by equation (2).
.. x = 2, y = 3.
x y 7x + 5y
19 ~ ^43 ~  82
29
= gg, by equation (2),
551 _ 1247
•*' X ~ 82 ,V ~ 82 "
551 1247
Hence x = 2, y = 3; or x  — , 11 = ^ •
Example 4. Solve 4# 3 + 3a; 2 f/ + ?/ 3 =8,
2z 3 2a; 2 ?/ + £?/ 2 = l.
Put y = mx, and substitute in both equations. Thus
z 3 (4 + 3m + m 3 ) = 8 (1).
z 3 (22m + m 2 ) = l (2).
4 + 3m + m z _
•*' 22m + m 2 ~ '
m 3 8»i 9 +19m 12 = 0;
that is, (/;il)(?/i3) (m4) = 0;
.*. m=l, or 3, or 4.
(i) Take m = l, and substitute in either (1) or (2).
From (2), # 3 = 1; .*. x = l;
and y=mx=x=l.
(ii) Take m = 3, and substitute in (2) ;
3 /l
thus 5:r 3 = l; .*. x = \/ k'>
3/1
and y = vix = 3x = 3 */ .
(iii) Take7?& = 4; we obtain
3 / 1
10.r 3 =l; .. x=^;
3 /I
and y = mx = 4x=4. /r^.
MISCELLANEOUS EQUATIONS. L05
Hence the complete solution is
* =1, V5' v To*
•" = 1 ' s \/l> 4 \/^*
Note. The ahove method of solution may always be used when the
equations are of the same degree and homogeneous.
Example 5. Solve 3 lx 2 y 2 7y 4  112^ + 64 = (1),
x 2 7xy + 4y 2 + 8 = ('2).
From (2) we have 8 = x 2  Ixy + 4//' ; and, substituting in (1),
3\x 2 y 2  7# 4 + Uxy {x 2  Ixy + Ay 2 ) + {x 2  Ixy + 4 j/") = ;
.. 31xy 2  7 j/ 4 + (x 2  Ixy f Ay 2 ) (Uxy + x 2  Ixy + Ay 2 ) = ;
.. Slx 2 y 2 7y* + (x 2 + 4y 2 ) 2 (7xy) 2 = 0;
that is, s*10sy+9y 4 =0 (3).
.'. (x 2 y 2 )(x 2 9y 2 ) = 0;
hence x=±y, ov x= ±3y.
Taking these cases in succession and substituting in (2), we obtain
x = y=±2;
x=y=± ^J 
x=±3, y=±l\
3 >/17'^ =T \/
 yj
Note. It should be observed that equation (3) is homogeneous. The
method here employed by which one equation is made homogeneous by a
suitable combination with the other is a valuable artifice. It is especially
useful in Analytical Geometry.
Example 6. Solve (x+yft+2 {x  ?/)* = 3 {x 2  y*fi (1).
3x2y=13 (2).
i i i
Divide each term of (1) by (x 2  y 2 ) , or {x +y)* (x  y) r  ;
i i
. ( x +y\ s +2 (~ y Y=3
\xyj \.v + gj
106 HIGHER ALGEBRA.
i
[ x + v\ a
This equation is a quadratic in ( 1 , from which we easily find,
i
( x ±yY = 2oTl; whence ^=8 or 1 ;
\xyj xy
.'. 7x = 9y, or y = 0.
Combining these equations with (2), we obtain
13
x=9, y = 7; or x = ^,y=0.
EXAMPLES. X. b.
Solve the following equations :
1. 3x2y = 7, 2. bx y = 3, 3. 4^3^ = 1,
xy = %). y 2  6# 2 = 25. 12^ + 13y 2 = 25.
4. a, 4 + #y+2/ 4 = 931, 5. x 2 + ocy +3/ 2 = 84,
x 2 — xy +y 2 = 19. x  *Jxy+y =6.
6. x + Jxy +y = 65, 7. x +y = 7 + \A?y,
# 2 + #y +y 2 =2275. x 2 +y 2 = l33xy.
8. 3# 2 5y 2 = 7, 9. 5y 2 7^ = l7, 10. 3.r 2 + 165 = 16.ry,
Zxy  4y 2 = 2. bxy  6x 2 = 6. 7^y + 3y 2 = 1 32.
11. 3x 2 +xy+y 2 = l5, 12. # 2 + y 2 3 = 3.zy,
Zlxy  3x 2 bf = 45. 2x 2  6 + y 2 = 0.
13. .r 4 +y 4 =706, 14. x A +y* = 272, 15. ^y 5 = 992,
x+y = 8. xy = 2. xy = 2.
16. ,r+i = l, 17. £+££, 18. +t =
y y x 2 2 5
5.
4 „ e 3 2 5 5
?/+=25. = 1.  +  = 7;
11 11
19. x +y = 1072, 20. xy^+yx^=20, 21. # 2 +y 2 = 5,
11 33 11
^3 +y 3 = ie. ^ 2 +y 2 =65. 6(.i? 2 +y 2 ) = 5.
MISCELLANEOUS EQUATIONS. 107
22. Jx+y+J7y = 4, 23. y + Jx 2  1 = 2,
24 . JZ+JZ^, 25. f~f + v^ = i_ 7 ,
26. tf* + 4y 2  15* = 10 (3y  8), xy = 6.
27. .r 2 y 2 + 400 = 41ay, y 2 = 5.ry  4.r 2 .
28. 4i 2 + 5y=6+2Qay25y a + 2.v, 7#lly = 17.
29. 9. c 2 + 33.r 12 = 1 2xy  4,y 2 + 22y , at?  ovy = 18.
30. (.v 2  y 2 ) (.r  y) = 1 6a^, (a 4  y 4 ) (a 3  y 2 ) = 64Ga?y.
31 . 2.v 2  xy +y 2 =2y, 2x 2 + 4xy = 5^.
32. y—% + , —% = o , 5.v  7y = 4.
(.r + y) 2 (#y) 2 8
33. y(y 2 3.r#.r 2 ) + 24 = 0, x(j/ 2 4xy + 2x 2 ) + 8 = 0.
34. 3a 3  8ay 2 + if + 2 1 = 0, a 2 (y  x) = 1 .
35. y 2 (4v 2  108) = x (x 3  9y 3 ), 2x 2 + 9xy + y 2 = 108.
36. 6x i + x 2 y 2 + l6 = 2x(\2x+y 3 ), x 2 + xyy 2 = 4.
37. x (a + x)=y(b+y), ax + by = (x + y) 2 .
38. xy + «Z> = 2ax, xhf + a 2 b 2 = 26 2 y 2 .
39. fir_ a + .lzi > = _J L L . =0 .
a 2 b 2 x — b y — a a — b
40. 6.v 3 = 10a 2 6.r + 3a 3 y, ay 3 = 10ab 2 y + 3b 3 x.
41. 2a('A+4a 2 = 4:X 2 +^t i
\y xj 2a a*
137. Equations involving three or more unknown quantities
can only be solved in special cases. We shall here consider some
of the most useful methods of solution.
Example 1. Solve x + y +z =13 (1),
.7^ + 2/2 + 22 = 65 (2),
xy = 10 (3).
From (2) and (3), (x + yf + * 2 = 85.
Put u for x + y ; then this equation becomes
u*+z*=85.
108 HIGHER ALGEBRA.
Also from (1), u +z =13;
whence we obtain u = l or 6; z = 6 or 7.
Thus we have x + y = 7,1 and £ + ?/ = 6,
#?/ = 10 \ acy = 10
Hence the solutions are
x=5, or 2,' .r = 3db\/l.,
y = 2, or 5,1 or y^W^T,
* = 6 5 J z=l.
Example 2. Solve (a; + y) {x + z) = 30,
{y + z)(y + x) = 15,
[z+x)(z+y) =18.
Write m, 1;, w for ?/ + 2, a + as, a; + y respectively ; thus
viv — 30, tvu = 15, mv = 18 (1).
Multiplying these equations together, we have
wVu> 2 _ 30 x 15 x 18 = 15* x 6 2 ;
.*. uvw = ±90.
Combining this result with each of the equations in (1), we have
u = 3, v = 6, w = 5\ or w = 3, v = 6, w=5;
.. y + z=3,\ y+z=S,\
z + x = $, > or z+x = d>,\
x + y = 5)) x + y = 5,i
whence ce=4, y = l, 2 = 2; or x=i, y=l, «=2.
Example 3. Solve y 2 + ys + 2 2 = 49 (1),
2 2 + z:r + a; 2 = 19 (2),
x* + xy + y 2 =39 : (3).
Subtracting (2) from (1)
y 2 x 2 + z{y «)=30;
that is, (yx){x + y + z) = 30 (4).
Similarly from (1) and (3)
[zx){x+y+z)*=10 (5).
Hence from (4) and (5), by division
y* 3 .
«« ■
whence y = 3z2x.
MISCELLANEOUS EQUATIONS. 10f)
Substituting in equation (3), we obtain
z*8xa+8z s =13.
From (2), x 2 + xz + z~ = 19.
Solving tbese homogeneous equations as in Example 4, Art. 130, we obtain
a;=±2, z = ± 3 ; and therefore y = ± 5 ;
or jc= ±ts, 2= ± t^ ; and therefore y= T — ,
Example 4. Solve .t 2 yz = a 2 , y^ zx = 6 2 , z 2 — xy = c 2 .
Multiply the equations by y, 2, a; respectively and add ; then
c 2 .r + « 2 // + & 2 z = (1).
Multiply the equations by z, x, y respectively and add ; then
b 2 x + cy + a*z = (2).
From (1) and (2), by cross multiplication,
~^¥c 2 = V^W = ^W 2 = k su PP° se '
Substitute in any one of the given equations ; then
k 2 (a 6 + b 6 + c 6  3a 2 Z> 2 c 2 ) = 1 ;
x 11 z 1
a 4_^2 c 2 ^4_ c 2 ft2 C 4_ a i 7/ j *J a *+b*+c*3a?tP<?
EXAMPLES. X. c.
Joh
r e the following equations :
1.
9#+y8z=0,
2.
3a?+y2s=0,
4a? 8y +7«=0,
4^7 y 32 = 0,
yz+zx + xy = 47.
. r 3 + ^3 + 2 3 = 467<
3.
xyz=2,
4.
#+2^3=11,
.<v2+f z 2 = 22,
. r 2_ 4<y 2 + (S 2 = 37)
xy = b.
ass = 24.
5,
x 2 +fz 2 = 2l,
6.
.r 2 + xy + .?£ = 18,
3xz + 3yz2xy=\8,
y 2 +yz+y.v+ 12 = 0,
x+yz = 5.
z 2 + zx + zy = 30.
7.
x*+2xy+3xz=50,
8.
(y*)(s+#)=22,
2y 2 + 3yz+yx=\0,
(*+*) (*y)=33,
+ z.u+2:y = \n.
(,y)( t yc) = G.
110 HIGHER ALGEBRA.
9. x*y*zhi=\% #VW=8, x*yz 2 u 2 = l, 3xy 2 z 2 u 2 = 4.
10. aPy*z=12 t ^ 3 =54, .*% 3 2 2 = 72.
11. ay+#+y=23, 12. 2^4?+2/ = 17,
xz+x + z = 4l, 3yz+y6z = 52,
yz + ij + z = 27. §xz + 3s + 2#= 29.
13. xz+y^lz, yz + x=8z, x + y + z = l2.
14. .r 3 +y 3 + ^ 3 =a 3 , ^ 2 +y 2 + 2 2 = a 2 , # + # + s = a.
15. ^ 2 +y 2 +2 2 =3/^ + 2^ + .«y = « 2 , 3.r# + 2 = a*/3.
16. # 2 +y 2 M 2 = 21a 2 , ys + ^.ry = 6a 2 , 3x+y2z = 3a.
Indeterminate Equations.
138. Suppose the following problem were proposed for solu
tion :
A person spends .£461 in buying horses and cows; if each
horse costs £23 and each cow £16, how many of each does he buy 1 ?
Let x, y be the number of horses and cows respectively ; then
23a; + 16^ = 461.
Here we have one equation involving two unknown quantities,
and it is clear that by ascribing any value we please to x, we can
obtain a corresponding value for y ; thus it would appear at first
sight that the problem admits of an infinite number of solutions.
But it is clear from the nature of the question that x and y must
be positive integers ; and with this restriction, as we shall see
later, the number of solutions is limited.
If the number of unknown quantities is greater than the
number of independent equations, there will be an unlimited
number of solutions, and the equations are said to be indeter
minate. In the present section we shall only discuss the simplest
kinds of indeterminate equations, confining our attention to posi
tive integral values of the unknown quantities ; it will be seen
that this restriction enables us to express the solutions in a very
simple form.
The general theory of indeterminate equations will be found
in Chap. xxvi.
INDETERMINATE EQUATIONS.
Ill
Example 1. Solve 7# + 12j/ = 220 in positive integers.
Divide throughout by 7, the smaller coefficient ; thus
x + y+^ =31 + ;
.. x + y+^~ =31 ...
Since x anil y are to be integers, we must have
5yS
(1)
and therefore
that is,
and therefore
7
l%9
= integer ;
= integer ;
w2
%l+ *== integer;
1/2
: integer =p suppose.
or
.. y2 = 7p,
y = lp + 2
Substituting this value of y in (1),
.r + 7p + 2 + 5> + l = 31;
that is, x = 2§l2p
(2).
.(3).
If in these results we give to p any integral value, we obtain corresponding
integral values of x and y; but if p > 2, we see from (3) that x is negative ;
and if p is a negative integer, y is negative. Thus the only positive integral
values of x and y are obtained by putting p = 0, 1, 2.
The complete solution may be exhibited as follows :
p= 0, 1, 2,
a: = 28, 16, 4,
y= 2, 9, 16.
Note. When we obtained
5yS
integer, we multiplied by 3 in order
to make the coefficient of y differ by unity from a multiple of 7. A similar
artifice should always be employed before introducing a symbol to denote
the integer.
Example 2. Solve in positive integers, 14x  11// = 29.
Divide by 11, the smaller coefficient; thus
(1).
x +
Sx
11
i/2 + ir ;
3x7
11
= 2  x + y = integer ;
112 HIGHER ALGEBRA.
12£  28 .
hence — ^ = m teg er *>
g* ^
that is, x  2 + —— = integer ;
Qfc ^
.. '— — = integer =_p suppose;
!
.*. X = \\p + §
and, from (1), y — 14p + 5
This is called the general solution of the equation, and by giving to p
any positive integral value or zero, we obtain positive integral values of x
and y ; thus we have
p = 0, 1, 2, 3,
.t = 6, 17, 28, 39,
y = 5, 19, 33, 47,
the number of solutions being infinite.
Example 3. In how many ways can £5 be paid in halfcrowns and florins?
Let x be the number of halfcrowns, y the number of florins ; then
5^ + 4y = 200;
••• x +y+\= 5 °;
x .
.' . 2 — integer = 2^ suppose ;
.*. x=4p,
and y = 505p.
Solutions are obtained by ascribing to p the values 1, 2, 3, ...9; and
therefore the number of ways is 9. If, however, the sum may be paid either
in halfcrowns or florins, p may also have the values and 10. If ^ = 0,
then x = 0, and the sum is paid entirely in florins ; if p = 10, then y = 0, and
the sum is paid entirely in halfcrowns. Thus if zero values of x and y are
admissible the number of ways is 11.
Example 4. The expenses of a party numbering 43 were £5. 14s. Qd. ; if
each man paid 5s., each woman 2s. 6d., and each child Is., how many were
there of each?
Let x, y, z denote the number of men, women, and children, respectively;
then we have
x + y + z= 43 (1),
10.r + 5?/ + 2z = 229.
Eliminating z, we obtain 8x + By = 143.
The general solution of this equation is
x=Sp + l,
y = 458p;
INDETERMINATE EQUATIONS. 113
Hence by substituting in (1), we obtain
z = 5p3.
Here p cannot be negative or zero, but may have positive integral values
from 1 to 5. Thus
p= 1, 2, 3, 4, 5;
x 4, 7, 10, 13, 16; *
y = 37, 29, 21, 13, 5;
2=2, 7, 12, 17, 22.
EXAMPLES. X. d.
Solve in positive integers :
1. 3.i + 8y = 103. 2. 5#+2y=53. 3. 7.>;+ 12y=152.
4. l&P+lly=414 5. 23a?+25y=915. 6. 4L>; + 47y = 2191.
Find the general solution in positive integers, and the least values
of x and y which satisfy the equations :
7. 5.v7y = 3. 8. 6a?13y=l. 9. 8#2ty=33.
10. I7y13#=0. 11. 19y23a?=7. 12. 77y3Qa?=295.
13. A farmer spends £752 in buying horses and cows ; if each horse
costs £37 and each cow £23, how many of each does he buy ?
14. In how many ways can £5 be paid in shillings and sixpences,
including zero solutions ?
15. Divide 81 into two parts so that one may be a multiple of 8
and the other of 5.
16. What is the simplest way for a person who has only guineas
to pay 105. 6d. to another who has only halfcrowns ?
17. Find a number which being divided by 39 gives a remainder 16,
and by 56 a remainder 27. How many such numbers are there ?
18. What is the smallest number of florins that must be given to
discharge a debt of £1. (5s. 6d., if the change is to be paid in halfcrowns
only?
19. Divide 136 into two parts one of which when divided by 5
leaves remainder 2, and the other divided by 8 leaves remainder 3.
20. I buy 40 animals consisting of rams at £4, pigs at £2, and oxen
at £17 : if I spend £301, how many of each do I buy ?
21. In my pocket I have 27 coins, which are sovereigns, halfcrowns
or shillings, and the amount I have is £5. 05. 6d. ; how many coins of
each sort have I ?
H. H. A. 8
CHAPTER XL
Permutations and Combinations.
139. Each of the arrangements which can be made by taking
some or all of a number of things is called a permutation.
Each of the groups or selections which can be made by taking
some or all of a number of things is called a combination.
Thus the •permutations which can be made by taking the
letters a, b, c, d two at a time are twelve in number, namely,
ab, ac, ad, be, bd, cd,
ba, ca, da, cb, db, dc ;
each of these presenting a different arrangement of two letters.
The combinations which can be made by taking the letters
a, b, c, d two at a time are six in number : namely,
ab, ac, ad, be, bd, cd;
each of these presenting a different selection of two letters.
From this it appears that in forming combinations we are only
concerned with the number of things each selection contains ;
whereas in forming permutations we have also to consider the
order of the things which make up each arrangement; for instance,
if from four letters a, b, c, d we make a selection of three, such
as abc, this single combination admits of being arranged in the
following ways :
abc, acb, bca, bac, cab, cba,
and so gives rise to six different permutations.
PERMUTATIONS AND COMBINATIONS. 115
140. Before discussing the general propositions of this
chapter there is an important principle which we proceed to
explain and illustrate by a few numerical examples.
If one operation can be performed in m ivays, and (when it
has been performed in any one of these ways) a second operation
can then be performed in n tvays ; the number of ways of per
forming the two operations ivill be m x n.
If the first operation be performed in any one way, we can
associate with this any of the n ways of performing the second
operation : and thus we shall have n ways of performing the two
operations without considering more than one way of performing
the first; and so, corresponding to each of the m ways of per
forming the first operation, we shall have n ways of performing
the two; hence altogether the number of ways in which the two
operations can be performed is represented by the product
m x n.
Example 1. There are 10 steamers plying between Liverpool and Dublin;
in how many ways can a man go from Liverpool to Dublin and return by a
different steamer?
There are ten ways of making the first passage ; and with each of these
there is a choice of nine ways of returning (since the man is not to come back
by the same steamer) ; hence the number of ways of making the two journeys
is 10 x 9, or 90.
This principle may easily be extended to the case in which
there are more than two operations each of which can be per
formed in a given number of ways.
Example 2. Three travellers arrive at a town where there are four
hotels; in how many ways can they take up their quarters, each at a
different hotel?
The first traveller has choice of four hotels, and when he has made his
selection in any one way, the second traveller has a choice of three ; there
fore the first two can make their choice in 4 x 3 ways ; and with any one such
choice the third traveller can select his hotel in 2 ways ; hence the required
number of ways is 4 x 3 x 2, or 24.
141. To find the number of permutations of \\ dissimilar things
taken r at a time.
This is the same thing as finding the number of ways in which
we can fill up r places when we have n different things at our
disposal.
The first place may be tilled up in n ways, for any one of the n
things may be taken ; when it has been filled up in any one of
8—2
116 HIGHER ALGEBRA.
these ways, the second place can then be filled up in n  1 ways ;
and since each way of filling up the first place can be associated
with each way of filling up the second, the number of ways in
which the first two places can be filled up is given by the product
n (n  1). And when the first two places have been filled up in
any way, the third place can be filled up in h — 2 ways. And
reasoning as before, the number of ways in which three places can
be filled up is n (n  1) (n  2).
Proceeding thus, and noticing that a new factor is introduced
with each new place filled up, and that at any stage the number
of factors is the same as the number of places filled up, we shall
have the number of ways in which r places can be filled up
equal to
n (n l)(n— 2) to r factors ;
and the r th factor is
n — (r— 1), or n — r+1.
Therefore the number of permutations of n things taken r at
a time is
n{n 1) (n 2) (nr + 1).
Cor. The number of permutations of n things taken all at
a time is
n (n  1) (?i  2) to n factors,
or n(n — Y)(n—2) 3.2.1.
It is usual to denote this product by the symbol \n, which is
read "factorial n." Also n\ is sometimes used for \n.
142. We shall in future denote the number of permutations
of n things taken r at a time by the symbol n P r , so that
"P r = w(wl)(w2) (nr + 1);
also "P = \n.
In working numerical examples it is useful to notice that the
suffix in the symbol n P r always denotes the number of factors in
the formula we are using.
143. The number of permutations of n things taken r at
a time may also be found in the following manner.
Let "P r represent the number of permutations of n things
taken r at a time.
PERMUTATIONS AND COMBINATIONS. 117
Suppose we form all the permutations of n things t;iken r — 1
at a time ; the number of these will be "P .
' r—l
With each of these put one of the remaining n — r + 1 tilings.
Each time we do this we shall get one permutation of u things
r at a time; and therefore the whole number of the permutations
of n things r at a time is n P r _ ] x (n  r + 1) ; that is,
By writing r—l for r in this formula, we obtain
"P_ 1 = '^ r _ 2 x(nrf2),
similarly, 'P = 'P r _ a x (n  r + 3),
"P^P.x (71 I),
"P x =7l.
Multiply together the vertical columns and cancel like factors
from each side, and we obtain
n P r = n(nl)(n2) (nr+l).
Example 1. Four persons enter a railway carriage in which there are six
seats ; in how many ways can they take their places ?
The first person may seat himself in 6 ways ; and then the second person
in 5 ; the third in 4 ; and the fourth in 3 ; and since each of these ways may
be associated with each of the others, the required answer is 6x5x4x3,
or 360.
Example 2. How many different numbers can be formed by using six out
of the nine digits 1, 2, 3, ...9?
Here we have 9 different things and we have to find the number of per
mutations of them taken 6 at a time ;
. * . the required result = 9 P 6
=9x8x7x6x5x4
= 60480.
144. To find the number of combinations of n dissimilar
tilings taken r at a time.
Let "C r denote the required number of combinations.
Then each of these combinations consists of a group of r
dissimilar things which can be arranged among themselves in
r ways. [Art. 142.]
118 HIGHER ALGEBRA.
Hence "C r x \r is equal to the number of arrangements of n
things taken rata time ; that is,
*C x\r = "P
r  r
= n (n — 1) (n — 2) . . . (n  r + 1) ;
_ tt(wl)(w2)...(wr+l)
r V '"
Cor. This formula for n C r may also be written in a different
form ; for if we multiply the numerator and the denominator by
\n — r we obtain
n
(n  1) (n  2) ... {n  r + 1) x \ n — r
\r n — r
The numerator now consists of the product of all the natural
numbers from n to 1 ;
\n
.'. "C r = . ~ (2).
It will be convenient to remember both these expressions for
n C r , using (1) in all cases where a numerical result is required,
and (2) when it is sufficient to leave it in an algebraical shaj)e.
Note. If in formula (2) we put r = n, we have
\n i
n ~jn_0" 0'
but n C n =l, so that if the formula is to be true for r = n, the symbol 10 must
be considered as equivalent to 1.
Example. From 12 books in how many ways can a selection of 5 be
made, (1) when one specified book is always included, (2) when one specified
book is always excluded ?
(1) Since the specified book is to be included in every selection, we
have only to choose 4 out of the remaining 11.
Hence the number of ways = n C 4
1 1x10 x9x8
~ 1x2x3x4
= 330.
PERMUTATIONS AND COMBINATIONS. 119
(2) Since the specified book is always to be excluded, we have to
select the 5 books out of the remaining 11.
Hence the number of ways = n C 6
_ 11x10x9x8x7
1x2x3x4x5
= 462.
145. The number of combinations of n things r at a time is
equal to the number of combinations of\\ things n — r at a time.
In making all the possible combinations of n things, to each
group of r things we select, there is left a corresponding group of
n  r things ; that is, the number of combinations of n things
r at a time is the same as the number of combinations of n things
n — r at a time ;
.. "C = n C .
r n — r
The proposition may also be proved as follows :
\n
"0 _ r = =— [Art. 144.1
n — r
n
— (n  r)
n
n — r r
Such combinations are called complementary.
Note. Put r=w, then tt C = n C n =l.
The result we have just proved is useful in enabling us to
abridge arithmetical work.
Example. Out of 14 men in how many ways can an eleven be chosen?
The required number = 14 C U
14 x 13 x 12
1x2x3
= 364.
If we had made use of the formula u C n , we should have had to reduce au
expression whose numerator and denominator each contained 11 factors.
120 HIGHER ALGEBRA.
146. Tojind the number of ways in which m + n things can be
divided into two groups containing in and n things respectively.
This is clearly equivalent to finding the number of combi
nations of ra + n things ra at a time, for every time we select
one group of ra things we leave a group of n things behind.
Ira + n
Thus the required number = h=
1 ra \7b
Note. If n = m, the groups are equal, and in this case the number of
\2m
different ways of subdivision is  — ~ — [9 ; for in any one way it is possible
to interchange the two groups without obtaining a new distribution.
147. To jind the number of ways in which m + n + p things can
be divided into three groups containing m, n, p things severally.
First divide ra + n + p things into two groups containing m
and n + p things respectively : the number of ways in which this
\m + n+p
can be done is r=
\m
n+p
Then the number of ways in which the group of n+p things
can be divided into two groups containing n and p things respec
\n+p
tively is
n p
Hence the number of ways in which the subdivision into three
groups containing m, n, p things can be made is
m + n+p n+p \m + n + ]>
x , ,  , or
in
n + p \n \p 5 Ira \n \p
J3wi
Note. If we put ?i=p = m. we obtain : — r= — ; hut this formula regards
as different all the possible orders in which ~th.e three groups can occur in
any one mode of subdivision. And since there are 13 such orders cor
responding to each mode of subdivision, the number of different ways in
3ot
which subdivision into three equal groups can be made is  — r^f — r^ •
771 771 m 3
Example. The number of ways in which 15 recruits can be divided into
115
three equal groups is ,   ; and the number of ways in which they
I 15
can be drafted into three different regiments, five into each, is _— Hr — .
[6 J 5 [6
PERMUTATIONS AND COMBINATIONS. 121
148. In the examples which follow it is important to notice
that the formula for 'permutations should not be used until the
suitable selections required by the question have been made.
Example 1. From 7 Englishmen and 4 Americans a committee of is to
be formed; in how many ways can this be done, (1) when the committee con
tains exactly 2 Americans, (2) at least 2 Americans ?
(1) "We have to choose 2 Americans and 4 Englishmen.
The number of ways in which the Americans can be chosen is 4 C, ; and
the number of ways in which the Englishmen can be chosen is 7 C 4 . Each of
the first groups can be associated with each of the second ; hence
the required number of ways = 4 C 2 x 7 C 4
li \1
= ~2"[2 X TTJ3
17
'J^ = 210.
223
(2) The committee may contain 2, 3, or 4 Americans.
"We shall exhaust all the suitable combinations by forming all the groups
containing 2 Americans and 4 Englishmen ; then 3 Americans and 3 English
men; and lastly 4 Americans and 2 Englishmen.
The sum of the three results will give the answer. Hence the required
number of ways = *C 2 x 7 C 4 + 4 C 3 x 7 (7 3 + 4 C 4 x 7 C,
17 4 17 17
X TTT^ + TK X rl  ^ + 1 X
[2 1 2 [4 j_3 j_3 34 [2)5
= 210 + 140 + 21 = 371.
In this Example we have only to make use of the suitable formulae for
combinations, for we are not concerned with the possible arrangements of the
members of the committee among themselves.
Example 2. Out of 7 consonants and 4 vowels, how many words can be
made each containing 3 consonants and 2 vowels?
The number of ways of choosing the three consonants is 7 C 3 , and the
number of ways of choosing the 2 vowels is *C a ; and since each of the first
groups can be associated with each of the second, the number of combined
groups, each containing 3 consonants and 2 vowels, is 7 C 3 x 4 C 2 .
Further, each of these groups contains 5 letters, which may be arranged
among themselves in [5 ways. Hence
the required number of words = 7 C 3 x 4 C 2 x Jo
~34 X [2]2 X "
= 5x7
r
= 25200.
122 HIGHER ALGEBRA.
Example 3. How many words can be formed out of the letters article, so
that the vowels occupy the even places?
Here we have to put the 3 vowels in 3 specified places, and the 4 conso
nants in the 4 remaining places ; the first operation can be done in 1 3 ways,
and the second in 1 4 . Hence
the required number of words =3x[4
= 144.
In this Example the formula for permutations is immediately applicable,
because by the statement of the question there is but one way of choosing the
vowels, and one way of choosing the consonants.
EXAMPLES XI. a.
1. In how many ways can a consonant and a vowel be chosen out of
the letters of the word courage?
2. There are 8 candidates for a Classical, 7 for a Mathematical, and
4 for a Natural Science Scholarship. In how many ways can the
Scholarships be awarded?
3. Find the value of 8 P 7 , 25 P 5 , 24 <7 4 , 19 C U .
4. How many different arrangements can be made by taking 5
of the letters of the word equation ?
5. If four times the number of permutations of n things 3 together
is equal to five times the number of permutations of n — 1 things
3 together, find n.
6. How many permutations can be made out of the letters of
the word triangle? How many of these will begin with t and end
with e ?
7. How many different selections can be made by taking four of
the digits 3, 4, 7, 5, 8, 1 ? How many different numbers can be formed
with four of these digits ?
8. If 2n C 3 : n Oj = 44 : 3, find n.
9. How many changes can be rung with a peal of 5 bells ?
10. How many changes can be rung with a peal of 7 bells, the tenor
always being last ?
11. On how many nights may a watch of 4 men be drafted from a
crew of 24, so that no two watches are identical ? On how many of these
would any one man be taken?
12. How many arrangements can be made out of the letters of the
w r ord draught, the vowels never being separated ?
PERMUTATIONS AND COMBINATIONS. 1 23
13. In a town council there are 25 councillors and 10 aldermen ;
how many committees can be formed each consisting of 5 councillors
and 3 aldermen ?
14. Out of the letters A, B, C, p, q, r how many arrangements can
be made (1) beginning with a capital, (2) beginning and ending with a
capital ]
15. Find the number of combinations of 50 things 4G at a time.
16. If n C 12 = n C s , find n C 17 , 22 <7 n .
17. In how many ways can the letters of the word vowels be
arranged, if the letters oe can only occupy odd places ]
18. From 4 officers and 8 privates, in how many ways can 6 be
chosen (1) to include exactly one officer, (2) to include at least one
officer?
19. In how many ways can a party of 4 or more be selected from
10 persons ?
20. If ™C r = ls C r + 2 , find'<7 5 .
21. Out of 25 consonants and 5 vowels how many words can be
formed each consisting of 2 consonants and 3 vowels ?
22. In a library there are 20 Latin and 6 Greek books; in how
many ways can a group of 5 consisting of 3 Latin and 2 Greek books be
placed on a shelf ?
23. In how many ways can 12 things be divided equally among 4
persons ?
24. From 3 capitals, 5 consonants, and 4 vowels, how many words
can be made, each containing 3 consonants and 2 vowels, and beginning
with a capital ?
25. At an election three districts are to be canvassed by 10, 15, and
20 men respectively. If 45 men volunteer, in how many ways can they
be allotted to the different districts ?
26. In how many ways can 4 Latin and 1 English book be placed
on a shelf so that the English book is always in the middle, the selec
tion being made from 7 Latin and 3 English books?
27. A boat is to be manned by eight men, of whom 2 can only row
on bow side and 1 can only row on stroke side; in how many ways can
the crew be arranged ?
28. There are two works each of 3 volumes, and two works each of
2 volumes ; in how many ways can the 10 books be placed on a shelf so
that volumes of the same work are not separated ?
29. In how many w r ays can 10 examination papers be arranged so
that the befit and worst papers never come together?
124 HIGHER ALGEBRA.
30. An eightoared boat is to be manned by a crew chosen from 11
men, of whom 3 can steer but cannot row, and the rest can row but can
not steer. In how many ways can the crew be arranged, if two of the
men can only row on bow side?
31. Prove that the number of ways in which p positive and n
negative signs may be placed in a row so that no two negative signs shall
be together is p + 1 C n .
32. If 56 P r + 6 : 54 P r + 3 = 30800 : 1, find r.
33. How many different signals can be made by hoisting 6 differ
ently coloured flags one above the other, when any number of them
may be hoisted at once ?
34. U^C 2r : 24 C 2r _ 4 = 225 : 11, find r.
149. Hitherto, in the formulae we have proved, the things
have been regarded as unlike. Before considering cases in which
some one or more sets of things may be like, it is necessary to
point out exactly in what sense the words like and unlike are
used. When we speak of things being dissimilar, different, un
like, we imply that the things are visibly unlike, so as to be
easily distinguishable from each other. On the other hand we
shall always use the term like things to denote such as are alike
to the eye and cannot be distinguished from each other. For
instance, in Ex. 2, Art. 148, the consonants and the vowels may
be said each to consist of a group of things united by a common
characteristic, and thus in a certain sense to be of the same kind;
but they cannot be regarded as like things, because there is an
individuality existing among the things of each group which
makes them easily distinguishable from each other. Hence, in
the final stage of the example we considered each group to
consist of five dissimilar things and therefore capable of [5
arrangements among themselves. [Art. 141 Cor.]
150. Suppose we have to find all the possible ways of arrang
ing 12 books on a shelf, 5 of them being Latin, 4 English, and
the remainder in different languages.
The books in each language may be regarded as belonging to
one class, united by a common characteristic ; but if they were
distinguishable from each other, the number of permutations
would be )12, since for the purpose of arrangement among them
selves they are essentially different.
PERMUTATIONS AND COMBINATIONS. 125
If, however, the books in the same language are not dis
tinguishable from each other, we should have to find the number
of ways in which 12 things can be arranged among themselves,
when 5 of them are exactly alike of one kind, and 4 exactly alike,
of a second kind : a problem which is not directly included in any
of the cases we have previously considered.
151. To find the number of ways in which n things may be
arranged among themselves, taking them all at a time, when p
of the things are exactly alike of one kind, q of them exactly
alike of another kind, r of them exactly alike of a third kind, and
the rest all different.
Let there be n letters ; suppose p of them to be a, q of them
to be b, r of them to be c, and the rest to be unlike.
Let x be the required number of permutations ; then if in
any one of these permutations the_p letters a were replaced by p
unlike letters different from any of the rest, from this single
permutation, without altering the position of any of the remaining
letters, we could form I p new permutations. Hence if this change
were made in each of the x permutations we should obtain x x \p
permutations.
Similarly, if the q letters b were replaced by q unlike letters,
the number of permutations would be
x x \p x <7.
In like manner, by replacing the r letters c by r unlike letters,
we should finally obtain x x \p x \q x \r permutations.
But the things are now all different, and therefore admit of \n
permutations among themselves. Hence
x x \p x \q x
r \n;
r
that is, x — ~. — ' •
\p \g p
which is the required number of permutations.
Any case in which the things are not all different may be
treated similarly.
126 HIGHER ALGEBRA.
Example 1. How many different permutations can be made out of the
letters of the word assassination taken all together ?
We have here 13 letters of which 4 are s, 3 are a, 2 are i, and 2 are n.
Hence the number of permutations
~^[32j£
= 13.11.10.9.8.7.3.5
= 1001 x 10800 = 10810800.
Example 2. How many numbers can be formed with the digits
1, 2, 3, 4, 3, 2, 1, so that the odd digits always occupy the odd places?
The odd digits 1, 3, 3, 1 can be arranged in their four places in
l^2 wa y s (1) 
The even digits 2, 4, 2 can be arranged in their three places in
13
y^ ways (2).
Each of the ways in (1) can be associated with each of the ways in (2).
14 13
Hence the required number = y^=x x j^ = 6 x 3 = 18.
152. To find the number of permutations of n things r at a
time, when each thing may be repeated once, twice, up to r
times in any arrangement.
Here we have to consider the number of ways in which r
places can be filled up when we have n different things at our
disposal, each of the n things being used as often as we please in
any arrangement.
The first place may be filled up in n ways, and, when it has
been filled up in any one way, the second place may also be filled
up in n ways, since we are not precluded from using the same
thing again. Therefore the number of ways in which the first
two places can be filled up iswxn or n 2 . The third place can
also be filled up in n ways, and therefore the first three places in
n 3 ways.
Proceeding in this manner, and noticing that at any stage the
index of n is always the same as the number of places filled up,
we shall have the number of ways in which the r places can be
filled up equal to n r .
PERMUTATIONS AND COMBINATIONS. 127
Example. In how many ways can 5 prizes be given away to 4 boys, when
each boy is eligible for all the prizes?
Any one of the prizes can be given in 4 ways; and then any one of the;
remaining prizes can also be given in 4 ways, since it may be obtained by the
boy who has already received a prize. Thus two prizes can be given away in
4 a ways, three prizes in 4 :! ways, and so on. Hence the 5 prizes can be given
away in 4 5 , or 1024 ways.
153. To find the total number of ways in which it is possible
to make a selection by taking some or all of \\ things.
Each tiling may be dealt with in two ways, for it may either
be taken or left; and since either way of dealing with any one
thing may be associated with either way of dealing with eacli one
of the others, the number of selections is
2x2x2x2 to n factors.
But this includes the case in which all the things are left,
therefore, rejecting this case, the total number of ways is 2"l.
This is often spoken of as "the total number of combinations"
of n things.
Example. A man has 6 friends ; in how many ways may he invite one or
more of them to dinner?
He has to select some or all of his 6 friends ; and therefore the number of
ways is 2 s  1, or 63.
This result can be verified in the following manner.
The guests may be invited singly, in twos, threes, ; therefore the
number of selections = 6 C 1 + 6 C 2 + 6 C 3 + 6 C 4 + 6 C 5 + <>C 6
= 6 + 15 + 20 + 15 + 6 + 1 = 63.
154. To find for what value of r the number of combinations
of n things r at a time is greatest.
Since "C = ^( ?l  1 )( n  2 ) (wr + 2)(nr + l)
1.2.3 (rl)r
, _ n(nl)(n2) (wr + 2)
1. 2.3 (r1)
"C = n C . x
n — r + 1
r
The multiplying factor may be written —   1,
which shews that it decreases as r increases. Hence as r receives
128 HIGHER ALGEBRA.
the values 1, 2, 3 in succession, n G r is continually increased
71 4 1
until 1 becomes equal to 1 or less than 1.
r
Now 1^1,
r
i 71+1 ^
so long as > z ;
r
that is, — —  > r.
We have to choose the greatest value of r consistent with
this inequality.
(1) Let n be even, and equal to 2m; then
n + 1 2m +1 1
2 2— — + s ;
and for all values of ?• up to ?n inclusive this is greater than r.
Hence by putting r — m = — , we find that the greatest number of
combinations is "C .
n
2
(2) Let n be odd, and equal to 2m + 1 ; then
n + 1 2m + 2 
— =5— » + li
and for all values of r up to m inclusive this is greater than r ;
but when r  m + 1 the multiplying factor becomes equal to 1, and
*C.= n C : that is, "C +  n C •
mi+I m ' ' n+ 1 7i—l J
2 2
and therefore the number of combinations is greatest when the
things are taken — — , or — ^— at a time; the result being the
same in the two cases.
155. The formula for the number of combinations of n things
r at a time may be found without assuming the formula for the
numbes of permutations.
Let "C r denote the number of combinations of n things taken
r at a time; and let the n things be denoted by the letters
a, b, c, d,
PERMUTATIONS AND COMBINATIONS. 120
Take away a; then with tin 1 remaining letters we cm form
"~ X C combinations of n— 1 letters taken r  1 at a time. With
eaeli of these write a; thus we see that of the combinations
of n tilings r at a time, the number of those which contain
a is w ~ l C x \ similarly the number of those which contain
b is n ~ x C , : and so for each of the n letters.
Tlierefore n x "~*C r _ l is equal to the number of combinations
r at a time which contain a, together with those that contain b,
those that contain c, and so on.
But by forming the combinations in this manner, each par
ticular one will be repeated r times. For instance, if r=3, the
combination abc will be found anions; those containing a, amonir
those containing b, and among those containing c. Hence
*c= n  x c r 1 x.
r r— i .,
By writing u — 1 and r — 1 instead of n and r respectively,
ni 1
rl°
»2
Similarly, V^ = ^G r _ z x 
n— r + 2/~1 _nr + \ri
U — T +
o
>.
2 ^i 2 ;
and finally, n  r+1 C 1 = »r + 1.
Multiply together the vertical columns and cancel like factors
from each side ; thus
"C . n (rcl)(n2) (nr+ l)
r(rl)(r2) 1
156. To find the total number of ways in which it is possible
to make a selection by taking some or all out qfip + c x +r +
tilings, ivJierenfp are alike of one kind, q alike of a second kind, r
alike of a third kind; and so on.
The p things may be disposed of in p + 1 ways ; for wo may
take 0, 1, 2, 3, p °f thorn. Similarly the q things may be
disposed of in q + \ ways; the r things in r+1 ways; and
so on.
H. II. A. 9
130 HIGHER ALGEBRA.
Hence the number of Avays in which all the tilings may be
disposed of is (^ + 1) (q + 1) (r + 1)
But this includes the case in which none of the things are
taken ; therefore, rejecting this case, the total number of
ways is
(jp + l)fe+l)(r + .l) 1.
157. A general formula expressing the number of permuta
tions, or combinations, of n things taken r at a time, when the
things are not all different, may be somewhat complicated ; but a
particular case may be solved in the following manner.
Example. Find the number of ways in which (1) a selection, (2) an ar
rangement, of four letters can be made from the letters of the word
proportion.
There are 10 letters of six different sorts, namely o, o,o; p,p; r, r; t; i; n.
In finding groups of four these may be classified as follows :
(1) Three alike, one different.
(2) Two alike, two others alike.
(3) Two alike, the other two different.
(4) All four different.
(1) The selection can be made in 5 ways ; for each of the five letters,
p, r, t, i s n, can be taken with the single group of the three like letters o.
(2) The selection can be made in 3 C 2 ways ; for we have to choose two out
of the three pairs o, o; p, p; r, r. This" gives 3 selections.
(3) This selection can be made in 3 x 10 ways ; for we select one of the
3 pairs, and then two from the remaining 5 letters. This gives 30 selections.
(1) This selection can be made in 6 C 4 ways, as we have to take 4 different
letters to choose from the six o, p, r, t, i, n. This gives 15 selections.
Thus the total number of selections is 5 + 3 + 30 + 15 ; that is, 53.
In finding the different arrangements of 4 letters we have to permute in
all possible ways each of the foregoing groups.
(1) gives rise to 5 x = , or 20 arrangements.
(2) gives rise to 3 x ^=^ , or 18 arrangements.
(3) gives rise to 30 x = , or 360 arrangements.
(4) gives rise to 15 x j4 , or 3G0 arrangements.
Thus the total number of arrangements is 20 + 18 + 360 + 360; that is, 758.
PERMUTATIONS AND COMBINATIONS. 131
EXAMPLES. XI. b.
1. Find the number of arrangements that can he made out of the
letters of the words
(1) independence, (2) superstitious,
(3) institutions.
2. In how many ways can 17 billiard balls be arranged, if 7 of
them are black, 6 red, and 4 white %
3. A room is to be decorated with fourteen flags ; if 2 of them are
blue, 3 red, 2 white, 3 green, 2 yellow, and 2 purple, in how many ways
can they be hung?
4. How many numbers greater than a million can be formed with
the digits 2, 3, 0, 3, 4, 2, 3?
5. Find the number of arrangements which can be made out of the
letters of the word algebra, without altering the relative positions of
vowels and consonants.
6. On three different days a man has to drive to a railway station,
and he can choose from 5 conveyances ; in how many ways can he make
the three journeys ?
7. I have counters of n different colours, red, white, blue, ; in
how many ways can I make an arrangement consisting of r counters,
supposing that there are at least r of each different colour ?
8. In a steamer there are stalls for 12 animals, and there are
cows, horses, and calves (not less than 12 of each) ready to be shipped;
in how many ways can the shipload be made?
9. In how many ways can n things be given to p persons, when
there is no restriction as to the number of things each may receive ?
10. In how many ways can five things be divided between two
persons ?
11. How many different arrangements can be made out of tl ie letters
in the expression a z b 2 c* when written at full length?
12. A letter lock consists of three rings each marked with fifteen
different letters ; find in how many ways it is possible to make an
unsuccessful attempt to open the lock.
13. Find the number of triangles which can be formed by joining
three angular points of a quindecagon.
14. A library has a copies of one book, b copies of each of two
books, c copies of each of three books, and single copies of d books. In
how many ways can these books be distributed, if all are out at once I
15. How many numbers less than 10000 can be made with the
eight digits 1, 2, 3, 0, 4, 5, 6, 7 ?
16. In how many ways can the following prizes be given away to a
class of 20 boys: first and second Classical, first and second Mathe
matical, first Science, and first French ?
9—2
132 HIGHER ALGEBRA.
17. A telegraph has 5 arms and each arm is capable of 4 distinct
positions, including the position of rest ; what is the total number of
signals that can be made ?
18. In how many ways can 7 persons form a ring? In how many
ways can 7 Englishmen and 7 Americans sit down at a round table, no
two Americans being together?
19. In how many ways is it possible to draw a sum of money from
a bag containing a sovereign, a halfsovereign, a crown, a florin, a shilling,
a penny, and a farthing?
20. From 3 cocoa nuts, 4 apples, and 2 oranges, how many selec
tions of fruit can be made, taking at least one of each kind ?
21. Find the number of different ways of dividing mn things into
n equal groups.
22. How many signals can be made by hoisting 4 flags of different
colours one above the other, when any number of them may be hoisted
at once ? How many with 5 flags ?
23. Find the number of permutations which can be formed out of
the letters of the word series taken three together ?
24. There are p points in a plane, no three of which are in the same
straight line with the exception of q, which are all in the same straight
line; find the number (1) of straight lines, (2) of triangles which result
from joining them.
25. There are p points in space, no four of which are in the same
plane with the exception of q, which are all in the same plane; find
how many planes there are each containing three of the points.
26. There are n different books, and p copies of each; find the
number of ways in which a selection can be made from them.
27. Find the number of selections and of arrangements that can be
made by taking 4 letters from the word expression.
28. How many permutations of 4 letters can be made out of the
letters of the word examination ?
29. Find the sum of all numbers greater than 10000 formed by
using the digits 1, 3, 5, 7, 9, no digit being repeated in any number.
30. Find the sum of all numbers greater than 10000 formed by
using the digits 0, 2, 4, 6, 8, no digit being repeated in any number.
31. If of p + q + r things p be alike, and q be alike, and the rest
different, shew that the total number of combinations is
(p + l)(q+l)2 r l.
32. Shew that the number of permutations which can be formed
from 2n letters which are either a's or 6's is greatest when the number
of a's is equal to the number of Z>'s.
33. If the n f 1 numbers a, b, c, d, be all different, and each of
them a prime number, prove that the number of different factors of the
expression a m bcd is (m + 1) 2 W — 1.
CHAPTER XIT.
Mathematical Induction.
158. Many important mathematical formula? are not easily
demonstrated by a direct mode of proof; in such cases we fre
quently find it convenient to employ a method of proof known as
mathematical induction, which we shall now illustrate.
Example 1. Suppose it is required to prove that the sum of the cubes
of the first n natural numbers is equal to < — ^— — 'J .
We can easily see by trial that the statement is true in simple cases, such
as when re=l, or 2, or 3 ; and from this we might be led to conjecture that
the formula was true in all cases. Assume that it is true when n terms are
taken ; that is, suppose
13 + 2 3 + 33 + to itteims= H ( ;t+1 )j 3 .
Add the («+ l) th term, that is, (n+ 1) 3 to each side ; then
13 + 2 3 + 3 3 + to n + 1 terms =j n ^ 2 +1 ^  +(n+iy
= {n + iy('j+n + l\
\
(n+l) 8 (n a +4n+4)
4
\
\ (n + l)(K + 2) )\
2 ! '
which is of the same form as the result we assumed to be true for n terms,
n + 1 taking the place of n ; in other words, if the result is true when we take
a certain number of terms, whatever that number may be, it is true when we
increase that number by one; but we see that it is true when 3 terms are
taken ; therefore it is true when 4 terms are taken ; it is therefore true when
5 terms are taken; and so on. Thus the result is true universally.
134 HIGHER ALGEBRA.
Example 2. To determine the product of n binomial factors of the form
x + a.
By actual multiplication we have
(x + a) (x + b) (x + c) = x 3 + (a + b + c) x 2 + (ab + bc + ca) x + abc ;
(x+a) (x + b) (x + c) (x + d) = x*+(a + b + c + d)x 3
+ (ab + ac+ ad + bc+ bd + cd) x~
+ (abc + abd + acd + bed) x + abed.
In these results we observe that the following laws hold :
1. The number of terms on the right is one more than the number of
binomial factors on the left.
2. The index of x in the first term is the same as the number of
binomial factors ; and in each of the other terms the index is one less than
that of the preceding term.
3. The coefficient of the first term is unity ; the coefficient of the second
term is the sum of the letters a, b, c, ; the coefficient of the third
term is the sum of the products of these letters taken two at a time;
the coefficient of the fourth term is the sum of their products taken three at
a time ; and so on ; the last term is the product of all the letters.
Assume that these laws hold in the case of n  1 factors ; that is, suppose
(x + a) (x+b)... (x + h) = x 71 ' 1 +p 1 x n ~ 2 +p. 2 x n ~ 3 +p. i x n  i + ... +p> n ^ ,
where p 1 = a + b + c+ ...h;
p. 2 = ab + ac + ... + ah + bc + bd+ ;
p 3 = abc + abd+ ;
p n _ x = abc...h.
Multiply both sides by another factor x + k ; thus
(x + a) (x + b) ... (x + h) (x + k)
= x n + (p x + k) x n ~ l + (p. 2 +p x k) x n ~* + (p 3 + pJc) x n ~ 3 +... +l^ n  x k.
Now ^i + A;:=(a + & + c + . ..+/*) + &
= sum of all the n letters a, b, c,...k;
p. 2 +p 1 k=p. 2 + k (a + b + ... + h)
= sum of the products taken two at a time of all the
n letters a, b, c, ... k;
p. A +p. 2 k =p 3 + k (ab + ac + . . . + ah + bc + . . .)
= sum of the products taken three at a time of all
the n letters a, b, c, ... k;
2? n _ 1 A* = product of all the n letters a, b, c, ... k.
MATHEMATICAL INDUCTION. 135
If therefore the laws hold when ?tl factors are multiplied together
they hold in the case of n factors. But we havo seen that they hold in the
case of 4 factors; therefore they hold for 5 factors; therefore also for 6
factors ; and so on ; thus they hold universally. Therefore
[x + a) (x + b) {x +c) ... (x + k) = x 11 + ,V U_1 + Stfp* + S. A x n ~* + . . . + 8 n
where S^the sum of all the n letters a, b, c ... Js;
<So = the sum of the products taken two at a time of these n letters.
S n =the product of all the n letters.
159. Theorems relating to divisibility may often be esta
blished by induction.
Example. Shew that .c u l is divisible by x1 for all positive integral
values of n.
x n l z' l1 l
By division = x n ~ l ^ — ;
J x1 x1 *
if therefore x n ~ l  1 is divisible by x  1, then x*  1 is also divisible by x  1.
But x' 1  1 is divisible by x  1 ; therefore x 3  1 is divisible by x  1 ; there
fore x 4,  1 is divisible by^r  1, and so on ; hence the proposition is established.
Other examples of the same kind will be found in the chapter on the
Theory of Numbers.
1G0. From the foregoing examples it will be seen that the
only theorems to which induction can be applied are those
which admit of successive cases corresponding to the order of
the natural numbers 1, 2, 3, n.
EXAMPLES. XII.
Prove by Induction :
1. 1+3 + 5+ + (2nl) = n 2 .
2. l 2 + 2 2 + 3 2 + + n 2 =i?i(n+l)(2tt+l).
3. 2 + 2 2 + 2 3 + + 2» = 2(2' l l).
4. T~o + o~q + q~T + ton terms = — ^ .
1.22.33.4 n+1
5. Prove by Induction that .r n — y n is divisible by x+y when n is
even.
CHAPTER XIII.
Binomial Theorem. Positive Integral Index.
161. It may be shewn by actual multiplication that
(x + a) (x + b) (x + c) {x + d)
= x 4 + (a + b + c + d) x 3 + (ab + ac + ad + bc + bd + cd) x*
+ (abc + abd + acd + bed) x + abed (1).
We may, however, write down this result by inspection ; for the
complete product consists of the sum of a number "of partial pro
ducts each of which is formed by multiplying together four
letters, one being taken from each of the four factors. If we
examine the way in which the various partial products are
formed, we see that
(1) the term x 4 is formed by taking the letter x out of each
of the factors.
(2) the terms involving x 3 are formed by taking the letter x
out of any three factors, in every way possible, and one of the
letters a, 6, c, d out of the remaining factor.
(3) the terms involving x 2 are formed by taking the letter x
out of any two factors, in every way possible, and two of the
letters a, b, c, d out of the remaining factors.
(4) the terms involving x are formed by taking the letter x
out of any one factor, and three of the letters a, b, c, d out of
the remaining factors.
(5) the term independent of x is the product of all the letters
«, b, c, d. .
Example 1. (x  2) (x + 3) (x  5) (x + 9)
= x 4 + ( 2 + 3  5 + 9) z 3 + ( 6 + 10 18 15 + 27 45) a 2
+ (30  54 + 90  135) x + 270
= x 4 + 5a; 3  47.<c 2  69z + 270.
BINOMIAL THEOREM. POSITIVE INTEGRAL INDEX. 137
Example 2. Find the coefficient of x* in the product
(x  3) (* + 5) [x  1) (x + 2) (x  8).
The terms involving x* are formed by multiplying together the x in any
three of the factors, and two of the numerical quantities out of the two re
maining factors ; hence the coefficient is equal to the sum of the products
of the quantities  3, 5, 1,2,  8 taken two at a time.
Thus the required coefficient
= 15 + 3 G + 21 5 + 1040 2 + 8 10
= 39.
1G2. If in equation (1) of the preceding article we suppose
b=c=d=a, we obtain
(x + a) 4 = x 4 + iax* + 6a V + 4a 3 as + a 4 .
The method here exemplified of deducing a particular case
from a more general result is one of frequent occurrence in
Mathematics ; for it often happens that it is more easy to prove
a general proposition than it is to prove a particular case of it.
We shall in the next article employ the same method to prove
a formula known as the Binomial Theorem, by which any binomial
of the form x + a can be raised to any assigned positive integral
power.
163. To find the expansion of (x + a) n ivhen n is a positive
integer.
Consider the expression
(x + a) (x + b) (x + c) (x + k),
the number of factors being n.
The expansion of this expression is the continued product of
the n factors, x + a, x + b, x + c, x + k, and every term in the
expansion is, of n dimensions, being a product formed by multi
plying together n letters, one taken from each of these n factors.
The highest power of x is x n , and is formed by taking the
letter x from each of the n factors.
The terms involving x n ~ l are formed by taking the letter x
from any n—\ of the factors, and one of the letters a, b, c, ... k
from the remaining factor ; thus the coefficient of x n ~ 1 in the
final product *is the sum of the letters a, b, c, k; denote it
by^.
The terms involving x n ~ 2 are formed by taking the letter x
from any n — 2 of. the factors, and two of the letters a, b, c, ... k
from the two remaining factors ; thus the coefficient of x n ~ in
the final product is the sum of the products of the letters
a, b, c, ... k taken two at a time; denote it by S 2 .
138 HIGHER ALGEBRA.
And, generally, the terms involving x n ~ r are formed by taking
the letter x from any n — r of the factors, and r of the letters
a, b, c, ... k from the r remaining factors ; thus the coefficient of
x"~ r in the final product is the sum of the products of the letters
a, b, c, ...k taken r at a time; denote it by S r .
The last term in the product is abc ... k; denote it by S n .
Hence (x + a)(x + b)(x + c) (x + k)
= x n + Sx n ~ l + SjxT* + ••• + £ x"~ r + ...+S ,x + S .
12 r n— 1 n
In $j the number of terms is n ; in S 2 the number of terms is
the same as the number of combinations of n things 2 at a time ;
that is, n C 2 ; in S 3 the number of terms is n C 3 ; and so on.
Now suppose b, c, ... k, each equal to a; then S l becomes
"Ca: S, becomes "C\a 2 : S becomes "Cjf: and so on: thus
(x + a) n = x n + n C l ax n  1 + n C 2 a 2 x n ~ 2 + "C^aV" 3 + . . . + "Ca" ;
substituting for *C lt n C 2 , ... we obtain
x» » «i n(n—l) „ „_„ n(n — \)(n—2) „ n „
(x+a) n = x"+nax n l +  \ — r^oV J + v 1 /v — l a 3 x n 3 +... + a n ,
the series containing n+ 1 terms.
This is the Binomial Theorem, and the expression on the right
is said to be the expansion of (x + a)*.
164. The Binomial Theorem may also be proved as follows :
By induction we can find the product of the n factors
x + a, x + b, x + c, ...x + k as explained in Art. 158, Ex. 2; we
can then deduce the expansion of (x + a) n as in Art. 163.
165. The coefficients in the expansion of (x + a)" are very
conveniently expressed by the symbols "C,, "C 2 , n C 3 , ... n C n .
We shall, however, sometimes further abbreviate them by omitting
n, and writing (7,, C 2 , C 3 , ... C n . With this notation we have
(x + a) n = x" + C x ax n ~ l + C 2 a 2 x n ~ 2 + C 3 a 3 x n ~ 3 + ... + Ca\
If we write — a in the place of a, we obtain
(x a) n = x" + C\( a) x n  l + C 2 (a) 2 x n  2 +C 3 (a) 3 x n  3 +... + C n (a) n
= x n  C,ax n ~ l + C„a 2 x n  2  C,a 3 x n ~ 3 + ... + ( IYG a\
1 2 3 \ / n
Thus the terms in the expansion of (x + a) n and (x — a) n are
numerically the same, but in (x  a)' 1 they are alternately positive
and negative, and the last term is positive or negative according
as n is even or odd.
BINOMIAL THEOREM. POSITIVE INTEGRAL INDEX. 1 39
Example 1. Find the expansion of {x + y) 6 .
By the formula,
{x + yf = x« + 8 tfi xhj + 6 C^xY + fi C 3 afy 8 + B C 4 .r 2 */ 4 + 8 <7 e x,f + »C e ,/'
= z? + 6.1V + loo; 4 ?/ 2 + 20a; 3 ?/ 3 + loxhf + Gxi/> + if',
on calculating the values of 6 C 1 , G C 2 , 6 C 3 ,
Example 2. Find the expansion of (a  2.r) 7 .
[a  2x) 7 = a 7  7 C X a c ' (2x) + 7 C 2 a 5 (2a;) 2  7 C 3 a 4 (2a;) 3 + to 8 terms.
Now remembering that n C r = n C n _ r , after calculating the coefficients up to
7 C 3 , the rest may be written down at once; for 7 G X = 7 C^ 7 Cr — 7 C< x \ and so on.
Hence
(a  2x) 7 = a 7  7a B {2x) + jp a 5 (2xf  \^\ « 4 (2a;) 3 +
= a 7  la 6 (2x) + 21a 5 (2a;) 2  35a 4 {2xf + 35a 3 (2a) 4
 21a 2 (2a;) 5 + la (2s) 6  (2.r)~
= a 7  Ua 6 x + 84a 5 a; 2  280a 4 .r 3 + 560a 3 a; 4
 672aV + USaafi  128a; 7 .
Example 3. Find the value of
(a + Jtf^ly + (a Ja  1)".
We have here the sum of two expansions whose terms are numerically
the same ; but in the second expansion the second, fourth, sixth, and eighth
terms are negative, and therefore destroy the corresponding terms of the first
expansion. Hence the value
= 2 {a 7 + 21a 5 (a 2  1) + 35a 3 (a 2  l) 2 + la (a 2  l) 3 }
= 2a (64a 6  112a 4 + 56a 2  7 ).
166. In the expansion of (x + a) n , the coefficient of the second
term is n C l ; of the third term is n G 2 ; of the fourth term is "C 3 ;
and so on ; the suffix in each term being one less than the
number of the term to which it applies ; hence "C r is the co
efficient of the (r + l) th terin. This is called the general term,
because by giving to r different numerical values any of the
coefficients may be found from n C r ; and by giving to x and a
their appropriate indices any assigned term may be obtained.
Thus the (r + l) th term may be written
■Cjrw, or »(»l)("2)(» — +»,,,„,.
t
In applying this formula to any particular case, it should 1><>
observed that the index of a is the same as the svffix of C, and
that the sum of the indices ofx and a is n.
140 HIGHER ALGEBRA.
Example 1. Find the fifth term of (a + 2a; 3 ) 17 .
The required term = 17 C 4 a 13 (2a; 3 ) 4
17.16.15.14
1.2.3.4
= 38080a 13 x 12 .
xl6ft 13 .T 12
Example 2. Find the fourteenth term of (3  a) 15 .
The required term = 15 C 13 (3) 2 (  a) 13
= 15 C 2 x(9a 13 ) [Art. 145.]
=  945a 13 .
167. The simplest form of the binomial theorem is the ex
pansion of (l+x) n . This is obtained from the general formula
of Art. 163, by writing 1 in the place of x, and x in the place
of a. Thus
(1 + x) n = l+ H C i x + "C 2 x 2 + . . . + "C r x r + ..+ "Cx n
1 n(n\) 2
1 + nx + — ^ — zr—i ar +
4 r" '
1.2 ~ + }
the general term being
n(n—l)(n—2) (nr+ 1) ,
tb .
The expansion of a binomial may always be made to depend
upon the case in which the first term is unity ; thus
{x + y yJ (X (i + l)J
V
= x n (l + z) n , where z =  .
x
Example 1. Find the coefficient of a; 16 in the expansion of (as 2  2a;) 10 .
We have (a; 2  2a;) 10 = a; 20 ( 1   V ;
/ 2\ 10
and, since a; 20 multiplies every term in the expansion of ( 1   J , we have in
this expansion to seek the coefficient of the term which contains — .
Hence the required coefficient = 10 C 4 (  2) 4
10 . 9 . 8 . 7
xl6
1.2.3.4
= 3360.
In some cases the following method is simpler.
BINOMIAL THEOEEM. POSITIVE INTEGRAL INDEX. 141
Example 2. Find the coefficient of x r in the expansion of ( ./• i  j .
Suppose that x r occurs in tlio (p + l) ,h term.
The (p + 1)°' term = *C P (x) n i> ( iY
= n C p x"' 1 *".
2*1 — r
But this term contains x r , and therefore 2n5p = r, or p = 
5
Thus the required eoellicient = n C iJ = n Co, l _,.
5
n
g(2nr)
= (3n + ?•)
2 n — i'
Unless —  — is a positive integer there will be no term containing x r in
the expansion.
1G8. In Art. 163 we deduced the expansion of (x + «)" from
the product of n factors (x + a) (x + b) ... (x + k), and the method
of proof there given is valuable in consequence of the wide gene
rality of the results obtained. But the following shorter proof of
the Binomial Theorem should be noticed.
It will be seen in Chap. xv. that a similar method is used
to obtain the general term of the expansion of
(a + b + c+ )".
161). To prove the Binomial Theorem.
The expansion of (x + a)' 1 is the product of n factors, each
equal to x + a, and every term in the expansion is of n dimen
sions, being a product formed by multiplying together n letters,
one taken from each of the n factors. Thus each term involving
x"~ r a r is obtained by taking a out of any r of the factors, and x
out of the remaining n — r factors. Therefore the number of
terms which involve x"~ r a r must be equal to the number of ways
in which r things can be selected out of n ; that is, the coellicient
of x n ~ r cC is "6' r , and by giving to r the values 0, 1, 2, 3, ... n in
succession we obtain the coefficients of all the terms. Hence
(x + a) n = x l 4 m C J X* 1 a + n C,,x n °a 2 + . . . + n C r x"a r + ...+ a",
since *C and "C n are each equal to unity.
142 HIGHER ALGEBRA.
EXAMPLES. XIII. a.
Expand the following binomials :
1. (#3) 5 . 2. (3^ + 2y) 4 . 3. {Zxyf.
4. (l3a 2 ) 6 . 5. {a?+x)\ 6. (1^j/) 7 .
">• g*js)' n  (H' 12  NT
Write clown and simplify :
13. The 4 th term of (a?  5) 13 . 14. The 10 th term of (1  2x) 12 .
15. The 12 th term of (2#  1) 13 . 16. The 28 th term of (5x + 8y) 30 .
(a \ 10
17. The 4 th term of U + 96 J .
/ b\ 8
18. The 5 th term of (2a  J .
19. The V th term of (^'  ^Y .
5. 8
20. The 5 th term of ( — x  V  %
Find the value of
21. (x + s/2y + (xj2)\ 22. (V^^+^CV^ 3 ^^) 5 
23. ( v /2 + l) 6 ( N /2l) 6 . 24. (2Vr^) 6 + (2 + v / I^^) 6 .
a cV\ 10
25. Find the middle term of f  + 
\x a
^»
26. Find the middle term of ( 1  "— j .
27. Find the coefficient of a. 18 in L'V 2 + — ] .
28. Find the coefficient of x 18 in (ax A  bx) 9 .
( 1\ 15
29. Find the coefficients of x 32 and #~ 17 in ( x A  g J
/ a 3 \ 9
30. Find the two middle terms of ( 3a  — ) .
BINOMIAL THEOREM. POSITIVE INTEGRAL INDEX. 1 b3
31. Find the term independent of x in ( x 2 — — ) .
32. Find the 13 th term of Ux  \\ .
33. If x* occurs in the expansion of lx+\ , find its coefficient.
/ 1 \ 3»i
34. Find the term independent uf ;/; in f x— ., j .
/ 1\'' 1
35. If x p occurs in the expansion of ( xr+ I , prove that its co
.... , . \2n
eihcient is  .
1
j3 (4 "^
\@n+p)
170. In the expansion of (1 4 x) u the coefficients of terms equi
distant from the beginning and end are equal.
The coefficient of the (r + l) th term from the beginning is
"C..
Tlie (r+l) th term from the end has n + 1— (r+1), or nr
terms before it; therefore counting from the beginning it is
the (n — r + l) th term, and its coefficient is "C n _ r , which has been
shewn to be equal to "C r . [Art. 145.] Hence the proposition
follows.
171. To find the greatest coefficient in the expansion of
(l + x)»
The coefficient of the general term of (1 +x)" is m C r j and we
have only to find for what value of r this is greatest.
By Art. 154, when n is even, the greatest coefficient is "C n ;
i
and when n is odd, it is "C ,, or "C , , ; these two coefficients
2 2
being equal.
172. To find the greatest term in the expansion of (x + a)".
We have (x + a)" = x" (l + Y ;
therefore, since x n multiplies every term in ( 1 + j , it will be
sufficient to find the greatest term in this latter expansion.
144 HIGHER ALGEBRA.
Let the r th and (r+l) th be any two consecutive terms.
The (r+l) th term is obtained by multiplying the r th term by
. — : that is, by ( 1 )  . [Art. 166.1
r x \ r J x L _ J
Vh + ]
The factor — 1 decreases as r increases ; hence the
r
(r+l) th term is not always greater than the r th term, but only
until ( 1 )  becomes equal to 1, or less than 1.
\ r J x
+ 1
Now — 1  > 1
/n + 1 1 \ a
\ r J x
•j
, n + 1 ., x
so long as 1 >  ;
a
.. n + 1 x
that is, >  + 1,
r a
or — — > r ( 1 ).
a
If — — be an integer, denote it by j> j then if r — ]) the
 + 1
a
multiplying factor becomes 1, and the (p + l) th term is equal to the
/> th ; and these are greater than any other term.
71+1
If — — be not an integer, denote its integral part by q ;
 + 1
a
then the greatest value of r consistent with (1) is q\ hence the
(q + 1 ) th term is the greatest.
Since we are only concerned with the numerically greatest
term, the investigation will be the same for (xa)"; therefore
in any numerical example it is unnecessary to consider the sign
of the second term of the binomial. Also it will be found best
to work each example independently of the general formula.
BINOMIAL THEOREM. POSITIVE INTEGRAL INDEX. 1 to
Example 1. If x = , find the greatest term in the expansion of (1+ I
Denote the ?•"' and (/ + l) tu terms by T r and T r ± 1 respectively; then
9r 4
hence T 7 ^. l > T r ,
. 9r 4
so long as x >1;
° r 3
that is 36  4r > 3r,
or 3G>7r.
The greatest value of r consistent with this is 5 ; hence the greatest term
is the sixth, and its value
3i4
243~
Example 2. Find the greatest term in the expansion of (3 2a:) 9 when
(3 2^ = 3^1  2 J;
(2rV
1  — J .
„ „ 9r+l 2* ...
Here *r+i = ~o~ x T r , numerically,
10  r 2
— X 3
X ^r,. ;
iience T r+1 > T r ,
i 10 ~ r 2 i
so long as x  > 1 ;
r 6
that is, 20>5r.
Hence for all values of r up to 3, we have T r+l >T r ; but if r=4, then
T r+x = T r> and these are the greatest terms. Thus the 4"' and 5 th terms are
numerically equal and greater than any other term, and their value
=3"x»C,x f J =3 6 x 84x8 =489888.
H. H. A. 10
146 HIGHER ALGEBRA.
173. To find the sum of the coefficients in the expansion
of (I +x)".
In the identity (1 + a?) n = 1 + C x x + G 2 x 2 + C 3 x 3 + . . . + C t af,
put x = 1 ; thus
2*=l + C x + C 2 + C 3 +... + C n
= sum of the coefficients.
Cor. C\ + C g + C q + ... + C =Tl;
12 3 n '
that is "the total number of combinations of n things" is 2" — 1.
[Art. 153.]
174. To prove that in the expansion of (1 + x) n , the sum of
the coefficients of the odd terms is equal to the sum of the coefficients
of the even terms.
In the identity ( 1 + x) n = 1 + C x x + C 2 x 2 + C 3 x 3 + ... + C x\
put x =  1 ; thus
= lC 1 + a 8 (7 a + (7 4 C 6 + ;
... i + 1+ c 4 + ;..... 0 1 +'C a + C.+
1
=  (sum of all the coefficients)
—
= 2
nl
175. The Binomial Theorem may also be applied to expand
expressions which contain more than two terms.
:
Example. Find the expansion of (x z + 2x l) 3 .
Regarding 2x  1 as a single term, the expansion
= (x) 3 + 3 (a 2 ) 2 (2x  1) + 3a; 2 (2x  l) 2 + (2x  l) 3
= x 6 + 6a; 3 + 9a; 4  4c 3  9a; 2 + 6x — l, on reduction.
176. The following example is instructive.
Example. If (1 + x) n = c + c x x + c#? + +c n x n ,
find the value of c + 2c 2 + 3c 2 + 4c 3 + + ( n +l)c n (1),
and c 1 2 + 2c 2 2 + 3c 3 2 + +nc n 2 (?).
The series (l) = {c + c 1 + c 2 + + c n ) + (c x + 2c 2 + 3c 3 + +nc n )
=2 w + ?i Jl + (/tl) + v J_L / + + il
= 2 n + n(l + l) n ~ 1
2 n +w.2» 1 .
BINOMIAL THEOREM. POSITIVE INTEGRAL INDEX. 1 IT
To find the value of the series (2), we proceed thus ;
c x x + 2c 2 x 2 + 3c 3 x* + + nc n x n
=»«{lHnl),+ <"^ 8 > *» + + *»}
= nx (1 + x)" 1 ;
hence, hv chauging x into  , we have
x
&+£+!*+ + ^=!(i + i)* 1 (s) .
X X' X s X n X \ xj W
Also c Q + c 1 x + c. 2 x 2 + + e n z % =(l+z) n (4).
If we multiply together the two series on the lefthand sides of (3) and (4),
we see that in the product the term independent of x is the series (2) ; hence
• n f l\ u_1
the series (2) = term independent of x in  (1 + x) n ( 1 +  I
)l
term independent of x in — (l + x)' 1 ' 1
= coefficient of x n in n (1 + .r)
2>ll
= ?ix 2n  1 <
i2nl
n1 In 1
EXAMPLES. XIII. b.
In the following expansions find which is the greatest term :
1. (x — y) 30 when #=11, y = 4
2. ( 2x  3y) 28 when x = 9, y = 4.
3. (2a + b) u when a =4, 6 = 5.
5
4. (3 + 2x) lb when x —  .
ss
In the following expansions find the value of the greatest term :
2
5. (1 + x) n when x =  , n = 6.
o
6. (« + #)* when a= s , .r = , ??=9.
10—2
148 HIGHER ALGEBRA.
7. Shew that the coefficient of the middle term of (l + x) 2n is
equal to the sum of the coefficients of the two middle terms of
(1+tf) 2 " 1 .
8. If A be the sum of the odd terms and B the sum of the
even terms in the expansion of (x + a) n , prove that A 2 B 2 = (x 2 a 2 ) n .
9. The 2 nd , 3 rd , 4 th terms in the expansion of (x+y) n are 240, 720,
1080 respectively ; find x, y, n.
10. Find the expansion of (1 + 2x  x 2 )\
11. Find the expansion of (Zx 2 2ax + 3a 2 ) 3 .
12. Find the r th term from the end in (x + a) n .
(]\ 2n + l
xj
14. In the expansion of (1 + #) 43 the coefficients of the (2r + l) th and
the (r + 2) th terms are equal; find r.
15. Find the relation between r and n in order that the coefficients
of the 3r th and (r + 2) th terms of (l+x) 2n may be equal.
16. Shew that the middle term in the expansion of (1 +x) 2n is
1 .3.5...(2nl) sn ^
hi
If c , Cj, c 2 , ... <? n denote the coefficients in the expansion of (1 +x) n ,
prove that
17. ^ + 2^ + 303 + +nc n =n.2 n  1 .
c, c, c n 2 n + 1 l
 4  4 . . H — = .
2 3 n+l n+\
18. c +i + o 2 4 +
io c.2c 2 3c 8 nc n n{n+\ )
iy. — i 1 r + ~ — O
c c x C 2 c n _ x A
v , N , . c,c, c n (n+l) H
20. (co+ej ( Cl + c 2 ) (c n . 1 + c n ) =  1  2 ^ '
M a 2 2 c, 2 3 c 2 2 4 c, 2 n + 1 c n 3' 1 + 1 1
21. 2c + — i+ 2 + — ? + + — p^ =..
2 3 4 n + l n + l
2w
22. c ( f+c 1 +c 2 + + c « = i 7i ]^ •
\2n
23. c c r + CjC r + j + c 2 c r + 2 + +c n _ r c n =  =~~ .
CHAPTER XIV.
Binomial Theorem. Any Index.
177. In the last chapter we investigated the Binomial
Theorem when the index was any positive integer; we shall now
consider whether the formula? there obtained hold in the case
of negative and fractional values of the index.
Since, by Art. 167, every binomial may be reduced to one
common type, it will be sufficient to confine our attention to
binomials of the form (1 +x) n .
By actual evolution, we have
(1 + xf = V 1 + X = 1 + ^ X   X 2 + yr. x 3  ;
and by actual division,
(1  x)~ 2 = 7^  x  a = 1 + 2x + 3x* + ix 3 + :
[Compare Ex. 1, Art. CO.]
and in each of these series the number of terms is unlimited.
In these cases we have by independent processes obtained an
i
expansion for each of the expressions (1 + x) 2 and (1 + x)~~. We
shall presently prove that they are only particular cases of the
general formula for the expansion of (1 + x) n , where it is any
rational quantity.
This formula was discovered by Newton.
178. Suppose we have two expressions arranged in ascending
powers of x, such as
, m (m  1 ) „ m (m  1 ) (m  2) ,
I + mx+ v 'x+  x /x 'a?+ (I ).
and l+n.v +  l g >x+± ] J K  a? + (2).
150 HIGHER ALGEBRA.
The product of these two expressions will be a series in as
cending powers of x\ denote it by
1+ Ax +Bx 2 + Cx 3 + Dx 4 + ;
then it is clear that A, B, C, are functions of m and n,
and therefore the actual values of A, B, C, in any particular
case will depend upon the values of m and n in that case. But
the way in which the coefficients of the powers of a; in (1) and (2)
combine to give A, B, C, is quite independent of m and n ;
in other words, whatever values in and n may have, A, B, C,
preserve the same invariable form. If therefore we can determine
the form of A, B, C, for any value of m and n, we conclude
that A, B, C, will have the same form for all values of m
and n.
The principle here explained is often referred to as an example
of "the permanence of equivalent forms ; " in the present case we
have only to recognise the fact that in any algebraical product the
form of the result will be the same whether the quantities in
volved are whole numbers, or fractions ; positive, or negative.
We shall make use of this principle in the general proof of
the Binomial Theorem for any index. The proof which Ave
give is due to Euler.
179. To prove the Binomial Theorem ivhen the index is a
positive fraction.
Wliatever be the value of m, positive or negative, integral or
fractional, let the symbol f(m) stand for the series
, m (m  1) „ m (mY) (m — 2) s
1 + mx + — y— ^ — x + — v ' v ' x 3 + ... ;
then.y(n) will stand for the series
 n(n — l)„ n(n — l)(n — 2) „
1 + nx + \ ' x 2 + v ' v ' x 3 + ....
If we multiply these two series together the product will be
another series in ascending powers of x, whose coefficients loill be
unaltered inform whatever m and n may be.
To determine this invariable form of the product we may give
to m and n any values that are most convenient ; for this purpose
suppose that m and n are positive integers. In this casey(m)
is the expanded form of (1 + x) m , andy*(?i) is the expanded form of
(1 +x) n ; and therefore
BINOMIAL THEOREM. ANY INDEX. 1.51
f(m) xf(n)  (1 + x) m x (1 + a?)" = (1 + x) m+ \
but when m and n are positive integers the expansion of (1 + x)" , + "
, / v (m + n) (m + n  1 ) .
I . —
This then is the form of the product of f(m) x/(><) in o#
cases, whatever tlie values of m and n may be; and in agreement
with our previous notation it may be denoted hyf(m + n) ; there
fore for all values ofm and n
/(m) xf(n)=f(m + n).
Also /(w) x/(n) x/(^) =/(w + ») x/( p)
=f(m + n +p), similarly.
Proceeding in tliis way we may shew that
f(m) xf(n) x/(j;)...to k factors =/(»» + n +p +...to k terms).
Let each of these quantities m, ?i, j), be equal to ■=■ ,
rC
where h and k are positive integers ;
but since h is a positive integer, f (h) = (1 + x) h ;
but y* ( y ) stands for the series
, h k\k J 2
,, vi , h k \k J ,
.*. ( 1 + a;) = 1 + T x + x / x + ,
« 1.2
which proves the Binomial Theorem for any positive fractional
index.
152 HIGHER ALGEBRA.
180. To prove the Binomial Theorem when the index is any
negative quantity.
It has been proved that
f(m) x/(w) =/(w* + n)
for all values of m and n. Replacing in by — n (wliere n is
positive), we have
f(n) •xf(n)=f(n + 7i)
=/(0)
=%
since all terms of the series except the first vanish ;
•'• /hr /(  n) '
but/(w) = (l + x)'\ for any positive value of n;
or (1 + *)"" =/(*)•
But f(—n) stands for the series
1 + ( n) x + ^ 'f—, = ar + ;
1 . L
... (1 + «.) = 1 + (_ W ) a. + (rg ) <" " " *> g» + ;
which proves the Binomial Theorem for any negative index.
Hence the theorem is completely established.
181. The proof contained in the two preceding articles may
not appear wholly satisfactory, and will probably present some dif
ficulties to the student. There is only one point to which we
shall now refer.
In the expression iov f(in) the number of terms is finite when
vi is a positive integer, and unlimited in all other cases. See
Art. 182. It is therefore necessary to enquire in what sense we
BINOMIAL T11EOUEM. ANY INDEX. 153
are to regard the statement thaty(m) x/(n) =f(m + n). It a\ ill
be seen in Chapter xxi., that when x< 1, each of the series/^/),
/( n )i/( m + n ) * s convergent, and/(m + «) is the true arithmetical
equivalent of f(m) *f(n). But when sol, all these series are
divergent, and we can only assert that if we multiply the series
denoted by/(m) by the series denoted by f(u), the first r terms
of the product will agree with the first r terms of f(m + n),
whatever finite value r may have. [8ee Art. 308.]
3
Example 1. Expand (1  xf 2 to four terms.
3
Id 1 ), ,.J(H(S)
Example 2. Expand (2 + 3a;) 4 to four terms.
(2 + 3z) 4 = 2<(l + ^)~ 4
182. In finding the general term we must now. use the
formula
m(w1)(w2) (nr + l) r
x
written in full ; for the symbol "C r can no longer be employed
when n is fractional or negative.
Also the coefficient of the general term can never vanish unless
one of the factors of its numerator is zero; the series will there
fore stop at the r th term, when n — r + 1 is zero ; that is, when
r=oi+ l ; but since r is a positive integer this equality can never
hold except when the index n is positive and integral. Thus the
expansion by the Binomial Theorem extends to w+1 terms when
n is a positive integer, and to an infinite number of terms in all
other cases.
154 HIGHER ALGEBRA.
1
Example 1. Find the general term in the expansion of (1 +x)'\
The (r+l) th term —  L±
r
5) (2r + 3)
2 r lr
af.
The number of factors in the numerator is r, and r  1 of these are nega
tive ; therefore, by taking  1 out of each of these negative factors, we may
write the above expression
(i)~ 1  8  6 <»V
i
Example 2. Find the general term in the expansion of (lnx) n .
The (r + 1)' term = » V " A " / M £ (  «»)r
E /
= !(!«) (lar.) (1F^Un) _ ^
w r I r
l(ln)(l2n) (1rl.n) ^
= ( _ i)r ( _ i)ri (nl)(2nl) (rl.n1) ^
(n  1) (2«  1) . . ....(^l.nl)
since (_1)» (_ l)ri = (_ i)2ri = _ 1#
Example 3. Find the general term in the expansion of (1  x)~ 3 .
The(r + irterm=< 3 '( 4 >'_ 5 )^( 3 '+ 1 ) ( ,)r
r
= (1)r 3.4.5 (r + 2) (1)ffa ,
~ [ } 1.2.3 r X
_ (r+l)(r+2) .
~ 1.2 *»
by removing like factors from the numerator and denominator.
BINOMIAL THEOREM. ANY INDEX. 155
EXAMPLES. XIV. a.
Expand to 4 terms the following expressions:
s
1
1. (l+xf.
2.
(1 + *)*".
3.
(ix)K
4. (l+.r 2 )"*.
5.
(l3.'\ ; .
6.
i
(l3ap.
7. (1+fce)"*.
8.
('♦r
9.
(?/■
io. (i +*•)"'
11.
(2 + .r)" 3 .
12.
i
(9 + 2.t,)' 2 .
13. (8+12a) § .
14.
3
(9fcr)~*.
15.
i
(4a8./;p
Write down and
simplify :
i
16. The 8 th term of (1 + 2.f)~ 2 .
17. The 11* ter
•m of (1 
ii
2s 3 ) 2 .
16
18. The 10 th term of (1 + 3a 2 ) 3 .
19. The 5 th term of (3a  26)  K
20. The (r + 1 ) th term of ( 1  x) ~ .
21. The (r + l) th term of (1  x)  4 .
i
22. The (r + 1 ) th term of ( 1 + xf.
n
23. The (r + 1 ) th term of ( 1 + x) 3 .
24. The 14 th term of (2 10  2\v) 2 .
n
25. The 7 th term of (3 8 + 6%) 4 .
183. If we expand (1 — x)~ 2 by the Binomial Theorem, we
obtain
(lx) 2 =l + 2x + 3x 2 +4x* + j
but, by referring to Art. 60, we see that this result is only true
when x is less than 1. This leads us to enquire whether we are
always justified in assuming the truth of the statement
(1 +x) n = 1 + nx+ = ar +
1 . 2i
156 HIGHER ALGEBRA.
and, if not, under what conditions the expansion of (1 + x) n may
be used as its true equivalent.
Suppose, for instance, that n — — l; then we have
(1 x)~ r = 1 + x + x 2 + x 3 + x* + (1);
in this equation put x = 2 ; we then obtain
(l)~ 1 =l+2 + 2 2 + 2 3 + 2 4 +
This contradictory result is sufficient to shew that we cannot
take
, n(nl)
l+nx+— \ — ~— ' x 2 +
as the true arithmetical equivalent of (1 + x) n in all cases.
Now from the formula for the sum of a geometrical pro
gression, we know that the sum of the first r terms of the
1  x r
series (1) = z
v ' I —x
1 x r
1  X 1  x y
and, when x is numerically less than 1, by taking r sufficiently
x r
large we can make ^ as small as we please ; that is, by taking
a sufficient number of terms the sum can be made to differ as
little as we please from ^ . But when x is numerically
x r
greater than 1, the value of ^ r increases with r. and therefore
1  x
no such approximation to the value of is obtained by taking
JL vC
any number of terms of the series
1 + X + X s + X 3 4
It will be seen in the chapter on Convergency and Diver
gency of Series that the expansion by the Binomial Theorem
of (1+x)" in ascending powers of a? is always arithmetically in
telligible when x is less than 1.
But if x is greater than 1, then since the general term of
the series
, n(n\) „
1 + nx H .j x" +
I . 
BINOMIAL THEOREM. ANY INDEX. 157
contains x r , it can be made greater than any Unite quantity by
taking r sufficiently large ; in which case there is no limit to the
value of the above series; and therefore the expansion of (1 + x) n
as an infinite series in ascending powers of x has no meaning
arithmetically intelligible when x is greater than 1.
184. We may remark that we can always expand (x + y)"
by the Binomial Theorem ; for we may write the expression in
either of the two following forms :
x"
('*!)'■ '(•♦ff.
and we obtain the expansion from the first or second of these
according as x is greater or less than y.
185. To find in its simplest form the general term in the
expansion of (1 — x) u .
The (r + l) th term
( n)(n I) (71 2)... (nr+1)
(*y
= ( iy »(* +1 H w + 2 ) (** + »• 1)
= (_ I)* ttv*+l)(tt+2)...ytt + rl ) x r
n (n + 1) (n + 2) ... (n + r  1) r
From this it appears that every term in the expansion of
(1 x)~* is positive.
Although the general term in the expansion of any binomial
may always be found as explained in Art. 182, it will be found
more expeditious in practice to use the above form of the general
term in all cases where the index is negative, retaining the
form
n(n l)(n2) ... ( n  r + 1 ) ,
i x
t
only in the case of positive indices.
158 HIGHER ALGEBRA.
Example. Find the general term in the expansion of  _ . .
1  1 
—  — = (l3x) 3 .
The (r + l) th term
1.4.7 (3r2) 3rrr
1.4.7 (Sr2) ^
^H : w •
r
_i
If the given expression had been (1 + Sx) 3 we should have used the same
formula for the general term, replacing Sx by  3x.
186. The following expansions should be remembered :
(1  x)' 1 = 1 + x + x 2 + x 3 + + x r +
(1  x)~ 2 = 1 + 2x + 3x 2 + ±x 3 + + (r + 1) x r +
(I  x)~ 3 =1 + 3x + 6x* + 10x 3 + + ( r+ l J% —K r +
expansion of (1 + x) n , when n is unrestricted in value, will be
found in Art. 189 ; but the student will have no difficulty in
applying to any numerical example the method explained in
Art. 172.
Example. Find the greatest term in the expansion of (l+a;)~ n when
2
x =  , and n — 20.
3
fi j_ <t' ^
We have ^V+i— ,xxT r , numerically,
 19+r ? r •
•"• ■'r+l > 'r»
2 (19 + r)
so long as — £ > 1 ;
that is, 38 >r.
Hence for all values of r up to 37, we have jr r+1 >T r ; but if r=38, then
I^k = T,. , and these are the greatest terms. Thus the 38 th and 39 th terms
are equal numerically and greater than any other term.
BINOMIAL THEOREM. ANY INDEX. 159
188. Some useful applications of the Binomial Theorem are
explained in the following examples.
Example 1. Find the first three terms in the expansion of
i _i
(l + 3*) r (l2x) 3.
Expanding the two binomials as far as the term containing x'\ we have
, /3 2\ /8 3 2
1 13 55 .
= 1 + Q X + 72 X "'
If in this Example ^='002, so that ar = 000004, we see that the third
term is a decimal fraction beginning with 5 ciphers. If therefore we were
required to find the numerical value of the given expression correct to 5 places
of decimals it would be sufficient to substitute *002 for x in 1 +  x, neglect
o
ing the term involving x 2 .
Example 2. When x is so small that its square and higher powers may
be neglected, find the value of
J(± + xJ*
Since x and the higher powers may be neglected, it will be sufficient to
retain the first two terms in the expansion of each binomial. Therefore
i
the expression
_tl±±±l
b(i+.)
KS).
the term involving x being neglected.
160 HIGHER ALGEBRA.
Example 3. Find the value of rj= to four places of decimals.
x/47
_i
1   1 / 2 \ 2
^ = (47) *=(7*2)*=(ln)
1/ 1 3 ^ 5 :L_
7^ + 72 + 3 74 + 2 7G+
7 + 73 + 2 * 7 5 + 2 *7 7+ ""
To obtain the values of the several terms we proceed as follows :
1)1 !
7 ) 142857 =t,
7 ) 020408 '
7 ) 002915 = 73,
7 ) 000416
•000059 = ^;
5 1
and we can see that the term  . = is a decimal fraction beginning with
5 ciphers.
.. i = 142857 + 002915 + 000088
\/47
= •14586,
and this result is correct to at least four places of decimals.
Example 4. Find the cube root of 126 to 5 places of decimals.
!
(126)3 = (5 3 + l) a
5
1
/ t 1 1 M 5 1 \
~ 5 V 3"5 :J 9'5« + 81*5 9 '")
1 1_ 1 J. _1 1
~ 3 ' 5 2 ~ 9 ' 55 + 81 *5 7 ■•"
1^1^ 1 W_
~ + 3*"l0 2 9'10 5 + 81 *10 7 "••
_ 04 00032 0000128
= 51 h —
^ 3 9 81
=5f 013333 ...  000035 ...+...
= 5 '01329, to five places of decimals.
BINOMIAL THEOREM. ANY INDEX. 101
EXAMPLES. XIV. b.
Find the (r+1)" 1 term in each of the following expansions :
I i
!• (l+#) 2 . 2. (l.t) 5 . 3. (l+3.e) :] .
J 3
4. (l+#) 3 . 5. (l+.r 2 )3. 6 . (i2.v)~*.
7. (a+fo?)" 1 . 8. (2.r)~ 2 . 9. tt{rfx*)\
10  7=A=. 11. 3/ * 12. , *
</T+2* N f/ (l3.^ V&Z^
Find the greatest term in each of the following expansions :
4
13. ( 1 + .v) ~ 7 when x=— .
lo
— 2
14. ( 1 + a?) 2 when a?= 5 .
 1  1 1
15. (1 — 74?) 4 wheu# = .
o
16. (2a? + 5J/) 12 when a? = 8 and y = 3.
17. (5  4.v) ~ 7 when t v= .
25
18. (3r 2 + 4/)  n when x = 9, y = 2, « = 1 5.
Find to five places of decimals the value of
19. v98. 20. 4/998. 21. \ 3/ 1003. 22. \ 4/ 2400.
1
1 3
23. ^=. 24. (1^)3. 25. (630) *. 26. tfilla
If x be so small that its square and higher powers may be neglected,
find the value of
1 3
27. (l7tf) s (l + 2a?)"*.
28.
V4*.(af) •
29.  < 8 + 3 f .
(2 + 3.r) V45./
30.
(l+*)"'x(4+3*)«
(1 +./•)
H. H.A.
1 1
162 HIGHER ALGEBRA.
31. V^+C+jj ' 32 . ^T^^1^
(1+5*)*+ (4+Y
33. Prove that the coefficient of sf in the expansion of (l4r) *
is
v2 '
31 Prove that (1 +*)*=2 l ■ — ^ + ^ (f^ ) f
35. Find the first three terms in the expansion of
1
(1 + x) 2 Vl + 4x '
36. Find the first three terms in the expansion of
3
(! + #)* + *Jl + bx
37. Shew that the n th coefficient in the expansion of (1  x)~ n is
double of the (nl) th .
189. To find the numerically greatest term in the expansion
of (1 + x) n , for any rational value of\\.
Since we are only concerned with the numerical value of the
greatest term, we shall consider x throughout as positive.
Case I. Let n be a positive integer.
The (r+l) th term is obtained by multiplying the r th term
by . x ; that is, by f 1 J x ; and therefore the
terms continue to increase so long as
'n+ 1
Or 1  1 ) 1 '
., (n+ l)x ,
that is. — > 1 + x,
r
(n + 1 ) x
or *— — >r.
1+02
BINOMIAL THEOREM. ANY INDEX. 103
(ll 4 1 ^ X
If — — be an integer, denote it by p; then if r=p, the
multiplying factor is 1, and the (;>+l) th term is equal to the
^> th , and these are greater than any other term.
( 71 4 1 ) X
If — , — be not an integer, denote its integral part by q ;
then the greatest value of r is 7, and the (q + l) th term is the
greatest.
Case II. Let n be a positive fraction.
As before, the (r+ l) th term is obtained by multiplying the
_. , (n + 1 t \
r m term by ( — ■ I )x.
(1) If x be greater than unity, by increasing r the above
multiplier can be made as near as we please to  x ; so that after
a certain term each term is nearly x times the preceding term
numerically, and thus the terms increase continually, and there
is no greatest term.
(2) If x be less than unity we see that the multiplying
factor continues positive, and decreases until r > n + 1 , and from
this point it becomes negative but always remains less than 1
numerically ; therefore there will be a greatest term.
As before, the multiplying factor will be greater than 1
(n + l)x
so Ions: as ^ — > r.
1 +x
( Jl 4 1 \ X
If ^ — be an integer, denote it by p ; then, as in Case I.,
the (p + l) th term is equal to the £> th , and these are greater than
any other term.
( 7t 4" 1 ) X
If ^p • be not an integer, let q be its integral part; then
the (q 4 l) th term is the greatest.
Case III. Let n be negative.
Let n  — in, so that m is positive ; then the numerical
nil _L f J
value of the multiplying factor is — . x ; that is
(
ml \
+ 1 ) x.
r J
11—2
164 HIGHER ALGEBRA.
(1) If x be greater than unity we may shew, as in Case II.,
that there is no greatest term.
(2) If x be less than unity, the multiplying factor will be
greater than 1, so long as
(m  1 ) x
that is, — > 1  x,
r
(m—\)x
or . — — > r.
I x
lyn. 1 ) CC
If ^— — be a positive integer, denote it by p \ tlien the
x — ■ x
(p + l) th term is equal to the p th term, and these are greater than
any other term.
(fjr 1 ) £C
If * ' — be positive but not an integer, let q be its inte
1 x
gral part ; then the (q + l) th term is the greatest.
If i ' — be negative, tlien m is less than unity ; and by
writing the multiplying factor in the form (1 — J x, we
see that it is always less than 1 : hence each term is less than
the preceding, and consequently the first term is the greatest.
190. To find the number of homogeneous products of v dimen
sions that can be formed out of the n letters a, b, c, and their
powers.
By division, or by the Binomial Theorem, we have
■= = 1 + ax + a 2 x 2 + a 3 x 3 + ,
1 — ax
1
1 — bx
1
1 — ex
= 1 + bx + b 2 x 2 + b 3 x 3 + ,
= 1 + ex + c 2 x 2 + c 3 x 3 + ,
BINOMIAL THEOREM. ANY INDEX. KJ5
Hence, by multiplication,
1 1 1
1 _ ax 1 — bx 1 — ex
= (1 + ax + aV + ...) (1 + bx + b*x* + ...) (1 + ex + c°x 2 + ...) ...
= 1 + x (a + b + c + ...) +x 2 (a 2 + ab + ac + b' 2 + bc±c 2 4 . . . ) + ...
= 1 + S t x + Sjfx? + S a x a + suppose ;
where S lt >S'.,, S aJ are the sums of the homogeneous pro
duets of one, two, three, dimensions that can be formed of
a, b, c, and their powers.
To obtain the number of these products, put a, b, c, each
equal to 1 ; each term in JS l9 S 2 , S :i , now becomes 1, and the
values of S l9 S 2 , S :i , so obtained give the number of the
homogeneous products of one, two, three, dimensions.
Also
1 1 1
1 — ax 1 — bx 1 — ex
becomes — or (1 — a;) ".
(1  x)
Hence S r = coefficient of x r in the expansion of (1 — x)~
n(n+ l)(n + 2) (n+r 1)
~ jr
n + r—1
\r \n—
1
191. To find the number of terms in the expansion of any
multinomial when the index is a positive integer.
In the expansion of
(a t + a B + a B + +a r )",
every term is of n dimensions; therefore the number of terms is
the same as the number of homogeneous products of n dimensions
that can be formed out of the r quantities a,, a , ... a r , and their
powers ; and therefore by the preceding article is equal to
I?' + n — 1
n r — 1
166 HIGHER ALGEBRA.
192. From the result of Art. 190 we may deduce a theorem
relating to the number of combinations of n things.
Consider n letters a, b, c, d, ; then if we were to write
down all the homogeneous products of r dimensions which can be
formed of these letters and their powers, every such product
would represent one of the combinations, r at a time, of the n
letters, when any one of the letters might occur once, twice,
thrice, ... up to r times.
Therefore the number of combinations of n things r at a time
when repetitions are allowed is equal to the number of homo
geneous products of r dimensions which can be formed out of n
\n + r — 1
letters, and therefore equal to ,  , or n+r *C .
\r n—\ T
That is, the number of combinations of it things r at a time
when repetitions are allowed is equal to the number of com
binations of n + r— 1 things r at a time when repetitions are
excluded.
193. We shall conclude this chapter with a few miscel
laneous examples.
(1  2a;) 2
Example 1. Find the coefficient of x r in the expansion of ~ .
The expression = (1  Ax + 4.x 2 ) (1 +PyC +p^xr + ... +p r x r + ...) suppose.
The coefficient of x r will be obtained by multiplying p r , p r  x , p r » by 1,
4,4 respectively, and adding the results ; hence
the required coefficient =p r  4p r _ x + 4p r _ 2 .
But p r =( iy fe±afc±9 . [Ex . 3 , Art. 182.]
Hence the required coefficient
= ( . 1) r (r+lHr + 2) _ 4( _ 1)r . 1 rJ^ + 4( _ ira ( I ^r
= ^[(r + l)(r + 2) + 4r(r + l)+4r(rl)]
fl) r
BINOMIAL TIIEOltEM. ANY INDEX. Hi?
Example 2. Find the value of the scries
„ , 5 5.7 5.7.!)
2 4 . 4
_2. 3 T 3.3 2 ^ 1 4 . 3 :J + •••
m U . 3 . 5 1 3.5.7 1 3.5.7.9 1
The expression = 2 + — — . — + . — + : —  . _ +
v [2 3 1 3 3 ;! 14 3 4
3 5 3 5 7 3 5 7 1)
 Q 2 Ll? 2 ~ 2 ' 2 ' 2 2J 2 ' 2 ' 2 ' 2 2 4
2 '3?*" 1 ~3~ '3 :i+ ]i~ *3 5+ ••'
3 3 5 3 5 7
2 2 2' 2 /2\ 2*2*2
1 *3 + "~J2~
z z 2 2 /2\ a 2 2 2 /2V
■o.rffl" 1
= 3 5 =V 3 
Example 3. If ?t is any positive integer, shew that the integral part of
(3 + Jl) n is an odd number.
Suppose I to denote the integral and/ the fractional part of (3 + a/7)' 1 .
Then I+f=3 n +C 1 S" i s /7 + a 2 S n ~ 2 . 7+(7 8 3*«^7) 8 + (1).
Now 3 N /7 is positive and less than 1, therefore (S^)' 1 is a proper
fraction; denote it by/';
.•./' = 3 n C , 1 3' l V7 + C' 2 3' l  2 .7+C 3 3' l 3( v /7) 3 + (2).
Add together (1) and (2) ; the irrational terms disappear, and we have
I+f+f = 2 (3» + C 2 3' 1 " 2 . 7 + . . . )
= an even integer.
But since/ and/' are proper fractions their sum must be 1 ;
:\ I=an odd integer.
EXAMPLES. XIV. c.
Find the coeflicient of
1. x m in the expansion of
2. a n in the expansion of
3. «* in the expansion of
X "T" X
(1  xf '
4 + 2a  a 2
(l+«) 3
3x 2  2
168 HIGHER ALGEBRA.
2 4 x + X 2
4. Find the coefficient of x n in the expansion of
( 1 + ^J
5. Prove that
1 1 1.3 1 1.3.5 2_ 13.5.7 1^
2 * 2 + 271 ' 2* 2.4.6'2 3 + 2.4.6.8'2*
6. Prove that
3 3^5 3.5.7
4 H ' 4. 8 H ~ 4.8.12
7. Prove that
V 3'
N /8 = 1 +  + t^~; + , \ \ n +
2n 2n(2n + 2) 2n(2n + 2) (2n + 4)
+ ~3 + ~ 3.6 " + 3.6.9 +
~ 2 V + 3 + _ 3T6~ + 3.6.9 + J
8. Prove that
7 h + ? i + ^ (^1) . n{n\)(n2) 1
' J 7 + 7.14 + 7.14.21 + J
±n Ji . % j. »(*+!) , n(» + l)(n + 2) \
(2 2.4 + " 2.4.6 + J •
9. Prove that approximately, when x is very small,
"7! 9 \ 2 256' '
2 ( 1+ r 6 'V
10. Shew that the integral part of (5 + 2 >JQ) n is odd, if n be a
positive integer.
11. Shew that the integral part of (8 + 3 V /7) H is odd, if n be a
positive integer.
12. Find the coefficient of x n in the expansion of
(l2.v + 3.v 2 4.v 3 + )*.
/ 1\ 4 ' 1
13. Shew that the middle term of ( x +  1 is equal to the coefficient
of x n in the expansion of (1 Ax) ^" 2 .
14. Prove that the expansion of (1 — x^) n may be put into the form
(1  xf n + 3nx (1  xf n ~ 2 + 3n @ n  3 ) x i (i _ x yn  4 +
BINOMIAL THEOREM. ANY INDEX. L69
15. Prove that the coefficient of at* ill the expansion — , is
1,0,  1 according as n is of the form 3m, 3m  1, or 3//<.+ 1.
16. In the expansion of (a + b + c) s find (1) the number of terms,
(2) the sum of the coefficients of the terms.
17. Prove that if n be an even integer,
111 1 2"" 1
l\ nl \'S \ n  3 \b\ n5 \ u\ ,1 rc '
18. If c , (',, C 2 , f n are the coefficients in the expansion of
(1 +.f) u , when n is a positive integer, prove that
I//1
a) c c l+ c 2 c 3+ +(mv(i)' 1/ ,  ;^r_ 1 .
(2) ^2^ + 3^4^+ + (_i)n (/i4 . 1)t . M = 0>
(3) c* c *+c£c*+ + (l)»c n 2 =0, or (1)^,
according as n is odd or even.
19. If *„ denote the sum of the first n natural numbers, prove that
(1) (l;r) 3 = ^ + %^ + ^.^+ +V»~ 1 +...
j2^ + 4
(2) 2 (*! *, B + 82*2, _j + + 8 n 8 n + l ) = — — ^ .
„ T . 1.3.5.7 (2)il)
20. If fr 2 .4. 6 , 8 2n . P*>™ that
(!) ?2n + l + <Mj» + Man  1 + + 2n \<ln + 2 + ?«?• +1 = 5
(2) 2 { ?2n  ?1 y, (l _ j + g^a. _ 2 + + (  1)"  1 tj n _ #„ + J
21. Find the sum of the products, two at a time, of the coefficients
in the expansion of (1 +x) n , when n is a positive integer.
22. If (7 +4 v /3) n =p + /3, where n and p are positive integers, and 9
a proper fraction, shew that (1 f3)(p + p) = l.
23. If c , <?!, c^, r n are the coefficients in the expansion of
(1 +#)*, where ?i is a positive integer, shew that
c 2 . c \ (I) n_1 fn ,11 1
2 3 n 2 3 n
CHAPTER XV.
Multinomial Theorem.
194. We have already seen in Art. 175, how we may
apply the Binomial Theorem to obtain the expansion of a multi
nomial expression. In the present chapter our object is not
so much to obtain the complete expansion of a multinomial as
to find the coefficient of any assigned term.
Example. Find the coefficient of a 4 b"c 3 d 5 in the expansion of
(a + b + c + d) u .
The expansion is the product of 14 factors each equal to a+b + c + d, and
every term in the expansion is of 14 dimensions, being a product formed by
taking one letter out of each of these factors. Thus to form the term a 4 b 2 c*d 5 ,
we take a out of any four of the fourteen factors, b out of any hco of the re
maining ten, c out of any three of the remaining eight. But the number of
ways in which this can be done is clearly equal to the number of ways of ar
ranging 14 letters when four of them must be a, two 6, three c, and five d ;
that is, equal to
114
A TT ralg . [Art. 151.]
412 3 5 L J
This is therefore the number of times in which the term a 4 b 2 c*d 5 appears
in the final product, and consequently the coefficient required is 2522520.
195. To find the coefficient of any assigned term in the ex
pansion of (a + b + c + cl + . ..) p , where p is a positive integer.
The expansion is the product of p factors each equal to
a + ft + c + cZ + ..., and every term in the expansion is formed by
taking one letter out of each of these p factors ; and therefore
the number of ways in which any term a a b^cyd 8 ... will appear
in the final product is equal to the number of ways of arranging
p letters when a of them must be a, (3 must be b, y must be c;
and so on. That is,
\p
the coefficient of a a bPcyd s ... is = — ~f~^ — ,
o p \y 6 ...
where a + j3 + y + S + ... =p.
MULTINOMIAL THEOREM. 171
Cok. Jn the expansion of
(a + bx + cx~ + da? + ... )'',
the term involving a"b&cyd 6 ... is
^L.a^^v^).
or i— T5Ht^t aWcyd 5 ... xfi+2y+M + ..
where a + /3 + y + $ + ... = p.
This may be eallecl tlte general term of the expansion.
Example. Find the coefficient of «* in the expansion of (a + i.c + ex 2 )'.
The general term of the expansion is
ii« a ^V +2 ? (i),
where a + p + y = \).
We have to obtain by trial all the positive integral values of /3 and 7
which satisfy the equation fi + 2y = 5; the values of a can then be found from
the equation a + /3 + 7 = 9.
Putting 7 = 2, we have /3 = 1, and a = G;
putting 7 = 1, we have /3 = 3, and a = 5;
putting 7 = 0, we have /3 = 5, and a = 4.
The required coefficient will be the sum of the corresponding values of the
expression (1).
Therefore the coefficient required
9 19 19
= 252a 6 6c 2 + 5Q4a*&c + 12Ga 4 b\
19G. To find the general term in the expansion of
(a + bx + ex 2 + clx 3 + . . .) n ,
vjhere n is any rational quantity.
By the Binomial Theorem, the general term is
n(nl)(n2)...(np + l) ( ,,_ v + rf + ^ +
where jp is a positive integer
172 HIGHER ALGEBRA.
And, by Art. 195, the general term of the expansion of
(6a; + ex' + dx 3 + ...)''
\P
\pjy_\o_—
where ft, y, 8 . . . are positive integers whose sum is p.
Hence the general term in the expansion of the given ex
pression is
where /? + y + S + ... = /?.
197. Since (a + bx + ex 2 + dx 3 + ..)" may be written in the
form
„A 6 c 2 a* 3 y
ail +x + x+ar+ ... ,
\ a a a J
it will be sufficient to consider the case in which the hrst term
of the multinomial is unity.
Thus the general term of
(1 + bx + ex 2 + dx 3 + . . .)"
n
is 
(nl)(n2). (np + l) bpcyd8 ^ +9f+u+
\p \v \ 8
where fi + y + &\...=p.
Example. Find the coefficient of x 3 in the expansion of
(l3z2.r 2 + 6x' 3 )3.
The general term is
S(S0(t»)...(* + o
V ,., ,, iOsA^e) 8 /^^ (i).
We have to obtain by trial all the positive integral values of /3, 7, 5 which
satisfy the equation j3 + Zy + 35 = 3 ; and then p is found from the equation
2>=/3 + 7 + 5. The required coefficient will be the sum of the corresponding
values of the expression (1).
MULTINOMIAL THEOREM. 173
In finding /3, 7, 5, ... it will be best to commence by giving to 5 successive
integral values beginning with the greatest admissible. In the present case
the values are found to be
8=1, 7 = 0, 18 = 0, p=l;
5 = 0, 7 =1, 0=1, p=2;
5 = 0, 7 = 0, 0=3, p = 3.
Substituting these values in (1) the required coefficient
^)<^)(>)< 3 " 2 >+^#^
(3)
s
4_4_4
3 3~3
198. Sometimes it is more expeditious to use the Binomial
Theorem.
Example. Find the coefficient of x 4 in the expansion of (1  2x + 3.r 2 ) 3 .
The required coefficient is found by picking out the coefficient of x x from
the first few terms of the expansion of (1  2x  Sx 2 )  * by the Binomial
Theorem ; that is, from
1 + 3 {2x  Sx 2 ) + 6 (2.r  3x 2 ) 2 + 10 {2x  3.r) :J + 15 (2.r  3.r 2 ) 4 ;
we stop at this term for all the other terms involve powers of x higher
than x*.
The required coefficient = 6 . 9 + 10 . 3 (2) 2 (  3) + 15 (2) 4
= 66.
EXAMPLES. XV.
Find the coefficient of
1. a 2 Pc 4 d in the expansion of (ab — c+d) w .
2. a 2 b ry d in the expansion of (a + b — c — d) s .
3. a?b s c in the expansion of (2a + 6f 3c) r .
4. x~yh A in the expansion of {cub  by + cz) 9 .
5. x 3 in the expansion of (l+3# — 2a 2 ) 3 .
6. x A in the expansion of (l + 2.r + 3.r 2 ) 10 .
7. .'•" in tlie expansion of (1 + 2.v  x 2 )'\
8. A" 8 in the expansion (if (1  2.r + 3# 2  4.r' ! ) 4 .
174 HIGHER ALGEBRA.
Find the coefficient of
9. .r 23 in the expansion of (1  2x + 3x 2  x 4  .i/') 5 .
i
10. x 5 in the expansion of (1 2x + 3x 2 ) 2 .
i
11. x 3 in the expansion of (1  2x + 3x 2  4a 3 ) 2 .
( X 2 X*\ ~ 2
12. x 8 in the expansion of ( 1  — + '» ) .
13. x* in the expansion of (2  4x + 3x 2 ) ~ 2 .
3
14. X s in the expansion of ( 1 + Ax 2 + 1 Ox 4 + 20^ G ) " * .
15. x 12 in the expansion of (3  15x* + 18^')  l .
i
16. Expand (1  2x  2x 2 )* as far as x 2 .
2
17. Expand (1 + 3x 2  6x*) 3 as far as x 5 .
4
18. Expand (8  9^ + 1 8a 4 ) 3 " as far as x 8 .
19. If (l+x + x 2 + +xP) n = a + a l x + a.^v 2 + a llf> x n r>,
prove that
(1) a +a 1 +a a + +a^=(p+l) n .
(2) a 1 +2a 2 +3a 8 + +«p.a«p=5»i>(p+l)*.
20. If a , a 15 a 2 > ft 3 ••• are the coefficients in order of the expansion
of (1 +x+x 2 ) n , prove that
a 2 a 2 + a 2 a 2 + + (l) n  1 aU 1 =^a n {l(l)^a n }.
21. If the expansion of (1 +x + x 2 ) n
be a + a l x+a 2 x 2 + ... +a r af r + ... +a 2n x 2n ,
shew that
«o + a 3 + a 6 + ... =a l + a 4 + a+ ... =« 2 + a 6 +a 8 + ... = 3 n_1 .
CHAPTER XVI.
Logarithms.
199. Definition. The logarithm of any number to a given
base is the index of the power to which the base must be raised
in order to equal the given number. Thus if a x = JV, x is called
the logarithm of N to the base a.
Examples. (1) Since 3 4 = 81, the logarithm of 81 to base 3 is 4.
(2) Since lO^lO, 10 2 = 100, 10 3 = 1000,
the natural numbers 1, 2, 3,... are respectively the logarithms of 10, 100,
1000, to base 10.
200. The logarithm of iV to base a is usually written log a jy,
so that the same meaning is expressed by the two equations
a x = N; x = \og a N.
From these equations we deduce
an identity which is sometimes useful.
Example. Find the logarithm of 32 £/■! to base 2 N /2.
Let x be the required logarithm; then,
by definition, (2 x /2)« = 32 4/4 ;
1 2
.. (2. 2*)* = 2 s . 2* ;
3 2
.. 2^ = 2 5 ^;
3 27
hence, by equating the indices,  x =  r ;
.'. x = — = 36.
o
176 HIGHER ALGEBRA.
201. When it is understood that a particular system of
logarithms is in use, the suffix denoting the base is omitted.
Thus in arithmetical calculations in which 10 is the base, we
usually write log 2, log 3, instead of log 10 2, log l0 3,
Any number might be taken as the base of logarithms, and
corresponding to any such base a system of logarithms of all
numbers could be found. But before discussing the logarithmic
systems commonly used, we shall prove some general propositions
which are true for all logarithms independently of any particular
base.
202. TJie logarithm of 1 is 0.
For a° = 1 for all values of a ; therefore log 10, whatever
the base may be.
203. The logarithm of the base itself is 1.
For a 1 = a ; therefore log a a = 1 .
201. To find the logarithm of a product.
Let MN be the product; let a be the base of the system, and
suppose
a: = log. J/, y = \og a J\T;
so that a* = M, a* = N.
Thus the product MN==a x x a y
= a x+y ;
whence, by definition, log a MN = x + y
= 100^1/"+ low N.
Similarly, \og a 3INP = \og a M+ log a iV+ log a P;
and so on for any number of factors.
Example. log 42 = log (2 x 3 x 7 )
= log2 + log3 + log7.
205. To find the logarithm of a fraction.
M
Let zz be the fraction, and suppose
x = \og a M i 2/ = log a iT;
so that a x = M t a y = N.
Thus the fraction
whence', by definition, log a *—=x — y
LOGARITHMS.
M
a*
N
a 7 '
a J ;
177
= log J/  Xos^N.
30
Example. log (if) = log —
= log 30 log 7
=log(2x3xo)log7
= log 2 + log 3 + log 5  log 7.
206. :Z'o find the logarithm of a number rained to any power,
integral or fractional.
Let log a (J/'') be required, and suppose
x = \og a M, so that a" — 21 ;
then M * = (aj
= aT;
whence, by definition, \og a (JP) — px\
that is, \og a (M>)=p\o% a M.
I 1
Similarly, log a (J/ r ) =  log tt J/.
207. It follows from the results we have proved that
(1) the logarithm of a product is equal to the sum of the
logarithms of its factors ;
(2) the logarithm of a fraction is equal to the logarithm of
the numerator diminished by the logarithm of the denominator ;
(3) the logarithm of the p*** power of a number is^> times the
logarithm of the number ;
(4) the logarithm of the r th root of a number is equal to th
of the logarithm of the number.
Also we see that by the use of logarithms the operations of
multiplication and division may be replaced by those of addition
and subtraction ; and the operations of involution and evolution
by those of multiplication and division.
H. II. A. 12
178 HIGHER ALGEBRA.
r a 3
Example 1. Express the logarithm of —^ m terms of log a, log b and
log c.
. Ja* a 2
3
= log a 2  log (c^ 2 )
3
= log«(logc 5 + logfc 2 )
3
= = log a  5 log c  2 log &.
Example 2. Find a; from the equation a x . c~ 2 *=& 3a!+1 .
Taking logarithms of both sides, we have
x log a  2x log c = (Sx + 1) log 6 ;
.. x (log a  2 log c  3 log b) = log b ;
_ lo 2 6
log a  2 log c  3 log b '
EXAMPLES. XVI. a.
Find the logarithms of
1. 16 to base J2, and 1728 to base 2 v '3.
2. 125 to base 5 v /5, and *25 to base 4.
3. stt. to base 2 x /2, and '3 to base 9.
256
4. '0625 to base 2, and 1000 to base 01.
5. 0001 to base '001, and i to base 9^/3.
4 /~*r i 3 r~^
6. kI gp , — j , */ a 2 to base a.
a?
7. Find the value of
l0g 8 128, l0g 6 ^, logfrgj, lo g 3 43 49 '
Express the following seven logarithms in terms of log a, logb, and
logo.
8. log(N^) fi . 9. log{Va 2 xyb s ). 10. logflcFW).
LOGARITHMS. 179
11. log^o^x^oJR). 12. log(^a V6 3 jVP7a).
13. log 14. logj^J + f ffi 3
15. Shew that log f ' ; £^ = 1 logo   log2  2 , log 3.
VW7J2 4 5 3
16. Simplify logV 72!) V 9" 1 . 27 " 3 .
75 5 3 9
17. Prove that log —  2 log  + 1< >g — = h .g 2.
Solve the following equations:
18. o«=c&* 19. a 2 ».6 3 *=c s .
90 °^  & 21  a ' 2 * • ^ = m6 l
U * &** c " a 3 *.6 2 »=m 10 J '
22. If \og(x' 2 y 3 ) = a } and log = 6, find log* and log//.
23. If a 3 " * . V> x = a x + \ b 3x , shew that x log (") = log a.
24. Solve the equation
(a*  2a*b* + b*) x  1 = (a  ft) 2 * (a + 6)  '.
Common Logarithms.
208. Logarithms to the base 10 are called Common Logar
ithms; this system was first introduced, in 1615, by Briggs, a
contemporary of Napier the inventor of logarithms.
From the equation 10 x  JV, it is evident that common logar
ithms "will not in general be integral, and that they will not
always be positive.
For instance 3154 > 10 ' and < 10 4 ;
ion.
12—2
.*. log 3151«=3 + a fraction.
180 HIGHER ALGEBRA.
Again, 06 > 10~ 2 and < 10 _l ;
.*. log *06 =  2 + a fraction.
209. Definition. The integral part of a logarithm is called
the characteristic, and the decimal part is called the mantissa.
The characteristic of the logarithm of any number to the
base 10 can be written down by inspection, as we shall now shew.
210. To determine the characteristic of the logarithm of any
number greater than unity.
Since 10 1  10,
10 2 =100,
10 3 1000,
it follows that a number with two digits in its integral part lies
between 10' and 10 2 ; a number with three digits in its integral
part lies between 10 2 and 10 3 ; and so on. Hence a number
with n digits in its integral part lies between 10" _I and 10".
Let N be a number whose integral part contains n digits;
then
J\T— in(ttl)+ a fraction .
.*. log iV= (n — 1) + a fraction.
Hence the characteristic is n — 1 \ that is, the characteristic of
the logarithm of a number greater than unity is less by one than
the number of digits in its integral part, and is positive.
211. To determine the characteristic of the logarithm of a
decimal fraction.
Since 10°= 1,
1(rs =iJcr 01 '
108 =i=' 001 >
LOGARITHMS. 181
it follows that a decimal with one cipher immediately after the
decimal point, such as 0324, being greater than 01 and less
than 1, lies between 10~ 2 and 10 1 ; a number with two ciphers
after the decimal point lies between 10 _:i and 10""; and so on.
Hence a decimal fraction with n ciphers immediately after the
decimal point lies between 10~ ( " + 1) and 10~".
Let D be a decimal beginning with n ciphers ; thou
/) _ 1 f)~(w + l) + ■ fraction.
.. log J) = — (n + l) + n fraction.
Hence the characteristic is  (n+ 1) ; that is, the characteristic
of the logarithm of a decimal fraction is greater by unity titan the
number of ciphers immediately after the decimal point, and is
negative.
212. The logarithms to base 10 of all integers from 1 to
200000 have been found and tabulated j in most Tables they are
given to seven places of decimals. This is the system in practical
use, and it has two great advantages :
(1) From the results already proved it is evident that the
characteristics can be written down by inspection, so that only
the mantissse have to be registered in the Tables.
(2) The mantissse are the same for the logarithms of all
numbers which have the same significant digits; so that it is
sufficient to tabulate the mantissse of the logarithms of integers.
This proposition we proceed to prove.
213. Let N be any number, then since multiplying or
dividing by a power of 10 merely alters the position of the
decimal point without changing the sequence of figures, it follows
that N x 10''. and N ~ 10 7 , where p and q are any integers, are
numbers whose significant digits are the same as those of N.
Now log (N x 10 p ) = log N+p log 10
= log J\ r +p (1 ).
Again, log (A T  1 9 )  log N  q log 1
= logiV7 (2).
In (1) an integer is added to logiV^, and in (2) an integer is
subtracted from log N ; that is, the mantissa or decimal portion
of the logarithm remains unaltered.
182 HIGHER ALGEBRA.
In this and the three preceding articles the mantissse have
been supposed positive. In order to secure the advantages of
Briggs' system, we arrange our work so as always to keep the
mantissa positive, so that when the mantissa of any logarithm
has been taken from the Tables the characteristic is prefixed
with its appropriate sign according to the rules already given.
214. In the case of a negative logarithm the minus sign is
written over the characteristic, and not before it, to indicate that
the characteristic alone is negative, and not the whole expression.
Thus 430103, the logarithm of 0002, is equivalent to 4 + 30103,
and must be distinguished from — 4*30103, an expression in which
both the integer and the decimal are negative. In working with
negative logarithms an arithmetical artifice will sometimes be
necessary in order to make the mantissa positive. For instance,
a result such as  3*69897, in which the whole expression is
negative, may be transformed by subtracting 1 from the
characteristic and adding 1 to the mantissa. Thus
 369897   4 + (1  69897) = 430103.
Other cases will be noticed in the Examples.
Example 1. Required the logarithm of 0002432.
In the Tables we find that 3859636 is the mantissa of log 2432 (the
decimal point as well as the characteristic being omitted) ; and, by Art. 211,
the characteristic of the logarithm of the given number is  4 ;
.. log 0002432 = 43859636.
Example 2. Find the value of ^00000165, given
log 165 = 22174839, log 697424=58434968.
Let x denote the value required ; then
I l
log a = l©g (00000165) 5 = = log (00000165)
o
= i (62174839) ;
the mantissa of log 00000165 being the same as that of log 165, and the
characteristic being prefixed by the rule.
Now  (62174839) =  (10 + 42174839)
= 28434968
LOGARITHMS. 183
and 8434908 is the mantissa of log 007424; hence x is a number consisting
of these same digits but with one cipher after the decimal point. [Art. 211. J
Thus a: = 0097424.
215. The method of calculating logarithms will be explained
in the next chapter, and it will there be seen that they are first
found to another base, and then transformed into common loga
rithms to base 10.
It will therefore be necessary to investigate a method for
transforming a system of logarithms having a given base to a
new system with a different base.
216. Suppose that the logarithms of all numbers to base a
are known and tabulated, it is required to find the logarithms
to base b.
Let N be any number whose logarithm to base b is re
quired.
Let y = log 6 iV, so that b y = N ;
••■ log. (&") = logJT;
that is, ylog£ = log,JV;
1/ — r X log N.
J log a 6 0u '
or k&^wK* 10 ^ C 1 )
1U Oa°
Now since N and b are given, los: N and log b are known
from the Tables, and thus log^V may be found.
Hence it appears that to transform logarithms from base a
to base b we have only to multiply them all by ; this is a
J r J J log b '
constant quantity and is given by the Tables; it is known as the
modulus.
217. Tn equation (1) of the preceding article put a for N\
thus
. i , i
On Oa
log/t x log 8 /j = 1 .
184 HIGHER ALGEBRA.
This result may also be proved directly as follows :
Let x = log/?, so that a* = b ;
then by taking logarithms to base b, we have
x \og b a = \og b b
.. log a 6xlog 4 a = l.
218. The following examples will illustrate the utility of
logarithms in facilitating arithmetical calculation ; but for in
formation as to the use of Logarithmic Tables the reader is
referred to works on Trigonometry.
4 5
Example 1. Given log 3 = 4771213, find log {(27) 3 x (81)»H90)*}.
27 4 81 5
The required value = 3 log j= +  log 1Q  = log 90
= 3(l<^3»l)+oog3*2)(log3*+l)
KM)"*aK + t)
= ^log35H
= 46280766585
= 27780766.
The student should notice that the logarithm of 5 and its
powers can always be obtained from log 2 ; thus
log 5 = log — = log 10  log 2 = 1  log 2.
Example 2. Find the number of digits in 875 1C , given
log 2 = 3010300, log 7 = 8450980.
log (875 16 ) = 16 log (7x125)
= 16 (log 7+ 3 log 5)
= 16(log7 + 331og2)
= 16x29420080
=47072128;
hence the number of digits is 48. [Art. 210.]
LOGARITHMS. 185
Example 3. Given log 2 and log 3, find to two places of decimals the
value of x from the equation
Taking logarithms of both sides, we have
(3  4a) log G + (x + 5) log 4 = log 8 ;
.. (3  4.r) (log 2 + log 3) + (x + 5) 2 log 2 = 3 log 2 ;
.. .r (  4 log 2  4 log 3 + 2 log 2) = 3 log 2  3 log 2  3 log 3  10 log 2 ;
10 log 2 + 3 log 3
.r =
2 log 2 + 4 log 3
_ 44416639
~2al054a2
= 177...
EXAMPLES. XVI. b.
1. Find, by inspection, the characteristics of the logarithms of
21735, 238, 350, '035, % 87, 875.
2. The mantissa of log 7623 is '8821259 ; write down the logarithms
of 7623, 7623, U07623, 762300, '000007623.
3. How many digits are there in the integral part of the numbers
whose logarithms are respectively
430103, 14771213, 369897, 56515 1
4. (Jive the position of the first significant figure in the numbers
whose logarithms are
27781513, 6910815, 54871384.
Given log 2 = 3010300, log 3 ='4771213, log 7 = 8450980, find the
value of
5. log 64. 6. log 84. 7. log 128.
8. log 0125. 9. log 144. 10. log 4^.
11. log^l2. 12. logW — . 13. logN 4 / : 0l05.
\ i
14. Find the seventh root of 00324, having given that
log 44092388 = 76443036.
15. Given log 194*8445 = 2'2896883, find the eleventh root of (392) 2 .
186 HIGHER ALGEBRA.
16. Find the product of 37203, 37203, 0037203, 372030, having
given that
log 37203 = 15705780, and log!915631 = 6:28231 20.
3 //3 2 5**\
17. Given log 2 and log 3, find log /( —y ) .
X
18. Given log 2 and log 3, find log (#48 x 108 1 f ^6).
19. Calculate to six decimal places the value of
V
/ 294 x 125 \ 2
V 42 x 32 J '
given log 2, log 3, log 7; also log 9076226 = 39570053.
20. Calculate to six places of decimals the value of
(330^49) 4 ^\ / 22x70;
given log 2, log 3, log 7 ; also
log 11 = 10413927, and logl78141516 = 42507651.
21. Find the number of digits in 3 12 x 2 8 .
/21\ 100
22. Shew that ( — J is greater than 100.
23. Determine how many ciphers there are between the decimal
/1\ 1000
point and the first significant digit in (  j
Solve the following equations, having given log 2, log 3, and log 7.
24. 3*~ 2 = 5. 25. 5* = 10l 26. 5 5 ~ 3 *=2* + 2 .
27. 2F = 2 2 * + 1 .5 3 . 28. 2*. 6* 2 =5 2 *. 7 1 "*.
29. 2 x + y = 6»
3* =3
' "I 30. 3 l ~ x  y =4y \
22, + ij 2 2x ~ l =3 3j/_a; J
31. Given log 10 2 = 30103, find log 25 200.
32. Given log 10 2 = .30103, log 10 7 = 84509, find log 7N /2 and logV27.
CHAPTER XVII.
EXPONENTIAL AND LOGARITHMIC SERIES.
219. In Chap. XVI. it was stated that the logarithms in
common use were not found directly, but that logarithms .are
first found to another base, and then transformed to base 10.
In the present chapter we shall prove certain formulae known
as the Exponential and Logarithmic Series, and give a brief ex
planation of the way in which they are used in constructing a
table of logarithms.
220. To expand a 1 in ascending powers of x.
By the Binomial Theorem, if n>l,
K)"
1 nx(nx—\) 1 nx (nx — 1) (nx 2) 1
= 1 + nx .  + — v — r . » + * ~± J  . s +
n 2 n" 3 n 6
x (x ) x (x ) (x — \
I? I 3 .
By putting sb=1, we obtain
(i).
(')•
188 HIGHER ALGEBRA.
hence the seizes (1) is the x ih power of the series (2); that is,
1 + x + ,„ + rz
3
and this is true however great n may be. If therefore n be
indefinitely increased we have
x 2 x 3 x A /'ill
1+ * + 2 + ]3 + _4 + = ( 1 + 1+ ^ + ^ + U +
1 1 1
The series 1 + 1 + — + — + — +
y
is usually denoted by e ; hence
■5 3 4
x X X
, = l + « + + + +
Write ex for x, then
cV cV
6** =  CX 4 tjj + ry +
Now let e e = «, so that c = log/* ; by substituting for c we
obtain
a' = l+x\og e a + — Vo + ,» +
If lr
This is the Exponential Theorem.
Cor. When n is infinite, the limit of ( 1 +  ) = e.
[See Art. 266.]
Also as in the preceding investigation, it may be shewn that
when n is indefinitely increased,
(, x\ n  X 2 x 3 x 4
1+ n) =1+X+ Y2 + ]3 + \i +
EXPONENTIAL AND LOGARITHM!*! SERIES. 180
tli.tt is, when n is infinite, the limit of ( 1 +  ) —
V nj
c T .
x 1
l>y putting — = , we have
H)K)~={K)7
Now m is infinite when n is infinite;
(x\ n
1 — J — e *.
Hence the limit of (1 ) = e~\
(■.')'
221. In the preceding article no restriction is placed upon
the value of x; also since  is less than unity, the expansions we
have used give results arithmetically intelligible. [Art. 183.]
But there is another point in the foregoing proof which
deserves notice. We have assumed that when n is infinite
/ 1\7 2\ / rV
:)•••(•
nj \ nj \ n J . x r
the limit or , is r
\r \r
for all values of r.
Let us denote the value of
iB ( a! "3( a, ~3 ■{ x  r ^r)
H(*^)
by u r .
™, u 1 / r— 1\ as 1 1
Then — z = lx ) = + —
u , r \ n J r n nr
Since n is infinite, we have
U X . . x
— — =  ; that is, u ~ — u r . .
u , r r l
a
It is clear that the limit of u is r^; hence the limit of u 3 is
ft> x x
,x; that of u A is .— r ; and generally that of u t is .— .
190 HIGHER ALGEBRA.
222. The series
ii 111
E + H + ~@ + '
which we have denoted by e, is very important as it is the base
to which logarithms are first calculated. Logarithms to this
base are known as the Napierian system, so named after Napier
their inventor. They are also called natural logarithms from the
fact that they are the first logarithms which naturally come into
consideration in algebraical investigations.
When logarithms are used in theoretical work it is to be
remembered that the base e is always understood, just as in
arithmetical work the base 10 is invariably employed.
From the series the approximate value of e can be determined
to any required degree of accuracy ; to 1 places of decimals it is
found to be 27182818284.
Example 1. Find the sum of the infinite series
, 1 1 1
We have e = l + l+  + — + rg + ;
and by putting x =  1 in the series for e x ,
e " 1=1  1+ i2i3 + n
hence the sum of the series is  (e + e~ x ).
a
Example 2. Find the coefficient of x r in the expansion of
e x
1  ax — x
o
= (1  ax  x 2 ) e~ x
n n fi «■ «* (l) r x r 1
= (la,^)l, + ^ + ... + L_L_ + j.
EXPONENTIAL AND LOGARITHMIC SERIES. 101
(!)>• (l)'ia (1)'
The coefficient required —
r
l) r
r1 r2
{l + arr(rl)},
223. 7V> expand log, (1 + tt) ira ascending powers of \.
From Art. 220,
/r (loge a) 8 ^ y 3 (log. 4 '
L 2
a" =• 1 + y 1< >g e r6 + •  + ^ v  c ' +
lii this series write 1 + x for a; thus
(1 +x)'J
= 1 + y log, (1 + *) + f 2 {log, (1 + *)}■ + £ {log e (1 + a;)} 3 + ... (1).
Also by the Binomial Theorem, when x < 1 we have
(i+«yi+» + g^*+ y fr 1 >fr 8) rf+ (2).
Now in (2) the coefficient of y is
,+ 1.3* + 1.2J + 1.2.3.4 + '
r**^ rp& •>••*
. m %K/ \K/ «C»
that is, £ —  + — —  +
2i o ±
Equate this to the coefficient of y in (1) ; thus we have
l0g t ,(l +Ct') = t 7J+  +
This is known as the Logarithmic Series.
Example. If x < 1, expand {log,, (1 + x)} 9 in ascending powers of .t.
By equating the coefficients of y 2 in the series (1) and (2), we see that the
required expansion is double the coefficient of y' 2 in
?/(!/!) r2 . y (y  1) (y  2) . , y(yi)(y2)(y3)
____. x + 1.2.8 + 1.2.3.4 ^ + '
that is, double the coefficient of y in
y1 (yl)(y2) (y  1) (y  2) (y  3)
1.2* + 1.2.3 * + "1.2.3.4 +
Thn 8 {log.(l + *)P=2{^i(l+l)^(l + l + l)«* }.
192 HIGHER ALGEBRA.
224. Except when x is very small the series for log e (l + x)
is of little use for numerical calculations. We can, however,
deduce from it other series by the aid of which Tables of Logar
ithms may be constructed.
By writing — for x we obtain log. : hence
1 1 1
■ lo S.( n+1 ) lo & w = S"2? + 3^" (1) 
1 n  1
By writing for x we obtain log e ; hence, by changing
signs on both sides of the equation,
log 8 n  log e (n  1 ) =  + s— a + #7 — 3 + (2).
From (1) and (2) by addition,
log.(" + l)log,(nl) = 2( +_+__+ ...J (3).
From this formula by putting n = 3 we obtain log e 4 — log e 2,
that is log e 2 ; and by effecting the calculation we find that the
value of log 6 2 69314718...; whence log e S is known.
Again by putting n = 9 we obtain log e 10 — log e 8; whence we
findlog e 10 = 230258509....
To convert Napierian logarithms into logarithms to base 10
we multiply by . =j= , which is the modulus [Art. 216] of the
1
common system, and its value is — , or '43429448...;
^' oOJjOoDk) J . . .
we shall denote this modulus by /x.
In the Proceedings of the Royal Society of London, Vol. xxvn.
page 88, Professor J. 0. Adams has given the values of e, /x,
log e 2, log e 3, log e 5 to more than 260 places of decimals.
225. If we multiply the above series throughout by /x, we
obtain formulae adapted to the calculation of common logarithms.
Thus from (1), /x log e (ra + 1)  /* log e ?i = £ _ ^ + JL . ... •
EXPONENTIAL AND LOGARITHMIC SERIES. 193
that is,
log I0 ( M + 1)  log, n = £  t + Jt. _ m
Similarly from (2),
l ^lo g] >l)^ + ^ + ^ + (2) .
From either of the above results we see that if the logarithm
ot one of two consecutive numbers be known, the logarithm of
the other may be found, and thus a table of logarithms can be
constructed. °
to J2 Sl ; " ld + ^ e «*"«*«* that the above formula are only needed
to calculate the logarithms of prime numbers, for the logarithm
ot a compose number may be obtained by adding together the
logarithms of its component factors.
In order to calculate the logarithm of any one of the smaller
prime numbers, we do not usually substitute the number in either
of the formula (1) or (2), but we endeavour to find some value
ot n by which division may be easily performed, and such that
either 7^+ 1 or nl contains the given number as a factor. We
then find log(n+l) or log( w l) and deduce the logarithm of
the given number. °
Example. Calculate log 2 and log 3, given ^=43429448.
By putting n = 10 in (2), we have the value of log 10 log 9; thus
1  2 log 3 = 043429448 + 002171472 + 000144765 + 000010857
+ 000000868 + 000000072 + 000000006 ;
12 log 3 =045757488,
log 3 = 477121256.
Putting M = 80 in (1), we obtain log 81 log 80; thus
4 log 3  3 log 2  1 = 005428681  000033929 + 000000283  000000003 ;
3 log 2 = 908485024  005395032,
log 2 = 301029997.
In the next article we shall give another series for
iog 9 {7i + l)\ g e n which is often useful in the construction of
Logarithmic Tables. For further information on the subject the
reader is referred to Mr Glaisher's article on Logarithms in the
hncyclopcvdia Britannica.
H. H. A. I •>
194 HIGHER ALGEBRA.
226. In Art. 223 we have proved that
2 3
log e (l + x) = x~2 + 3"~"'
changing x into  cc, we have
2 3
log.(l «)=■*— 2 ~ J" —
By subtraction,
. 1 + x { a / x 3 x 5 \
Put = — ■ = , so that x — x = ; we thus obtain
lx n Zn + 1
los* (n + \) — log. n = 2< 5 7 + 777^ = va + jt= ^r. + ...}.
oeV ; &e (2w + l 3(2?i + l) 3 5(2w + l) 5 J
Note. This series converges very rapidly, but in practice is not always
so convenient as the series in Art. 224.
227. The following examples illustrate the subject of the
chapter.
Example 1. If a, 8 are the roots of the equation ax 2 + bx + c = 0, shew
a 2 ,02 a 3 . ffi
that \og(abx + cx' 1 ) = loga+(a. + p)x  J —x 2 + —^ x 3 ...
 o
Since a + 8 = — , a/3 =  , we have
a a
a  bx + cx 2 =a {l + {a + B)x + aBx 2 )
= a (1 + cur) (1 + px).
.. log (a  bx + ex 2 ) = log a + log (1 + ax) + log (1 + Bx)
ax' 2 a 3 x 3 a B 2 x 2 B 3 x 3 .
= loga + ax _ + _... + Bx'^+^...
= \oga + {a + B)x a ^^ x 2 + a *+^ x 3  ...
Example 2. Prove that the coefficient of x n in the expansion of
2 1
log (1 + x + x 2 ) is — or  according as n is or is not a multiple of 3.
n n
1x 3
log (1 + x + x 2 ) = log — — =log (1  x' 3 )  log (1  x)
Q X 6 X 9 X 3r ( X 2 X 3 x r \
EXPONENTIAL AND LOGARITHMIC SERIES. 1 93
If n is a multiple of 3, denote it by 3,; then the eoeffieient of *» j, _ I
from the first series, together with g I fr om the second 3eries; ^ J
coefficient is   +  or _ ?
n n ' n '
If . is not a multiple „f 3, *» does not ocour in the first series, therefore
the required coefficient is  .
n
228. To prove that e is incommensurable.
For if not, let e = ™ where m and n are positive integers;
then ^ui.. 1 1 . 1 1
7i If 1 3 (w n+l "
multiply both sides by \n;
• '• m ^irJ = integer + i + ___J___ * 1
w+1 (n+l)(n+2) (n+])(jw.2)(n+S) + "
But — L + _____J_____ 1
n + 1 (n + l)( n+ 3) (n + 1 )~(^T2)^T3) + * ' '
is a proper fraction, for it is greater than * and less than the
geometrical progression
_ + _i i__
\3 + ••• 'j
n+\ (n+1) 2 ' (n+iy
that is, less than I; hence an integer is equal to an integer plus
a fraction, which is absurd; therefore e is incommensurable.
EXAMPLES. XVIL
1. Find the value of
0 + 57 + ? ■.+...
2^3 4 + 5~6
2. Find the value of
2 2 . 2 2 ^ 3 . 2 3 i72 4 +
5 . 2 5 " ' *
15 9
19G HIGHER ALGEBRA.
3. Shew that
a a? a?
h ge (n+a)lo gt (na)=2l + ^ .+_+ \
*\*& />»o O"**
»</ %A/ *t/
4. if y=*'2 +3  4 +•••»
2 3
shew that x =y +^ + ^ + . . . .
5. Shew that
+ ~ ( : )+^P ' ) +... = log e alog e Z>.
a 2 V~^" / + 3 V a
6. Find the Napierian logarithm of — — correct to sixteen places
of decimals.
/ 1 2 3 \
7. Prove that e" 1 = 2 ( . + .=■ + nr +....) .
8. Prove that
iog,d + xr« ( i »•)''=« (£ + o + o +••••) •
9. Find the value of
H '' 2  f + j2 (' 4  # 4 ) + i ^   //,m + • • ■ ■
10. Find the numerical values of the common logarithms of 7, 1 1
and 13; given ^ = 43429448, log 2 = '30103000.
11. Shew that if ax 2 and — 2 are each less unity
12. Prove that
log c (l + 3a+2^ 2 ) = 3.r — + — — + ... ;
and find the general term of the series.
13. Prove that
, 1 + 3.? „ 5x 2 S5X 3 65.iT 4
and find the general term of the series.
14. Expand — ^— in a series of ascending powers of x.
EXPONENTIAL AND LOGARITHMIC SERIES. 197
15. Express  (e ix + e~ ix ) in ascending powers of .r, where i'= </ — 1.
25
16. Shew that
17. If a and /3 be the roots of x 2 jtxr + tf = 0, shew that
a 2 4 ft 2 n 3 4 ft 3
18. If .r<l, find the Bum of the series
1 „ 2 , 3 4 .
19. Shew that
i A i\ n , i i
log, (1 + ) =1
nj 2(» + l) 2.3(>4l) 2 3.4(>t + l) 3 ""
l+.r + .^ + .t 3
1
20. If log, ., x ^,^^,2^ , 3 De expanded in a series of ascending
powers of #, shew that the coefficient of o: n is — if n be odd, or of
3
the form 4m + 2. and  if n be of the form 4m.
21. Shew that
22. Prove that
2 3 , 3 3 4 3
1 + ]2 + J3 + (4 +  =5e 
2 log, n  log, (« + 1)  log, (» _ 1)= + — + _ +
23. Shew that — 1 — — +
ft + 1 2(?i+l) 2 ' 3(7i+I) 3 "
1 1_ 1
n 2?i 2 3n 3
<) 24 81
24. If log, Yq =  «, log e 25 = ~ ?; ' 1() S« go = C ' sheW that
log, 2 = 7a  26 + 3c, log, 3 = 1 1 a  36 + 5c, log, 5 = 1 6a  Ah + Vc ;
and calculate log, 2, log, 3, k>g e 5 to 8 places of decimals.
CHAPTER XVIII.
INTEREST AND ANNUITIES.
229. In this chapter we shall explain how the solution of
questions connected with Interest and Discount may be simplified
by the use of algebraical formulae.
"We shall use the terms Interest, Discount, Present Value in
their ordinary arithmetical sense ; but instead of taking as the
rate of interest the interest on ,£100 for one year, we shall find it
more convenient to take the interest on £1 for one year.
230. To find the interest and amount of a given sum in a
given time at simple interest.
Let P be the principal in pounds, r the interest of £1 for one
year, n the number of years, I the interest, and M the amount.
The interest of P for one year is Pr, and therefore for n years
is Pnr ; that is,
/ =Pnr (1).
Also M = P + I;
that is, M=P(l+nr) (2).
From (1) and (2) we see that if of the quantities P, n, r, 7,
or P, ?i, r, M, any three be given the fourth may be found.
231. To find the present value and discount of a given sum
due in a given time, allowing simple interest.
Let P be the given sum, V the present value, D the discount,
r the interest of £1 for one year, n the number of years.
I
INTEREST AND ANNUITIES. 199
Since V is the sum which put out to interest at the present
time will in u years amount to P, we have
P= V(\+nr);
1 + nr
P
And D = P 
1 + nr '
Pnr
1 + nr '
Note. The value of D given by this equation is called the true discount.
But in practice when a sum of money is paid before it is due, it is customary
to deduct the interest on the debt instead of the true discount, and the
money so deducted is called the banker's discount; so that
Banker's Discount = Pnr.
Pnr
True Discount =
1 + nr'
Example. The difference between the true discount and the banker's
discount on £1900 paid 4 months before it is due is 6s. 8d.; find the rate
per cent., simple interest being allowed.
Let r denote the interest on £1 for one year; then the banker's discount
1900r
• 1900 '' i «. * a * ■ ~~ ^~
is —  — , and the true discount is ■
i*
1900r
1900r ~3~ 1
""• ~3~ 7~T~3 ;
l +i r
whence 1900r 2 =3 + >;
1 ±Jl + 22800 _ 1±151
*'• r ~ 3800 ~ 3800 '
t, • • . , 152 1
Rejecting the negative value, we nave f—aSui = o? «
.. rate per cent. = 100r = 4.
232. To find the interest and amount of a given sum in a
given time at compound interest.
Let P denote the principal, 7? the amount of £1 in one year,
n the number of years, I the interest, and M the amount.
200 HIGHER ALGEBRA.
The amount of P at the end of the first year is PR ; and, since
this is the principal for the second year, the amount at the end of
the second year is PR x R or PR 2 . Similarly the amount at the
end of the third year is PR 3 , and so on ; hence the amount in
n years is PR" \ that is,
M=PR";
.'. I=P(R n l).
Note. If r denote the interest on £1 for one year, we have
R = l+r.
233. In business transactions when the time contains a
fraction of a year it is usual to allow simjyle interest for the
fraction of the year. Thus the amount of ,£1 in ^ year is
v
reckoned 1 +  ; and the amount of P in 4f years at compound
interest is PR* (1 + ^ r J . Similarly the amount of P in
n + — years is PR" (In — ) .
m \ m/
If the interest is payable more than once a year there is a
distinction between the nominal annual rate of interest and that
actually received, which may be called the true annual rate ; thus
if the interest is payable twice a year, and if r is the nominal
r
annual rate of interest, the amount of £1 in half a year is 1 +^ ,
and therefore in the whole year the amount of <£1 is (1 + J,
r 2
or 1 + r + — ; so that the true annual rate of interest is
4
r 2
234. If the interest is payable q times a year, and if r is
the nominal annual rate, the interest on .£1 for each interval is
r
 , and therefore the amount of P in n years, or qn intervals, is
In this case the interest is said to be "converted into principal"
(f times a year.
INTEREST AND ANNUITIES. 201
If the interest is convertible into principal every moment,
then q becomes infinitely great. To find the value of the amount,
r 1
put  = — , so that q  rx : thus
q x
the amount = P (l +Y =P(l + Y" = P {(l + i)T'
= Pe nr , [Art. 220, Cor.,]
since x is infinite when q is infinite.
235. To find the present value and discount of a given stun
due in a given time, allowing comjwund interest.
Let P be the given sum, V the present value, D the discount,
R the amount of £1 for one year, n the number of years.
Since V is the sum which, put out to interest at the present
time, will in n years amount to P, we have
P=VR n ',
it
and D = P(lR).
Example. The present value of £672 due in a certain time is £126; if
compound interest at 4£ per cent, be allowed, find the time; having given
log 2 = 30103, log 3 = 47712.
Here Hol) = iI' *»**=!'
Let n be the number of years ; then
672=126 y ;
. 25 . 672
•'• ?ll0g 24 = 1 ° g i26'
. 100 . 16
or ?ilog 96 =logy ;
.. n (log 100  log 96) = log 16  log 3,
4 log 2  log 3
n =
2  5 log 2  log 3
•72700
" = 01773 = ' Veiy nea y '
thus the time is very nearly 41 years.
202 HIGHER ALGEBRA.
EXAMPLES. XVIII. a.
When required the following logarithms may be used,
log 2 = 3010300, log 3 = '4771 213,
log 7 = 8450980, log 11 = 10413927.
1. Find the amount of £100 in 50 years, at 5 per cent, compound
interest; given log 114674 = 20594650.
2. At simple interest the interest on a certain sum of money is
,£90, and the discount on the same sum for the same time and at the
same rate is £80 ; find the sum.
3. In how many years will a sum of money double itself at 5 per
cent, compound interest ?
4. Find, correct to a farthing, the present value of £10000 due
8 years hence at 5 per cent, compound interest ; given
log 6768394 = 48304856.
5. In how many years will £1000 become £2500 at 10 per cent,
compound interest ?
6. Shew that at simple interest the discount is half the harmonic
mean between the sum due and the interest on it.
7. Shew that money will increase more than a hundredfold in
a century at 5 per cent, compound interest.
8. What sum of money at 6 per cent, compound interest will
amount to £1000 in 12 years ? Given
log 106 = 20253059, log 49697 = 46963292.
9. A man borrows £600 from a moneylender, and the bill is
renewed every halfyear at an increase of 1 8 per cent. : what time will
elapse before it reaches £6000 1 Given log 1 18 = 2071882.
10. "What is the amount of a farthing in 200 years at 6 per cent,
compound interest? Given log 106 = 20253059, log 11 50270 = 20611800.
Annuities.
*
236. An annuity is a fixed sum paid periodically under
certain stated conditions ; the payment may be made either once
a year or at more frequent intervals. Unless it is otherwise
stated we shall suppose the payments annual.
An annuity certain is an annuity payable for a fixed term of
years independent of any contingency ; a life annuity is an
annuity which is payable during the lifetime of a person, or of
the survivor of a number of persons.
INTEREST AND ANNUITIES. 203
A deferred annuity, or reversion, is an annuity which does
not begin until after the lapse of a certain number of years ; and
when the annuity is deferred for n years, it is said to commence
after n years, and the first payment is made at the end of n + 1
years.
If the annuity is to continue for ever it is called a perpetuity ;
if it does not commence at once it is called a deferred perpetuity.
An annuity left unpaid for a certain number of years is said
to be forborne for that number of years.
237. To find the amount of an annuity left unpaid for a given
number of years, allowing simple interest.
Let A be the annuity, r the interest of £1 for one year, n the
number of years, M the amount.
At the end of the first year A is due, and the amount of this
sum in the remaining n  1 years is A + (n — 1) rA ; at the end of
the second year another A is due, and the amount of this sum in
the remaining (a— 2) years is A + (ii — 2) rA • and so on. Now
M is the sum of all these amounts ;
.. M={A + (nl)rA} + {A + (n2)rA} + + (A + rA) + A,
the series consisting of n terms ;
.. J/=Wil+(l + 2 + 3+ +nl)rA
= nA + — K — — ' rA.
238. To find the amount of an annuity left unpaid for a
given number of years, allowing compound interest.
Let A be the annuity, R the amount of <£1 for one year, n
the number of years, M the amount.
At the end of the first year A is due, and the amount of this
sum in the remaining n— 1 years is AR n ~ x ; at the end of the
second year another A is due, and the amount of this sum in the
remaining n  2 years is AR n ~ 2 ; and so on.
.. M = AR n ~ x + AR" 2 + +AR 2 + AR + A
= A(l +R + R 2 + to n terms)
. R m  1
= A R^l
204 HIGHER ALGEBRA.
239. In finding the present value of annuities it is always
customary to reckon compound interest; the results obtained
when simple interest is reckoned being contradictory and un
trustworthy. On this point and for further information on the
subject of annuities the reader may consult Jones on the Value
of Annuities and Reversionary Payments, and the article Annuities
in the Encyclopaedia Britannica.
240. To find t/ie present value of an annuity to continue for
a (jiven number of years, allowing compound interest.
Let A be the annuity, R the amount of £>\ in one year, n
the number of years, V the required present value.
The present value of A due in 1 year is AR~ l ;
the present value of A due in 2 years is AR~' J ;
the present value of A due in 3 years is AR~ 3 ;
and so on. [Art. 235.]
Now V is the sum of the present values of the different
payments ;
.. V=AR 1 +AR 3 + AK 3 + tow terms
1  R~"
= AR~ l
= A
lR 1
1R"
Rl
Note. This result may also be obtained by dividing the value of M,
given in Art. 238, by R n . [Art. 232.]
Cor. If we make n infinite we obtain for the present value
of a perpetuity
RV r '
241. If mA is the present value of an annuity A, the annuity
is said to be worth m years' purchase.
In the case of a perpetual annuity mA — — ; hence
1 100
m =  =
r rate per cent.
INTEREST AND ANNUITIES. 205
that is, the number of years' purchase of a perpetual annuity is
obtained by dividing 100 by the rate per cent.
As instances of perpetual annuities we may mention the
income arising from investments in irredeemable Stocks such as
many Government Securities, Corporation Stocks, and Railway
Debentures. A good test of the credit of a Government is fur
nished by the number of years' purchase of its Stocks ; thus the
2 p. c. Consols at 96} are worth 35 years' purchase ; Egyptian
4 p. c. Stock at 96 is worth 24 years' purchase ; while Austrian
5 p. c. Stock at 80 is only worth 16 years' purchase.
242. To find the present value of a deferred annuity to
commence at the end of p years and to continue for n years, allow
ing compound interest.
Let A be the annuity, R the amount of £1 in one year, V the
present value.
The first payment is made at the end of (;> + l) years.
[Art. 236.]
Hence the present values of the first, second, third... pay
ments are respectively
AR { * +l \ AR (p+2 \ AR (p+3 \ ...
.'. V=AR (p+l) + AR {p+ »+AR (1,+3 >+ ton terms
1 7?~"
= AR~ (p+1)  — —
AR~ V A R p "
Cor. The present value of a deferred perpetuity to commence
after p years is given by the formula
V ~RV
243. A freehold estate is an estate which yields a perpetual
annuity called the rent ; and thus the value of the estate is equal
to the present value of a perpetuity equal to the rent.
It follows from Art. 241 that if we know the number of years'
purchase that a tenant pays in order to buy his farm, we obtain
the rate per cent, at which interest is reckoned by dividing 100
by the number of years' purchase.
206 HIGHER ALGEBRA.
Example. The reversion after 6 years of a freehold estate is bought for
£20000; what rent ought the purchaser to receive, reckoning compound
interest at 5 per cent. ? Given log 105 = 0211893, log 1340096 = 1271358.
The rent is equal to the annual value of the perpetuity, deferred for 6
years, which may be purchased for £20000. '
Let £A be the value of the annuity; then since .R = l05, we have
20000^* ^° 5 ' 6 ;
•0o
.. A x (105) ~ 6 = 1000;
log A 6 log 105 = 3,
log A = 31271358 = log 134009G.
.. A = 1310096, and the rent is £1340. Is. lid.
244. Suppose that a tenant by paying down a certain sum
lias obtained a lease of an estate for p + q years, and that when
q years have elapsed he wishes to renew the lease for a term
p + n years ; the sum that he must pay is called the fine for
renewing n years of the lease.
Let A be the annual value of the estate ; then since the
tenant has paid for p of the p + n years, the fine must be equal
to the present value of a deferred annuity A, to commence after
p years and to continue for n years ; that is,
. n AR* AR p ~ n r 4
the fine = = — . [Art. 242.1
A — 1 A — 1
EXAMPLES. XVIII. b.
The interest is supposed compound unless the contrary is stated.
1. A person borrows ,£672 to be repaid in 5 years by annual in
stalments of ,£120; find the rate of interest, reckoning simple interest.
2. Find the amount of an annuity of ,£100 in 20 years, allowing
compound interest at 4 per cent. Given
log 1045 = 0191163, log24117 = 13823260.
3. A freehold estate is bought for £2750 ; at what rent should it
be let so that the owner may receive 4 per cent, on the purchase money ?
4. A freehold estate worth £120 a year is sold for £4000; find the
rate of interest.
INTEREST AND ANNUITIES. 207
5. How many years' purchase should be given for a freehold
estate, interest being calculated at 3i per cent.?
6. If a perpetual annuity is worth 25 years' purchase, find the
amount of an annuity of £625 to continue for 2 years.
7. If a perpetual annuity is worth 20 years' purchase, find the
annuity to continue for 3 years which can be purchased for £2522.
8. When the rate of interest is 4 per cent., find what sum must
be paid now to receive a freehold estate of £400 a year 10 years hence;
having given log 104 = 20170333, log 675565 8296670.
9. Find what sum will amount to £500 in 50 years at 2 per cent.,
interest being payable every moment; given e _1 = '3678.
10. If 25 years' purchase must be paid for an annuity to continue
n years, and 30 years' purchase for an annuity to continue 2?i years,
find the rate per cent.
11. A man borrows £5000 at 4 per cent, compound interest ; if the
principal and interest are to be repaid by 10 equal annual instalments,
find the amount of each instalment ; having given
log 104 =01 70333 and log 675565 = 5829667.
12. A man has a capital of £20000 for which he receives interest
at 5 per cent. ; if he spends £1800 every year, shew that he will be
ruined before the end of the 17 th year; having given
log 2 = '3010300, log 3 = '4771213, log 7 = '8450980.
13. The annual rent of an estate is £500 ; if it is let on a lease
of 20 years, calculate the fine to be paid to renew the lease when 7 years
have elapsed allowing interest at 6 per cent. ; having given
logl06 = 20253059, log4688385 = '6710233, log3'118042 = '4938820.
14. If a, b, c years' purchase must be paid for an annuity to con
tinue n, 2/i, 3?i years respectively; shew that
a 2 — ab + b 2 = ac.
15. What is the present worth of a perpetual annuity of £10
payable at the end of the first year, £20 at the end of the second,
£30 at the end of the third, and so on, increasing £10 each year;
interest being taken at 5 per cent, per annum ?
CHAPTER XIX.
INEQUALITIES.
245. Any quantity a is said to be greater than another
quantity b when a b is positive; thus 2 is greater than 3,
because 2  ( 3), or 5 is positive. Also b is said to be less
than a when b a is negative; thus 5 is less than 2, because
— 5— (— 2), or  3 is negative.
In accordance with this definition, zero must be regarded as
greater than any negative quantity.
In the present chapter we shall suppose (unless the contrary
is directly stated) that the letters always denote real and positive
quantities.
246. If a > b, then it is evident that
a + c > b + c ;
a — c > b — c ;
ac > be ;
a b
that is, an inequality will still hold after each side has been
increased, diminished, multiplied, or divided by the same positive
quantity.
247. If aob,
by adding c to each side,
a>b+ c;
which shews that in an inequality any term may be transposed
from one side to the other if its sign be changed.
If a > b, then evidently b < a ;
that is, if the sides of an inequality be transposed, the sign of
inequality must be reversed.
INEQUALITIES. 209
If a > b, then a  b is positive, and ba is negative; that
is, — a — (— b) is negative, and therefore
— a < — b ;
hence, if the signs of all the terms of an inequality be changed,
the sign of inequality must be reversed.
Again, if a > b, then —a < — b, and therefore
— ac < — be ;
that is, if the sides of an inequality be multiplied by the same
negative quantity, the sign of inequality must be reversed.
248. If a.>b,, a. > b oi a^>b.,, a >b , it is clear
that
a l + a 2 + a 3 +...+ a m > 6, + b^+b a + ... + b m ;
and a : a 2 a ,' a ,n >h AK' b , > r
249. If a>b, and if p, q are positive integers, then ^/a>^Jb,
11 V V
or a 1 > b 9 ; and therefore a' 1 > b' ; that is, a' 1 > b'\ where n is any
positive quantity.
Further, — < = ; that is a~ n < b~".
250. The square of every real quantity is positive, and
therefore greater than zero. Thus (a  b) 2 is positive ;
. ■ . a 2 — 2ab + b 2 > ;
. • . a 2 + b 2 > 2ab.
Similarly — ^ > Jxy ;
that is, the arithmetic mean of tivo positive quantities is greater
than their geometric mean.
The inequality becomes an equality when the quantities are
equal.
251. The results of the preceding article will be found very
useful, especially in the case of inequalities in which the letters
are involved symmetrically.
H. H. A. li
210 HIGHER ALGEBRA.
Example 1. If a, b, c denote positive quantities, prove that
a 2 + b 2 +c 2 >bc + ca + ab;
and 2 (a 3 + b 3 + c 3 )>bc (b + c) + ca(c + a) + ab (a + b).
For & 2 + c 2 >2bc (1);
c 2 + a 2 >2c«;
a 2 + b 2 >2al);
whence by addition a 2 + b 2 + c 2 > be + ca + a&.
It may be noticed that this result is true for any real values of a, b, c.
Again, from (1) b 2 bc + c 2 >bc (2);
.. b 3 + c 3 >bc(b+c) (3).
By writing down the two similar inequalities and adding, we obtain
2 (a 3 + b 3 + c 3 ) > be (b + c) + ca [c + a) + ab{a+b).
It should be observed that (3) is obtained from (2) by introducing the
factor b + c, and that if this factor be negative the inequality (3) will no
longer hold.
Example 2. If x may have any real value find which is the greater,
.r 3 +l or x 2 + x.
x 3 +l (x 2 + x) =x 3  x 2  (x  1)
= (x 2 l)(xl)
= (.rl) 2 (* + l).
Now [x  l) 2 is positive, hence
x 3 + 1 > or < x 2 + x
according as x + 1 is positive or negative; that is, according as x > or <  1.
If x—  1, the inequality becomes an equality.
252. Let a and b be two positive quantities, $ their sum
and P their product ; then from the identity
4a6 = (a + bf  (a  b)\
we have
iP = S 2  (a  b) 2 , and S 2 = ±P+(a b) 2 .
Hence, if S is given, P is greatest when a — b\ and if P is
given, S is least when
a= b;
that is, if the sum of two positive quantities is given, their product
is greatest when they are equal ; and if the product of two positive
quantities is given, their sum is least when they are equal.
INEQUALITIES. 211
253. To find the greatest value of a product the sum of whose
factors is constant.
Let there be n factors a, b, c, ... k, and suppose that their
sum is constant and equal to s.
Consider the product abc ... k, and suppose that a and b are
any two unequal factors. If we replace the two unequal factors
Tii ii> a + b a + b , , .. ,
a, b by the two equal factors — — , — — the product is increased
while the sum remains unaltered ; hence so long as the product
contains two unequal factors it can be increased tvithout altering
the sum of the factors ; therefore the product is greatest when all
the factors are equal. In this case the value of each of the n
s /s\"
factors is  , and the greatest value of the product is (  ) ,
n \nj
/a + b + c + ... +k\
\ n )
or
\n/ '
a + b + c+ ... +k\"
Cor. If «, b, c, ... k are uneqiud,
/a + b + c + ... +k\ n 7 7
( ) > abc ... k ;
\ n J
that is,
a + b + c+ ... + k
n
> (abc ... k)".
By an extension of the meaning of the terms arithmetic mean
and geometric mean this result is usually quoted as follows :
the arithmetic mean of any number of positive quantities is greater
than the geometric mean.
Example. Shew that (l r + 2 r + S r + . . . + n r ) n > n n ( \nY ;
where r is any real quantity.
c . lr + 2 r+Zr + +n r 1
Since >(l r .2 r .3 r « r )' 1 ;
n
.'. ( ) >l r .2 r .3 r n r , that is, >(») r ;
whence wo obtain the result required.
14—2
212 HIGHER ALGEBRA.
254. To find the greatest value q/'a m b"c p . . . when a + b + c + ...
is constant; m, n, p, ... being positive integers.
Since m, n, p,... are constants, the expression a m b n c p ... will
be greatest when ( — ) () (  ) ... is greatest. But this last
& \mj \nj \pj &
expression is the product of m + n + p + ... factors whose sum is
m ( — )+ n (—) + p ( — ) + .. ., or a + b + c + . . ., and therefore con
\mj \nj * \pj
stant. Hence a m b n c p ... will be greatest when the factors
a o c
ni n p
are all equal, that is, when
a b c a + b + c +
m n p m+n+p +
Thus the greatest value is
/a + b + c+ ...\ M4 * 4 * + "
m m n n p p . . . ( )
Example. Find the greatest value of (a + x) s (a a:) 4 for any real value
of x numerically less than a.
The given expression is greatest when ( — — J ( — j— j is greatest ; but
the sum of the factors of this expression is 3 ( —^ J + 4 I — ^— ) , or 2a;
hence {a + x) 3 (a  x)* is greatest when — ^— = — ^— , or x=   .
6 3 . 8 4
Thus the greatest value is * a 7 .
255. The determination of maximum and minimum values
may often be more simply effected by the solution of a quad
ratic equation than by the foregoing methods. Instances of
this have already occurred in Chap. ix. ; we add a further
illustration.
Example. Divide an odd integer into two integral parts whose product
is a maximum.
Denote the integer by 2/i + 1 ; the two parts by x and 2n + 1  x ; and
the product by y ; then (2n + 1) x  x* = y ; whence
2x = (2n + 1) ± V^h + I) 2 ^ ;
INEQUALITIES. 213
but the quantity under the radical must be positive, and therefore y cannot
11
be greater than  (2/t + l) 2 , or n' 2 + n +  ; and since y is integral its greatest
value must be n + n\ in which case x = n+ 1, or n ; thus the two parts are n
and n+1.
256. Sometimes we may use the following method.
Example. Find the minimum value of ' — ' — ' .
c + x
Put c+x=y ; then
.. . (ac + y){bc + y)
the expression = ^^ 
y
_ (a  c) (b  c)
+y+ac+bc
( a ~ C Jy b ~ C) ^yy + ac + bc + 2j (ac)(bc) .
Hence the expression is a minimum when the square term is zero ; that
is when y=J(a c)(b c).
Thus the minimum value is
ac + bc + 2 *J(a  c) (b  c) ;
and the corresponding value of x is */(«  c) {b  c) c.
EXAMPLES. XIX. a.
1. Prove that (ab + xy) (ax + by) > 4abxy.
2. Prove that (b + c) (c + a) (a + b) > 8abc.
3. Shew that the sum of any real positive quantity and its
reciprocal is never less than 2.
4. If a 2 + b 2 = l, and x 2 +y 2 = l, shew that ax + by<\.
5. If « 2 + 6 2 + c 2 =l, and x 2 +y 2 + z 2 = l, shew that
ax + by + cz < 1.
6. If a > b, shew that a a b b > a b b a , and loe  < losr = .
7. Shew that (.r 2 ^ + y 2 z + z 2 x) (xy 2 +yz 2 + zx 2 ) > D.''V  2 .
8. Find which is the greater 3«6 2 or aP+26 3 .
9. Prove that a 3 6 + ab 3 < « 4 + 6 4 .
10. Prove that 6abc < be (b + c) + ca(c + a) + ab (a + b).
11. Shew that b 2 c° + c 2 a 2 + a 2 b 2 > abc (a + b + c).
214 HIGHER ALGEBRA.
12. Which is the greater x 3 or x 2 +x + 2 for positive values of x%
13. Shew that x 3 + lSa 2 x > hax* + 9a 3 , if x > a.
14. Find the greatest value of x in order that 7x 2 + 11 may be
greater than x^ + Hx.
15. Find the minimum value of x 2  12#+40, and the maximum
value of 24?  8  9x 2 .
16. Shew that ( \nf > »• and 2 . 4 . 6. . . 2?i< (w + l) n .
17. Shew that (x +y + ,s) 3 > 27^^.
18. Shew that n* > 1 . 3 . 5 . . .(2n  1 ).
19. If ?i be a positive integer greater than 2, shew that
2 ft >l+?iV2 ,7_1 .
21. Shew that
(1) (x+y +z) 3 > 27 (y + z x) (z + x  y) (x+y  z).
(2) xyz>(y+zx)(z + xy)(x+yz).
22. Find the maximum value of (7  x) A (2 + #) 5 when # lies between
7 and  2.
no T7 v xu • • 1 f (5 + x)(2 + x)
23. Find the minimum value of =** .
1+*
*257. To prove that if a and b are positive and unequal,
a m +b m / a + b\ m " .
— > ( — — ) , except when m xs a positive proper jr action.
We have a 1 " + 6 m = f y + ^J + ( ^ gJ ; and
since — ~— is less than — — , we may expand each of these
L 
expressions in ascending powers of — — . [Art. 184.]
2
a" + 6 m /a + b\ m m (m  1) (a + b\"— f a  b \*
•'•"~2~ = v~2J + 1.2 \~r) \~r)
m (m  \)(m  2)(m3) fa + bV" 4 fa  b\ 4
+ 1.2.3.4 "A 2 J 12 j + '"
INEQUALITIES. 215
(1) If m is a positive integer, or any negative quantity,
all the terms on the right are positive, and therefore
a" + b' n fa + b s
= >
fa + 6 V"
\2~) '
(2) If m is positive and less than 1, all the terms on
the right after the first are negative, and therefore
a m + b m fa + fr
—  — <
fa + b\ m
(3) If on > 1 and positive, put m =  where n < 1 ; then
76
i i i
+ b m \ m fd" + b 7l \ H
fa m + b m \ m fa n + b n \ n
{■2 ) = (2) ;
1 1 1
'a m + b m \ m (a*) m + (b») H , /ox
— o— > o i ]j y ( 2 )
i
> + b'"\" 1 a + b
> — ~ —
.*. — = — >
fi7
Hence the proposition is established. If m = 0, or 1, the
inequality becomes an equality.
*25&. If there are n 'positive quantities a, b, c, ...k, then
a m +b m + c m + ... + k m /a + b + c+...+k
>
)■
n \ n
unless m & rt positive proper fraction.
Suppose on to have any value not lying between and 1.
Consider the expression a m + b m + c" 1 + ... + k"\ and suppose
that a and b are unequal ; if we replace a and b by the two equal
.... a + b a + b .. , „ , 7
quantities — — , —  , the value or a + + c+...+fc remains un
_ i j
altered, but the value of a" 1 + b m + c m + ... + k'" is diminished, since
216 HIGHER ALGEBRA.
Hence so long as any two of the quantities a, b, c,...&are unequal
the expression a m + b m + c m + ... + k m can be diminished without
altering the value of a + b + c + ...+k; and therefore the value
of a" 1 + b' n + c m + . . . + k m will be least when all the quantities
a, b, c,...k are equal. In this case each of the quantities is equal
a + b + c + ... + k
to ;
n
and the value of a m + b m + c m + ... +k m then becomes
fa + b + c + . . . + k\ m
n { » ) •
Hence when a, b, c,...k are unequal,
fa + b + c + ... + k\ m
a m + b m + c n +...+le m /a + b + c+ ...+fc
>
n
If ?n lies between and 1 we may in a similar manner prove
that the sign of inequality in the above result must be reversed.
The proposition may be stated verbally as follows :
The arithmetic mean of the m th powers of n positive quantities
is greater than the m th power of their arithmetic mean in all cases
except when m lies between and 1.
*259. If a and h
> are positive integers, and a > b, and if ^ x be a
positive quantity,
(• * 3'  (' * iT
For
(l + Jf = l + , +
(•3S*(«3(3S«
the series consisting of a + 1 terms ; and
('^•♦■•(■^♦(•{)(«S)jS*
the series consisting of b + 1 terms.
After the second term, each term of (1) is greater than the
corresponding term of (2) ; moreover the number of terms in (1)
is greater than the number of terms in (2) ; hence the proposition
is established.
INEQUALITIES. 217
*2G0. To prove that 'I±^ / lh V
X
/I +
>
ix' v iy 5
if'x and y are proper fractions and ])ositive, and x > y.
x /l+x
For .71±f >flr< //g* 
,. 1 . 1 + a 1 1 + y
according as lo<? n > or <  log
lx y °l2/'
But S^lzS" 2 ^ +? + ?+•••)» t AAM ^
and Ilog^2(l + !\
2/ & 12/ \ 3
5 7 '
1 . 1 4 a; 1 . \+y
 log >  log  —  ,
x * lx y ° 1  y '
and thus the proposition is proved.
•261. To prove that (1 + x) ,+x (1 x) 1_x >l, if x<l, and to
777 , k /a + b\ a+b
d i' < luce that a a b D > ( _ J
Denote (1 +jb) 1+ * (1 a) 1 * by P; then
logP = (l+a)log(l + x) + (l aj)log(la;)
= x {log (1 +x) log (1  x)) + log (1 + x) + log (l—x)
r\ I **s JO \ _ / *C i// SI/ \
= 2x^ + 3 + ^ + ...)2^ + 1 +  6 + ...)
. / •*/ JC Jit \
Hence log P is positive, and therefore P> 1 ;
that is, (1 +*) 1 + r (l *)''>!.
218 HIGHER ALGEBRA.
f9
In this result put x = — , where u > z ; then
u
Z Z
sY + w/, z\ l ~u
{ 1 + u) ^i) %1 ;
'u + z\
y^Y >r>rl .
\ u J \ u J
.'. (u + z) u+ *(uz) u  z >u 2u .
Now put u + z = a, u — z = b, so that w = — — — ;
(TJ
* EXAMPLES. XIX. b.
1. Shew that 27 (a 4 + & 4 + c 4 ) > (a + b + c)\
2. Shew that n (n + l) 3 < 8 (l 3 + 2 3 + 3 3 + ... + n 3 ).
3. Shew that the sum of the m ih powers of the first n even num
bers is greater than n (n + l) m , if m > 1.
4. If a and /3 are positive quantities, and a > /3, shew that
(
■SMjr
Hence shew that if n > 1 the value of ( 1 +  ) lies between 2 and
2718...
5. If a, b, c are in descending order of magnitude, shew that
/a + c\ a ^. fb + c\ b
\ac) \bcj
'a + b + c+...+k\ a + b + c +  + ,i
6. Shew that ( a ' ~ v ' v ~ ) < a a b b <*. . .>&*.
7. Prove that  log (1 + a m ) <  log (1 + a n ), if m > n.
lib Ih
8. If ii is a positive integer and x < 1, shew that
1 _ #n + 1 J _ A .n
<
?& + 1 n
INEQUALITIES. 219
9. If a, b, c are in H. P. and »> 1, shew that n n \c n > 2b n .
10. Find the maximum value of x 3 (4a  .r) 5 if x is positive and less
i i
than 4a; and the maximum value of x*(\—xf when x i.s a proper
fraction.
11. If x is positive, shew that log (1 +.r) < x and
1+.?; "
12. If x + y + z=l, shew that the least value of  \ h is i) :
x y z
and that (1  x) (1  y) (1  z) > 8xyz.
13. Shew that (a+b+c+d) (a 3 + 6 3 + c 3 + a 73 ) > (a 2 + 6* + c 2 + cl 2 f.
14. Shew that the expressions
a(ab)(ac) + b (b c) (ba) + c (ca) (c  b)
and cfi{ab)(ac) + b 2 (bc){ba) + c 2 (ca){cb)
arc both positive.
15. Shew that (x m + y m ) n < (.t' n +y n ) MI , if m > n.
fa 4h\ a + b
16. Shew that a b fr < (^p) .
17. If at, 6, c denote the sides of a triangle, shew that
(1 ) a 2 (pq)(p r) + b 2 (q  r) (q p) + c 2 (r p) (r  q)
cannot be negative; p, q, r being any real quantities;
(2) ah/z + b 2 zx + c 2 xy cannot be positive, if x + y + z = 0.
18. Shew that [1 j3 15 \2nl > (\n) n
19. If a,b,c, d, are p positive integers, whose sum is equal
to n, shew that the least value of
\a\bJ±\ c L » (g) P " r (g+ 1 ) r >
where q is the quotient and r the remainder when n is divided by £>.
CHAPTER XX.
LIMITING VALUES AND VANISHING FRACTIONS.
a
262. If a be a constant finite quantity, the fraction  can
x
be made as small as we please by sufficiently increasing x ; that
a
is, we can make  approximate to zero as nearly as we please
by taking x large enough ; this is usually abbreviated by saying,
" when x is infinite the limit of  is zero."
x
Again, the fraction  increases as x decreases, and by making
x
x as small as we please we can make ■ as large as we please ;
x
thus when x is zero  has no finite limit; this is usually ex
JO
pressed by saying, " when x is zero the limit of  is infinite."
263. When we say that a quantity increases without limit
or is infinite, we mean that we can suppose the quantity to become
greater than any quantity we can name.
Similarly when we say that a quantity decreases without
limit, we mean that we can suppose the quantity to become
smaller than any quantity we can name.
The symbol go is used to denote the value of any quantity
which is indefinitely increased, and the symbol is used to
denote the value of any quantity which is indefinitely dimi
nished.
LIMITING VALUES. 221
204. The two statements of Art. 2G2 may now be written
symbolically as follows :
if x is co , then  is ;
x
if x is , then is co .
x
But in making use of such concise modes of expression, it
must be remembered that they are only convenient abbreviations
of fuller verbal statements.
26~>. The student will have had no difficulty in understanding
the use of the word limit, wherever we have already employed it;
but as a clear conception of the ideas conveyed by the words
limit and limiting value is necessary in the higher branches of
Mathematics we proceed to explain more precisely their use and
meaning.
266. Definition. If y =f(x), and if when x approaches a
value a, the function f(x) can be made to differ by as little as
we please from a fixed quantity b, then b is called the limit of
y when x — a.
For instance, if S denote the sum of n terms of the series
1+2 + 2 2 + 2~ J + '" ; then ' S ' = 2 ~2^ t Art 56 *1
Here S is a function of n, and ^— , can be made as small
as we please by increasing n ; that is, the limit of S is 2 when
n is infinite.
267. We shall often have occasion to deal with expressions
consisting of a series of terms arranged according to powers of
some common letter, such as
a + a x x + a a x" + a 3 x 3 +
where the coefficients o , a,, a 2 , a 3 , ... are finite quantities
independent of x, and the number of terms may be limited or
unlimited.
It will therefore be convenient to discuss some propositions
connected with the limiting values of such expressions under
certain conditions.
222 HIGHER ALGEBRA.
268. The limit of the series
a + a x x + a 2 x 2 + a 3 x 3 +
when x is indefinitely diminished is a .
Suppose that the series consists of an infinite number of terms.
Let b be the greatest of the coefficients a lf a si o 3 , ... ; and
let us denote the given series by a + S ; then
S<bx + bx 2 + bx 3 + ... ;
bx
and if x < 1 , we have S < = .
1 — x
Thus when x is indefinitely diminished, S can be made as
small as we please ; hence the limit of the given series is a .
If the series consists of a finite number of terms, S is less
than in the case we have considered, hence a fortiori the pro
position is true.
269. In the series
,3
a + a,x + a 2 x + a 3 x + . . . ,
by taking x small enough ive may make any term as large as we
please compared with the sum of all that follow it ; and by taking
x large enough we may make any term as large as we please
compared with the sum of all that precede it.
The ratio of the term a x n to the sum of all that follow
n
it is
a x n a
or
a n+1 x n+l +a n+2 x" + ' 2 + ... '■ a n+1 x + a u+2 x*+..
When x is indefinitely small the denominator can be made
as small as we please ; that is, the fraction can be made as large
as we please.
Again, the ratio of the term a n x n to the sum of all that
precede it is
a x n a
or
a ,cc n l +a »x n 2 +...' a ,y + a a y 2 +...'
n—l n — 2 n — lts n—2<J
where y =  .
u x
LIMITING VALUES. 223
When x is indefinitely largo, y is indefinitely small ; hence,
as in the previous case, the fraction can be made as large as
we please.
270. The following particular form of the foregoing pro
position is very useful.
In the expression
Hi
a x + a ,x + + a.x + a ,
ii H — 1 1 »
consisting of a finite number of terms in descending powers of x,
by taking x small enough the last term a can be made as large
as we please compared with the sum of all the terms that precede
it, and by taking x large enough the first term ax* can be made
as large as we please compared with the sum of all that follow it.
Example 1. By taking n large enough we can make the first term of
n 4  5/i 3 7/i + 9 as large as we please compared with the sum of all the other
terms ; that is, we may take the first term ?i 4 as the equivalent of the whole
expression, with an error as small as we please provided n be taken large
enough.
3.t 3 — 2x' 2 — 4
Example 2. Find the limit of == — : — when (1) x is infinite : (2) x is
zero.
(1) In the numerator and denominator we may disregard all terms but
3a; 3 3
the first ; hence the limit is ^s , or ^ .
OXr O
4 1
(2) When x is indefinitely small the limit is — , or   .
8 2
* / 1 + x
Example 3. Find the limit of ^ /  —  when
V 1 — x
x is zero.
Let P denote the value of the given expression ; by taking logarithms we
have
log P=i {log (1+ x) log (1x)}
X
^(l + ^' + '^+..V [Art. 226.]
Hence the limit of log P is 2, and therefore the value of the limit
required is e' 2 .
224 HIGHER ALGEBRA.
VANISHING FRACTIONS.
271. Suppose it is required to find the limit of
x 2 + ax — 2a 2
x 2  a 2
when x = a.
If we put x = a + h, then h will approach the value zero as x
approaches the value a.
Substituting a + h for x,
x 2 + ax— 2a 2 3ah + h 2 3a + h
x 2 a 2 = 2ah + h 2 = 2a~+h'
and when h is indefinitely small the limit of this expression
. 3
is .
a
There is however another way of regarding the question; for
x 2 + ax  2a 2 (x  a) (x + 2a) x + 2a
x 2 — a 2 (x — a)(x + a) x + a '
and if we now put x = a the value of the expression is
^r , as before.
j
2 . fi 2
If in the given expression ^ ~ — we pat x = a before
x — a
simplification it will be found that it assumes the form  , the
value of which is indeterminate ; also Ave see that it has this
form in consequence of the factor x a appearing in both
numerator and denominator. Now we cannot divide by a zero
factor, but as long as x is not absolutely equal to a the factor
x  a may be removed, and we then find that the nearer x
approaches to the value «, the nearer does the value of the
3
fraction approximate to ^ , or in accordance with the definition of
Art. 266,
i , ! i . . , n x t ax — Jia . o
when x = a, the limit of ^ . — is  .
x~ ^a" 2
VANISHING FRACTIONS. 225
272. If f(x) and <f> (x) are two functions of x, each of which
becomes equal to zero for some particular value a of x, the
fraction ^~ takes the form Ki and is called a Vanishing
<f) (a) v
Fraction.
Example 1. If x = S, find the limit of
■T 3 5:r 2 + 733
.T 3  x — ox— '6
When x = 3, the expression reduces to the indeterminate form ^; but by
removing the factor x3 from numerator and denominator, the fraction
becomes ^ ~ 2x+1 , When x = S this reduces to  , which is therefore the
x 2 + 2x + 1 4
required limit.
Example 2. The fraction J'^aJx + a becomeg when % _ a
xa
To find its limit, multiply numerator and denominator by the surd con
jugate to J'dxa Jx + a; the fraction then becomes
(Sxa)(x + a) ^ or 2 ,
[x a )(Jdxa + Jx + a)' J'6xa+ >Jx + a
whence by putting x = a we find that the limit is —j= .
1  21 x
Example 3. The fraction 1 _% x becomes ^ when x=l.
To find its limit, put x = l + h and expand by the Binomial Theorem.
Thus the fraction
1  (1 + fe)* _ V 3 9
l(l + /0i l(l+J*^»F+.)
1 1,
3 + 9 ;< 
1 2 7
5 + 25 /l 
5
Now h = when *ael; hence the required limit is  .
273. Sometimes the roots of an equation assume an in
determinate form in consequence of some relation subsisting
between the coefficients of the equation.
H. H. A. lo
226 HIGHER ALGEBRA.
For example, if ax + b = ex + d,
(a — c)x = d — b,
db
x =
a — c
But if c = a, then x becomes — j — , or go ; that is, the root of
a simple equation is indefinitely great if the coefficient of x is
indefinitely small.
274. The solution of the equations
ax + by + c = 0, a'x + b'y + c = 0,
be'  b'e ca' — c'a
ab' — ab 1 ab'a'b '
If ab' — a'b = 0, then x and y are both infinite. In this case
— = — = m suppose ; by substituting for a', b\ the second
c
equation becomes ax + by + — = 0.
c
If — is not equal to c, the two equations ax + by + c = and
m
c'
ax + b ii H = differ only in their absolute terms, and being
J Ml J
inconsistent cannot be satisfied by any finite values of x and y.
If — is equal to c. we have ==, and the two equations
m ^ a b c
are now identical.
Here, since be — b'e = and ca' — c'a — the values of x and y
Q
each assume the form  , and the solution is indeterminate. In
fact, in the present case we have really only one equation
involving two unknowns, and such an equation may be satisfied
by an unlimited number of values. [Art. 138.]
The reader who is acquainted with Analytical Geometry will
have no difficulty in interpreting these results in connection with
the geometry of the straight line.
VANISHING FHACTIONS. 227
275. We shall now discuss some peculiarities which may
arise in the solution of a quadratic equation.
Let the equation be
ax 2 + bx + c  0.
If c = 0, then
ax 2 + bx = 0;
whence x = 0, or — :
a
that is, one of the roots is zero and the other is finite.
If 6 = 0, the roots are equal in magnitude and opposite in
sign. [Art. 118.]
If a = 0, the equation reduces to bx + c = ; and it appears
that in this case the quadratic furnishes only one root,
namely — = . But every quadratic equation has two roots, and in
order to discuss the value of the other root we proceed as follows.
Write — for x in the original equation and clear of fractions ;
*J
thus
cy 2 + by + a = 0.
Now put a = 0, and we have
cy 2 + by = 0;
b c
the solution of which is y — 0, or — ; that is, x = oo, or — T .
J c> ' b
Hence, in any quadratic equation one root will become infinite
if the coefficient ofx 2 becomes zero.
This is the form in which the result will be most frequently
met with in other branches of higher Mathematics, but the
student should notice that it is merely a convenient abbreviation
of the following fuller statement :
In the equation ax 2 + bx + c = 0, if a is very small one root is
very large, and as a is indefinitely diminished this root becomes
indefinitely great. In this case the finite root approximates
to y as its limit.
o
The cases in which more than one of the coefficients vanish
may be discussed in a similar manner.
15—2
228 HIGHER ALGEBRA.
EXAMPLES. XX.
Find the limits of the following expressions,
(1) when #=oo, (2) when x = 0.
 (2s 3) (3 5*) (3r? I)*
7# 2 
6# + 4
(3 + 2# 3 )(#5)
(4s 3 
9)(l + #)
l# 5
! l#
•
y^r^^ j \f »j ^
# 4 + 9 '
(a?3)(25.v)(3a7+l)
(2a? I) 3
_^_^ :L ^ r (3  *?)(* + 5) (2 7*)
' 2^1 ' 2# 2 ' (7.rl)(# + l) 3 '
Find the limits of
7. 'a —  , when x— — 1. 8. , when # = 0.
# 2 l #
gjc _ g  x ginx _ Qtna
9. . — / n , . when # = 0. 10. , when #=a.
log(l+#) .r«
v/#\/2a + V^2a .
11. . , when a?=2a.
V # 2  4a 2
log(l+# 2 +# 4 )
12  l^(isto) ' whena,=ft
la?+loga? ,
13. , when a?=l.
1  \/2x — x 2
1 3
,. (a 2 x 2 ) 2 + (axf ,
14. ^ — ^, when x=a,
(a?afi)*+(axf
\fa 2 + ax + x 2 — *Ja 2 — ax+x 2 .
15. . , when x = 0.
\Ja + x — \j a — x
16. { y , when n =
17. w log i_ 1 , when w = cc .
18. A / , when x = 0.
A' a — x
00
)
CHAPTER XXI.
NVERGENCY AND DIVERGENCY OF SERIES.
076 Ax expression in which the successive terms are formed
^ ! lr law is called a series ; if the series terminate at
by some reguJ« ^ w ^e ^ number q{
some assigiMgl t^ m lt: lb ca'iea a B ^J.
terms is unfimitfd, it is called an infinite series.
In the preset chapter we shall usually denote a series by
an expression c ° ue form
u x + n 2 + i/ 3
+ + u +
/
Suppose that we have a series consisting of w terms.
The sum of the series will be a function of n; if n increases
indetinitely, the sum either tends to become equal to a certain
finite fcmi*, or else it becomes infinitely great.
An infinite series is said to be convergent when the sum
of the first n terms cannot numerically exceed some finite
quantity however great n may be.
An infinite series is said to be divergent when the sum of
the first n terms can be made numerically greater than any finite
quantity by taking n sufficiently great.
978 If we can find the sum of the first n terms of a given
series we may ascertain whether it is convergent or divergent
W examining whether the series remains finite, or becomes in
finite, when n is made indefinitely great.
For example, the sum of the first n terms of the series
. 1*"
1 + x + x 2 + x* + ... is ■. _ a . •
230 HIGHER ALGEBRA.
If x is numerically less than 1, the sum appro. ; , the
finite limit j—  , and the series is therefore converge,^!
If x is numerically greater than 1, the sum of t„ first
n terms is — y , and by taking n sufficiently great, tk can
cUve?ge d n e t. greater ""* ™ Y &n * e l""*^ thus «» *>•» is
serie's r d i^ent Um * ^ ** " ^ * "» *" d **■*» ^
If x=  1, the series becomes
11+11+1 1+
The sum of an even number of terms is I ,;L the sum
of an odd number of terms is 1 • and thiw +\L •„ T
between the values and 1. Thi t,fe X Ac ^ ™ 1
which may be called oiBo^ or A^c^ 5 ^ ^
f /^' ?i here are , many 0ases in which we'haic ,„. h od
of finding the. sum of the first n terms of a series. We p"
therefore to investigate rules by which we can test the cTt
S: T ° f a ^ «*• *Hout effecting its
280. 4n tra/mfe series w* taAicA *Ae ferw are alternately
Let the series be denoted by
M ,  % + %  u, + u  M +
where w 1 >^>^ a >w,> M ....
* o 4 5
for,™ 6 giVe " SerfeS may be Written in each of the following
K«,)+(«,«0 +(».«,) + ^
».K«J(« 4 «,)K« r ) ( 2 ).
From (1) we see that the sum of any number of terms is
a positive quantity; and from (2) that the sum of any nnmber
of terms is less than «, ; hence the series is convergent.
CONVEltGENCY AND DIVERGENCE OF SERIES. 231
281. For example, the series
, 11111
1 f 1 h
2 3 4 5 6
is convergent. By putting x • 1 in Art. 223, we see that its
sum is log e 2.
Again, in the series
23 4 _5 6 _7
T~2 + 3"4 + 5 ~6 + '
each term is numerically less than the preceding term, and the
series is therefore convergent. But the given series is the sum of
i 11111 m
1 2 + 34 + 5"6 + ' (1) '
and 11+11 + 11 + , (2).
Now (1) is equal to log e 2, and (2) is equal to or 1 according'
as the number of terms is even or odd. Hence the given series
is convergent, and its sum continually approximates towards
log,, 2 if an even number of terms is taken, and towards 1 + log 8 2
if an odd number is taken.
282. An infinite seizes in which all the terms are of the same
sign is divergent \f each term is greater than some finite quantity
however small.
For if each term is greater than some finite quantity a,
the sum of the first n terms is greater than na ; and this, by
taking n sufficiently great, can be made to exceed any finite
quantity.
283. Before proceeding to investigate further tests of con
vergency and divergency, we shall lay down two important
principles, which may almost be regarded as axioms.
I. If a series is convergent it will remain convergent, and
if divergent it will remain divergent, when we add or remove
any finite number of its terms ; for the sum of these terms is
a finite quantity.
II. If a series in which all the terms are positive is con
vergent, then the series is convergent when some or all of the
terms are negative ; for the sum is clearly greatest when all
the terms have the same sign.
We shall suppose that all the terms are positive, unless the
contrary is stated.
232 HIGHER ALGEBRA.
284. An infinite series is convergent if from and after some
fixed term the ratio of each term to the preceding term is numerically
less than some quantity zuhich is itself numerically less than unity.
Let the series beginning from the fixed term be denoted by
u, + u c> + u^ + u^ +
12 3 4
U U 1 U A
and let — < r, — < r, * < r
U : U 2 U s
where r < 1.
Then u, +u r , + u+u A +
12 3 4
/_. u 9 u. u a n u u
1 V ^ <f a u x u 3 u 2 Ul
1
< it, (1 + r + r 2 + r 3 + ) ;
tliat is, < ~ 1 , since r < 1.
1  r
u.
Hence the given series is convergent.
285. In the enunciation of the preceding article the student
should notice the significance of the words " from and after a
fixed term."
Consider the
1
series
+ 2x +
u
n
3x 2 + 4:X 3
nx
n—1
+
+ nx' 1 '
1 >
n ly
"» +
Here
f
+
\x;
and by taking n sufiiciently large we can make this ratio ap
proximate to x as nearly as we please, and the ratio of each term
to the preceding term will ultimately be x. Hence if x < 1 the
series is convergent.
But the ratio — — will not be less than 1, until = < 1:
u , n—\
n—1
that is, until n > ^ .
1 — x
Here we have a case of a convergent series in which the terms
may increase up to a certain point and then begin to decrease.
99 1
For example, if x^^r—, then = 100, and the terms do not
100 1 — x
begin to decrease until after the 100 th term.
CONVERGENCY AND DIVERGENCY OF SERIES. '2Xi
286. An infinite series in which all the terms are of the same
sign is diverge) it if from and after some fixed term the ratio of ea<li
term to the 'preceding term is greater than unify, or equal to unify.
Let the fixed term be denoted by t* . If the ratio is equal to
unity, each of the succeeding terms is equal to u , and the sum
of n terms is equal to nu l ; hence the series is divergent.
If the ratio is greater than unity, each of the terms after the
fixed term is greater than u x , and the sum of n terms is greater
than nu } ; hence the series is divergent.
287. In the practical application of these tests, to avoid
having to ascertain the particular term after which each term is
greater or less than the preceding term, it is convenient to find
the limit of —  when n is indefinitely increased; let this limit
n—\
be denoted by A.
If X< 1, the series is convergent. [Art. 284.]
If \> 1, the series is divergent. [Art. 286.]
If X=l, the series may be either convergent or divergent,
and a further test will be required ; for it may happen that
— — < 1 but continually approaching to 1 as its limit ivhen n is
n — 1 m
indefinitely increased. In this case we cannot name any finite
quantity r which is itself less than 1 and yet greater than X.
u
Hence the test of Art. 284 fails. If, however, — — > 1 but con
u
H — I
tinually approaching to 1 as its limit, the series is divergent by
Art. 286.
We shall use " Liin — — " as an abbreviation of the words
u ,
n — 1
U
"the limit of — — when n is infinite."
u .
n — 1
Example 1. Find whether the series whose n lh term is —  ., — is con
1 di
vergent or divergent.
„ ?/ n (n + l)a: n ru^ 1 (n + l)(nl)
('„_! n 2 {h  1) 2 n*
him — — —x\
"n I
234 HIGHER ALGEBRA.
hence if x < 1 the series is convergent ;
if x > 1 the series is divergent.
u
If x = l, then Lim — — =1, and a further test is required.
Example 2. Is the series
l 2 + 2 2 x + 3 2 x 2 + 4?x s +
convergent or divergent?
_ T . u n n 2 x n ~ l
Here Lim — 7jL =Lim. —. „ 9 =x.
u ni (nl)x n 2
Hence if x < 1 the series is convergent ;
if x> 1 the series is divergent.
If x = 1 the series becomes l 2 + 2 2 + 3 2 + 4 2 + . . . , and is obviously divergent.
Example 3. In the series
a+(a + d)r+{a + 2d)r 2 +... + (a + n1 . d)r n ~ 1 + ...,
, . w n T . a t (n  1) d
Lim — —=Lim — —. ^.r = r;
»*i a + (n2)d
thus if r< 1 the series is convergent, and the sum is finite. [See Art. 60, Cor.]
288. If there are two infinite series in each of which all the
terms are ])Ositive, and if the ratio of the corresponding terms in
the two series is always finite, the two series are both convergent,
or both divergent.
Let the two infinite series be denoted by
u x + u a + u a + w 4 + ,
and v, + v, + v, + v. +
12 3 4
The value of the fraction
u i + u , + u a^ +n n
lies between the greatest and least of the fractions
\ *, », [Art. 14.1
and is therefore a, finite quantity, L say ;
Hence if one series is finite in value, so is the other; if one
series is infinite in value, so is the other; which proves the
proposition.
CONVERGENCY AND DIVERGENCY OF SERIES. 235
289. The application of this principle is very important, for
by means of it we can compare a given series with an auxiliary
series whose convergency or divergency has been already esta
blished. The series discussed in the next article will frequently
be found useful as an auxiliary series.
290. The infinite series
1111
y T 2 P 3 P 4.1'
is always divergent except when p is positive and greater than 1.
Case I. Let;? > 1.
The first term is 1 ; the next two terms together are less than
2 . 4
j—; the following four terms together are less than—; the fol
Z 4
lowing eight terms together are less than — ; and so on. Hence
o
2 4 8
the series is less than I + t^+th+ttt, +•••;
2 P 4' o 1
that is, less than a geometrical progression whose common ratio
2
~j is less than 1, since p > 1 ; hence the series is convergent.
Case II. Let_p=l.
The series now becomes 1 + ^ + ^ +  + ■=;+ ...
2 3 4 5
2 1
The third and fourth terms together are greater than  or ^ ;
t —
4 1
the following four terms together are greater thau ^ or  ; the
o 2
8 1
following eight terms together are greater than — or  ; and so
on. Hence the series is greater than
1111
2 + 2 + 2 + 2 + '"'
and is therefore divergent. [Art. 2^6.]
Case III. Let p<\, or negative.
Each term is now greater than the corresponding term in
Case II., therefore the series is divergent.
Hence the series is always divergent except in the case when
p is positive and greater than unity.
236 HIGHER ALGEBRA.
Example. Prove that the series
2 3 4 n+1
is divergent.
Compare the given series with 1 + « + « + v ■ "^ — •"••••
Thus if «* n and v n denote the n th terms of the given series and the
auxiliary series respectively, we have
u n _n + l . 1 _ w + 1
i' n n 2 ' re ?i
7/
hence Zim, — =1, and therefore the two series are both convergent or both
divergent. But the auxiliary series is divergent, therefore also the given
series is divergent.
This completes the solution of Example 1. Art. 287.
291. In the application of Art. 288 it is necessary that the
limit of — should be finite ; this will be the case if we find our
auxiliary series in the following way :
Take u , the n th term of the given series and retain only the
highest powers of n. Denote the result by v n ) then the limit of
u
 is finite by Art. 270, and v may be taken as the 7i th term of
the auxiliary series.
3/2n 2  1
Example 1. Shew that the series whose n th term is ,, = is
r Z/S?v i + 2n+5
divergent.
As n increases, u n approximates to the value
l/w ' or 4/3 * i
n 12
1 u 3 /2
Hence, if v„= r ,we have Lim — = ^r, which is a finite quantity;
~ v n v/ 3
n 1 
1
therefore the series whose n th term is — may be taken as the auxiliary
series. But this series is divergent [Art. 290] ; therefore the given series is
divergent.
CONVERGENCY AND DIVERGENCY OF SERIES. 237
Example 2. Find whether the series in which
v n = ^/;< :} +l n
is convergent or divergent.
Here "»=« \\/ * + tf ~ *J
//
( 1 + »» + ;" 1 )
~3n 2 9>< 5 +
If we take v n = = , we have
v M 3 9n' J
N
Luti — =x.
v„ 3
n
But the auxiliary series
JL JL Jl l
P + 22 + 3 2+ '" n 1+ " 
is convergent, therefore the given series is convergent.
292. To shew that the expansion of (1 + x) n by the Binomial
T/teorem is convergent when x < 1.
Let u r , u r+l represent the ?* th and (?+l) th terms of the ex
pansion ; then
u . , nr+1
w r
r
When r>?6+l, this ratio is negative; that is, from this
point the terms are alternately positive and negative when x
is positive, and always of the same sign when x is negative.
7/
Now when r is infinite, Lim — — = x numerically ; therefore
since x < 1 the series is convergent if all the terms are of the
same sign; and therefore a fortiori it is convergent when some of
the terms are positive and some negative. [Art. 283.]
293. To shew that the expansio?i of a x in ascending p owers
of x is convergent for every value of x.
W # 1°#« ^ 1 1 • 7 • U 11 1
Here — * = —  — ; and therefore Lim — = < 1 whatever be
«„_, n1 «*__,
the value of x; hence tlie series is convergent.
238 HIGHER ALGEBRA.
294. To shew that the expansion of log (1 + x) in ascending
powers of x is convergent when x is numerically less than 1.
i
ni n't I
Here the numerical value of — — = x. which in the limit
u , n
is equal to x \ hence the series is convergent when x is less than 1.
If a5 = l, the series becomes 1— k + 77t+> an ^ is con "
2 3 4
vergent. [Art. 280.]
If x~ — 1, the series becomes — 1 — — q _ t" •••> an( ^ * s
a O 4:
divergent. [Art. 290.] This shews that the logarithm of zero is
infinite and negative, as is otherwise evident from the equation
e°°=0.
295. The results of the two following examples are important,
and will be required in the course of the present chapter.
\q<j x
Example 1. Find the limit of — 2  when x is infinite.
Put x = ev; then
logs y y
X ~ eV y* yi
i  y y' 2
y \2 ^ 3 + "
also when x is infinite y is infinite ; hence the value of the fraction is zero.
Example 2. Shew that when n is infinite the limit of nx n = 0, when x<l.
Let x= , so that y>l;
if
also let y n =z, so that n\ogy = logz; then
fu^=— = i ^^ = — logz .
y n z'logy logy' z
Now when n is infinite z is infinite, and — s_ = 0; also logy is finite;
z
therefore Lim nx n = 0.
296. It is sometimes necessary to determine whether the
product of an infinite number of factors is finite or not.
Suppose the product to consist of n factors and to be denoted by
uMAia io ;
then if as n increases indefinitely u <<1, the product will ulti
mately be zero, and if u n > 1 the product will be infinite ; hence in
order that the product may be finite, u must tend to the limit 1 .
CONVERGENCY AND DIVERGENCY OF SERIES. 239
Writing 1 + v n for u n , the product becomes
(l+* 1 )(l+*,)(l+* 8 ) (l+O
Denote the product by P and take logarithms j then
logP = log(l+v 1 ) + log(l+v 8 ) +...+ log(l + vJ (1),
and in order that tlie product may be finite this series must be
convergent.
Choose as an auxiliary series
v,+v 2 + v 3 + +v n (2).
/ _1 .
r . log(l + t;) _. r 2** + "
Now Lim2l ^ = Lim\ /==1,
v \ v I
n n
since the limit of v is when the limit of u is 1 .
n n
Hence if (2) is convergent, (1) is convergent, and the given
product finite.
Example. Shew that the limit, when n is infinite, of
13 3 5 5 7 2nl 2n + l
2' 2 '4*1*6' 6 ~JT~'~2ir
is finite.
The product consists of 2n factors; denoting the successive pairs by
Uj, m 2 , Ug,... and the product by P, we have
P = u x v 2 u 3 u
n>
2nl 2«+l , 1
where ** n = — s — • 5 — = 1  t  ?;
■ 2m 2n 4«
but logP = logM 1 + log« 2 + logM 3 + ...+logM n (1),
and we have to shew that this series is finite.
Now log« n = log (l ^)= ~ ^
32/i 1 '•'
therefore as in Ex. 2, Art. 291 the series is convergent, and the given product
is finite.
297. In mathematical investigations infinite series occur so
frequently that the necessity of determining their convergency or
divergency is very important ; and unless we take care that the
series we use are convergent, we may be led to absurd conclusions.
[See Art. 183.]
240 HIGHER ALGEBRA.
For example, if we expand (l—x)~ 2 by the Binomial Theorem,
we find
( 1  a;) 2 = 1 + 2x + 3ar + 4a 3 +
But if we obtain the sum of n terms of this series as ex
plained in Art. 60, it appears that
in O 9 nl *■ ^ ^^
1 + 2.*; + 3ar + ... + nx = t= r a — ■= :
(1  x) 1  x
whence
I +'2x+ 3x~ + ... + nx 4 7 z ^ +
(lx) 2 '—'■—  (1a?) 9 1*
i
By making n infinite, we see that z  a can only be re
J ° (lx) 2
garded as the true equivalent of the infinite series
1 + 2x + 3x 2 + ix 3 +
x nx
when rz ri + =— vanishes.
(1 x)~ lx
If n is infinite, this quantity becomes infinite when x=l,
or aj>l, and diminishes indefinitely when a,*<l, [Art. 295], so
that it is only when x < 1 that we can assert that
\
Ta =* 1 + 2x + 3x 2 + 4# 3 + to inf. j
and we should be led to erroneous conclusions if we were to use
the expansion of (1  x)~ 2 by the Binomial Theorem as if it were
true for all values of x. In other words, we can introduce the
infinite series 1 + 2x + 3x 2 + ... into our reasoning without error
if the series is convergent, but we cannot do so when the series
is divergent.
The difficulties of divergent series have compelled a distinction
to be made between a series and its algebraical equivalent. For
example, if we divide 1 by (1  x) 2 , we can always obtain as
many terms as we please of the series
l + 2a;+3£ 2 +4a; 3 +
whatever x may be, and so in a certain sense p. ^ niay be
called its algebraical equivalent ; yet, as we have seen, the equi
valence does not really exist except when the series is con
CONVERGENCY AND DIVERGENCY OF SERIES. 241
vergent. It is therefore more appropriate to speak of — —
(l — X)
as the generating function of the series
1 +2a, + 3a 2 +
being that function which Avhen developed by ordinary alge
braical rules will give the series in question.
The use of the term generating function will be more fully
explained in the chapter on Recurring Series.
EXAMPLES. XXI. a.
Find whether the following series are convergent or divergent.:
. Ill 1
1 # 1 _ 4.
x x + a x^2a .v + 3a
x and a being positive quantities.
1 1 1 1
1.2 + 273 + 371 + 475 +
_1_ 1 1 1
6  xy (*+i)(y+i) 4 >+a)(y+*) (*+3)(y+3) + '
x and y being positive quantities.
x x 2 x 3 x*
4 1 1 1 h .
1.2^2.3^3.4^4.5
/>» /)»2 o»o o***
tf \Mj \Mj %A/
T72 + 3T4 + 576 + 778 +
n , 2 2 3 2 4 2
6  1+ I + I + I +
7  \/l + \/i + \/f + \/1 +
8. 1 + toe + bx 2 + la? + 9af* +
2 __ 1 i. A
y * "i^ + 2/' + 3p + 4p +
ia 1 + 2 + 5 + Ib +  + ,^TT +
3 ., 8 , 15 n 2 \
11. x +  x 2 + x* + — x A + . . . + .,—.. x n +
5 10 17 n l + 1
H. H. A. 16
242 HIGHER ALGEBRA.
.n , 2 6 , 14 , 2' l 2
12. l + g * +5 *» +I ^P + „. + — *pi+
1 — — —
14. 2.r+— +  + . ..+ ,— +
8 2 7 ?t 3
1C /2 2 2\! /3 3 3\ 2 /4 4 4\ 3
15 ' (pl) + (2 3 2) +«=■ ="
/4_ 4 _4\
\3 4 3^
2 2 3 3 4 4
16. 1 +  +  +  +  +
17. Test the series whose general terms are
(1) Jn*+ln. (2) jtF+l Jnt^i.
18. Test the series
/1N 1 1 1 1
.r A+l a+2 x+3
/on 11 1 1 1
(2) + r+— —+ 5+— T5+
A # 1 A+l ^ 2 a + 2
x being a positive fraction.
19. Shew that the series
2" 3^ 4"
1+ I + I + E +
is convergent for all values of p.
20. Shew that the infinite series
u x + 2i 2 + u 3 + u± +
is convergent or divergent according as Lim^fu n iti <1, or >1.
21. Shew that the product
2 2 4 4 6 2ti2 2tt2 2n
• 1 * 3' 3' 5 ' 5 2713' 2»l"S^Ti
is finite when n is infinite.
22. Shew that when x=\, no term in the expansion of (1 +#)" is
infinite, except when n is negative and numerically greater than unity.
CONVERGENCY AND DIVERGENCY OF SERIES. 243
*298. The tests of convergency and divenrencV wp i.™
sriven m Art* 9x7 ogi 11 «» . & cxlt v we nave
proved in the next article enables n/^T of ^ b ST^
I .1 1 1
l» + 2» + 3? + ••• + ,7 + 
venter 6 " ddit£0U,a ** wUch ^ S01 » eti »' es "» fad con
tergent when the vservs is convergent if after some particular term
— < ^ ; onrf Me „*»*» «,<« J, AWjori «,/ t(!re the vsertes is
divergent if — 5 > _ n
U ni V., '
Let us suppose that Wj and », are the particular terms.
Case I. Let *■ < Ei &<! . then
2
w, + w fl + u 3 +
= 2t
that is,
V w i ** 2 w, y
< — (v. f y + v + )
Hence, if the ^series is^convergent the wseries is also con
vergent.
Case II. Let  2 > 3» ^^
**, v, tt, «
; then
i a 2
M i + ^ 9 + U, +
V V, V a 27, J
16—2
244 HIGHER ALGEBRA.
that is, > — (v l + v 2 + v 3 + ...).
Hence, if the ^series is divergent the itseries is also di
vergent.
*300. We have seen in Art. 287 that a series is convergent
or divergent according as the limit of the ratio of the ?i th term
to the 'preceding term is less than 1, or greater than 1. In the
remainder of the chapter we shall find it more convenient to use
this test in the equivalent form :
A series is convergent or divergent according as the limit of
the ratio of the n th term to the succeeding term is greater than 1,
or less than 1 ; that is, according as Lim — — > 1, or < 1.
Similarly the theorem of the preceding article may be
enunciated :
The wseries will be convergent when the vseries is convergent
u v
provided that Lim — — > Lim — — ; and the itseries will be di
vergent when the vseries is divergent provided that
Lim ^^ Lim ^.
*301. The series whose general term is u n is convergent or di
vergent according as Lim \ n ( —  — 1 \ >> 1, or < 1.
Let us compare the given series with the auxiliary series
whose general term v is — .
"When p > 1 the auxiliary series is convergent, and in this
case the given series is convergent if
u n (n+iy
U n + l n?
, or (l + iy.
thatis,if JS B . > l + g + J > CPlV + „
u n+i n 2n~
/ u ,\ p (pl)
n KCr l r p+ ^ +
that is, if Lim \n ( —  1 ) 1 >■)>.
I Wh J)
CONVERGENCY AND DIVERGENCY OF SERIES. 245
But the auxiliary series is convergent if y; is greater than 1
by a Unite quantity however small ; hence the first part of the
proposition is established.
When p< 1 the auxiliary series is divergent, and by proceed
ing as before we may prove the second part of the proposition.
Example. Find whether the series
a; 1 ^ L3 x= 1.3 .5 x[
l + 2* 3 + 2.4* 5 + 2~i.d'T + '"
is convergent or divergent.
it 1
Here Lim — — =; hence if x<l the series is convergent, and if x>l
u n+l x "
the series is divergent.
u
If x= 1, Lim — — = 1. In this case
u n+l
and
_ 1  3  5 (2w 3) 1
M " ~ 2 . 4 . 6 ...... (2n  2) ' 2~/T=T '
w n 2n(2n+l)
u n+1 (2n  1) (2n  1) '
'• "Urn J" (2nl) 2 '
hence when a; = 1 the series is convergent.
*302. T/ie series whose general term is u n is convergent or di
vergent, according as Lim ( n log —  j > 1, or < 1.
Let us compare the given series with the series whose general
term is —  .
n l
When p > 1 the auxiliary series is convergent, and in this
case the given series is convergent if
u /„ lv
s >
n+ I
1 + ij ; [Art. 300.]
that is, if log — — > p log (1 h ) :
! ** P 7 J
or if log — " > ' ^5 +
'u ., « 2n 2 " 3
71+ 1
246 HIGHER ALGEBRA.
that is, if Lim In log — — ) >p.
Hence the first part of the proposition is established.
When p < 1 we proceed in a similar manner ; in this case the
auxiliary series is divergent.
Example. Find whether the series
2 2 z 2 3 s x 3 4*r 4 5 5 x 5
is convergent or divergent.
„ u n n n x n (n + l)
Here — * = — ' v .
n+l /pW+1 7^
w n+l
n_ ' [n + l (n+l)** A lyV
H)'
.. Lim 3l = 1 . [Art. 220 Cor.l.
w n+1 <?*
Hence if a?< the series is convergent, if #> the series is divergent.
If:r=,then ^St—
e u n+l
•.log  n — = logewlog( 1 +  )
_1 J_
~2n 3n 8+ " '
. u n 1 1
■. Lim [ n log — — 1 =  :
hence when x =  the series is divergent.
*303. If Lim ^ = 1, and also Liminf^  l)) = 1, the
w n+1 ) \ u n+ i J)
tests given in Arts. 300, 301 are not applicable.
To discover a further test we shall make use of the auxiliary
series whose general term is —  r . In order to establish
n (log n) p
the convergency or divergency of this series we need the theorem
proved in the next article.
CONVERGENCY AND DIVERGENCY OF SERIES. 247
*304. If $ (n) is positive for all positive integral values of n
and continually diminishes as n increases, and if a be any posit ive
integer, then the two infinite series
</>(l) + <£(2) + </>(3) + ... + </>(n) + ...,
and a<£ (a) + a 2 <£(a 2 ) + a 3 </> (a 3 ) + . . . + a n <£ (a n ) + . . . ,
are both convergent, or both divergent.
In the first series let us consider the terms
</>(«*+ 1), <f>(a k + 2), <f>(a k + S), <M« i+1 ) 0)
beginning with the term which follows </>(«*).
The number of these terms is a k+l  a k , or a k (a 1), and each
of them is greater than <£(a* +1 ); hence their sum is greater than
1
a k (a 1) <f>(a k+1 ); that is, greater than x a k+l cf> (a k+1 ).
By giving to k in succession the values 0, 1, 2, 3,... we have
4>(2) + 4>(3)f<M4) + ++W>^x«*W;
Co
<]>(a + 1) + <£(« + 2) + <f>(a+ 3)+ + <£(«*)> x a 2 <f>(a 2 ) ;
therefore, by addition, $! — <£(1) > S 2 ,
ct
where £, , S 2 denote the sums of the first and second series respec
tively; therefore if the second series is divergent so also is the
first.
Again, each term of (1) is less than <£(«*), and therefore the
sum of the series is less than (a— 1) x a k <j>(a k ).
By giving to k in succession the values 0, 1, 2, 3... we have
<j>{2) + <£(3) + 4>(4) + + <£(«) < (a 1) x <£(1);
<f>(a + I) + <f>(a + 2) + <f>(a + 3) + +<f>(a 2 )<(a 1) x a<f>(a);
therefore, by addition
4+(l)<(«l){4 + *(l)};
hence if the second series is convergent so also is the first.
»'
Note. To obtain the general term of the second series we take </>(») the
general term of the first series, write a n instead of n and multiply by a n .
248 HIGHER ALGEBRA.
*305. The series whose general term is — ^ r— is convergent
n(logn) p u
if p > 1, and divergent if p = 1, or p < 1.
By the preceding article the series will be convergent or
divergent for the same values of p as the series whose general
term is
1 1 11
ft" \l c\y (\V X
a"(loga") p ' (n\oga) p ' (log a)' n p '
The constant factor 7= r _ is common to every term ; there
fore the given series will be convergent or divergent for the same
values of p as the series whose general term is —  . Hence the
required result follows. [Art. 290.]
*306. The series whose general term is u n is convergent or di
vergent according as Lim \\\ ( —  — 1 ] — 1 > log n > 1, or < 1.
Let us compare the given series with the series whose general
term is — .
n (log ny
When j) > 1 the auxiliary series is convergent, and in this
case the given series is convergent by Art. 299, if
u n (w + l){log(n+l)}'
M ,+i n {log n) v
Now when n is very large,
log (n + l) = log n + log ( 1 +  J = log n +  , nearly;
Hence the condition (1) becomes
■(!)•
u . , V nj V n log n ,
n + l N ' N O '
thatis, ^>(l + l)(l+P
u H+l \ nj \ nlog
n log n) '
u i 1 P
that is, —  > 1 +  +
u ., ?i wlogw
n + l o
CONVERGENCY AND DIVERGENCY OF SERIES. 240
1 ) > 1 + . P ;
l0g?4
or <n
CSr 1 ) 1 } 10 *"^
Hence the first part of the proposition is established. The
second part may be proved in the manner indicated in Art. 301.
Example. Is the series
2 2 2 2 .4 2 2 2 .4 2 .6 2
^3 2 ^3 2 .5 2 ^3 2 .5 2 .7 2
convergent or divergent?
Here A. = *£* . 1 + I + * (1).
ti
.. Lt«i — * =1, and we proceed to the next test.
Fromfl), »fel)=l+5 < 2 >
.. Lim In (  1  1)1=1, and we pass to the next test.
*» ffe 1 ) 1 } 108 ^'^
•••^"[ffe 1 ) 1 } 108 "] 30 '
since Lt/u — ^— = [Art. 295]; hence the given series is divergent.
n
*307. We have shewn in Art. 183 that the use of divergent
series in mathematical reasoning may lead to erroneous results.
But even when the infinite series are convergent it is necessary to
exercise caution in using them.
For instance, the series
 JC %)C Ou Jb
+ 4/2~J/3 + 474~^5 + '"■
is convergent when x=l. [Art. 280.] But if we multiply the
series by itself, the coefficient of x 2n in the product is
1 1 1 + 1 + 1
250 HIGHER ALGEBRA.
Denote this by a 2n ; then since
1 1 J^
a„ > — ; — , and is therefore infinite when n is infinite.
2 " Jn '
If x=l, the product becomes
%a x +a 3  a B f ... + a gJ1  a 2n+1 + a 2>I+a  ...,
and since the terms ol, a .,. a ffl ._ ... are infinite, the series has
2h' 2/i+ 1' 2;i+2 '
no arithmetical meaning.
This leads us to enquire under what conditions the product
of two infinite convergent series is also convergent.
*308. Let us denote the two infinite series
a + a x x 4 a 2 x + a 3 x + . . . + a 2 x + . . .,
b Q + b^x + b 2 x 2 + b 3 x 3 + . . . + b 2n x 2n + . . .
by A and B respectively.
If we multiply these series together we obtain a result of
the form
a <A + ( a A + a (A) x + ( a J>o + a fii + afiz) x 2 + ...
Suppose this series to be continued to infinity and let us
denote it by G ; then we have to examine under what conditions
C may be regarded as the true arithmetical equivalent of the
product AB.
First suppose that all the terms in A and B are positive.
Let A„ , B„ , C„ denote the series formed by taking the first
2/1 » 2« ' 2« JO
2w + 1 terms of A, B, C respectively.
If we multiply together the two series A 2ai B 2ni the coefficient
of each power of x in their product is equal to the coefficient of
the like power of x in C as far as the term x 2 " ; but in A 2n B, n
there are terms containing powers of x higher than x 2n , whilst
x 2n is the highest power of x in C 0n ; hence
^o B* > C 2 .
2/i 2/1 2/1
If we form the product A B the last term is a b x 2n ; but
C 2n includes all the terms in the product and some other terms
besides ; hence
C. >A B .
%n ii ii
CONVERGENCY AND DIVERGENCY OF SERIES. 251
Thus C is intermediate in value between A B and A B ,
T 2 " ■• 1 /. B B 2/» 2/i'
whatever be the value or n.
Let 4 and B be convergent series ; put
A = AX, B =B Y.
where X and Y are the remainders after n terms of the series
have been taken; then when n is infinite X and Y are both
indefinitely small.
.. A n B H = (AX)(BY) = ABBXAY+XY' }
therefore the limit of A B is AB. since A and B are botli finite.
Similarly, the limit of A 2n B„ a is AB.
Therefore C which is the limit of C 2n must be equal to AB
since it lies between the limits of A B and A n B„ .
B B 2« 2;«
Next suppose the terms in A and B are not all of the same
sign.
In this case the inequalities A n B n > C„ > A B are not
■I 2n 2b 2b b b
necessarily true, and we cannot reason as in the former case.
Let us denote the aggregates of the positive terms in the
two series by P t P' respectively, and the aggregates of the
negative terms by iV, N'; so that
A = PN, B^FN'.
Then if each of the expressions P, P\ JV, N' represents a con
vergent series, the equation
AB = PF NF PN' + NN\
has a meaning perfectly intelligible, for each of the expressions
PP\ NF, PN\ NN' is a convergent series, by the former part
of the proposition ; and thus the product of the two series A and
B is a convergent series.
Hence the product of two series will be convergent provided
that the sum of all the terms of the same sign in each is a con
vergent series.
But if each of the expressions P, N y P', N' represents a
divergent series (as in the preceding article, where also F = P
and N' = N), then all the expressions PF, NF, PN\ NN' are
divergent series. When this is the case, a careful investiga
tion is necessary in each particular example in order to ascertain
whether the product is convergent or not.
252 HIGHER ALGEBRA.
^EXAMPLES. XXI. b.
Find whether the following series are convergent or divergent
1 .r 2 1.3.5 #* 1.3.5.7.9 £«
1. 1+ 2*4 + 2.4.6'8 + 2.4.6.8.10' 12 +
3 3.6 2 3.6.9 ^ 3.6.9.12
2. 1 + ^+7. 10 ^ +7.10.13* + 7.10.13.16 A +
o 2 22 a 2 2 .4 2 G 2 2 . 4 2 . 6 2 o
3. ^+374^+3.4.5.6^+3.4.5.6.7.8*°+ —
2# 3 2 .? 2 4 3 ^ 5 4 ^
4 n 1 1 1 h
*' 2 ^ 3 ^ 4 ,5
1 12 13 14
l 2 1 2 .3 2 1 2 .3 2 .5 2 2
* 2 2 + 2 2 .4 2 ' r + 2 2 .4 2 .6 2 ^ + *
7 i , g(la) , ( l + a)«(lg)(2 g)
'• X "T 12 ~ "*" l 2 . 2 2
a being a proper fraction.
(2 + q)(l+a)q(la)(2,a)(3q)
I 2 . 2 2 . 3 2
a+x (a + 2#) 2 (a + 3ai) 3
8 * IT*— 12~ + "13"" +
9 . 1+ ^ + ^MM,
1 . y 1 . Z . y (y+1)
a(a + l)(a + 2)/30+l)(/3 + 2)
1.2.3.y(y+l)(y + 2) ' "*"
10. x 1 (log 2)* + a? 3 (log 3)i + a? 4 (log 4)* +
11. i +a+ __^+— _^ +
12  If ^; = '^r^w^S^' ■ where * is a positive \
integer, shew that the series w 1 + ?^ 2 + « 3 + is convergent if
^ _ a _ i j s positive, and divergent if A  a  1 is negative or zero
CHAPTER XXII.
Undetermined Coefficients.
309. In Art. 230 of the Elementary Algebra, it Avas proved
that if any rational integral function of x vanishes when x = a,
it is divisible by x — a. [See also Art. 514. Cor.]
Let p x n + p x x n " ' + pjf " 2 + +p n
be a rational integral function of x of n dimensions, which
vanishes when x is equal to each of the unequal quantities
«!> «*, %i «„•
Denote the function hy f(x); tlien since f(x) is divisible
by x  a l , we have
f(x)=:(xa l )(p x" i + ),
the quotient being of n — 1 dimensions.
Similarly, since f(x) is divisible by x a, 7 , we have
2W n ~ X + = (xaj(pjf + )«
the quotient being of n — 2 dimensions; and
Proceeding in this way, we shall finally obtain after n di
visions
f(x) =p (x  a) (xa}(xa a ) (x a H ).
310. If a rational integral function of\\ dimensions vanishes
for more than n values of the variable, the coefficient of each power
of the variable must be zero.
Let the function be denoted hyf(x), where
f(x) !>x" +p ) x"~ x +p,c'' + +p n ;
254 HIGHER ALGEBRA.
and suppose tha,tf(x) vanishes when x is equal to each of the
unequal values a lt a 2i a 3 a n ; then
f( x ) =Po ( x  a i) ( x ~ a 2 ) ( x ~ °0 ( x ~ a ,)
Let c be another value of x which makes f(x) vanish ; then
since f(c) = 0, we have
Po ( c ~ a i) (° ~ a *) ( G ~ a s) (c«J = 0;
and therefore p = 0, since, by hypothesis, none of the other
factors is equal to zero. Hence f (x) reduces to
2\x n  x +p 2 x n  2 + 2 ) 3 X "~ 3+ +P n 
By hypothesis this expression vanishes for more than n values
of x, and therefore p x = 0.
In a similar manner we may shew that each of the coefficients
2>o, P 3 , Vn mus t be equal to zero.
This result may also be enunciated as follows :
If a rational integral function of n dimensions vanishes for
more than n values of the variable, it must vanish for every value
of the variable.
Cor. If the function f(x) vanishes for more than n values
of x, the equation f (x) — has more than n roots.
Hence also, if an equation of n dimensions has more than n
roots it is an identity.
Example. Prove that
(x  b) (x  c) (x  c) (x  a) (x  a) (x — b) _ 1
(a b) (a c) {b c) (6  a) (ca) (cb)~
This equation is of tivo dimensions, and it is evidently satisfied by each
of the three values a, 6, c ; hence it is an identity.
311. If two rational integral functions of n dimensions are
equal for more than n values of the variable, they are equal for
every value of the variable.
Suppose that the two functions
2 ) x n +p 1 x n  1 +2> 2 x" 2 + +p H ,
q o x n + q^" 1 + q 2 x 2 + + q mt
are equal for more than n values of x ; then the expression
U>»  %) x ' 1 + (Pi  ?i) x "~ l + (p»  ad x "~ 2 + + (p*  ?.)
UNDETERMINED COEFFICIENTS. 255
vanishes for more than n values of x; and therefore, by the
preceding article,
that is,
2\ = %> Pi=9li> Pi^Vv l> n = <l n > *
Hence the two expressions are identical, and therefore are
equal for every value of the variable. Thus
if two rational integral functions are identically equal, we may
equate the coefficients of the like powers of the variable.
This is the principle we assumed in the Elementary Algebra,
Art. 227.
Cor. This proposition still holds if one of the functions is
of lower dimensions than the other. For instance, if
p x" + pff~ l + pjf~ 2 + pjf~* + +p n
= q 2 x n ~ 2 + q 3 x n ~ 3 + +q n ,
we have only to suppose that in the above investigation q o = 0,
q = 0, and then Ave obtain
^o=°> Pi=°> P2=v s > Ps=q 3 > p,, = q»
312. The theorem of the preceding article is usually referred
to as the Principle of Undetermined Coefficients. The application
of this principle is illustrated in the following examples.
Example, 1. Find the sum of the series
1.2 + 2.3 + 3.4+ +n(n+l).
Assume that
1.2 + 2. 3 + 3. 4 + ... + n(n + l)=A + Bn+Cn 2 + Dn 3 + En i +...,
where A, B, C, D, E,... are quantities independent of n, whose values have
to be determined.
Change n into n + 1 ; then
1. 2 + 2.3+...+?i(;i + l) + (?t + l) (n + 2)
= A+B(n + l) + C(n+l)* + D(n + l)3 + E(n + iy+....
By subtraction,
(n + 1) [n+2) = B+C {2n + l) + D (3}v> + 3}i + l) + E {±n* + 6ri + ±n + l)+ .. .
This equation being true for all integral values of n, the coefficients of the
respective powers of n on each side must be equal ; thus E and all succeeding
coefficients must be equal to zero, and
3D = 1; 3D + 2C = 3; D + C + B = 2;
1 2
whence 1) =  • , (7=1, B =  .
o o
256 HIGHER ALGEBRA.
Hence the sum =A + — + n 2 +  n 3 .
o o
To find A, put n = l; the series then reduces to its first term, and
2 = A + 2, or A = 0.
Hence 1 .2 + 2 . 3 + 3. 4 + ... + n(;i + l) =  n (n + 1) (n + 2).
Note. It will be seen from this example that when the n lh term is a
rational integral function of n, it is sufficient to assume for the sum a
function of n which is of one dimension higher than the w th term of the
series.
Example 2. Find the conditions that x 3 +px 2 + qx + r may be divisible by
x 2 + ax + b.
Assume x 3 +px 2 + qx + r=(x + k) (x 2 + ax + 6).
Equating the coefficients of the like powers of x, we have
k + a=p, ak + b = q, kb = r.
From the last equation k =  ; hence by substitution we obtain
b
r n ar ,
r + a=p, and — +b = q;
that is, r = b (pa), and ar = b (qb);
which are the conditions required.
EXAMPLES. XXII. a.
Find by the method of Undetermined Coefficients the ,sum of
1. l 2 +3* + 5*+7*+...to n terms.
2. 1.2. 3 + 2. 3. 4 + 3. 4. 5 + .. .ton terms.
3. 1. 2 2 + 2.3 2 + 3.4 2 + 4.5 2 +... to n terms.
4. I 3 + 3 3 + 5 3 + V 3 + . . .to n terms.
5. l 4 + 2* + 3 4 + 4 4 + ...to?i terms.
6. Find the condition that x 3 3px + 2q may be divisible by a
factor of the form a?+%ax + a 2 .
7. Find the conditions that ax 3 + hv 2 \cx + d may be a perfect cube.
8. Find the conditions that a 2 A A + bx 3 +cx 2 + dx+f 2 may be a
perfect square.
9. Prove that ax 2 + 2bxy + cif + 2tlv + 2ey +/ is a perfect square,
if b' 1 = ac, d = a/, e 2 = cf.
UNDETERMINED COEFFICIENTS. 257
10. If a.< :i + bx 2 + cx + d is divisible by x 2 + h 2 , prove that <id = bc.
11. If 3tP — f>qx+4r is divisible by (x — c) 2 , shew that g*=r*,
12. Trove the identities :
a 2 (xb)(x— c) b 2 (xc)(x — a) c 2 (x  a) (x  b) _ 2
( } (a6)(«c) + ~(bc){ba)~ + ~Jc^aJ(cb) "
/ n (■y^>)(^c)<.yc Q (ffc)(# eg) (.?«)
w (rt6)(ac)(aJ)" t " (bc)(bd)(ba)
(x  d) (x  a) (x ~b) (x  a) (x  b) (x  c)
+ {cd){ca)(cb) + \da){db)\dc)** '
13. Find the condition that
ax 2 + 2/ixy + by 2 f 2gx + 2fy + c
may be the product of two factors of the form
jfctf+gy+r, jt/.t' + ^'y + r'.
14. If £ = lx + my + nz, r) = nx + ly + mz, £=mx + n// + l~, and if the
same equations are true for all values of x, y, z when £, 77, £ are inter
changed with x t y, 2 respectively, shew that
l 2 +2mn = l, m 2 + 2ln = 0, n 2 + 2lm=0.
15. Shew that the sum of the products //  /• together of the n
quantities a, a 2 , a 3 , ,..a n is
( « y + 1 l)(tt* + a l)...(a»l) i(„r)(»r+l).
(a 1) (a 2  1).. .(a*' 1) a
313. If the infinite series a + a 2 x + a.,x 2 + a 3 x 3 + is equal
to zero for every finite value of x for which the series is convergent,
tit en each, coefficient must be equal to zero identically.
Let the series be denoted by S, and let S\ stand for the ex
pression a l + a 2 x + a :i x 2 + ; then S = a + xS t , and therefore,
l>y hypothesis, a + xS t = for all finite values of x. But since S
is convergent, #, cannot exceed some finite limit; tlierefore by
taking x small enough xS x may be made as small as we please.
In this case the limit of & is a ; but S is always zero, therefore
a Q must be equal to zero identically.
Removing the term a , we have xS x = for all finite values of
x; that is, a x + a 2 x + ajc 2 + vanishes for all finite values of x.
Similarly, we may prove in succession that each of the
coefficients a n a.,, a is equal to zero identically.
H. ir.A. 17
258 HIGHER ALGEBRA.
314. If tivo infinite series are equal to one another for every
finite value of the variable for which both series are convergent, the
coefficients of like powers of the variable in the two series are equal.
Suppose that the two series are denoted by
a + a x x + a x 2 + a 3 x 3 +
and A + A x x + A 2 x 2 + Aj? + ;
then the expression
«o " A o + («i " A i ) x + ( a 2  A 2 ) °° 2 + («■  A a) °f +
vanishes for all values of x within the assigned limits ; therefore
by the last article
a A = O t a.A^Q, a B A a = 0, a 3 A a = 0,
that is, « = ^ > °i^n a 2 = A 2 , a. d = A 3 , ;
which proves the proposition.
2 + x 2
Example 1. Expand = „ in a series of ascending powers of x as far
as the term involving x 5 .
2 + X 2
Let r— — ' — 2 = a + a x x + a 2 x' 2 + a.jx? + ...,
where a , fl a , a. 2 , a :i ,... are constants whose values are to be determined; then
2 + x 2 — (1 + x  x 2 ) (a Q + Oj a; + a 2 ar + o a re 3 + . . . ) .
In this equation we may equate the coefficients of like powers of x on
each side. On the righthand side the coefficient of x n is a n + a u _ 1  a n _ 2 ,
and therefore, since x 2 is the highest power of x on the left, for all values of
?t>2 we have
this will suffice to find the successive coefficients after the first three have
been obtained. To determine these we have the equations
a = 2, a 1 + a = 0, a. 2 + a 1 a = l;
whence a = 2, ^=2, a 2 =5.
Also a 3 + a 2 a 1 = 0, whence a 3 = 7;
a 4 + a 3 ~~ a 2 = 0, whence a 4 = 12 ;
and a 5 + a±a 3 = 0, whence a 5 =  19 ;
thus , 2 + X ~ „ = 2  2x + 5x 2  7.t 3 + 12x 4  19a 5 + . . .
l + ica; 2
UNDETERMINED COEFFICIENTS. 250
Example 2. Prove that if n and r are positive integers
*.fry+«fez3 ( .y fr?)fr«> fr ~ + .„
£ I 3
is equal to if r be less than n, and to w if r = n
We have
= x n + terms containing higher powers of x. . .(1).
Again, by the Binomial Theorem,
(g*l)n =c »w ne (»l)ai + ^_(±Ll) e (n2)*_ j ( 2 ).
By expanding each of the terms e nx , e (n ~ l)X y ... we find that the coefficient
of x r in (2) is
n
r
(»l) r n(nl) (n2) r w(m1)(w2) (n3)* 1
r [r j2 r 3 r
and by equating the coefficients of x r in (1) and (2) the result follows.
Example 3. If y = ax + bx 2 + ex 3 + ,
express x in ascending powers of y as far as the term involving y 3 .
Assume x=py + qy 2 + ry 3 + ,
and substitute in the given series ; thus
y = a{py + qy* + ry 3 +...) + b(py + qy 2 +...y 2 + c{2>y + qif+...y t +....
Equating coefficients of like powers of y, we have
an = 1 ; whence p =  .
a
aq + bp = ; whence q = — 5 .
a 6
a r + 2bpq + cp 3 — ; whence r = — =■ , .
a 5 a 1
m, V &'V" (2&   ac) y
Thus # = •'4 +  ?— —
This is an example of Reversion of Series.
Cor. If the series for ?/ be given in the form
y = k + ax + bx 2 + ex? + ...
put yk = z;
then z — ax + bx + ex 3 + . . . ;
from which x may be expanded in ascending powers of z, that is of y  k.
17—2
260 HIGHER ALGEBRA.
EXAMPLES. XXII. b.
Expand the following expressions in ascending powers of x as far
1+2^; l&g l+x
1_^_^.2 "' i_#_6#2 ■* 2+.r + .r2'
4 3 + * 5 l  .
' 2  x — x 2 1 + ax — ax 2 — .r 3
a 4 b v
6 Find « and b so that the n th term in the expansion of 7 ,
(l.r)
may be (Sn2)x n ~ 1 .
7. Find a, b, c so that the coefficient of x n in the expansion of
a + bx + cx 2 , , ,
— I ^ — may be n + l.
(l.r) 3 J
8. If y 2 + % =# (3/ + 1), shew that one value of y is
.r + s .rjs S A +
9. If cx z + ax y = 0, shew that one value of x i «
y e?/ 3 3c 2 ;/ 5 12c 3 ;/ 7
a a 4 a 7 «'°
Hence shew that x= 00999999 is an approximate solution of the
equation x 3 + 100.? 1 = 0. To how many places of decimals is the
result correct ?
10. In the expansion of ( 1 + x) ( 1 + ax) ( 1 + a\c) ( 1 + a\v) , the
number of factors being infinite, and a < 1, shew that the coefficient of
r . I \ n hr(rl)
X 1S (la)(la 2 )(l« 3 ) (lO
11. When a < 1, find the coefficient of x n in the expansion of
(1  ax) (1 —a 2 x) (1 — d A x) to inf. '
12. If n is a positive integer, shew that
(1) n n+1 n(nl) n+1 + n ^~ 1 ' (n2)* +1  =jn\ n+.l ;
(2) n n (n+l)(nl) n + K — — ^ (n2)« =1;
the series in each case being extended to n terms ; and
(3) l"»2»+ 7t ^~ 1 < 3 a  =(l) w \n;
(4) (n+p) n n(n+pl) n + — ^— — ' (n+p2) n  = \n;
l± '—
the series in the last two cases being extended to n + 1 terms.
CHAPTER XXIII.
Partial Fractions.
315. In elementary Algebra, a group of fractions connected
by the signs of addition and subtraction is reduced to a more
simple form by being collected into one single fraction whose
denominator is the lowest common denominator of the given
fractions. But the converse process of separating a fraction into
a group of simpler, or jwtial, fractions is often required. For
3 — 5a;
example, if we wish to expand ^ „ in a series of ascend
1 — iX r OXT
ing powers of x, we might use the method of Art. 314, Ex. 1, and
so obtain as many terms as we please. But if we wish to find the
general term of the series this method is inapplicable, and it is
simpler to express the given fraction in the equivalent form
1 2
1 — . Each of the expressions (1 —a;) 1 and (1 — 3aj) 1
I — x l — ox
can now be expanded by the Binomial Theorem, and the general
term obtained.
316. In the present chapter we shall give some examples
illustrating the decomposition of a rational fraction into partial
fractions. For a fuller discussion of the subject the reader is
referred to Serret's Cours d'Algebre Superieure, or to treatises on
the Integral Calculus. In these works it is proved that any
rational fraction may be resolved into a series of partial fractions;
and that to any linear factor x — a in the denominator there cor
responds a partial fraction of the form —  ; to any linear
X — cc
factor x  b occurring twice in the denominator there correspond
7? 7?
two partial fractions, — l j and . — *__ . If x — b occurs three
x — b (x — by
times, there is an additional fraction . hnl au d so on  To
(xb)"
262 HIGHER ALGEBRA.
any quadratic factor x 2 +px + q there corresponds a partial
Px + Q
fraction of the form — : if the factor x 2 + vx + q occurs
x' +])x + q
P x + Q
twice, there is a second partial fraction y—~ L — s ; and so on.
' r (x+2)x + q)
Here the quantities A v B lt P 2 , B 3 , P, Q, P v Q x are all
independent of x.
We shall make use of these results in the examples that
follow.
5x — 11
Example 1. Separate =^ ^ into partial fractions.
Since the denominator 2.r 2 + x  6 = (x + 2) (2x  3), we assume
5.rll A B
+
2x 2 + x§ x + 2 2.c3'
where A and B are quantities independent of x whose values have to be
determined.
Clearing of fractions,
5xll = A (2xS) + B(x + 2).
Since this equation is identically true, we may equate coefficients of like
powers of x ; thus
2A+B = 5, ~SA + 2B=U;
whence A = 3, B= 1.
5.rll 3 1
'"' 2x 2 + x6~ x + 2 2xB'
Example 2. Resolve ■; r~. r, into partial fractions.
. (x  a) (x + b)
mx + n A B
Assume . z. =r = h
(xa)(x + b) xa x + b'
.' . mx + n = A {x + b) +B (xa) (1).
We might now equate coefficients and find the values of A and B, but it
is simpler to proceed in the following manner.
Since A and B are independent of x, we may give to x any value we please.
In (1) put xa = 0, ov x = a; then
ma + n
A = r ;
a + b
t n , r, nibn
putting x + b = 0, or x—  b, B— — .
CI "T*
mx + n
(x  a) (x + b)
1 /ma + n mbii\
~ a + b \ xa x + b J
PARTIAL FRACTIONS. 203
23 v  11 r' 2
Example 3. Resolve 7^7— .rrx — ^ mto partial fractions.
y£x — I) (J — x j
23xll.r 2 ABC
Assume = ^^ rm r = n r+ 5 H 5 i 1 ) 5
(2.cl) (3 + x)(3.r) 2.cl 3 + x 3x w
. • . 23x  lLc 2 = .1 (3 + x) (3 x)+B (2x  1) (3  x) + G (2x  1) (3 + x).
By putting in succession 2^1 = 0, 3 + x— 0, 3  # = 0, we find that
4 = 1, B = i, C= 1.
23.c  lis 2 1 4 1_
•'• (2.cl)(9x 2 )~2xl + 3+x 3 a:'
3.t 2 + x — 2
Example 4. Resolve . — ' , ' — ^— : into partial fractions.
[x — &)" (J. — &x\
3s 2 + s2 ^ B G
Assume ■z^n — s7 — ~k — k~ H « +
(x  2) (1  2x) " 1  2x x2 (x  2) 2 '
.• . %x + x2 = A (x  2) 2 + £ (1  2x) {x  2) + C (1  2x).
Let 1  2x = 0, then A =   ;
o
let a; 2 = 0, then C=4.
To find B, equate the coefficients of x 2 ; thus
3 = A  2B ; whence B =  ^ .
o
3.r + x  2
' ' (x 2) a (1  2x) 3(1 2x) 3 {x  2) (x  2) 2 '
42  19a;
Example 5. Resolve — — r, r into partial fractions.
1 [x 2 +l)(x4] L
4219.C Ax + B C
Assume 73 — tt, r; = —3 — r  +
(.^ + 1)^4) x+l .i4'
.. 42  19.r = (Ax +B) (x 4) + C (x* + l).
Let x = 4, then C=2;
equating coefficients of x' 2 , = A + C, and .4=2;
equating the absolute terms, 42 =  4Z? + C, and B =  11,
42  19a 2s 11 2
•'" p+l)(x4)"^TT *4*
317. The artifice employed in the following example will
sometimes be found useful.
2G4 HIGHER ALGEBRA.
9x* — 24rc 2 + 48#
Example. Resolve ; 7 ^t~. =rr into partial fractions.
r (a:2) 4 (a:+l)
9z 3 24;r 2 +48;c A f(x)
Assume —. ^tx~? — rrv — — n +
(x2)*(x + l) x+1 {x2) 4 '
where A is some constant, and / (x) a function of x whose value remains to
he determined.
.. 9x 3 24£ 2 + 48x = ,l (x2)*+(x + l)f{x).
Let x=  1, then A =  1.
Substituting for ^4 and transposing,
(x + 1) / (a) = [x  2) 4 + 9a 3  24s 3 + 48x = x 4 + x* + 16* + 16 ;
.'./(*) = £ 3 + 16.
•r 3 + 16
To determine the partial fractions corresponding to  — — , put x2 = 2;
\x—2)
.r 3 +16 (2 + 2) 3 + 16 2 3 + 6^ 2 +122 + 24
then
(x2) 4 Z* z*
1 6 12 24
~z + z^ + ~z^ + ~z 4 ~
1 6 12 24
+ /. nva + /_ n\n +
~ x2^(x2)' 2 ' (x2f^ (x2f
9x 3 24j; 2 + 48* 1,1, 6 , 12 24
= i \ « + / Z Svi + TZ ^J +
" (x2) 4 (x + l) " x + 1 x2 (x2)" (x2f (x2) 4 *
318. In all the preceding examples the numerator has been
of lower dimensions than the denominator ; if this is not the case,
we divide the numerator by the denominator until a remainder is
obtained which is of lower dimensions than the denominator.
6r 3 + 5# 2 7
Example. Resolve zr= —  = into partial fractions.
* ox  2x  1
By division,
v = 2x + 3 + ,
3a: 2  2.x  1 Sx 2  2x  1
8a; 4 5 1
and ^= — pr ^ = s =■ +
3x 2 2xl 3.c + l x1'
6^ + 5^7 5 1
= 2.r + 3 +   +
'* 3x 2 2xl 3.T+1 *l'
319. We shall now explain how resolution into partial
fractions may be used to facilitate the expansion of a rational
fraction in ascending powers of x.
PARTIAL FRACTIONS. 265
;; ,■'  ;• _ 2
Example 1. Find the general term of —  — — — — . when expanded in a
(* 2)(l  2x) l
series of ascending powers of x.
By Ex. 4, Art. 316, we have
3.r 2 + .r2 15 4
(*2) 2 (l2*) 3(12*) 3(*2) (*2) 2
15 4
+
3(12*) 3(2*) (2a?) a
Hence the general term of the expansion is
/ r 6 1 r+l\
V 3 + 6 • 2 r sr y
7 + *
Example 2. Expand r^r— . in ascending powers of * and find
(1 + *) (I + *~)
the general term.
. 7 + * .4 JB* + C
Assume H —  =  — +  ;
(1 + *) (1+* J ) 1+x 1 + * 2
.\ 7 + * = J(l + * 2 ) + (E*+C)(l + *).
Lctl + *=:0, then A = 3;
equating the absolute terms, 7 = A + C, whence C = i ;
equating the coefficients of * 2 , = A + B, whence B—  3.
7 + * 3 43*
+
(1 + *)(1+* 2 ) _ 1 + *^ 1+* 2
= 3(1 + .r) 1 + (4  3*) (1 + x 2 )~l
= 3{l* + * 2  + (_l)P; C P + ...j
+ (43*) {l.r 2 + *» + (1)p*'^+...}.
To find the coefficient of x r :
r
(1) If /• is even, the coefficient of * r in the second series is 4(l) 2 ;
r
therefore in the expansion the coefficient of x r is 3 + 4 (  1) 2 .
rl
(2) If r is odd, the coefficient of * r in the second series is  3 (  1) '
r+l
and the required coefficient is 3 (  1) 2  3.
EXAMPLES. XXIII.
Resolve into partial fractions :
, lx\ 46+13.r l+3. r + 2 .r 2
lbj; + 6jf ' 12.t 2 lU15' (1 2.r) (1 .//')'
266 HIGHER ALGEBRA.
.y 2  10a; +13 2x*+x 2 x3
' (xl)(x 2 5x+6)' x(xl)(2x + 3)'
9 „ x* 3x* 3a; 2 + 10
6 * (a;l)(^ + 2) 2 ' 7 * (#+l) 2 (#3)
26^ 2 + 208o; Q 2^ 2 lLr + 5
(a; 2 + l)(^ + 5)' (^3) (^ 2 + 2^ 5)
3^8x 2 +10 ,, 5^ + 6.r 2 + 5.r
(071)* * (^ 2 l)(^+l) 3 '
Find the general term of the following expressions when expanded
in ascending powers of x.
l + 3# 5a; + 6 u # 2 + 7;f + 3
12# l + llo; + 28^* ' (2+a?)(l#)' tf 2 + 7a + uy
2#4 .« 4 + 3^+2a' 2
15. t^ 5tt^ — ^ • 16.
(1  x 2 ) (1  2.r) * (1  x) ( 1 + x  2x 2 )
3 + 2xx 2 no 4 + 7x
17. 7 , . w , — ttx* • 18.
(l+a?)(l4a?) 2 * (2 + 3a;)(l+.r) 2 '
19 . *"* 20. 1 * +i *
3
(^1)(^ 2 +1)" (1tf)
21 . „ 1 22 . »«•■
(1  cw?) (1  te) (1  co;) * ' (2  3.r + a 2 ) 2 '
23. Find the sum of n terms of the series
(l) I +  + — +
[ } (i+^)(i+^ 2 ) (i+^ 2 )(n^) (i+^)(i+^ 4 )
. . x (1  ax) ax (1  a 2 x)
^ ' (1 +x) (l + ax) (1 + a%) + (1 + ax) (1 +a%) (1 + a 3 .r) +
24. When a? < 1, find the sum of the infinite series
1 x 2 x A
(lx) (lx 3 ) + (1 a?) (1 .r 5 ) + (1tf 5 ) (1 ^) +
25. Sum to n terms the series whose p th term is
xp(1+xp + 1 )
(l^)(l.^ + 1 )(l^ + 2 )'
26. Prove that the sum of the homogeneous products of n dimen
sions which can be formed of the letters a, b, c and their powers is
a n + 2 (b c) + b n + 2 (c a) + c n + 2 (ab)
a 2 (bc) + b 2 (ca) + c 2 (ab) *
CHAPTER XXIV.
Recurring Series.
320. A series u + u l + u 2 + u 3 +
in which from and after a certain term each term is equal to the
sum of a fixed number of the preceding terms multiplied respec
tively by certain constants is called a recurring series.
321. In the series
1 + 2x + 3ar + 4a? + 5a; 4 + ,
each term after the second is equal to the sum of the two
preceding terms multiplied respectively by the constants 2x, and
 x 2 j these quantities being called constants because they are
the same for all values of n. Thus
5x 4 = 2x . 4a; 3 + ( x 2 ) . 3a; 2 ;
that is,
u 4 = 2xn 3 — x 2 u 2 ;
and generally when n is greater than 1, each term is connected
with the two that immediately precede it by the equation
u — 2xii , — x 2 u . ,
h n— 1 n—2*
or u — 2xu , + x 2 u „ = 0.
H n — 1 ii — 2
In this equation the coefficients of u n , «*,_,, and l*,_ a , taken
with their proper signs, form what is called the scale of relation.
Thus the series
1 + 2x + 3a; 2 + 4a; 3 + 5x 4 +
is a recurring series in which the scale of relation is
1  2x + x 2 .
322. If the scale of relation of a recurring series is given,
any term can be found when a sufficient number of the preceding
268 HIGHER ALGEBRA.
terms are known. As the method of procedure is the same
however many terms the scale of relation may consist of, the
following illustration will be sufficient.
If 1  px  qx 2  rx 3
is the scale of relation of the series
a + a<x + a a x 2 + ajc 3 +
we have
a n x*=px  a n i x "~ 1 + <l x "  a n  2 x ' l ~ 2 + rx3 • a n  3 x ' l ~ 3 i
or a m =pa n _, + &».■ + m « 3 5
thus any coefficient can be found when the coefficients of the
three preceding terms are known.
323. Conversely, if a sufficient number of the terms of a
series be given, the scale of relation may be found.
Example. Find the scale of relation of the recurring series
2 + 5x + 13x 2 + 35x 3 +
Let the scale of relation be 1 px  qx*, then to obtain p and q we have
the equations 13  5p  2q = 0, and 35  13p  5q = ;
whence p = 5, and q=  6, thus the scale of relation is
1  5x + 6a; 2 .
324. If the scale of relation consists of 3 terms it involves
2 constants, p and q ; and we must have 2 equations to de
termine p and q. To obtain the first of these we must know
at least 3 terms of the series, and to obtain the second we
must have one more term given. Thus to obtain a scale of
relation involving two constants we must have at least 4 terms
'O
given.
If the scale of relation be 1 — px — qx 2  rx 3 , to find the
3 constants we must have 3 equations. To obtain the first of
these we must know at least 4 terms of the series, and to obtain
the other two we must have two more terms given ; hence to find
a scale of relation involving 3 constants, at least G terms of the
series must be given.
Generally, to find a scale of relation involving m constants,
we must know at least 2m consecutive terms.
Conversely, if 2m consecutive terms are given, we may assume
for the scale of relation
1 ~ l\ x ~ l\ x * ~ lh x * ~ PJ**
RECURRING SERIES. 2G9
325. To find the sum ofn terms of a recurring series.
The method of finding the sum is the same whatever be the
scale of relation ; for simplicity we shall suppose it to contain
only two constants.
Let the series be
a u + a x x + a 2 x 2 + aj£ + (1)
and let the sum be S ; let the scale of relation be 1 — px — qx* ;
so that for every value of n greater than 1, we have
Now S—a it + a.x + a,x 2 + ...+ a ,x"~\
1 2 /»— 1 '
— px S= — pa x — pa x x* — ... — 2^>ci H _ 2 x n ~ l — pa x* t
 qa? S=  qajt?  ... qa H _ 3 x* 1 qa H _ i x n qa H _ l x u + \
... (i  px _ qtf) S  a + {a x pa ) x  {pa n _ x + qa n _ a ) x n  qa^x** 1 ,
for the coefficient of every other power of x is zero in consequence
of the relation
a nP a nl<2 a «2= '
. s _ % + («, P<Q x (P a , t , + qa n  3 ) x" + qa n _ } x H+l
1 px— qx 2 1  px — qx 2
Thus the sum of a recurring series is a fraction whose de
nominator is the scale of relation.
32G. If the second fraction in the result of the last article
decreases indefinitely as n increases indefinitely, the sum of an
infinite number of terms reduces to — \ — — ! —  — ^ — .
1 — px — qx"
If we develop this fraction in ascending powers of x as
explained in Art. 314, we shall obtain as many terms of the
original series as we please; for this reason the expression
1 —px — qx 2
is called the generating function of the series.
327. From the result of Art. 325, we obtain
a n + ( a , —P a .) X o .xi
 °, v ' — £ —%— = a lt + a.x + ax + ... +a x n + l
1 px — qx' ° ' 2 " 1
1  px— qx
270 HIGHER ALGEBRA.
from which we see that although the generating function
1 — px — qx 2
may be used to obtain as many terms of the series as we please,
it can be regarded as the true equivalent of the infinite series
a + a l x + a 2 x 2 + ,
only if the remainder
( I**,., +q a n 2 ) xn + ( 2 a n^" +l
1 — poj — qx 2
vanishes when n is indefinitely increased ; in other words only
when the series is convergent.
o v
328. When the generating function can be expressed as a
group of partial fractions the general term of a recurring series
may be easily found. Thus, suppose the generating function
can be decomposed into the partial fractions
ABC
h h
1— ax 1 + bx (I— ex) 2 '
Then the general term is
&
{Aa r + ( l) r M r + (r + 1) Cc r } x\
In this case the sum of n terms may be found without using
the method of Art. 325.
Example. Find the generating function, the general term, and the sum
to n terms of the recurring series
1  Ix  x 2  43.C 3 
Let the scale of relation be 1 px  </.r 2 ; then
l + 7j><z = 0, 43 + 2> + 7</ = 0;
whence p = l, 5 = 6; and the scale of relation is
1  x  6.r 2 .
Let S denote the sum of the series ; then
S = llx x 2 4Sx s 
xS=  x + 7x 2 + x*+
Qx 2 S= 6x 2 + 42.r 3 +
.. (lx6x 2 )S = l8x,
s J" 8 * 
which is the generating function.
RECURRING SERIES. 271
1  8.r 2 1
If we separate ^„ into partial fractions, we obtain  —  ;
1xU.r ± 1 + 2a; 1305'
whence the (r+ l) tU or general term is
{(lyw^v ].<■>■.
Putting r = 0, 1, 2,...n 1,
the sum to ?i terms
= { 2  2 2 x + 2% 2 ... + ( I)"" 1 2" a;' 1 " 1 }  (1 + 3a + 3%* + . . . + 3" 1 x n ~ l )
_ 2 + (  I)' 1 " 1 2 n+1 x n _ 1_ 3* x n
~ l+lte 1  3x~ '
329. To find the general term and sum of n terms of the
recurring series a + a i + a_,+ , we have only to find the
general term and sum of the series a + a l x + a 2 x 2 + , and put
x — 1 in the results.
Example. Find the general term and sum of n terms of the series
1 + 6 + 24 + 84+
The scale of relation of the series 1 + 6.r + 24x 2 + 84x 3 + . . . is 1  ox + Ooj 2 ,
1 + x
and the generating function is —  — *— — .
1 — OX + OX"
This expression is equivalent to the partial fractions
4 3
1  Sx 1  2a; '
If these expressions be expanded in ascending powers of x the general
term is (4 . 3 r  3 . 2 r ) x r .
Hence the general term of the given series is 4 . 3 r 3. 2 r ; and the sum
of n terms is 2 (3' 1  1)  3 (2' 1  1).
330. We may remind the student that in the preceding
article the generating function cannot be taken as the sum of
the series
1 +6x + 24:x 2 +8±x 3 +
except when x has such a value as to make the series convergent.
Hence when x = 1 (in which case the series is obviously divergent)
the generating function is not a true equivalent of the series.
But the general term of
1 + 6 + 24 + 84 +
is independent qfx, and whatever value x may have it will always
be the coefficient of x" in
1 + Gx + 24* 2 + 84a 3 +
We therefore treat this as a convergent series and find its
general term in the usual way, and then put x = 1.
272 HIGHER ALGEBRA.
EXAMPLES. XXIV.
Find the generating function and the general term of the following
series :
1. l + 5.r + 9.r 2 +13.r 3 + 2. 2.v + 5.r 2 7.r 3 +
3. 2 + 3x + 5x 2 + 9x 3 + 4. 7 6x + 9x 2 + 27x 4 +
5. 3 + 6a? + Ux 2 + 36.r* + 98.^ + 276.1 5 +
Find the n th term and the sum to n terms of the following series :
6. 2 + 5 + 13 + 35+ 7. l+6.v 2 + 30.v 3 +
8. 2 + 7^ + 25^ + 91^+
9. 1 + 2.v + 6x 2 + 20# 3 + 66x* + 212^ +
10. ^ + 2 + + 8+
11. Shew that the series
1 2 + 2 2 + 3 2 + 4 2 + + n 2 ,
1 3 + 2 3 + 3 3 + 4 3 + +n 3 ,
are recurring series, and find their scales of relation.
12. Shew how to deduce the sum of the first n terms of the re
curring series
a + a x x + a 2 x 2 + a^v 3 +
from the sum to infinity.
13. Find the sum of 2n + 1 terms of the series
31 + 139 + 4153+
14. The scales of the recurring series
a + a v v+ a^x 2 + a 3 .r 3 + ,
b + b 1 x+b^c 2 {b 3 .v 3 + ,
are 1 +px+qx 2 , l + rx + sx 2 , respectively; shew that the series whose
general term is (<x n +6 n )^" is a recurring series whose scale is
l + (p + r)x + (q + s +pr) x 2 + (qr +ps) x 3 + qsx*.
15. If a series be formed having for its n ih term the sum of n terms
of a given recurring series, shew that it will also form a recurring
series whose scale of relation will consist of one more term than that
of the given series.
CHAPTER XXV.
CONTINUED FllACTIONS.
331. All expression of the form a + is called a
a
c + 
e + ...
continued fraction ; here the letters a, b, c, may denote any
quantities whatever, but for the present we shall only consider
the simpler form a x + , where a n a 2i « 3 ,... are positive
2 a 3 + ...
integers. This will be usually written in the more compact form
1 1
a, +
a 2 + a 3 +
332. When the number of quotients a a , « 3 ,... is finite the
continued fraction is said to be terminating ; if the number of
quotients is unlimited the fraction is called an infinite contirmed
fraction.
It is possible to reduce every terminating continued fraction
to an ordinary fraction by simplifying the fractions in succession
beginning from the lowest.
333. To convert a given fraction into a continued fraction.
tn
Let — be the tnven fraction ; divide in by n, let a be the
quotient and j> the remainder ; thus
m p 1
— — a. + =a, + — :
n n n
P
si. H. A 18
274
HIGHER ALGEBRA.
divide n by ^», ^ «„ be the quotient and q the remainder ; thus
n q 1
 = a . +  = a s +  ;
V ' V ' P
9.
divide p by q, let a. 6 be the quotient and r the remainder ; and .so
on. Tims
1 1 1
rn,
— = a. +
n
i
1
= a. +
«o +
a 2 + a 3 +
a 3 +.
If m is less than ?t, the first quotient is zero, and we put
7)1 1
n ti
■m
and proceed as before.
It will be observed that the above process is the same as that
of finding the greatest common measure of m and n ; hence if m
and n are commensurable we shall at length arrive at a stage
where the division is exact and the process terminates. Thus
every fraction whose numerator and denominator are positive
integers can be converted into a terminating continued fraction.
251
Example. Reduce ^^ to a continued fraction.
Finding the greatest common measure of 251 and 802 by the usual
process, we have
5
251
802
3
G
G
49
1
8
and the successive quotients are 3, 5, 8, G; hence
251 1 1 1 1
802 ~ 3+ 5+ 8+ 6'
334. The fractions obtained by stopping at the first, second,
third, quotients of a continued fraction are called the first,
second, third, convergents, because, as will be shewn in
Art. 339, each successive convergent is a nearer approximation
to the true value of the continued fraction than any of the
preceding convergents.
CONTINUED FRACTIONS. 275
335. To shew that the convergents ewe alternately less and
greater than the continued fraction.
1 1
Let the continued fraction be a l +
a 2 + a 3 +
The first convergent is «,, and is too small because the part
is omitted. The second convergent is a i — , and is
a a +a 3 + ° l a k
too great, because the denominator a a is too small. The third
convergent is a, \ , and is too small because a \ — is too
a 2+ CC 3 %
great ; and so on.
When the given fraction is a proper fraction a t = ; if in this
case we agree to consider zero as the first convergent, we may
enunciate the above results as follows :
The convergents of an odd order are all less, and the convergents
of an even order are all greater, than the continued fraction.
336. To establish the law of formation of the successive con
vergents.
Let the continued fraction be denoted by
1 1 1
a x +
a 2 + a 3 + a 4 +
then the first three convergents are
a.
a x a 3 + 1 o, (a, a, + !) + «,
1 a 2 a 3 . a 2 + 1
and we see that the numerator of the third convergent may be
formed by multiplying the numerator of the second convergent
by the third quotient, and adding the numerator of the first con
vergent ; also that the denominator may be formed in a similar
manner.
Suppose that the successive convergents are formed, in a
similar way; let the numerators be denoted by^,^.,, p 3 ,..., and
the denominators by q lt q , q 3 ,...
Assume that the law of formation holds for the » tt convergent;
that is, suppose
1\ = »J».i +P»i In = <* n ?., + Q„ 2 
18—2
27g HIGHER ALGEBRA.
The (*+ l) th convergent differs from the ft* only in having
the quotient a n + ± in the place of aj hence the (» + 1) «*
vergent
^nn ^» + ^»i ? by supposition.
« n+1 ?„ + ?„_!'
If therefore we put
co. th«t the numerator and denominator of the (» + l) th con
we ^ ^* ^^ which was supposed to hold in the case of
^ f tttVs hold in P the case of the third con
vergent, hence it holds for the fourth, and so on; therefore *
holds universally.
337. It will be convenient to call a H the n* partial quotient;
the complete quotient at this stage being a n +
a„ +1 + «« +2 4
We shall usually denote the complete quotient at'any stage by ft.
We have seen that
let the continued fraction be denoted by m ; then x differs from
& only in taking the complete quotient ft instead of the partial
quotient a„ ; thus
_ ft j^il + ff»2
X ~kq n _ x + q n  2 '
338 // Eb 6 e tf l6 n th convergent to a continued fraction, then
Q
Let the continued fraction be denoted by
111
a, +
1 a Q + a 3 + a 4 +
CONTINUED FRACTIONS. 277
then
= (" 1 ) 2 (P. 2 9«a iV, ^2)1 similarly,
But p 2 q x ]\ q, = (<h % + 1)  «x • a, = 1 = ( l) 2 J
hence />„ g^, #,_, g, = ( 1)".
When the continued fraction is less than unity, this result will
still hold if we suppose that a x = 0, and that the first convergent
is zero.
Note. When we are calculating the numerical value of the successive
convergents, the above theorem furnishes an easy test of the accuracy of the
work.
Cor. 1. Each convergent is in its lowest terms ; for iip n and
q n had a common divisor it would divide p n q nl — p n _ l q ni or unity ;
which is impossible.
Cor. 2. The difference between two successive convergents is
a fraction whose numerator is unity ; for
q» ?«_i q n q n ^ q,,q n i'
EXAMPLES. XXV. a.
Calculate the successive convergents to
1. 2 + l * l ' l
2.
6+ 1+ 1+ 11+ 2
1111111
3. 3 +
2+ 2+ 3+ 1+ 44 2+ 6
111111
3+ 1+ 2+ 2+ 1+ 9"
Express the following quantities as continued fractions and find the
fourth convergent to each.
729
4.
253
5.
832
6.
1189
179'
159'
3927 "
8.
•37.
9.
1139.
10.
•3029.
7.
2318'
11. 4310.
278 HIGHER ALGEBRA.
12. A metre is 39*37079 inches, shew by the theory of continued
fractions that 32 metres is nearly equal to 35 yards.
13. Find a series of fractions converging to "24226, the excess in
days of the true tropical year over 365 days.
14. A kilometre is very nearly equal to "62138 miles; shew that
A , . .. 5 18 23 64 . •■*•■* «.
the fractions , ^ , == , ^z are successive approximations to the
ratio of a kilometre to a mile.
15. Two scales of equal length are divided into 162 and 209 equal
parts respectively; if their zero points be coincident shew that the
31 st division of one nearly coincides with the 40 th division of the other.
16. If — s is converted into a continued fraction, shew
n 3 + n u + n + l
that the quotients are n — 1 and n+l alternately, and find the suc
cessive convergents.
17. Shew that
Pn + \~Pn  1 _ Pn
(!)
2n + 1 9.n  1 9.n
(2) (^O^fHvr 2  1
\ Pn / \ Pn + U \ c Jn ,
X g«l
18. If — is the n th convergent to a continued fraction, and a n the
corresponding quotient, shew that
339. Each convergent is nearer to the continued fraction than
any of the 'preceding convergents.
Let x denote the continued fraction, and *— " , ^*±J —"±2
9* ?« + ! ^+2
three consecutive convergents; then x differs from *a±l only in
taking the complete (n + 2) th quotient in the place of a ; denote
this by k: thus x = ? n+l +Pn ;
and ^^ ~ a; =
Pn + 1 Pn^l^nPn^l 1
& + 1 ?„ + , (%» +l + 7.) y. + , (%„ + , + ?„) "
CONTINUED FRACTIONS. 270
Now k is greater than unity, and q m is less than q ; lience on
botli accounts the difference between " ' ' and x is less than the
difference between — " and x: that is, every convergent is nearer
to tlie continued fraction than the next preceding convergent,
and therefore a fortiori than any preceding convergent.
Combining the result of this article with that of Art. .°>3.>, it
follows that
tli^ convergent of an odd order continually increase, hat are
always less than the continued fraction ;
tin' covrergents of an even order continually decrease, hut are
always greater than the continued fraction.
340. To find limits to the error made in taking any convergent
for the continued fraction.
p p p
Let — , Y ^ 1 r_n±2 ] )0 three consecutive convorgents, and let
k denote the complete (n + 2) th quotient;
then x = ^^ t
p k 1
<ln <ln( k <ln + >+ n J
'.(*•« +9 i)
Now k is greater than 1, therefore the difference between x and
p.. . • i .. i
— is less than , and greater than 
p
Again, since <7, 1 + l ><7„, the error in taking " instead of x is
1 1
less than — 5 and greater than 770— .
?. v. + ,
341. From the last article it appears that the error in
p 1
taking — instead of the continued fraction is less than  ,
q m ?.?.+,
or — ; ; ; that is, less than 3 : hence the larger
a (a ., 7 +q ,) « . .7 "
/ (I V II +1 ill 2 Ft— 1/ 11+ I ili
a i+l is, the nearer does £2 approximate to the continued fraction;
/ «
280 HIGHER ALGEBRA.
therefore, any convergent which immediately precedes a large
quotient is a near approximation to the continued fraction.
Again, since the error is less than — g , it follows that in order
to find a convergent which will differ from the continued fraction
by less than a given quantity  , we have only to calculate the
a
successive convergents up to — , where q n 2 is greater than a.
342. The properties of continued fractions enable us to find
two small integers whose ratio closely approximates to that of
two incommensurable quantities, or to that of two quantities
whose exact ratio can only be expressed by large integers.
Example. Find a series of fractions approximating to 3* 14159.
In the process of finding the greatest common measure of 14159 and
100000, the successive quotients are 7, 15, 1, 25, 1, 7, 4. Thus
314159 = 3+1 1 1 1 111
7+ 15+ 1+ 25+ 1+ 7+ 4
The successive convergents are
3 22 333 355
1 ' 7 ' 106 ' 113 '
this last convergent which precedes the large quotient 25 is a very near
approximation, the error being less than ^ , and therefore less than
25TP5) » •° 00004 
343. Any convergent is nearer to the continued fraction than
any other fraction whose denominator is less than that of the
convergent.
V P
Let x be the continued fraction, — , '*=*■ two consecutive
°n ?.,
r
convergents,  a fraction whose denominator s is less than q .
° 8 "
r v r
If possible, let  be nearer to x than — , then  must be
« ?» s
7) . P
nearer to x than  Ji ^ 1 [Art. 339] ; and since x lies between  and
In  J I"
£5=? it follows that  must lie between — and — ' .
9.X S % ?»!
CONTINUED FRACTIONS. 281
Hence
r P»*P. P n i fWi<5 ^ l ■
.'. rq n _ x ~ sp n _ x < £ ;
that is, an integer less than a fraction ; which is impossible.
p r
Therefore — must be nearer to the continued fraction than  .
& *
P P'
344. If  , — be two consecutive conver gents to a continued
fraction x, then — , is greater or less than x 2 , according as  is
greater or less than — , .
q
Let k be the complete quotient corresponding to the con
vergent immediately succeeding — , ; then x — f—. — ,
° J ° q" lcq ' + q
' '' 5 " * = WW^YY w {hq ' + qY " "' w + pY]
= (tfp'q , pq)(pq'2>'q)
qq'(kq' + q) 2
The factor ky'q'  pq is positive, since p' >p, q' >q, and k> I ;
pp'
lience — , > or < x 2 , according as ]iq' —p'q is positive or negative ;
that is, according as  > or < — , .
Cor. It follows from the above investigation that the ex
pressions ]iq'—2 )/ q i VP ~ c L c L^> p 2  q 2 M 2 , q' 2 x 2 —p' 2 have the same
sign.
EXAMPLES. XXV. b.
222
1. Find limits to the error in taking — yards as equivalent to
a metre, given that a metre is equal to 10936 yards.
282 HIGHER ALGEBRA.
2. Find an approximation to
JL J_ J JL JL
+ 3+ 5+ 7+ 9+ 11+
which differs from the true value by less than 0001.
99
3. Shew by the theory of continued fractions that = differs from
1*41421 by a quantity less than .
„ a 3 + 6a 2 + 13a+10 ,. , , ,. n
4. Express , \ . , A « , 1K rs as a continued fraction, and
1 a 4 +6a 3 + 14a+15a + 7
find the third convergent.
5. Shew that the difference between the first and n th convergent
is numerically ecpial to
1 1 1 (l) n
+ ...+
Mi Ms Wh 9nl2n
p
6. Shew that if a n is the quotient corresponding to s 5 ,
^ ' Pn1~ a " «nl+ «u 2 + 0»3+ '" «3+ a 2+ «1 '
(2 ) _i =an+ _L_ ^2 L_ ... J_ 1. ,
qn1 a nl+ «»2+ tt n3+ «3+ «2
1111
7. In the continued fraction — , shew that
«+ a+ « + « +
( 1 ) Pn +P\ + 1 =Pn  lPn + 1 + £>„£>„ + 2 >
( 2 ) Pn = q n l
8. If — is the ?i th convergent to the continued fraction
111111
«+ b+ a+ b+ a+ 6 +
a
b r  n 
a
shew that q 2n =p 2n + u q 2n  1 = r #».. •
9. In the continued fraction
1111
a+ 6+ «+ 6+ '
shew that
Pn + 2~ ( ah + 2 ) P n +Pn2 = °i 9n + 2 ~ ( ab + 2 ) ?u + •?•* 2 = °
CONTINUED FRACTIONS. 283
10. Shew that
/ 111 x . L \
a[a\ +  to 2/i quotients
\ x a.v.j,+ ar 3 + oa; 4 + /
= «.v,H to 2/i quotients.
:r.,+ oa? 3 + .v,+
11. If r; ,  , ■ , are the n tU , (n — l) th . (?i2) th convergent* to the
iV (^ A3
continued fractions
111 111 111
a i+ (( 2+ a S+ ' tt *+ a B + W 4+ ' ( ':5+ "4+ a 5+ '
respectively, shew that
J/= OjP + 5, iV T = (a^ + 1) P + aJL
12. If — is the n th convergent to
j. i i_
a + « + a + " ' '
shew that p n and q n are respectively the coefficients of x n in the
expansions of
# . «.# + x 2
and
1 — ax — x 2 1 — «r — x 2 '
a n _ Qn
Hence shew that p n —<Ini = i > where a, /3 are the roots of the
equation t 2  at  1 = 0.
13. If — is the n th convergent to
9n
_l 1 1 1_
a + b\ a+ bt " ' '
shew that p n and q n are respectively the coefficients of x n in the
expansions of
x + bx 2 — ^ . ax+(ab + l)x 2 — x A
and
1  (ab + 2) x 2 + x* 1  (aft + 2) x 2 + x A '
Hence shew that
op, n = bq 2n _ x = ab a ,
where a, /3 are the values of x 2 found from the equation
l(ab + 2)x 2 + x A = 0.
CHAPTER XXVI.
INDETERMINATE EQUATIONS OF THE FIRST DEGREE.
345. In Chap. X. we have shewn how to obtain the positive
integral solutions of indeterminate equations with numerical co
efficients; we shall now apply the properties of continued fractions
to obtain the general solution of any indeterminate equation of
the first degree.
346. Any equation of the first degree involving two un
knowns x and y can be reduced to the form ax±by = ± c, where
a, 6, c are positive integers. This equation admits of an unlimited
number of solutions ; but if the conditions of the problem require
x and y to be positive integers, the number of solutions may be
limited.
It is clear that the equation ax + by = — c has no positive
integral solution ; and that the equation ax — by = — c is equivalent
to by — ax — c) hence it will be sufficient to consider the equations
ax ±by — c.
If a and b have a factor m which does not divide c, neither of
the equations ax±by = c can be satisfied by integral values of x
and y ; for ax ± by is divisible by m, whereas c is not.
If a, b, c have a common factor it can be removed by division;
so that we shall suppose a, b, c to have no common factor, and
that a and b are prime to each other.
347. To find the general solution in positive integers of the
equation ax — by — c.
Let  be converted into a continued fraction, and let — denote
6 q
the convergent just preceding j ; then aq—bp = ±l. [Art. 338.]
INDETERMINATE EQUATIONS OF THE FIRST DEGREE. 285
I. If aq — bj) — 1, the given equation may l»e written
ax — by — c (aq — b]j) ;
.. a(x — cq) b (y — c/>).
Now since a and b have no common factor, x — cq must be
divisible by b ; hence x — cq = bt, where t is an integer,
x cq y — c P.
b a
that is, x = bt + cq, y — at + cj) \
from which positive integral solutions may be obtained by giving
to t any positive integral value, or any negative integral value
en cd
numerically smaller than the less of the two quantities j , — \
also t may be zero; thus the number of solutions is unlimited.
II. If aq — bp — — 1, we have
ax — by — — c (aq — bji) ;
.'. a(x + cq) = b (y + cj)) ',
x + cq y + cp
. • . — =—  = — = t, an integer ;
o a
lience x = bt — cq, y — at — cp;
from which positive integral solutions may be obtained by giving
to t any positive integral value which exceeds the greater of the
CO CD
two quantities =,—; thus the number of solutions is unlimited.
o a
III. If either a or b is unity, the fraction j cannot be con
verted into a continued fraction with unit numerators, and the
investigation fails. In these cases, however, the solutions may be
written down by inspection; thus if 6 = 1, the equation becomes
ax — y = c; whence y = ax—c, and the solutions may be found by
ascribing to x any positive integral value greater than  .
a
Note. It should be observed that the series of values for x and y form
two arithmetical progressions in which the common differences are b and a
respectively.
286 HIGHER ALGEBRA.
Example. Find the general solution in positive integers of 29.r  42*/ = 5.
In converting — into a continued fraction the convergent just before —
13
is jr ; we have therefore
29xl342x9 = l;
.. 29x6542x45 =  5;
combining this with the given equation, we obtain
29 (* + 65) =42(# + 45);
x + 65 u + 45
•*• £j = 29" = *» an mte 8 er 5
hence the general solution is
a: = 42«65, ij = 20t~4o.
348. Given one solution in positive integers of the equation
ax — by = c, to jind the general solution.
Let h, k be a solution of axby = c; then ah — bk = c.
.'. ax — by = ah  bk ;
.'. a (x — h) — b(y — k);
x—h y—k
.'. — z — = = t. an integer ;
b a
.'. x = h + bt, y — k + at ;
which is the general solution.
349. To Jind the general solution in positive integers of the
equation ax + by = c.
a D
Let t be converted into a continued fraction, and let — be the
b q
convergent just preceding j ; then aq — bp = ± 1.
I. If aq — bp=l, we have
ax +by = c (aq — bp);
.'. a(cq — x) = b(y + c2));
cq — x y + cp
.' . == — = — — — = L an integer ;
b a ° '
. ' . x = cq — bt, y ' — at  cp ;
INDETERMINATE EQUATIONS OF THE FIRST DEGREE. 287
from which positive integral solutions may be obtained by giving
CI) CO
to t positive integral values greater than — and less than j .
Thus the number of solutions is limited, and if there is no integer
fulfilling these conditions there is no solution.
II. If aq — bp =  1, we have
ax + by = — c (aq — bp) ;
.. a(x + cq) = b(cpy);
x + co en  y
• —  — = — = t. an integer ;
JL « ' 7
o a
. • . x=bt — cq, y = cj) — at ;
from which positive integral solutions may be obtained by giving
co c P
to t positive integral values greater than ~ and less than
As before, the number of solutions is limited, and there may be
no solution.
III. If either a or b is equal to unity, the solution may be
found by inspection as in Art. 317.
350. Given one solution in positive integers of Ike equation
ax + by = c, to find the general solution.
Let A, k be a solution of ax f by — c ; then ah + bk = c.
. ' . ax + by — ah + bk ;
.'. a (x — h) — b (k  y) ;
x —hk—y
.'. — 7 — = — — t, an integer ;
o a
.'. x = h + bt, y  k — at ;
which is the general solution.
351. To find the number of solutions in positive integers of the
equation ax + by = c.
Let T be converted into a continued fraction, and let  be the
b q
convergent just preceding j ; then aq — bp = at 1.
288 HIGHER ALGEBRA.
I. Let aq bp = l ; then the general solution is
x = cqbt, y = at ep. [Art. 349.]
Positive integral solutions will be obtained by giving to t
positive integral values not greater than °f , and not less
o
than — .
a
c c
(i) Suppose that  and T are not integers.
a b °
Let — = m+f. ± = n + a.
a J b y '
where m, n are positive integers and J\ g proper fractions ; then
the least value t can have is m+ 1, and the greatest value is n;
therefore the number of solutions is
cq cp . c j.
nvi = ± — +f g=—+fg.
b a J J ab J J
Now this is an integer, and may be written — + a fraction, or
ab
—r a fraction, according as /is greater or less than g. Thus the
number of solutions is the integer nearest to — , greater or less
according as/ or g is the greater.
(ii) Suppose that z is an integer.
In this case g  0, and one value of x is zero. If we include
c
this, the number of solutions is r+f, which must be an in
ao
teger. Hence the number of solutions is the greatest integer in
C C
^7+1 or j , according as we include or exclude the zero solution.
c
(iii) Suppose that  is an integer.
cc
In this case/=0, and one value of y is zero. If we include
this, the least value of t is m and the greatest is n; hence
the number of solutions is 71 — m + l. or —r  q + 1. Thus the
ab
INDETERMINATE EQUATIONS OF TJIE FIRST DEGREE. 289
c c
number of solutions is the greatest integer in 7 + 1 or —=■. ae
ab ab
cording as we include or exclude the zero solution.
c c
(iv) Suppose that  and 7 are both integers.
In this case f— and y = 0, and both x and y have a zero
value. If we include these, the least value t can have is m, and
the greatest is n ; hence the number of solutions is 11111+ 1, or
y + 1. If we exclude the zero values the number of solutions is
ab
4i.
ab
II. If aq bp=  1, the general solution is
x = bt — cq } y— cp — at,
and similar results will be obtained.
352. To find the solutions in positive integers of the equa
tion ax + by + cz — d, we may proceed as follows.
By transposition ax + by = d — cz ; from which by giving to z
in succession the values 0, 1, 2, 3, we obtain equations of
the form ax + by = c, which may be solved as already explained.
353. If we have two simultaneous equations
ax + by + cz=d, ax + b'y + cz = d\
by eliminating one of the unknowns, z say, we obtain an equation
of the form Ax + By = C. Suppose that x —f, y — g is a solution,
then the general solution can be written
x=f+Bs, y = gAs,
where s is an integer.
Substituting these values of x and y in either of the given
equations, we obtain an equation of the form Fs + Gz = II, of
which the general solution is
8 = h + Gt, z = k  Ft say.
Substituting for s, we obtain
x=f+Bh + BGt, y = gAhAGt;
and the values of x, y, z are obtained by giving to t suitable
integral values.
H. H. A. 19
290 HIGHER ALGEBRA.
354. If one solution in positive integers of the equations
ax + by + cz = d, ax + b'y + c'z = d',
can be found, the general solution may be obtained as follows.
Lety, g, h be the particular solution ; then
af+ bg + ch = d, a'f+ b'g + ch = d'.
By subtraction,
a(xf) + b(yg)+c(z h) = 0, 1
a'(x/) + b'(yg) + c'(zh) = 0; j
whence
xf = yg _ zh _ t
be — b'c ca — c'a ab' — a'b k '
where t is an integer and k is the H.C.F. of the denominators
be — b'c, ca — c'a, ab' — a'b. Thus the general solution is
x =f+ (be' — b'c) j , y — g + (ca' — c'a) =• , z = h + (ab'  a'b) ?.
/c fc fc
EXAMPLES. XXVI. j
Find the general solution and the least positive integral solution, of
1. 775.r711y = l. 2. 455#519y=l. 3. 436#393y = 5.
4. In how many ways can ,£1. 19s. 6d. be paid in florins and half
crowns ?
5. Find the number of solutions in positive integers of
lLe+15y=1031.
6. Find two fractions having 7 and 9 for their denominators, and
such that their sum is 1 £§.
7. Find two proper fractions in their lowest terms having 12
and 8 for their denominators and such that their difference is — .
24
8. A certain sum consists of x pounds y shillings, and it is half
of y pounds x shillings ; find the sum.
Solve in positive integers :
9. 6#+ty + 4s=122\ 10. 12.rlly + 4^=22
lhr + 8y 6^=145
21 10. 1 2x  1 \y + 4z = 221
5J ' 4.v+ 5y+ z=ll)
INDETERMINATE EQUATIONS OF THE FIRST DEGREE. 291
11. 20^21^=381 12. 13^ + 1 Is =103)
3y+ 4s =34/ ' 7z  by= 4J "
13. 7.r + 4y + 19^ = 84. 14. 23.r+17.y + lU = 130.
15. Find the general form of all positive integers which divided
by 5, 7, 8 leave remainders 3, 2, 5 respectively.
16. Find the two smallest integers which divided by 3, 7, 11 leave
remainders 1, 6, 5 respectively.
17. A number of three digits in the septenary scale is represented
in the nonary scale by the same three digits in reverse order ; if the
middle digit in each case is zero, find the value of the number in the
denary scale.
18. If the integers 6, «, b are in harmonic progression, find all the
possible values of a and b.
19. Two rods of equal length are divided into 250 and 243 equal
parts respectively ; if their ends be coincident, find the divisions which
are the nearest together.
20. Three bells commenced to toll at the same time, and tolled at
intervals of 23, 29, 34 seconds respectively. The second and third
bells tolled 39 and 40 seconds respectively longer than the first ; how
many times did each bell toll if they all ceased in less than 20 minutes?
21. Find the greatest value of c in order that the equation
7.r + 9y = c may have exactly six solutions in positive integers.
22. Find the greatest value of c in order that the equation
14r + lly=c may have exactly five solutions in positive integers.
23. Find the limits within which c must lie in order that the
equation 19x + 14y = c may have six solutions, zero solutions being
excluded.
24. Shew that the greatest value of c in order that the equation
ax + by = c may have exactly n solutions in positive integers is
(n + l)abab, and that the least value of c is (nl)ab + a + b } zero
solutions being excluded.
in o
CHAPTER XXVII.
RECURRING CONTINUED FRACTIONS.
355. We have seen in Chap. XXV. that a terminating con
tinued fraction with rational quotients can be reduced to an
ordinary fraction with integral numerator and denominator, and
therefore cannot be equal to a surd ; but we shall prove that a
quadratic surd can be expressed as an infinite continued fraction
whose quotients recur. We shall first consider a numerical
example.
Example. Express ^19 as a continued fraction, and find a series of
fractions approximating to its value.
x /19 = 4 + ( v /194) = 4+ Tl9 3  + ;
v
v /19 + 4_ 2 ,x/19 z _2_ 5 ,
3 + 3 V19 + 2 '
N /19 + 2_ 1j _ ^193 , . 2
=1+^ — = 1 +
5 5 \/19 + 3'
,/19 + S ^£98 5 .
2 2 \/19 + 3'
v /19 + 3_ 1 1 V 1 92_ 1 , 3
1 + 5 ~ 1 + N /
2 +
/L9 + 2 n iN /194 0i 1
3 ~ \/19 + 4'
N /19 + 4 = 8 + ( N /194) = 8 +
after this the quotients 2, 1, 3, 1, 2, 8 recur; hence
1 1 Jl_£ 1 2.
V 19  4 + 2+ 1+ 3+ 1+ 2+8+ •■••
It will be noticed that the quotients recur as soon as we come to a
quotient which is double of the first. In Art. 361 we shall prove that this is
always the case.
RECURRING CONTINUED FRACTIONS, 293
[Explanation. In each of the lines above we perform the same series of
operations. For example, consider the second line : we first find the
greatest integer in  —  — ; this is 2, and the remainder is  — 2, that
o 6
is ^— ^ — . We then multiply numerator and denominator by the surd
o
5
conjugate to ^192, so that after inverting the result . , we begin a
new line with a rational denominator.]
The first seven convergents formed as explained in Art. 336 are
4 9 13 48 61 170 1421
1 ' 2 ' 3 ' 11 ' 14 ' 39 ' 326 '
The eiTor in taking the last of these is less than , ' _ , and is therefore
less than . —  , or , and a fortiori less than 00001. Thus the
seventh convergent gives the value to at least four places of decimals.
356. Every periodic continued fraction is equal to one of the
roots of a quadratic equation of which the coefficients are rational.
Let x denote the continued fraction, and y the periodic part,
and suppose that
1 1 »
x = a+ z , ,
b + c +
1 1 1
h+ k+ y 1
1
1 1 1
and y — m +
n + u + v + y'
where a, b, c,...h, k, m, n,...u, v are positive integers.
p p
Let  , , be the convergents to x corresponding to the
quotients h, k respectively; then since y is the complete quotient,
p'y+p . pqx
we have x— , : wlience y = V — £ — , .
qy+q qxp
r r
Let  , — be tlie convergents to y corresponding to the
s s
r i/ + r
quotients u, v respectively ; then y= ~ .
Substituting for y in terms of x and simplifying we obtain a
quadratic of which the coefficients are rational.
294 HIGHER ALGEBRA.
The equation s'y 2 + {s — r) y — r = 0, which gives the value of
y, has its roots real and of opposite signs ; if the positive value of
v'y + p
ii be substituted in x = , — — , on rationalising the denominator
qy+q
the value of x is of the form ~ — , where A, B, C are integers,
G
B being positive since the value of y is real.
,1111
Example. Express l + s — 5 — ~— ^— ... as a surd.
1 1
Let x be the value of the continued fraction ; then x  1 = = — — — ;
£> + O + \X — 1)
whence 2x 2 + 2x  7 = 0.
The continued fraction is equal to the positive root of this equation, and
is therefore equal to ^—  — .
EXAMPLES. XXVII. a.
Express the following surds as continued fractions, and hnd the
sixth convergent to each :
1. v /3. 2. ^5. 3. y/6. 4. s/8.
5. v/11. 6. x /13. 7. x/14. 8. V22.
9. 2^3. 10. 4 v /2. 11. 3^5. 12. 4 N /10.
13  j& 14  V33 15  \/s 16  \/n
268
17. Find limits of the error when — — is taken for N /17.
65
916
18. Find limits of the error w T hen '— is taken for v/23.
19. Find the first convergent to N /101 that is correct to five places
of decimals.
20. Find the first convergent to VI 5 that is correct to five places
of decimals.
Express as a continued fraction the positive root of each of the
following equations :
21. x* + 2xl = 0. 22. a 8 4*? 3=0. 23. la? 8x 3=0.
24. Express each root of x 2  5^ + 3 = as a continued fraction.
Ill
25. Find the value of 3 + 5— x— x 
6+ 6+ 6 +
26. Find the value of , —  —
1+ 3+ 1+ 3 +
RECURRING CONTINUED FRACTIONS. 295
111111
1+ 2 + 3+ 1+ 2 + 3 +
1111
27. Find the value of 3+
28. Find the value of 5 + ,
1+ 1+ 1+ 10 +
29. Shew that
* + i+6+ i+ e+""~*\ 1+ a+ 2+ 3+ 2+ ; ■
30. Find the difference between the infinite continued fractions
111111 111111
1+ 3+ 5+ 1+ 3+ 5+ •"' 3+ 1+ 5+ 3+ 1+ 5+ ""
*357. To convert a quadratic surd into a continued fraction.
Let N be a positive integer which is not an exact square,
and let a x be the greatest integer contained in J N j then
N /iV = «, + (Jff a,) = «, + j£— , if r, = W »,\
Let b be the greatest integer contained in — ' ; then
JM+a l = b  JNb x r x + a x ^ h , JNa 2 ^ h
r i
+
where « 2 = b i r 1 — a x and r x r 2 = N — a„ 2 .
Similarly
r 2 2 »* 2  JN + a./
where « 3 = bf 2 — a s and r 2 r 3 — N — a 3 2 ;
and so on ; and generally
JN+a , . JNa . r
 —  s=i = b , + v — " = b , + , "
"' jy + a '
> it
Ml
H— 1 ' (11
where a n = &„_,/•„_,  a„_ 1 and ?•„_,*•„ = N  a/.
1111
Hence *JN= a, +
and thus JN can be expressed as an infinite continued fraction.
"We shall presently prove that this fraction consists of re
curring periods ; it is evident that the period will begin when
ever any complete quotient is first repeated.
296 HIGHER ALGEBRA.
We shall call the series of quotients
JA r + a, JN + a 2 JJST+ a
JA T .
r x r 2 r 3
3
)
the first, second, third, fourth complete quotients.
*358. From the preceding article it appears that the quan
tities a v r v b v b , b 3 are positive integers; we shall now prove
that the quantities a 2 , a 3 , a 4 , , r 8 , r 3 , r 4 , . . . are also positive in
tegers.
p p p
Let — , —.. —r. be three consecutive convergents to JN. and
q q q ° x
P"
let — be the convergent corresponding to the partial quotient b n .
The complete quotient at this stage is — ; hence
v^=
— p + p
r , t = P JW+a„P+r v p
Clearing of fractions and equating rational and irrational
parts, we have
«y + r nP = ^Y> c k<l + r n q =p ;
whence a n ( pq  pq) =pp* ~ <Z<7 '^j r n {ptf —p<i) = A T q' 2 —p' 2 .
But pq' —p'qssdslf and pq —pq, pp' —qq'N, Nq 2 — p 2 have
the same sign [Art. 344] ; hence a n and r n are positive integers.
Since two convergents precede the complete quotient 
r*
this investigation holds for all values of n greater than 1.
*359. To prove that the complete and partial quotients recur.
In Art. 357 we have proved that r n r n _ l = N—a 2 . Also r n and
r n _ l are positive integers ; hence a n must be less than ^/JV, thus
a n cannot be greater than a v and therefore it cannot have any
values except 1, 2, 3, ...a x ' } that is, the number of different values of
a n cannot exceed a x .
Again, a n+1 =r v b u a h , that is r n b n = a n + a n+v and therefore
r n b n cannot be greater than 2a l ; also b n is a positive integer ;
hence r n cannot be greater than 2a v Thus r n cannot have any
values except 1, 2, 3,...2a 1 ; that is, the number of different values
ofi\ cannot exceed 2a r
RECURRING CONTINUED FRACTIONS. 297
Thus the complete quotient — — cannot have more than
r n
2a* different values ; that is, some one complete quotient, and
therefore all subsequent ones, must recur.
Also b n is the greatest integer in — — ; hence the partial
r
n
quotients must also recur, and the number of partial quotients in
each cycle cannot be greater than 2a
2
']
*3G0. To prove that a, < a u + r n .
We have «,_, + a n = b H _ 1 r n _ l ;
«»_i + a «= or >?, t ,i 5
since 6„_ l is a posit ive integer ;
But N"a;=r n r n _ l  i
a i ~ a n < r n ,
which proves the proposition.
*361. To shew that the period begins loith the second partial
quotient and terminates ivith a partial quotient double of the first.
Since, as we have seen in Art. 359, a recurrence must take
place, let us suppose that the (n+ l) th complete quotient recurs at
the (*+ l) th ; then
a. = a , r, = r , and b. = b ;
we shall prove that
a, . =a , . r m , = r , , b, = b , .
4 — 1 n — 1' »— 1 ii — 1' *— 1 ii — 1
We have
r. , r = N a, 2 ' — iV — a 2 = r ,r —r . r, ;
* — i * » it H — i ii H — l * '
v = r
Again,
a , _ a * 1 7 7
.. " ' ~ b , — o m . = zero, or an integer.
n1
298 HIGHER ALGEBRA.
But, by Art. 360, a l a n _ l <.r ii _ x , and a l a s _ l <r s _ i ; that is
a. a. , < r ' , : therefore a ,  a a , < r . ; hence ^ — is less
n 1
than unity, and therefore must be zero.
Thus «,_! = «„_!, and also 6 # _ 1 = 6 fl _i.
Hence if the (n + l) th complete quotient recurs, the ?^ th com
plete quotient must also recur; therefore the (n l) th complete
quotient must also recur; and so on.
This proof holds as long as n is not less than 2 [Art. 358],
hence the complete quotients recur, beginning with the second
quotient —  . It follows therefore that the recurrence
1 r x
begins with the second partial quotient b x ; we shall now shew
that it terminates with a partial quotient 2a x .
Let  "be the complete quotient which just precedes the
second complete quotient  when it recurs ; then — — a
1 H
anc j v l ar e two consecutive complete quotients ; therefore
»",
but N a* = r, ; hence r n = 1.
Again, a y — a H < ?'„, that is < 1 ; hence a x  a n  0, that is
«« = «,•
Also a n + a = r n b n — b n ; hence b n = 2a i ; which establishes the
proposition.
*362. To shew that in any period the partial quotients equi
distant from the beginning and end are equal, the last partial
quotient being excluded.
Let the last complete quotient be denoted by * — ; then
r n =l, ci n = a x , b n =2a r
We shall prove that
^ 2 =^ 2) «h2=« 3 > ^.2= & 2^
r.
RECURRING CONTINUED FRACTIONS. 299
We have
r«x = r* r H _ t =N a,; N  a;  r, .
Also
»„_, + a x = «„_, + a H = r m _, &„_, = r, &„_, ;
and «,+«., = ?•,&,;
. . =  = 0, — o M1 = zero, or an integer,
i
But "' I ^ < CT i ~ a "~ l , that is < a '~ a ' 1 , which is less than
unity ; thus a 2  a n _ x = • hence a,,.! = « 2 , and o,,^ = b l .
Similarly r n _ 2 = r 2 , «„_ 2 = « 3 , 6„_ 2  b 2 ; and so on.
*363. From the results of Arts. 3G1, 362, it appears that
when a quadratic surd v /iV r is converted into a continued fraction,
it must take the following form
J_ J_ J_ J_ J_ J 1_
1 &i + & 3 +6 3 + ° 3 + & 2 + °i +2a,+
*364. To obtain the penultimate convergents of the recurring
periods.
Let n be the number of partial quotients in the recurring
period ; then the penultimate convergents of the recurring periods
are the ?i ih , 2n th , 3n th , convergents ; let these be denoted by
V \ ^=, ^, respectively.
xt /v 111 11
Now JiV = a l + i — j — —  — — 
v b i + b 2 + b 3 + b 7l _ l + 2a l +
7)
so that the partial quotient corresponding to — +1 is 2a t ; hence
Pn+X = ^Pn+Pnl
SWl " 2 «1 9n + ?nl '
Tlie complete quotient at the same stage consists of the period
2«,+T r : ,
b i + K + 6 «i +
300 HIGHER ALGEBRA,
and is therefore equal to a x + J 'N ; hence
Clearing of fractions and equating rational and irrational
parts, we obtain
*iP.+JVi = jfy«i a i9 , H + 9 r ni=^« (!)•
Again — can be obtained from — and ^ by taking for the
quotient
1 1 1
2 * 1 + V?V^ C
which is equal to rtj + — . Thus
in
& = 1 2^ = & , from (1);
** U+%)q n + q^ P« + %.q n
? 2 „
. l(A + *&) (2)>
In like manner we may prove that if —  is the penultimate
icn
convergent in the c th recurring period,
«i ^c« +Fcni = Nq mi a, q cn + q m _ l =#*,
and by using these equations, we may obtain £— , — , suc
cessively.
It should be noticed that equation (2) holds for all multiples
of n ; thus
Ol
the proof being similar to that already given.
*365. In Art. 356, we have seen that a periodic continued
fraction can be expressed as the root of a quadratic equation
with rational coefficients.
RECURRING CONTINUED FRACTIONS. 301
Conversely, we might prove by the method of Art. 357 that
an expression of the form — tt~~ > where <4> B, C are positive
integers, and B not a perfect square, can be converted into a
recurring continued fraction. In this case the periodic part will
not usually begin with the second partial quotient, nor will
the last partial quotient be double the first.
For further information on the subject of recurring continued
fractions we refer the student to Serret's Cours cVAlgebre Supe
rieure, and to a pamphlet on The Expression of a Quadratic Surd
as a Continued Fraction, by Thomas Muir, M.A., F.R.S.E.
^EXAMPLES. XXVII. b.
Express the following surds as continued fractious, and find the
fourth convergent to each :
1. N /a 2 + l. 2. J a* a. 3. N /«l.
4. V /T7T. 5 . y«"^f . 6 . ^l
7. Prove that
J9a* + 3 = 3a+ — — i
2a + 6a + 2a + 6a +
and find the fifth convergent.
8. Shew that
2 1111 i—» — r
p i+ p+ i+ p
9. Shew that
/111 \ 111
V a \\ =P a \ + — —
\ P9 Cl 2 + tt 3 + P9 Ct 4 + / aCL l + P a Z + OCl \ +
10. If Ja' 2 + 1 be expressed as a continued fraction, shew that
2(a*+l)q n =p n _ 1 +p n + 1 , 2p n = q n _ l + q n + l .
11 Tf 1111
11. If .%'= — ...,
a x + « 2 + a i+ a 2 +
1111
• ?/ ~2a 1 + 2« 2 + 2a x + 2a 2 + ""'
1111
~3tf 1 +~3a 2 + 3^ + 3« 2 + '"'
shew that x {f  z) + 2y (z 2  .r 2 ) + 3z {a?  y 1 ) = 0.
302 HIGHER ALGEBRA.
12. Prove that
(
JL J_ Jl J_ V———— ^ = 
b + a+ b + a+ '")\b+ a+ b+ a+ '")"b
1 J_ J_ J_
13. U X ~ a+ b+ b+ a+ a+ ••'
J_ J_ J_ J_
y b + a+ a+ i+ b+ •••'
shew that (ab 2 + a + b)x (a 2 b + a + b)y = a 2 b 2 .
14. If — be the n th convergent to Ja 2 + l. shew that
P2 2 +P 3 2 +'"+P 2 n + l = Pn + lPn + 2PlP2
15. Shew that
1 1 1 \ l + bc
(— —
\a+ b +
c +
v a+ b + c+ ' "/ \ ^+ a + c + / l+ctb'
16. If — denote the r ih convergent to ^— — — , shew that
q r & 2
Pi+Pi>+ >~+P°.nl=P2np<L, ? 3 + ?5 + ■  + &»  1 = ?8» ~ ft.
17. Prove that the difference of the infinite continued fractions
_i_j_2_ i i i
a+ b+ c+ •' b+ a+ c+ '••'
is equal to =■ .
1 + ao
18. If s/JV is converted into a continued fraction, and if n is the
number of quotients in the period, shew that
19. If \/^ De converted into a continued fraction, and if the pen
ultimate convergents in the first, second, ...k th recurring periods be
denoted by n lt n 2i ...n k respectively, shew that
*CHAPTER XXVIII.
INDETERMINATE EQUATIONS OF THE SECOND DEGREE.
*366. The solution in positive integers of indeterminate
equations of a degree higher than the first, though not of much
practical importance, is interesting because of its connection with
the Theory of Numbers. In the present chapter we shall confine
our attention to equations of the second degree involving two
variables.
*367. To shew Iww to obtain the positive integral values of
x arid y which satisfy the equation
ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0,
a, b, c, f, g, h being integers.
Solving this equation as a quadratic in x, as in Art. 127, we
have
ax + hy+g = ±J(h 2 ab) y 2 + 2 (hg  af)y+(g 2 ac)...(l).
Now in order that the values of x and y may be positive
integers, the expression under the radical, which we may denote
kv py 2 + 2gy + r, must be a perfect square ; that is
py 2 + 2qy + r = z 2 , suppose.
Solving this equation as a quadratic in y, we have
py + q = ± Jq 2 jjr+pz 2 ;
and, as before, the expression under the radical must be a perfect
square ; suppose that it is equal to t 2 ; then
t 2  pz 2 = q 2  pr^
where t and z are variables, and j\ q, r are constants.
304 HIGHER ALGEBRA.
Unless this equation can be solved in positive integers, the
original equation does not admit of a positive integral solution.
We shall return to this point in Art. 374.
If a, b, h are all positive, it is clear that the number of
solutions is limited, because for large values of x and y the sign
of the expression on the left depends upon that of ax 2 + 2hxy + by 2
[Art. 2G9], and thus cannot be zero for large positive integral
values of x and y.
Again, if h* — ab is negative, the coefficient of y 2 in (1) is
negative, and by similar reasoning we see that the number of
solutions is limited.
Example. Solve in positive integers the equation
a; 2  4xy + &y*  2x  20*/ = 29.
Solving as a quadratic in x, we have
x = 2y + 1 ± ^30 + 24//  2y\
But 30 + 24?/  2j/ 2 = 102  2 (y  G) 2 ; hence (y  6) 2 cannot be greater than
51. By trial we find that the expression under the radical becomes a
perfect square when (y6) 8 =l or 49; thus the positive integral values of y
are 5, 7, 13.
When ?/ = 5, x = 21 or 1; when y = 7, x = 25 or 5; when y = 13,
x = 29 or 25.
*3G8. We have seen that the solution in positive integers
of the equation
ax 2 + 2hxy + by 2 + 2gx + 2fy + c =
can be made to depend upon the solution of an equation of the
form
x 2 ± Ny 2 = ± a,
where iV* and a are positive integers.
The equation x 2 + Ny* = — a has no real roots, whilst the
equation x 2 + Ny 2 = a has a limited number of solutions, which
may be found by trial ; we shall therefore confine our attention
to equations of the form x 2  Ny 2 = ± a.
*369. To sJiew that the equation x 2 Ny 2 =l can always be
solved in positive integers.
Let JN be converted hito a continued fraction, and let
2_ l '— be any three consecutive convergents; suppose that
q q 9
INDETERMINATE EQUATIONS OF THE SECOND DEGREE. 305
t 17" a,/ ,n
is the complete quotient corresponding to „ ; then
*. (/"/ ~ V'q) W P" 2 [Art. 358].
But r„ = 1 at the end of any period [Art. .'3(51] ■
.. ]> — JVq " ] } ( 1 — ] )( 2 )
/
, being the penultimate convergent of any recurring period.
If the number of quotients in the period is even, , i.s an even
convergent, and is therefore greater than v /iV, and therefore
P
greater than ; thus pq —pq = 1. Jn this case p'* — N"q' a = J,
and therefore x=]>\ y = q is a solution of the equation xr — Ny* = 1.
p
Since — is the penultimate convergent of any recurring
period, the number of solutions is unlimited.
If the number of quotients in the period is odd, the penultima 1 1
convergent in the tirst period is an odd convergent, but the
penultimate convergent in the second period is an even convergent.
Thus integral solutions will be obtained by putting x=p', y — q\
where — is the penultimate convergent in the second, fourth,
q
sixth, recurring periods. Hence also in this case the number
of solutions is unlimited.
*370. To obtain a solution, in positive inte<iers of the equation
As in the preceding article, we have
f£ ~KT '2 ' I
v Jq =pqpq>
If the number of quotients in the period is odd, and if
<1
>
is an odd penultimate convergent in any recurring period, ,<,
and therefore pq —pq ' — — 1.
In this case p' 2 — Nq 2  \, and integral solutions of the
equation x 2 — X y 1 = — 1 will be obtained by putting x =p\ y — q ',
where — is the penultimate convergent in the first, third, fifth...
q
recurring periods.
,, „ nappeu i.H.10 we can discover , irv
11. n. A. 1:0
2fc — U
306 HIGHER ALGEBRA.
Example. Solve in positive integers x 2  13y 2 = ±1.
We can shew that
11111
^133 + 1+ 1+ITl+ 6+
Here the number of quotients in the period is odd ; the penultimate con
18
vergent in the first period is = ; hence a; = 18, y = o is a solution of
x 2 13y 2 =l.
By Art. 364, the penultimate convergent in the second recurring period is
1 / 18 5 io\ ,u * • 649
2U + 18 Xl3 J' thatlS '180'
hence # = 649, y = 180 is a solution of x 2  13y 2 =l.
By forming the successive penultimate convei'gents of the recurring
periods we can obtain any number of solutions of the equations
x 2  13?/ 2 =  1, and x 2  lSy 2 = + 1.
*371. When one solution in positive integers of x 2 — Nif = 1
lias been found, we may obtain as many as w r e please by th
following method.
Suppose that x = h, y = k is a solution, h and k being positive
integers; then (A 2 — Nk 2 ) n = 1 , where n is any positive integer.
Thus x 2 Ni/= (h*  m?y .
. •. (x + yjN) (x  yJJST) = (h + kJN) n (h  k s !X)\
Put x + yJN = (h + kJX)' 1 , xy JN = (h  kJN)" ;
.. 2x = (h + kJN) n + (hkJJYy;
2 Us in = (h + kjiry  (h  kjNy.
Tlie values of x and y so found are positive integers, and by
ascribing to n the values 1, 2, 3,..., as many solutions as we please
can be obtained.
Similarly if x = h i y = k is a solution of the equation
x 2 — Xy 2 = — 1, and if n is any odd positive integer,
x*  Ntf = (h 2  Nk 2 )\
Thus the values of x and y are the same as already found, but
n is restricted to the values 1, 3, 5,
*372. By putting x = ax', y = ay the equations x 2 — Ny 2 — ± a 3
become of 2 — IFy f * = d= 1, which we have already shewn how to
solve.
INDETERMINATE EQUATIONS OF THE SECOND DEGREE 307
*373. We have seen in Art. 3G9 that
P n ~ N( f~ = ~ r n ( V<1 ~ V<l) = * r..
Hence if a is a denominator of any complete quotient which
occurs in converting JX into a continued fraction, and if — is
i • q
the convergent obtained by stopping short of this complete
quotient, one of the equations x~ — Xy 2 — ±a is satisfied by the
values x =p\ y = q'
Again, the odd convergents are all less than JN, and the
even convergents are all greater than JX ; hence if — , is an even
<2
t
convergent, x=p, y = q is a solution of x* Xif ' — a: and if —.
q
is an odd convergent, x =p , y — q is a solution of X s — Xy 2 = — a.
*374. Tlie method explained in the preceding article enables
us to find a solution of one of the equations x 2 — Xy 2 — ±a only
when a is one of the denominators which occurs in the process of
converting JX into a continued fraction. For example, if Ave
convert J7 into a continued fraction, we shall find that
and that the denominators of the complete quotients are 3, 2, 3, 1.
The successive convergents are
2 3 5 8 37 45 82 127
1' 1' 2' 3' IT 17' 31' 18 ' ;
and if we take the cycle of equations
2 I" 2 O 2 *" 2 CI o f o o o h' 2 1
x  ty = — 3, ar — Iff = 2, ar — 7^ = — 3, cc  / y = 1 ,
we shall find that they are satisfied by taking
for x the values 2, 3, 5, 8, 37, 45, 82, 127,
and iovy the values 1, 1, 2, 3, 14, 17, 31, 48,
*375. It thus appears that the number of cases in which solu
tions in integers of the equations x 2 — Xy 2 = ± a can be obtained
with certainty is very limited. In a numerical example it may,
however, sometimes happen that we can discover by trial a
20—2
308 HIGHER ALGEBRA.
positive integral solution of the equations x 2 — Ny 2 = =*= a, when a
is not one of the above mentioned denominators ; thus we easily
find that the equation # 2 7?/ 2 = 53 is satisfied by y=2, # = 9.
When one solution in integers has been found, any number of
solutions may be obtained as explained in the next article.
*376. Suppose that x =f, y = g is a solution of the equation
x 2 _ Ny 2 = a ; and let x = h, y  k be any solution of the equation
x 2  JSfy 2 = 1 ; then
x* ~ Ny 2 = (f 2  Kg 2 ) (h 2  Nk 2 )
= (fh±Ngk) 2 N{fk±gh)\
By putting x fh ± Kgk, y fk ± gh,
and ascribing to h, k their values found as explained in Art. 371,
we may obtain any number of solutions.
*377. Hitherto it has been supposed that N is not a perfect
square ; if, however, N is a perfect square the equation takes the
form x 2  n 2 y 2 = a, which may be readily solved as follows.
Suppose that a = be, where b and c are two positive integers,
of which b is the greater ; then
(x + ny) (x — ny) = be.
Put x + ny = b, x  ny = c ; if the values of x and y found
from these equations are integers we have obtained one solution
of the equation ; the remaining solutions may be obtained by
ascribing to b and c all their possible values.
Example. Find two positive integers the difference of whose squares is
equal to 60.
Let x, y be the two integers ; then ,xr  y 2 = 60 ; that is, (as + y) (x  y) = 60.
Now 60 is the product of any of the pair of factors
1,60; 2,30; 3,20; 4,15; 5,12; 6,10;
and the values required are obtained from the equations
ic + y = 30, # + y = 10,
xy= 2; xy= 6;
the other equations giving fractional values of x and y.
Thus the numbers are 16, 14; or 8, 2.
INDETERMINATE EQUATIONS OF THE SECOND DEGREE. 309
Cor. In like manner we may obtain the solution in positive
integers of
ax' + 2hxy + by 1 + %jx + *2/'y + c = k,
if the lefthand member can be resolved into two rational linear
factors.
*378. If in the general equation a, or b, or both, are zero,
instead of employing the method explained in Art. 3G7 it is
simpler to proceed as in the following example.
Example. Solve in positive integers
2.ry  4a 2 + V2x  5y = 11 .
Expressing y in terms of x, we have
4a; 2  12* + 11 „ , 6
V= — ^rr — =2tfl+;
2x  5 2x  5
n
In order that y may be an integer = must be an integer ; hence 2.r  5
2iX — O
must be equal to ± 1, or ± 2, or ± 3, or ± G.
The cases ±2, ±6 may clearly be rejected; hence the admissible values
of x are obtained from 2x  5 = ± 1, 2x  5 = ± 3 ;
whence the values of .x are 3, 2, 4, 1.
Taking these values in succession we obtain the solutions
x = S, y = ll; s=2, y = 3; # = 4, ?/ = 9; ar=l, y= 1;
and therefore the admissible solutions are
a; = 3, y = 11; x = 4, y = 9.
*379. The principles already explained enable us to discover
for what values of the variables given linear or quadratic
functions of x and y become perfect squares. Problems of this
kind are sometimes called DiopJiantine Problems because they
were first investigated by the Greek mathematician Diophantus
about the middle of the fourth century.
Example 1. Find the general expressions for two positive integers which
are such that if their product is taken from the sum of their squares the
difference is a perfect square.
Denote the integers by x and y ; then
x xy + y' 2 = z' 2 suppose ;
.. x(xy) = z 2 y.
This equation is satisfied by the suppositions
mx= n {z + y), n (x y) = m (z  y),
where m and n are positive integers.
310 HIGHER ALGEBRA.
Hence mx  ny  nz = 0, nx + (m n)y mz  0.
From these equations we obtain by cross multiplication
x _ V 2 .
2mn  ri 2 m 2  n 2 m 2  mn + ri 1 '
and since the given equation is homogeneous we may take for the general
solution
x = 2mnn 2 , y = m 2 n 2 , z = m 2 vin + n 2 .
Here m and n are any two positive integers, »i being the greater; thus if
?n = 7, n = 4, we have
x = ±0, y = SS, 3 = 37.
Example 2. Find the general expression for three positive integers in
arithmetic progression, and such that the sum of every two is a perfect
square.
Denote the integers by xy, x, x + y; and let
2xy=p 2 , 2x = q 2 , 2x + y = r 2 ;
then p 2 + r 2 = 2q 2 ,
or i*q*=q*p>.
This equation is satisfied by the suppositions,
m (r  q) = n (q  j>), n (r + q) = m (q +p),
where m and n are positive integers.
From these equations we obtain by cross multiplication
V = <l _ r
w 2 + 2mnm 2 m 2 + n 2 m 2 + 2mn n 2 '
Hence we may take for the general solution
p=n* + 2mnm*, q = m 2 + n 2 , r = )u 2 + 2mnu 2 ;
whence x = = {m 2 + w 2 ) 2 , y = inin (m 2  w 2 ) ,
and the three integers can be found.
From the value of x it is clear that m and n are either both even or both
odd ; also their values must be such that x is greater than y, that is,
(m 2 + n 2 ) 2 >8mn{m 2 n 2 ),
or m z (m  Sn) + 2inn 2 + 8m n* + n 4 > ;
which condition is satisfied if m>Sn.
If m = 9, w=l, then a = 3362, y =2880, and the numbers are 482, 33G2,
6242. The sums of these taken in pairs are 3844, 6724, 9604, which are the
squares of 62, 82, 98 respectively.
INDETERMINATE EQUATIONS OF THE SECOND DEGREE. 311
*EXAMPLES. XXVIII.
Solve in positive integers :
1. 5a 2 10.iv/ + 7?/ 2 = 77. 2. 7^2^+3y 2 =27.
3. y 2 4.ry + 5.r 2 10.i = 4. 4. xy  2.v  y = S.
5. 3.y + 3.ry4j/ = 14. 6. 4^ 2 y 2 =315.
Find the smallest solution in positive integers of
7. .r 2 14y 2 =l. 8. ^19^=1. 9. .t 2 = 4iy 2 l.
10. x 2  61/ + 5 = 0. 11. x 2 7y 2 9 = 0.
Find the general solution in positive integers of
12. .r 2 3/=l. 13. x 2 5y 2 =l. 14. .v 2  17y 2 =  1.
Find the general values of x and y which make each of the following
expressions a perfect square :
15. x 2 3xy + 3y 2 . 16. afi+2xy + 2f. 17. 5^+y 2 .
18. Find two positive integers such that the square of one exceeds
the square of the other by 105.
19. Find a general formula for three integers which may be taken
to represent the lengths of the sides of a rightangled triangle.
20. Find a general formula to express two positive integers which
are such that the result obtained by adding their product to the sum
of their squares is a perfect square.
21. " There came three Dutchmen of my acquaintance to see me,
being lately married ; they brought their wives with them. The men's
names were Hendriek, Claas, and Cornelius; the women's Geertruij,
Catriin, and Anna : but I forgot the name of each man's wife. They
told me they had been at market to buy hogs ; each person bought as
many hogs as they gave shillings for one hog; Hendriek bought 23 hogs
more than Catriin; and Claas bought 11 more than Geertruij ; likewise,
each man laid out 3 guineas more than his wife. I desire to know the
name of each man's wife." (Miscellany of Mathematical Problems, 1743.)
22. Shew that the sum of the first n natural numbers is a perfect
square, if n is equal to k 2 or k' 2  1, where k is the numerator of an odd,
and k' the numerator of an even convergent to N ^2.
CHAPTER XXIX.
SUMMATION OF SERIES.
380. Examples of summation of certain series have occurred
in previous chapters ; it will be convenient here to give a
synopsis of the methods of summation which have already been
explained.
(i) Arithmetical Progression, Chap. IV.
(ii) Geometrical Progression, Chap. Y.
(iii) Series which are partly arithmetical and partly geo
metrical, Art. 60.
(iv) Sums of the powers of the Natural Numbers and allied
Series, Arts. 68 to 75.
(v) Summation by means of Undetermined Coefficients,
Art. 312.
(vi) Recurring Series, Chap. XXIY.
We now proceed to discuss methods of greater generality ;
but in the course of the present chapter it will be seen that some
of the foregoing methods may still be usefully employed.
381. If the r th term of a series can be expressed as the dif
ference of two quantities one of which is the same function of r
that the other is of r  1 , the sum of the series may be readily
found.
For let the series be denoted by
and its sum by S , and suppose that any term u r can be put in
the form v r v r _ 1 ; then
^.=(«i0+( w .«i)+(*.«f) + » + ( w .i0 ! + (*. ,, i)
= v  v n .
SUMMATION OF SERIES. 313
Example. Sum to n terms the series
1 1 1
+ U . »„v„ . n T + „ . , . +
(l + s)(l+2s) (l + 2u)(l + 3.r) (l + 3j)(l + 4ar)
If we denote the series by
*±(——\
 a;\l + 2x 1 + 3*/'
_!/ 1 1 \
Ws a;\l + 3# 1 + 4*,/ '
■ x\l + nx i+ n+ i.x/
b}' addition, SL= I ^ — I
ar\l + a; l +w + l.a?/
n
(1 + x) (1 + n + l ..r)
382. Sometimes a suitable transformation may be obtained
by separating u into partial fractions by the methods explained
in Chap. XXIII.
Example. Find the sum of
1 a a 2
+ n, x~n z~r + t, ., > ,, 5— v + . . . to n terms.
(l + x)(l + ax) (l + ax)(l+a*x) (1 + ax) (1 + a 3 x)
nu, ft* « n_1 A B
The n th term= — n . w , — = —. t + — suppose:
{l + a n  1 x)(l + a n x) l + a" 1 ^ l + a n x **
.'. a n ~ x A (1 + a**) + B (1 + a 7 ' 1 *).
By putting 1 + «" _1 .r, 1 + a u x equal to zero in succession, we obtain
a nl n n
A= , B= —
1 a' 1 a'
1/1 a
Hence u, = —
1 1  a \1 + a: 1 + ax J
.... 1 / a a 2 \
similarly, t**=; , — 5 — 5 .
1 / a" 1 a w \
Wn ~la Vl+a*" 1 * l + a n j/
'"• *~lo\l + * l + a n x)'
314 HIGHER ALGEBRA.
383. To jind the sum of\\ terms of a series each term of which
is composed of r factors in arithmetical progression, the first factors
of the several terms being in the same arithmetical progression.
Let the series be denoted by u x + u 2 + u s + + u n ,
where
u„  (a + nb) (a + n + 1 . b) (a + n + 2 . b) ... (a + n + r — 1 . b).
Replacing n by n— 1, we have
«„ ! = (a + n — 1 . &) (a + nb) (a + n + 1 . 6) . .. (a + n + r — 2 . b) ;
k »i
.'. (« + ?i — 1 . b) u n = (a + n + /• — 1 . b) ?.«.„_! = v n , say.
Replacing n by n + 1 we have
(a + w + r. 6)tf» = « Il+1 j
therefore, by subtraction,
(r+l)b . u n = v n+ iv n .
Similarly, (7+1)6. w B _, = r /(  /<„_,,
(r+ 1) b .u. 2 = v s v 2f
(r + 1) b . «*! = w 2 — ^i
By addition,
(r + l)&.#, y ft+ i»
that is, S n
(r+l)6
1 >
(a 4 n + r . £A u n ~
= (r+l)i +6 ' S ' ly;
where C is a quantity independent of n, which may be found by
ascribing to n some particular value.
The above result gives us the following convenient rule :
Write down the n th term, affix the next factor at the end, divide
by the number of factors thus increased and by the common differ
ence, and add a constant.
It may be noticed that C = — , \ — T = — , 5tt m, ; it is
J (r+l)b (r+l)b lf
however better not to quote this result, but to obtain C as above
indicated.
SUMMATION OF SERIES. 315
Example. Find the sum of n terms of the series
1.3.5+3.5.7+5.7.9+
The n th term is {2n  1) (2« + 1) (2n + 3) ; hence by the rule
 _ (2wl)(2n + l)(2n + 3)(2n+5) , n
>\ ^ +C7.
To determine C, put n = 1 ; then the series reduces to its first term, and
we have 15 = —  — j — : — f C ; whence C = — ;
8 8
(2nl)(2n+l)(2n+3)(2n+5) 15
•'• S *~ 8 + 8"
= n (2n 3 + 8« 2 + In  2), after reduction.
384. The sum of the series in the preceding article may
also be found either by the method of Undetermined Coefficients
[Art. 312] or in the following manner.
We have u n = (2w  1) (2w + 1) (2w + 3) = $n 3 + 12>i 2  2m  3;
. \ S m = 82?i 3 + 122^ 2  22m  3m,
using the notation of Art. 70 ;
. \S U = 2m 2 (m + l) 2 + 2m (n + 1) (2n + 1)  n (n + 1)  3m
= w(2m 3 + 8m 2 + 7?i2).
385. It should be noticed that the rule given in Art. 383 is
only applicable to cases in which the factors of each term form an
arithmetical progression, and the first factors of the several terms
are in the same arithmetical progression.
Thus the sum of the series
1.3. 5 + 2.4. 6 + 3.5. 7 + to n terms,
may be found by either of the methods suggested in the preceding
article, but not directly by the rule of Art. 383. Here
u n = n (m + 2) (m + 4)  n (m +1 + 1 ) (m + 2 + 2)
= n{n+ l)(™ + 2) + 2n(n+ \) + u(a + '2) + 2/4
= n (m + 1) (m + 2) + 3m (n + 1) + 3m.
The rule can now be applied to each term ; thus
S n = \n (m+ 1) (m + 2)(m + 3) +n (n+ 1) (m + 2) + « (»+ 1) + C
\ )> (r h l)(/r»l) (» + 5), the constant being zero.
316 HIGHER ALGEBRA.
386. To find the sum of\\ terms of a series each term of which
is composed of the reciprocal of the product of r factors in arith
metical progression, the first factors of the several terms being in
tlie same arithmetical progression.
Let the series be denoted by u x + it, + u :i + + u n ,
where
— = (a + nb) (a + n + 1 . b) (a + n + 2 . b) (a + n + rl .b).
4i
U n
M„_i
Replacing n by n  1,
— = (a + n  1 . b) (a + nb) (a + n + 1 . b) ...(a + n + r—2 . b) ;
l nl
.'. (a + n + rl . b) u n = (a + n  1 . b) u n _ x = v ni say.
Replacing n by n + 1, we have
(a+nb)u n = v n+1 ;
therefore, by subtraction,
(rl)b. u n = v n v n+1 ,
Similarly (r  1) b . u n _ x = v ll _ l  v n ,
(r — l)b . u. 2 = v. 2 — v St
(r— 1) b . Wj = v x — v. 2 .
By addition, (r — 1) b . S n = v x — v n+1 ;
, , , . Q ^i  ?W _ r _ ( a + nb) u n
tnatis *"(rl)6~ U (rl)6 '
where C is a quantity independent of n, which may be found by
ascribing to n some particular value.
Thus S n = C , * . =  , .
(rl)6 (a + n+L.b)... (a + n + rl. b)
Hence the sum may be found by the following rule :
Write doivn the n th term, strike off a factor from the beginning,
divide by the number of factors so diminished and by the common
difference, change the sign and add a constant.
The value of C= , Vv~7 = t tti u i '■> but ** i s advisable in
(r — 1) o (r — 1) 6
each case to determine C by ascribing to n some particular value.
SUMMATION OF SERIES. 317
Example 1. Find the sum of re terms of the series
The re' 1 ' term is 
1.2.3.4 + 2.3.4.5 + 3.4.5.6 +
1
»(n + l)(n + 2)(n + 3)'
hence, hy the rule, we have
3(n+l)(w + 2)(» + 3)
rut »=1, then ^=03^; whence (7=1;
• 5 X *
" 18 3(re + l)(re + 2)(re + 3)
By making n indefinitely great, we obtain fi^ = — .
Example 2. Find the sum to n terms of the series
3 4 5
+ a— T7—  + r t — r— 5 +
1.2.42.3.5 3.4.6
Here the rule is not directly applicable, because although 1,2,3, ,
the first factors of the several denominators, are in arithmetical progression,
the factors of any one denominator are not. In this example we may
proceed as follows :
n + 2 (n+2) 2
" re(re+l)(re + 3) n{n+l) (n + 2) (re + 3)
re (re 41) + 3re + 4
: re(re + l)(re + 2)(re + 3)"
1 3
(re + 2)(w + 3) (re + l)(re + 2)(re + 3) w(re+l)(n + 2)(re + 3)'
Each of these expressions may now be taken as the ?i th term of a series
to which the rule is applicable.
• S c l 3 4
n + S 2(re + 2)(re + 3) 3 (re+ 1) (re + 2) (re + 3) '
put re=l, then
313 4 29
17271= C "4 " 27174 " 372.3 .4' Whence C = 36 ;
_29 1 3 4
n 36 re + 3 2 (re + 2) (re + 3) 3 (re + 1) (re + 2) (re + 3)'
318 HIGHER ALGEBRA.
387. In cases where the methods of Arts. 383, 386 are directly
applicable, instead of quoting the rules we may always effect the
summation in the following way, which is sometimes called ' the
Method of Subtraction.'
Example. Find the sum of n terms of the series
2.5 + 5.8 + 8.11 + 11.14+
The arithmetical progression in this case is
2, 5,8, 11, 14,
In each term of the given series introduce as a new factor the next term
of the arithmetical progression ; denote this series by &", and the given series
by S; then
S' = 2. 5. 8 + 5. 8. 11 + 8. 11. 14+ +(3wl)(3n+2)(3»+5);
.. £'2.5.8 = 5.8.11 + 8. 11.14 + 11. 14.17+... to (u1) terms.
By subtraction,
_2.5.8=9[5.8 + 8.11 + 11.14+...to(»l)terms](3nl)(3n+2)(3n+5),
 2 . 5 . 8 = 9 [S  2 . 5]  (3/i  1) (3n+2) (3n+5),
9S = (3/i  1) (3/i + 2) (3/i + 5) 2. 5. 8 + 2, 5.0,
fif=n(3n 3 +6n+l).
388. When the n th term of a series is a rational integral
function of n it can be expressed in a form which will enable us
readily to apply the method given in Art. 383.
For suppose <j> (n) is a rational integral function of n of p
dimensions, and assume
cf)(n) = A +Bn+ Cti(n + 1) +B)i(u+ l)(n + 2)+ ,
where A, JB, C, D, are undetermined constants p + l in
number.
This identity being true for all values of n, we may equate
the coefficients of like powers of n; we thus obtain ^> + 1 simple
equations to determine the p + 1 constants.
Example. Find the sum of n terms of the series whose general term is
n*+6n 3 + 5w 2 .
Assume
7i 4 + 6/i 3 + 5/t 2 = A + Bn+ Gn [n + 1) + Dn [n + 1) (n + 2) + En (n + 1) (n + 2) (w + 3) ;
it is at once obvious that ,4=0, 2? = 0, E = 1 ; and by putting n =  2, n =  3
successively, we obtain C =  6, J) = 0. Thus
« 4 + 6» 3 + 5/< 2 =//(n + l) (n+2) (?i + 3)6/t(/i + l).
SUMMATION OF SERIES. 310
Hence S n = s n (/t + l)(» + 2)(n + 3)(?i + 4)  2n(n + l)(n + 2)
o
= \n(n+l)(n+2){n' i + 7n + 2).
o
Polygonal and Figurate Numbers.
•
389. If in the expression n+ \n(n— l)b, which is the sum
of n terms of an arithmetical progression whose first term is 1
and common difference b, we give to b the values 0, 1, 2, 3,
we get
•>
n j
u, \n (n + 1), n* s \n (Bn — 1)
which are the u ih terms of the Polygonal Numbers of the second,
third, fourth, fifth, orders; the first order being that in which
each term is unity. The polygonal numbers of the second, third,
fourth, fifth, orders are sometimes called linear, triangular
square, pentagonal
)
390. To find the sum of the first n terms of the r th order of
j>olygonal numbers.
The n ih term of the r tb order is n + \n (n  1) (r — 2);
.. $ i =$n + l(r2)%(nl)u
= \n (n + 1) + 1 (r 2)(nl) n (n + 1) [Art. 383]
= in(n + l){(r2)(nl) + $}.
391. If the sum of n terms of the series
1, 1, 1, 1,1, ,
be taken as the ?* th term of a new series, we obtain
1,2,3,4,5,
n in + 1 )
If again we take — , which is the sum of n terms of the
j
last series, as the ?t th term of a new series, we obtain
1, 3, 6, 10, 15,
By proceeding in this way, we obtain a succession of series
such that in any one, the n ih term is the sum of n terms of the
preceding series. The successive series thus formed are known
as Figurate Numbers of the first, second, third, ... orders.
320
HIGHER ALGEBRA.
392. To find the n th term and the sum of n terms of ilie r th
order offigurate numbers.
The n ih term of the first order is 1; the n th term of the
second order is n; the n th term of the third order is Hn, that is
\n (n + 1); the n tYl term of the fourth order is 2 " V > tnat is
1 . 2
n(n+l)(n+2) .. tIl , , . . ~ M , . ^ n(n+l) (n+2)
— L± ' • the u th term of the fifth order is 2, — *= — ^5 ■ ,
1.2.3 1 . 2 . o
xl . w(w+l)(n + 2)(M + 3)
that is — 2 ^n — — ; and so on.
4
Tims it is easy to see that the n th term of the r th order is
w(?*+l )(w + 2)...(n + r2) i rc+r2
. 01
r1
n — 1 I r — 1
A«rain, the sum of n terms of the r th order is
n (n + 1) (n + 2) . . . (w + r  1)
which is the w th term of tlie (r + l) th order.
Note. In applying the rule of Art. 383 to find the sum of n terms of
any order of figurate numbers, it will be found that the constant is always
zero.
393. The properties of figurate numbers are historically
interesting on account of the use made of them by Pascal in
his Traite du triangle arithmetique, published in 1665.
The following table exhibits the Arithmetical Triangle in its
simplest form
1 ...
1 1
1
1
1 1
1
1
1
1 2
3
4
5 6
8
9
1 3
6
10
15 21
28
36 .
1 4
10
20
35 56
84
1 5
15
35
70 126
1 6
21
56
126 ...
1 7
28
84
. . .
1 8
36
. ■ .
1 9
• •
1
SUMMATION OF SERIES. 321
Pascal constructed the numbers in the triangle by the follow
ing rule :
Each number is the sum of that immediately above it and that
immediately to the left of it;
thus 15 = 5 + 10, 28 = 7 + 21, 126 = 56 + 70.
From the mode of construction, it follows that the numbers in
the successive horizontal rows, or vertical columns, are the hgurate
numbers of the first, second, third, . . . orders.
A line drawn so as to cut off an equal number of units from
the top row and the lefthand column is called a base, and the
bases are numbered beginning from the top lefthand corner.
Thus the 6th base is a line drawn through the numbers 1, 5, 10,
10, 5, 1 ; and it will be observed that there are six of these num
bers, and that they are the coefficients of the terms in the ex
pansion of (1 + x) 5 .
The properties of these numbers were discussed by Pascal
with great skill : in particular he used his Arithmetical Trianyle
to develop the theory of Combinations, and to establish some
interesting propositions in Probability. The subject is fully
treated in Todhunter's History of Probability, Chapter n.
304. "Where no ambiguity exists as to the number of terms
in a series, we have used the symbol % to indicate summation ;
but in some cases the following modified notation, which indicates
the limits between which the summation is to be effected, will be
found more convenient.
Let cf> (x) be any function of x, then 2 <f> (x) denotes the sum
x=l
of the series of terms obtained from <f> (x) by giving to x all posi
tive integral values from I to m inclusive.
'a 1
For instance, suppose it is required to find the sum of all the
terms of the series obtained from the expression
(pl)(p2)...(pr)
by giving to p all integral values from r + 1 to j> inclusive.
H.H. A. 21
322 HIGHER ALGEBRA.
Writing the factors of the numerator in ascending order,
. *=* (p  r) (p  r + 1) ... (p  1)
the required sum = 2 — — —
= i{1.2.3.. ..r+2.SA....(r+l)+...+(pr)(pr+l)...(pl)}
= l (pr)(pr + l) „(pl)p [Art. 383.]
\r r + 1 L J
= y~(l)(y2)...(^r)
jr+1
i
Since the given expression is zero for all values of p from 1 to
r inclusive, we may write the result in the form
% p (pl) ( p2) •■■(p r) _ p(pl) ( ff2) ...(pr)
v\ \r  r + 1
EXAMPLES. XXIX. a.
Sum the following series to n terms :
1. 1.2.3 + 2.3.4 + 3.4.5 +
2. 1.2.3.4 + 2.3.4.5 + 3.4.5.6 +
3. 1.4.7 + 4.7.10 + 7.10.13 +
4. 1.4.7 + 2.5.8 + 3.6.9 +
5. 1.5.9 + 2.6.10 + 3.7.11 +
Sum the following series to n terms and to infinity :
I 1 1
1.2^2.3 3.4^
II 1
7 ' 174 + 4. 7 + 77l0 +
1 1 1_
1.3.5 + 3.5.7 + 5.7.9 +
1 1_ 1
1.4.7 + 4.7.10 + 7.10.13 +
4 5 6
10 1 1  4
1.2.3^2.3.4 .3.4.5^
11 J_ _1_ _JL
* 3.4.5 + 4.5.6 + 5.6.7 + *""'
io 1 3 5 7
\9, — — — l — i u
1.2.3 2.3.4 3.4.5 4.5.6
SUMMATION OF SERIES. 323
Find the sum of n terms of the series :
13. 1 ,3.2 2 +2.4.3 a +3.5.4*+
14. (?i 2 l 2 ) + 2<> 2 2 2 ) + 3(> 2 3 2 ) +
Find the sum of n terms of the series whose n a term is
15. »*(»* 1). 16. (n* + DR + 4)(n 2 + 5n + H).
?i 2 (?i' 2 l) ?* 4 + 2/> 3 + h 2 1
17. A 9 i • *&• v, •
4w 2 l u + /i
1Q n*+3n ?+2n+2 7i*+n 2 + l
iy. ., _ . zu. ,
n* + 2)i iv + n
21. Shew that the ?i th term of the r th order of figurate numbers is
equal to the r th term of the n tXx order.
22. If the n th term of the r th order of figurate numbers is equal to
the (n + 2) th term of the (>2) th order, shew that r=n+%
23. Shew that the sum of the first n of all the sets of polygonal
numbers from the linear to that of the ? ,th order inclusive is
{r\)n(n + \), „ oN
►Summation by the Method of Differences.
395. Let u n denote some rational integral function of », and
let Mj, u. 2 , w 3 , tt 4 ,... denote the values of u n when for n the values
1 , 2, 3, 4, . . . are written successively.
We proceed to investigate a method of finding u n when a
certain number of the terms u x , u. 2 , w 3 , u 4 ,... are given.
From the series u x , u 2 , u 3 , u A , u 5 ,... obtain a second series
by subtracting each term from the term which immediately
follows it.
The series
u. 2 — w,, u s u. 2 , u 4 — u 3 , u 5 u 4 ,...
thus found is called the series of the first order of differences, and
may be conveniently denoted by
Aw,, &u~ &u A , At* 4 ,...
By subtracting each term of this series from the term that
immediately follows it, w r e have
Am., A?*,, Attg — Awg, Aw 4 — Awj,...
which may be called the series of the second order of differences,
and denoted by
A../',, A.,?'.,, AjWg,...
324 HIGHER ALGEBRA.
From this series we may proceed to form the series of the
third, fourth, fifth,... orders of differences, the general terms of
these series being A 3 u r , A A u r , A 5 ?t r) ... respectively.
From the law of formation of the series
Uj t u. 2 , u 3 , u±, u 5 , u 6 ,
Attj, Au. 2 , Au 3 , Aw 4 , Au 5 ,
A.y^ , A 2 w 2 , A.m 3 , A 2 u i ,
. A 3 Wj, A 3 u,, A 3 u 3 ,
it appears that any term in any series is equal to the term
immediately preceding it added to the term below it on the left.
Thus u. 2 = «j 4 Au ly and Ait. 2 = Au i + A.m^ .
By addition, since u. 2 + Au. 2 = u 3 we have
ii.j = t^ + 2Au ± + A. 2 u x .
In an exactly similar manner by using the second, third, and
fourth series in place of the first, second, and third, we obtain
Au 3 = Au x + 2A. 2 u 1 + A^.
By addition, since u 3 + Au 3 = u 4i we have
?f 4  u x + ZAu x + SA^ + A^ .
So far as we have proceeded, the numerical coefficients follow
the same law as those of the Binomial theorem. We shall now
prove by induction that this will always be the case. For sup
pose that
u n+i = «i + mAmj + v 9 A,u l + ... + "CVA^j + + A n Wj i
X ■ J
then by using the second to the (n + 2) th series in the place of the
first to the (n + l) th series, we have
it (11 — 1 )
Au n+1 = A%! + nA. 2 u } + A. — jrf A 3 Wj + . . . + B C f r _ 1 A^w 1 + . . . + A n ^u Y .
By addition, since u n+l + Au n+1 = u n+2i we obtain
M»+a = Mj + fa + 1) Awj + . . . + ( n C r + *G r _j) A r u x + ...+ A„ +1 «, .
SUMMATION OF SERIES. 32.",
But *C r + HJ r i  (— ^ + l) x »C r _ x = ?i±i x "C,,.,
(n + l)w(wl) ...(w+lr+1) _
1.2. 3... (rl)r
Hence if tlie law of formation holds for u n+l it also holds for
f£ n+8 , hut it is true in the case of w 4 , therefore it holds for u rn and
therefore universally. Hence
, 1X . (wl)(w2) .
"„ = Ui + (n  1) A?^ + — ■£*_ *■ A 2 ?^ + ... + An.iWi.
39G. To find the sum of w terms of the series
in terms of the differences of u l .
Suppose the series u^, u. 2 , u 3 ,... is the first order of differences
of the series
Vl, v. 2) v 3 , v 4 ,...,
then v n+1 = (v n+1  v n ) + (v n  v n _ t ) + ... + (v 2  v x ) + v x identically ;
• '• ^»+l = u a + u nl + •■• + u 2 + u \ + v l •
Hence in the series
0, v a1 v 3i v 4 , v 5
1 ) 2 ) 3 J 4 J
Aw 1? Aw 2 , Aw 3
the law of formation is the same as in the preceding article;
•'■ «»+i = + «Wi + 4 — s— Aw x + . . . + A„?^ ;
that is, Wj + w 3 + u z + ... + u n
n (n—\) t n(n—l)(n2) >
 nu x + — y —r — AWj + — ^ A 2 M! + . . . + A„?f , .
The formula) of this and the preceding article may be ex
pressed in a slightly different form, as follows : if a is the first
term of a given series, (I x , d 2 , d 3 ,... the first terms of the suc
cessive orders of differences, the n th term of the given series is
obtained from the formula
326 HIGHER ALGEBRA,
and the sum of n terras is
^i"^" 1 )^ , »(»l )(" 2 ) f/ n(»l)(tta)(n3)
2 j3 4
Example. Find the general term and the sum of ?* terms of the series
12, 40,90, 168, 280, 432,
The successive orders of difference are
28, 50, 78, 112, 152,
22, 28, 34, 40,
6, 6, 6,
0, 0,...
,c ™, ,x 22(rel)(re2) 6 (re 1) (re 2) (re 3)
Hence the n th term = 12 + 28 (re  1) + — K  ~P ' +  v M M  '
l± II
= ?i 3 + 5re 2 + 6>t.
The sum of n terms may now be found by writing down the value of
2re 3 + 52re 2 + 62re. Or we may use the formula of the present article and
obtain S^ia^ 28 "'" 1 ' + 22 "'" 1)( " 2 » + «M»D (»2) (8)
= ^(3re 2 + 26re + 69re + 46),
= in(re+l)(3n 2 + 23re + 46).
397. It will be seen that this method of summation will only
succeed when the series is such that in forming the orders of
differences we eventually come to a series in which all the terms
are equal. This will always be the case if the n th term of the
series is a rational integral function of n.
^» j
For simplicity we will consider a function of three dimensions;
the method of proof, however, is perfectly general.
Let the series be
u. + u a + u a + + u +u .,+u . „ + u , „ +
1 2 3 ii n + \ n+2 u + 3
where u = An 3 + Bn 2 + Cn + D.
"
and let v , w , % denote the ?i th term of the first, second, third
n' ii* ii * *
orders of differences;
SUMMATION OF SERIES. 327
then v h  m m+1 — u n = A(3n* + 3n+ l) + 2?(2» + 1) +C:
that is, v n = 3Au 2 + (3A + 2B) n + A + 11 + C ;
Similarly w = v . , — v = 3A (2n + I) + 3A + 211
and z =w .— iv =64.
H »ti ii
Thus the terms in the third order of differences are equal;
and generally, if the n ih term of the given series is of p dimensions,
the terms in the p th order of differences will be equal.
Conversely, if the terms in the ]j th order of differences are
equal, the u tu term of the series is a rational integral function of
ii of p dimensions.
Example. Find the « th term of the series 1,  3, 3, 23, G3, 129,
The successive orders of differences are
2, 6, 20, 40, GO,
8, 14, 20,20,
6, 6, 0,
Thus the terms in the third order of differeLces are equal ; hence we may
assume it H = A+Bn+Cn 2 + Dn 3 ,
where A, B, G, D have to be determined.
Putting 1, 2, 3, 4 for 7i in succession, we have four simultaneous
equations, from which we obtain A =3, B = 3, C =  2, D — \ ;
hence the general term of the series is 3  3n  2n 2 + n 9 .
398. If a ri is a rational integral function of p dimensions
in n, the series
a, + ax + ajx 2 + ... + a x n
12 »i
is a recurring series, ivhose scale of relation is (I — x) p+1 .
Let S denote the sum of the series ; then
S (1  x)  a o + (a x  a )x + {a,  ajx* + . .. + (a,  a ,_>"  ax" + l
= a + b t x + bjc 2 + ... + bx"  ax" + \ say;
here b =a —a , , so that 6 n is of p  1 dimensions in n.
n h it — 1 ' "• x
Multiplying this last series by 1  x, we have
S(ixy
=s+(^a„)*+(^^K+..+(6n6„iK(« J ,+6>" +l +«X +a
= c^+{ba )x+c^ 2 + c i x^...+cX{a i +b i y , ^ + a i :c ,+ % say;
here c n b n b u u so that c n is of p  2 dimensions in n.
328 HIGHER ALGEBRA.
Hence it follows that after the successive multiplications by
1 — x, the coefficients of x n in the first, second, third, . . . products
are general terms in the first, second, third, . . . orders of differences
of the coefficients.
By hypothesis a n is a rational integral function of n of p
dimensions ; therefore after p multiplications by 1  x we shall
arrive at a series the terms of which, with the exception of p
terms at the beginning, and p terms at the end of the series, form
a geometrical progression, each of whose coefficients is the same.
[Art. 397.]
Thus S (1  xf = k(x p + x>' +1 + ...+ x") +/(a?),
where k is a constant, and f (x) stands for the p terms at
the beginning and p terms at the end of the product.
r.Silxy J^l^ K/ix);
k x»(lx"^) + (lx)f(x) ^
that is, a = (1 x) p+l '
thus the series is a recurring series whose scale of relation is
(lx) p+1 . [Art. 325.]
If the general term is not given, the dimensions of a n are
readily found by the method explained in Art. 397.
Example. Find the generating function of the series
3 + 5a; + 9a; 2 +15a; 3 + 23a; 4 + 33a; 5 +
Forming the successive orders of differences of the coefficients, we have
the series
2, 4, G, 8, 10,
2, 2, 2, 2, ;
thus the terms in the second order of differences are equal ; hence a n is a
rational integral function of n of two dimensions ; and therefore the scale
of relation is (1  a;) 3 . We have
S = 3 + 5x + 9a; 2 + 15.r 3 + 23a; 4 + 33a; 5 +
 SxS =  9.r  15a; 2  27.x 3  45a; 4  69^ 
Sx 2 S = 9a; 2 + 15a; 3 + 27.r 4 + 45a; 5 +
x s S=  3^ 5a; 4  9a; 5 
By addition, ( 1  a;) 3 S = 3  4a; + 3a; 2 ;
34.r + 3a; 2
•*• b ~ (1a;) 3 *
SUMMATION OF SERIES. 329
399. We have seen in Chap, xxiv. that the generating
function of a recurring series is a rational fraction whose denomi
nator is the scale of relation. Suppose that this denominator can
be resolved into the factors (1 — ax) (1 — bx) (1 — ex) ; then the
generating function can be separated into partial fractions of the
 ABC
to rm , 1
1  ax 1 — bx 1  ex
Each of these fractions can be expanded by the Binomial Theorem
in the form of a geometrical series; hence in this case the re
curring series can be expressed as the sum of a number of
geometrical series.
If however the scale of relation contains any factor 1  ax
more than once, corresponding to this repeated factor there will be
A A
partial fractions of the form ^ — 7, ... — r=, : which
(1 axy (1  ax)
when expanded by the Binomial Theorem do not form geometrical
series; hence in this case the recurring series cannot be expressed
as the sum of a number of geometrical series.
400. The successive orders of differences of the geometrical
progression
a, ar, ar 2 , ar 3 , ar\ ar n ,
are «(rl), a(r—l)r, a(rl)r 2 , a(r—\)r ? '
a(rl) 2 , a(rl) 2 r, a(r\fr 2 ,
which are themselves geometrical progressions having the same
common ratio r as the original series.
401. Let us consider the series in which
where </>(rc) is a rational integral function of n of p dimensions,
and from this series let us form the successive orders of differences.
Each term in any of these orders is the sum of two parts, one
arising from terms of the form ar n ~\ and the other from terms of
the form <£(?i) in the original series. Now since <f>(n) is of ;;
dimensions, the part arising from <f>(n) will be zero in the (p + l) th
and succeeding orders of differences, and therefore these series
will be geometrical progressions whose common ratio is r.
[Art. 400.]
330 HIGHER ALGEBRA.
Hence if the first few terms of a series are given, and if the
p th order of differences of these terms form a geometrical pro
gression whose common ratio is r, then we may assume that the
general term of the given series is ar"" 1 +f(n), where f(n) is a
rational integral function of n of p  1 dimensions.
Example. Find the n th term of the series
10, 23, 60, 169, 494,
The successive orders of differences are
13, 37, 109, 335,
24, 72, 216,
Thus the second order of differences is a geometrical progression in which
the common ratio is 3 ; hence we may assume for the general term
u n —a . S n ^ + bn + c.
To determine the constants a, b, c, make n equal to 1, 2, 3 successively;
then a + b + c=10, 3a + 2b+c = 23, 9a + 3b + c = 60;
whence a = 6, 6=1, c = S.
Thus u n = 6 . 3' 1 " 1 + n + 3 = 2 . 3» + n + 3.
402. In each of the examples on recurring series that we
have just given, on forming the successive orders of differences
we have obtained a series the law of which is obvious on inspec
tion, and we have thus been enabled to find a general expression
for the ?4 th term of the original series.
If, however, the recurring series is equal to the sum of a
number of geometrical progressions whose common ratios are
«, b, c, ..., its general term is of the form Aa"' 1 + Bb n ~ l + Cc n ~\
and therefore the general term in the successive orders of
differences is of the same form ; that is, all the orders of differ
ences follow the same law as the original series. In this case to
find the general term of the series we must have recourse to the
more general method explained in Chap. xxiv. But when the
coefficients are large the scale of relation is not found without
considerable arithmetical labour ; hence it is generally worth
while to write down a few of the orders of differences to see
whether we shall arrive at a series the law of whose terms is
evident.
403. We add some examples in further illustration of the
preceding principles.
SUMMATION OF SERIES. 33]
Example 1. Find the sum of n terras of the series
1.2'3 + 2.3'3 3+ 3.4'3 5 + 4.5 *3^ +
„ 2« + 3 1
" 1l(ll + l) 3"
2n + 3 .4 7?
Assuming — = . =  + = ,
n(u+l) n n + 1
we find A =3, B= 1.
tt /3 1 \ 1 1 1 11
Hence t/,. = ( ) — =  .  — , . — .
" \n n + 1) 3" n 3" 1 n + 1 3"'
and therefore #,, = 1 .  .
n n + 1 3' 1
Example 2. Find the sum of n terms of the series
1 _3_ _5 7
3 + 3. 7 + 3. 7. 11 + 3. 7. 11. 15 +
The ri h term is .,„■,., r, kt: •, •
3.7 . 11 (An 5) (4/il)
. 2nl A (n + 1) + B An + B
ssume 3 7 (4n _ 5) (4 n _i) ~ 3 . 7 ......4»l " 3.7 (4„  5) '
.. 2rcl = ,4n + (J+I> > )(.t» + .B)(4?il).
On equating coefficients we have three equations involving the two
unknowns A and B, and our assumption will be correct if values of A and B
can be found to satisfy all three.
Equating coefficients of n 2 , we obtain ^1=0.
Equating the absolute terms, 1 = 2B; that is B = %; and it will be
found that these values of A and B satisfy the third equation.
1 1 1 1
""' V,l ~2 *3.7 (4»5) 2'3.7 (4»5)(4»l) ;
hence S„ = . —
" 2 2 3.7.11 (4»l)
Example 3. Sum to n terms the series
G. 9 + 12. 21 + 20. 37 + 30. 57 + 42. 81 +
By the method of Art. 396, or that of Art. 397, we find that the ;t th terra
of the series 6, 12, 20, 30, 42, is ?r + 3» + 2,
and the ;« th term of the series
9, 21, 37, 57, 81, is2n*+6n+l.
3.32 HIGHER ALGEBRA.
Hence »„=(« + 1) (a + 2) {2m (m+3) + 1}
= 2m [n + 1) (?i + 2) (»+ 3) + (n + 1) (m + 2) ;
•'• S«=ln(»+l)(»+2)(n+3)(n+4)+(n+i)(n+2)(n+8)2.
Example 4. Find the sum of ??. terms of the series
2.2 + 6.4 + 12.8 + 20.16 + 30.32+
In the series 2, 6, 12, 20, 30, the ?i th term is n 2 + n ;
hence u n = {n 2 + n) 2 n .
Assume (rc 2 + m) 2' 1 = (An 2 + Bn+ C)2 n  {A (nl) 2 + B (n  1) + C\ 2" 1 ;
dividing out by 2' 1_1 and equating coefficients of like powers of n, we have
2 = A t 2 = 2A+B, 0=CA + B;
whence A=2, B= 2, 0=4.
.. w n = (2?i 2  2n + 4) 2 n  { 2 (n  l) 2  2 (n  1) + 4 } 2" 1 j
and S n = (2m 2  2m + 4) 2 n  4 = (n a  n + 2) 2*«  4.
EXAMPLES. XXIX. b.
Find the n th term and the sum of n terms of the series
1. 4, 14, 30, 52, 80, 114,
2. 8, 26, 54, 92, 140, 198,
3. 2, 12, 36, 80, 150, 252,
4. 8, 16, 0, 64, 200, 432,
5. 30, 144, 420, 960, 1890, 3360,
Find the generating functions of the series :
6. 1 + 3x + 7x 2 +13.^ + 21a 4 + 31a 6 +
7. 1 + 2a + 9a 2 + 20a 3 + 35a 4 + 54a 3 +
8. 2 + 5a + 10a 2 + 1 7a 3 + 26a 4 + 37a 5 +
9. 1  3a + 5a 2  7 X s + 9a 4  11a 6 +
10. I 4 + 2% + 3 4 a 2 + 4 4 ^ + 5 4 a 4 +
Find the sum of the infinite series :
11. 3 + 32 + 33 + g4 +
12 i 2 _? 2 + ??_iV 2 _ 62 +
1Z> * 5 + 5 2 53 + 5« 5* + "
SUMMATION OF SERIES. 333
Find the general term and the sum of n terms of the series :
13. 9, 16, 29, 54, 103,
14. 3, 1, 11, 39, 89, 167,
15. 2, 5, 12, 31, 8(i,
16. 1, 0, 1, 8, 29, 80, 193,
17. 4, 13, 35, 94, 262, Tr»5
Find the sum of n terms of the series :
18. 1 + 8* + 3.>/ + 4./,' ; + 5.t 1 +
19. 1+ 3.i + 6x 2 + lO.f' 5 + 1 5.r* +
onJLi 4 1 5 1 6 1
1.2*2 + 2.3 "2 :2 ' f 3.4'2 ! + 4.5'2 4 +
21 ' 2T3 4+ i£ i 4S+ 4^ 5 4 ' + 0 44+
22. 3.4 + 8. 11 + 15.20 + 24.31+35.44+
23. 1.3 + 4.7 + 9.13 + 16.21+25.31 +
24. 1.5 + 2.15 + 3.31+4.53 + 5.81 +
o C 1 2 3 4
25 1 A k — 4
' 1.3^1.3.5 1.3.5.7 1.3.5.7.9
nn 1.2 2.2' 3.2 3 4.2 4
26 ' ^ + 14 + T5 + T6 +
27. 2.2 + 4.4 + 7.8 + 11.16+16.32 +
28. 1 . 3 + 3 . 3 2 + 5 . 3 3 + 7 . 3 4 +9. 3> + ...
rtr . 1 1.3 1.3.5 1.3.5.7
' 2. 42. 4. 62.4. 6. 82. 4. 6. 8. 10
30 ± +— 2+i5L 92, l l 2 3+
^ 1.2 + 2.3' 2+ 3.4' 2 + 4.5" 2 +
_4_ 1 _5_ 1 (J 1
1.2.3*3 2.3.4' 3 2 + 3.4. 5" 3 3 +
32 ±+A + H + ^ +
(3^ 4 5 6
33 19 I 28 1 _39_ J_ 52 1
' 1 . 2 . 3 ' 4 + 2 . 3 . 4 " 8 + 3 . 4 . 5 * 16 + 4 . 5 . 6 ' 32 +
334 HIGHER ALGEBRA.
404. There are many series the summation of which can be
brought under no general rule. In some cases a skilful modifi
cation of the foregoing methods may be necessary ; in others it
Avill be found that the summation depends on the properties of
certain known expansions, such as those obtained by the Binomial,
Logarithmic, and Exponential Theorems.
Example 1. Find the sum of the infinite series
2 12 28 50 78
[I + 2 + 3_ + I + 5 +
term of the series 2, 12, 28, 50, 78. .. v . is 3n + n 2; hence
3h 2 +j«2 3h(h1)+4»2
a
»" ]n n
+
2
;i2 n1 In"
Put n equal to 1, 2, 3, 4,... in succession ; then we have
2 4 2 3 4 2
", = 4; „ 2 = 3 + ri  r2 ; « 3 =j i + ^  gj
i
and so on.
Whence ,Sf„ = Se + 4e  2 {e  1) = 5e + 2.
Example 2. If (1 + a;) n = c + c r r + c 2 .r 2 + . . . + c n x n , find the value of
lc 1 + 2 2 c 2 + 3 2 c 3 +... + n\v
As in Art. 398 we may easily shew that
l 2 + 2 2 .r + 3 2 .r 2 + &x 3 +...+ nx n ~ l + . . . =
Also c n + c n _ x x + . . .c. 2 £ n  2 + c^ 11 ' 1 + c x n = (1 + .r) n .
Multiply together these two results; then the given series is equal to
(l + .r) n+1 . . (2  1  x) n+1
the coefficient of x 11 x in ,., .„ , that is, in  — 7  J= — .
(1  x) A (1  x) 3
The only terms containing jc n1 in this expansion arise from
2"+! (1  .r) 3  (n + 1) 2> l (1  .t)" 2 + \!l±Jl l %*i (i _ ^l.
.. the given series = f L^+3 2 »+i _ „ („ + 1) gn + ?ii"±: l ) 2 hi
n(w+l)2« ! .
MISCELLANEOUS METHODS OF SUMMATION. 335
Example 3. If b = a + l, and )i is a positive integer, find the value of
IP _ (n _ 1) „,,.  + »»>(»«> rfj. . _ C3)(»4)(»5) ^ +
2 \6
By the Binomial Theorem, we see that
(n8) (n2) (n5)(n4)(»3)
are the coefficients of x n , .r' 12 , .r n_4 , .r' 1 * 5 , in the expansions of (1 x) ',
(1.t) 2 , (1x)*, (l.r) 4 , respectively. Hence the sum required is
e<pial to the coefficient of x* in the expansion of the scries
1 ax* ax 4 a*x 6
+
1bx {1bx) 3 ' (1fcc) 8 (1  bx)* '
and although the given expression consists only of a finite number of terms,
this series may be considered to extend to infinity.
But the sum of the series = , — ; — • ( 1 + , ) = z — z
1bx \ 1bx J 1bx + ax"
i
, since b — a+1.
1  (a + l)x + ax
Hence the given series = coefficient of x n in
(lx)(lax)
= coefficient of x n in = (   ~  }
a  1 \1 ax 1—x)
a
H+l _ 1
a1 ■
Example 4. If the series
, x 3 x e X* X 7 x' 2 X 5 X 8
1 + J3 + JG + ' • r + ]5 + 7 + ' 2_ + 5 + 8_ +
are denoted by a, b, c respectively, shew that a 8 + 6 3 + c 8 3o6c=l.
If w is an imaginary cube root of unity,
a 3 + b 3 + c 3  Sabc = {a + b + c) (a + wb + w'c ) (a + wb + ojc) .
.t 2 x z .t 4 X s
Now ' lA ~ h + c = 1+x + ~\9 + \3 + Tl + ~\5 +
and
w.r ur\r w 4 .c 4 w'.r'
>/ + lob + OJC1+ C0X+ — + r^ + —T + r=~
\ \ \ \
I
= e
similarly a + io'b + wc = c
bc =
1, since l + w + ur = 0.
0)=X
•. , , ., , , o , X uX co 2 X (l+u> + w ! )x
336 HIGHER ALGEBRA.
405. To find the sum of the r th powers of the first n natural
numbers.
Let the sum be denoted by S n ; then
S H =V+2 r + 3 r + ... + n r .
Assume that
S =A n n r+i +An r + A n r  1 +An r ~ 2 + ... +An + A + 1 (1),
n 1 2 3 r r + 1 \ / '
where A , A^ A 2 , A 3 , ... are quantities whose values have to be
determined.
Write n + 1 in the place of n and subtract; thus
(n + l) r = A {(n + l) r+1  n r+1 ] + A x {{n + 1)'  n r ]
+ A 2 {(n+ l) 1 n' 1 } + A 3 {(n + iy~ 2 n r  2 } + ... +A r ...(2).
Expand (?i+l) r+ \ (n + l) r , (n+l) r_1 , ... and equate the co
efficients of like powers of n. By equating the coefficients of n r ,
we have
1
l=A. (r + 1), so that A a = T .
By equating the coefficients of n r ! , we have
A (r+ l)r 1
r = ° — — + A x r ; whence A x = ^ .
Equate the coefficients of n r p , substitute for A and A Jf and
multiply both sides of the equation by
\P
r(rl)(r2) ... {r 2 )+ 1) ;
we thus obtain
i ~p + l + 2 + A 'r + A 'r(rl) + ^ r(r  1) (r2) + "^
In (1) write w — 1 in the place of n and subtract; thus
n r =A {n r+l (niy +i }+A l {?i r (nl) r } + A 2 {n r ' 1 (niy 1 } + ...
Equate the coefficients of n r ~ p , and substitute for A , A 1 ; thus
o '4 + ^^gzi) +i /^;);^) .... w
p + 1 2 2 r 3 r (?•  1 ) 4 v (?•  1) (r  2)
MISCELLANEOUS METHODS OF SUMMATION. 337
From (3) and (4), by .addition .and subtraction,
2 p + 1 "r * r(rl)(r2)
o^/_^)^/ 0'i)(pg(^3) + (6) .
3 r (r  1) ■ r (?•  1) (r  2) (r  3) w
By ascribing to }> i n succession the values 2, 4, 6, . . . , we see
from (G) that each of the coefficients A.^ A 5> A.,... is equal
to zero; and from (5) we obtain
1 r ___1_ r (rl)(r2) ,
6 " 1^ J 30 ' li
. _J_ r(rl)(r2)(r3)(r4)
8 ~42" 6 ;
By equating the absolute terms in (2), we obtain
\=A^A X + A % + A Z + +A r 
and by putting n= 1 in equation (1), we have
1 = A + A l + A a + A 9 + +A r + A r+l ;
thus A r+1 = 0.
406. The result of the preceding article is most conveniently
expressed by the formula,
„ n r+x 1 , „ r r _ x _ r(rl)(r2) r _ 3
" r+1 2 l 2 3 4
r(rl)(r2)(r3)(r4) ^ +
'6
w } lprp 7? ■ i 7? 1 7? i 7? i 7? — 5
The quantities B x , B 3 , 2? 5 , ... are known as Bernoulli's Numbers;
for examples of their application to the summation of other series
the advanced student may consult Boole's Finite Differences.
Example. Find the value of l 5 + 2 5 + 3 5 f + n 5 .
ttt , r. n6 n5 ^,5 . _5 . 4 . 3 „ _
We have S„ = ^ + ^ + ^ ^ n*  B a — j— n* + C,
_?t 6 n 5 5?i 4 n 2
~6 + "2 + l2~r2'
the constant being zero.
II. ii. A. 22
HIGHER ALGEBRA.
EXAMPLES. XXIX. c.
Find the sum of the following series: ^ ^ ^
JL + A
5. l + ^+\T'T 1 2 ii 3
3 r/3
6. *rz
p r p ri q £± £ + f^l.2 + to r + 1 terms.
(1 + .r) _ »^"2) _ X + 2 ^
7 TX^" "' 12 "•(1+^)"
1+tm;
?i(?il)(™ 2 ) 1+3a?  to n terms.
2n + l , K / 2/i + x Y + ... to n terms.
o,2 W 2 ( 7l 2 _ 12) 7i 2 (ft 2  1 2 )^ 2 ^ 2 ) + to w + 1 terms.
9. ij[i+ii7? 1 2 .2*.3 2
1 + 2 3
1L r2T3 + 3^T5 + 5T677 +
2 3^6 11 18
12. ji + ] + [3 + 4 + 5 +
2a 8
^ W 23s 5 121s 6 _
is. 1 +J23 + "[7'r 16
14 Without fuming the formula, find the sum of the series:
W !«+*+*+ +»«• « 17 + 2; + 3? + + "
SUMMATION OF SERIES. 339
33 43 53
15. Find the sum of l 3 + 2 3 +  + — + _+
B I* I*
16. Shew that the coefficient of x n in the expansion of ., is
(lX) 2 r.r
fl 1 "' 1 / 1 (" 2 l)(« 2 4) , 1 (n *lKn*4)(n*9) )
Y + if" n c+ [7 * + }'
17. If n is a positive integer, find the value of
8 . ( »i)^ + e» g )(' t  8 ) 2 ^ (» 8 )(» 4 )» 6 ) g ^ +
\ 2 \o
and if 11 is a multiple of 3, shew that
1  ( » 1)+ (»»H»3) _(»8)(»4)(»6) + =( _ 1)n
18. If ?i is a positive integer greater than 3, shew that
rf+ «flga ( . y+ "(»i)<««)(— *) ( ,l 4 y + ...
=»»*(» + 3) SP*" 4 .
19. Find the sum of ??. terms of the series :
1 2 3
W i + i2 + i"4 + l + 2 2 + 2 4+ l+3 2 + 3 4 +
(2) _5__J_+JL__L+i3 _JLL+ 17
2.3 3.4 4.5 5.6 6.7 7.8
(l) n + 1 x n
20. Sum to infinity the series whose n th term is
?i(n+l)(n + 2)
21. If (1 + x) n — Cq + c v v + c^v 2 + CyV 3 + + c n # n , n being a positive
integer, find the value of
(n  \)\ + (n  3) 2 c 3 + (?i  5) 2 c 5 +
22. Find the sum of n terms of the series :
„ N 2 4 8 16 32
1.5 5.7 7.17 17.31 31.65
7 17 31 49 _71_
^ 1.2.3 2.3.4 + 3.4.5 4.5.6 5.6.7
23. Prove that, if a < 1, ( 1 + or) ( 1 + A) ( 1 + a?x) ....
ax a*x 2 aPx 3
= 1 + 5 ,+T: =5wi — =K +
1a 2 ' (l« 2 )(la 4 ) ' (l« 2 )(la 4 )(la' ! ) "
22—2
340 HIGHER ALGEBRA.
24. If A r is the coefficient of x r in the expansion of
2/ *A2 / ^.\2
2~ 3 J '
(i + ,f(i + ) 2 (i + ) 2 (i + ;
2 s ,, , v j , 1072
prove that ^l r = 2^ (^4 r i + ^r 2 ) > and ^4 = "3^5 ■
25. If n is a multiple of 6, shew that each of the series
n ^~\i —  3+ [5  3 "
w(wl)(w2) 1 , n(nl)(n2)(nS)(n4) 1
11 [3 *3 + 5 ""'32 •
is equal to zero.
26. If n is a positive integer, shew that
pti + 1 _ qn + 1
is equal to .
27. If P r =(wr)(»r+l)(nr+2) (nr+^1),
&=r(r+l)(r+2) (r+^1),
shew that
ho k \nl+p + q
P& + P2Q2 + P3Q3+ + P »i^i = > + g +ln2
28. If ?i is a multiple of 3, shew that
, »3 (m4)(w5) (w5)(w6)(w7)
1 "^" + "~ 3 H
^ (nrl)(wr2)...(tt2r + l) ,
+ (!) u. "'"•••'
3 1
is equal to  or — , according as n is odd or even.
u n n
29. If x is a proper fraction, shew that
x x z x 5 x x 3 X s
1_^2 l_#6 T l_a?io 1 +.v2^1+^ ' l+.r 10
CHAPTER XXX.
Theory of Numbers.
407. In this chapter we shall use the word number as equi
valent in meaning to positive integer.
A number which is not exactly divisible by any number
except itself and unity is called a prime number, or a prime; a
number which is divisible by other numbers besides itself and
unity is called a composite number \ thus 53 is a prime number,
and 35 is a composite number. Two numbers which have no
common factor except unity are said to be prime to each other ;
thus 24 is prime to 77.
408. We shall make frequent use of the following elementary
propositions, some of which arise so naturally out of the definition
of a prime that they may be regarded as axioms.
(i) If a number a divides a product be and is prime to one
factor b, it must divide the other factor c.
For since a divides be, every factor of a is found in be; but
since a is prime to b, no factor of a is found in b; therefore all
the factors of a are found in c ; that is, a divides c.
(ii) If a prime number a divides a product bed..., it must
divide one of the factors of that product ; and therefore if a
prime number a divides b", where n is any positive integer, it
must divide b.
(iii) If a is prime to each of the numbers b and c, it is prime
to the product be. For no factor of a can divide b or c ; there
fore the product be is not divisible by any factor of a, that is, a
is prime to be. Conversely if a is prime to be, it is prime to eacli
of the numbers b and c.
Also if a is prime to each of the numbers b, c, d, ..., it is
prime to the product bed... ; and conversely if a is prime to any
number, it is prime to every factor of that number.
342 HIGHER ALGEBRA.
(iv) If a and b are prime to each other, every positive
integral power of a is prime to every positive integral power of b.
This follows at once from (iii).
(v) If a is prime to b, the fractions = and j are in their
bo
ft
lowest terms, n and m being any positive integers. Also if j and
 are any two equal fractions, and j is in its lowest terms, then
c and d must be equimultiples of a and b respectively.
409. The number of primes is infinite.
For if not, let p be the greatest prime number; then the
product 2 . 3 . 5 . 7 . 11 . . .p, in which each factor is a prime num
ber, is divisible by each of the factors 2, 3, 5, . . .p ; and therefore
the number formed by adding unity to their product is not
divisible by any of these factors ; hence it is either a prime
number itself or is divisible by some prime number greater than
p : in either case p is not the greatest prime number, and there
fore the number of primes is not limited.
410. No rational algebraical formula can represent prime
numbers only.
If possible, let the formula a + bx + ex 2 + dx 3 + ... represent
prime numbers only, and suppose that when x = m the value of
the expression is ]), so that
p — a + bm + cm 2 + dm 3 + ;
when x = m + np the expression becomes
a + b (m + np) + c {m + np) 2 + d (m + np) 3 + ...,
that is, a + bm + cm 2 + dm 3 + . . . + a multiple of p,
or p + a multiple of p,
thus the expression is divisible by £>, and is therefore not a prime
number.
411. A number can be resolved into prime factors , in only one
way.
Let N denote the number; suppose N = abed..., where
a, b, c, d, ... are prime numbers. Suppose also that JV = a/3yS...,
where a, /3, y, 8, ... are other prime numbers. Then
abed... = a/3yS... ;
THEORY OF NUMHEHS. 343
hence a must divide; abed... ; but eacli of the factors of this pro
duct is a prime, therefore a must divide one of them, a suppose.
But a and a are both prime ; therefore a must be equal to a.
Hence bed. . . =/3yS. . . ; and as before, /? must be equal to one of the
factors of bed... J and so on. Hence the factors in a/3y<$... are
equal to those in abed..., and therefore iV can only be resolved
into prime factors in one way.
412. To find the number of divisors of a composite number.
Let N denote the number, and suppose N"=a p b g <f..., where
a, b, c, ... are different prime numbers and p, q, r, ... are positive
integers. Then it is clear that each term of the product
(l+a + a' + ...+a'')(l+b + b 2 + ... + V) (I + c + c 2 + ...+c r )...
is a divisor of the given number, and that no other number is a
divisor ; hence the number of divisors is the number of terms in
the product, that is,
(f>+l)fe+l)(r + l)
This includes as divisors, both unity and the number itself.
413. To find the number of ways in which a composite number
can be resolved into two factors.
Let N" denote the number, and suppose N = a'tyc' . . . , where
a, b, c... are different prime numbers and ]), q, r... are positive
integers. Then each term of the product
(I + a + a 2 + ... + of) (1 + b + b 2 + . . . + b' 1 ) (1 + c + c 2 + . . . + c r ) . . .
is a divisor of iV; but there are two divisors corresponding to
each way in which iV can be resolved into two factors ; hence the
required number is
}(!>+l)& + l)(r + l)
This supposes N not a perfect square, so that one at least of the
quantities^, q, r, ... is an odd number.
If N is a perfect square, one way of resolution into factors
is x /iVx JNj and to this way there corresponds only one divisor
JX. If we exclude this, the number of ways of resolution is
!{(p+l)(? + l)(r + l)...l},
and to this we must add the one way JN x N /iV; thus we obtain
for the required number
\{(P + !)(</+ !)(<•+ l) + lj
344 HIGHER ALGEBRA.
414. To find the number of ways in which a composite
number can be resolved into two factors which are prime to each
other.
As before, let the number N = a v b q c r .... Of the two factors
one must contain a p , for otherwise there would be some power of
a in one factor and some power of a in the other factor, and thus
the two factors would not be prime to each other. Similarly b q
must occur in one of the factors only ; and so on. Hence the
required number is equal to the number of ways in which the
product abc... can be resolved into two factors; that is, the
number of ways is (1 + 1)(1 + 1)(1 + 1)... or 2"" 1 , where n is
the number of different prime factors in N.
415. To find the sum of the divisors of a number.
Let the number be denoted by a p b q c r ..., as before. Then each
term of the product
(1 +a + a 2 + ...+a r )(l+b + b 2 + ... + b' 1 ) (1 + c + c 2 + ...+c r )...
is a divisor, and therefore the sum of the divisors is equal to this
product ) that is,
the sum required =
a . ■_ i &»+'_! c r+1 l
a
1 * bl " c1
Example 1. Consider the number 21600.
Since 21600 = 6 3 . 10 2 = 2 3 . 3 3 . 2 2 . 5 2 = 2 3 . 3 3 . 5 2 ,
the number of divisors = (5 + 1) (3 + 1) (2 + 1) = 72 ;
.. ... ,. . 261 3*l 5 3 l
the sum of the divisors = — — ? . 5 — — .  — 
2 — 1 o — 1 5 — 1
= 63x40x31
= 78120.
Also 21600 can be resolved into two factors prime to each other in 2 3_1 ,
or 4 ways.
Example 2. If n is odd shew that n (n 2  1) is divisible by 24.
We have n(n 2  l) = 7i {n 1) (n+1).
Since n is odd, n  1 and n+1 are two consecutive even numbers ; hence
one of them is divisible by 2 and the other by 4.
Again n  1, n, n + 1 are three consecutive numbers ; hence one of them
is divisible by 3. Thus the given expression is divisible by the product of 2,
3, and 4, that is, by 24.
THEORY OF NUMBERS. 34".
Example 3. Find the highest power of 3 which is contained in J 100.
Of the first 100 integers, as many are divisible by 3 as the number of
times that 3 is contained in 100, that is, 33 ; and the integers are 3, G, 9,... 99.
Of these, some contain the factor 3 again, namely 9, 18, 27,... 99, and their
number is the quotient of 100 divided by 9. Some again of these last
integers contain the factor 3 a third time, namely 27, 54, 81, the number of
them being the quotient of 100 by 27. One number only, 81, contains the
factor 3 four times.
Hence the highest power required = 33 + 11 + 3 + 1 = 48.
This example is a particular case of the theorem investigated in the next
article.
416. To find the highest 'power of a prime number a which is
contained in In.
n iii n
Let the greatest integer contained in , — 2 , — tJ ... respectively
Cv Ct CL
be denoted by / (  ] , /(,], /(§),... Then among the numbers
1,2, 3, ... n. there are / (  j which contain a at least once, namely
the numbers a, 2a, 3a, 4a, ... Similarly there are I[A which
contain a 2 at least once, and I ( — g ) which contain « 3 at least once;
and so on. Hence the highest power of a contained in \n is
'©♦'©)*'6) + ~
417. In the remainder of this chapter we shall find it con
venient to express a multiple of n by the symbol Jl(n).
418. To prove that the prodicct of r consecutive integers is
divisible by r.
Let P n stand for the product of r consecutive integers, the
least of which is n ; then
P n = n(n+l)(n + 2) ... (u + rl),
and P n+l = (n+l)(n + 2)(n+3) ...(n+r);
• \ nP m+i = (n + r) P = nP n + rP H ;
p
.. 1> P =lsxr
= r times the product of r — 1 consecul ive integer.
346 HIGHER ALGEBRA.
Hence if the product of r — 1 consecutive integers is divisible by
\r — 1, we have
P m+1 P m = rM(\rl)
= M(\r).
Now P, = ?', and therefore P 2 is a multiple of \r \ therefore
also P. , P , . . . are multiples of (r. We have thus proved that if
the product of r— 1 consecutive integers is divisible by \r — 1, the
product of r consecutive integers is divisible by \r ; but the
product of every two consecutive integers is divisible by 1 2 ;
therefore the product of every three consecutive integers is divisible
by 1 3 ; and so on generally.
This proposition may also be proved thus :
By means of Art. 416, we can shew that every prime factor
is contained in \n + r as often at least as it is contained in \n \r.
This we leave as an exercise to the student.
419. If p is a prime number, the coefficient of every term in
the expansion q/*(a + b) p , except the first and last, is divisible by p.
"With the exception of the first and last, every term has a co
efficient of the form
p(pl)(p2)...(pr + l)
'
where r may have any integral value not exceeding p — 1. Now
this expression is an integer; also since p is prime no factor of \ r
is a divisor of it, and since p is greater than r it cannot divide
any factor of \r ; that is, (p — 1) (p — 2)... (p  r + 1) must be
divisible by r. Hence every coefficient except the first and
the last is divisible by p.
420. If p is a prime number, to prove that
(a + b + c + d + ...) p = a 5 + b 1 ' + c p + d p + . . . + M(p).
Write ft for b + c + . . . ; then by the preceding article
(a + py = a* + p' + M(p).
Again J3 p = (b + c + d+ . . . ) p = (b + y) p suppose ;
= b p + y + M{p).
By proceeding in this way we may establish the required result.
THEORY OF NUMBERS. 347
421 [Fermat'a Theorem.] If p is a prime number and N is
prime to p, then N"" 1 lis a multiple of p.
We have proved that
(a + b + c + d+ ...y^a' + V+c* + d"+ ... + M (p);
let each of the quantities «, 6, Cj 4 ... be equal to unity, and sun
pose they are N in number ; then J ' P
But ,V is prime top, and therefore iV'  1 is a multiple of p.
' °°\ } Si T ^ is P l ™°> Pli* an even number except when
/>=J. lherefore r
Hence either 2^ + 1 or S^  1 « a multiple of ft
that is .V • = 7^ ± 1, where K is some positive integer.
422. It should be noticed that in the course of Art. 421 it
tins result is sometimes more useful than Fermat's theorem.
Example 1. Shew that n 7  n is divisible by 42.
Since 7 is a prime, n 7  n = M (7) ;
a T 1S ° n?  n:=n (» G l) = >i(n + l)(nl)(n* + nS + l).
Now (n  1) n (n + 1) is divisible by 3 ; hence n?  n is divisible by 6 x 7, or 42.
Dowfra of^J; tSL* iS \ pHme DU ^ ber ' shew that the difference of the p"
mXpleof^ 7 mimbeiS GXCeedS thG dlfference of the numbers b/a
Let .r, y be the numbers ; then
* p x=M(p) and y»y=M (p),
thatls > * p y p (*y)=^(p);
whence we obtain the required result.
Example 3. Prove that every square number is of the form Sn or on ± 1.
Tf v  V iS  UOt ? r L m l t0 5 x'J e have AT2 = 5;i where » w some positive integer
S£r^PTi?5n 1 l G • ~ l iS i* n i ultipIe 0f 5 ^ Fermat '« theorem ; thus
eitner n» 1 or N*+l is a multiple of 5 ; that is, tf»=5w ± 1.
348 HIGHER ALGEBRA.
EXAMPLES. XXX. a. i
1. Find the least multipliers of the numbers
3675, 4374, 18375, 74088
respectively, which will make the products perfect squares.
2. Find the least multipliers of the numbers
7623, 109350, 539539
respectively, which will make the products perfect cubes.
3. If x and y are positive integers, aud if x—y is even, shew that
a?—y 2 is divisible by 4.
4. Shew that the difference between any number and its square
is even.
5. If ixy is a multiple of 3, shew that 4x 2 + 7xy  2y 2 is divisible
by 9.
6. Find the number of divisors of 8064.
7. In how many ways can the number 7056 be resolved into
two factors ?
8. Prove that 2 4 ' 1  1 is divisible by 15.
9. Prove that n (?i + 1) (n + 5) is a multiple of 6.
10. Shew that every number and its cube when divided by 6 leave
the same remainder.
11. If n is even, shew that n (;i 2 + 20) is divisible by 48.
12. Shew that n (?i 2  1) (Sn + 2) is divisible by 24.
13. If n is greater than 2, shew that n 5 — 5n 3 f 4?i is divisible by
120.
14. Prove that 3 2n + 7 is a multiple of 8.
15. If n is a prime number greater than 3, shew that ?i 2  1 is
a multiple of 24.
16. Shew that n 5 — n is divisible by 30 for all values of n, and by
240 if n is odd.
17. Shew that the difference of the squares of any two prime
numbers greater than 6 is divisible by 24.
18. Shew that no square number is of the form 3?i — 1.
19. Shew that every cube number is of the form 9?i or 9n±L
THEORY OF NUMBERS. 349
20. Shew that if a cube number is divided by 7, the remainder
is 0, 1 or 6.
21. If a number is both square and cube, shew that it is of the
form 7n or 7?t+l.
22. Shew that no triangular number can be of the form 3u  1.
23. If 2» 4 1 is a prime number, shew that l 2 , 2 2 , 3 2 ,...n 2 when
divided by 2>i+l leave different remainders.
24. Shew that a x + a and a*  a are always even, whatever a and x
may be.
25. Prove that every even power of every odd number is of the
form 8r + l.
26. Prove that the 12 th power of any number is of the form I3)i
or 13ft+l.
27. Prove that the 8 th power of any number is of the form I7n
or I7n±l.
28. If n is a prime number greater than 5, shew that n 4 — 1 is
divisible by 240
29. If n is any prime number greater than 3, except 7, shew that
n G — 1 is divisible by 168.
30. Show that ?i 36  1 is divisible by 33744 if n is prime to 2, 3, 19
and 37.
31. When p + l and 2p + l are both prime numbers, shew that
«** — 1 is divisible by 8(p + l)(2/) + l), if x is prime to 2, £> + l,and
2p+h
32. If p is a prime, and X prime to p, shew that x p1 ~ pV  1 is
divisible by p r .
33. If m is a prime number, and a, b two numbers less than m,
prove that
a m  2 + a m ~ 3 b + a m  4 b' i + + b m ~ 2
is a multiple of m.
423. If a is any number, then any other number N may
be expressed in the form N = aq + r, where q is the integral
quotient when N is divided by a, and r is a remainder less than a.
The number a, to which the other is referred, is sometimes called
the modulus ; and to any given modulus a there are a different
350 HIGHER ALGEBRA.
forms of a number iV, each form corresponding to a different
value of r. Thus to modulus 3, we have numbers of the form
3<7, 3q + l, 3q + 2; or, more simply, 3q, 3q±l, since 3q + 2 is
equal to 3 (q+ 1)  1. In like manner to modulus 5 any numbe^
will be one of the five forms 5q, 5q ± 1, 5q ± 2.
424. If 6, c are two integers, which when divided by a
leave the same remainder, they are said to be congruent with
respect to the modulus a. In this case b — c is a multiple of a, and
following the notation of Gauss we shall sometimes express this
as follows :
b = c (mod. a), or b  c=0 (mod. a).
Either of these formulae is called a congruence.
425. If b, c are congruent with respect to modulus a, then
pb and pc are congruent, p being any integer.
For, by supposition, b  c = ?za, where w is some integer ;
therefore ])b — pc — pna ; which proves the proposition.
426. If a is prime to b, and the quantities
a, 2a, 3a, (b — 1 ) a
are divided by b, the remainders are all different.
For if possible, suppose that two of the quantities ma and
ma when divided by b leave the same remainder r, so that
ma = qb + r, m'a = q'b + r ;
then (m  7/1') a = (qq')b ;
therefore b divides (m — m') a ; hence it must divide m — m', since
it is prime to a ; but this is impossible since m and m' are each
less than b.
Thus the remainders are all different, and since none of the
quantities is exactly divisible by b, the remainders must be the
terms of the series 1, 2, 3, b — 1, but not necessarily in this
order.
Cor. If a is prime to b, and c is any number, the b terms
of the a. p.
c, c + a, c + 2a, c + (b — 1) a,
THEORY OP NUMBERS. 351
when divided by b will leave the same remainders as the terms
of the series
c, c+ 1, c+ 2, c + (b 1),
though not necessarily in this order ; and therefore the re
mainders will be 0, 1, 2, b 1.
427. (/"b., b.j b 3 , ... are respectively congruent to c n c„, c } , ...
wn7A regard to modulus a, £/te?i //te products b,b a b a ..., ^c.c.^ ...
(otf also congruent.
For by supposition,
b 1 c l = n x a, b 2 c 2 = u 2 a, b 3 c :i u./i, ...
where n lt n 2 , n 3 ... are integers;
. •. b x b a b a ... =(<?! + »,») (c a + rc 2 «) (c a + w :j a) . . .
= c,c 2 c 3 ... + M (a),
which proves the proposition.
428. We can now give another proof of Fermat's Theorem.
If p be a prime number and N prime to p, then N 1 ' 1 — 1 is
a multiple of p.
Since JV and p are prime to each other, the numbers
if, 2tf, 3.V, (pl)iV (1),
when divided by p leave the remainders
1, 2, 3, (p1) (2),
though not necessarily in this order. Therefore the product of
all the terms in (1) is congruent to the product of all the terms
in (2), p being the modulus.
That is, ^—1 N''~ i and \p  1 leave the same remainder when
divided by p ; hence
but i^—l is prime to p ; therefore it follows that
JP" 1  1 = M (j>).
429. We shall denote the number of integers less than a
number a and prime to it by the symbol <f> (a) ; thus <f>(2) = 1 ;
<£(13) = 12; </>(18) = G; the integers less than 18 and prime to
it being 1, 5, 7, 11, 13, 17. It will be seen that we here
consider unity as prime to all numbers.
352 HIGHER ALGEBRA.
430. To shew that if the numbers a, b, c, d, ... are prime to
each other,
(f> (abccl . . .) = <£ (a) . </> (b) . <£ (c) . . . .
Consider the product ab ; then the first ab numbers can be
written in b lines, each line containing a numbers ; thus
1, 2, k, a,
a+l, a + 2, a + k, a + a,
2a +1, 2a + 2, 2a + k, 2a + a,
(&_ \) a + 1, (6 1) a + 2, ... (bl)a + k, ... (b  1) a + a.
Let us consider the vertical column which begins with h ; if
k is prime to a all the terms of this column will be prime to a ;
but if k and a have a common divisor, no number in the column
will be prime to a. Now the first row contains <£ (a) numbers
prime to a \ therefore there are <£ (a) vertical columns in each
of which every term is prime to a ; let us suppose that the
vertical column which begins with k is one of these. This column
is an A. p., the terms of which when divided by b leave remainders
0, 1, 2, 3, ... 6 — 1 [Art. 426 Cor.]; hence the column contains
<£ (b) integers prime to b.
Similarly, each of the </> (a) vertical columns in which every
term is prime to a contain <£ (b) integers prime to b ; hence in the
table there are <f> (a) . cj> (b) integers which are prime to a and
also to by and therefore to ab ; that is
<£ (ab)  <£ (a) . <£ (6).
Therefore cf> (abed ...) = <f> (a) . <j> (bed . . .)
= cj> (a) . (f)(b) . <j> (cd ...)
= <f>(a).<t>(b).<t>(c).<}>(d)....
431. To find the number of positive integers less than a
given number, and prime to it.
Let JV denote the number, and suppose that JV = a p b q c r ... ,
where a, b, c, ... are different prime numbers, and p, q, r ...
positive integers. Consider the factor a 1 ' ; of the natural num
bers 1, 2, 3, ... a p — 1, a p , the only ones not prime to a are
a, 2a, 3a, ... (a* 1  I) a, (a 1 " 1 ) a,
THEORY OF NUMBERS. 353
and the number of these is a''~ i ; hence
4> (a v ) = a"  a'' =a?(l ^ .
Now all the factors a p , b'\ c\ ... are prime to each other ;
. \ </> (a)Vc r . . .) = <j> (a 1 ') . </> (b 1 ) . (c r ) . . .
H)H)H)
^••K)H)H)
that is, ^ W = iir(ii)(iJ)(iI)....
Example. Shew that the sum of all the integers which are less than N
and prime to it is ^N<p (N).
If x is any integer less than N and prime to it, then Nx is also an
integer less than N and prime to it.
Denote the integers by 1, p, q, r, ... , and their sum by S; then
S = l+p + q + r+... + (Nr) + {Nq) + (Np) + {Nl),
the series consisting of (N) terms.
Writing the series in the reverse order,
S = {Nl) + (Np) + (Nq) + (Nr)+...+r + q+p + l;
.. by addition, 2S = N + N + N+ ... to <p (N) terms;
.. S = $N<p(N).
432. From the last article it follows that the number of
integers which are less than J¥ and not prime to it is
''(■.4>(>i)(» : 3("i)'
tli at is,
N N N N N N N
_++_+..._ . . . + + ....
a b c ao ac be abc
N
Here the term — gives the number of the integers
a, la, da, ... — .a
(t
N
which contain a as a factor; the term — = gives the number of
ao
H. II. A. 23
354 HIGHER ALGEBRA.
N . .
the integers ab, 2ab, Sab, ... j ab, which contain ab as a factor,
ao
and so on. Further, every integer is reckoned once, and once
only ; thus, each multiple of ab will appear once among the
multiples of a, once among the multiples of b, and once negatively
among the multiples of ab, and is thus reckoned once only.
iV N N
Again, each multiple of abc will appear among the — , j , —
a o c
terms which are multiples of a, b, c respectively; among the
JV & iV
— , — , = terms which are multiples of ab, ac, be respectively ;
ab' ac' be r ' r J '
and among the j multiples of abc; that is, since 33+1 = 1,
each multiple of abc occurs once, and once only. Similarly, other
cases may be discussed.
433. [Wilson's Theorem.] If p be a prime number, 1 + \p  1
» is divisible by p.
By Ex. 2, Art. 314 we have
Ijp1 = (P~ i)"" 1  (P  i) (P  2r ' + ^z^ipD (p  $y>
Jpl)(p 2)(p3) {p _ irl+ top _ lterms .
and by Fermat's Theorem each of tlie expressions (j)  l) p ~ l ,
(p2) p ~\ (p2>y~\ ... is of the form 1 +M(p), thus
pl = M(p) +fl(pl) + (P~ l )(P~ 2 ) ...top I terms!
=M(p) + {(iiy>(iy>}
= M(p) — 1, since p — 1 is even.
Therefore 1 + I p  1 = M (p).
This theorem is only true when p is prime. For suppose p
has a factor q; then q is less than p and must divide \p — 1 ; hence
1 + \p — 1 is not a multiple of q, and therefore not a multiple of p.
Wilson's Theorem may also be proved without using the
result quoted from Art. 314, as in the following article.
THEORY OF NUMBERS. ,'}:>5
434. [Wilson's Theorem.] If p be a prime number, 1 + lp— 1
is divisible by p.
Let a denote any one of the numbers
1, 2, 3, 4 5 ... (p1) (1),
then a is prime to p, and if the products
\.a, 2. a, 3. a, (^; — 1 ) a
are divided by p, one and only one of them leaves the re
mainder 1. [Art. 426.]
Let this be the product ma; then we can shew that the
numbers m and a are different unless a=j)~ 1 or 1 For if a 2
were to give remainder 1 on division by^>, we should have
a~  1 = (mod. p),
and since p is prime, this can only be the case when a + 1  p,
or a — 1 — 0; that is, when a=p— 1 or 1.
Hence one and only one of the products 2a, 3a, ... (p — 2) a
gives remainder 1 when divided by p ; that is, for any one of the
series of numbers in (1), excluding the first and last, it is
possible to find one other, such that the product of the pair is of
the form M (p) + 1 .
Therefore the integers 2, 3, 4, ... (p2), the number of
which is even, can be associated in pairs such that the product of
each pair is of the form M (j?) + 1 .
Therefore by multiplying all these pairs together, we have
2.3.4 ... (p2) = M(p) + l;
thatis, 1.2.3.4 ... (pl) = (pl){M(p) + l} ;
whence \p  1 = M (p) +p  1 j
or 1 + 1^ — 1 is a multiple of p.
Cor. If 2p + l is a prime number /jp\ 2 + ( iy i s divisible
by 2p + l.
For by Wilson's Theorem 1 + \2p is divisible by 2p + 1 . Put
n = 2p + 1, so that p+ 1 = n —p ; then
\2p = 1.2.3.4 p(p+l)(p + 2) (n1)
= 1 (w1) 2(n2) 3(»3) ... p (np)
= a multiple of n + ( iy (\p) 2 .
Therefore 1 + (  l) p (\p) 2 is divisible by n or 2p + 1, and
therefore (^) 2 + (— l/ is divisible by 2;;+l.
23—2
356 HIGHER ALGEBRA.
435. Many theorems relating to the properties of numbers
can be proved by induction.
Example 1 . If p is a prime number, x p  x is divisible by p.
Let x p  x be denoted by f(x) ; then
/ (x + 1) / (x) = {x + 1)p  (x + 1)  {xP  x)
=px p ~ l + P< f~ 1) * p " 2 + . . . +px
J. • a
= a multiple of p, if p is prime [Art. 419.]
.. f(x + 1) =f(x) + a multiple of p.
If therefore/^) is divisible by^i, so also is/(.r + l); but
/(2) = 2* 5 ~ 2 = (1 + 1)^2,
and this is a multiple of p when p is prime [Art. 419] ; therefore / (3) is divisible
by p, therefore /(4) is divisible by^, and so on; thus the proposition is true
universally.
This furnishes another proof of Fermat's theorem, for if x is prime to p,
it follows that x p ~ J  1 is a multiple of p.
Example 2. Prove that 5 2,l + 2  24/i  25 is divisible by 576.
Let 5 2n+2  24?i  25 be denoted by f(n) ;
then /(?i+l) = 5 2n+4 24(w + l)25
= 5 2 .5 2w + 2 24n49;
.. f(n+l)  25/ (n) =25 (24n + 25)  24u  49
= 576 (n + 1).
Therefore if f(n) is divisible by 576, so also is /(u + 1); but by trial we
see that the theorem is true when n = l, therefore it is true when n=2, there
fore it is true when ?i = 3, and so on; thus it is true universally.
The above result may also be proved as follows :
52n+2 _ 2in  25 = 25 M + 1  24;i  25
= 25 (1 + 24)" 24rc25
= 25 + 25 . n . 24 + M (24 2 )  24n  25
= 576n + iW(576)
= i)/(576).
EXAMPLES. XXX. b.
1. Shew that 10 n + 3 . 4" + 2 + 5 is divisible by 9.
2. Shew that 2 . 7 n + 3 . 5 H  5 is a multiple of 24.
3. Shew that 4 . 6 n + 5 n + x when divided by 20 leaves remainder 9.
4. Shew that 8 . 7 n + 4" + 2 is of the form 24 (2r  1).
THEORY OF NUMBERS. 357
5. If p is prime, shew that 2 \p3 + l is a multiple of p.
6. Shew that a v, + l a is divisible 1>y 30.
7. Shew that the highest power of 2 contained in 2 r 1 is
2'';l.
8. Shew that 3 4 ' 1 +  + 5 2 ' 1 +* is a multiple of 14.
9. Shew that 3** +6 +160» a  56n 243 is divisible l»y 512.
10. Prove that the sum of the coefficients of the odd powers of x
in the expansion of (l+ < r+# 2 + # 3 + .r 4 ) n "" 1 , when n is a prime number
other than 5, is divisible by n.
11. If n is a prime number greater than 7, shew that n°l is
divisible by 504.
12. If n is an odd number, prove that ?i* 5 + 3>i 4 + 7>i 2  11 is a
multiple of 128.
13. If p is a prime number, shew that the coefficients of the terms
of (Ha?)** are alternately greater and less by unity than some mul
tiple of p.
14. If p is a prime, shew that the sum of the (pl) th powers of
any p numbers in arithmetical progression, wherein the common differ
ence is not divisible by p, is less by 1 than a multiple of p.
15. Shew that a 12  b 12 is divisible by 91, if a and b are both prime
to 91.
16. If p is a prime, shew that \p 2r 1 2r  1  1 is divisible by p.
17. If n— 1, n + 1 are both prime numbers greater than 5, shew
that n(?i 2 4) is divisible by 120, and ?i 2 (>i 2 + 16) by 720. Also shew
that n must be of the form 30£ or 30^ + 12.
18. Shew that the highest power of n which is contained in \n r ~ 1
, . n r — nr + r 1
is equal to .
n 1
19. If p is a prime number, and a prime to ]), and if a square
number c 2 can be found such that c 2 — a is divisible by p t shew that
l(pD
a 2  1 is divisible by p.
20. Find the general solution of the congruence
98a; 1=0 (mod. 139).
358 HIGHER ALGEBRA.
21. Shew that the sum of the squares of all the numbers less than
a given number N and prime to it is
?(i30J)09 + ?ci.»a.)a*.
and the sum of the cubes is
?(i3(x])(t9 + ?a^ci«a*.,
a,b,c... being the different prime factors of iV.
22. If jt? and q are any two positive integers, shew that \pq is
divisible by (£>)«. # and by (\q) p . \p.
23. Shew that the square numbers which are also triangular are
given by the squares of the coefficients of the powers of x in the ex
pansion of r 2> an< ^ ^hat the square numbers which are also
L — \)X f" X
pentagonal by the coefficients of the powers of x in the expansion of
1_
24. Shew that the sum of the fourth powers of all the numbers
less than N and prime to it is
5 \ a
gg(l<*)(li»)(l «*)...,
a, 6, c,... being the different prime factors of A".
25. If (f> (iV) is the number of integers which are less than JV and
prime to it, and if x is prime to JV, shew that
^^ 1 = (mod. JV).
26. If d v d 2 , d s , ... denote the divisors of a number JV, then
(f> (dj + (c? 2 ) + <£ (d 3 ) + ... =iV.
Shew also that
<t> (!) r  ; — 9~0( 3 ) r 1 ; — fi + 0( 5 )T^ — ™   odinf. =  ? ~ l — kJ.
* CHAPTER XXXI.
The General Theory of Continued Fractions.
*436. In Chap. xxv. we have investigated the properties of
Continued Fractions of the form a, +  —  . . . , where a , a , . . .
«,+ %+ 2 ' 3 '
are positive integers, and a^ is either a positive integer or zero.
We shall now consider continued fractions of a more general
type.
*i37. The most general form of a continued fraction is
~' „ "i rZl ' where a i> a 2> a 3> ■••> *,» K K ••• represent
a 1 =*= « 2 * a 3 =*=
any quantities whatever.
The fractions — , — , — , . . . are called components of the
a, a 2 a 3
continued fraction. We shall confine our attention to two cases;
(i) that in which the sign before each component is positive ;
(ii) that in which the sign is negative.
*438. To investigate the law of formation of the successive
convergents to the continued fraction
b i b 2 b 3
ergents are
6, a 2 b x a 3 .a 2 b i +b 3 .b l
The first three convergents are
AVe see that the numerator of the third convergent may be
formed by multiplying the numerator of the second convergent by
a 3 , and the numerator of the first by b 3 and adding the results
together ; also that the denominator may be formed in like
manner.
360 HIGHER ALGEBRA.
Suppose that the successive convergents are formed in a
similar way; let the numerators be denoted by p it p 2 , p 3 ...,
and the denominators by q lt q 2 , q 3 , ...
Assume that the law of formation holds for the n th con
vergent ; that is, suppose
p a p , + b p „, q =a q , + bq _.
In nl n — \ nl n — 2> J n nLn — 1 hIh—2
The (n+l) th convergent differs from the w th only in having
a h — — in the place of a ; hence
a
n + \
the (n+ l) tb convergent
If therefore we put
?> ^,=a nP +b ,,p ,, q + , = a +,q +b ,,q ,,
we see that the numerator and denominator of the (u + l) th con
vergent follow the law which was supposed to hold in case of the
?t th . But the law does hold in the case of the third convergent ;
hence it holds for the fourth ; and so on ; therefore it holds
universally.
*439. In the case of the continued fraction
b, b 2 b 3
«1  a 2 ~ CC 3 ~
we may prove that
Vn = a nPnl ~ kPn* , Qn = »«?«! ~ k<ln2 ',
a result which may be deduced from that of the preceding article
by changing the sign of b n .
*440. In the continued fraction
h K K
a 1 + a 2 + « 3 +
we have seen that
p =a p ,+bp „, q ^a q ,+bq a .
J n nl ii — \ nL n — 2' Jn nln — 1 nJn—2
but
GENERAL THEOHY OF CONTINUED FRACTIONS. Ml
?„ + , \9n QnJ'
and is therefore a proper fraction: hence " + 1 — — is numerically
&.+1 ft
less than — — — — , and is of opposite sign.
In In}
By reasoning as in Art. 335, we may shew that every con
vergent of an odd order is greater than the continued fraction,
and every convergent of an even order is less than the continued
fraction ; hence every convergent of an odd order is greater than
every convergent of an even order.
Thus 2 '^  ^ is positive and less than £ss=l  ^ 2 " ; hence
32/1+] 9 an 2ft*J 2*2/1
2 2/1 + 1 1 2/1 — 1
2 2/1 + 1 2 2/11
Also ?*=i  2  1 is positive and less than ^=*  &=s ■ hence
22/1 — 1 22/1 22/1 — 1 22/12
2 2/1 2 2/1 — 2
2 2/1 2 2/1 — 2
Hence the convergents of an odd order are all greater than
the continued fraction but continually decrease, and the con
vergents of an even order are all less than the continued fraction
but continually increase.
Suppose now that the number of components is infinite, then
the convergents of an odd order must tend to some finite limit,
and the convergents of an even order must also tend to some
finite limit ; if these limits are equal the continued fraction tends
to one definite limit ; if they are not equal, the odd convergents
tend to one limit, and the even convergents tend to a different
limit, and the continued fraction may be said to be oscillating; in
this case the continued fraction is the symbolical representation of
two quantities, one of which is the limit of the odd, and the other
that of the even convergents.
362
HIGHER ALGEBRA.
*441. To shew that the continued fraction
a x + a 2 + a 3 +
has a definite value if the limit of rf"" n+I when n is infinite is
greater than zero.
The continued fraction will have a definite value when n is
infinite if the difference of the limits of ^ and — is equal to zero.
9n+l ?«
Now
whence we obtain
n+lffnl fPn _ Pn1
?«+! W« £«!>
Pjt+X
<2n+l
_ £» = (_ 1 )l 6 "+^»l & ng»3 KV* KVl (P* _ Pl\
But
k.i?
k.M?
n+lin1
li+1 !7u + ^B+lS'nl a n+l Qn + J
k^g
an(
a n+l q n a.
■i K?i + &«£«*) _ «*»«+! + ^n+A^2 .
"n+lSnl ^n+lSnl ^n+1
b^ x q K
also neither of these terms can be negative; hence if the limit of
" n+1 is greater than zero so also is the limit of " + ; in which
case the limit of J**^ i s less than 1 : and therefore ^±i_ ^ £ a
Qn+i m t q„+i q n
the limit of the product of an infinite number of proper fractions,
and must therefore be equal to zero : that is, ^ and ■ tend to
?«+: q n
the same limit ; which proves the proposition.
.For example, in the continued fraction
V T 3
n~
3+5
a. a
2n+l +
• • »
Lim f^ 1 = Lim . 7 \, 2 ' = 4 :
*«+i (n+iy
and therefore the continued fraction tends to a definite limit.
GENERAL THEORY OF CONTINUED FRACTIONS. 3G3
*442. In the continued fraction
c\i, cl H;„
1 2 A
if the denominator of every component exceeds the numerator by
unity at least, the convergents are positive fractions in ascending
order of magnitude.
By supposition — ' , * , — 3 , . . . are positive proper fractions
a i a 2 a ?
in each of which the denominator exceeds the numerator by
unity at least. The second convergent is 1 , and since a i
a   2
exceeds ft t by unity at least, and * is a proper fraction, it follows
tliat «, 2 is greater than ft,; that is, the second convergent is
a positive proper fraction. In like manner it may be shewn
that 2 , is a positive proper fraction ; denote it by f, then
the third convergent is — ! . , and is therefore a positive proper
fraction. Similarly we may shew that — — — — — is a positive
proper fraction ; hence also the fourth convergent
ft, K K K
«1  CV 2 ~ %~ a ,
is a positive proper fraction ; and so on.
Again, p — a p , — ft p ., q—aa , — ft q „ ;
O ' i ii nl ii — 1 »/ n — 2' 2 ii iuii I niii 2 '
^
2
ie same sign.
hence ^s±J  ^ and ;  n  ^=* have tl
But P °~  ^ = g «\   1 = h ^ , and is therefore positive ;
9 2 <?i <V* 2 ft 2 r 'i <M 2
hence ^ 2 >^ , ^ > ^ 2 , ^>^ ?3 ; and so on; which proves the
9jt 7, ( l, 7, <7 4 %
proposition.
364 HIGHER ALGEBRA.
Cor. If the number of the components is infinite, the con
vergents form an infinite series of proper fractions in ascending
order of magnitude ; and in this case the continued fraction must
tend to a definite limit which cannot exceed unity.
*443. From the formula
Pn = <*nPni + hPnt* 9n = a &n, + &.&_«>
we may always determine in succession as many of the con
vergent^ as we please. In certain cases, however, a general
expression can be found for the 11 th convergent.
c c c
Example. To find the w th convergent to  —
5 5 5
We have p n = Sp,^  6p n _ 2 ; hence the numerators form a recurring series
any three consecutive terms of which are connected by the relation
Pn " 5p«! + §Pn2'
Let S =p 1 +p. 2 x +Prf? + • • • +p n x n ~ 1 + . . . ;
Pl + (P2~ 5 Pl) X
then, as in Art. 325, we have S=
1  5x + 6a; 2
But the first two convergents are  , =^ ;
6 18 12
1  hx + Qx l3x 1  2x '
whence p n = 18 . 3" 1  12 . 2' 1 " 1 = 6 (3"  2").
Similarly if S' = q x + q& + q 3 x* +...+ q n x n ~ x + . . . ,
we find ^ = ___ = i ___ r _ ;J
whence g n = 9 . 3* 1  4 . 2* 1 = S'^ 1  2' l + ] .
y w _ 6(3"2 w )
""" ffn ~ 3n+1  2?l+1 '
This method will only succeed when a„ and b n are constant
for all values of n. Thus in the case of the continued fraction
... , we may shew that the numerators of the
a + a+ a+
successive convergents are the coefficients of the powers of x in
the expansion of ^ 72 , and the denominators are the
ft I ti / Y*
coefficients of the powers of x in the expansion of ^ 72 .
1 ~ ~ OjX — ox
GENERAL THEORY OF CONTINUED FRACTIONS.
305
*444. For the investigation of the general values of p n and q n
the student is referred to works on Finite Differences ; it is only
in special cases that these values can be found by Algebra. The
following method will sometimes be found useful.
12 3
Example. Find the value of — =— 5—
1 + z + o +
The same law of formation holds for p n and q n ; let us tal<e i* n to denote
either of them ; then u n = nu n _ x + nu n _ 2 ,
u n  (n + 1) «„_! =  (u B _,  ?w n _ 2 ).
Similarly, i^j  RU tt _ s =  (« u _ 2 ra 1 « n _ 3 ).
or
whence by multiplication, we obtain
u n (n + l)u n _ 1 = (iy^(u 2 3u i ).
1 2
The first two convergents are  , T ; hence
1 A
p n (n + l) Pn ^=(l) n \ q n (n + l)q n i = ( I)"" 2 
Tims
^n Pn1 (" !)
?ll
0n
gn1 _ (~ I)"" 2
7i + l m lra+1 iw+l to j» + l '
At! _ Art = (~ l)" 8
In nl In
ffn1 9»2
In
n  1 In
Ps_Pi
13 2'
2
1
2'
?3 _ <h
3 2
1
[3'
'2 2 2'
3i
whence, by addition
n + l
12 3 + 4
+
7 a
, 1 1 1
= 17^+1 i "77 +
n+1 2 3 1
Lit 1 .
j n+1 '
, ( l) n ~ a
n + 1
By making n infinite, we obtain
e) eV
2n e
which is therefore the value of the given expression
366 HIGHER ALGEBRA.
*445. If every component of — — — — — — ... is a proper
3j j + a o + a 3 +
fraction with integral numerator and denominator, the continued
fraction is incommensurable.
For if possible, suppose that the continued fraction is com
mensurable and equal to ^ , where A and B are positive integers ;
XL
then 7 = *—m , where f, denotes the infinite continued fraction
— * ^ ... ; hence f = — *— —  = ^ suppose. Now A, B, « x , 6 X
are integers and f is positive, therefore C is a positive integer.
C b
Similarly = = — *= , where f a denotes the infinite continued
J B a 2 +f
fraction — ■* — * . . . ; hence / = — S^ * = 7* suppose ; and as
a 3 + a +
before, it follows that D is a positive integer ; and so on.
. . B C D , .. , 5 . ,
Again, 7 , ■= , jy , ... are proper tractions ; tor j is less
than — , which is a proper fraction : ■= is less than * ; ^ is
a x B a s O
less than — ; and so on.
Thus A, B, C, D, ... form an infinite series of positive integers
in descending order of magnitude ; which is absurd. Hence the
given fraction cannot be commensurable.
The above result still holds if some of the components are
not proper fractions, provided that from and after a fixed com
ponent all the others are proper fractions.
For suppose that — and all the succeeding components are
n
proper fractions ; thus, as we have just proved, the infinite con
tinued fraction beginning with s is incommensurable ; denote
a
n
7) k
it by k, then the complete quotient corresponding to — n is ^
in
and therefore the value of the continued fraction is ^^ — / "~ 2 .
9nl + hn2
GENERAL THEORY OF CONTINUED FRACTIONS. 3G7
V P
This cannot be commensurable unless — n_1 = l ^^ • and this
tfnl ?««
condition cannot hold unless ?2=a = *=2 , Pn = B = P^* t ;ultl
?n a ?n3 Qnz 9n4
P P
finally H= — ; that is £>6 = 0, which is impossible ; hence the
% . ?,
given fraction must be incommensurable.
1 1 K
*446. //* eirary component of —  — « — * ... ?'s a proper
a i "" a a ~ a 3 ""
fraction with integral numerator and denominator, and if the
value of the infinite continued fraction beginning with any com
ponent is less than unity, the fraction is incommensurable.
The demonstration is similar to that of the preceding article.
* EXAMPLES. XXXI. a.
1. Shew that in the continued fraction
\ _h h_
a x  a 2  a 3  '
Pn = a nPn  1 ~ ^nPn  2 J Qn = a n9.n \~ "?i?n  2 ■
" m
2. Convert  '^  ) into a continued fraction with unit nume
rators.
3. Shew that
« V*+6=«+^ ^
(2 ) V^=« 2 _ ±r. .......
4. In the continued fraction — — — — — — .... if the denominator
«1~ a %~ a 3~
of every component exceed the numerator by unity at least, shew that
p n and q n increase with n.
5. If a lf a,, rtg,..^,, are in harmonical progression, shew that
a n 1 1 1 1 «2
^"2^ 2^ 2~=" 2^ SJ'
368 HIGHER ALGEBRA.
6. Shew that
cc+  —   —  ... + [a  ... = 2a 2 ,
V 2a+ 2a + J V 2a 2a J
\/ 1 1 \ ,
■7T o^)= a ~
and ( a + s ■ — : ... ) ( a 
2«+ 2a+"7\ 2a 2a "7 2a 2  2a' 2
7. In the continued fraction
b b b
a+ a+ a+
shew that p n + x = 6a n , bq n + 1  ap n + 1 = % n _ x .
b b b a x — 8P
8. Shew that —  = 6.^7 — ^n i
a+ a+ a+ a x + 1 — p x + 1
.v being the number of components, and a, /3 the roots of the equation
k 2 — ak — b = 0.
9. Prove that the product of the continued fractions
J_ L _L A_ ,7 _1_ J_ J_ _ x _
6+ c+ d+ a+ '"' c + 6+ a+ 0?+ '"'
is equal to — 1 .
Shew that
1 4 9 64 (?i 2 l) 2 (n + l)(w + 2)(2/i + 3)
10.
11.
12.
1 5 13 25 » 2 +(»+l) 2 6
L JL _§_ ^ 2 ~1 _ ?*fo+ 3 )
1 5 7 2» + l~ 2 '
£ i ji £±L 5± 1+ i+i« + . + ...+«.
2 3 4 ?i + l— ?i + 2 I— ! — L_
13 . !> «* 2=1 =.1.
1 3 4 5 w+l
14.
15.
4 6 8 2/i + 2 2(e 2 l)
1+ 2+ 3+ n+ e 2 +l
3.3 3.4 3.5 3(n + 2) _'6(2 e3+l)
1+ 2+ 3+ n+ " 5e 3 2
16. If u, = v, Ucy = f , Uo = —f , , each successive fraction
1 a a + e> •* a + 26
bein» formed by taking the denominator and the sum of the numerator
and denominator of the preceding fraction for its numerator and denomi
nator respectively, shew that u „=**—= — .
CONVERSION OF SERIES INTO CONTINUED FRACTIONS. 309
17. Prove that the n ih convergent to the continued fraction
J' f ■)' y'H +1 y.
is
r+l r+l r+l »* +1 l '
18. Find the value of — \ % %
cij + l a.,+ l a 3 +l '
a 19 a 2 > a 3v being positive and greater than unity.
19. Shew that the n lh convergent to 1   —  is equal to
the (2m — 1 ) lh convergent to n —  — , —  —
v ; 6 1+2+1+ 2 +
20. Shew that the 3n th convergent to
1111111 n
is
21. Shew that
5 2 1 5 2 1 5 3>i + l
1 2 3 3d
2+ 3+ 4+ e2 '
hence shew that e lies between 2§ and 2 r 8 T .
Conversion of Series into Continued Fractions.
*417. It will be convenient here to write the series in the form
Put
1 1 1
— +— + — +..
^6 1 u 2 u 3
1 1
• —
1
.... + —
u
n
1
u u ., u
r r+l r
+ x '
r
then (u r + x r ) (u r+ , + lb) = uu r+ x ,
u
u + u , ,
r r+ 1
1 1
— + — =
1
i
<
LC/lH/t?
or
»i
u { + u ?
10
H. H. A.
24
370 HIGHER ALGEBRA.
Similarly,
2
11111 1 u y
— + — + —=— + = — — — 
u x u 2 u 3 u x u s + x 2 u x  u x + u 2 + x a
12 2
u x U Q
u x  u x + u 2 — u 2 + u 3 '
and so on ; hence generally
1 1 1
1
— +— + — + .
.. + —
U l U 2 U 3
U n
1
11*
U 2
<,,
u x —
u x
+ u 2 
U 2 + U 3~
«*«_!+«*,
Example 1. Express as a continued fraction the series
1 x x 2 , ,._ x n
+ + (!) n
Put
a a a x a^a.^ a a 1 a 2 ...a n
1 x 1
a n a n a n+l a n + Vn
then (a n + y n ) (a n+l  x) = a n a n+1 ;
Hence
a„x
•'• Un= * — •
a n+1  x
# 1 1 a u x
a a a x a + y a + a^x
. 1 X X 2 1 X ( 1 X \ 1 X
Again, + = ) =  —
a a^ fl <V2 a a \aj a x aj a a (a 1 
+ Vi)
a x
«o+ a 1 + y 1 x
rt .r a x x
a + a x  x + a 2 x'
X X 2 ^'^
and generally : 1 ■ ...+( l) n
a o,q(Ii a^a^ac, a a 1 a.2...a n
a x a^ a n \ x
a o + <h ~ x + a 2 ~ x + a n~ x
Example 2. Express log(l+£) as a continued fraction.
/y»2 />»3 <j*4
We have log(l + a;) = .T + —  — +
The required expression is most simply deduced from the continued
fraction equivalent to the series
x x 2 x s x x
+ +
«! « 2 a 3 a 4
CONVERSION OF SERIES INTO CONTINUED FRACTIONS. .',71
1 x 1
By putting ■ — = ,
, . • "n x
we obtain V,l= ;
a n+l  a n x
hence we have
X .T 3 .T 3 X A X (IfX <i 2 2 x <i.."x
<l l rt 2 (l 3 (J i ( 'l + a 2~ a l X + a 3~ U, 2 X + a ,~ a i x +
x Vx 2°x 3 2 .r
.. log (! + *) =
1+ 2x+ 32x+ ~iSx +
*44S. In certain cases we may simplify the components of the
continued fraction by the help of the following proposition :
The continued fraction
&. K h K
a x + a 2 + a 3 + a 4 +
is equal to the continued fraction
C A 5 C A ^3 C 3 C A ... •
c x a x + c 2 a 2 + c 3 a 3 + c/c 4 +
where c n c 2 , c 3 , c 4 , are any quantities whatever.
7 7
Let /" denote — — — — ; tlien
the continued fraction =
«i +/i c i«i + c i/i
Let /„ denote 3  — — ... ; then
«2 + / 2 C 2<\. + C 2 X '
£ Q Q
Similarly, c f =   '^~ ; and so on; whence the proposition
C 3 rt 3 + G J 3
is established.
24—2
372 HIGHER ALGEBBA..
^EXAMPLES. XXXI. b.
Shew that
1111 , ,. 1
1. + + + (l) Jl —
U Q Wj u 2 m 3 u n
1 ic 2 u t 2 u 2 n _
n1
« + Mj  U Q + U. 2  ?<<! + U n ~ u
n lv nl
1 X X 2 a?
2. + — + + +
3.
&n ClfiCCt Cl^QztCto CIqCC iCtv • • • &JI
J. Ctry.0 iXxJu ^n 1 *^
r—\ r r+1 ?' + 2
?•  2 ~~ »~ r + 1 r + 2
.2^ 1111 , ...
4. — ^—^ — . — , — . — to n quotients.
n+l 141—4
r , 11 1114 9 n 2
5. l + s + « + +
2 3 ?i+l 1 3 5 7 2n+l
11 111 4 ?i 4
6. i»+oa+ +
l 2 ' 2 2 (n + 1) 2 1 l 2 + 2 2  n*+(n+Yf'
x x 2x 3x
7. e x = l +
1 x + 2 x + S x+4
 1111 la
8. r + i 5—, + ...= — 
a ab abc abed a+ bl+ c— 1+ d—l +
 , 1 1 1 1 1 r 73 r 5
9. l+ + i + a + iB+ ...=i+ rn — ,5 , , yrr •••
r ?** ? ,J r lb r — r 3 + 1 — r 5 + 1 — r + 1 —
a x + a 2 + a 3 + ' a n 1+ « x + « 2 f « 3 + ' ' <(„_
Hl
u. if p=4 ,4 4 ■ q
a+ b+ c+ ' c b+ c+ d+ '
shew that P (a+ 1 + Q) = a + Q.
o
12. Shew that 1 \ ... is equal to the con
9i M2 Ms Mt
9* />» />»
tinued fraction — — .... where #., q„, a*. ... are the
denominators of the successive convergents.
CHAPTER XXXII.
PROBABILITY.
449. Definition. If an event can happen in a ways and fail in
b ways, and each of these ways is equally likely, the probability,
or the chance, of its happening is z , and that of its failing is
r r ° a + b G
a + b '
For instance, if in a lottery there are 7 prizes and 25 blanks,
. 7
the chance that a person holding 1 ticket will win a prize is — ,
25
and his chance of not winning is — .
Oa
450. The reason for the mathematical definition of pro
bability may be made clear by the following considerations :
If an event can happen in a ways and fail to happen in b
ways, and all these ways are equally likely, we can assert that the
chance of its happening is to the chance of its failing as a to b.
Thus if the chance of its happening is represented by ka, where
k is an undetermined constant, then the chance of its failing
will be represented by kb.
.. chance of happening + chance of failing = k (a + b)
Now the event is certain to happen or to fail ; therefore the sum
of the chances of happening and failing must represent certainty.
If therefore we agree to take certainty as our unit, we have
1 = k (a + b), or k — T :
v ' a + b
.. the chance that the event will happen is
and the chance that the event will not happen is
a + b
b
a + b
Cor. If p is the probability of the happening of an event,
the probability of its not happening is 1 — p.
374 HIGHER ALGEBRA.
451. Instead of saying that the chance of the happening of
an event is T , it is sometimes stated that the odds are a to b
a + o
in favour of the event , or b to a against the event.
452. The definition of probability in Art. 449 may be given
in a slightly different form which is sometimes Useful. If c is the
total number of cases, each being equally likely to occur, and of
these a are favourable to the event, then the probability that the
event will happen is  , and the probability that it will not
c
happen is 1 .
Example 1. What is the chance of throwing a number greater than 4
with an ordinary die whose faces are numbered from 1 to 6?
There are 6 possible ways in which the die can fall, and of these two
are favourable to the event required ;
therefore the required chance =  =  .
Example 2. From a bag containing 4 white and 5 black balls a man
draws 3 at random ; what are the odds against these being all black ?
The total number of ways in which 3 balls can be drawn is 9 <7 3 , and
the number of ways of drawing 3 black balls is 5 C 3 ; therefore the chance
of drawing 3 black balls
~*C % ~ 9.8.7 = 42 '
Thus the odds against the event are 37 to 5.
Example 3. Find the chance of throwing at least one ace in a single
throw with two dice.
The possible number of cases is 6 x 6, or 36.
An ace on one die may be associated with any of the 6 numbers on the
other die, and the remaining 5 numbers on the first die may each be asso
ciated with the ace on the second die ; thus the number of favourable cases
is 11.
Therefore the required chance is — .
3b
Or we may reason as follows :
There are 5 ways in which each die can be thrown so as not to give an
ace ; hence 25 throws of the two dice will exclude aces. That is, the chance
25
of not throwing one or more aces is ^ ; so that the chance of throwing one
36
ace at least is 1  ^ , or ^,
do oo
PROBABILITY. 375
Example 4. Find the chance of throwing more than 15 in one throw with
3 dice.
A throw amounting to 18 must be made up of 6, G, G, and this can occur
in 1 way; 17 can be made up of G, G, 5 which can occur in 3 ways; 16 may
be made up of G, G, 4 and 6, 5, 5, each of which arrangements can occur in
3 ways.
Thereforo the number of favourable cases is
1 + 3 + 3 + 3, or 10.
And the total number of cases is 6 3 , or 21G;
therefore the required chance =^ =
21G 108
Example 5. A has 3 shares in a lottery in which there are 3 prizes and
6 blanks ; B has 1 share in a lottery in which there is 1 prize and 2 blanks :
shew that A's chance of success is to ZJ's as 1G to 7.
A may draw 3 prizes in 1 way ;
3 2
he may draw 2 prizes and 1 blank in —^— x 6 ways :
JL • m
6 . 5
he may draw 1 prize and 2 blanks in 3 x r^r ways ;
JL • m
the sum of these numbers is 64, which is the number of ways in which A can
9.8.7
win a prize. Also he can draw 3 tickets in ' ' , or 84 ways ;
therefore 4's chance of success = — r = —  .
84 21
Z»"s chance of success is clearly  ;
o
1 C 1
therefore A 's chance : B's chancer— : 
— L O
= 16 : 7.
6.5.4
Or we might have reasoned thus: A will get all blanks in * ' , or
20 5
20 ways ; the chance of which is —. , or — ;
J ' 84 21
therefore A's chance of success = 1 — = —.
— 1 ZL
453. Suppose that there are a number of events A, B, C,...,
of which one must, and only one can, occur ; also suppose that
a, b, c, ... are the numbers of ways respectively in which these
events can happen, and that each of these ways is equally likely
to occur ; it is required to find the chance of eacli event.
The total number of equally possible ways is a + b + c+ ...,
and of these the number favourable to A is a; hence the chance
376 HIGHER ALGEBRA.
that A will happen is = . Similarly the chance that B
rr a + b + c+ ... J
will happen is . : and so on.
ri a + b + c + ...
454. From the examples we have given it will be seen that
the solution of the easier kinds of questions in Probability requires
nothing more than a knowledge of the definition of Probability,
and the application of the laws of Permutations and Combina
tions.
EXAMPLES. XXXII. a.
1. In a single throw with two dice find the chances of throwing
(1) five, (2) six.
2. Prom a pack of 52 cards two are drawn at random ; find the
chance that one is a knave and the other a queen.
3. A bag contains 5 white, 7 black, and 4 red balls: find the
chance that three balls drawn at random are all white.
4. If four coins are tossed, find the chance that there should be
two heads and two tails.
5. One of two events must happen : given that the chance of the
one is twothirds that of the other, find the odds in favour of the other.
6. If from a pack four cards are drawn, find the chance that they
will be the four honours of the same suit.
7. Thirteen persons take their places at a round table, shew that
it is five to one against two particular persons sitting together.
8. There are three events A, B, C, one of which must, and only
one can, happen; the odds are 8 to 3 against A, 5 to 2 against B: find
the odds against C.
9. Compare the chances of throwing 4 with one die, 8 with two
dice, and 12 with three dice.
10. In shuffling a pack of cards, four are accidentally dropped ; find
the chance that the missing cards should be one from each suit.
11. A has 3 shares in a lottery containing 3 prizes and 9 blanks ;
B has 2 shares in a lottery containing 2 prizes and 6 blanks : compare
their chances of success.
12. Shew that the chances of throwing six with 4, 3, or 2 dice
respectively are as 1 ; 6 ; 18,
PROBABILITY. 377
13. There are three works, one consisting of 3 volumes, one of 4,
and the other of 1 volume. They are placed on a shelf at random ;
prove that the chance that volumes of the same works are all together
3
18 140 '
14. 1 and B throw with two dice ; if A throws 9, find i>'s chance
of throwing a higher number.
15. The letters forming the word Clifton are placed at random in
a row : what is the chance that the two vowels come together ?
16. In a hand at whist what is the chance that the 4 kings are
held by a specified player ]
17. There are 4 shillings and 3 halfcrowns placed at random in
a line : shew that the chance of the extreme coins being both half
crowns is  . Generalize this result in the case of m shillings and
n halfcrowns.
455. We have hitherto considered only those occurrences
which in the language of Probability are called Single events.
When two or more of these occur in connection with each other,
the joint occurrence is called a Confound event.
For example, suppose we have a bag containing 5 white
and 8 black balls, and two drawings, each of three balls, are
made from it successively. If we wish to estimate the chance
of chawing first 3 white and then 3 black balls, w^e should be
dealing with a compound event.
In such a case the result of the second drawing might or
might not be dependent on the result of the first. If the balls
are not replaced after being drawn, then if the first drawing gives
3 white balls, the ratio of the black to the white balls remaining
is greater than if the first drawing had not given three white;
thus the chance of drawing 3 black balls at the second trial
is affected by the result of the first. But if the balls are re
placed after being drawn, it is clear that the result of the second
drawing is not in any way affected by the result of the first.
We are thus led to the following definition :
Events are said to be dependent or independent according as
the occurrence of one does or does not affect the occurrence of the
others. Dependent events are sometimes said to be contingent.
378 HIGHER ALGEBRA.
456. If there are two independent events the respective pro
babilities of which are known, to find the probability that both will
happen.
Suppose that the first event may happen in a ways and fail
in b ways, all these cases being equally likely ; and suppose that
the second event may happen in a' ways and fail in b' ways,
all these ways being equally likely. Each of the a + b cases may
be associated with each of the a + b' cases, to form (a + b) (a! + b')
compound cases all equally likely to occur.
In aa' of these both events happen, in bb' of them both fail,
in ab' of them the first happens and the second fails, and in a'b
of them the first fails and the second happens. Thus
aa
(a + b){a'+b')
bb'
(a + b)(a+b')
ab'
(a + b)(a'+b')
a'b
(a + b)(a'+b')
is the chance that both events happen ;
is the chance that both events fail ;
is the chance that the first happens and the second
fails ;
is the chance that the first fails and the second
happens.
Thus if the respective chances of two independent events are
p and p\ the chance that both will happen is pp'. Similar
reasoning will apply in the case of any number of independent
events. Hence it is easy to see that if p lf p 2 , p 3 , ... are the
respective chances that a number of independent events will
separately happen, the chance that they will all happen is
p x p 2 p 3 ... ; the chance that the two first will happen and the rest
fail is 2\Po (1 — P 3 ) (1 —pj' > an d similarly for any other par
ticular case.
457. If p is the chance that an event will happen in
one trial, the chance that it will happen in any assigned suc
cession of r trials is p r ' ; this follows from the preceding article
by supposing
P 1 =P 2 =P 3 = =P
To find the chance that some one at least of the events will
happen we proceed thus : the chance that all the events fail
is (1 p x ) (1 ]).,) (1 p 3 ) .j and except in this case some one
of the events must happen ; hence the required chance is
PROBABILITY. 379
Example 1. Two drawings, each of 3 balls, arc made from a bag con
taining 5 wbitc and 8 black balls, the balls being replaced before tbe second
trial : find the chance that the first drawing will give 3 white, and the second
3 black balls.
The number of ways in which 3 balls may be drawn is 13 C 3 ;
3white 5C.,;
3black *C Z .
Therefore the chance of 3 white at the first trial = •— ^f
and the chance of 3 black at the second trial =
1.2" 1.2.3 143 '
8.7.6 13. 12. 11 _ 28
1.2.3 : 1.2.3 = 143*
•j >
therefore the chance of the compound event = x " = < .
14o 143 20449
Example 2. In tossing a coin, find the chance of throwing head and tail
alternately in 3 successive trials.
Here the first throw must give either head or tail ; the chance that the
second gives the opposite to the first is  , and the chance that the third throw
is the same as the first is ^ .
Therefore the chance of the compound event = x = = j .
2 2 4
Example 3. Supposing that it is 9 to 7 against a person A who is now
35 years of age living till he is 65, and 3 to 2 against a person B now 45
living till he is 75 ; find the chance that one at least of these persons will be
alive 30 years hence.
9
The chance that A will die within 30 years is — ;
3
the chance that B will die within 30 years is ^ ;
9 3 27
therefore the chance that both will die is ^ x ' , or —  ;
lb o 8U
therefore the chance that both will not be dead, that is that one at least will
. .. . , 27 53
be alive, >sl8o> or.
458. By a slight modification of the meaning of the symbols
in Art. 45G, we are enabled to estimate the probability of the
concurrence of two dependent events. For suppose that when the
first event has happened^ a denotes the number of ways in which
the second event can follow, and b' the number of ways in which
it will not follow ; then the number of ways in which the two
380 HIGHER ALGEBRA.
events can happen together is aa\ and the probability of their
aa
concurrence is . 7 , , , — tr •
(« + o) (a + o )
Thus if p is the probability of the first event, and p' the
contingent probability that the second will follow, the probability
of the concurrence of the two events is pp .
Example 1. In a hand at whist find the chance that a specified player
holds both the king and queen of trumps.
Denote the player by A ; then the chance that A has the king is clearly
13
^; for this particular card can be dealt in 52 different ways, 13 of which fall
to A. The chance that, when he has the king, he can also hold the queen is
12
then — : for the queen can be dealt in 51 ways, 12 of which fall to A.
ol
m * « , . n 13 12 1
Therefore the chance required = — x ^ =  .
u 52 ol 17
Or we might reason as follows :
The number of ways in which the king and the queen can be dealt to A is
equal to the number of permutations of 13 things 2 at a time, or 13 . 12.
And similarly the total number of ways in which the king and queen can be
dealt is 52 . 51.
13 . 12 1
Therefore the chance = „ * „„ = — , as before.
52.51 17
Example 2. Two drawings, each of 3 balls, are made from a bag con
taining 5 white and 8 black balls, the balls not being replaced before the
second trial: find the chance that the first drawing will give 3 white and
the second 3 black balls.
At the first trial, 3 balls may be drawn in 13 C 3 ways ;
and 3 white balls may be drawn in 5 C 3 ways;
5.4 13 . 12 . 11 5
therefore the chance of 3 white at first trial :
1.2" 1.2.3 143
When 3 white balls have been drawn and removed, the bag contains
2 white and 8 black balls ;
therefore at the second trial 3 balls may be drawn in 10 C 3 ways ;
and 3 black balls may be drawn in 8 C 3 ways ;
therefore the chance of 3 black at the second trial
8.7.6 . 10.9.8 _ 1_ m
"1.2.3 ' 1.2.3 ~ 15 '
therefore the chance of the compound event
5 7 7
x == . =
143 15 429
The student should compare this solution with that of Ex. 1, Art. 457.
PROBABILITY. 381
459. If an event can happen in ttvo or more different ways
which are mutually exclusive, the chance that it wilt happen is
the sum of the chances of its happening in these different ways.
This is sometimes regarded as a selfevident proposition arising
immediately out of the definition of probability. It may, how
ever, be proved as follows :
Suppose the event can happen in two Avays which cannot
concur ; and let ^ , ■=* be the chances of the happening of the
event in these two ways respectively. Then out of bfi 2 cases
there are a x b 2 in which the event may happen in the first way,
and a b J ways in which the event may happen in the second;
and tliese ivays cannot concur. Therefore in all, out of b l b 2 cases
there are a,b„ + a k ,b, cases favourable to the event: hence the
chance that the event will happen in one or other of the two
ways is
a x b 2 + a 2 b x a_ x a,
bh 6, bf
12
Similar reasoning will apply whatever be the number of ex
clusive ways in which the event can happen.
Hence if an event can happen in n ways which are mutually
exclusive, and if p lt p a , p^ Pn are the probabilities that the
event will happen in these different ways respectively, the pro
bability that it will happen in some one of these ways is
Pi+Pl+Pa* +Pn
Example 1. Find the chance of throwing 9 at least in a single throw
with two dice.
4
9 can be made up in 4 ways, and thus the chance of throwing 9 is , .
3
10 can be made up in 3 ways, and thus the chance of throwing 10 is ^ .
2
11 can be made up in 2 ways, and thus the chance of throwing 11 is .
12 can be made up in 1 way, and thus the chance of throwing 12 is ^ .
Now the chance of throwing a number not less than 9 is the sum of these
separate chances ;
.*. the required chance = <V/ . = n .
ou lb
382 HIGHER ALGEBRA.
Example 2. One purse contains 1 sovereign and 3 shillings, a second
purse contains 2 sovereigns and 4 shillings, and a third contains 3 sovereigns
and 1 shilling. If a coin is taken out of one of the purses selected at
random, find the chance that it is a sovereign.
Since each purse is equally likely to be taken, the chance of selecting
the first is  ; and the chance of then drawing a sovereign is  ; hence the
chance of drawing a sovereign so far as it depends upon the first purse is
 x j , or =^ . Similarly the chance of drawing a sovereign so far as it
3 4 12
12 1
depends on the second purse is  x  , or  ; and from the third purse the
3 6 9
13 1
chance of drawing a sovereign is  x  , or  ;
.. the required chance = — +  + . =  .
x*5 y tc o
460. In the preceding article we have seen that the pro
bability of an event may sometimes be considered as the sum of
the probabilities of two or more separate events ; but it is very
important to notice that the probability of one or other of
a series of events is the sum of the probabilities of the separate
events only when the events are mutually exclusive, that is, when
the occurrence of one is incompatible with the occurrence of any
of the others.
Example. From 20 tickets marked with the first 20 numerals, one is
drawn at random : find the chance that it is a multiple of 3 or of 7.
The chance that the number is a multiple of 3 is — , and the chance that
2
it is a multiple of 7 is — ; and these events are mutually exclusive, hence the
. , , 6 2 2
required chance is — + ^ , or  .
But if the question had been: find the chance that the number is a
multiple of 3 or of 5, it would have been incorrect to reason as follows :
Because the chance that the number is a multiple of 3 is — , and the
4
chance that the number is a multiple of 5 is — , therefore the chance that
6 4 1
it is a multiple of 3 or 5 is ^ + ^ , or  . For the number on the ticket
might be a multiple both of 3 and of 5, so that the two events considered
are not mutually exclusive.
461. It should be observed that the distinction between
simple and compound events is in many cases a purely artificial
PROBABILITY. 383
one ; in fact it often amounts to nothing more than a distinction
between two different modes of viewing the same occurrence.
Example. A bag contains 5 white and 7 black balls; if two balls arc
drawn what is the chance that one is white and the other black?
(i) Regarding the occurrence as a simple event, the chance
= (5*7H.=C 2 = 6  6 .
(ii) The occurrence may be regarded as the happening of one or other
of the two following compound events :
(1) drawing a white and then a black ball, the chance of which is
12 * 11 ° r 132 ;
(2) drawing a black and then a white ball, the chance of which is
7 5 35
i2 X ir 0r 132'
And since these events are mutually exclusive, the required chance
j$5 ^5_ _35
 132 + 132~66'
It will be noticed that we have here assumed that the chance of drawing
two specified balls successively is the same as if they were drawn simul
taneously. A little consideration will shew that this must be the case.
EXAMPLES. XXXII. b.
1. What is the chance of throwing an ace in the first only of two
successive throws with an ordinary die ?
2. Three cards are drawn at random from an ordinary pack : find
the chance that they will consist of a knave, a queen, and a king.
3. The odds against a certain event are 5 to 2, and the odds in
favour of another event independent of the former are 6 to 5 ; find the
chance that one at least of the events will happen.
4. The odds against A solving a certain problem are 4 to 3, and
the odds in favour of B solving the same problem are 7 to 5 : what is
the chance that the problem will be solved if they both try 1
5. What is the chance of drawing a sovereign from a purse one
compartment of which contains 3 shillings and 2 sovereigns, and the
other 2 sovereigns and 1 shilling ?
6. A bag contains 17 counters marked with the numbers 1 to 17.
A counter is drawn and replaced; a second drawing is then made:
what is the chance that the first number drawn is even and the second
odd?
384 HIGHER ALGEBRA.
7. Four persons draw each a card from an ordinary pack: find
the chance (1) that a card is of each suit, (2) that no two cards are of
equal value.
8. Find the chance of throwing six with a single die at least once
in five trials.
9. The odds that a book will be favourably reviewed by three
independent critics are 5 to 2, 4 to 3, and 3 to 4 respectively ; what is
the probability that of the three reviews a majority will be favourable ?
10. A bag contains 5 white and 3 black balls, and 4 are successively
drawn out and not replaced ; what is the chance that they are alternately
of different colours %
11. In three throws with a pair of dice, find the chance of throwing
doublets at least once.
12. If 4 whole numbers taken at random are multiplied together
shew that the chance that the last digit in the product is 1, 3, 7, or 9
. 16
1S 625'
13. In a purse are 10 coins, all shillings except one which is a
sovereign ; in another are ten coins all shillings. Nine coins are taken
from the former purse and put into the latter, and then nine coins are
taken from the latter and put into the former : find the chance that
the sovereign is still in the first purse.
14. If two coins are tossed 5 times, what is the chance that there
will be 5 heads and 5 tails \
15. If 8 coins are tossed, what is the chance that one and only
one will turn up head?
16. A, B, C in order cut a pack of cards, replacing them after each
cut, on condition that the first who cuts a spade shall win a prize : find
their respective chances.
17. A and B draw from a purse containing 3 sovereigns and
4 shillings : find their respective chances of first drawing a sovereign,
the coins when drawn not being replaced.
18. A party of n ^persons sit at a round table, find the odds against
two specified individuals sitting next to each other.
19. A is one of 6 horses entered for a race, and is to be ridden by
one of two jockeys B and C. It is 2 to 1 that B rides A, in which
case all the horses are equally likely to win ; if C rides A, his chance
is trebled : what are the odds against his winning ?
20. If on an average 1 vessel in every 10 is wrecked, find the chance
that out of 5 vessels expected 4 at least will arrive safely.
PROBABILITY. 385
462. The probability of the happening of an event in one
trial being known, required the probability of its happening once,
twice, three times, ... exactly in n trials.
Let p be the probability of the happening of the event in
a single trial, and let q = 1 p\ then the probability that the
event will happen exactly r times in n trials is the (r + l) th term
in the expansion of (q + p)*.
For if we select any particular set of r trials out of the total
number n, the chance that the event will happen in every one of
these r trials and fail in all the rest is p r q"~ [Art. 456], and as
a set of r trials can be selected in n C r ways, all of which are
equally applicable to the case in point, the required chance is
C r p q .
If we expand (/; + q)" by the Binomial Theorem, we have
2f + "C 1 2) n  1 q + n C jS' 2 q 2 + ... +"C n _ r p r q" r + ... + q n ;
thus the terms of this series will represent respectively the
probabilities of the happening of the event exactly n times, n — 1
times, n — 2 times, ... inn trials.
463. If the event happens n times, or fails only once,
twice, ... (n — r) times, it happens r times or more ; therefore the
chance that it happens at least r times in n trials is
P n + "Cy*q + "C aP n Y+ ... tv^r.
or the sum of the first n — r + 1 terms of the expansion of
Example 1. In four throws with a pair of dice, what is the chanco of
throwing doublets twice at least ?
c i
In a single throw the chance of doublets is ^ , or ^ ; and the chance of
do o
5
failing to throw doublets is ^ . Now the required event follows if doublets
are thrown four times, three times, or twice ; therefore the required chanco
/l 5\ 4
is the sum of the first three terms of the expansion of h+d .
1 19
Thus the chance = — (1 + 4.5 + 6.5)= ^ .
H. H. A. 25
386 HIGHER ALGEBRA.
Example 2. A bag contains a certain number of balls, some of which are
white; a ball is drawn and replaced, another is then drawn and replaced;
and so on : if p is the chance of drawing a white ball in a single trial, find
the number of white balls that is most likely to have been drawn in n trials.
The chance of drawing exactly r white balls is n C r p r q n  r , and we have to
find for what value of r this expression is greatest.
Now n C r p r q n  r > n C r  l p r  l q n  (r ~ l \
so long as (nr + l)p>rq,
or (n + l)p>(p + q)r.
But p + 5 = 1; hence the required value of r is the greatest integer in
p[n + l).
If n is such that pn is an integer, the most likely case is that of pn
successes and qn failures.
464. Suppose that there are n tickets in a lottery for a prize
of £x; then since each ticket is equally likely to win the prize, and
a person who possessed all the tickets must win, the money value of
x
each ticket is £  : in other words this would be a fair sum to
n
pay for each ticket; hence a person who possessed r tickets might
TX
reasonably expect £ — as the price to be paid for his tickets by
any one who wished to buy them; that is, he would estimate
£ x as the worth of his chance. It is convenient then to in
n
troduce the following definition :
If p represents a person's chance of success in any venture
and M the sum of money which he will receive in case of success,
the sum of money denoted by pM is called his expectation.
465. In the same way that expectation is used in reference
to a person, we may conveniently use the phrase probable value
applied to things.
Example 1. One purse contains 5 shillings and 1 sovereign : a second
purse contains 6 shillings. Two coins are taken from the first and placed in
the second ; then 2 are taken from the second and placed in the first :
find the probable value of the contents of each purse.
The chance that the sovereign is in the first purse is equal to the sum of
the chances that it has moved twice and that it has not moved at all ;
PROBABILITY. ,387
112 8
that is, the chance =  .  + 5 .1=t.
6 4 3 4 .
.*. the chance that the sovereign is in the second purse =r.
Hence the probahle value of the first purse
3 1
= T of 25*. + . of 6*.=£1. O.s. 3r/.
4 4
.*. the probable value of the second purse
=31*.2Q£«.=10*. <></.
Or the problem may be solved as follows :
The probable value of the coins removed
= s of 25s. = 8^s.;
the probable value of the coins brought back
=^of (Gs.+S : V>\)=3 r W,
.\ the probable value of the first purse
= (2581 + 3^) shillings = £1. Ck 3d., as before.
Example 2. A and B throw with one die for a stake of £11 which is to
be won by the player who first throws 6. If A has the first throw, what are
their respective expectations?
1 5 5 1
In his first throw A' a chance is  ; in his second it is ^ x  x  , because
b o 6
each player must have failed once before A can have a second throw ; in his
/5\ 4 1
third throw his chance is (  J x ^ because each player must have failed
twice; and so on.
Thus A's chance is the sum of the infinite series
5MKi© 4+ }•
Similarly #'s chance is the sum of the infinite series
WM» ,+ G) 4 * J
.. A' a chance is to 7>"s as G is to 5; their respective chances are therefore
 and =y, and their expectations are £6 and £5 respectively.
26— 8
388 HIGHER ALGEBRA.
466. We shall now give two problems which lead to useful
and interesting results.
Example 1. Two players A and B want respectively m and n points of
winning a set of games ; their chances of winning a single game are p and q
respectively, where the sum of p and q is unity ; the stake is to belong to
the player who first makes up his set : determine the probabilities in favour
of each player.
Suppose that A wins in exactly m + r games; to do this he must win the
last game and m1 out of the preceding m + r1 games. The chance of
this is ™+^ 1 m _ 1 p™ 1 q r 2 h or m ^~ 1 C m  1 p m q r .
Now the set will necessarily be decided in m + n  1 games, and A may
win his m games in exactly m games, or m+ 1 games, ... , or m + n  1 games;
therefore we shall obtain the chance that A wins the set by giving to r the
values 0, 1, 2, ... n  1 in the expression m+r  1 C m _ 1 p m q r . Thus A J s chance is
similarly B's chance is
n(nAl\ \m + n2 )
1.2 * jm1
ii
This question is known as the " Problem of Points," and has
engaged the attention of many of the most eminent mathematicians
since the time of Pascal. It was originally proposed to Pascal by
the Chevalier de Mere in 1654, and was discussed by Pascal and
Fermat, but they confined themselves to the case in which the
players were supposed to be of equal skill : their results were also
exhibited in a different form. The formulae we have given are
assigned to Montmort, as they appear for the first time in a work
of his published in 1714. The same result was afterwards ob
tained in different ways by Lagrange and Laplace, and by the
latter the problem was treated very fully under various modi
fications.
Example 2. There are n dice with / faces marked from 1 to /; if these
are thrown at random, what is the chance that the sum of the numbers
exhibited shall be equal to p?
Since any one of the / faces may be exposed on any one of the n dice,
the number of ways in which the dice may fall is / n .
Also the number of ways in which the numbers thrown will have p for
their sum is equal to the coefficient of x p in the expansion of
{x l + x* + x 3 + ... + x f ) n \
for this coefficient arises out of the different ways in which n of the indices
1, 2, 3, .../can be taken so as to form p by addition.
PROBABILITY. 3S!)
Now the above expression = x 11 (l + x + x 2 + ... + x f ')"
(£?)"■
We have therefore to find the coefficient of x p ~ n in the expansion of
(I  x') n (I  x)~ n .
. t n(nl) .,. n(nl)(n2) .,,
v « ■, »(n+l) „ ?t(»+l)(w + 2) _
and <1  .r) " = 1 + nx + * ' a; 2 + . 1 ^ 3 — x 3 +...
Multiply these series together and pick out the coefficient of x p ~ n in the
product ; we thus obtain
n(n+l)...{pl) n(n+l)...(pfl)
it
\ P n \ p  n f
n (n  1) M(;t + l)...(j>2/l)
+ 1.2 •" \ pn2f
where the series is to continue so long as no negative factors appear. The
required probability is obtained by dividing this series by/ n .
This problem is due to De Moivre and was published by him
in 1730 j it illustrates a method of frequent utility.
Laplace afterwards obtained the same formula, but in a much
more laborious manner ; he applied it in an attempt to demon
strate the existence of a primitive cause which has made the
planets to move in orbits close to the ecliptic, and in the same
direction as the earth round the sun. On this point the reader
may consult Todhunter's History of Probability, Art. 987.
EXAMPLES. XXXII. c.
1. In a certain game A'a skill is to 2>'s as 3 to 2 : find the chance
of .1 winning 3 games at least out of 5.
2. A coin whose faces are marked 2, 3 is thrown 5 times : what
is the chance of obtaining a total of 12 ?
3. In each of a set of games it is 2 to 1 in favour of the winner
of the previous game : what is the chance that the player who wins
the first game shall win three at least of the next four ?
4. There are 9 coins in a bag, 5 of which are sovereigns and
the rest are unknown coins of equal value ; find what they must be if
the probable value of a draw is 12 shillings.
390 HIGHER ALGEBRA.
5. A coin is tossed n times, what is the chance that the head will
present itself an odd number of times ?
6. From a bag containing 2 sovereigns and 3 shillings a person
is allowed to draw 2 coins indiscriminately; find the value of his ex
pectation.
7. Six persons throw for a stake, which is to be won by the one
who first throws head with a penny ; if they throw in succession, find
the chance of the fourth person.
8. Counters marked 1, 2, 3 are placed in a bag, and one is with
drawn and replaced. The operation being repeated three times, what
is the chance of obtaining a total of 6 ?
9. A coin whose faces are marked 3 and 5 is tossed 4 times : what
are the odds against the sum of the numbers thrown being less than 15?
10. Find the chance of throwing 10 exactly in one throw with
3 dice.
11. Two players of equal skill, A and B, are playing a set of
games ; they leave off playing when A wants 3 points and B wants 2.
If the stake is £16, what share ought each to take \
12. A and B throw with 3 dice : if A throws 8, what is Z?'s chance
of throwing a higher number ?
13. A had in his pocket a sovereign and four shillings ; taking out
two coins at random he promises to give them to B and C. What is
the worth of (7's expectation ?
14. In five throws with a single die what is the chance of throwing
(1) three aces exactly, (2) three aces at least.
15. A makes a bet with B of 5s. to 2s. that in a single throw with
two dice he .will throw seven before B throws four. Each has a pair
of dice and they throw simultaneously until one of them wins : find B's
expectation.
16. A person throws two dice, one the common cube, and the other
a regular tetrahedron, the number on the lowest face being taken in the
case of the tetrahedron; what is the chance that the sum of the
numbers thrown is not less than 5 ?
17. A bag contains a coin of value J/, and a number of other coins
whose aggregate value is m. A person draws one at a time till he
draws the coin 31 : find the value of his expectation.
18. If 6n tickets numbered 0, 1, 2, 6n 1 are placed in a bag,
and three are drawn out, shew that the chance that the sum of the
numbers on them is equal to 6?i is
3?&
(6nl)(6n2)'
PROBABILITY. 3Dl
*Inverse Probability.
*467. In all the cases we have hitherto considered it lias been
supposed that our knowledge of the causes which may produce a
certain event is sucli as to enable us to determine the chance of
the happening of the event. We have now to consider problems
of a different character. For example, if it is known that an
event has happened in consequence of some one of a certain
number of causes, it may be required to estimate the probability
of each cause being the true one, and thence to deduce the pro
bability of future events occurring under the operation of the
same causes.
*468. Before discussing the general case we shall give a
numerical illustration.
Suppose there are two purses, one containing 5 sovereigns
and 3 shillings, the other containing 3 sovereigns and 1 shilling,
and suppose that a sovereign lias been drawn : it is required to
find the chance that it came from the first or second purse.
Consider a very large number iV of trials ; then, since before
the event eacli of the purses is equally likely to be taken, we may
assume that the first purse would be chosen in ^ iV of the trials,
5
and in  of these a sovereign would be drawn ; thus a sovereign
8 °
5 1 5
would be drawn  x ~iV, or —N times from the first purse.
o 1 lb
The second purse would be chosen in  N of the trials, and in
3
j of these a sovereign would be drawn ; thus a sovereign would
3
be drawn JV times from the second purse.
Now JV is very large but is otherwise an arbitrary number ;
let us put iV16n; thus a sovereign would be drawn 5 n times
from the first purse, and Qn times from the second purse; that is,
out of the lln times in which a sovereign is drawn it comes
from the first purse bn times, and from the second purse 0?i
392 HIGHER ALGEBRA.
times. Hence the probability that the sovereign came from the
5
first purse is — , and the probability that it came from the
A' 6
second is rr.
*469. It is important that the student's attention should be
directed to the nature of the assumption that has been made in
the preceding article. Thus, to take a particular instance,
although in 60 throws with a perfectly symmetrical die it may
not happen that ace is thrown exactly 10 times, yet it will
doubtless be at once admitted that if the number of throws is
continually increased the ratio of the number of aces to the
number of throws will tend more and more nearly to the limit
— . There is no reason why one face should appear oftener than
6
another ; hence in the long run the number of times that each of
the six faces will have appeared will be approximately equal.
The above instance is a particular case of a general theorem
which is due to James Bernoulli, and was first given in the Ars
Conjectandi, published in 1713, eight years after the author's
death. Bernoulli's theorem may be enunciated as follows :
If p is the probability that an event happens in a single trial,
then if the number of trials is indefinitely increased, it becomes a
certainty that the limit of the ratio of the number of successes to the
number of trials is equal to p ; in other words, if the number of
trials is N, the number of successes may be taken to be pN.
See Todhunter's History of Probability, Chapter vn. A proof
of Bernoulli's theorem is given in the article Probability in the
Encyclopaedia Britannica.
*470. An observed event has Jiappened through, some one of a
number of mutually exclusive causes : required to find the . pro
bability of any assigned cause being the true one.
Let there be n causes, and before the event took place ' suppose
that the probability of the existence of these causes was estimated
at P x , P 2 , P 3 , ... P n . Let p r denote the probability that when the
r* 1 * cause exists the event will follow : after the event has occurred
it is required to find the probability that the r th cause was the
true one.
PROBABILITY. 393
Consider a \ cry great number JV of trials ; then the first cause
exists in P X N of these, and out of this number the event follows
in p x P x N j similarly there are p^^N trials in which the event
follows from the second cause; and so on for each of the other
causes. Hence the number of trials in which the event follows is
;md the number in which the event was due to the r th cause is
'P,.I\N ', lience after the event the probability that the r th cause
was the true one is
pJPjr+NUpP);
tli at is, the probability that the event was produced by the r"'
PrK
cause is
Mvn
*471. It is necessary to distinguish clearly between the pro
bability of the existence of the several causes estimated before
the event, and the probability after the event has happened of any
assigned cause being the true one. The former are usually called
a priori probabilities and are represented by P x , P , P . ... P n \
the latter are called a posteriori probabilities, and if we denote
them by Q t1 Q„, Q 3 , ... Q Hf we have proved that
Qr
2 ( P P) '
where p r denotes the probability of the event on the hypothesis
of the existence of the r th cause.
From this result it appears that S (Q) = lj which is other
wise evident as the event has happened from one and only one
of the causes.
We shall now give another proof of the theorem of the pre
ceding article which does not depend on the principle enunciated
in Art. 469.
*472. An observed event has happened through some one of a
member of mutually exclusive causes : required to find the pro
bability of any assigned cause being the true one.
Let there be n causes, and before the event took place suppose that
the probability of the existence of these causes was estimated at
P t , P 2 , P z , ... P n . Let p r denote the probability that when the
? th cause exists the event will follow ; then the antecedent proba
bility that the event would follow from the r th cause is p r P r .
394 HIGHER ALGEBRA.
Let Q r be the a posteriori probability that the r th cause was the
true one; then the probability that the r th cause was the true one
is proportional to the probability that, if in existence, this cause
would produce the event ;
. A._JL = _ <?» . s(<?) _ i .
" pA pA '" p. p , Hp p ) s( P P)'
Pr p r
<?,=
2 (pP) ■
Hence it appears that in the present class of problems the
product P r p r , will have to be correctly estimated as a first step;
in many cases, however, it will be found that P lt P 2 , P 3 , ... are
all equal, and the work is thereby much simplified.
Example. There are 3 bags each containing 5 white balls and 2 black
balls, and 2 bags each containing 1 white ball and 4 black balls : a black ball
having been drawn, find the chance that it came from the first group.
Of the five bags, 3 belong to the first group and 2 to the second ; hence
If a bag is selected from the first group the chance of drawing a black
2 4 2 4
ball is  ; if from the second group the chance is  ; thus p x =  , p. 2 = ;
7 o / o
p 6 ^
" lh 1_ 3o' lh 2 ~~25'
Hence the chance that the black ball came from one of the first group is
JL^/A 8\15
35 ' \35" h 25/ 43*
*473. When an event has been observed, we are able by
the method of Art. 472 to estimate the probability of any
particular cause being the true one ; we may then estimate
the probability of the event happening in a second trial, or
we may find the probability of the occurrence of some other
event.
For example, p r is the chance that the event will happen
from the r th cause if in existence, and the chance that the r th
cause is the true one is Q r ; hence on a second trial the chance
that the event will happen from the r th cause is p r Q r . Therefore
the chance that the event will happen from some one of the
causes on a second trial is 2 (2 } Q)'
PROBABILITY. 395
Example. A purse contains 4 coins which arc either sovereigns or
shillings; 2 coins are drawn and found to be shillings: if these are replaced
what is the chance that another drawing will give a sovereign?
This question may be interpreted in two ways, which we shall discuss
separately.
I. If we consider that all numbers of shillings are a priori equally likely,
we shall have three hypotheses; for (i) all the coins may be shillings, (ii)
three of them may be shillings, (iii) only two of them may be shillings.
Here P^P.^P.^;
also ^ = 1, J> a =g, P»=q
Hence probability of iirst hypothesis = 15 (1 + o + r) — tTv^ Qi>
probability of second hypothesis = o ^ ( * + 2 + f ) = To ~ ^' J '
probability of third hypothesis —  f ( 1 + ~ + . ) =T7\=Qy
Therefore the probability that another drawing will give a sovereign
1 3_ 2 1_ 5^ 1
~4 *10 + 4 'To~40  8*
II. If each coin is equally likely to be a shilling or a sovereign, by taking
/l IV
the terms in the expansion of I  +  J , we see that the chance of four
1 . 4 6
shillings is r^ , of three shillings is 77: , of two shillings is — . ; thus
lb 10 10
P_l P ± P _A.
il_ 16' 2 ~1G' ^»16'
also, as before, i J i = l> Pi— a* Ps—r'
Qi_Q2_Q*_ Qi + Q2+Q* _ 1
6 ' 12 ' 6 24 24'
Hence
Therefore the probability that another drawing will give a sovereign
= (<2ix0)+(q,x^ + ((? :j x)
~ 8 + 16 4
396 HIGHER ALGEBRA.
*474. We shall now shew how the theory of probability may
be applied to estimate the truth of statements attested by wit
nesses whose credibility is assumed to be known. We shall
suppose that each witness states what he believes to be the truth,
whether his statement is the result of observation, or deduction,
or experiment; so that any mistake or falsehood must be
attributed to errors of judgment and not to wilful deceit.
The class of problems we shall discuss furnishes a useful
intellectual exercise, and although the results cannot be regarded
as of any practical importance, it will be found that they confirm
the verdict of common sense.
*475. When it is asserted that the probability that a person
speaks the truth is p, it is meant that a large number of state
ments made by him has been examined, and that p is the ratio
of those which are true to the whole number.
*476. Two independent witnesses, A and B, whose proba
bilities of speaking the truth are p and p' respectively, agree in
making a certain statement : what is the probability that the
statement is true %
Here the observed event is the fact that A and B make the
same statement. Before the event there are four hypotheses ; for
A and B may both speak truly ; or A may speak truly, B falsely;
or A may speak falsely, B truly ; or A and B may both speak
falsely. The probabilities of these four hypotheses are
PP\ p( l ~P\ P'QP)* ( 1 P)( 1 ~P') respectively.
Hence after the observed event, in which A and B make the
same statement, the probability that the statement is true is to
the probability that it is false as pp to (1  p) (1 p') ; that
is, the probability that the joint statement is true is
pp'
statement is true is
./„//
ppp
ppY + {1p){ip')(ip") }
and so on for any number of persons.
pp' + (lp)(lp')'
Similarly if a third person, whose probability of speaking the
truth is p", makes the same statement, the probability that the
PROBABILITY. 397
*477. In the preceding article it lias been supposed that we
have no knowledge of the event except the statement made by A
and B ; if we have information from other sources as to the
probability of the truth or falsity of the statement, this must be
taken into account in estimating the probability of the various
hypotheses.
For instance, if A and B agree in stating a fact, of which
the a priori probability is P, then we should estimate the pro
bability of the truth and falsity of the statement by
Ppp* and (1  P) (1 — p>) (1 — p') respectively.
Example. There is a raffle with 12 tickets and two prizes of £9 and £3.
A, B, C, whose probabilities of speaking the truth are ^, §, f respectively,
report the result to D, who holds one ticket. A and B assert that he has
won the £9 prize, and C asserts that he has won the £3 prize; what is D's
expectation?
Three cases are possible; D may have won £9, £3, or nothing, for A, B,
C may all have spoken falsely.
Now with the notation of Art. 472, we have the a priori probabilities
Pi PA P™.
*i 12 « *al2' ^ 3 ~12'
12 24 1133 1 1 2_ 2
also Pi~2 X 3 X 530> **~~2 X 3 X 5 ~ 30 ' A_ 2 * 3 X 5~3() ;
" 4 3 20 27'
4 3
hence D's expectation =— of £9 + — of £3 =£1. 13s. id.
*478. With respect to the results proved in Art. 47G, it
should be noticed that it was assumed that the statement can be
made in two ways only, so that if all the witnesses tell falsehoods
they agree in telling the same falsehood.
If this is not the case, let us suppose that c is the chance
that the two witnesses A and B will agree in telling the same
falsehood ; then the probability that the statement is true is to
the probability that it is false as pp' to c (1 —p) (1 — p').
As a general rule, it is extremely improbable that two
independent witnesses will tell the same falsehood, so that c is
usually very small; also it is obvious that the quantity c becomes
smaller as the number of witnesses becomes greater. These con
siderations increase the probability that a statement asserted by
two or more independent witnesses is true, even though the
credibility of each witness is small.
398 HIGHER ALGEBRA.
Example. A speaks truth 3 times out of 4, and B 7 times out of 10; they
both assert that a white ball has been drawn from a bag containing 6 balls
all of different colours : find the probability of the truth of the assertion.
There are two hypotheses ; (i) their coincident testimony is true, (ii) it is
false.
•1 P 5 
6' 2 ~6'
Here P x = ~ , P 2
^ 1 4 X 10' • P2 ~25 X 4 X 10 ;
for in estimating p. 2 we must take into account the chance that A and B will
both select the white ball when it has not been drawn ; this chance is
11 1
5 X 5 ° r 25 *
Now the probabilities of the two hypotheses are as P^ to P 2 po, and
35
therefore as 35 to 1; thus the probability that the statement is true is — .
*479. The cases we have considered relate to the probability
of the truth of concurrent testimony; the following is a case of
traditionary testimony.
If A states that a certain event took place, having received an
account of its occurrence or nonoccurrence from B, what is the
probability that the event did take place 1
The event happened (1) if they both spoke the truth, (2) if
they both spoke falsely ; and the event did not happen if only
one of them spoke the truth.
Let p, p denote the probabilities that A and B speak the
truth ; then the probability that the event did take place is
pp' + (lp)(lp) }
and the probability that it did not take place is
p(l 2 ))+p'(lp).
*480. The solution of the preceding article is that which has
usually been given in textbooks; but it is" open to serious objec
tions, for the assertion that the given event happened if both A
and B spoke falsely is not correct except on the supposition that
the statement can be made only in two ways. Moreover,
although it is expressly stated that A receives his account from
B, this cannot generally be taken for granted as it rests on
A'& testimony.
PROBABILITY. 399
A full discussion of the different ways of interpreting the
question, and of the different solutions to which they lead, will be
found in the Educational Times Reprint, Yols. XXVII. and XXXII.
^EXAMPLES. XXXII. d.
1. There are four balls in a bag, but it is not known of what
colours they are ; one ball is drawn and found to be white : find the
chance that all the balls are white.
2. In a bag there are six balls of unknown colours; three balls
are drawn and found to be black; find the chance that no black ball
is left in the bag.
3. A letter is known to have come either from London or Clifton ;
on the postmark only the two consecutive letters ON are legible ; what
is the chance that it came from London ?
4. Before a race the chances of three runners, A, B, C, were
estimated to be proportional to 5, 3, 2 ; but during the race A meets
with an accident which reduces his chance to onethird. What are now
the respective chances of B and C ?
5. A purse contains n coins of unknown value ; a coin drawn at
random is found to be a sovereign; what is the chance that it is the
only sovereign in the bag ?
6. A man has 10 shillings and one of them is fcnown to have two
heads. He takes one at random and tosses it 5 times and it always
falls head : what is the chance that it is the shilling with two heads ?
7. A bag contains 5 balls of unknown colour; a ball is drawn
and replaced twice, and in each case is found to be red : if two balls
are now drawn simultaneously find the chance that both are red.
8. A purse contains five coins, each of which may be a shilling
or a sixpence ; two are drawn and found to be shillings : find the prob
able value of the remaining coins.
9. A die is thrown three times, and the sum of the three numbers
thrown is 15 : find the chance that the first throw was a four.
10. A speaks the truth 3 out of 4 times, and B 5 out of 6 times :
what is the probability that they will contradict each other in .stating
the same fact ?
400 HIGHER ALGEBRA.
11. A speaks the truth 2 out of 3 times, and B 4 times out of 5 ;
they agree in the assertion that from a bag containing 6 balls of different
colours a red ball has been drawn : find the probability that the state
ment is true.
12. One of a pack of 52 cards has been lost ; from the remainder
of the pack two cards are drawn and are found to be spades ; find the
chance that the missing card is a spade.
13. There is a raffle with 10 tickets and two prizes of value £5
and £1 respectively. A holds one ticket and is informed by B that
he has won the £b prize, while C asserts that he has won the ,£1 prize :
what is A's expectation, if the credibility of B is denoted by §, and
that of C by f ?
14. A purse contains four coins ; two coins having been drawn are
found to be sovereigns : find the chance (1) that all the coins are
sovereigns, (2) that if the coins are replaced another drawing will give
a sovereign.
15. P makes a bet with Q of ,£8 to £120 that three races will be
won by the three horses A, B, C, against which the betting is 3 to 2,
4 to 1, and 2 to 1 respectively. The first race having been won by A,
and it being known that the second race was won either by B, or by
a horse D against which the betting was 2 to 1, find the value of P's
expectation.
16. From a bag containing n balls, all either white or black, all
numbers of each being equally likely, a ball is drawn which turns out
to be white; this is replaced, and another ball is drawn, which also
turns out to be white. If this ball is replaced, prove that the chance
of the next draw giving a black ball is  (n — 1) (2n + l)~ l .
17. If mn coins have been distributed into m purses, n into each,
find (1) the chance that two specified coins will be found in the same
purse; and (2) what the chance becomes when r purses have been
examined and found not to contain either of the specified coins.
18. A, B are two inaccurate arithmeticians whose chance of solving
a given question correctly are £ and y 1 ^ respectively ; if they obtain the
same result, and if it is 1000 to 1 against their making the same
mistake, find the chance that the result is correct.
19. Ten witnesses, each of whom makes but one false statement in
six, agree in asserting that a certain event took place ; shew that the
odds are five to one in favour of the truth of their statement, even
although the a 'priori probability of the event is as small as ^9 — r •
PRoUAl'.ILITY.
41 1 1
Local Probability. Geometrical Methods.
*481. The application of Geometry to questions of Pro
bability requires, in general, the aid of the Integral Calculus;
there are, however, many easy questions which can be solved by
Elementary Geometry.
Example 1. From each of two equal lines of length I a portion is cut
off at random, and removed : what is the chance that the sum of the
remainders is less than I?
Place the lines parallel to one another, and suppose that after cutting,
the righthand portions are removed. Then the question is equivalent to
asking what is the chance that the sum of the righthand portions is greater
than the sum of the lefthand portions. It is clear that the first sum is
equally likely to be greater or less than the second; thus the required
probability is  .
a
Cor. Each of two lines is known to be of length not exceeding I: the
chance that their sum is not greater than Z is  .
a
Example 2. If three lines are chosen at random, prove that they are
just as likely as not to denote the sides of a possible triangle.
Of three lines one must be equal to or greater than each of the other
two ; denote its length by I. Then all we know of the other two lines is that
the length of each lies between and /. But if each of two lines is known to
be of random length between and 1, it is an even chance that their sum
is greater than /. [Ex. 1, Cor.]
Thus the required result follows.
Example 3. Three tangents are drawn at random to a given circle :
shew that the odds are 3 to 1 against the circle being inscribed in the triangle
formed by them.
P
O
Draw three random lines P, (), 11, in the same plane as the circle, and
draw to the circle the six tangents parallel to these lines.
H. H.A. 2G
402
HIGHER ALGEBRA.
Then of the 8 triangles so formed it is evident that the circle will be
escribed to 6 and inscribed in 2 ; and as this is true whatever be the original
directions of P, Q, R, the required result follows.
*4:82. Questions in Probability may sometimes be con
veniently solved by the aid of coordinate Geometry.
Example. On a rod of length a + b+c, lengths a, b are measured at
random: find the probability that no point of the measured lines will
coincide.
Let AB be the line, and suppose AP = x and PQ = a; also let a be
measured from P towards B, so that x must be less than b + c. Again let
AP' = y, P'Q' = b, and suppose P'Q' measured from P' towards B, then y must
be less than a + c.
Now in favourable cases we must have AP'>AQ, or else AP>AQ\
hence y>a + x, or x>b + y (1).
Again for all the cases possible, we must have
x>0, and <& + c)
2/>0, and <a + c)
Take a pair of rectangular axes and make OX equal to b + c, and OY
equal to a + c.
Draw the line y = a + x, represented by TML in the figure; and the line
x = b + y represented by KB.
Q
n
•A P' W B f 0.—Q b  K
Then YM, EX are each equal to c, 031, OT are each equal to a.
X
The conditions (1) are only satisfied by points in the triangles MYL and
ItXR, while the conditions (2) are satisfied by any points within the rect
angle OX, OY;
c 2
.*. the required chance = — .
{a + c)(b + c)
*483. We shall close this chapter with some Miscellaneous
Examples.
Example 1. A box is divided into m equal compartments into which n
balls are thrown at random ; find the probability that there will be p com
partments each containing a balls, q compartments each containing b balls,
r compartments each containing c balls, and so on, where
Z>a+qb + rc + =n.
PROBABILITY. 403
Since each of the n halls can fall into any one of the m compartments
the total number of cases which can occur is m n , and these are all equally
likely. To determine the number of favourable cases we must find the
number of ways in which the n balls can be divided into p, <1, r, ... parcels
containing a, b, c, ... balls respectively.
First choose any g of the compartments, where s stands for p + q + r + ... ;
\m
the number of ways in which this can be done is . — — — (1).
\s \ms v '
Next subdivide the s compartments into groups containing p t q, r, ...
severally; by Art. 147, the number of ways in which this can be done is
\»\1
r ..
(2).
Lastly, distribute the n balls into the compartments, putting a into each
of the group of p, then b into each of the group of q, c into each of the
group of r, and so on. The number of ways in which this can be done is
In
 (3).
(\a)*(\b)«(\c_)
Hence the number of ways in which the balls can be arranged to satisfy
the required conditions is given by the product of the expressions (1), (2), (3).
Therefore the required probability is
\m
t
m" (\a)>> (\b)i ([£)• pjr. \ mpqr
Example 2. A bag contains n balls ; k drawings are made in succession,
and the ball on each occasion is found to be white : find the chance that the
next drawing will give a white ball'; (i) when the balls are replaced after
each drawing ; (ii) when they are not replaced.
(i) Before the observed event there are n + 1 hypotheses, equally likely;
for the bag may contain 0, 1, 2, 3, ... n white balls. Hence following the
notation of Art. 471,
1 = Pj — P2 = P 3 = . . . = P n ;
Hence after the observed event,
Qr =
7*
thus the required chance =
l»+2*+ 3*+. ..+n*
xt drawing will giv<
Now the chance that the next drawing will give a white ball =2  Q r \
n p + 2* + 3* + ...+n*
and the value of numerator and denominator may be found by Art. 405.
26—2
404 HIGHER ALGEBRA.
In the particular case when k = 2,
the required chance = <— *= — S 4 ^
3 (n + 1)
~2(2n+l)'
If n is indefinitely large, the chance is equal to the limit, when n is in
1 V^ 2 «*+!
finite, of
and thus the chance is
n ' k + 2 ' fc + 1.'
fe + 1
&+2*
(ii) If the halls are not replaced,
and Q r =i
it
r r  1 r  2 r  A; + 1
r — n " n — 1 ' n  2 " ' » — k + 1 '
p, (rk + l)(rk + 2) (rl)r
r=K
r y
r=0
(>£+l)(rifc + 2) (rl)r
(uifc + l)(nJfe + 2) (nl)n (n+1)
The chance that the next drawing will give a white ball= 2 . Q r
r =0 U — Ii
s"(rfc)(rfc+l) (rl)r
(;i  A) (u  /c + 1) ?i (n + 1) r=0
fc + 1 (/iA)(n/v + l) n(»i + l)
_ (nk)(nk + l) n(n + l) * k~+ 2~
Jfc+1
~k + 2'
which is independent of the number of balls in the bag at first.
Example 3. A person writes n letters and addresses n envelopes ; if the
letters are placed in the envelopes at random, what is the probability that
every letter goes wrong ?
Let u n denote the number of ways in which all the letters go wrong, and
let abed . . . represent that arrangement in which all the letters are in their
own envelopes. Now if a in any other arrangement occupies the place of an
assigned letter b, this letter must either occupy a's place or some other.
(i) Suppose b occupies a's place. Then the number of ways in which
all the remaining n  2 letters can be displaced is u n _ 2 , and therefore the
numbers of ways in which a may be displaced by interchange with some one
of the other n 1 letters, and the rest be all displaced is (n  1) «„_ 2 .
PROBABILITY. 405
(ii) Suppose a occupies i>'s place, and b does not occupy a's. Then in
arrangements satisfying the required conditions, since a is fixed in &'s place,
the letters b, c, d, ... must be all displaced, which can be done in h__j ways;
therefore the number of ways in which a occupies the place of another letter
but not by interchange with that letter is (n  1) u n  l ;
.. v n = (nl) (M n _! + «„_„);
from which, by the method of Art. 444, we find u n  nu n _ 1 = (  l) n (ttj  Uj).
Also n 1 = 0, tig = 1 ; thus we finally obtain
, f 1 i i ( 1 )' 1 !
Now the total number of ways in which the n things can be put in n
places is In ; therefore the required chance is
11 1 _ ( 1)"
[2 £ + 4 '•• + in '
Tlie problem liere involved is of considerable interest, and in
some of its many modifications lias maintained a permanent place
in works on the Theory of Probability. It was first discussed
by Montmort, and it was generalised by De Moivre, Euler, and
Laplace.
*484. The subject of Probability is so extensive that it is
impossible here to give more than a sketch of the principal
algebraical methods. An admirable collection of problems, illus
trating every algebraical process, will be found in "NVliitworth's
Choice and Chance; and the reader who is acquainted with the
Integral Calculus may consult Professor Crofton's article Proba
bility in the Encyclopcedia JJritannica. A complete account of
the origin and development of the subject is given in Todhunter's
History of the Theory of Probability from the time of Pascal to
that of Laplace.
The practical applications of the theory of Probability to
commercial transactions are beyond the scope of an elementary
treatise ; for these we may refer to the articles Annuities and
Insurance in the JEncyclopcedia Britannica.
^EXAMPLES. XXXII. e.
L What are the odds in favour of throwing at lea.st 7 in a single
throw with two dice ?
2. In a purse there are 5 sovereigns and 4 shillings. If they are
drawn out one by one, what is the chance that they come out sovereigns
und shillings alternately, beginning with ;t sovereign?
406 HIGHER ALGEBRA.
3. If on an average 9 ships out of 10 return safe to port, what
is the chance that out of 5 ships expected at least 3 will arrive 1
4. In a lottery all the tickets are blanks but one; each person
draws a ticket, and retains it : shew that each person has an equal
chance of drawing the prize.
5. One bag contains 5 white and 3 red balls, and a second bag
contains 4 white and 5 red balls. From one of them, chosen at random,
two balls are drawn : find the chance that they are of different colours.
6. Five persons A, B, C, B, E throw a die in the order named
until one of them throws an ace : find their relative chances of winning,
supposing the throws to continue till an ace appears.
7. Three squares of a chess board being chosen at random, what
is the chance that two are of one colour and one of another 1
8. A person throws two dice, one the common cube, and the other
a regular tetrahedron, the number on the lowest face being taken in
the case of the tetrahedron ; find the average value of the throw, and
compare the chances of throwing 5, 6, 7.
9. A's skill is to 2?'s as 1 : 3 ; to Cs as 3 : 2 ; and to Z)'s as 4 : 3 :
find the chance that A in three trials, one with each person, will succeed
twice at least.
10. A certain stake is to be won by the first person who throws
an ace with an octahedral die : if there are 4 persons what is the
chance of the last ?
11. Two players A, B of equal skill are playing a set of games ; A
wants 2 games to complete the set, and B wants 3 games: compare
their chances of winning.
12. A purse contains 3 sovereigns and two shillings : a person
draws one coin in each hand and looks at one of them, which proves
to be a sovereign ; shew that the other is equally likely to be a sovereign
or a shilling.
13. A and B play for a prize ; A is to throw a die first, and is to
win if he throws 6. If he fails B is to throw, and to win if he throws
6 or 5. If he fails, A is to throw again and to win with 6 or 5 or 4,
and so on : find the chance of each player.
14. Seven persons draw lots for the occupancy of the six seats in
a first class railway compartment : find the chance (1) that two specified
persons obtain opposite seats, (2) that they obtain adjacent seats on
the same side.
15. A number consists of 7 digits whose sum is 59 ; prove that the
.4
chance of its being divisible by 11 is — .
16. Find the chance of throwing 12 in a single throw with 3 dice.
PROBABILITY. 407
17. A bag contains 7 tickets marked with the numbers 0, 1, 2, ...G
respectively. A ticket is drawn and replaced ; find the chance that
after 4 drawings the sum of the numbers drawn is 8.
18. There are 10 tickets, 5 of w T hich are blanks, and the others are
marked with the numbers 1, 2, 3, 4, 5 : what is the probability of
drawing 10 in three trials, (1) when the tickets are replaced at every
trial, (2) if the tickets are not replaced ?
19. If n integers taken at random are multiplied together, shew
that the chance that the last digit of the product is 1, 3, V, or 9 is — ;
o
An _ 9>i Kn .pi
the chance of its being 2, 4, 6, or 8 is — =— — ; of its being 5 is
and of its beinc: is
10 H 8' l 5 n + 4 n
10*
20. A purse contains two sovereigns, two shillings and a metal
dummy of the same form and size ; a person is allowed to draw out one
at a time till he draws the dummy : find the value of his expectation.
21. A certain sum of money is to be given to the one of three
persons A, B, C who first throws 10 with three dice; supposing them
to throw in the order named until the event happens, prove that their
chances are respectively
/8\ 2 56 . /7\ 2
(ja)' W> and [&)'
22. Two persons, whose probabilities of speaking the truth are
2 5
 and  respectively, assert that a specified ticket has been drawn out
of a bag containing 15 tickets: what is the probability of the truth of
the assertion ?
23. A bag contains —  counters, of which one is marked 1,
two are marked 4, three are marked 9, and so on ; a person puts in his
hand and draws out a counter at random, and is to receive as many
shillings as the number marked upon it : find the value of his ex
pectation.
24. If 10 things are distributed among 3 persons, the chance of
a particular person having more than 5 of them is _ .... .
25. If a rod is marked at random in n points and divided at
those points, the chance that none of the parts shall be greater than
— th of the rod is — .
n a n
408 HIGHER ALGEBRA.
26. There are two purses, one containing three sovereigns and a
shilling, and the other containing three shillings and a sovereign. A coin
is taken from one (it is not known which) and dropped into the other ;
and then on drawing a coin from each purse, they are found to be two
shillings. What are the odds against this happening again if two more
are drawn, one from each purse 1
27. If a triangle is formed by joining three points taken at random
in the circumference of a circle, prove that the odds are 3 to 1 against
its being acuteangled.
28. Three points are taken at random on the circumference of a
circle: what is the chance that the sum of any two of the arcs so
determined is greater than the third ?
29. A line is divided at random into three parts, what is the chance
that they form the sides of a possible triangle ?
30. Of two purses one originally contained 25 sovereigns, and the
other 10 sovereigns and 15 shillings. One purse is taken by chance
and 4 coins drawn out, which prove to be all sovereigns : what is the
chance that this purse contains only sovereigns, and what is the prob
able value of the next draw from it?
31. On a straight line of length a two points are taken at random ;
find the chance that the distance between them is greater than b.
32. A straight line of length a is divided into three parts by two
points taken at random ; find the chance that no part is greater than b.
33. If on a straight line of length a + b two lengths a, b are
measured at random, the chance that the common part of these lengths
c 2
shall not exceed c is — r , where c is less than a or b ; also the chance
ab
that the smaller length b lies entirely within the larger a is — . .
(Jj
34. If on a straight line of length a + b + c two lengths a, b are
measured at random, the chance of their having a common part which
shall not exceed d is T — . . 7 . , where d is less than either a or b.
(c + a)(c+6)'
35. Four passengers, A, B, C, D, entire strangers to each other, are
travelling in a railway train which contains I firstclass, m secondclass,
and n thirdclass compartments. A and B are gentlemen whose re
spective a priori chances of travelling first, second, or third class are
represented in each instance by X, fi, v, C and D are ladies whose
similar a priori chances are each represented by I, m, n. Prove
that, for all values of X, fi, v (except in the particular case when
X : p : v=l. : m : oi), A and B are more likely to be found both in the
company of the same lady than each with a different one.
CHAPTER XXXIII.
Determinants.
485. The present chapter is devoted to a brief discussion of
determinants and their more elementary properties. The slight
introductory sketch here given will enable a student to avail
himself of the advantages of determinant notation in Analytical
Geometry, and in some other parts of Higher Mathematics ;
fuller information on this branch of Analysis may be obtained
from Dr Salmon's Lessons Introductory to the Modern Higher
Algebra, and Muir's Theory of Determinants.
48G. Consider the two homogeneous linear equations
a ] x + b l y = 0,
a 2 x + b 2 y = 0;
multiplying the first equation by b si the second by 6 , sub
tracting and dividing by x, we obtain
This result is sometimes written
a x b x
a„ b„
0,
and the expression on the left is called a determinant. It consists
of two rows and two columns, and in its expanded form each
term is the product of two quantities; it is therefore said to be
of the second order.
The letters « , b a a , b 2 are called the constituents of the
determinant, and tile terms «,/>,,, ab. are called the elements,
410
HIGHER ALGEBRA.
487. Since
a.
a.
a A  «A =
a.
a„
it follows that the value of the determinant is not altered by chang
ing the rows into columns, and the columns into rows.
488. Again, it is easily seen that
«■
\
— —
h
a x
, and
« 2
K
K
a 2
a,
a„
6„
a 2
a.
K
b,
that is, if we interchange two rows or two columns of the deter
minant, ive obtain a determinant ivhich differs from it only in sign.
489. Let us now consider the homogeneous linear equations
a x x + b x y + c x z = 0,
a 2 X + b 2 V + c 2 z = o,
a 3 x + b 3 y + c 3 z = 0.
By eliminating x, y, z, we obtain as in Ex. 2, Art. 16,
a > (K c s  K c 2 ) + b i (<v* 3  vO + c , ("A  a A) = °»
or
a.
K
c s
+ *.
C 2
a 2
+ C ,
a 2
K
K
C 3
C s
%
"'I
K
= 0.
This eliminant is usually written
a
K c 2
h C 3
o,
and the expression on the left being a determinant which consists
of three rows and three columns is called a determinant of the
third order.
490. By a rearrangement of terms the expanded form of
the above determinant may be written
a X vV 8 ~ h S C 2 ) + « 2 (&3 C 1 " K C 3 ) + «a( 6 l C .  h 2^\
or
a.
K
K
+ (L 2
h
h
+ %
K
K
c.
C 3
C 3
C l
<V
c 2
DETERMINANTS.
411
hence
a
*i
^3 C 3
(I. «.,
ft..
*, K \
<\ C 2 C 3
that is, the value of the determinant is not altered by changing the
rows into column*, and the columns into rov)S.
a.,
491.
6.
From the preceding article,
c 2
=«,
* a
C i
+ « a
6.
C 3
+ «3
6,
<\
»,
C 3
K
C .
K
c.
— a.
\
c a
~ a *
h
<",
+ «a
»,
c >
K
C z
\
C 3
K
C 2
■(!)•
Also from Art. 489,
a t
a.
a„
h
c,
= «,
ft
h.
Co
ft
K
C 3
ft.
rt,
«,
c 8
+ C.
ft..
(2).
We shall now explain a simple method of writing down the
expansion of a determinant of the third order, and it should be
noticed that it is immaterial whether we develop it from the first
row or the first column.
From equation (1) we see that the coefficient of any one of
the constituents «,, a 2i a 3 is that determinant of the second order
which is obtained by omitting the row and column in which
it occurs. These determinants are called the Minors of the
original determinant, and the lefthand side of equation (1) may
be written
« 1 ^iM 2 + M 3J
where A t , A , A. s are the minors of a,, a^ a 3 respectively.
Again, from equation (2), the determinant is equal to
a l A l b l B x +0,(7,,
where A lt />',, 0, are the minors of a lt ft,, c, respectively.
412
HIGHER ALGEBRA.
492. The determinant a x b l c x
<** h 2 C 2
% b 3 C 3
= «, (K C 3 ~ K C 2) + h l ( C 2«3  C /h) + C , (« A " «3 6 2 )
= " b l( a 2 C B ~ «« C ») ~ Cl l ( C 2 b 3  C J>>)  C l (K a 3 ~ 6 A) >
hence
«,
K
a 2 b 2
c.
b x a x
a
(i
3 C 3
Thus it appears that if two adjacent columns, or rows, of the
determinant are interchanged, the sign of the determinant is
changed, but its value remains unaltered.
If for the sake of brevity we denote the determinant
a x b t c x
(l 2 h C 2
«3 K °3
by (a x b 2 c 3 ), then the result we have just obtained may be written
(VsO =  (»A c s)
Similarly we may shew that
( c i«A) =  ( a A b 3 ) = + («A C 3 )
493. If two rows or two columns of tlie determinant are
identical the determinant vanishes.
For let D be the value of the determinant, then by inter
changing two rows or two columns we obtain a determinant
whose value is — D; but the determinant is unaltered; hence
J) = — D, that is D = 0. Thus we have the following equations,
a A, — aA n + aJL m = D.
1 1
2 2
3 3
494. If each constituent in any row, or in any column, is
multiplied by the same factor, then the determinant is multiplied
by that factor.
DETERMINANTS.
413
For
tna x
ma 2
ma
a
by
c,
K
c 2
K
C 3
— ma x . A x — ma 2 . A 2 + ma A . A 3
= m(a l A l a 2 A 2 + a. i A,y,
which proves the proposition.
Cor. If each constituent of one row, or column, is the same
multiple of the corresponding constituent of another row, or
column, the determinant vanishes.
495. If each constituent in any row, or column, consists of tivo
terms, then the determinant can he expressed as the sum of tvio
other determinants.
Thus we have
a v + a,
h
c,
=
a,
*.
Ci
+
a,
*t
Ci
a 2 + a 2
K
C 2
Ct 2
K
c 2
a 2
b.
c 2
«3 + a 3
K
C 3
^
b
3
C 3
a .
K
C 3
for the expression on the left
= (a x + a^ A ,  (a 2 + a 2 ) ^1 2 + (a 3 + a 3 ) y1 2
= (V,  M* + M 3 ) + (<v*i  M 2 + a A) ;
which proves the proposition.
In like manner if each constituent in any one row, or column,
consists of m terms, the determinant can he expressed as the
sum of m other determinants.
Similarly, we may shew that
a x + a x
*i+A
C l
a 2 + a 2
&*+& c 2
«3 + «3
*.+A C 3
a l fr,
c ,
+
«,
ft
c >
+
B 
6,
C ,
+
a .
A
c ,
»1 \
c 2
"2
A
c a
a 2
&,
c,
a 2
ft
c 2
«3 6 3
C 3
««
ft
C 3
tt 3
^3
c >
a 3
ft
<*3
414
HIGHER ALGEBRA.
These results may easily be generalised; thus if the con
stituents of the three columns consist of m, n, p terms respec
tively, the determinant can be expressed as the sum of mnp
determinants.
Example 1. Shew that
b + c
c + a
a + b
a b
bc
c — a
a
b
c
— Babe a 3  b 3 — c 3 .
The given determinant
b
a
a

c
b
b
a
c
c
b
c
b
c
a
b
a a
+
c
a
b
a
a

b
b
c
c
c
a
b
b
c
a
a
b
c
Of these four determinants the first three vanish, Art. 493; thus the ex
pression reduces to the last of the four determinants ; hence its value
= _ < c ( c 2 ab)b (ac  & 2 ) + a (a 2  be)}
= Babe  a 3  b 3  c 3 .
Example 2. Find the value of
67 19
39 13
81 24
21
14
26
We have
67
39
81
19 21
=
13 14
24 26
10 + 57 19 21
+ 39 13 14
9 + 72 24 26
10 19
21
+
13
14
9 24
26
57 19 21
10 19 21
13 14
9 24 26
10 19 19 + 2
13 13 + 1
9 24 24 + 2
10 19
13
9 24
39
72
2
1
i
13 14
24 26
= 10
13
24
1
2
+ 9
19
13
2
1
= 2063= 43.
496.
Consider the determinant
as in the last article we can shew that it is equal to
6*
+
+
qc t
9 C 3
C 2
c„
DETERMINANTS.
415
and the last two of these determinants vanish [Art. 494 Cor.].
Tims we see that the given determinant is equal to a new one whose
first column is obtained by subtracting from the constituents of
the first column of the original determinant equimultiples of the
corresponding constituents of the other columns, while the second
and third columns remain unaltered.
Conversely,
a
C 2
c.
a { + j)b x + qc } b {
a ., + PK + <7 C 2 ^ 2
c„
and what has been here proved with reference to the first column
is equally true for any of the columns or rows ; hence it appears
that in reducing a determinant we may replace any one of the
rows or columns by a new row or column formed in the following
way :
Take the constituents of the row or column to be replaced,
and increase or diminish them by any equimidtij)les of the cor
responding constituents of one or more of the other rows or
columns.
After a little practice it will be found that determinants
may often be quickly simplified by replacing two or more rows
or columns simultaneously : for example, it is easy to see
that
a i + 2 } b } b }  qc x c,
% + P h 2 K ~ C 1 C 2 °2
a 3 +2 jb 3 KQ c s c 3
€l l
Ct 2
a..
b 2
b„
c.,
but in any modification of the rule as above enunciated, care
must be taken to leave one row or column unaltered.
Thus, if on the lefthand side of the last identity the con
stituents of the third column were replaced by c l +ra li c 2 + ra^
c, + ra respectively, we should have the former value in
creased by
a x + 2>b t b x — qc x ra x
« a + i'K K  v c i
ra„
ra..
416
HIGHER ALGEBRA.
and of the four determinants into which this may be resolved
there is one which does not vanish, namely
ra,
pb s  qc 2 . „ 9
ra
Example 1, Find the value of I 29 26 22
25 31 27
! 63 54 46
The given determinant
3 264l = 3x4x
6 31 4
9 54 8
1
26
1
=  12 x
1
26
1
2
31
1
o
— O
5
3
54
2
1
2
=: 12
1 26
12 I 3
1
o
2
= 132.
[Explanation. In the first step of the reduction keep the second column
unaltered; for the first new column diminish each constituent of the first
column by the corresponding constituent of the second ; for the third new
column diminish each constituent of the third column by the corresponding
constituent of the second. In the second step take out the factors 3 and
 4. In the third step keep the first row unaltered ; for the second new row
diminish the constituents of the second by the corresponding ones of the
first ; for the third new row diminish the constituents of the third by twice
the corresponding constituents of the first. The remaining steps will be
easily seen.]
Example 2. Shew that
The given determinant
abc 2a 2a
2b bca 2b
2c 2c cab
= (a + b + cf.
a + b + c a + b + c a+b+c
2b bca 2b
2c 2c ca— b
= (a + b + c) x
11 1
2b bca 2b
2c 2c cab
(a + b + c) x 1
2b bca
2c cab
(a + b + c) x I bca I = (a + b + c) 3 .
c— a b I
DETERMINANTS.
417
[Explanation. In the first new determinant the first row is the sum of
the constituents of the three rows of the original determinant, the second
and third rows being unaltered. In the third of the new determinants the
first column remains unaltered, while the second and third columns are
obtained by subtracting the constituents of the first column from those of
the second and third respectively. The remaining transformations are suffi
ciently obvious.]
497. Before shewing how to express the product of two de
terminants as a determinant, we shall investigate the value of
«i«i + & A + ^7, »A + h A + c i?2 a i a 3 + b A + r i7s
Vi + KPt + c *y, %% + h A + c *y 2 « 2 <* 3 + h A + c a y a
a 3 a i + h A + Wi %% + b A + ^ 3 y 2 «3 a 3 + &A + c 3 y 3
From Art. 495, we know that the above determinant can be
expressed as the sum of 27 determinants, of which it will be
sufficient to give the following specimens :
a 3 a 2
*1<S
« 2 ft 3
«3 a 3
a x a {
V.
^3
?
a ! a ,
c iy 2
*J8
V:
KA
c 2 y 3
v»
C V
2 / 2
V
«3 a !
b A
c 3 y 3
Vl
<V/,.
I'fi
these are respectively equal to
a i a 2 a 3
a 2
a.
a.
ct„
» "Ay,
a.
«„
a„
3 C 3
a Ay 2
«.
a„
ct„
"3 *,
the first of which vanishes; similarly it will be found that 21
out of the 27 determinants vanish. The six determinants that
remain are equal to
( a Ay 3  a Ay 2 + a 2&yi  a Ay 3 + a Ay 2 ~ a Ayd x
ft
a>„
cc„
b.
that is,
a.
<x„
a„
ft 7i
X
ft y.
ft r 3
«,
«„
«„
», <3
hence the given determinant can be expressed as the product of
two other determinants.
498. The product of two determinants is a determinant.
Consider the two linear equations
a x X x + b x X a = 0)
ag X x + b^ 9 = o]
H. H. A. 27
(1),
418
HIGHER ALGEBRA.
.(2).
(3).
where X l = a^ + a 2 x 2 \
^=A*i+/W
Substituting for X, and X 2 in (1), we have
(a^ + 6^) x, + (a^+bfij x 2 = 0\
(a t a x + b 2 P } ) x x + (a 2 a, + bfi 2 ) x 2 = Oj
In order that equations (3) may simultaneously hold for
values of x x and x 2 other than zero, we must have
a^ + bfr a x a 2 + bfi 2 = (4).
But equations (3) will hold if equations (1) hold, and this
will be the case either if
a a b 2
(5),
or if X l = and X 2 = 0;
which last condition requires that
a.
a.
ft
ft
= o
.(6).
Hence if equations (5) and (6) hold, equation (4) must also
hold ; and therefore the determinant in (4) must contain as
factors the determinants in (5) and (6) ; and a consideration of
the dimensions of the determinants shews that the remaining
factor of (4) must be numerical ; hence
«i a i + & A »i a , + h A
the numerical factor, by comparing the coefficients of afyafl,
on the two sides of the equations, being seen to be unity.
a l
\
X
a l
ft
==
«.
K
tt 2
ft
Cor.
a.
«■,
a* + b*
a i a 2 + h h
a,a 2 + bfi 2
a' + b:
The above method of proof is perfectly general, and holds
whatever be the order of the determinants.
Since the value of a determinant is not altered when we
write the rows as columns, and the columns as rows, the product
of two determinants may be expressed as a determinant in
several ways ; but these will all give the same result on ex
pansion.
DETERMINANTS.
419
4
B l
Ci
=z
a l
h
*i
A.,
B S
C 2
a 2
b..
C 2
^
~B,
c*
a 3
h
C i
Example. Shew that
the capital letters denoting the minors of the corresponding small letters in
the determinant on the right.
Let D, D' denote the determinants on the right and lefthand sides
respectively; then
DD'.
a 1 A l  b l B 1 + c 1 C 1
— a x A. 2 + b x B%  c x C 2
a 2 A x  b. 2 B x + c 2 C x
a 2 A 2 + b 2 B 2  c 2 C 2
 a 3 A 2 + b.jB.2  c 3 C 2
^ 3 ^ 3 t 3 2?3 + c 3 C 3
D
D
D
[Art. 493.]
thus DD' = D 3 , and therefore D' = D*.
EXAMPLES. XXXIII. a.
Calculate the values of the determinants
1.
1
1
1
35
37
34
23
26
25
13
16
19
14
17
20
15
18
21
3.
4.
7.
a h
h b
9 f
ab
bc
c — a
9 5.
/
c
b—c ca
c — a, a—b
b bc
y
l
— X
y
X
1
6.
a
8.
b + c
b
c
a
c + a
c
13 3 23
30
53
39 9 70
1 1
1
1 1+i
v 1
1 1
l+y
a
•
b
a + b
If <o is one of the imaginary cube roots of unity, find the value of
9.
1 CO
9
CO CO"
2 1
U)
co
1
CO
10.
CO'
co
co
1
CO
co
co
1
11. Eliminate I, m } n from the equations
al + cm + bn = 0, cl 4 bm + an = 0. bl + am + en = 0.
and express the result in the simplest form.
27—2
420
HIGHER ALGEBRA.
12. Without expanding the determinants, prove that
a
x
P
b
y
c
z
r
x
z
b
a
c
1
P
r
x
P
a
y
b
z
r
c
13. Solve the equations
14.
15.
16.
17.
1
1
1
1
a
a?
x
x
■2
a
b
c i
1
b
y
y 2
zx
a"
b 2
1
c
c 3
= (bc)(c a) (a  b).
= (b  c) (c  a) (a  b) (a + b + c).
6
*2
18.
19.
yz zx xy
la a + b
b + a 26
c + a c+b
(b+cY
6 2
^2
= (yz)(zx) (x y)(yz + zx + xy).
a + c
b + c
2c
4(b + c)(c + a)(« + b).
{c + af
a
6 2
(a+bf
2dbc{a+b+cf.
20. Express as a determinant
f
c
b
c
b
1)
a a x
=
0.
(2)
15
2x
11
10
m m m
11 3x
17
16
b x b
1x
14
13
the following identities :
b+c c + a a+b
= 2
a b c
•
q + r r+p p + q
p q r
y+
z z + x x\
y
x y z
0.
^r
21. Find the condition that the equation lx + my + nz=0 may be
satisfied by the three sets of values (a u b lt c{) (a. 2 , 6 2 , c 2 ) (a 3 , b 3 , c 3 ) ;
and shew that it is the same as the condition that the three equations
a l x + b 1 y + c i z = 0, a 2 x + b 2 y + c 2 z = 0, a 3 x + b 2 y + c 3 z =
may be simultaneously satisfied by I, m, n.
DETERMINANTS.
421
22.
Fi
nd the value of
a' 1 f X 2 ab + cX ca  bX
X
X c b
ab  cX b 2 + X 2 be + aX
— e X a
ca + bX be  aX c 2 + X 2
b —a X
Prove that
a + ib c + id
X
aifi y id
— c + id a ib
\ yid a + ifi
23.
where i = *j — 1, can be written in the form
AiB CiD
CiB A + iB
hence deduce the following theorem, due to Euler :
The product of two sums each of four squares can be expressed as the
sum of four squares.
Prove the following identities :
24.  1 bc + ad b 2 e 2 + a 2 d 2
1 ca + bd c 2 a 2 + b 2 d 2
1 ab + cd a 2 b 2 + c 2 d 2
=  (bc) (e  a) (ab) (a d) (b  d) (cd).
25.
be — a 2
— be + ca + ab
(a + b) (a + c)
26.
ca  b 2 ab  c 2
bc — ca + ab be + ca  ab
(b + c)(b + a) (c + a)(c + b)
= 3(bc)(c a) (a b)(a + b + c) (be + ca + ab).
(a  x) 2 (a  y) 2 (a — z) 2
(hxf {by) 2 (bzf
ioxf (ey) 2 (cz) 2
= 2 (bc) (e  a) (a b){yz)(z x) (x y).
27. Find in the form of a determinant the condition that the
expression
Ua? + V@ 2 + Wy 2 + 2u'Py + 2 c'ya + 2 ic'aft
may be the product of two factors of the first degree in a, ft, y
28. Solve the equation :
u + a 2 x w' + abx v' + acx
w' + abx v + b 2 x u' + bcx
c + acx u' + bcx w + c 2 x
expressing the result by means of determinants.
= 0,
422
HIGHER ALGEBRA.
499. The properties of determinants may be usefully em
ployed in solving simultaneous linear equations.
Let the equations be
a x x + h x y + c x z + d x = 0,
a 2 X + b 2 y + c 2 z + d 2 = 0,
aja + bg + cji + d^O;
multiply them by A lt A s , A. A respectively and add the results,
A , A j A 3 being minors of a lt a 2i a a in the determinant
D =
a,
a„
*,
«3 ^3 ^
The coefficients of y and z vanish in virtue of the relations proved
in Art. 493, and we obtain
(Mi  M* + M 3 ) * + (M i  M 2 + Mb) = °
Similarly we may shew that
(6,5,  6 A + 6 A) 2/ + (<*A  <*A + <*A) = 0,
and
fcff,  e,C, + c 3 C 3 ) ■ + (dfi,  dfi a + d 3 C,) = 0.
Now «A  a t A, + «A   (6,5,  6,5, + 6 A)
hence the solution may be written
x —y z
<*,
*.
«■
<*,
J a
c s
d 3
* 3
C 3
^
«l
c l
^
<* 2
C 2
^3
a z
C B
rf,
°l
6,
<*,
°J
* 2
^3
°a
6,
1
«I
^
C !
«2
^
C 2
«■
^3
C 3
or more symmetrically
x y
*,
c ,
<h
K
c,
*,
h
C 3
^
^
C !
*.
a 2
C 2
d 2
%
C s
d 3
<*,
6,
*,
«.
b.
J,
«„
K
*
a.
3 K C 3
500. Suppose we have the system of four homogeneous linear
equations :
DETERMINANTS.
423
a x x + b x y + c,s + d x u — 0,
a 2 x + b 2 y + c 2 z + d 2 u = 0,
a 3 x + b 3 y + c 3 z + d 3 u = 0,
ax + b A y + c A z + du = 0.
4 W 4 4
From the last three of these, we have as in the preceding article
x —y z — u
K
« s
<*,
K
C 3
d.
h
C 4
d
4
Cl 2
C 2
rf,
%
C 3
^3
a *
C 4
^
«. ^2
% \
»4 6 4
^
4
£
«2
K
C 2
^3
K
C ,
»4
K
C 4
Substituting in the first equation, the eliminant is
«.
K C 2 d 2
"ft,
K c 3 d z
J \ C 4 d *
«. C 2 ^2
+ C i
«a c 3 <*a
«4 C 4 ^4
(l ° h 2 d 2
~d.
a s K c h
a 4 b 4 d *
a 2
K
C 2
%
K
C 3
a 4
h
C 4
0.
This may be more concisely written in the form
a 2
a.
b x c, d x
h 2 C 2 ( h
b.
C 3
c.
<*3
d,
= 0;
the expression on the left being a determinant of the fourth order.
Also we see that the coefficients of <z,, b x , c,, d x taken with
their proper signs are the minors obtained by omitting the row
and column which respectively contain these constituents.
501. More generally, if we have n homogeneous linear
equations
a x x x + b x x 2 + c x x 3 + + k x x n = 0,
a 2 x x + b 2 x 2 + c 2 x 3 + + k 2 x n = 0,
a x, + b x a + ex, + + kx= 0,
« 1 « 2 n 3 « n '
involving u unknown quantities x x , x 2 , x 3 , ...as,,, these quantities
can be eliminated and the result expressed in the form
a.
a.
k
a
> c
ii n
0.
424 HIGHER ALGEBRA.
The lefthand member of this equation is a determinant which
consists of n rows and n columns, and is called a determinant of
the II th  order.
The discussion of this more general form of determinant is
beyond the scope of the present work ; it will be sufficient here
to remark that the properties which have been established in the
case of determinants of the second and third orders are quite
general, and are capable of being extended to determinants of
any order.
For example, the above determinant of the n th order is
equal to
a 1 A 1 b 1 B l + c 1 C 1 d 1 D 1+ ... + (l)" 1 k 1 K l ,
or ct l A 1 a 2 A 2 + a B A 3 a 4 A 4 + ... + (1)" 1 a n A n ,
according as we develop it from the first row or the first column.
Here the capital letters stand for the minors of the constituents
denoted by the corresponding small letters, and are themselves
determinants of the (nl) th order. Each of these may be ex
pressed as the sum of a number of determinants of the (n — 2) th
order ; and so on ; and thus the expanded form of the deter
minant may be obtained.
Although we may always develop a determinant by means of
the process described above, it is not always the simplest method,
especially when our object is not so much to find the value of
the whole determinant, as to find the signs of its several
elements.
502. The expanded form of the determinant
a l
*,
c i
a 2
\
C 2
%
K
C 3
= afi 2 G 3 ~ afi 3 C 2 + a A G i ~ a 2 K C 3 + a * h i c 2 ~ % h 2 G x ;
and it appears that each element is the product of three factors,
one taken from each row, and one from each column; also the
signs of half the terms are + and of the other half  . The signs
of the several elements may be obtained as follows. The first
element ci x b 2 c^ in which the suffixes follow the arithmetical order,
is positive ; we shall call this the leading element ; every other
element may be obtained from it by suitably interchanging the
suffixes. The sign + or  is to be prefixed to any element ac
1
DETERMINANTS. 425
cording as it can be deduced from the leading element by an
even or odd number of permutations of two suffixes ; for instance,
the element a 3 b 2 c 1 is obtained by interchanging the suffixes 1 and
3, therefore its sign is negative ; the element ajb 1 c 2 is obtained
by first interchanging the suffixes 1 and 3, and then the suffixes
1 and 2, hence its sign is positive.
503. The determinant whose leading element is a x b 2 c 3 d A ...
may thus be expressed by the notation
%^aJ> a e B d A ,
the 2 * placed before the leading element indicating the aggregate
of all the elements which can be obtained from it by suitable
interchanges of suffixes and adjustment of signs.
Sometimes the determinant is still more simply expressed by
enclosing the leading element within brackets; thus (a^crf ...)
is used as an abbreviation of 5 ± a,b„c„d A ....
Example. In the determinant (a^c^e^ what sign is to be prefixed to
the element a^c^e.,1
From the leading element by permuting the suffixes of a and d we get
a 4 b 2 c 3 d x e 5 ; from this by permuting the suffixes of b and c we have a 4 b 3 c 2 d 1 e 5 ;
by permuting the suffixes of c and d we have a i b. i c 1 d 2 e 5 ; finally by permuting
the suffixes of d and e we obtain the required element rt 4 & 3 c 1 rf 5 ? 2 ; and since
we have made four permutations the sign of the element is positive.
504. If in Art. 501, each of the constituents 6 , c , ... k is
equal to zero the determinant reduces to a A ; in other words
it is equal to the product of a y and a determinant of the (n — l) th
order, and we easily infer the following general theorem.
If each of the constituents of the first row or column of a
determinant is zero except the first, and if this constituent is equal
to m, the determinant is equal to m times that determinant of lower
order ivhich is obtained by omitting the first column and first
row,
Also since by suitable interchange of rows and columns any
constituent can be brought into the first place, it follows that if
any row or column has all its constituents except one equal to
zero, the determinant can immediately be expressed as a deter
minant of lower order.
This is sometimes useful in the reduction and simplification
of determinants.
426
HIGHER ALGEBRA.
Example.
Find the value of
30
11
20
38
6
3
9
11
2
36
3
19
6
17
22
Diminish each constituent of the first column by twice the corresponding
constituent in the second column, and each constituent of the fourth column
by three times the corresponding constituent in the second column, and
we obtain
8 20
5
15 36
9
7 17
4
8 11
20
5
>
3
15 2 36 9
7 16 17 4
row has three zero constituents 1
= 3
8 20 5
= 3
10
8 19 5
8 19 5
7
17 4
7
17 4
= 3
8
7
5
4
505. The following examples shew artifices which are oc
casionally useful.
Example 1. Prove that
= (a + b + c + d) (a  b + c  d) (a  b  c + d) (a + b  c  d).
a
b
c
d
b
a
d
c
c
d
a
b
d
c
b
a
By adding together all the rows we see that a + b + c + d is a factor of the
determinant; by adding together the first and third rows and subtracting
from the result the sum of the second and fourth rows we see that
ab + c  d is also a factor ; similarly it can be shewn that abc + d and
a + bc d are factors ; the remaining factor is numerical, and, from a com
parison of the terms involving a 4 on each side, is easily seen to be unity ;
hence we have the required result.
Example 2. Prove that
1
1
1
1
a
b
c
d
« 2
6 2
c 2
d
a*
b*
d*
: (a  6) (a  c) (a  d) (b c)(b d) (c  d).
The given determinant vanishes when b — a, for then the first and second
columns are identical ; hence a  b is a factor of the determinant [Art. 514].
Similarly each of the expressions a  c, a  d, b — c, b  d, c  d is a factor of
the determinant; the determinant being of six dimensions, the remaining
factor must be numerical ; and, from a comparison of the terms involving
bc 2 d 3 on each side, it is easily seen to be unity ; hence we obtain the required
result.
DETERMINANTS.
427
EXAMPLES. XXXIII. b.
Calculate the values of the determinants
1.
1 1
I 2
3.
1
1
a
1
1
3
4
1
1
3
4
6
10
10
20
2.
7 13 10 6
5
9
8 12 11 7
4 10 6 3
1 1 1
a 1 1
1 a 1
1 a
4.
5.
7.
9.
1 1
3 2 14
15 29 2 14
16 19 3 17
33 39 8 38
x y z
x z y
y z x
z y x
a b c
a a + b a + b + c
a 2a + b 3a + 2b + c
1
1
1
b + c
a
1
b
c+a
1
c
c
1
a
b
a + b
6.
8.
\\a
1
1
1
— x
y
— z
1
l+b
1
1
X
— c
b
1
1
1+c
1
1
1
1
l+d
y
s
c
b
a
a
d
a + b + c + d
4a + 3b + 2c + d
1 a 3a + b 6a + 36 + c \0a + 6b + 3c + d
10. If o> is one of the imaginary cube roots of unity, shew that
the square of
CO
CO
V
2 "?
co" or
CO'
CO"'
CO 3
=
CO 3
1
1
CO
CO
CO''
1
1
2
1
1
1
2
1
1
1 2
1
2
1
1
hence shew that the value of the determinant on the left is 3 v / — 3.
11. If {Pbc)x+{chfg)y + {bghf)z = 0,
(ch fy) x + (g* ca) y + (af gh) z=0,
(bghf)x+(afgh)y + (h* ab)z=0,
shew that abc + 2fgh  a/ 2  bg 2  ch 2 = 0.
428
HIGHER ALGEBRA.
Solve the equations :
12. x+ y+ 0=1,
ax + by + cz=k,
a 2 x + b 2 y + c 2 z = l 2 .
14.
15.
where
13. ax + by + cz=k,
a 2 x + b 2 y + c 2 z = k 2 ,
a?x + b 3 y + c?z= P.
x + y+ z+ u=l,
ax + by + cz+ du=k,
a 2 x + b 2 y + c 2 z + d 2 u = k 2 ,
cfix + b s y + c 3 z + d 3 u = P.
Prove that
b+c—a—d
c+a—b—d
a+bcd
16. Prove that
be — ad be (a + d)ad(btc)
ca — bd ca(b\d) — bd(e + a)
ab — cd ab (c + d) cd{a + b)
= 2 (b e) (ca) (ab) (ad) (b  d) (cd). \
a 2 a 2 (b — c) 2
b 2 b 2 (ca) 2
c 2 c 2  (a  b) 2
be
ca
ab
= (bc)(ca){ab)(a + b + c)(a 2 + b 2 + c 2 ).
17. Shew that
a b
f «
e f
d e
c d
b c
c
b
a
d
c
b
f cc
e f
d e
e
d
c
b
a
f
f
e
d
c
b
a
ABC
CAB
B C A
A=a 2 d 2 + 2ce 2bf,
B=e 2 b 2 +2ac2df,
C=e 2 f 2 + 2ae2bd.
18. If a determinant is of the ?i th order, and if the constituents
of its first, second, third, ...n th rows are the first n figurate numbers of
the first, second, third, ...n th orders, shew that its value is unity,.
CHAPTER XXXIV.
MISCELLANEOUS THEOREMS AND EXAMPLES.
506. We shall begin this chapter with some remarks on the
permanence of algebraical form, briefly reviewing the fundamental
laws which have been established in the course of the work.
507. In the exposition of algebraical principles we proceed
analytically : at the outset we do not lay down new names and
new ideas, but we begin from our knowledge of abstract
Arithmetic ; we prove certain laws of operation which are capable
of verification in every particular case, and the general theory of
these operations constitutes the science of Algebra.
Hence it is usual to speak of Arithmetical Algebra and Sym
bolical Algebra., and to make a distinction between them. In the
former we define our symbols in a sense arithmetically intelligible,
and thence deduce fundamental laws of operation ; in the latter
we assume the laws of Arithmetical Algebra to be true in all
cases, whatever the nature of the symbols may be, and so find
out what meaning must be attached to the symbols in order that
they may obey these laws. Thus gradually, as we transcend the
limits of ordinary Arithmetic, new results spring up, new lan
guage has to be employed, and interpretations given to symbols
which were not contemplated in the original definitions. At the
same time, from the way in which the general laws of Algebra
are established, we are assured of their permanence and uni
versality, even when they are applied to quantities not arithmeti
cally intelligible.
508. Confining our attention to positive integral values of
the symbols, the following laws are easily established from a priori
arithmetical definitions.
430 HIGHER ALGEBRA.
I. The Law of Commutation, which we enunciate as follows :
(i) Additions and subtractions may be made in any order.
Thus a + bc = ac + b = bc + a.
(ii) Multiplications and divisions may be made in any order.
Thus axb=bxa;
axbxc = bxcxa = axcxb' } and so on.
ab± c = a x b = c = (a f c) x b = (b + c) xa.
II. The Law of Distribution, which we enunciate as follows : 
Multiplications and divisions may be distributed over additions
and subtractions.
Thus {a  b + c) m = am —bm + cm,
(a — b)(c — d) = ac — ad — bc + bd.
[See Elementary Algebra, Arts. 33, 35.]
And since division is the reverse of multiplication, the distri
butive law for division requires no separate discussion.
III. The Laws of Indices.
(i) a m xa n = a m+n
3
a m + a H = a m ".
(n) [a ) = a .
[See Elementary Algebra, Art. 233 to 235.]
These laws are laid down as fundamental to our subject, having
been proved on the supposition that the symbols employed are
positive and integral, and that they are restricted in such a way
that the operations above indicated are arithmetically intelligible.
If these conditions do not hold, by the principles of Symbolical
Algebra we assume the laws of Arithmetical Algebra to be true
in every case and accept the interpretation to which this assump
tion leads us. By this course we are assured that the laws of
Algebraical operation are selfconsistent, and that they include in
their generality the particular cases of ordinary Arithmetic.
509. From the law of commutation we deduce the rules
for the removal and insertion of brackets [Elementary Algebra,
Arts. 21, 22] ; and by the aid of these rules we establish the law
MISCELLANEOUS THEOREMS AND EXAMPLES. 431
of distribution as in Art. 35. For example, it is proved that
(a b)(c — d)~ac — ad—bc + bd,
with the restriction that a, b, c, d are positive integers, and a
greater than b, and c greater than d. Now it is the province of
Symbolical Algebra to interpret results like this when all restric
tions are removed. Hence by putting a = and c = 0, we obtain
(— b) x (— d) = bd, or the product of two negative quantities is
positive. Again by putting 6 = and c= 0, we obtain a x (—d) =—a<I,
or the product of two quantities of opposite signs is negative.
"We are thus led to the Rule of Signs as a direct consequence
of the law of distribution, and henceforth the rule of signs is
included in our fundamental laws of operation.
510. For the way in which the fundamental laws are applied
to establish the properties of algebraical fractions, the reader is
referred to Chapters xix., xxi., and xxn. of the Elementary Algebra ;
it will there be seen that symbols and operations to which we
cannot give any a priori definition are always interpreted so as
to make them conform to the laws of Arithmetical Algebra.
511. The laws of indices are fully discussed in Chapter xxx.
of the Elementary Algebra. When m and n are positive integers
and m > n, we prove directly from the definition of an index that
a m xa n = a m+n ; a m r a' 1 = a m ~ n j (a m ) n = a m ".
We then assume the first of these to be true when the indices
are free from all restriction, and in this way we determine mean
ings for symbols to which our original definition does not apply.
p
The interpretations for a\ a , a~" thus derived from the first law
are found to be in strict conformity with the other two laws ;
and henceforth the laws of indices can be applied consistently and
with perfect generality.
512. In Chapter vill. we defined the symbol i or J— 1 as
obeying the relation i 2 = — 1 . From this definition, and by
making i subject to the general laws of Algebra we are enabled
to discuss the properties of expressions of the form a + ib, in
which real and imaginary quantities are combined. Such forms
are sometimes called complex numbers, and it will be seen by
reference to Articles 92 to 105 that if we perform on a complex
number the operations of addition, subtraction, multiplication,
and division, the result is in general itself a complex number,
432 HIGHER ALGEBRA.
Also since every rational function involves no operations but
those above mentioned, it follows that a rational function of a
complex number is in general a complex number.
Expressions of the form a x+ly , \og(x±iy) cannot be fully
treated without Trigonometry; but by the aid of De Moivre's
theorem, it is easy to shew that such functions can be reduced to
complex numbers of the form A + iB.
The expression e x+iy is of course included in the more general
form a x+i \ but another mode of treating it is worthy of attention.
We have seen in Art. 220 that
(x\ n
1 H — ) , when n is infinite,
nj
x being any real quantity ; the quantity e x+i!/ may be similarly
defined by means of the equation
e *+iy
= Lini (1 H ) , when n is infinite,
\ n J
x and y being any real quantities.
The development of the theory of complex numbers will be
found fully discussed in Chapters x. and XI. of Schlomilch's
Handbuch der algebraischen Analysis.
513. We shall now give some theorems and examples illus
trating methods which will often be found useful in proving
identities, and in the Theory of Equations.
514. To find the remainder ivhen any rational integral function
of x is divided by x  a.
Let fix) denote any rational integral function of x ; divide
f(x) hyxa until a remainder is obtained which does not involve
x ; let Q be the quotient, and R the remainder ; then
f(x) = Q(xa) + R.
Since R does not involve x it will remain unaltered whatever
value we give to x ; put x = a, then
f(a) = QxO + R;
now Q is finite for finite values of x, hence
.MISCELLANEOUS THEOREMS AND EXAMPLES. 433
Cor. If f{x) is exactly divisible by x  a, then R == 0, that is
f(a) = ; hence if a rational integral function of x vanishes when
x — a, it is divisible by x  a.
515. The proposition contained in the preceding article is so
useful that we give another proof of it which has the advantage
of exhibiting the form of the quotient.
Suppose that the function is of n dimensions, and let it be
denoted by
p x n +2\ x "~ }+ P^"~ 2 +P^"" 3+ +P»>
then the quotient will be of n  1 dimensions ; denote it by
q{f c" x + qi x n  2 +q 2 x n  3 + ... +q H _ l ;
let R be the remainder not containing x ; then
pjf +p 1 x 1 +2> 2 x' 2 +p. a x"~ 3 + ••• +P„
= (xa) (q X^ + qi x"~ 2 + q 2 x"~ 3 + ... + q a _ x ) + R
Multiplying out and equating the coefficients of like powers of x,
we have
9. 2  a Qi=P 2 > or qs = a 4i+P a '>
q 3  n 2 = ihi or & = «<i 2 + ih ;
R  oq n  % =P n , or R = aq n _ l +p n ;
thus each successive coefficient in the quotient is formed by
multiplying by a the coefficient last formed, and adding the
next coefficient in the dividend. The process of finding the
successive terms of the quotient and the remainder may be
arranged thus :
Po Pi P 2 P 3 PaX Pa
«?0 Ct( lx Cl( l 2 Cl( In2 (l( 2nl
% q x % v, ?., x
Thus R = aq^ +p n  « («<?„ +#.i) + P* =
=P<P* +P^"~ 1 +P/<< n ~ 2 + ••• +P,r
If tlie divisor is x + a the same method can be used, only in
this case the multiplier is  a.
H. H. A. 28
434 HIGHER ALGEBRA.
Example. Find the quotient and remainder when 3a; 7  a; 6 + 31a: 4 + 21a; + 5
is divided by x + 2.
Here the multiplier is  2, and we have
310 31 00 21 5
6 14 28 6 12 24 6
3 7 14 3 6 12  3 11
Thus the quotient is 3.r 6  7a; 5 + 14a; 4 + 3a; 3  6a; 2 +12a;3, and the re
mainder is 11.
516. In the preceding example the work has been abridged
by writing down only the coefficients of the several terms, zero
coefficients being used to represent terms corresponding to powers
of x which are absent. This method of Detached Coefficients may
frequently be used to save labour in elementary algebraical
processes, particularly when the functions we are dealing with
are rational and integral. The following is another illustration.
Example. Divide 3a; 5  8a; 4  5a; 3 + 26a; 2  33a; + 26 by a; 3  2a; 2  4a; + 8.
1 + 2 + 48)38 5 + 2633 + 26(32 + 3
3 + 6 + 1224
2 +
2
7 +
4
233
8 + 16
3
3 +
617 + 26
6 + 1224
5+2
Thus the quotient is 3a; 2  2;r + 3 and the remainder is  5a; + 2.
It should be noticed that in writing down the divisor, the sign of every
term except the first has been changed ; this enables us to replace the process
of subtraction by that of addition at each successive stage of the work.
517. The work may be still further abridged by the following
arrangement, which is known as Horner's Method of Synthetic
Division.
1
3_8 5 + 2633 + 26
2
6 + 1224
4
 4 8 + 16
8
6 + 1224
32+ 3+ 0 5+ 2
[Explanation. The column of figures to the left of the vertical line
consists of the coefficients of the divisor, the sign of each after the first being
changed; the second horizontal line is obtained by multiplying 2, 4, 8
by 3, the first term of the quotient. We then add the terms in the second
column to the right of the vertical line ; this gives  2, which is the coeffi
cient of the second term of the quotient. With the coefficient thus obtained
MISCELLANEOUS THEOREMS AND EXAMPLES. 435
we form the next horizontal line, and add the terms in the third column;
this gives 3, which is the coefficient of the third term of the quotient.
By adding up the other columns we get the coefficients of the terms in
the remainder. ]
Example. Divide 6a 5 + ba*b  8a?b 2  6a 2 b 3  6a ¥ by 2a 3 + 3a 2 6  b z
to four terms in the quotient.
2
6+5866
3
9+0+3
6 + 02
1
3 +
 1
12 + 0
321+04  +11+04
Thus the quotient is 3a 2  2ab  b 2  4a~ 2 6 4 , and llb 5 4a~ 2 b 7 is the
remainder.
Here we add the terms in the several columns as before, but each sum has
to be divided by 2, the first coefficient in the divisor. When the requisite
number of terms in the quotient has been so obtained, the remainder is
found by merely adding up the rest of the columns, and setting down the
results without division.
The student may easily verify this rule by working the division by
detached coefficients.
518. The principle of Art. 514 is often useful in proving
algebraical identities; but before giving any illustrations of it
we shall make some remarks upon Symmetrical and Alternating
Functions.
A function is said to be symmetrical with respect to its vari
ables when its value is unaltered by the interchange of any pair
of them ; thus x + y + z, be + ca + ab, x 3 + y 3 + z 3 — xyz are sym
metrical functions of the first, second, and third degrees respec
tively.
It is worthy of notice that the only symmetrical function of
the first degree in x, y, z is of the form M (x + y + z), where M is
independent of x, y, z.
519. It easily follows from the definition that the sum,
difference, product, and quotient of any two symmetrical expres
sions must also be symmetrical expressions. The recognition of
this principle is of great use in checking the accuracy of alge
braical work, and in some cases enables us to dispense with much
of the labour of calculation.
For example, we know that the expansion of (as + y + z) 3 must
be a homogeneous function of three dimensions, and therefore
of the form x 3 + y 3 + z 3 + A (x 2 y + xy 2 + y 2 z + yz 2 + z 2 x + zx 2 ) + Bxyz,
where A and B are quantities independent of x, y, z.
28—2
436 HIGHER ALGEBRA.
Put z = 0, then A = 3, being the coefficient of x 2 y in the ex
pansion of (x + y) 3 .
Put x = y = z = l, and we get 27 = 3 + (3 x 6) + B ; whence
B = 6.
Thus (x + y + z) 3
= x 3 + y 3 + z 3 + 3x 2 y + 3xy 2 + 3y 2 z + 3yz 2 + 3z 3 x + 3zx 2 + 6xyz.
520. A function is said to be alternating with respect to its
variables, when its sign but not its value is altered by the inter
change of any pair of them. Thus x — y and
a 2 (bc) + b 2 (ca) + c 2 (a  b)
are alternating functions.
It is evident that there can be no linear alternating function
involving more than two variables, and also that the product of
a symmetrical function and an alternating function must be an
alternating function.
521. Symmetrical and alternating functions may be con
cisely denoted by writing down one of the terms and prefixing
the symbol % ; thus %a stands for the sum of all the terms of which
a is the type, %ab stands for the sum of all the terms of which
ab is the type; and so on. For instance, if the function involves
four letters a, b, c, d }
^aa + b + c + d;
%ab = ab + ac + ad +bc + bd+ cd;
and so on.
Similarly if the function involves three letters a, b, c,
$a 2 (b c) = a 2 (bc)± b 2 (c  a) + c 2 (a  b) •
%a 2 bc = a 2 bc + b 2 ca + c 2 ab;
and so on.
It should be noticed that when there are three letters involved
%a 2 b does not consist of three terms, but of six : thus
2<a 2 b = a 2 b + a 2 c + b 2 c + b 2 a + c 2 a + c 2 b.
The symbol 2 may also be used to imply summation with
regard to two or more sets of letters; thus
%yz (bc) = yz (b~c) + zx (ca) + xy (a  b).
MISCELLANEOUS THEOREMS AND EXAMPLES. 437
522. The above notation enables us to express in an abridged
form the products and powers of symmetrical expressions : thus
(a+b + c) 3 = %a 3 + 32a 2 b + Gabc j
(a + b + c + df = 2« 3 + 3$a 2 b + Gtabc;
(a + b + c) 4 = %a A + i%cfb + 6Sa"6 fl + 1 2%a 2 bc;
%a x 2« 2 = 2a 3 + %a 2 b .
Example 1. Prove that
(a + b) 5  a 5  b* = 5ab (a + b) (a 2 + ab + b 2 ).
Denote the expression on the left by E ; then E is a function of a which
vanishes when a = ; hence a is a factor of E ; similarly 6 is a factor of E.
Again E vanishes when a—  b, that is a + b is a factor of E; and therefore
E contains ab(a + b) as a factor. The remaining factor must be of two
dimensions, and, since it is symmetrical with respect to a and b, it must be
of the form Act? + Bab + Ab' z ; thus
(a + b) 5  a 5  b 5 = ab (a + 6) (Aa* + Bab + A b~),
where A and B are independent of a and b.
Putting a = 1, b = 1, we have 15 = 2A + B ;
putting a = 2, b =  1, we have 15 = 5A  2B ;
whence A = o, J5 = 5; and thus the required result at once follows.
Example 2. Find the factors of
(&3 + c 3) (bc) + (c 3 + a 3 ) (ca) + (a 3 + b 3 ) (a  b).
Denote the expression by E ; then E is a function of a which vanishes
when a = b, and therefore contains a  b as a factor [Art. 514]. Similarly it
contains the factors bc and ca; thus E contains (b  c) (c  a) (a  b) as a
factor.
Also since E is of the fourth degree the remaining factor must be of the
first degree; and since it is a sj^mmetrical function of a, b, c, it must be of
the form M{a + b + c). [Art. 518];
.. E = M (bc) (ca) (ab)(a + b + c).
To obtain M we may give to a, b, c any values that we find most con
venient; thus by putting a = 0, 6 = 1, c = 2, we find M=l, and we have the
required result.
Example 3. Shew that
(x + y + zjtx 5 y b  z? = 5 (y + z) (z + x) (x + y) (x 2 + y 2 + z~ + yz + zx+ xy).
Denote the expression on the left by E ; then E vanishes when y=z,
and therefore y + z is a factor of E; similarly z + x and x + y are factors;
therefore E contains (y + z) (z + x) [x + (/)asa factor. Also since E is of the
438 HIGHER ALGEBRA.
fifth degree the remaining factor is of the second degree, and, since it is
symmetrical in x, y, z, it must be of the form
A (x 2 + y 2 + z 2 ) + B (yz + zx + xy) .
Put»=2/=z=l; thus 10=^1+5;
put x=2, y=l, 2 = 0; thus 35 = 5A + IB ;
whence A=B = 5,
and we have the required result.
523. We collect here for reference a list of identities which
are useful in the transformation of algebraical expressions; many
of these have occurred in Chap. xxix. of the Elementary Algebra.
^bc (b — c) = —(bc)(c a) (a  b).
$a 2 (bc) = (bc)(ca)(ab).
$a(b 2 c 2 ) = (bc)(ca)(ab).
2a 3 (bc) = (bc) (ca) (ab) (a + b + c).
a s + b 3 + c 3  3abc = (a + b + c)(a 2 +b 2 + c 2  bcca ab).
This identity may be given in another form,
a 3 + b 3 + c 3 3abc = l(a + b + c){(bc) 2 + (ca) 2 + (ab) 2 }.
(bc) 3 + (ca) 3 + (ab) 3 = 3(bc)(ca)(ab).
(a + b + c) 3 a 3 b 3 c 3 = 3(b + c)(c + a)(a + b).
Hbc (b + c) + 2abc = (b + c)(c + a)(a + b).
%a 2 {b + c) + 2abc =(b + c)(c + a) (a + b).
(a + b + c) (be + ca + ab)  abc =(b + c)(c + a) (a + b).
2b 2 c 2 + 2c V + 2a 2 b 2 tffrc*
= (a + b + c)(b + ca)(c + ab)(a+bc).
EXAMPLES. XXXIV. a.
1. Find the remainder when 3^ + 1 1^ + 90# 2  19# + 53 is divided
by x + 5.
2. Find the equation connecting a and b in order that
2x i 7x 3 +ax + b
may be divisible by x  3.
MISCELLANEOUS THEOREMS AND EXAMPLES. 439
3. Find the quotient and remainder when
jfi _ 5#4 + 9 iV 3 _ q x i _ iq v + 13 j >s divided by x 2  3v + 2.
4. Find a in order that x 3 7x + 5 may be a factor of
tf _ 2x A  4^ + 19.V 2  Six + 12 + a.
5. Expand ^.^^.g ^ descending powers of x to four
terms, and find the remainder.
Find the factors of
6. a(6c) 3 + 6(ca) 3 + c(a6)3.
7. a 4 (6 2  c 2 ) + 6 4 (c 2  a 2 ) + c 4 (a 2  6 2 ).
8. (a + 6 + c) 3 (6 + ca) 3 (c+a6) 3 (a + 6c) 3 .
9. a (6  cf + & (c  af + c(a 6) 2 + 8a6c.
10. a (6 4  c 4 ) + b (c 4  a 4 ) + c(a i  6 4 ).
11. (6c + ca + a6) 3  J 3 ** 3  c% 3  a 3 6 3 .
12. (a + 6 + c) 4 (6 + c) 4 (c + a) 4 (a + 6) 4 + a 4 + 6 4 + c 4 .
13. (a + 6 + c) 5 (6 + ca) 5 (c + a6) 5 (a + 6c) 5 .
14. (tf  a) 3 (6  cf + (x  b) s (c  af + (x  c) 3 (a  6) 3 .
Prove the following identities :
15. 2 (6 + c  2a) 3 = 3(6 + c 2a) (c + a 26) (a + 6 2c).
a(bcf He*)* c{abf _ flM . g
i0, (ca)(a6r (a6)(6cr (6c)(ca)
17 J^ _?L 2c (6c)(ca)(a6)_ 3
'" a + 6 6 + c c + a (6 + c)(c + a)(a+6)
18. 2 a 2(& + c)2a 3 2a&c=^& + ca)(c + a6)(a + &c)
iy * (a6)(ac)^(6c)(6a)^(ca)(c6)
20. 42(6c)(6 + c2a) 2 = 92(6c)(6 + ca) 2 .
21. ty+z)*(e+x)*(x+y)*=tx*(y+zY+2(^z) 3 2^ 2 z*'
22. ^ («6  c 2 ) (ac 6 2 ) = (26c) (26c  2a 2 ).
23. «6c (2a) 3  (26c) 3 = abc 2a 3  26 3 c 3 = (a 2  6c) (6 2  ca) (c 2  a6).
24. 5(6 c) 3 (6 + c  2a) = ; hence deduce 2  y) (£ + 7  2a ) 3 = °
440 HIGHER ALGEBRA.
25. (b + cf+(c + af+(a + bfZ(b + c)(c + a)(a + b)
= 2(a 3 \b 3 + c 3 3abc).
26. If x=b+ca, y = c + ab, z = a + bc, shew that
#3 + ^3 + # _ g^g = 4 (a 3 + 6 3 + c 3  3a6c).
27. Prove that the value of a 3 + b 3 + c 3  3a6c is unaltered if we
substitute s«. s b, sc for a, 6, c respectively, where
3s = 2(a + 6 + c).
Find the value of
28 . , , w « w i + /1 »_ w _ » +
(ab)(ac)(xa) (bc)(ba)(xb) (ca) (cb) (xc)'
a 2 — b 2 — c 2 b 2 — c z — a 2 c 2 — a 2  Z> 2
29. 7 ft7 ;+77 wt x +
(a b)(a c) (b c)(b — a) (c a) (cb) '
30. ( a +P)( a + <l) , (b+p)(b + q) + ( c +p)(c + (J )
(ab)(ac)(a+x) (bc)(ba) (b + x) (ca) (cb)(c + x) '
31. 3 __ w ^ w ^ . 32. s
(a b) (a c) (a — d)' (a b) (a — c) (a d) '
33. If x + y + z = s, and #yz =£< 2 , shew that
'jp _y\(p__z\ + fp__ A /£. _ #\ + /£ _ A /> _ y\ 4
,y« p)\zs p) \zs pj\xs p) \xs pj\ys p) ' " 8
Miscellaneous Identities.
524. Many identities can be readily established by making
use of the properties of the cube roots of unity; as usual these
will be denoted by 1, w, o> 2 .
Example. Shew that
(x + yf x 7 y 7 = Ixy (x + y) (x 2 + xy + y 2 ) 2 .
The expression, E, on the left vanishes when x = 0, y = 0, x + y = 0;
hence it must contain xy (x + y) as a factor.
Putting x = coy, we have
E = {(1 + ta)7  W 7  1} y7= {(_ w 2)7 _ w 7 _ !} y i
= (_ w 2  w l)y 7 = 0;
hence E contains x  wy as a factor ; and similarly we may shew that it con
tains x  ury as a factor; that is, E is divisible by
(x ury) (x  to 2 ?/), or x^ + xy + y 2 .
MISCELLANEOUS IDENTITIES. 441
Further, E being of seven, and xy(x + y) (x 2 + xy + y 2 ) of five dimensions,
the remaining factor must be of the form A (x 2 + y' 2 ) + Bxy ; thus
(x + y) 7  x 7  y 7 = xy {x + y) (x 2 + xy + y 2 ) (Ax 2 + Bxy + Ay 2 ).
Putting a; = l, y = l, we have 21 = 2^+5;
putting x = 2, y= 1, we have 21 = 5^1 2B;
whence A = 7, B = 7 ;
.. (x + y) 7  x 7  y 7 = Ixy (x + y)(x 2 + xy + y 2 ) 2 .
525. We know from elementary Algebra that
a a + b 3 + c 3  3abc = (a+ b + c) (a 2 + b 2 + c 2 be ca — ab) ;
also we have seen in Ex. 3, Art. 110, that
a* + b 2 + c 2 — be — ca — ab = (a + ub + ore) (a + <a 2 b + wc) ;
hence a 3 + b 3 + c 3 — 3abc can be resolved into three linear factors;
thus
a 3 + b 3 + c 3  3abc = (a + b +c) (a + mb + arc) (a + <D 2 b + wc).
Example. Shew that the product of
a 3 + b 3 + c 3  dabc and a; 3 + y 3 + z 3  Sxyz
can be put into the form A 3 + B 3 + C 3  SABC.
The product = [a + b + c) (a + wb + ore) (a + w 2 & + wc)
x (x + y + z) (x + uy + urz) (x + w 2 y + uz).
By taking these six factors in the pairs (a + b + c) (x + y + z);
(a + u>b + w 2 c) (x + cry + uz) ; and (a + urb + uc) (x + wy + urz),
we obtain the three partial products
A + B + C, A + wB + uC, A+u 2 B + u)C,
where A = ax + by + cz, B — bx + cy + az, C = cx + ay + bz.
Thus the product = (A + B + C) [A + wB + u 2 C) (A + orB + «C)
= A 3 + B 3 +C 3 SABC.
526. In order to find the values of expressions involving
a, b, c when these quantities are connected by the equation
a + b + c = 0, we might employ the substitution
a — h + k, b = ioh + (x> 2 k, c = ufh + u>k.
If however the expressions involve a, b, c symmetrically the
method exhibited in the following example is preferable.
442 HIGHER ALGEBRA.
Example. If a + b + c = 0, shew that
6 (a 5 + b 5 + c 5 ) = 5 (a 3 + 6 3 + c 3 ) (a 2 + 6 2 + c 2 ).
We have identically
(1 + ax) (1 + bx) (1 + cz) = 1 +px + qx 2 +rx 3 ,
w here p — a + b + c, q = bc + ca + ab, r — abc.
Hence, using the condition given,
(1 + ax) (1 t bx)(l + cx) = l + qx 2 + rx 3 .
Taking logarithms and equating the coefficients of x n , we have
(~ ' (a n + b n + c n ) = coefficient of x n in the expansion of log(l + qx 2 + rx 3 )
n
= coefficient of x n in [qx 2 + rx 3 )  ^ {qx 2 + rx 3 ) 2 + ^ (qx 2 + rx 3 ) 3  . . .
By putting rc = 2, 3, 5 we obtain
a 2 + b 2 + c 2 a 3 + b 3 + c 3 a 5 + b 5 +c 5
j— =*• 3— =r ' T =  <?r;
whence » — = ~ • « '
and the required result at once follows.
If a=fiy, 6 = 7 a, c = a/3, the given condition is satisfied; hence
we have identically for all values of a, /3, y
6{(iS7) 5 + (7«) 5 +("/3) 5 }
= 5{(/3 7 ) 3 + (Ya) 3 + (a/3) 3 } {{§ y?+ (y a) 2 + (a/S) 2 }
that is,
(/37) 5 + (7a) 5 + (a^) 5 =5( J 87)( 7 a)(a i 3)(a 2 + ^ + 7 2 /377aa^;
compare Ex. 3, Art. 522.
EXAMPLES. XXXIV. b.
1. If (a + b + cf = a 3 + b z + c 3 , shew that when n is a positive
integer (a + b + cf n + 1 = a 2n + 1 + b 2n + 1 + c 2n + K
2. Shew that
(a + <ob + a) 2 c) 3 + (a + a> 2 b + a>c) 3 = (2a  b  c) (2b  c  a) (2c  a  b).
3. Shew that (x+y) n x 1l y n is divisible by xy(x 2 + xy+y 2 ), if
n is an odd positive integer not a multiple of 3.
4. Shew that
a 3 (bz  cy) 3 + b 3 (ex  azf + c 3 (ay  bx) 3 = Sabc (bz  cy) (ex  az) (ay  bx).
MISCELLANEOUS IDENTITIES. 443
5. Find the value of
(6 c)(c — a) (ab) + (b a>c) (c  a>a) (a  cob) + (6  eo 2 c) (c  arc<) (a  a> 2 6).
6. Shew that (a 2 + b 2 + c 2  be — ca  ab) (x 2 f y 2 + z 2 — yz  zx  xy)
may be put into the form A 2 + B 2 + C 2 BCCA AB.
7. Shew that (a 2 + ab + b 2 ) (x 2 + xy + y 2 ) can be put into the form
A 2 + AB + B 2 , and find the values of A and B.
Shew that
8. 2 (« a + 26c) 3  3 (a 2 + 26c) (6 2 + 2ca) (c 2 + 2ab) = (a 3 + 6 3 + c 3  3a6c) 2 .
9. 2 (a 2  fc) 3  3 (a 2  be) (b 2  ca) (c 2  ab) = (a 3 + 6 3 + c 3  3a6c) 2 .
10. (« 2 + 6 2 + c 2 ) 3 +2(6c + ca + a6) 3 3(a 2 + 6 2 + c 2 )(6c + ca + a6) 2
= (a?+b 3 + c 3 3abc) 2 .
If a + 6 + c = 0, prove the identities in questions 11 — 17.
11. 2(a 4 + 6 4 + c 4 ) = (a 2 + 6 2 + c 2 ) 2 .
12. a 5 + 6 5 + c 5 =  5a6c (6c + ca + ab).
13. a 6 + 6 6 + c 6 = 3a W  2 (6c + ca + a6) 3 .
14. 3(a 2 + 6 2 + c 2 )(a 5 + 6 6 + c 5 ) = 5(a 3 + 6 3 + c 3 )(« 4 + 6 4 + c 4 ).
,_ a 7 + b 7 + c 7 a 5 + 6 5 + c 5 a0 + 6 2 + c 2
15. m = !■ • R •
/6c ca ab\ ( a b c \
16. +— J =9
a b c J\b — c c — a
a
17. (6 2 c + c 2 a + a 2 6  3a6c) (6c 2 + ca 2 + ab 2  3abc)
= (be + ca + ab) 3 + 27a 2 6 2 c 2 .
18. 25 {Q,  zf + (z  x) 7 + (x  y) 7 } {{y  zf + (z xf + (x yf)
= 21 {(y  zf + (z  xf + (x yf] 2 .
19. {(yz) 2 + (zxf + (xy) 2 } 3 54:(i/z) 2 (zx) 2 (xyf
= 2(y + z2x) 2 (z+x 2y) 2 (x +y 2z) 2 .
20. (6  cf + (e  a) 6 + (a  6) 6  3 (6  c) 2 (c  a) 2 (a  bf
= 2 (a 2 + b 2 + c 2 be ca ab) 3 .
21. (6c) 7 + (ca) 7 + (a6) 7
= 7(6c)(ca)(a6)(a 2 + 6 2 + c 2 6ccaa6) 2 .
22. If a + 6 + c = 0, and x+y + z = 0, shew that
4 (ax + by + czf 3(ax + by + cz) (a 2 + 6 2 + c 2 ) (x 2 +y 2 + z 2 )
2(bc)(ca)(ab)(yz)(z x) (xy) = 54abcxyz.
444 HIGHER ALGEBRA.
If a + b + c + d=0, shew that
a 5 + &5 +c 5 + c #> g? + b 3 + (? + d 3 a 2 + b 2 + c 2 + d 2
23. g 3— • 2
24. (a 3 + Z> 3 + c 3 + d 3 ) 2 = 9 (M + cda + da& + abc) 2
= 9 (be ad) (ca  bd) (ab  cd).
25. If 2s = a + b + c and 2o 2 = a 2 + 6 2 + c 2 , prove that
5 (s  b) (s  c) (a 2  a 2 ) + 5a6cs = {s 2  cr 2 ) (4s 2 + a 2 ).
26. Shew that (a? + 6.2% + 3.vy 2  y 3 ) 3 + (v 3 + 6xy 2 + 3x 2 y ~ x 3 ) 3
= Zlxy (x+y) (x 2 + xy +y 2 ) 3 .
27. Shew that 2
a 5
(a — b)(a — c) (a  d)
= a 2 + b 2 + c 2 + d 2 + ab + ac + ad+bc + bd+cd.
28. Resolve into factors
2« 2 6 2 c 2 + (a 3 + b 3 + c 3 ) abc + Z> 3 c 3 + c% 3 + a 3 Z> 3 .
Elimination.
527. In Chapter xxxiii. we have seen that the eliminant of
a system of linear equations may at once be written down in the
form of a determinant. General methods of elimination ap
plicable to equations of any degree will be found discussed in
treatises on the Theory of Equations ; in particular we may refer
the student to Chapters iv. and VI. of Dr Salmon's Lessons Intro
ductory to the Modern Higher Algebra, and to Chap. xm. of
Burnside and Panton's Theory of Equations.
These methods, though theoretically complete, are not always
the most convenient in practice. We shall therefore only give a
brief explanation of the general theory, and shall then illustrate
by examples some methods of elimination that are more practi
cally useful.
528. Let us first consider the elimination of one unknown
quantity between two equations.
Denote the equations by f(x) = Q and <£ (x) = 0, and suppose
that, if necessary, the equations have been reduced to a form in
which f(x) and <£ (x) represent rational integral functions of x.
Since these two functions vanish simultaneously there must be
some value of x which satisfies both the given equations ; hence
ELIMINATION. 445
the eliminant expresses the condition that must hold between the
coefficients in order that the equations may have a common root.
Suppose that x = a, x = J3, x = y,... are the roots of f(x) = 0,
then one at least of the quantities <f> (a), <f> (/?), <f> (y), must
be equal to zero ; hence the eliminant is
4> (a) <f> tf) <f> (y) =0.
The expression on the left is a symmetrical function of the
roots of the equation fix) = 0, and its value can be found by the
methods explained in treatises on the Theory of Equations.
529. We shall now explain three general methods of elimina
tion : it will be sufficient for our purpose to take a simple
example, but it will be seen that in each case the process is
applicable to equations of any degree.
The principle illustrated in the following example is due to
Euler.
Example. Eliminate x between the equations
ax* + bx 2 + cx + d = 0, fx 2 + gx + h = 0.
Let x + k be the factor corresponding to the root common to both equa
tions, and suppose that
ax 3 + bx 2 + ex + d = (x + k) (ax 2 + lx + m),
and fx 2 + gx + h = (x + k) (fx + n) ,
k, I, m, n being unknown quantities.
From these equations, we have identically
(ax s + bx 2 + cx + d)(fx + n) = (ax 2 + Ix + m) (fx 2 + gx + h).
Equating coefficients of like powers of x, we obtain
fl an + agbf=0,
gl +fm bn + ah cf= 0,
Jd + gm en  df= 0,
hmdn =0.
From these linear equations by eliminating the unknown quantities I, in,
n, we obtain the determinant
/ a agbf
g f b ahcf
h g c df
h d
= 0.
446
HIGHER ALGEBRA.
530. The eliminant of the equations f(x) = 0, <f> (x) = can
be very easily expressed as a determinant by Sylvester's Dialytic
Method of Elimination. We shall take the same example as
before.
Example. Eliminate x between the equations
ax s + bx 2 + cx + d = 0, fx 2 +gx + h = 0.
Multiply the first equation by x, and the second equation by x and x 2 in
succession ; we thus have 5 equations between which we can eliminate the 4
quantities x 4 , x z , x 2 , x regarded as distinct variables. The equations are
ax* + bx 2 +cx + d=0,
ax i + bx 3 +cx 2 + dx =0,
fx 2 + gx + h = 0,
fx s + gx 2 + hx = 0,
fx 4 + gx 3 + ltx 2 =0.
Hence the eliminant is
a
/
a
b
/
9
b
c
f
9
h
c
d
9
h
d
h
= 0.
531. The principle of the following method is due to Bezout;
it has the advantage of expressing the result as a determinant of
lower order than either of the determinants obtained by the pre
ceding methods. We shall choose the same example as before,
and give Cauchy's mode of conducting the elimination.
Example. Eliminate x between the equations
ax 3 + bx 2 + cx + d=0, fx 2 + gx + h = 0.
From these equations, we have
a _ bx 2 + ex + d
f gx 2 +hx '
ax + b cx + d
fx +g~ Jix '
(ag  bf) x 1 + {ah ~cf)x df= 0,
(ah  cf) x 2 +(bh  eg  df) x  dg = 0.
Combining these two equations with
fx 2 +gx + h = 0,
whence
and
ELIMINATION. 447
and regarding x' z and x as distinct variables, we obtain for the eliminant
f g h =o.
ag  bf ah cf  df
a h  cf bh  eg df  dg
532. If we have two equations of the form <£, (x, y) — 0,
<£.,(#, 2/)=0, then y may be eliminated by any of the methods
already explained; in this case the eliminant will be a function of x.
If we have three equations of the form
0, (*» y> z ) = °> 2 (^ y> z ) = °> 03 ( a; > y» *) = °>
l>y eliminating z between the first and second equations, and then
between the first and third, we obtain two equations of the form
•A, (»> V) = °> ^ ( x > y) = °
If we eliminate y from these equations we have a result of
the form/* (a:) = 0.
By reasoning in this manner it follows that we can eliminate
n variables between n + 1 equations.
533. The general methods of elimination already explained
may occasionally be employed with advantage, but the eliminants
so obtained are rarely in a simple form, and it will often happen
that the equations themselves suggest some special mode of
elimination. This will be illustrated in the following examples.
Example 1. Eliminate Z, m between the equations
lx + my = a, vixly = b, Z 2 + m 2 =l.
By squaring the first two equations and adding,
7.c 2 + mx 2 + »»V + *V = a2 + & 2 >
that is, (Z 2 + /» 2 ) (.t 2 + y*) = a 2 + Z, 2 ;
hence the eliminant is .t 2 + ?/ 2 = a 2 + ZA
If Z = cos0, m= sin $, the third equation is satisfied identically; that is,
the eliminant of
x cos 6 + y sin 6 = a , x sin 6  y cos = Z>
is x 2 + y* = a° + b*.
448 HIGHER ALGEBRA.
Example 2. Eliminate x, y, z between the equations
y*+z*=zayz, z 2 + x 2 =bzx, x 2 + y* = cxy.
v z z x . x , y
We have *■ + =a,  + ~ = h +" = c 5
z y x z y x
by multiplying together these three equations we obtain,
w 2 z 2 z 2 z 2 z 2 y 2 .
z 2 t/ 2 a; 2 s a 2/ 2 X'
hence 2 + (a 2  2) + (6 2  2) + (c 2 2) = abc ;
.. a 2 + & 2 +c 2 4 = a&c.
Example 3. Eliminate #, ?/ between the equations
x 2 y 2 =pxqy, ±xy = qx+py, x 2 + y 2 =l.
Multiplying the first equation by x, and the second by y, we obtain
x s + Sxy~=p {x 2 + y 2 )\
hence, by the third equation,
p = x 3 + Sxy 2 .
Similarly q = Bx 2 y + y s .
Thus p + q={x+y) 3 > pq={xy) 3 \
.: (p + q)* + (p q)* = {x + y)* + {x  yf
= 2(x 2 + y 2 );
Example 4. Eliminate x, y, z between the equations
v z z x T x y
? = a, = b, ^ = c.
z y x z y x
x(y 2 z 2 )+y(z 2 x 2 )+z(x 2 y 2 )
We have a + o + c =
xyz
_{ yz){zx) (xy)
xyz
If we change the sign of x, the signs of b and c are changed, while the
sign of a remains unaltered ;
(yz){z + x)(x + y)
hence abc—
Similarly, bca =
and cab =
xyz
(y + z){zx)(x + y)
xyz
(y + z)(z + x){x^y)
xyz
ELIMINATION. 441)
.. {a^b+c)(b + ca){c + ah){a + bc) =  {ul ~ Z " )2 ^~f J^ztl
\z y) \x z) \y x)
= a?b 2 c 2 .
.• . 26V + 2c 2 a 2 + 2a 2 6 2  a 4  i 4  c 4 + a 2 6 2 c 2 = 0.
EXAMPLES. XXXIV. c.
1. Eliminate m from the equations
m 2 x — ?ny + a=0 } my + x=Q.
2. Eliminate m, n from the equations
m\v — my + a = 0, n 2 x — ny + a = 0, mn + 1 = 0.
3. Eliminate m, n between the equations
mx — ny — a (m 2 — n 2 ), nx + my = 2amu, m 2 + n 2 = 1 .
4. Eliminate p, q, r from the equations
p + q + r — Of a(qr+rp+pq) = 2ax,
apqr=y, qr= — 1.
5. Eliminate x from the equations
ax 2  2a 2 x + 1=0, a 2 + x 2  3ax = 0.
6. Eliminate m from the equations
y + mx=a (1 + ??i), wy  x— a (1  m).
7. Eliminate a:, y, z from the equations
yz = a 2 , zx=b 2 , xy = c 2 , x 2 +y 2 + z 2 = d 2 .
8. Eliminate p, q from the equations
x(p + q)=y, pq = k(l+pq), xpq = a.
9. Eliminate x, y from the equations
x — y = a, x 2 — y 2 = b 2 , x 3 —y 3 = c 3 .
10. Eliminate x, y from the equations
x+y = a, x 2 +y 2 = b 2 , #*+#*=c*.
11. Eliminate x, y, z, u from the equations
x = by + cz + rfw } y=cz + cfo + a#,
2 = cfti + a# + fry, w = cu; + by + cs.
12. Eliminate x, y, z from the equations
x+y + z = 0, x 2 +y 2 + z 2 = a 2 ,
aP+ff+sP^fc, ^ ,5 +y 5 + 2 5 = c 5 .
n. h. a. 29
450 HIGHER ALGEBRA.
13. Eliminate #, y, z from the equations
y s # ' z^x^y ' \y zj\z xj\x y)
14. Eliminate #, y, z from the equations
ff 2 (y+z) = y 2 (z + x) = g 2 fo+ff) a ^ a!l(
a 3 b 3 c 3 abc
15. Eliminate x, y from the equations
4 (.r 2 + y 2 ) = ax + Z>y, 2(x 2 y 2 ) = ax  by, xy = c 2 .
16. Eliminate #, y, z from the equations
(y f z) 2 = 4a 2 yz, (2; + #) 2 = 46^, (a; + y ) 2 = 4c 2 #y.
17. Eliminate x, y, z from the equations
(x+y  z) (xy + z) = ayz, (y + z  x) (y  z + x) = 6s#,
(z+x—y) (z — x + y) = cxy.
18. Eliminate a?, y from the equations
x 2 y=a, x(x{y) = b, 2x\y = c.
19. Shew that (a+6 + c) 3 4 (b + c) (c + a) (a + 6) + 5a&c=0
is the eliminant of
cm; 2 + fry 2 + cz 2 = ax + by + cz =yz + zx + xy = 0.
20. Eliminate #, y from the equations
ax 2 tby 2 =ax+by = — — =c.
21. Shew that & 3 c 3 + c% 3 + a 3 P= 5a 2 b 2 c 2
is the eliminant of
ax+yz = bc, by + zx=ca i cz + xy = ab, xyz=abc.
22. Eliminate x, y, z from
x 2 +y 2 +z 2 =x + y + z=l,
^(xp)=(yq)= C {zr).
23. Employ Bezout's method to eliminate x, y from
ax 3 + bx 2 y + cxy 2 + dy 3 = 0, a'x 3 + b'x 2 y + c'xy 2 + d'y 3 = 0.
CHAPTER XXXV.
THEORY OF EQUATIONS.
534. Ix Chap. ix. we have established certain relations be
tween the roots and the coefficients of quadratic equations. We
shall now investigate similar relations which hold in the case of
equations of the n th degree, and we shall then discuss some of the
more elementary properties in the general theory of equations.
535. Let 2? x" + 2) 1 x"~ 1 + 2 ) 2 X "~ 2 + + Pni x+ Pn ^ e a ra tional
integral function of x of n dimensions, and let us denote it by
f(x); then y (a?) = is the general type of a rational integral equa
tion of the n th degree. Dividing throughout by^> , we see that
without any loss of generality we may take
x n +2) i x"~ 1 +2 , o^"~ 2 + + 2 :> ni x ' ] 2 ) n = Q
as the type of a rational integral equation of any degree.
Unless otherwise stated the coefficients £>, , ^> , . . . p n will always
be supposed rational.
536. Any value of x which makes f(x) vanish is called a
root of the equation f(x) = 0.
In Art. 514 it was proved that when f(x) is divided by
xa, the remainder is f(a) ; hence if f (x) is divisible by x — a
without remainder, a is a root of the equation f{x) = 0.
537. We shall assume that every equation of the form f(x) =
has a root, real or imaginary. The proof of this proposition will
be found in treatises on the Theory of Equations ; it is beyond
the range of the present work.
29—2
452 HIGHER ALGEBHA.
538. Every equation of the n th degree has n roots, and no more.
Denote the given equation by/(a;) = 0, where
f(x) =p Q «? +PJXT 1 +2> 2 x n ~°' + + P*'
The equation f(x) = has a root, real or imaginary; let this be
denoted by a,; then/(a) is divisible by xa } , so that
f(x) = (xa i )<f> l (x),
where <t> (x) is a rational integral function of n1 dimensions
Igain, the equation *» = has a root real or "0^^**
this be denoted by a 2 ; then <£» is divisible by xa 2 , so that
fa^^ixaj^x),
where <f> a (x) is a rational integral function of n  2 dimensions.
Thus /(a>) = (« ~ «,) (*  O *b(*>
Proceeding in this way, we obtain, as in Art. 309,
/(«) = PoO* ~ °i) ( a " ^) (*  a )*
Hence the equation f(x)= has n roots, since f(x) vanishes
when sc has any of the values a x , a 2 , a 3 ,...a n .
Also the equation cannot have more than n roots; for if x has
any value different from any of the quantities a xi a 2 , a ...«„, all
the factors on the right are different from zero, and therefore
f(x) cannot vanish for that value of x.
In the above investigation some of the quantities a l ,a 2 ,a 3 ,...a n j
may be equal; in this case, however, we shall suppose that the
equation has still n roots, although these are not all different.
539. To investigate the relations between the roots and the J
coefficients in any equation.
Let us denote the equation by
x n +p 1 x n  l +2> 2 x n ~ 2 + +Pnl X + P» = > j
and the roots by a, b, c, k; then we have identically
x"+p 1 x n  l +2) x n ~'+ +P n i x+ P»
= (xa) (xb)(xc) (xk)' }
hence, with the notation of Art. 163, we have
x n +p l x H ~ l +p a x n * + +P n .^+P m
 wT  S x x n ~ l + S s x*~*  + ( iy%^ + (" !)"£,■
THE011Y OF EQUATIONS. 453
Equating coefficients of like powers of x in this identity,
 %> x — S l ~ sum of the roots ;
p a = S„ = sum of the products of the roots taken two at a
time;
—p B — S a sum of the products of the roots taken three at a
time ;
(— \)*p **S U = product of the roots.
If the coefficient of x" is p oi then on dividing each term by
]? ui the equation becomes
aj" + ;  1 x n  l + ] ^x" 2 + + P»=1 X + P» =
Po Po Pu Po
and, with the notation of Art. 521, we have
& = &, 2aft=&, %abe = %>. , abc...k = ( 1)"^ .
Po V, P Po
Example 1. Solve the equations
x + ay + a 2 z = a 3 1 x + by + b 2 z = b*, x + cy + c 2 z = c*.
From these equations we see that a, &, e are the values of t which
satisfy the cubic equation
t 3 zt 2 ytx = 0;
hence z = a + b + c, y= (bc + ca + ab), x = abc.
Example 2. If a, b, c are the roots of the equation x 3 +p 1 x 3 +PzC+p a =0,
form the equation whose roots are a 2 , 6 2 , c 2 .
The required equation is (y  a 2 ) (y  b 2 ) (y  c 2 )=0,
or (x 2  a 2 ) (a; 2  6 2 ) (x 2  c 2 ) = 0, if y = x 2 ;
that is, (x  a) (x  b) (x  c) (x + a) (x + b) (x + c) = 0.
But (x  a) (x  b) (x  c) = x 3 +p 1 x 2 +p 2 x +p t ;
hence {x + a)(x + b) (x + c) = x 3  p x x 2 +p.&  p v
Thus the required equation is
(x 3 +p 1 x' z +p 2 x +p 3 ) (x 3 p x x 2 +p*x p 3 ) = 0,
or (x 3 +p&) 2  (p x x 2 +p 3 ) 2 = 0,
or x 6 + (2ft  j^ 2 ) x 4 + (p. 2  2p 1 p. i ) x 2  p. 2 = ;
and if we replace x 2 by y, we obtain
f + (2p,p 2 ) y 2 + (p.?  2p dh ) y p : 2 =0.
454 HIGHER ALGEBRA.
540. The student might suppose that the relations established
in the preceding article would enable him to solve any proposed
equation; for the number of the relations is equal to the number
of the roots. A little reflection will shew that is this not the
case • for suppose we eliminate any n  1 of the quantities
a, b,c,...k and so obtain an equation to determine the remaining
one; then since these quantities are involved symmetrically in
each of the equations, it is clear that we shall always obtain an
equation having the same coefficients; this equation is therefore
the original equation with some one of the roots a, b, c,...k sub
stituted for x.
Let us take for example the equation
x 3 + p x x 2 + p 2 x + p 3 = ;
and let a, b, c be the roots; then
a + b + c = —] :> x >
ab + ac + bc= p s)
abc= — p 3 .
Multiply these equations by a 2 ,  a, I respectively and add ; thus
rf = l\rfl\u>p 3i
that is, a 3 + p x a? + p 2 a + p 3 = 0,
which is the original equation with a in the place of x.
The above process of elimination is quite general, and is
applicable to equations of any degree.
541. If two or more of the roots of an equation are con
nected by an assigned relation, the properties proved in Art. 539
will sometimes enable us to obtain the complete solution.
Example 1. Solve the equation 4.r 3  24a; 2 + 23x + 18 = 0, having given
that the roots are in arithmetical progression.
Denote the roots by a  b, a, a + b ; then the sum of the roots is 3a ; the
sum of the products of the roots two at a time is 3a 2  6 2 ; and the product
of the roots is a (a 2  6 2 ) ; hence we have the equations
3a = 6, 3a 2 Z> 2 = ^, a( a 2 6 2 )=;
5
from the first equation we find a = 2, and from the second 6=±, and
a
since these values satisfy the third, the three equations are consistent.
1 9
Thus the roots are   , 2,  .
2 £
THEORY OF EQUATIONS. 455
Example 2. Solve the equation 24a; 3  14.r 2  (ftx + 45 = 0, one root being
double another.
Denote the roots by a, 2a, b\ then we have
Sa + b = ^, 2a 2 + 3a&=^, 2a6=^.
From the first two equations, we obtain
8a 2 2a3 = 0;
3 1 , , 5 25
.. a = orand&=or.
1 25
It will be found on trial that the values a= , 6 = ^ do not satisfy
15
the third equation 2a 2 6 = ~ —  ; hence we are restricted to the values
o
3 5
a =  v b=.
Thus the roots are 7 , ~ >  o •
542. Although we may not be able to find the roots of an
equation, we can make use of the relations proved in Art. 539
to determine the values of symmetrical functions of the roots.
Example 1. Find the sum of the squares and of the cubes of the roots
of the equation x 3 px 2 + qxr — 0.
Denote the roots by a, b, c ; then
a + b + c=p, bc + ca + ab = q.
Now a 2 + b 2 + c 2 = (a + b + c) 2  2 (bc + ca + ab)
—p 2  2q.
Again, substitute «, b, c for x in the given equation and add; thus
a 3 + b 3 + <?p{a 2 + b 2 + c 2 ) + q{a + b + c)Sr = 0;
.. a 3 + b 3 + c 3 =p(p 2 2q) pq + Sr
=p 3  Spq 4 dr.
Example 2. If a, 6, c, d are the roots of
x A +px* + qx 2 + rx + s = 0,
find the value of Ha 2 b.
We have a + b + c + d =  p (1),
ab + ac + ad + bc + bd + cd = q (2),
abc + abd + acd + bcd — r (3).
456 HIGHER ALGEBRA.
From these equations we have
pq = Sa 2 6 + 3 (dbc + aid + acd + bed)
= Sa 2 63r;
.. 2ab = 3rpq.
EXAMPLES. XXXV. a.
Form the equation whose roots are
1. , , ±V& 2. 0, 0, 2, 2, 3, 3.
3. 2, 2, 2, 2, 0, 5. 4. a + b, ab, a + b, ab.
Solve the equations :
5. a? 16x3+ 86x 2  1 76# + 105 = 0, two roots being 1 and 7.
6. 4r 3 + 16.r 2  9x  36 = 0, the sum of two of the roots being zero.
7. 4^ + 2(Xr 2  23.r + 6 = 0, two of the roots being equal.
8. Sx 3 — 26x 2 + 52.27 — 24 = 0, the roots being in geometrical pro
gression.
9. 2a? — x 2 — 22# 24 = 0, two of the roots being in the ratio of
3: 4.
10. 24x* + 46.2? 2 + 9# — 9 = 0, one root being double another of the
roots.
11. &r 4  2^ 27^ ,2 + 6# + 9 = 0, two of the roots being equal but
opposite in sign.
12. 54^ 39^ 2 26^7+ 16 = 0, the roots being in geometrical pro
gression.
13. 32^348^ 2 +22^3 = 0, the roots being in arithmetical pro
gression.
14. 6#*  29^ + 40a 3  1x 12 = 0, the product of two of the roots
being 2.
15. #*  2x* 21.r 2 + 22.27 + 40 = 0, the roots being in arithmetical
progression.
16. 27.27 4 195.27 3 + 494.r 2  520.27 + 192 = 0, the roots being in geo
metrical progression.
17. 18a 3 + 8U 2 + 121.37 + 60 = 0, one root being half the sum of the
other two,
THEORY OF EQUATIONS. 457
18. If a, b, c are the roots of the equation X s paP+qx  r = 0, find
the value of
(1) ^+1 + 1 (2) i + i,+
a 2 ' 6 2 c 2 ' w W r c 2 rt 2 t a aja ■
19. If a, £>, c are the roots of .r 3 + g'.r + r=0, find the value of
(1) {hcy + (ca)* + (ab)\ (2) (& + c )i + ( c + «)i + ( a + ?,)i.
20. Find the sum of the squares and of the cubes of the roots of
#* + qx 2 + rx + s = 0.
21. Find the sum of the fourth powers of the roots of
x 3 +qx+r=Q.
543. 7?i an equation with real coefficients imaginary roots
occur in pairs.
Suppose that f(x) = is an equation with real coefficients,
and suppose that it has an imaginary root a + ib ; we shall shew
that a — ib is also a root.
The factor of f(x) corresponding to these two roots is
(x — a — ib) (x — a + ib), or (x — a) 2 + b 2 .
Let f(x) be divided by (x — a) 2 + b 2 ; denote the quotient by
Q, and the remainder, if any, by Rx + E ', then
f(x) = Q{(x a) 2 + b 2 } + Rx + E.
In this identity put x = a + ib, then f(x) — by hypothesis ; also
(x  a) 2 + b 2 = ; hence R (a + ib) + E = 0.
Equating to zero the real and imaginary parts,
Ra + E=Q, Rb = 0;
and b by hypothesis is not zero,
.. R = and # = 0.
Hence f (x) is exactly divisible by (x — a) 2 + b 2 , that is, by
(x — a  ib) (x — a + ib) \
hence x = aib is also a root.
544. In the preceding article we have seen that if the equa
tion f{x) = has a pair of imaginary roots a ± ib, then (x — a) 2 + b 2
is a factor of the expression f(x).
458 HIGHER ALGEBRA.
Suppose that a±ib, c±id, e±ig,... are the imaginary roots
of the equation f(x)  0, and that cf> (x) is the product of the
quadratic factors corresponding to these imaginary roots; then
<j>(x) = {(xa) 2 + b 2 }{(xc) 2 + d 2 }{(xe) 2 + /}....
Now each of these factors is positive for every real value of x;
hence <f> (x) is always positive for real values of x.
545. As in Art. 543 we may shew that in an equation with
rational coefficients, surd roots enter in pairs; that is, if a + Jb is
a root then a Jb is also a root.
Example 1. Solve the equation 6z 4  13x 3  35z 2  x + 3 = 0, having given
that one root is 2  ^3.
Since 2^/3 is a root, we know that 2+^/3 is also a root, and corre
sponding to this pair of roots we have the quadratic factor x 2  4# + 1.
Also 6z 4  13z 3  35a: 2  x + 3 = {x 2  4x + 1) (6a; 2 + 11* + 3) ;
hence the other roots are obtained from
6a; 2 + 11a; + 3 = 0, or (3a; + 1) (2z + 3)=0;
1 3
thus the roots are   ,   , 2 + ^/3, 2^3.
Example 2. Form the equation of the fourth degree with rational
coefficients, one of whose roots is ,J2 + sj  3.
Here we must have /J2 + /J3, J2J 3 as one pair of roots, and
 >/2 + ^/  3,  J2  J  3 as another pair.
Corresponding to the first pair we have the quadratic factor x 2  2 /v /2x + 5,
and corresponding to the second pair we have the quadratic factor
x 2 + 2 f J2x + 5.
Thus the required equation is
(x 2 + 2,J2x + 5) (x 2 2 f J2x + 5) = 0,
or (x 2 + 5) 2 8a 2 = 0,
or a^ + 2x 2 + 25 = 0.
Example 3. Shew that the equation
A 2 B 2 C 2 H 2 ,
+ 7 + + ... + — ,=&,
xa xb xc ' xh
has no imaginary roots.
If possible let p + iq be a root ; then p  iq is also a root. Substitute
these values for x and subtract the first result from the second ; thus
{(pa
A 2 B 2 C 2 H*
i) 2 + q 2 (pb) 2 + q 2 ^(pc) 2 + q 2 ' "" ' (ph) 2 + q
which is impossible unless q =
THEORY OF EQUATIONS. 459
546. To determine the nature of some of the roots of an
equation it is not always necessary to solve it ; for instance, the
truth of the following statements will be readily admitted.
(i) If the coefficients are all positive, the equation has no
positive root ; thus the equation # 5 + x 3 + 2x + 1 = cannot have a
positive root.
(ii) If the coefficients of the even powers of x are all of one
sign, and the coefficients of the odd powers are all of the contrary
sign, the equation has no negative root; thus the equation
x 7 + x 5  2x 4 + x 3  3x 3 + 7x 5 =0
cannot have a negative root.
(iii) If the equation contains only even powers of x and the
coefficients are all of the same sign, the equation has no real
root ; thus the equation 2x 8 + 3x* + x 2 + 7 = cannot have a real
root.
(iv) If the equation contains only odd powers of x, and the
coefficients are all of the same sign, the equation has no real root
except x = ; thus the equation x 9 + 2x 5 + 3x 3 + x = has no real
root except x = 0.
All the foregoing results are included in the theorem of the
next article, which is known as Descartes' Rule of Signs.
547. An equation f(x) = cannot have more positive roots
than there are changes of sign in f (x), and cannot have more
negative roots than there are changes of sign in f (x).
Suppose that the signs of the terms in a polynomial are
+ H 1 1 1 — ; we shall shew that if this polynomial
is multiplied by a binomial whose signs are A — , there will be at
least one more change of sign in the product than in the original
polynomial.
Writing down only the signs of the terms in the multiplica
tion, we have
+
+
—
—
+
—
—
—
+
—
+
—
+
—
+
+
—
—
+
—
—
—
+
—
+
—
—
—
+
+
—
+
+
+
—
+
—
+
+ ± — =F \ =F=FH + h
460 HIGHER ALGEBRA.
Hence we see that in the product
(i) an ambiguity replaces each continuation of sign in the
original polynomial;
(ii) the signs before and after an ambiguity or set of am
biguities are unlike;
(iii) a change of sign is introduced at the end.
Let us take the most unfavourable case and suppose that all
the ambiguities are replaced by continuations; from (ii) we see
that the number of changes of sign will be the same whether we
take the upper or the lower signs; let us take the upper; thus
the number of changes of sign cannot be less than in
+ + + +  + +,
and this series of signs is the same as in the original polynomial
with an additional change of sign at the end.
If then we suppose the factors corresponding to the negative
and imaginary roots to be already multiplied together, each factor
x — a corresponding to a positive root introduces at least one
change of sign; therefore no equation can have more positive
roots than it has changes of sign.
Again, the roots of the equation f(—x) = are equal to those
of /(^) = but opposite to them in sign; therefore the negative
roots of f(x)0 are the positive roots of /*(#) = 0; but the
number of these positive roots cannot exceed the number of
changes of sign in f{— x) ; that is, the number of negative roots
of f(x) = cannot exceed the number of changes of sign in
/( ■>
Example. Consider the equation a; 9 + 5x 8  x* + Ix + 2 = 0.
Here there are two changes of sign, therefore there are at most two
positive roots.
Again /( x)= — x 9 + 5x 8 + x 3 7x + 2, and here there are three changes
of sign, therefore the given equation has at most three negative roots, and
therefore it must have at least four imaginary roots.
EXAMPLES. XXXV. b.
Solve the equations :
1. 3x A — lO.'o" 3 + 4x 2  a — 6 = 0, one root being ^ .
2. 6s 4  l&e 3  35# 2  x + 3 = 0, one root being 2  N /3.
3. x A + 4r 3 + 5x 2 + %x 2 = 0, one root being  1 + ,J ^1 .
THEORY OF EQUATIONS. 4G L
4. X* + 4./," + G.f 2 + 4x + 5 = 0, one root being «/l.
5. Solve the equation x 5 x A + 8x 2 9x — 15 = 0, one root being
^3 and another 1 —2J 1.
Form the equation of lowest dimensions with rational coefficients,
one of whose roots is
6. s iZ+J^2. 7. J^l+J5.
8. J2J^2. 9. N /5 + 2 x /6.
10. Form the equation whose roots are ± 4 a/3, 5 =l 2 */  1.
11. Form the equation whose roots are 1± >/ 2, 2± J 3.
12. Fomi the equation of the eighth degree with rational co
efficients one of whose roots is »J2 + J3 + x / — 1.
13. Find the nature of the roots of the equation
3x i + l2x 2 + bx4 = 0.
14. Shew that the equation 2.v 7  x A + 4.V 3  5 = has at least four
imaginary roots.
15. What may be inferred respecting the roots of the equation
a 10 4a 6 + x A 2.y3=0?
16. Find the least possible number of imaginary roots of the
equation x° — o?> + x A + x 2 + 1 = 0.
17. Find the condition that x 3 px 2 + qx  r = may have
(1) two roots equal but of opposite sign ;
(2) the roots in geometrical progression.
18. If the roots of the equation x l +p.v 3 \qx 2 + rx + s = are in
arithmetical progression, shew that p 3 — 4pq + 8r=0; and if they arc
in geometrical progression, shew that p 2 s = r 2 .
19. If the roots of the equation x n  1 = are 1, a, /3, y, . . ., shew that
(la)(l/3)(l 7 ) =n.
If a, b, c are the roots of the equation x 3 px 2 + qx r = 0, find the
value of
20. Za 2 b 2 . 21. (b + c)(c + a)(a + b).
22. S (* + !)• 23. $a 2 b.
If a, b, c, d are the roots of x A +px 3 + qx + rx + s = 0, find the value of
24. %a*b& 25. $a\
462 HIGHER ALGEBRA.
548. To find the value of f (x + h), when f (x) is a rationed
integral Junction ofx.
Let f(x) =p Q x n +p l x n ~ l +P 2 X "~ 2 + +Pni x + Pn i then
fix + h) = p (x + h) n +2\ (x + h) H ~ } +2> 2 (x + h)"* +
+2\A x + h ) + Pn I
Expanding each term and arranging the result in ascending
powers of h, we have
Po x n \2\x n ~ l +2^x n ~ 2 + ... +p n . l x+p n
+ h {np^ 1 + (n ljjyrf 1 + (n2) p 2 x n ~ 3 + ...+ p^}
+ ^{n(nl)p x" 2 + (nl)(n2) Pl x'> 3 +... + 2p n _ 2 }
+ .
+ ^{n(nl)(n2)...2.1 2 > }
\n l
This result is usually written in the form
/(» + *)=/(*) + hf{x) + *J/» + *j/» + ... + *i/»,
and the functions f (x), f"(x), f"(x),... are called the first,
second, third,... derived functions oifix).
The student who knows the elements of the Differential Cal
culus will see that the above expansion of f(x + h) is only a
particular case of Taylor's Theorem; the functions f (x), f" (x),
f'"{x) may therefore be written down at once by the ordinary
rules for differentiation: thus to obtain f'(x) £romf(x) we multiply
each term in f(x) by the index of x in that term and then
diminish the index by unity.
Similarly by successive differentiations we obtain fix),
J \X), ....
By writing — h in the place of h, we have
f(xh)=f(x)h/'(x) + h 'f" (x )Jff'"( x)+ ... + {  I) %f{x).
The function f(x + h) is evidently symmetrical with respect
to x and h; hence
i
,n
fix + h) =/(h) + xf (h) + *r (h) + ... 4 f/* (h).
£ \n
THEORY OF EQUATIONS. 463
Here the expressions f'(h) i f"{Ji),f , "{1b) i ... denote the results
obtained by writing h in the place of x in the successive derived
functions f'(x), /"(#), f"(x),....
Example. If / {x) = 2x*  x s  2x z + 5x  1 , find the value of / (x + 3).
Here / (x) = 2x*  x*  2x* + 5x  1, so that / (3) = 131 ;
/' (x) = 8z 3  Sx°  4x + 5, ana /' (3) = 182 ;
^ ) =12a»3aj2, and ^ = 97;
QSte1, and / ^ 3) = 23;
ii
Thus / (x + 3) = 2s 4 + 23r* + 97x 2 + 182* + 131 .
The calculation may, however, be effected more systematically by Horner's
process, as explained in the next article.
549. Let f{x) =p x n +p 1 x n ~ 1 + p 2 x"~ 2 + ... +p n _ l x + p n ;
put x — y + h, and suppose that f [x) then becomes
Now y = x — h; hence we have the identity
p x" +p 1 x n ~ l + p 2 x n ~ 2 +... +p n _ 1 x + p n
= q o (x h) n +q x (x h) 1 + . . . + q n _ x (x  h) + q n ;
therefore q n is the remainder found by dividing f(x) by xh;
also the quotient arising from the division is
q (xh)* l +q l {xhy'+...+q H _ i .
Similarly q n _ l is the remainder found by dividing the last
expression by x h, and the quotient arising from the division is
9o( x  h T' 2 + QA x  h T~ 3 +  + Qn 2 ' }
and so on. Thus q n , q n _ 1 , q n _ a , ••• may be found by the rule ex
plained in Art. 515. The last quotient is q , and is obviously
equal to j) 
464
HIGHER ALGEBRA.
Example. Find the result of changing x into x + 3 in the expression
2a;4_^_2x 2 +5xl.
Or more briefly thus :
Here we divide successively by x  3.
212 5 1
6 15 39 132
5 13 44
6 33 138
131 4
11 46
6 51
182 =
=&
17
6
23~=
97 =
= <Z 2
2
1
2
5
1
2
5
13
44
131
2
11
46 
182
2
17
97
2
23
2
Hence the result is 2x i + 23a 3 + 97sc 2 + 182x + 131. Compare Art. 548.
It may be remarked that Horner's process is chiefly useful in numerical
work.
550. If the variable x changes continuously from a to b the
function f (x) will change continuously from f (a) to f (b).
Let c and c + h be any two values of x lying between a and b.
We have
/(«+*)/(•)= vw+§/"W+ +/"( e );
and by taking A small enough the difference between/(c + /i) and
f(c) can be made as small as we please; hence to a small change
in the variable x there corresponds a small change in the function
f (x), and therefore as x changes gradually from a to b, the func
tion/*^) changes gradually from /(a) to f(b).
551. It is important to notice that we have not proved that
f(x) always increases from /(a) to fib), or decreases from f(a)
to fib), but that it passes from one value to the other without
any sudden change; sometimes it may be increasing and at other
•times it may be decreasing.
The student who has a knowledge of the elements of Curve
tracing will in any particular example find it easy to follow the
gradual changes of value oif(x) by drawing the curve y =f(x).
552. If f (a) and f (b) are of contrary signs then one root of
the equation f (x) = must lie between a and b.
As x changes gradually from a to b, the function f(x) changes
gradually from f(a) to f(b), and therefore must pass through all
THEORY OF EQUATIONS. 4G5
intermediate values; but since f(a) and f(b) have contrary signs
the value zero must lie between them; that is, f(x) = for some
value of x between a and b.
It does not follow that f(x) = has only one root between a
and b; neither does it follow that if f(a) and /(b) have the same
signf(x) = has no root between a and b.
553. Every equation of an odd degree has at least one real
root whose sign is opposite to that of its last term.
In the function f(x) substitute for x the values + co , 0, co
successively, then
/( + oo) = + co, f(0)=p nJ /(oo) = oo.
If p n is positive, then f(x) = has a root lying between and
— oo , and if p n is negative f(x) = has a root lying between
and + co .
554. Every equation which is of an even degree and has its
last term negative has at least two real roots, one positive and one
negative.
For in this case
/(+co) = +co, f(0)=p n , f(co) = + co;
but p n is negative; hence f(x) = has a root lying between
and + co , and a root lying between and  co .
555. If the expressions f (a) and f (b) have contrary signs,
an odd number of roots of f (x) = will lie between a and b; and
«/*f(a) andi(h) have the same sign, either no root or an even number
of roots will lie between a and b.
Suppose that a is greater than b, and that a, /3, y, . . . k
represent all the roots of f(x) = which lie between a and b.
Let <f> (x) be the quotient when f(x) is divided by the product
(x — a) (x — /3) (x — y) ... (x — k ) ; then
f(x) — (x — a)(x—/3)(x y) ... (x — k ) <£ (x).
Hence f (a) = (a — a) (a — /3) (a — y) ... (a  k) </> («)•
/(8)=(5a)(6 J 3)(6 r )...(ftK)*(5).
Now <{>(a) and <f>(b) must be of the same sign, for otherwise a
root of the equation <j£>(.x') = 0, and therefore of f (x) = 0, would
lie between a and b [Art. 552], which is contrary to the hypo
H. H. A. 30
466 HIGHER ALGEBRA.
thesis. Hence if /(a) and /(b) have contrary signs, the ex
pressions
(a  a) (a fi)(ay) ... (a  k),
(ba)(bP)(by)...(b K )
must have contrary signs. Also the factors in the first expression
are all positive, and the factors in the second are all negative;
hence the number of factors must be odd, that is the number of
roots a, /?, y, ... k must be odd.
Similarly if /(a) and /(b) have the same sign the number of
factors must be even. In this case the given condition is satisfied
if a, /?, y, . . . k are all greater than a, or less than b ; thus it does
not necessarily follow that y* (as) = has a root between a and b.
556. If a, b, c, ...k are the roots of the equation /(x) = 0, then
/ (x) = j? (x — a)(xb)(x — c) ... (x — k).
Here the quantities a, b, c, ... k are not necessarily unequal.
If r of them are equal to a, s to b, t to c, . . . , then
/(x) = p (x — a) r (x  b) s (x — c)'
In this case it is convenient still to speak of the equation
/{x) — as having n roots, each of the equal roots being considered
a distinct root.
557. 1/ the equation f(x) = has r roots equal to a, then the j
equation f (x) = will have r — 1 roots equal to a.
Let <£(#) be the quotient when /(x) is divided by (x — a) r ;
then /(x) = (x — a) r $>{x).
Write x + h in the place of x; thus
/(x + h) = (xa + h) r 4>(x+ h) ;
.../(») + ¥'(x) + %/"(x)+.. .
= Uxa) r + r(xa) r  x h + ...\U(x) + hcf>'(x)+~ <}>"(x)+ ...] .
In this identity, by equating the coefficients of A, we have
/'(x)=r(x  ay'^x) + (x  a) r $ (x).
Thus/'(aj) contains the factor xa repeated r\ times; that
is, the equation /' (x) = has ?•  1 roots equal to a.
THEORY OF EQUATIONS. 4G7
Similarly we may shew that if the equation f (x) = has s
roots equal to b, the equation f (x) = has s — 1 roots equal to b ;
and so on.
558. From the foregoing proof we see that if f{x) contains
a factor (x — a)\ then f (x) contains a factor (x — a)* 1 ; and thus
f{x) and f'(x) have a common factor (x — a) r ~\ Therefore if
f(x) and fix) have no common factor, no factor in f(x) will be
repeated ; hence the equation f (x) = has or has not equal roots,
according as f (x) and f (x) have or have not a common factor
involving x.
559. From the preceding article it follows that in order to
obtain the equal roots of the equation f(x) = 0, we must first find
the highest common factor of f(x) and /*'(#).
Example 1. Solve the equation ar 1  llx 3 + 44ar 76x+ 48 = 0, which has
equal roots.
Here / {x) = x 4  lis 3 + 44a?  76a + 48,
/' {x) = 4x 3  33a 2 + 88x  76 ;
and by the ordinary rule we find that the highest common faetor of f(x) and
/' (a;) is x  2 ; hence (x  2) 2 is a factor of f(x) ; and
/(a;) = (a:2) 2 (a: 2 7x+12)
= (a;2) 2 (a;3)(a:4);
thus the roots are 2, 2, 3, 4.
Example 2. Find the condition that the equation aar 3 + 36a; 2 + 3ca; + d —
may have two roots equal.
In this case the equations f(x) = 0, and /' (x) = 0, that is
ax s + Sbx 2 + 3cx + d = (1),
ax 2 f 2bx + c = Q (2)
must have a common root, and the condition required will be obtained by
eliminating x between these two equations.
By combining (1) and (2), we have
bx 2 + 2cx + d = (3).
From (2) and (3), we obtain
a; 2 _ x _ 1
2 (bd  c 2 ) ~ beall ~ 2{ac  ft 2 ) '
thus the required condition is
(6c  adf =4 (ac  b) (bd  <■).
30—2
4fi8 HIGHER ALGEBRA.
560 We have seen that if the equation/^) = has r roots
equal to a, the equation /(a) = has r  1 roots equal to a. But
r(x) is the first derived function of f (x); hence the equation
f'\x) = must have r2 roots equal to a; similarly the equation
f>t x \ = o must have r  3 roots equal to a; and so on. IheseL
considerations will sometimes enable us to discover the equal
roots of f(x) = with less trouble than the method ot Art. 559.
561. If a, b, c, ...k are the roots of the equation f (x) = 0, to
prove that
f(x) = M + M + M + ... + lW k .
v ' xa xb xc xk
We have fix) = (x a) (x b) (xc) ... (xk);
writing x + h in the place of x,
f (x + h) = (x  a + h)(x  b 4 h)(x c + h) ... (xk + h) ... (1).
But f(x + h ) =/(«) .+ hf (x) + r^ /" (»)+... ',
hence f(x) is equal to the coefficient of h in the righthand
member of (1); therefore, as in Art. 163,
f(x)= (xb) (xc) ... (xk) + (x a)(x c) ... (xk)+ ...;
, • *,, n A x ) A x ) A x ) f( x )
that is, f'(x) = J ^^ +^{ + ^ J + ... +  / ^/ .
w x — a xb x — c x — k
562. The result of the preceding article enables us very easily '"
to find the sum of an assigned power of the roots of an equation.
Example. If S k denote the sum of the fc th powers of the roots of the
equation x 5 +px* + qx 2 + t = 0,
find the value of S 4 , S 6 and S_ 4 .
Let f(x)=x 5 +px li + qx* + t',
then /' ( x ) = 5x* + 4ps 3 + 2qx.
fix)
Now Z^=rf+(a+P)x 3 + (a 2 + ap)x 2 +(a* + a 2 p + q)x + a 4 + a?p + aq,
so a
and similar expressions hold for
/(*) fw />) /w
xb' xc' xd' xe'
THEUltY OF EQUATIONS. 409
Hence by addition,
5ar* + 4px* + 2qx =5x* + (S t + 5p)x? + (S 2 +pSJ x 2
+ {Si +pS 2 + 5q) x + {S 4 +p8 3 + qSJ.
By equating coefficients,
S 1 + 5p = ±p, whence 8 X .—  p ;
S 2 +pS l = 0, whence S 2 =p z ;
S 3 +pS 2 + oq = 2q, whence S 3 =  p % Sq;
S 4 +pS 3 + qS x = 0, whence S 4 =p i + 4j)q.
To find the value of S k for other values of k, we proceed as follows.
Multiplying the given equation by #*~ 5 ,
x* +px k ~ 1 + qx*~ 3 + to* 5 = 0.
Substituting for x in succession the values a, b, c, d, e and adding the
results, we obtain S k +pS k _ x + qS k _ 3 + tS k _ 5 = 0.
Put k = 5 ; thus S 5 +pS 4 + qS 2 + 5t = 0,
whence S 5 = p 5  op 2 q  bt.
Put k = 6 ; thus S 6 +pS 5 + qS 3 + tS x = 0,
whence S 6 =p 6 + 6p'*q + Sq 2 + bpt.
To find 5_ 4 , put k = 4, 3, 2, 1 in succession; then
S i +pS 3 + qS 1 + *#_! = (), whence S_ x = 0;
2,/
S 3 +pS 2 + 5q + tS_ 2 = 0, whence S_. 2 =  — ;
S. 2 +pS 1 + qS 1 + tS_ 3 = 0, whence S_ 3 = 0;
S 1 + 5p + qS_ 2 + tS_ 4 = 0, whence £_ 4 = %  Ap .
563. When the coefficients are numerical we may also pro
ceed as in the following example. 
Example. Find the sum of the fourth powers of the roots of
x*2x 2 + xl = 0.
Here f(x) = x*2x 2 + xl,
f'(x) = Sx 2 ±x + l.
Also * / .' = + +
f(x) x a x—b xc
/l a a 2 a3 \
= 2  +_+ + —+...
\x x X s x* J
O i>, Oo o«
= +— H + " +
X X~ X 3 X*
470 HIGHER ALGEBRA.
hence # 4 is equal to the coefficient of ^ in the quotient of f'{x) by f(x),
which is very conveniently obtained by the method of synthetic division as
follows :
1
2
1
1
34 + 1
63 + 3
42 + 2
4 2 + 2
105 + 5
3 + 2 + 2 + 5 + 10 +
3 2 2 5 10
Hence the quotient is  +  2 + 3 +  4 + g +
thus S. = 10.
EXAMPLES. XXXV. c.
1. If f{x) = x A + 10^ + 39# 2 + 76o; + 65, find the value of f(x  4).
2. If f(x)=x i  I2x 3 + 11x 2 9x + 1, find the value of/(# + 3).
3. If /(#) = 2# 4  13# 2 + 10a;  19, find the value of f(x + 1).
4. If f(x) =x* + 16^ + 12x 2 + 64a;  129, find the value of f(x  4).
5. If f(x) = ax 9 + bx 5 + ex + c£, find the value of /(# + /i) — f(co  It).
6. Shew that the equation 10a* 3  17# 2 +#+6=0 has a root
between and  1.
7. Shew that the equation x*  5x 3 + Sx 2 + 35#  70 = has a root
between 2 and 3 and one between  2 and  3.
8. Shew that the equation x*  l<2x 2 + I2x  3 =0 has a root
between  3 and  4 and another between 2 and 3.
9. Shew that x 5 + 5x*  20x 2  19a;  2 = has a root between 2 and
3, and a root between  4 and  5.
Solve the following equations which have equal roots :
10. a; 4 9a; 2 + 4a;+12 = 0. 11. ^6^ + 12^_ 1007 + 3 = 0.
12. a, 5  13#*+67# 3  17la;2 + 216^ 108=0.
13. x 5 x 3 + 4x 2 3x + 2 = 0. 14. 8^ + 4^318^+11^2=0.
15. x G 3x 5 + 6x 3 3x 2 3x + 2 = 0.
16. x 6  2x*  4x A + 12a,* 3  Sx 2  18a; + 18 = 0.
17. x i (a + b)x*a(ab)x 2 + ai(a + b)xa3b = Q.
TRANSFORMATION OF EQUATIONS. 471
Find the solutions of the following equations which have common
roots :
18. 2s*  2a 3 + x 2 + 3x 6 = 0, 4#*  2s 3 + 3x 9 = 0.
19. 4#* + 1 2x*  # 2 15.*; = 0, 6^ 4 + 1 3a; 3  4a; 2  1 bx = 0.
20. Find the condition that x 1l px 2 + r=0 may have equal roots.
21. Shew that x i + qx 2 + s = cannot have three equal roots.
22. Find the ratio of b to a in order that the equations
ax 2 + bx + a = and x s 2x 2 + 2xl =0
may have (1) one, (2) two roots in common.
23. Shew that the equation
x n + nx n ~ 1 + n (n  1) x n ~ 2 + ... + \n =
cannot have equal roots.
24. If the equation x 5  l0a 3 x 2 + b i x + c 5 = Q has three equal roots,
shew that ab*  9a 5 + c 5 = 0.
25. If the equation x A + ax 3 + bx 2 + cx + d=0 has three equal roots,
shew that each of them is equal to — ^ — ^r .
26. If x 5 hqx 3 + rx 2 + t = has two equal roots, prove that one of
them will be a root of the quadratic
15rx 2  6q 2 x + 25*  4qr = 0.
27. In the equation x 3  x  1 =0, find the value of S 6 .
28. In the equation x i  x 3 1x 2 + x + 6 = 0, find the values of £ 4
and S 6 .
Transformation of Equations.
564. The discussion of an equation is sometimes simplified
by transforming it into another equation whose roots bear some
assigned relation to those of the one proposed. Such transforma
tions are especially useful in the solution of cubic equations.
565. To transform an equation into another ivhose roots are
those of the proposed equation with contrary signs.
Let f(x) = be the proposed equation.
Put y for x; then the equation f(—y)  is satisfied by
every root of f(x)  with its sign changed ; thus the required
equation is f(—y) = 0.
472 HIGHER ALGEBRA.
If the proposed equation is
then it is evident that the required equation will be
p y n py~ l + p 2 f~ 2  + ( ir x A,y + ( W. = o,
which is obtained from the original equation by changing the
sign of every alternate term beginning with the second.
566. To transform an equation into another whose roots are
equal to those of the proposed equation multiplied by a given
quantity.
Let f{x) = be the proposed equation, and let q denote the
given quantity. Put y — qx, so that x —  , then the required
equation is f (  J = 0.
The chief use of this transformation is to clear an equation of
fractional coefficients.
Example. Remove fractional coefficients from the equation
*»•** .+10.
Put x =  and multiply each term by q 3 ; thus
3 13
By putting q = 4 all the terms become integral, and on dividing by 2,
we obtain
y s Sy 2 y + Q = 0.
567. To transform an equation into another whose roots are
the reciprocals of the roots of the proposed equation.
Let f(x) = be the proposed equation ; put y =  , so that
X —
y
 ; then the required equation isy(  ) = 0.
One of the chief uses of this transformation is to obtain the
values of expressions which involve symmetrical functions of
negative powers of the roots.
TRANSFORMATION OF EQUATIONS. 473
Example 1. If a, b, c are the roots of the equation
X s px 2 + qx — r = 0,
find the value of  + r ., + s .
a 2 b c
Write  for x, multiply by y'\ and change all the signs; then the re
y
suiting equation ry 9  qy 2 +py 1 = 0,
has for its roots
111
a' b' c ''
hence 2 = ^, S= =:
a r ab r
1 q 2  2pr
V
a 2 r
Example 2. If a, b, c are the roots of
« 3 + 2x 2 3xl = 0,
find the value of a 3 + b~ 3 + c~ 3 .
Writing  for x, the transformed equation is
y* + Sy 2 2yl = 0;
and the given expression is equal to the value of S s in this equation.
Here S 1= 3;
£ 2 =(3) 2 2(2) = 13;
and S 3 + 3S. 2 2S 1 3 = 0;
whence we obtain S,= 42.
1
568. If an equation is unaltered by changing x into — , it
is called a reciprocal equation.
If the given equation is
x n + Pl x n  l +p 2 x" 2 + + 1 > n _ 2 x i +Pn _ iX + ^=0,
the equation obtained by writing  for x, and clearing of fractions
is
V£? + l\p n ~ x + p n X~ 2 + • • • +l\n 2 +PF +1 = 0.
If these two equations are the same, we must have
Fl p ' ? ' 2 * ' '••' V »*~ p ' *»« ' P "p >
from the last result we have p =*fc 1, and thus we have two
classes of reciprocal equations.
474 HIGHER ALGEBRA.
(i) If p n =l t then
Px=Pn^ P*=P*» Pb=P*B> '>
that is, the coefficients of terms equidistant from the beginning
and end are equal.
(ii) If 2> n = ~ 1) then
p,=p s _ 1 , p 2 =p n  2 , Ps=p n  3 , ;
hence if the equation is of 2m dimensions p m = — p mi or £> TO =0.
In this case tlie coefficients of terms equidistant from the begin
ning and end are equal in magnitude and opposite in sign, and
if the equation is of an even degree the middle term is wanting.
569. Suppose that f (x) = is a reciprocal equation.
If f (x) = is of the first class and of an odd degree it has a
root —1; so that f (x) is divisible by aj + 1. If <f>(x) is the
quotient, then <f>(x) =0 is a reciprocal equation of the first class
and of an even degree.
If f(x) = is of the second class and of an odd degree, it
has a root + 1 ; in this case f(x) is divisible by as— 1, and as
before <j> (x) = is a reciprocal equation of the first class and of
an even degree.
If f(x) = is of the second class and of an even degree, it
has a root + 1 and a root  1 ; in this case f{x) is divisible by
x 2 — 1, and as before <f>(x) = Q is a reciprocal equation of the first
class and of an even degree.
Hence any reciprocal equation is of an even degree with
its last term positive, or can be reduced to this form; which may
therefore be considered as the standard form of reciprocal
equations.
570. A reciprocal equation of the standard form can be re
duced to an equation of half its dimensions.
Let the equation be
ax 2m + bx 2m  ] + cx 2m ~ 2 + ... + kx m + ... + ex 2 + bx + a = 0;
dividing by x m and rearranging the terms, we have
i) + 6(^' +5 L) + .(.r' +; l,)+... + *=a.
a x m +
x
TRANSFORMATION OF EQUATIONS. 475
Now
* +,+ ^^4)K)(* ,+ ^);
hence writing % for x +  , and giving to p in succession the values
1, 2, 3,... we obtain
x 2 +  2 = s*  2,
a 3 +  j = z (z 2 2) z = z 3  3z ;
x
x 4 +  4 = z (z 3  2>z)  (z 2  2) = z*  iz 2 + 2 ;
and so on; and generally x m + — is of m dimensions in z, and
therefore the equation in z is of m dimensions.
571. To find the equation whose roots are the squares of those
of a proposed equation.
Let f(x) = be the given equation ; putting y = x 2 , we have
x— Jy\ hence the required equation is f(Jy) — 0.
Example. Find the equation whose roots are the squares of those of the
equation vP+p^+ptfc+p^Q.
Putting x=Jy, and transposing, we have
(y+P2)Jy= (PiU+2h)>
whence {y 2 + 2p 2 y +pf) y =p x hj 2 + 2p x p z y + p 3 2 ,
or 2/ 3 + (3p a  Pl *) tf + (i> 2 2  2 Pnh ) yp. A 2 = 0.
Compare the solution given in Ex. 2, Art. 539.
572. To transform an equation into another whose roots
exceed those of the proposed equation by a given quantity.
Let f (x) = be the proposed equation, and let h be the given
quantity ; put y = x + h, so that x = y — h; then the required
equation is f(y — h) — 0.
Similarly f(y + h) = is an equation whose roots are less by
h than those oif(x) = 0.
476 HIGHER ALGEBRA.
Example. Find the equation whose roots exceed by 2 the roots of the
equation 4x* + 32a; 3 + 83a; 2 + 76a; + 21 = 0.
The required equation will be obtained by substituting x  2 for a; in the
proposed equation ; hence in Horner's process we employ x + 2 as divisor,
and the calculation is performed as follows :
4
32
83
76 21
4
24
35
6  9
4
16
3
0
4
8 1
13
4
0
4
Thus the transformed e
quation is
4a; 4 13a; 2
+ 9 = 0,
or
(4a; 2 9)(a; 2 l) =
The roots of this equal
;ion are
3
+ 2
, , +1, 1; h
the proposed equation are
1
"2'
7
2'
1, 3.
573. The chief use of the substitution in the preceding
article is to remove some assigned term from an equation.
Let the given equation be
2? Q x" + p 1 x"~ l + p 2 x n ~ 2 + ... +p n _ x x+p n = 0;
then if y = x — h, we obtain the new equation
A^+*)"+ip 1 6r+*r , +A(y+*r i ++p.=0i
which, when arranged in descending powers of y, becomes
(/w (77, *— 1 ) ^
■ 2 pft + (»  i) M +i\yf~*+ ■ =o.
If the term to be removed is the second, we put npji + ]) l = 0,
so that h = — — : if the term to be removed is the third we put
np
n(n\)
^2 — " Po h + ( n  l )l\ h +P* = °>
and so obtain a quadratic to find h \ and similarly we may remove
any other assigned term.
TRANSFORMATION OF EQUATIONS 477
Sometimes it will be more convenient to proceed as in the
following example.
Example. Remove the second term from the equation
px? + qx 2 + rx + s = 0.
Let a, /9, 7 be the roots, so that a + p + y= . Then if we increase
each of the roots by £ , in the transformed equation the sum of the roots
dp
will be equal to   +  ; that is, the coefficient of the second term will
p p
be zero.
Hence the required transformation will be effected by substituting x~
6p
for x in the given equation.
574. From the equation f(x) = we may form an equation
whose roots are connected with those of the given equation by
some assigned relation.
Let y be a root of the required equation and let cf>(x, y) =
denote the assigned relation; then the transformed equation can
be obtained either by expressing a; as a function of y by means
of the equation <£ (x, y) = and substituting this value of x in
f(x) = §; or by eliminating x between the equations f(x) = Q
and <f> (x, y) = 0.
Example 1. If a, b, c are the roots of the equation x 3 +p>x 2 + qx + r=0,
form the equation whose roots are
11 1
a  — , b , c — r .
be ca ab
When x = a in the given equation, y = a = in the transformed equation j
, , 1 a a
but a— = a — — =a + ;
be abc r
and therefore the transformed equation will be obtained by the substitution
x ry
y = x +  , or x = ~ ;
v r 1 + r
thus the required equation is
r 2 y 3 +Pr{l + r)y 2 + q{l + r) 2 y + (l + r)* = 0.
Example 2. Form the equation whose roots are the squares of the
differences of the roots of the cubic
x 3 + qx + r = Q.
Let a, b, c be the roots of the cubic ; then the roots of the required
equation are (b  c) 2 , (c  a) 2 , (a  b) 2 .
478 HIGHER ALGEBRA.
2a6c
2abc
Now (bc) 2 = fc 2 + c 2 2fcc = a 2 + & 2 + c 2 a 2 
a
a
= (a + b + c) 2  2 (be + ca + ah)  a 2
= 2 5 a 2 + ^;
a
also when x = a in the given equation, y = (bc) 2 in the transformed
equation ;
. 2r
.*. ?/ =  2o a; J H .
x
Thus we have to eliminate x between the equations
x s + qx + r=0,
and .r 3 + (2# + y) x  2r = 0.
By subtraction (#+?/)# = 3r ; or a: = .
Substituting and reducing, we obtain
y s + 6 (2 ^2 + 9^ + 27,2 + 4^3 _ .
Cor. If a, &, c are real, (&  c) 2 , (c  a) 2 , (a  6) 2 are all positive ; therefore
27r 2 + 4g 3 is negative.
Hence in order that the equation x z + qx + r = may have all its roots
real 27r 2 + 4<7 3 must be negative, that is () +(f) must be negative.
If 27r 2 + 4# 3 =0 the transformed equation has one root zero, therefore
the original equation has two equal roots.
If 27r 2 + 4g 3 is positive, the transformed equation has a negative root
[Art. 553], therefore the original equation must have two imaginary roots,
since it is only such a pair of roots which can produce a negative root in
the transformed equation.
EXAMPLES. XXXV. d.
1. Transform the equation x 3  4# 2 +  x —  = into another with
integral coefficients, and unity for the coefficient of the first term.
2. Transform the equation 3x A  5x 3 + x 2  x + 1 = into another
the coefficient of whose first term is unity.
Solve the equations :
3. 2x 4 + x 3 6x 2 + x + 2 = 0.
4. ^10^ + 26^10^+1 = 0.
5. x*5x i + 9x 3 9x 2 + 5.^1 = 0.
6. 4#fi  24^ + 57x A  Idx 3 + 57 x 2  Mx + 4 = 0.
TRANSFORMATION OF EQUATIONS. 479
7. Solve the equation 3./, 3  22.r 2 + 48.r  32 = 0, the roots of which
are in harmonica! progression.
8. The roots of x 3  lLr 2 + 36#36 = are in harmonica! pro
gression ; find them.
9. If the roots of the equation x 3  ax 2 +x—b=0 are in harmonica!
progression, shew that the mean root is 3b.
10. Solve the equation 4(Xr 4 22^2Lr 2 + 2.t*+l =0, the roots of
which are in harmonica! progression.
Remove the second term from the equations :
11. a? 8  &c*+ 10a? 3=0.
12. x*+4o?+ 2a 2 4# 2=0.
13. afi + 5x A + 3X 3 + x 2 + x  1 = 0.
14. afi  12s 5 + 3.v 2  1 7.v + 300 = 0.
x 3
15. Transform the equation a^j — 7=0 mto one whose roots
3
exceed by  the corresponding roots of the given equation.
22
16. Diminish by 3 the roots of the equation
17. Find the equation each of whose roots is greater by unity
than a root of the equation x 3  bx 2 + 6x  3 = 0.
18. Find the equation whose roots are the squares of the roots of
x* + x 3 + 2x 2 + x+ 1 = 0.
19. Form the equation whose roots are the cubes of the roots of
x 3 + 3x 2 + 2 = 0.
If a, b, c are the roots of x 3 + qx + r0, form the equation whose
roots are
20. ka~\ hb\ hr\ 21. b 2 c\ c 2 a 2 , a 2 b 2 .
b + c c + a a + b , 1 1 b 1
24. «(6 + c), &(c+a), c(a + b). 25. « 3 , 6 3 , c 3 .
~ n b c c a a b
26.  + r ,  + , T + 
c b a c b a
27. Shew that the cubes of the roots of x 3 + ax 2 +bx + ab=0 are
given by the equation x 3 + a 3 x 2 + b 3 x + a 3 b 3 = 0.
28. Solve the equation x*  bx*  bx 3 + 2bx 2 + 4a  20 = 0, whose
roots are of the form «, —a,b, — b, c.
29. If the roots of x 3 + 3px 2 + 3qx + r = are in harmonica] pro
gression, shew that 2g s =r(3pg r — r).
480 HIGHER ALGEBRA.
Cubic Equations.
575. The general type of a cubic equation is
x 3 +Px 2 + Qx + fi=0,
but as explained in Art. 573 this equation can be reduced to the
simpler form x 3 + qx + r = 0,
which we shall take as the standard form of a cubic equation.
576. To solve the equation x 3 + qx + r = 0.
Let x = y + z ; then
x 3 =y 3 + z 3 + 3yz (y + z) = y 3 + z 3 + 3yzx,
and the given equation becomes
y 3 + z 3 + (3yz + q)x+r = 0.
At present y, z are any two quantities subject to the con
dition that their sum is equal to one of the roots of the given
equation ; if we further suppose that they satisfy the equation
3yz + q = 0, they are completely determinate. We thus obtain
o 3
y 3 + z 3 =r, y 3 z 3 = ^;
hence y 3 , z 3 are the roots of the quadratic
Solving this equation, and putting
y z =\ + J r i + it »
sr ~ 2 V 4 27 ™
we obtain the value of x from the relation x = y + z; thus
1
r fr 2 q 3 Y f r 11
2 + V4 + 27} + HV"
r 2 q 3 ^
' 4 + 97
The above solution is generally known as Cardan's Solution,
as it was first published by him in the Ars Magna, in 1545. Cardan
obtained the solution from Tartaglia; but the solution of the
cubic seems to have been due originally to Scipio Ferreo, about
CU13IC EQUATIONS. 481
1505. An interesting historical note on this subject will be
found at the end of Burnside and Panton's Theory of Equations.
577. By Art. 110, each of the quantities on the righthand
side of equations (1) and (2) of the preceding article lias three
cube roots, hence it would appear that x has nine values ; this,
however, is not the case. For since yz = — ^, the cube roots are
to be taken in pairs so that the product of each pair is rational.
Hence if y, z denote the values of any pair of cube roots which
fulfil this condition, the only other admissible pairs will be
wy, ta*z and ii> 2 y, a)Z, where co, or are the imaginary cube roots of
unity. Hence the roots of the equation are
y + z, wy + w 2 z, <x) 2 y + wz.
Example. Solve the equation x 3  15.r = 126.
Put y + z for x, then
y"* + z* + {3yz15)x = 126;
put 3f/215 = 0,
then y^z 3 = 126;
also y*z s = 125 ;
hence y' s , z :i are the roots of the equation
i 2 126£ + 125 = 0;
.. 2/3 = 125, sfc=l;
y = 5, 2 = 1.
Thus j/ + 2 = 5 + l = 6;
u,y + u~z = ^— 5 +  2
= 3 + 2^/^3;
wy + wz =  32^/^3;
and the roots are 6, 3 + 2*73, 32 J 3.
578. To explain the reason why we apparently obtain nine
values for x in Art. 576, we observe that y and z are to be found
from the equations y 3 + z 3 + r = 0, yz = — \ j but in the process of
o
q 3
solution the second of these was changed into y J z 3 =  ^ , which
H.H. A. 31
482 HIGHER ALGEBRA.
2
would also hold if yz = — ~ , or yz = ^ ; hence the other six
values of x are solutions of the cubics
x 3 + wqx + r — 0, x 3 + ou 2 <p; + r = 0.
579. We proceed to consider more fully the roots of the
equation x 3 + qx + r = 0.
2 3
v q .
(i) If r + ~ h positive, then y 3 and z 3 are both real; let
y and # represent their arithmetical cube roots, then the roots
are y + z, wy + oy 2 z, (o 2 y + wz.
The first of these is real, and by substituting for a> and w 2 the
other two become
r 2 <7 3 .
(ii) If j + ^= is zero, then y 3 — z 3 \ in this case ?/ = z, and
the roots become 2y, 2/(w + co 2 ), 2/(00 + to 2 ), or 2y, —3/, — ?/.
r 2 <7 3
(iii) If — + ~ is negative, then ?/ 3 and 2 3 are imaginary ex
pressions of the form a + ib and a — ib. Suppose that the cube
roots of these quantities are m + in and m — in; then the roots of
the cubic become
m + in + m — in, or 2m ;
(m + m) o> + (m — in) <o 2 , or — m — w ^/3 ;
(m + m) co 2 + (m — in) <o, or —m + n ^3 ;
which are all real quantities. As however there is no general
arithmetical or algebraical method of finding the exact value of
the cube root of imaginary quantities [Compare Art. 89], the
solution obtained in Art. 576 is of little practical use when the
roots of the cubic are all real and unequal.
This case is sometimes called the Irreducible Case of Cardan's
solution.
580. In the irreducible case just mentioned the solution may
be completed by Trigonometry as follows. Let the solution be
1 1
x = (a + ib) 3 + (a  ib) 3 ;
BIQUADHATIC EQUATIONS. 483
put a = r cos 6, b = r sin 0, so that r 2 = a 2 + b\ tan =  ■
then (« + ^) 3 = {r (cos + i sin 6)f.
Now by De Moivre's theorem the three values of this ex
pression are
, ;.,: *\ J/ + 2tt . . 0+27A
r 3 (^cos  + * sin \ , H f cos l^fl + i sin
<9 + 4tt . . + 4
+ 1 sin
and r3 (cos
1
where r 3 denotes the arithmetical cube root of r, and the
smallest angle found from the equation tan =  .
a'
1
The three values of (a  ibf are obtained by changing the sign
ot 1 in the above results ; hence the roots are
2r!co8' 24cos^T, 2,icos^ti^ I
o 3
Biquadratic Equations.
581 We shall now give a brief discussion of some of the
methods which are employed to obtain the general solution of a
biquadratic equation. It will be found that in each of the
methods we have first to solve an auxiliary cubic equation ; and
thus it will be seen that as in the case of the cubic, the general
solution is not adapted for writing clown the solution of a
given numerical equation.
• 58 i 2 ; J he solution of a biquadratic equation was first ob
tained by Ferrari, a pupil of Cardan, as follows.
Denote the equation by
x 4 + 2px 3 + qx 2 + 2rx + s=0;
add to each side (ax + b) 2 , the quantities a and b being determined
so as to make the left side a perfect square; then
x 4 + 2px 3 +(q + a 2 )x 2 + 2(r + ab)x + s + b 2 = (ax + b) 2 .
Suppose that the left side of the equation is equal to (rf+px+k)*
then by comparing the coefficients, we have
p* + 2k  q + a 2 , pk = r + ab, If = s + b' 2 ;
31—2
484 HIGHER ALGEBRA.
by eliminating a and b from these equations, we obtain
(pk  r) 2 = (21c +p*  q) (k 2  s),
or 2k 3 qk 2 + 2 ( pr  s) k + p 2 s  qs  r 2 = 0.
From this cubic equation one real value of k can always be
found [Art. 553]; thus a and b are known. Also
(x 2 +px + W = ( ax + W'>
.'. x 2 +px + k = ±(ax + b);
and the values of x are to be obtained from the two quadratics
x 2 + (p — a) x + (k — b) = t
and x 2 + (2) + a)x+ (k + b) = 0.
Example. Solve the equation
x i  2.x 3  5x + Hh;  3 = 0.
Add a 2 x 2 + 2abx + b 2 to each side of the equation, and assume
x*  2x* + (a 2  5) x 2 + 2 {ab + 5) x + 6 2  3 = (x 2  x + k) 2 •
then by equating coefficients, we have
a 2 =2fc + 6, ab = k5, b 2 =k 2 + 3;
.. (2fc + 6)(fc 2 + 3) = (7c + 5) 2 ;
.. 2k* + 5k 2 M 7 = 0.
By trial, we find that k=  1 ; hence a 2 = 4, & 2 = 4, ab=  4.
But from the assumption, it follows that
(x 2 x + k) 2 =(ax + b) 2 .
Substituting the values of k } a and b, we have the two equations
x 2  x  1 = ± (2x  2) ;
that is, z 2 3.r + l = 0, and x 2 + x3 = Q;
whence the roots are — ~— , ^ — .
a 
583. The following solution was given by Descartes in 1637.
Suppose that the biquadratic equation is reduced to the form
x 1 + qx 2 + rx + s = ;
assume x 4 + qx 2 + rx + s = (x 2 + kx + 1) (x 2  kx + m) ;
BIQUADRATIC EQUATIONS. 485
then by equating coefficients, we have
I + vi — k 2 — q, k (m — l) = r, Im = s.
From the first two of these equations, we obtain
v v
2m = AT + q + t , 2l=k 2 + q j;
hence substituting in the third equation,
(k 3 + qk + r) (k 3 + qk  r) = 4sk 2 ,
or ¥ + 2qk 4 + (q 2  4s) k 2  r 2 = 0.
This is a cubic in k 2 wliich always has one real positive solu
tion [Art. 553]; thus when k 2 is known the values of I and m
are determined, and the solution of the biquadratic is obtained
by solving the two quadratics
x 2 + kx + I = 0, and x 2 — kx + m=0.
Example. Solve the equation
z 4 2a; 2 + 8j;3 = 0.
Assume x 4  2a; 2 + 8#  3 = (x 2 + kx + 1) (x 2  kx + m) ;
then by equating coefficients, we have
l + mk 2 = 2, k(ml) = 8, lm = Z;
whence we obtain {k 3  2k + 8) (A; 3  2k  8) =  12& 2 ,
or fc 6 4fc 4 + 16fc 2 64 = 0.
This equation is clearly satisfied when A; 2 4 = 0, or k— ±2. It will be
sufficient to consider one of the values of k ; putting k = 2, we have
m+l = 2, 7nl = 4] that is, l = l, m = 3.
Thus as*  2.r 2 + 8.r  3 = (.r 2 + 2x  1 ) {x 2  2x + 3) ;
hence a; 2 + 2.rl = 0, and x2x + 3 = ;
and therefore the roots are  1 ± J2, 1± J^2.
584. The general algebraical solution of equations of a
degree higher than the fourth has not been obtained, and Abel's
demonstration of the impossibility of such a solution is generally
accepted by Mathematicians. If, however, the coefficients of an
equation are numerical, the value of any real root may be found
to any required degree of accuracy by Horner's Method of ap
proximation, a full account of which will be found in treatises on
the Theory of Equations.
486 HIGHER ALGEBRA.
585. We shall conclude with the discussion of some miscella
neous equations.
Example 1. Solve the equations :
x + y + z + u = 0,
ax + by +cz + du = 0,
a 2 x + b 2 y + c 2 z + d 2 u = 0,
a?x + b 3 y + c 3 z + d 3 u = k.
Multiply these equations, beginning from the lowest, by 1, p, q, r re
spectively ; p, q, r being quantities which are at present undetermined.
Assume that they are such that the coefficients of y, z, u vanish ; then
x (a 3 +pa 2 + qa + r) = k,
whilst b, c, d are the roots of the equation
t 3 +pt 2 + qt + r = 0.
Hence a 3 +pa 2 + qa + r = (ab){ac){a d) ;
and therefore (a b)(a c) {a d)x = h.
Thus the value x is found, and the values of y, z, u can be written down
by symmetry.
Cor. If the equations are
x + y + z + u = l,
ax + by + cz + du = I;
a 2 x + b 2 y + c*z + d 2 u = k 2 ,
a 3 x + b s y + c 3 z + dhi = A; 3 ,
by proceeding as before, we have
x (a 3 +pa 2 + qa + r) = k 3 +pk 2 + qk + r;
.'. (ab)(ac)(ad) x = (k b)(kc)(kd).
Thus the value of x is found, and the values of y, z, u can be written
down by symmetry.
The solution of the above equations has been facilitated by the use of
Undetermined Multipliers.
Example 2. Shew that the roots of the equation
{x a){x b) (x  c) f 2 (x a)g 2 (xb) h 2 {x c) + 2fgh =
are all real.
From the given equation, we have
{xa){(xb)(xc)f*}{g*{xb) + h*(xc)2fgh}=0.
Let p, q be the roots of the quadratic
{xb)(xe)f*=0,
BIQUADRATIC EQUATIONS. 487
and suppose ^ to be not less than q. By solving the quadratic, we have
2x = b + c±J(bc)* + tf :i (1);
now the value of the surd is greater than b ~ c, so that p is greater than h
or c, and q is less than b or c.
In the given equation substitute for x successively the values
+ °°» v, q>  30 ;
the results are respectively
+ °° , fajp^bh Jp ~ c ) 2 > +{<J Jbq h Jc  q)~,  cc ,
since {p b)(p c) =f* = (b  q) (c  q).
Thus the given equation has three real roots, one greater than 2', one
between _p and q, and one less than q.
If p = q, then from (1) we have (6c) 2 + 4/ 2 = and therefore b = c,f=0.
In this case the given equation becomes
(x b){{x a) (x b)g* lr} =0 ;
thus the roots are all real.
If p is a root of the given equation, the above investigation fails ; for it
only shews that there is one root between q and + oo , namely p. But as
before, there is a second real root less than q ; bence the third root must also
be real. Similarly if q is a root of the given equation we can shew that all
the roots are real.
The equation here discussed is of considerable importance ; it occurs
frequently in Solid Geometry, and is there known as the Discriminating
Cubic.
586. The following system of equations occurs in many
branches of Applied Mathematics.
Example. Solve the equations :
x y z ,
a + \ b + \ c + \
x y z
a \fji. b + /j, c + fx
x y z ,
— + J —+ = 1.
a+f b+ v c+v
Consider the following equation in 6,
x y z (0X)(gft)(gy) .
a + + b + d + c + (a + e)(b + 6){c + 0y
x, y, z being for the present regarded as known quantities.
488 HIGHER ALGEBRA.
This equation when cleared of fractions is of the second degree in 6, and
is satisfied by the three values 8 = \ 6 = p., d = v, in virtue of the given
equations ; hence it must be an identity. [Art. 310.]
To find the value of #, multiply up by a+0, and then put a + = 0;
thus ..  (X)(^)("'0 .
(b  a) (c  a)
that is, . = fe+£Lfe+i4!ttd.
(a b) (ac)
By symmetry, we have
(b + \){b + fx)(b + v)
y=
and
{bc)(ba)
{c + \){c + fi)(c + v)
(c a) (c  b)
EXAMPLES. XXXV. e.
Solve the following equations :
1. a 3 18a = 35. 2. a?+ 7207 1720=0.
3. a 3 + 63a 316 = 0. 4. ff 3 + 21# + 342 = 0.
5. 28^9^+1=0. 6. & s 15# 8 33ar+ 847=0.
7. 2a 3 + 3a 2 + 3a + 1=0.
8. Prove that the real root of the equation a 3 + 12a 12 =
is 2^/2^4.
Solve the following equations :
9. a 4 3a 2 42 a 40 = 0. 10. a 4  10a 2  20a 16 = 0.
11. a 4 + 83? + 9a 2 8a 10 = 0.
12. a 4 + 2a 3  7 a 2  8a + 1 2 = 0.
13. ** 3^6^2=0. 14. a*23?12afi+10x + 3=0.
15. 4a 4  20a 3 + 33^ 2  20a + 4 = 0.
16. a 6 6a 4 17a 3 + 17a 2 + 6a1 = 0.
17. a 4 + 9a 3 + 1 2a 2  80a  1 92 = 0, which has equal roots.
18. Find the relation between q and r in order that the equation
A 3 + ^A + r=0 may be put into the form a 4 = (a 2 + «a+&) 2 .
Hence solve the equation
8a 3 36a + 27 = 0.
BIQUADRATIC EQUATIONS. 489
19. If jfi+3pafl+3qx+r and x*+2px+q
have a common factor, shew that
4(p 2 q) (q 2 —pr)  (pqr) 2 = 0.
If they have two common factors, shew that
p 2 q=0, q 2 pr=0.
20. If the equation ax s + 3bx 2 + 3cx + d=() has two equal roots,
shew that each of them is equal to —. rs? .
1 2 (etc  b 2 )
21. Shew that the equation x 4 +PX 3 + qx 2 + rx + s = may be solved
as a quadratic if r 2 =p 2 s.
22. Solve the equation
gfl  1 Sx A + 1 6.1* 3 + 28x 2  S2x + 8 = 0,
one of whose roots is J6 — 2.
23. If a, /3, y, 5 are the roots of the equation
x A + qx 2 + r.t 4 s = 0,
find the equation whose roots are /3+y + d + (/3y§) _1 , &c.
24. In the equation x 4 — px 3 + qx 2  rx + s = 0, prove that if the sum
of two of the roots is equal to the sum of the other two p 3  4pq + 8r = ;
and that if the product of two of the roots is equal to the product of
the other two r 2 =p 2 s.
25. The equation x°  209.£ + 56 = has two roots whose product is
unity : determine them.
26. Find the two roots of ^ — 409^ + 285 = whose sum is 5.
27. If a, b, c,...k are the roots of
X n +p 1 X n ~ 1 +p2X n ~ 2 + +Pnl$ +Pn = °>
shew that
(l+a 2 )(l+b 2 ) { l + k 2 ) = (lp,+p± ...) 2 + ( Pl p,+p, ...) 2 .
28. The sum of two roots of the equation
.r 4  8.r> + 21^ 2  20a + 5 =
is 4 ; explain why on attempting to solve the equation from the kuow
led^e of this fact the method fails.
MISCELLANEOUS EXAMPLES.
1. If s l , s 2i * 3 are the sums of n, 2n, Sn terms respectively of an
arithmetical progression, shew that s 3 = 3 (s 2 — sj.
2. Find two numbers such that their difference, sum and product,
are to one another as 1, 7, 24.
3. In what scale of notation is 25 doubled by reversing the digits?
4. Solve the equations :
(1) (#+2)(#+3)(a;4)(#5)=44.
(2) x(y + z) + 2 = 0, y(z2x) + 2l=Q, z(2xy) = b.
5. In an A. P., of which a is the first term, if the sum of the
first p terms = 0, shew that the sum of the next q terms
a{p + q)q ^
p—l
[R. M. A. Woolwich.]
6. Solve the equations :
( 1 ) (a + b) (ax + b)(a bx) = (a 2 x  b 2 ) (a + bx).
11 i
(2) x* + (2xZf={l2(xl)Y. [India Civil Service.]
7. Find an arithmetical progression whose first term is unity
such that the second, tenth and thirtyfourth terms form a geometric
series.
8. If a, fi are the roots of x+px+q = 0, find the values of
a 2 + a/3 + /3 2 , a 3 + /3 3 , a 4 + a 2 /3 2 + 4 .
9. If 2x — a + a~ 1 and 2y = b + b~ 1 , find the value of
xy + *J(x 2 \)(y 2  1).
10. Find the value of
3 3
(4 + Vl5)" 2 + (4Vi5)' 2
_ 3 3"
(6 + V35)" 2 (6\/35)'
[R. M. A. Woolwich.]
11. If a and /3 are the imaginary cube roots of unity, shew that
a 4 + ^ 4 + a 1 ^ 1 = 0.
MISCELLANEOUS EXAMPLES. 401
12. Shew that in any scale, whose radix is greater than 4, the
number 12432 is divisible by 111 and also by 112.
13. A and B run a mile race. In the first heat A gives B a start
of 11 yards and beats him by 57 seconds ; in the second heat A gives
B a start of 81 seconds and is beaten by 88 yards : in what time could
each run a mile ?
14. Eliminate x, y, z between the equations :
x 2 —yz— a 2 , y 2  zx = b 2 , z 2  xy = c 2 , x f y + z <= 0.
[R. M. A. Woolwich.]
15. Solve the equations :
ax 2 + bxy + ey 2 = bx 2 + cxy + ay 2 = d.
[Math. Tripos.]
16. A waterman rows to a place 48 miles distant and back in
14 hours: he finds that he can row 4 miles with the stream in the
same time as 3 miles against the stream : find the rate of the stream.
17. Extract the square root of
(1) (a 2 + ab + be + ea) {be + ca + ab + b 2 ) (be + ca + ab + c 2 ).
(2) l.r+\/22^158^ 2 .
10
18. Find the coefficient of x G in the expansion of (1  Sx) :i , and the
term independent of x in ( x 2  — ) .
\^S AX J
19. Solve the equations :
/1N 2.r3 3^8 ff+3 n
(2) x 2 y 2 = xy — ab, (x + y) (ax + by) = 2ab(a + b).
[Trin. Coll. Camb.]
20. Shew that if a(bc) x 2 + b (c a) xy + c(ab)y 2 is a perfect
square, the quantities a, b, c are in harmonica! progression.
[St Cath. Coll. Camb.]
21. If
(yz) 2 + (zx) 2 + (xy) 2 = (y + z2x) 2 + (z + x2y) 2 + (x+y2z) 2 ,
and x, y, z are real, shew that x=y = z. St Cath. Coll. Camb.]
22. Extract the square root of 3e582Gl in the scale of twelve, and
find in what scale the fraction  would be represented by 17.
o
492 HIGHER ALGEBRA.
23. Find the sum of the products of the integers 1, 2, 3, ... n taken
two at a time, and shew that it is equal to half the excess of the sum of
the cubes of the given integers over the sum of their squares.
24. A man and his family consume 20 loaves of bread in a week.
If his wages were raised 5 per cent., and the price of bread were raised
2\ per cent., he would gain 6d. a week. But if his wages were lowered
7^ per cent., and bread fell 10 per cent., then he would lose \\d.
a week : find his weekly wages and the price of a loaf.
25. The sum of four numbers in arithmetical progression is 48 and
the product of the extremes is to the product of the means as 27 to 35 :
find the numbers.
26. Solve the equations :
(1) a{bc)x 2 + b(ca)x+c(ab) = 0.
fr% . (xa)(xb) (xc)(xd) r , r m ..
(2) b ^ — — l — \ 1± — —£ . [Math. Tripos.]
v ' xa — b xcd L J
27. If /s/ax + ^/bx+\/cx=0 i shew that
(a + b + c + 3x) (a + b + cx) = 4(bc + ca + ab),
and if ^a + 4/6+4/c = 0, shew that (a + b + c) 3 = 27abc.
28. A train, an hour after starting, meets with an accident which
detains it an hour, after which it proceeds at threefifths of its former
rate and arrives 3 hours after time : but had the accident happened 50
miles farther on the line, it would have arrived l£ hrs. sooner : find the
length of the journey.
29. Solve the equations :
2x+y = 2z, 9z7x=6y, x 3 +f + z 3 =2l6.
[R. M. A. Woolwich.]
30. Six papers are set in examination, two of them in mathematics :
in how many different orders can the papers be given, provided only that
the two mathematical papers are not successive ?
31. In how many ways can £5. 4s. 2d. be paid in exactly 60 coins,
consisting of halfcrowns, shillings and fourpennypieces ?
32. Find a and b so that x 3 + ax 2 + llx + 6 and x 3 + bx i + l4x + 8
may have a common factor of the form x 2 \px + q.
[London University.]
33. In what time would A,B,C together do a work if A alone could
do it in six hours more, B alone in one hour more, and C alone in twice
the time 1
MISCELLANEOUS EXAMPLES. 493
34. If the equations ax + by = \, ex 2 + dy % = 1 have only i >ne solution
., . a 2 b 2 , , a b ___ „,
prove that — +7 = 1, and x =  , y = , . [Math. Tiuros.]
35. Find by the Binomial Theorem the first five terms in the expan
sion of (l2x + 2x 2 )~' 2 '
36. If one of the roots of x 2 f px + q — is the square of the other,
shew that p 3  q (3p  1 ) + q 2 = 0.
[Pemb. Coll. Camb.]
37. Solve the equation
x i 5x^6xb = 0.
[Queen's Coll. Ox.]
38. Find the value of a for which the fraction
x 3  ax 2 + 19.27  a — 4
x?(a + l) x 2 + 23xa~7
admits of reduction. Eeduce it to its lowest terms. [Math. Tripos.]
39. If a, b, c, x, y, z are real quantities, and
(a + b + c) 2 = 3 (be + ca + ab x 2 y 2  z 2 ),
shew that a = b = c, and x = 0, y = 0, 2 = 0.
[Christ's Coll. Camb.]
i
40. What is the greatest term in the expansion of ( 1   x ] when
the value of x is  ? [Emm. Coll. Camb.]
41. Find two numbers such that their sum multiplied by the sum
of their squares is 5500, and their difference multiplied by the difference
of their squares is 352. [Christ's Coll. Camb.]
1 __ b 2 + 3c 2
42. If x = \a, y = (kl)b, s = (\3)c, X= — z — , 2 ' , , express
Qj "T" 0" ~p C
x 2 +y 2 + z 2 in its simplest form in terms of a, b, c.
[Sidney Coll. Camb.]
43. Solve the equations :
(1) x a + 3j*=16x + 60.
(2) y 2 + z 2 x = z 2 + x' i y = x 2 +y 2 z = \.
[CoRrus Coll. Ox.]
44. If x, y, z are in harmonical progression, shew that
log (x + z) + log {x 2y + z) = 2 log (x  z).
494 HIGHER ALGEBRA.
45. Shew that
1 1.3/1\ 1.3.5 /lV , 4 ._ /oN ,_
[Emm. Coll. Camb.]
3a26~362c~3c2a'
then will b(x+y + z) (5c + 46  3a) = (9x + 83/ + 13^) (a + b + c).
[Christ's Coll. Camb.]
47. With 17 consonants and 5 vowels, how many words of four
letters can be formed having 2 different vowels in the middle and 1
consonant (repeated or different) at each end?
48. A question was lost on which 600 persons had voted ; the same
persons having voted again on the same question, it was carried by twice
as many as it was before lost by, and the new majority was to the former
as 8 to 7 : how many changed their minds? [St John's Coll. Camb.]
49. Shew that
lx
(l+x) 2 5x* 9^5 13^7
l+£?^ + 2.3 + 4.5 + 6.7 + "'
[Christ's Coll. Camb.]
50. A body of men were formed into a hollow square, three deep,
when it was observed, that with the addition of 25 to their number a
solid square might be formed, of which the number of men in each side
would be greater by 22 than the square root of the number of men in
each side of the hollow square : required the number of men.
51. Solve the equations :
(1) V (a + x) 2 + 2 V(a^0 2 = 3 \/a 2 ^ 2 .
(2) (x  a)* (x  6)2  {x  c)i (x  d)% = (a  c)% (6  d)K
52. Prove that
3/, t , 2 2  5 2.5.8
v/4 = H 1 v — — +
N ^6 6.12 6.12.18
[Sidney Coll. Camb.]
53. Solve $6(5a? + 6)^5(6#ll)=l.
[Queens' Coll. Camb.]
MISCELLANEOUS EXAMPLES. 405
54. A vessel contains a gallons of wine, and another vessel con
tains b gallons of water: c gallons are taken out of eaeh vessel and
transferred to the other; this operation is repeated any number of
times : shew that if c(a + b) = ab, the quantity of wine in each vessel
will always remain the same after the first operation.
55. The arithmetic mean between m and n and the geometric
mean between a and b are each equal to : find m and n in terms
m + n
of a and b.
56. If x, y, z are such that their sum is constant, and if
(z+x2y)(x+y2z)
varies as yz, prove that 2 (y + z)  x varies as yz.
[Emm. Coll. Camb.]
57. Prove that, if n is greater than 3,
1.2. M CV2.3.' l C_ 1 + 3.4.«<X_ 2  + (l)'(r+l)(/+2)=2.» 3 C r .
[Christ's Coll. Camb.]
53. Solve the equations :
(1) *J'2x  1 + */&v  2 = *J~4x  3 + *Jbx^~i.
3 I
(2) 4{(s a 16)*+8}=# 8 +16(# a 16)*
[St John's Coll. Camb.]
59. Prove that two of the quantities x, y, z must be equal to one
., .j. y  z z  x x — y n
another, if f h  — — + 2 = 0.
l+yz l+zx l+xy
60. In a certain community consisting of p persons, a percent, can
read and write ; of the males alone b per cent., and of the females alone
c per cent, can read and write : find the number of males and females in
the community.
61. If !•=?•'"
[Emm. Coll. Camb.]
62. Shew that the coefficient of x 4n in the expansion of
(1 — x + x 2 — x 3 )' 1 is unity.
63. Solve the equation
xa xb b a
+ = +
a x —a xb'
[London University.]
64. Find (1) the arithmetical series, (2) the harmonical series of
n terms of which a and b are the first and last terms ; and shew that
the product of the r* term of the first series and the {n — r+ l) tb term of
the second scries is ab.
496 HIGHER ALGEBRA.
65. If the roots of the equation
1 " q+ ^J * 2+p (1 +q) x+q ( q ~ 1) + f =0
are equal, shew that p 2 = 4q. [R. M. A. Woolwich.]
66. If a 2 + b 2 = lab, shew that
l°g jg (« + V) } = g ( lo S a + lo S h )
[Queen's Coll. Ox.]
67. If n is a root of the equation
x (1  ac)  x (a 2 + c 2 )  (1 + ac) = 0,
and if n harmonic means are inserted between a and c, shew that the
difference between the first and last mean is equal to ac {a — c).
[Wadham Coll. Ox.]
68. If n + 2 8 : W " 2 P 4 = 57 : 16, find n.
69. A person invests a certain sum in a 6rr per cent. Government
loan : if the price had been £3 less he would have received \ per cent,
more interest on his money ; at what price was the loan issued ?
70. Solve the equation :
{(^ 2 + ^ + l) 3 (^ 2 + l) 3 ^ 3 }{(^ 2 ^ + l) 3 (^ 2 + l) 3 +^ 3 }
= 3 {(^ 4 + x 2 + 1) 3  (#*+ If  a 6 } .
[Merton Coll. Ox.]
71. If by eliminating x between the equations
x 2 + ax + b = an d xy + 1 (x + y) + m = 0,
a quadratic in y is formed whose roots are the same as those of the
original quadratic in x, then either a =21, and 6 = m, or b + m=al.
[R. M. A. Woolwich.]
72. Given log 2 = '30103, and log 3 = 47712, solve the equations :
(1) 6*=y6«. (2) V5MV5*=q.
73. Find two numbers such that their sum is 9, and the sum of
their fourth powers 2417. [London University.]
74. A set out to walk at the rate of 4 miles an hour ; after he had
been walking 2 hours, B set out to overtake him and went 4£ miles
the first hour,4 miles the second, 5 the third, and so gaining a quarter
of a mile every hour. In how many hours would he overtake A l
75. Prove that the integer next above (^3 + l) 2m contains 2 m + 1 as
a factor.
MISCELLANEOUS EXAMPLES. 407
76. The series of natural numbers is divided into groups 1 ; 2, 3, 4 ;
5, 6, 7, 8, 9 ; and so on : prove that the sum of the numbers in the
?i th group is (?i l) 3 + n 3 .
77. Shew that the sum of n terms of the series
2 + 2_W + [3 \2/ + 4 \2J +
,, , 1.3.5.7 (2nl)
is equal to 1 =— : —  .
1 2'* \n
[R. M. A. Woolwich.]
1 + 2x
78. Shew that the coefficient of x n in the expansion of j— 2 is
n n1 w2
(l)S 3(l)3, 2(l)3,
according as n is of the form 3m, 3m + 1, 3«i + 2.
79. Solve the equations :
(1 ) £ = ^_ 2 _ yy z
a b c x+y + z
.„. x ii z v z x
y z x x y z
[Univ. Coll. Ox.]
80. The value of xyz is 7£ or 3f according as the series a, x, y, z,
b is arithmetic or harmonic : find the values of a and b assuming them
to be positive integers. [Merton Coll. Ox.]
81. If aybx=c \/(x a) 2 + (y b) 2 , shew that no real values of x
and y will satisfy the equation unless c 2 < a 2 + b 2 .
82. If (#+l) 2 is greater than 5x  1 and less than 7#3, find the
integral value of x.
83. If P is the number of integers whose logarithms have the
characteristic p, and Q the number of integers the logarithms of whose
reciprocals have the characteristic  q, shew that
log 10 Plog 10 # = p2 + l.
84. In how many ways may 20 shillings be given to 5 persons so
t lat no person may receive less than 3 shillings ?
85. A man wishing his two daughters to receive equal portions
• rilen they came of age bequeathed to the elder the accumulated interest
of a certain sum of money invested at the time of his death in 4 per
cent, stock at 88 ; and to the younger he bequeathed the accumulated
interest of a sum less than the former by £3500 invested at the same
time in the 3 per cents, at 63. Supposing their ages at the time of
their father's death to have been 17 and 14, what was the sum invested
in each case, and what was each daughter's fortune ?
11. 11. A 32
498 HIGHER ALGEBRA.
86. A number of three digits in scale 7 when expressed in scale 9
has its digits reversed in order : find the number.
[St John's Coll. Camb.]
87. If the sum of m terms of an arithmetical progression is equal
to the sum of the next n terms, and also to the sum of the next p
terms ; prove that (m + n)( i = (wi +p)( ) •
[St John's Coll. Camb.]
88. Prove that
1 1 1 / 1 1 1 V
+ 7—4 + 7—. vi = 7— + — 7 +
(yz? {zxf (xy) 2 \yz zx xy)
[R. M. A. Woolwich.]
89. If m is negative, or positive and greater than 1, shew that
l m + 3™ + 5 m + + (2nl) m >n m + 1 .
[Emm. Coll. Camb.]
90. If each pair of the three equations
x 2 p 1 x + q l = 0, aPptfC+q^Q, x 2 p 3 x+q 3 =0,
have a common root, prove that
Pi 2 +P? + P 3 2 + 4 (?i + ft + ft) = 2 (P2P2 +P?,Pi +PiP<J
[St John's Coll. Camb.]
91. A and B travelled on the same road and at the same rate from
Huntingdon to London. At the 50 th milestone fioin London, A over
took a drove of geese which were proceeding at the rate of 3 miles in 2
hours ; and two hours afterwards met a waggon, which was moving at
the rate of 9 miles in 4 hours. B overtook the same drove of geese at
the 45 th milestone, and met the waggon exactly 40 minutes before he
came to the 31 st milestone. Where was B when ^4 reached London ?
[St John's Coll. Camb.]
92. Ifa + 5 + c + c?=0, prove that
abc + bed + cda + dab = *J(bc ad) (ca  bd) {ah — cd).
[R. M. A. Woolwich.]
93. An A. P., a G. P., and an H. P. have a and b for their first two
terms : shew that their (?i + 2) th terms will be in G. P. if
1 — 770 «tn = . [Math. Tripos.]
ba(b 2n a 2n ) n L J
x
94. Shew that the coefficient of x n in the expansion of , r, r v
(x — a) (x  0)
a n — b n I
in ascending power of x is — ^ . — 7 ; and that the coefficient of x 2n
01 ab a n b n '
in the expansion of ,, L is 2 n_1 hi 2 + 4w + 2l __ r , r , .
r (l#) 3 » ' [Emm. Coll. Camb.]
MISCELLANEOUS EXAMPLES. 499
95. Solve the equations ;
, sF+y* : ay =34 : 15.
[St John's Coll. Camb.]
Till
96. Find the value of 1 +
ratio surd.
/ 1 / #1
*/x  y
1 1 1
1
... in the form of a quad
[R. M. A. Woolwich.]
97. Prove that the cube of an integer may be expressed as the
difference of two squares ; that the cube of every odd integer may be
so expressed in two ways ; and that the difference of the cubes of any
two consecutive integers may be expressed as the difference of two
squares. [Jesus Coll. Camb.]
98. Find the value of the infinite series
1 1 J 3 i
13 + 5 + 7 + 9 + '"
[Emm. Coll. Camb.]
99. If
x —
"
a
b + d+ b+ d +
and
then
y
a
a
d + b + d+ b+ '
bxdy=ac. [Christ's Coll. Camb.]
100. Find the generating function, the sum to n terms, and the
n th term of the recurring series 1 + 5# + 7x 2 + 1 7.V 3 + 31. z 4 +
101. If a, 6, o are in H. P., then
a+b c+b
(1) 2^> + 27^ >4 
(2) b 2 (ac) 2 =2{c 2 (ba) 2 + a 2 (cb) 2 }. [Pemb. Coll. Camb.]
102. If a, 6, c are all real quantities, and x 3  3b 2 x + 2c 3 is divisible
by x  a and also by x  b ; prove that either a = b = c, or a = — 26 = — 2<\
[Jesus Coll. Ox.]
103. Shew that the sum of the squares of three consecutive odd
i umbers increased by 1 is divisible by 12, but not by 24.
104. Shew that is the greatest or least value of ax 2 + 2bx + c,
according as a is negative or positive.
If x*+y A + z i +y 2 z 1 + z 2 x 2 + x 2 y 2 = Zxyz (x+g + z), and x, y, z arc all
real, shew that x=y=z. [St John's Coll. Camb.]
32—2
500 HIGHER ALGEBRA.
105. Shew that the expansion of
/ lVla; 2 "
V 2~
x JL3 ^ 1.3.5.7 a»
1S 2 + 2?4' 6 + 2. 4. 6.8* 10 +
106. If a, /3 are roots of the equations
x 2 +px + q = 0, x 2n +p n x n + q n = 0,
where n is an even integer, shew that ~ , — are roots of
P a
.r» + l + (#+l) n = 0. [Pemb. Coll. Camb.]
107. Find the difference between the squares of the infinite
continued fractions
b b b , d d d
a+ —  ^—, ^.. j aD0  c +
2a+ 2a + 2a+ " "' 2c+ 2c + 2c +
[Christ's Coll. Camb.]
108. A sum of money is distributed amongst a certain number of
persons. The second receives Is. more than the first, the third 2s.
more than the second, the fourth 3s. more than the third, and so on.
If the first person gets Is. and the last person £3. 7s., what is the
number of persons and the sum distributed 1
109. Solve the equations :
K ' a b + c b c + a c a + b
(2) ~ 2 + x *+f= l3 i> &&+"»=*&■
110. If a and b are positive and unequal, prove that
a*b n > n («  b) (ab) 2 .
[St Cath. Coll. Camb.]
111. Express ^r^ as a continued fraction; hence find the least
values of x and y which satisfy the equation 396.t'— 763y = 12.
112. To complete a certain work, a workman A alone would take
m times as many days as B and C working together ; B alone would
take n times as many days as A and C together ; C alone would take
p times as many days as A and B together : shew that the numbers of
days in which each would do it alone are as m + 1 : »+l : jp + 1.
Prove also H + ^— = 2. m ,, . ,_ ,
m+l n + l p + l [R. M. A. Woolwich.]
MISCELLANEOUS EXAMPLES. 501
113. The expenses of a hydropathic establishment are partly con
stant and partly vary with the number of boarders. Each boarder
pays £65 a year, and the annual profits are £9 a head when there are
50 boarders, and £10. 13s. 4d. when there are 60: what is the profit on
each boarder when there are 80 ?
114. If x 2 y = 2x — y, and x 2 is not greater than 1, shew that
[Peterhouse, Camb.]
X V
115. If s — h — —sr — i = Ti and xv — c 2 . shew that when a and c
a l y a 2 x 2 o °
are unequal,
(a 2 c 2 ) 2 b 2 c 2 = 0, or a 2 + c 2 b 2 = Q.
116. If (1 + x + x 2 f r = 1 + k\x + l' 2 x 2 + . . .,
and (x  1 ) 3r = a*"  c^' ~ 1 + c^s* " 2  . . . ;
prove that (1) \—k x + k 2  = 1,
!3r
(2) lk^ + hc., — = ±
\r\2r
[R. M. A. Woolwich.]
117. Solve the equations :
(1) {x — y) 2 + 2ab = ax+by, xy + ab = bx + ay.
(2) x 2 y 2 + z 2 = 6, 2yzzx + 2xy = 13, xy + z = 2.
118. If there are n positive quantities a lt a 2 ,... a n , and if the
square roots of all their products taken two together be found, prove
that
/ — / n — \ , N
Vaia 2 + V«i«3+ <— «— («i + « 2 + + a n);
hence prove that the arithmetic mean of the square roots of the
products two together is less than the arithmetic mean of the given
quantities. [R. M. A. Woolwich.]
119. If 6¥ + «V=a 2 6' i , and d 2 + V = x 2 +y 2 = \, prove that
Wx 6 + a*y G = (b 2 x A f a 2 y 4 ) 2 . [India Civil Service.]
120. Find the sum of the first n terms of the series whose r th terms
[St John's Coll. Camb.]
(1) ~ r ~_, (2) (a+r*6)*'
x+ 2
121. Find the greatest value of o . ~a
2iX" t~ *iX +
502 HIGHER ALGEBRA.
122. Solve the equations :
(1) l+^ 4 = 7(l+#) 4 .
(2) 3#y+20=ff0+6y=2^s+3d?=O.
123. If «x, a 2 y a 3> a i are an y f° ur consecutive coefficients of an
expanded binomial, prove that
— I — I 3 — = £ . [Queens' Coll. Camb.1
124. Separate ' \ , / „ =r into partial fractions ; and
3x — 8
find the general term when 2 is expanded in ascending powers
of X.
125. In the recurring series
5  lx + 2x* + lx 3 + bx 4 + 7x : > +
4 2
the scale of relation is a quadratic expression ; determine the unknown
coefficient of the fourth term and the scale of relation, and give the
general term of the series. [R. M. A. Woolwich.]
126. If x, y, z are unequal, and if
2a3v = ( ^2,and 2a3z^^^ ,
9 y z
(v  *) 2
then will 2a 3.?=— , and x+y + z = a. [Math. Tripos.]
Ob
127. Solve the equations :
(1) xy + 6 = 2xx 2 , xy9 = 2yy 2 .
(2) {ax)^ a = {by)^ h , b Xo & x = a lo %y.
128. Find the limiting values of
(1) x \fx 2 + « 2  *Jx A + a 4 , when x = oc .
, . \fa + 2x—\/3x , rr tt T
(2) — — — — ?—— , when x—a. [London University.]
\/Za + x 2sjx
129. There are two numbers whose product is 192, and the quotient
of the arithmetical by the harmonical mean of their greatest common
measure and least common multiple is 3f  : find the numbers.
[R. M. A. Woolwich.]
MISCELLANEOUS EXAMPLES. 503
130. Solve the following equations :
(1) yiar + 37 J/l3.r37= J/2.
(2) 6Vl2 2 + c\/ly 2 = «,
c \/l  # 2 + « Vl  2 2 = 6,
a*Jly 2 + b*Jlx 2 =c. _
131. Prove that the sum to infinity of the series
1 1.3 1.3.5 .23 2 , n
2^3 " 24)4 + ~Wb * * " 1S 24 ~ 3 * [Math. Tripos.]
132. A number consisting of three digits is doubled by reversing
the digits; prove that the same will hold for the number formed by
the first and last digits, and also that such a number can be found in
only one scale of notation out of every three. [Math. Tripos.]
133. Find the coefficients of x 12 and x r in the product of
1+x 3
n_ 2ui_ \ an( * 1 *+*"■ [R M. A. Woolwich.]
134. A purchaser is to take a plot of land fronting a street ; the
plot is to be rectangular, and three times its frontage added to twice
its depth is to be 96 yards. What is the greatest number of square
yards he may take ? [London University.]
135. Prove that
(a + b + c + dy + (a + bcdy + (ab + cdy + (abc + d)*
 (a + b + c  d)*  (a + b  c + d) A  (a  b+c+df  ( a + b + c + d)*
= 192 abed.
[Trin. Coll. Camb.]
136. Find the values of a, b, c which will make each of the ex
pressions xt + aaP + bx'Z + cx+l and x A + 2ax 3 + 2bx 2 + 2cx + 1 a perfect
square. [London University.]
137. Solve the equations :
f (1) 4^S = 3( ^=65.
(2) \j2x 2 +\ + \l&  1 =
V3  2
j
138. A farmer sold 10 sheep at a certain price and 5 others at 10*.
less per head; the sum he received for each lot was expressed in pounds
by the same two digits : find the price per sheep.
504 HIGHER ALGEBRA.
139. Sum to n terms :
(1) (2»~l)+2(2»3)+3(2»5)+....
(2) The squares of the terms of the series 1, 3, 6, 10, 15
(3) The odd terms of the series in (2). [Trin. Coll. Camb.]
140. If a, /3, y are the roots of the equation x 3 + qx + r=0 prove
that 3 (a 2 + /3 2 + y 2 ) (a 5 + /3 5 + y 5 ) = 5 (a 3 + /3 3 + y 3 ) (a 4 + £ 4 + y 4 ).
[St John's Coll. Camb.]
141. Solve the equations :
(1) a?(%5)= 41 (2) A 3 +y 3 + z 3 = 495)
y(2A + 7) = 27J' ar+y+*=15V.
Ay2=105 )
[Trin. Coll. Camb.]
142. If a, b, c are the roots of the equation x 3 + qx 2 +r = 0, form the
equation whose roots are a + bc, b + c — a, c + ab.
143. Sum the series :
(1) n + (nl)x + (?i2)x 2 +...+2z n  2 + x n  1 ;
(2) 3  x  2x 2  1 6a 3  28^ 4  676a 5 + ... to infinity ;
(3) 6 + 9 + 14 + 23 + 40 + .. . to n terms.
[Oxford Mods.]
144. Eliminate a, y, z from the equations
xi+yi + z 1 = a~ 1 , x+y + z=b.
.v 2 + y 2 + z 2 = c 2 , A 3 +3/ 3 + £r 3 = c? 3 ,
and shew that if a, y, z are all finite and numerically unequal, b cannot
be equal to d. [R. M. A. Woolwich.]
145. The roots of the equation 3a 2 (a 2 + 8) + 16(a 3  1) = are not
all unequal : find them. [R. M. A. Woolwich.]
146. A traveller set out from a certain place, and went 1 mile the
first day, 3 the second, 5 the next, and so on, going every day 2 miles
more than he had gone the preceding day. After he had been gone
three days, a second sets out, and travels 12 miles the first day, 13 the
second, and so on. In how many days will the second overtake the
first? Explain the double answer.
147. Find the value of
11111 1
3+ 2+ 1+ 3+ 2+ 1 + ""
MISCELLANEOUS EXAMPLES. 505
148. Solve the equation
x 3 + 3ax 2 + 3 (a 2  be) as + a 3 + b 3 + c 3  Zabc = 0.
[India Civil Service.]
149. If n is a prime number which will divide neither «, b, nor
a + b y prove that a n ~ 2 b — a n ~ :i b 2 + a n ~ i b 3 — ...+ab n ~ 2 exceeds by 1 a
multiple of n. [St John's Coll. Camb.]
150. Find the ?t th term and the sum to n terms of the series whose
sum to infinity is (1  abx 2 ){\ — ax)~ 2 (l — bx)~ 2 .
[Oxford Mods.]
151. If a, b, c are the roots of the equation x 3 + px + q = 0, find the
b 2 + c 2 c 2 + a 2 a 2 + b 2
equation whose roots are , — , — , .
1 a b c
[Trin. Coll. Camb.]
152. Prove that
(y + z 2xY + (z + x2y) i + (x+i/2z) i = 18 (x 2 + y 2 + z 2  yz  zx  xy) 2 .
[Clare Coll. Camb.]
153. Solve the equations :
( 1 ) x 3  20x 4133 = 0, by Cardan's method.
(2) x 5  4t 4  KU 3 + 40.i' 2 + 9x 36 = 0, having roots of the form
+ a, ±b, c.
154. It is found that the quantity of work done by a man in an
hour varies directly as his pay per hour and inversely as the square
root of the number of hours he works per day. He can finish a piece
of work in six days when working 9 hours a day at Is. per hour. How
many days will he take to finish the same piece of work when working
16 hours a day at Is. 6d. per hour ?
155. If s n denote the sum to n terms of the series
1.2 + 2.3 + 3.4+...,
and o^! that to n — 1 terms of the series
1 1 1
1.2.3.4 + 2.3.4.5 + 3.4.5.6 +  "'
shew that 1 8s n cr n _ x  s n + 2 = 0.
[Magd. Coll. Ox.]
156. Solve the equations :
(1) (12a?l)(&pl)(4a?l)(&el)=5.
(2) I fo+^ foS) 1 ( x+3)(x5) _2_ (a?+5)(a?7) 92
^ ; 5 (x + 2)(x  4) + 9 (x + 4) (x 6) "" 1 3 {x + 6)(*  8) ~ 585 *
[St John's Coll. Camb.]
506 HIGHER ALGEBRA.
157. A cottage at the beginning of a year was worth £250, but it
was found that by dilapidations at the end of each year it lost ten per
cent, of the value it had at the beginning of each year : after what
number of years would the value of the cottage be reduced below £25 ?
Given log 10 3 = 4771213. [R. M. A. Woolwich.]
158. Shew that the infinite series
1 1.4 1.4.7 1.4.7.10
+ 4 4.8 + 4.8.12 + 4.8.12.16 + '"'
i+?_l ll? 2.5.8 2.5.8. 11
+ 6 + 6 . 12 + 6 . 12 . 18 + 6 . 12 . 18 . 24 + '•• '
are equal. [Peterhouse, Camb.]
159. Prove the identity
H
x(x  a) x{x  a) (x  /3)
a/3 ~aPy~ +
r  X [ x(x+ a)  x(x + a)(x + p)  \
\ a a(5 a/3y J
_x> x 2 (x 2  a 2 ) _ x 2 (x 2 a 2 )(x 2 ^)
a
„2R2 „2/92„2 +••••
a 2 ^ 2 a 2 (3 2 y
[Trin. Coll. Camb.]
160. If n is a positive integer greater than 1, shew that
n*57i 3 + 60n 2 56n
is a multiple of 120. [Wadham Coll. Ox.]
161. A number of persons were engaged to do a piece of work
which would have occupied them 24 hours if they had commenced at
the same time; but instead of doing so, they commenced at equal
intervals and then continued to work till the whole was finished, the
payment being proportional to the work done by each : the first comer
received eleven times as much as the last ; find the time occupied.
162. Solve the equations :
x y 7
(1)
y 2 3 x 2 S x 3 +f
(2) y2 + z 2_ x{]/ + z) = a ^
z 2 + x 2 — y (z+x) = b 2 ,
x' 2 ty 2  z (x +3/) = c 2 . [Pemb. Coll. Camb.]
MISCELLANEOUS EXAMPLES. 507
163. Solve the equation
a 3 (6  c) {x  b) (x c) + b 3 (c a) {x  c) (x  a) + c 3 (a  b) (x  a) (x  b) = ;
also shew that if the two roots are equal
__+_ + j = 0. [St John's Coll. Camb.1
s]a — s /b~ sfc L
164. Sum the series :
(1) 1.2.4 + 2.3.5 + 3. 4.6+... to n terms.
(2) S + il + 5 +  toiuf 
165. Shew that, if a, 6, c, d be four positive unequal quantities and
s = a + b + c + d, then
(s  a) (s — b)(s — c) (s d)> 8labcd.
[Peterhouse, Camb.]
166. Solve the equations :
(1) \/x + a — \J y  a =  v /a, \/x—a\!i/ J ta=Ja.
(2) x + i/ + z = x 2 +f + z* = ^(x3 + i/ + z 5 ) = Z.
[Math. Tripos.]
167. Eliminate I, m, n from the equations :
lx+ my + nz = rax + ny + lz = nx + ly + mz = Jc 1 {I' 1 + m 2 + n 2 ) = 1 .
168. Simplify
a (b + c  a) 2 + . . . + . . . + (b + c  a) (c + a  b) {a + b  c)
a 2 (b + ca) + ... + ... (6 + ca)(c + a6)(a + 6c) '
[Math. Tripos.]
169. Shew that the expression
(x 2  yz) 3 + (y 2  zx) 3 + (z 2  xy) 3  3 (x 2  yz) (y 2  zx) (z 2 — xy)
is a perfect square, and find its square root. [London University.]
170. There are three towns A, B, and C; a person by walking
from A to B, driving from B to C, and riding from C to A makes the
journey in 15^ hours ; by driving from A to B, riding from B to C, and
walking from C to A lie could make the journey in 12 hours. On foot
he could make the journey in 22 hours, on horseback in 8 hours, and
driving in 11 hours. To walk a mile, ride a mile, and drive a mile he
takes altogether half an hour: find the rates at which he travels, and
the distances between the towns.
508 HIGHER ALGEBRA.
171. Shew that ?t 7 7n 5 +14?i 3 8?i is divisible by 840, if n is an
integer not less than 3.
172. Solve the equations :
(1) six 1 + 1 2y + *Jy 2 + I2x= 33, x+y=2S.
,~x u(yx) z(yx) , y(u — z) x(uz) 7
(2) — ^ =a, ^ ' = &, —  = c, — J = d.
w 2 — W 2W # — # ^y
[Math. Tripos.]
173. If s be the sum of n positive unequal quantities a, b,c..., then
+ — =■ + — + ... >  . [Math. Tripos.]
174. A merchant bought a quantity of cotton ; this he exchanged
for oil which he sold. He observed that the number of cwt. of cotton,
the number of gallons of oil obtained for each cwt., and the number of
shillings for which he sold each gallon formed a descending geometrical
progression. He calculated that if he had obtained one cwt. more of
cotton, one gallon more of oil for each cwt., and Is. more for each
gallon, he would have obtained £508. 9s. more ; whereas if he had
obtained one cwt. less of cotton, one gallon less of oil for each cwt., and
Is. less for each gallon, he would have obtained .£483. 13s. less : how
much did he actually receive ?
175. Prove that
2 (b + c  a  x)*(b  c) (ax) = 16 (b c)(c a) (a b)(x a) (x  b) (x  c).
[Jesus Coll. Camp,.]
176. If a, /3, y are the roots of the equation st?— paP+r =0, find the
equation whose roots are — — , ^~ ,  — . TR. M. A. Woolwich.]
a p y
177. If any number of factors of the form a 2 + b 2 are multiplied
together, shew that the product can be expressed as the sum of two
squares.
Given that (a 2 + b 2 )(c 2 + d 2 )(e 2 +f 2 )(c/ 2 + h' 2 )=p 2 + q 2 , find p and q in
terms of a, 6, c, d, e,f, g, h. [London University.]
178. Solve the equations
x 2 +y 2 =6l, a*y*=91. [R. M. A. Woolwich.]
179. A man goes in for an Examination in which there are four
papers with a maximum of m marks for each paper; shew that the
number of ways of getting 2m marks on the whole is
 (m + 1 ) (2m 2 + Am + 3). [Math. Tripos.]
MISCELLANEOUS EXAMPLES. 509
180. If a, j3 are the roots of « 8 +jw?+l=0, and y, S .are the roots
of x 2 + qx+l=0; shew that (a  y)(/3  y)(a + 8)(/3 + 8) = J 2  jo 2 .
[R. M. A. Woolwich.]
181. Shew that if a m be the coefficient of x m in the expansion of
(1 +#)*, then whatever n be,
/ ,s i (n~l)(n — 2)...(nm + l). , >
« « 1 + «.,...+(l) 1 « m _ 1 = ^ A w ; _^ 2U(i)i.
[New Coll. Ox.]
182. A certain number is the product of three prime factors, the
sum of whose squares is 2331. There are 7560 numbers (including
unity) which are less than the number and prime to it. The sum of
its divisors (including unity and the number itself) is 10560. Find the
number. [Corpus Coll. Camb.]
183. Form an equation whose roots shall be the products of every
two of the roots of the equation x 3  ax 2 + hx + c = 0.
Solve completely the equation
2afi + x A + x + 2 = 1 2x* + 1 2x 2 .
[R. M. A. Woolwich.]
184. Prove that if n is a positive integer,
n n n(n2) n + \ '(n4) n  = 2 B [w.
185. If (6V6 + 14) 2n + 1 =:jr, and if F be the fractional part of N,
prove that NF =20 2>t + 1 . [Emm. Coll. Camb.]
186. Solve the equations :
(1) x+y+z = 2, x 2 +y 2 + z 2 = 0, x 3 +y 3 + z 3 =  1.
(2) x*(yz) 2 = a 2 , y 2 (zx) 2 = b 2 , z 2 {xy) 2 =cK
[Christ's Coll. Camb.]
187. At a general election the whole number of Liberals returned
was 15 more than the number of English Conservatives, the whole
number of Conservatives was 5 more than twice the number of English
Liberals. The number of Scotch Conservatives was the same as the
number of Welsh Liberals, and the Scotch Liberal majority was equal
to twice the number of Welsh Conservatives, and was to the Irish
Liberal majority as 2 : 3. The English Conservative majority was 10
more than the whole number of Irish members. The whole number of
members was 652, of whom 60 were returned by Scotch constituencies.
Find the numbers of each party returned by England, Scotland, Ire
land, and Wales, respectively. [St John's Coll. Camb.]
188. Shew that a 5 (c  b) + b 5 (a  c) + & (b  a)
= (b c){c  a)(a  b) (2a 3 + 2a*b + abc).
510
HIGHER ALGEBRA.
189. Prove that
a 3 3c* 2 3a 1
a 2 a 2 + 2a 2a+l 1
a 2a+l « + 2 1
13 3 1
= (al)«
[Ball. Coll. Ox.]
190. If —  1 j H ? =0, prove that a, b, c are in harmonical
a c a—b c—b
progression, unless b = a + c. [Trin. Coll. Camb.]
191. Solve the equations :
(1) .r 3  13# 2 +1 5x + 189 = 0, having given that one root ex
ceeds another root by 2.
(2) .r 4  Ax 2 + 8x f 35 = 0, having given that one root is
2 + \/~3. [R. M. A. Woolwich.]
192. Two numbers a and b are given ; two others a v b ± are formed
by the relations 3a 1 = 2<x+6, 3b l = a + 2b; two more a 2 , b 2 are formed
from a lf b x in the same manner, and so on ; find a n , b n in terms of a and
b, and prove that when n is infinite, a n —b n . [R. M. A. Woolwich.]
193. If x +y + 2 + w = 0, shew that
mr (w + a;) 2 + yz (w — x) 2 + wy(w+y) 2
+ zx(io  yf + wz(w + z) 2 + xy (w  z) 2 + 4xyzw = 0.
[Math. Tripos.]
be a 2
194. If a +
be not altered in value by interchanging a
a 2 + fc2 + c 2
pair of the letters a, b, c not equal to each other, it will not be altered
by interchanging any other pair; and it will vanish if a + b + c=\.
[Math. Tripos.]
195. On a quadruple line of rails between two termini A and B y
two down trains start at 6.0 and 6.45, and two up trains at 7.15 and
8.30. If the four trains (regarded as points) all pass one another
simultaneously, find the following equations between x lt x 2 , x 3i x 4 , their
rates in miles per hour,
*53/i)
Am + 5#o Am + 1 Ox,
«VO Jb t
where m is the number of miles in AB. [Trin. Coll. Camb.]
196. Prove that, rejecting terms of the third and higher orders,
^4 * + (1 ~ y) 2 = l + ^+y) + ^(3.* 2 + ^ + 3y 2 ).
i+V(i #) (i y) 2 8
[Trin. Coll. Camb.]
MISCELLANEOUS EXAMPLES. 511
197. Shew that the sum of the products of the series
a, a — b, a 2b, , a — {n l)b,
taken two and two together vanishes when n is of the form 3m 8 — 1,
and 2a = (3m  2) (m + 1)6.
198. If n is even, and a + /3, a/3 are the middle pair of terms,
shew that the sum of the cubes of an arithmetical progression is
na{a 2 + (w 2 l)/3 2 }.
199. If «, b, c are real positive cpiantities, shew that
111 g8 + 68 + C 8
a b c a 3 b 3 c 3
[Trin. Coll. Camb.]
200. A, B, and C start at the same time for a town a miles distant ;
A walks at a uniform rate of u miles an hour, and B and C drive at a
uniform rate of v miles an hour. After a certain time B dismounts
and walks forward at the same pace as A, while C drives back to meet
A J A gets into the carriage with C and they drive after B entering the
town at the same time that he does : shew that the whole time occupied
a 3v + u . rT . r . .
was  .  hours. [Peterhouse, Camb.]
v 3u+v L ' J
201. The streets of a city are arranged like the lines of a chess
board. There are m streets running north and south, and n east and
west. Find the number of ways in which a man can travel from the
N.W. to the S.E. corner, going the shortest possible distance.
[Oxford Mods.]
202. Solve the equation */ x + 27 + v 55  x— 4.
[Ball. Coll. Ox.]
203. Shew that in the series
ab + (a + x) (b + x) + (a + 2x) (b + 2x) + to 2 n terms,
the excess of the sum of the last n terms over the sum of the first n
terms is to the excess of the last term over the first as ri l to 2n — 1 .
204. Find the n th convergent to
(1)
1 1
2 2
1
2
(2)
4 4
3+ 3 +
4
3 +
Pro 1
7e that
205.
{ax)Hyzy + {ayf{zxY + {azY{xyY
= 2{{ayf{azf{xyf(xzf+{az)' i {axf{yzf{yxY
^(axf{ayf{zxf{zyf}.
[Peterhouse, Camb.]
512 HIGHER ALGEBRA.
206. If a, #, y are the roots of x 3 + qx+r = 0, find the value of
ma + n m{3 + n my+n
ma — n m(3  n my — n
in terms of m, n, q, r. [Queens' Coll. Camb.]
207. In England one person out of 46 is said to die every year,
and one out of 33 to be born. If there were no emigration, in how
many years would the population double itself at this rate ? Given
log 2 = '3010300, log 1531 = 31849752, log 1518 = 31812718.
208. If (1 + x + x 2 ) n = a + a x x + a^c 2 + , prove that
7i (n — 1 ) n '
« P wa r i + y72~ «r2 + ( 1 ) r r! (n r ) \ a « = >
unless r is a multiple of 3. What is its value in this case 1
[St John's Coll. Camb.]
209. In a mixed company consisting of Poles, Turks, Greeks,
Germans and Italians, the Poles are one less than onethird of the
number of Germans, and three less than half the number of Italians.
The Turks and Germans outnumber the Greeks and Italians by 3;
the Greeks and Germans form one less than half the company ; while
the Italians and Greeks form sevensixteenths of the company : deter
mine the number of each nation.
210. Find the sum to infinity of the series whose n th term is
(n + l)n 1 (?i+2) 1 (x) n+1 . [Oxford Mods.]
211. If n is a positive integer, prove that
n(n 2 l) n(n 2 l)(n 2 2 2 )
n [2 + 2J_3
n{n 2 l){n 2 V) (n 2 r 2 )
* K ; \r \r + l ~ k ; '
[Pemb. Coll. Camb.]
212. Find the sum of the series :
(1) 6, 24, 60, 120, 210, 336, to n terms.
(2) 4  9x + 16x 2  25^ 3 + 36^  49^ + to inf.
1.3 3. 55. 7 7. 9 .
( 3 ) x + ^r + ^3+^r + tomf 
213. Solve the equation
Ax Qx + 2 8#+l
6x + 2 9.r + 3 12# =0.
8.r+l 12.r l6x + 2
[King's Coll. Camb.]
MISCELLANEOUS EXAMPLES. 513
214. Shew that
(1) a 2 (l + ^ 2 ) + ^ 2 (l+c 2 ) + c 2 (l+rt 2 )>6«6^
(2) ?2(rtP + « + Z)P + « + ^ + «+...)>(« ,> + 6' , + C"+...)(^+^ + C^+...),
the number of quantities a, 6, c,... being n.
215. Solve the equations
yz = a{y + z) + a\
zx=a(z+x) +/3>.
xi/ = a(x+y) + y\ [Trin. Coll. Camb.]
216. If n be a prime number, prove that
l(2»^l) + 2^ 1 +^+3f4«^!U...+(»l)^^^
is divisible by n. [Queen's Coll. Ox.]
217. In a shooting competition a man can score 5, 4, 3, 2, or
points for each shot: find the number of different ways in which he
can score 30 in 7 shots. [Pemb. Coll. Camb.]
218. Prove that the expression x>  bx 3 + ex 2 + dx  e will be the
product of a complete square and a complete cube if
126_9^_5e_^
5 " b ~ c ~ c 2 *
219. A bag contains 6 black balls and an unknown number, not
greater than six, of white balls ; three are drawn successively and not
replaced and are all found to be white; prove that the chance that
ft*7*7
a black ball will be drawn next is jr— r . [Jesus Coll. Camb.]
220. Shew that the sum of the products of every pair of the
squares of the first n whole numbers is —  n(n 2 — l)(4?i 2 — l)(5?i + G).
[Caius Coll. Camb.]
221. If + — — '4.£_i ^ = o has equal roots, prove
x — a xb xc
that a(bc) ±/3 (c a)±y (a b) = 0.
222. Prove that when n is a positive integer,
». 2 ...^ y . + < "»X»4) 8 ..,
(n4)(»5)(»6) «,_,. ,
j3 " +•
[Clare Coll. Camb.]
H. H.A. 33
514 HIGHER ALGEBRA.
223. Solve the equations :
(1 ) .r 2 + 2yz = if + 2z.r =g»+ %xy + 3 = 7G.
(2) .v+y + z = a + b + c
? + f + S  = 3
a b c
ax + by + cz = bc + ca + ab 
[Christ's Coll. Camb.]
224. Prove that if each of m points in one straight line be joined
to each of n in another by straight lines terminated by the points, then,
excluding the given points, the lines will intersect mn{m\){n—\)
times. [Math. Tripos.]
225. Having given y = x + x 1 + r>, expand x in the form
y + ay 2 + by 3 + ey i + dy s + ;
and shew that a 2 d 3abc + 2b 3 =  1. [Ball. Coll. Ox.]
226. A farmer spent three equal sums of money in buying calves,
pigs, and sheep. Each calf cost £1 more than a pig and £2 more
than a sheep ; altogether he bought 47 animals. The number of pigs
exceeded that of the calves by as many sheep as he could have bought
for £9 : find the number of animals of each kind.
227. Express log 2 in the form of the infinite continued fraction
1 1 2 2 3 2 n 2
1+ 1+ 1+ 1+ 1 +
[Euler.]
228. In a certain examination six papers are set, and to each are
assigned 100 marks as a maximum. Shew that the number of ways
in which a candidate may obtain forty per cent, of the whole number
of marks is
II (1245 1144 143) r ~ , f ,
!— ; ' _ r ' . i \ = > . [Oxford Mods.]
[5 {[240 6< [139 +ll> 38j L J
229. Test for convergency
x 1JJ x* 1.3.5.7 x*_ 1.3.5.7.9.11 x*_
2 + 2.4"6" + 2.4.6.8 , 10 + 2.4.G.8.10.12* 14 +
230. Find the scale of relation, the n th term, and the sum of n
terms of the recurring series 1 + 6 + 40 + 288 +
Shew also that the sum of n terms of the series formed by taking
for its r th term the sum of r terms of this series is
4 (2*  1) + i (2*  1)  ^ . [Caius Coll. Camb. ]
[Emm. Coll, Camb.]
MISCELLANEOUS EXAMPLES. 515
231. It is known that at noon at a certain place the sun is hidden
by clouds on an average two days out of every three : find the chance
that at noon on at least four out of five specified future days the sun
will be shining. [Queen's Coll. Ox.]
232. Solve the equations
x 2 + (j/ z) 2 = a 2 ^
y 2 + (z — x) 2 — b 2
z 2 + {x  i/) 2 = c 2
233. Eliminate x, ?/, z from the equations :
x 2 — x<i—xz y 2 yz — yx z 2 zx — zii . 7
2 =s* S — ^— = k , and ax + by + cz = 0.
a b c J
[Math. Tmros.]
234. Tf two roots of the equation .v 5 + px 2 + qx + r = be equal and
of opposite signs, shew that pq = r. [Queens' Coll. Camb.]
235. Sum the series :
( 1 ) 1 + 2\v + 3 V 2 + + ?ih; n ~ \
25 52 5?i 2 +12/i + 8
\ ) 12 o.{ o:? ~i" o3 o3 Tx* "•
is.23.33 ' 2 2 .3 3 .4 3 ' w 2 (w+1)3(tH2) 3 *
[Emm. Coll. Camb.]
236. If (1 +«V) (1 + a\i*)(l + a°x lc >)(\ +a*x**)
= l+A i x 4 + A 8 x 8 + A l2 x l2 +
prove that A gn + i = (rA Sn} &ndA 8n = a 2n A in ; and find the first ten terms
of the expansion. [Corpus Coll. Camb.]
237. On a sheet of water there is no current from A to B but a
current from B to C ; a man rows down stream from A to C in 3 hours,
and up stream from C to A in 3^ hours ; had there been the same cur
rent all the way as from B to C, his journey down stream would have
occupied 2 1 hours ; find the length of time his return journey would
r ave taken under the same circumstances.
238. Prove that the ?i th convergent to the continued fraction
3 3 3 . 3» +1 + 3(l)" +1
is
2+ 2+ 2+ 3» +1 (l)* +1 *
[Emm. Coll. Camb.]
239. If all the coefficients in the equation
x n + p x x n ~ 1 +p 2 x n ~ 2 + +p n =f(x) = 0,
be whole numbers, and if/(0) and/(l) be each odd integers, prove
that the equation cannot have a commensurable root.
[London University.]
516 HIGHER ALGEBRA.
240. Shew that the equation
is] ax + a + \]bx + /3 + *J ex + y =
reduces to a simple equation if f Ja± s /b± f Jc = 0.
Solve the equation
\f6x 2  1 5.i  7 + V4.r 2  8x  1 1  \/2x 2  5#+ 5 = 2#  3.
241. A bag contains 3 red and 3 green balls, and a person draws
out 3 at random. He then drops 3 blue balls into the bag, and again
draws out 3 at random. Shew that he may just lay 8 to 3 with
advantage to himself against the 3 latter balls being all of different
colours. [Pemb. Coll. Camb.]
242. Find the sum of the fifth powers of the roots of the equation
at  lx 2 + 4x  3 = 0. [London University.]
243. A Geometrical and Harmonica! Progression have the same
p tYl , q th , r th terms a, b, c respectively : shew that
a(bc)\oga + b (ea) log b + c(ab)\ogc = 0.
[Christ's Coll. Camb.]
244. Find four numbers such that the sum of the first, third and
fourth exceeds that of the second by 8 ; the sum of the squares of the
first and second exceeds the sum of the squares of the third and fourth
by 36; the sum of the products of the first and second, and of the
third and fourth is 42 ; the cube of the first is equal to the sum of the
cubes of the second, third, and fourth.
245. If T w T n + l , T n+2 be 3 consecutive terms of a recurring series
connected by the relation T n + fi = aT n + l — bT n , prove that
1 {T\ + 1 aT n T n + 1 + bT n *} =a constant.
246. Eliminate x, y, z from the equations :
1 ■++ =, .r*+y2 + 2 = Z> 2
x y z a
X s + y 3 + z 3 = c 3 , xyz = d 3 . i
[Emm. Coll. Camb.]
247. Shew that the roots of the equation
x* — px 3 + ox 2  rx + — „ =
are in proportion. Hence solve .r 4 — 1 2.r 3 + 47.^ 2 — 72.r + 36 = 0.
MISCELLANEOUS EXAMPLES. 517
248. A can hit a target four times in 5 shots; U three times in 1
shots; and twice in 3 shots. They fire a volley: what is the pro
liability that two shots at least hit? And if two hit what is the pro
bility that it is C who has missed? [St Cath. Coll. Camb.]
249. Sum each of the following series to n terms:
(1) 1+01+0 + 7 + 28 + 70+ ;
(2) 22 , l» , 6 23 , »■» . .
I.2.3.4 T 2.3.4.5 T 3.4.5.6 T 4.5.6.7 '
(3) 3 + x + 9x* + x 3 + 33x* + a* + 1 29^ ; +
[Second Public Exam. Ox.]
250. Solve the equations :
(1) y 2 +yz + z 2 =ax,\ (2) x(g + zx) = a,
z 2 + zx + x* = ay, I y(z + x y) = b,\
x 2 + xy+y 2 = az.) z (.</ +y  z) = c.
[Peterhoisk, Camb.]
251. If — h t + = 1 — , and a is an odd integer, shew that
a b c a+b+c J °
111 1
+ 7„ +  =
a H b n c n a n + b n + c n '
If u 6  v G + 5 tt¥(« 2  v 2 ) + 4md (1  u*v *) = 0, prove that
( w 2v 2 )6=16^V(lw 8 )(l— p 8 ). [Pemb. Coll. Cai ..
252. If x+y\z=3p J yz + zx + xy = 3q, xyz = r, prove that
(y + z  x) (z + x  y) (x +y  z) =  27js 3 + 36pg  8r,
and (.'/ + 2 ~" x ) 3 + (s + # — y) 3 + (#+#  *') 3 = 27j9 3  24/.
253. Find the factors, linear in x, y, z, of
{a (b + c) x 2 + b(c + a)y 2 + c(a + b) z 2 } 2  Aabc (x 2 +y 2 + z 2 )(ax 2 + by 2 + cz 2 ).
[Caius Coll. Camb.]
254. Shew that ( — J I >.r*yy.s»>( ^ )
\ x+y+z J J \ 3 J
[St John's Coll. Camb.]
255. By means of the identity \l  , ' ,„  " = = —  , prove that
r=n
* r=1 < 1; r!(rl)!(»r)! "
[Pemb. Coll. Camb.]
518 HIGHER ALGEBRA.
256. Solve the equations :
(1) ax\by+z=zx\ay\b=yz + bj. + a = 0.
(2) x fy +z ~u ■= 12,\
X 1 J^yl _ 2 2 _ u 2 _ g ?
x 3 +y 3 z 3 + u^ = 218,
>■
xy + zu = 45. J
257.
li p = q nearly, and n > 1, shew that
(?il)jt? + (
n+l)q ~\q)
If — agree with unity as far as the r th decimal place, to how many
places will this approximation in general be correct ? [Math. Tripos.]
258. A lady bought 54 lbs. of tea and coffee ; if she had bought
fivesixths of the quantity of tea and fourfifths of the quantity of
coffee she would have spent nineelevenths of what she had actually
spent ; and if she had bought as much tea as she did coffee and vice
versa, she would have spent 5s. more than she did. Tea is more ex
pensive than coffee, and the price of 6 lbs. of coffee exceeds that of
2 lbs. of tea by 5s. ; find the price of each.
, 259. If s n represent the sum of the products of the first n natural
numbers taken two at a time, then
J2 11 v_i 11
3! + 4! + + n\ + ~2l 6 '
[Caius Coll. Camb.]
260. If P « *
pa 2 + 2qab + rb 2 pac + q (be — a 2 ) — rab pc 2 — 2qca + ra 2 '
prove that P, p ; Q, q ; and R, r may be interchanged without altering
the equalities. [Math. Tripos.]
261. If a + j8 + y = 0, shew that
a n + 3 + j8 n + 3 + y M+3 = a/3y(a n + j8 B + 7 n ) + ^(a 2 + /3 2 + > 8 )(a w + 1 +/3« + 1 + y» + 1 ).
[Caius Coll. Camb.]
262. If a, /3, y, 8 be the roots of the equation
x* +pa? + qx 2 + rx + s = 0,
find in terms of the coefficients the value of 2(a/3) 2 (yS) 2 .
[London University.]
MISCELLANEOUS EXAMPLES. 519
263. A farmer bought a certain number of turkeys, geese, and
ducks, giving for each bird as many shillings as there were birds of
that kind; altogether he bought 23 birds and spent £10. 11*.; find
the number of each kind that he bought.
*o*
264. Prove that the equation
(y+z8xfi+(z+x  §y)i+(#+y  8^ = 0,
is equivalent to the equation
[St John's Coll. Camb.]
265. If the equation H . = 1 , have a pair of
1 x + a x+b x + c x + d L
equal roots, then either one of the quantities a or b is equal to one of
the quantities c or d, or else  + r =  +  Prove also that the roots
abed
are then  a, — a, :  b, — b, ; or 0, 0, , .
' ' ' ' a + b
[Math. Tripos.]
266. Solve the equations :
(1) x + y + z = ab, x l + y 1 + z l = a 1 b, xyz=a z .
(2) ay z + by + cz = bzx + cz + ax = cxy + a. >; + by = a + b + c.
[Second Public Exam. Oxford.]
267. Find the simplest form of the expression
+ >„ „„ * ^ — , + ...
(aj8)(ay)(a*)(a*) (0 a)((3 y)(/3  S)(/3  c)
^_
+ (*«.)(« /3)(e 7 )(e 8) '
[London University.]
268. In a company of Clergymen, Doctors, and Lawyers it is
fcund that the sum of the ages of all present is 2160; their average
a;e is 36; the average age of the Clergymen and Doctors is 39; of the
1 octors and Lawyers 32^; of the Clergymen and Lawyers 36f. If
each Clergyman had been 1 year, each Lawyer 7 years, and each
Doctor 6 years older, their average age would have been greater by
5 years : find the number of each profession present and their average
ages.
269. Find the condition, among its coefficients, that the expression
ciyX* + Aa^xhf + Ga. s vy + i't...ry 3 + « 4< y 4
should be reducible to the sum of the fourth powers of two linear
expressions in x and y. [London University.]
520
HIGHER ALGEBRA.
270. Find the real roots of the equations
x 2 + v 2 \w 2 =a 2 , vwhu{y + z)=^bc,
y 2 f w 2 + u 2 = b 2 , wu + v (z+x)=ca,
z 2 +u 2 +v 2 =c 2 , uv + w(x+y)=ab.
[Math. Tripos.]
271. It is a rule in Gaelic that no consonant or group of consonants
can stand immediately between a strong and a weak vowel ; the strong
vowels being a, o, u ; and the weak vowels e and i. Shew that the
whole number of Gaelic words of n + 3 letters each, which can be formed
2 1 ft + 3
of n consonants and the vowels aeo is —  — — where no letter is re
ft +2
peated in the same word. [Caius Coll. Camb.]
272. Shew that if x 2 +y 2 = 2z 2 , where x, y, z are integers, then
2x = r{l 2 + 2lkk 2 ), 2y = r(k 2 + 2lkl 2 ), 2z=r(l 2 + k 2 )
where r, I, and k are integers. [Caius Coll. Camb.]
273. Find the value of
274. Sum the series :
112 4 6
to inf.
1+ 1+ 3+ 5+ 7+ "
[Christ's Coll. Camb.]
(1)
»2
2.1 3 3.^ . B
+  — +  —  + to inf.
2.3 3.4 4.5
1 [2
(2) ^ +
+ +
[ft
(a + l)(a + 2)...(a + n)
a+l (a + l)(a + 2)
275. Solve the equations :
(1) 2^ + 3 = (2^l)(3y + l)(42l) + 12
= (2x+l)(3y l)(4g + l) + 80 = 0.
(2) 3ux 2oy = vx + uy = 3u 2 + 2v 2 = 14 ; xy = 10«v.
276. Shew that
a 2 + \ ab ac ad
ab b 2 + X be bd
ac be c 2 + X cd
ad bd cd d 2 + \
is divisible by X 3 and find the other factor. [Corpus Coll. Camb.]
MISCELLANEOUS EXAMPLES. 521
277. If c, b, c,... are the roots of the equation
find the sum of o s +6 s +c 8 +..., and shew that
a" b' 2 a* c 2 I/ 2 c 2 PniOr 2/*.,)
£> (t c a c b J 2> n
[St John's Coll. Camb.]
1 + 2a'
278. Hy the expansion of j , or otherwise, prove that
(3m 1) (3m 2) (3/t2)(3/t3)(3w4)
l3»+ jg lT 273
+ " 1.2.3.4 cVo.(l),
wlien n is an integer, and the series stops at the first term that vanishes.
[Math. Tripos.]
279. Two sportsmen A and B went out shooting and brought
home 10 birds. The sum of the squares of the number of shots was
2880, and the product of the numbers of shots fired by each was 48
times the product of the numbers of birds killed by each. If A had
fired as often as B and B as often as A, then B would have killed 5
more birds than A : find the number of birds killed by each.
280. Prove that 8 (« 3 + 6 s + c 3 ) 2 > 9 (a 2 + be) (b 2 + ca) (e* + ab).
[Pemb. Coll. Cams.]
281. Shew that the n lh convergent to
2 4 6 . _ 2» +1
... is 2
3 4 5 '" 2»2 r (nr)\ '
What is the limit of this when n is infinite? [Kino's Coll. Camb.]
282. If — is the ?i th convergent to the continued fraction
111111
a+ b+ c+ a+ b+ c\
shew that p 3n + 3 = bp 3n + (bc+l)q 3n . [Queens' Coll. Camb.]
283. Out of n straight lines whose lengths are 1, 2, 3, ...n inches
respectively, the number of ways in which four may be chosen which
will form a quadrilateral in which a circle may be inscribed is
L {2n (/i  2) (2*  5)  3 + 3 (  1 )"} . [Math. Tripos.]
522
HIGHER ALGEBRA.
284. If u 2 , u 3 are respectively the arithmetic means of the squares
and cubes of all numbers less than n and prime to it, prove that
?i 3 — 6nu 2 + 4m 3 = 0, unity being counted as a prime.
[St John's Coll. Camb.]
285. If n is of the form &m  1 shew that {y  z) n + (z x) n + (x  y) n
is divisible by x 2 +y 2 + z 2 yz — zxxy; and if n is of the form 6m +1,
shew that it is divisible by
(x 2 + y 2 + z 2 yz — zx — xy ) 2 .
286. If S is the sum of the m th powers, P the sum of the products
m together of the n quantities a lt a 2 , a 3 , ... a n , shew that
\ n 1 . S > ! n  m . \jm . P.
[Gaius Coll. Camb.]
287. Prove that if the equations
x 3 + qxr = and rx 3 — 2q 2 x 2 — 5qrx — 2q 3 — ?' 2 =Q
have a common root, the first equation will have a pair of equal roots ;
and if each of these is a, find all the roots of the second equation.
[India Civil Service.]
288. If x V2a 2  Sx 2 +y */2a 2 Sy 2 + z \/2a 2  3z 2 = 0,
where a 2 stands for x 2 +y' + z 2 , prove that
(x+y + z)(x+y + z)(xy + z)(x+yz) = Q.
[Thin. Coll. Camb.]
289. Find the values of x { , x 2 , ...x n which satisfy the following
system of simultaneous equations :
til /C\)
a x  b x Oj — b 2
x\
+
X.,
a 1 ~ W Cl 2 ~~ ^2
a x b n ~
 +...+
X r ,
a 2  b n
h
OC%
+
a n  b x a n  bo
+ ...+
x,
a n ~ K
[London University.]
290. Shew that yz  x l zx  y L xy  z
zx  y 2 xy z 2 yz x*
xy — z 2 yz — x 2 zx  y 2
where r 2 = x 2 +y 2 + z 2 , and u 2 =yz + zx + xy.
r 2
u 2
ll 2 1
u 2
r 2
u 2
u 2
it'
r 2
[Trin. Coll. Camb.]
MISCELLANEOUS EXAMPLES. 523
291. A piece of work was done by A, B, C\ at first A worked alone,
lmt after some days was joined by />', and these two after sonic days
were joined by C. The whole work could have been done by II and (",
if they had each worked twice the number of days that they actually
did. The work could also have been completed without B'h help if A
had worked twothirds and ('four times the number of days they actually
did; or if A and B had worked together for 40 days without C; or if
all three had worked together for the time that B had worked. The
number of days that elapsed before B began to work was to the
number that elapsed before C began to work as 3 to 5 : find the
number of days that each man worked.
292. Shew that if >S' r is the sum of the products r together of
l
then o H _ r = /6 ( . . .'
[St John's Coll. Camb.]
293. If a, b, c are positive and the sum of any two greater than
the third, prove that
'^'T(»t)'('*""i'«'
[St John's Coll. Camb.]
294. Resolve into factors
(a + b +c) (6+ e  a) (c + a  b) {a + bc) (a 2 + ¥ + c 8 )  8a 2 bc 2 .
Prove that
4{a 4 + /y 4 + y 1 + (a + ^ry) t }=(/3 + y) 4 + (y + a) l + (a + i3) 4
+ 6(^ + y) 2 (y + a) 2 + 6(y + a) 2 (a + /i{) 2 + 6(a + ^(^ + y).
[Jesus Coll. Camb.]
295. Prove that the sum of the homogeneous products of r dimen
sions of the numbers 1, 2, 3, ... w, and their powers is
L^L.M_^, 2 ,,.^(^;)(;^3^M,,,, t0ittcn J
[Emm. Coll. Camb.]
296. Prove that, if n be a positive integer
ifc+»?yq* , <» t r?? t ~ B) +""»(»>'
[Oxford Mods.]
297. If x(2a .»/)=.?/ (2az) = z (2au)=n (2a#) = 6*, shew that
x=y = z = u unless o 2 =2a 2 , and that if this condition is satisfied the
equations are not independent. [Math. Tkipos.]
524 HIGHER ALGEBRA.
298. Shew that if a, b, c are positive and unequal, the equations
ax+yz + z = 0, zx+by+z=Q, yz+zx+c=0,
give three distinct triads of real values for x, y, z ; and the ratio of the
products of the three values of x and y is b (b  c) : a {c  a).
[Oxford Mods.]
299. If A = ax bycz, D=bz + cy,
B=bycz ax, E= ex + az,
C = cz ax  by, F= ay + bx,
prove that ABC AD 2  BE?  CF 2 + 2DEF
= (a 2 + b 2 + c 2 ) (ax + by + cz) (x 2 +y 2 + z 2 ).
[Second Public Exam. Oxford.]
300. A certain student found it necessary to decipher an old
manuscript. During previous experiences of the same kind he had
observed that the number of words he could read daily varied jointly
as the number of miles he walked and the number of hours he worked
during the day. He therefore gradually increased the amount of daily
exercise and daily work at the rate of 1 mile and 1 hour per day
respectively, beginning the first day with his usual quantity. He found
that the manuscript contained 232000 words, that he counted 12000
on the first day, and 72000 on the last day ; and that by the end of half
the time he had counted 62000 words : find his usual amount of daily
exercise and work.
ANSWERS.
I. Paces 10—12.
1. (1) 546 : a. (2) 9 : 7. (3) bx : ay. 2. 18. 3. 385, 600.
4 11, 5. 5 : 13. 6. 5 : 6 or  3 : 5.
10. * = y = *,or^± i = Z Q. 17. abc+2fghaf*bg*cti>=0.
20. 3,4,1. 21. 3,4,1. 22. 7,3,2. 23. 3,4,1.
25. ± a (6 2  c 2 ), ± b (e 2  a 2 ), ± c (a 2  ft 2 ).
26. be(bc) t ca(ca),ab(ab).
II. Pages 19, 20.
z 3
' i 45 2. (1) 12. (2) 300a 3 6. 3. . , «,. .
13 5 14. 0, 3, 8. 15. — . 7T— •
li. u, o, . « j > cm bm 2aii
18 8 19 . g, 9, 10, 15. 20. 3 gallons from A ; 8 gallons from B.
21. 45 gallons. 23. 17:3. 24. a = ±b.
25. 64 per cent, copper and 36 per cent. zinc. 3 parts of brass are taken to
5 parts of bronze. 26. 63 or 12 minutes.
III. Packs
20,
27.
1.
5*. 2. 9.
3. lfc.
4.
2.
7. 60.
9.
„ 8
y = 2x — .
3D
r 3G
10. y = 5x + $
•
11.
4.
12.
22 2
1 = ^2 + .— .
15 loz
14. 36.
15.
1610 feet; 3059
16.
224^ cubic feet.
17.
4
:3.
18. Tbe regatta lasted 6 days; 4 lh , 5 th , 6 th days.
20. 16, 25 years; £200, £250. 21. 1 day 18 hours 28 minutes.
22. The cost is least when the rate is 12 miles an hour; and then the cost
per mile is £ : /ly, and for the journey is £9. 7*. <><'•
1. 277£. 2. 153. 3. 0. 4. v ' . 5. 30.
o
526 HIGHER ALGEBRA.
IV. a. Pages 31, 32.
2. 153. 3. 0. 4. n ( 10 *>
6. 42. 7. 185. 8. 1325^/3. 9. 75^/5.
21
10. 820a 16806. 11. n {n + l)an 2 b. 12. —(11a 96).
a
1 3
•13. j, ,..., 9. 14. 1, l£,...,39. 15. 33a;, 31z, ...,a%
16. z 2 a:+l, a; 2 2a; + 2, ..., x. 17. ?i 2 . 18. 3. 19. 5.
20. 612. 21. 4, 9, 14. 22. 1, 4, 7. 23. 495. 24. 160.
25. ^ ( o +1 Uff5. 26. n(»+l)a.
2a x a
IV. b. Pages 35, 36.
1. 10 or 8. 2. 8 or 13. 3. 2,5,8,...
4. First term 8, number of terms 59.
5. First term 1\, number of terms 54.
6. Instalments £51, £53, £55,... 7. 12. 8. 25.
fi
9 " 2(l.r) (2 + rc~ 8 •^ r) • 10# %2 * 12, te + 2)
13. 3, 5, 7, 9. [Assume for tbe numbers a  M, ad, a + d, a + 3d.]
14. 2,4,6,8. 15. p + qvi. 16. 12 or  17. 17. 6rl.
20. 10p8. 21. 8 terms. Series 1, 3, 4^,.
22. 3, 5, 7; 4,5,6. 23. ry = (n + lr)x.
V. a. Pages 41, 42.
1.
2059 1281
1458' 2 ' 512 '
3.
191J.
4.
682.
5.
1Q 93 1 ttfm ,
45 ' 6  I^^'
7.
?i*G)T
8.
364(^/3 + 1).
9.
^(585^2292).
10.
463
192*
11.
» 1?
2' '3'
12.
y,8,...,27. 13. 7,g,...
7
' 32'
14.
64
65*
15.
27
~. 16. 999.
08
17.
1
2'
18.
3(3 + ^/3)
2
19.
7(7 + N /42). 20. 2.
21.
16,24,36,...
22.
2.
23.
2. 24. 8, 12, 18.
25.
2, 6, 18.
28.
6, 3,1J,
ANSWERS. 527
V. b. Pages 46, 40.
1g" mi" 8 1 H
* (la)2 1a' Zl 3' 3  (i
9  (lr)flH ' 10> 10 ' 2ft ' 10  ^ ^i'"
s(.r»l) n(n+l) a . r=(.r 5 »l) a 7/ (x"//" 1 )
*l 2 xl xy1 *
2 / l'\ 23
14. 4/ 2 a + 9 f 1pJ. 15. li. 16. ^r. 19. B.2"+ f 2*« + 2.
2o (i + .)i«yd , 21 . « S'":"" 1 '"!
«c  1 r  1 ( ?•*  1 1
VI. a. Pages 52, 53,
2
11
1. (1) 5. (2) 3*. (3) 3H 2. 0^, 79 3. , J, I
4. Gaud 24. 5. 4:9. 10. >r (» 1 1).
11. ^w(?t + l)(?i 2 + ?i + 3). 12. gn(n+l)(2»+7).
13. n(n + l){n 2 + 3/t + l). 14. J (3 ,l + 1 + 1)  2' 1 + 1 .
15. 4" +1 4w(tt + l)(n 2 «l).
18. The n th term = fr + c (2n — 1), for all values of n greater than 1. Tho first
term is a + b + c; the other terms form the A. P. b + 3c, b f 5c, fr + 7c,....
22. I (2a + n^ld) la* + (n  1) ad +  ( " 1} d* [ .
VI. b. Page 56.
19. v\
1. 12,0. . 2. 1140. 3. 16646. 4. 2170. 5. 21321.
6. 52. 7. 11879. 8. 1840. 9. 11940. 10. 190.
11. 300. 12. 18296. 14. Triangular 364; Square 4900.
15. 120. 16. n1.
VII. a. Tack 59.
1. 333244. 2. 728G26. 3. 1710137. 4. *7074. 5. 112022.
6. 334345. 7. 1783212G. 8. 1625. 9. 2012. 10. 842.
11. M90001. 12. 231. 13. 1456. 14. 7071. 15. 
16. (1) 121. (2) 122000.
VII. b. Pages 65, 66.
1. 20305. 2. 4444. 3. 11001110. 4. 2000000. 5. i
6. 34402. 7. 6587. 8. 8978. 9. 26011. 10. 87214.
528 HIGHER ALGEBRA.
11. 30034342. 12. 710^3. 13. 2714687. 14. 2046. 15. 151*6.
5 2 5
16. 2073. 17. 1250125. 18. = . 19. 5 , 6 .
o o o
20. Nine. 21. Four. 22. Twelve. 23. Eight. 24. Eleven.
25. Twelve. 26. Ten. 30. 2 n + 2 7 + 2 6 .
31. 3 9 3 8 3 7 3 6 3 5 + 3 3 + 3 2 + l.
VIII. a. Pages 72, 73.
2 + ^/2 + ^/6 3 + ^6 + ^15
l. 4 z. 6 .
a*/ 6 + & \/ a  \/ a6 (« + 6) . a  1 + y/a 2  1 + N /2a (a  1)
3 * ~~ ~~2^ "' a1 *
3^/30 + 5^/15  12  10^/2 ^2 + ^/3 + „/5
5. ^ • 6. g .
5 4 1 2 3 1 5
7. 33 + 33 . 2^+3 . 2 + 3 3 . 22 + 3 :1 . 2 2 + 22.
541322 145
8. 56  56 . 23+ 5 6 . 23 56 . 2 + 56 . 2 3  2 3 ".
11 10 19 1 110 11 21
9. a 6 a 6 6*+a% 2 "...+a«6*6 4 . 10. 3*+3»+l.
1 13
11. 2 3 2 2 .7 4 +2.7 2 "7 4 ~.
11 10 1 2 1 10 11 21
12. 53+53 . 3 4 + 5 3 . 3 4 +. ..+53.34 +3*. 133 + 33
5 3 4 5 2 17 2
14. 1733.22 + 33.2 2 3. 22 + 33.2333.22.
15 4 3 2 5 1
15. 32 . 22 33 . 2 + 33 . 22 3 . 22 + 33 . 22 33 . 23.
S&4 33 1\ 31262116 11
16 . i(36_36 + 36_ 3 6 +3 6_ 1 \ 17# 2 5 +26+2 6 +26+26+2 6 + l
2 \ J gj .
3 5 1
18. 32 + 8 Q 6 + 86 . 19. v/5 + ^72.
o
20. V 5 \/7 + 2^3. 21. 1+^/3^/2. 22. * + aA " 4/^ •
23. 2 + JaJSb. 24. 3  ^7 + ^2  <J3. 25. 1 + ^/3.
26. 2 + ^/5. 27. 32^/2. 28. JU2J2.
29. 2^/3 + ^/5. 30. 3^/3^/6. 31. a/^^+a/
35. ll + 56 x /3. 36. 289. 37. 5v / 3 
ANSWERS. 529
38. 3^3 + 5. 39. 3. 40. 8^3.
41. 3 + ^/5 = 523007. 42. a^ + l+^+a x^/2 s 2.
43. Sa+Jlr^^r.
1. 02^/6.
4. .i  x + 1 .
_8_
7  29*
10.
3a 2  1
44.
1
2 '
VIII. b. PAGEfi
HI,
82.
2.  13.
3.
# .'V 1 C 2V/1.
5. 3 + ^ rT2 .
11
6.
196«/10.
4a.r \/  1
8  a 2 + x 2 '
9.
2 (3^  1 ) N / =T
.r + 1
11. Jl.
12.
100.
2a
13. ±(2 + 3 N / 3 l). 14. ±(56 V^T). 15. ±(1 + 4 J^Z).
16. ±2(1 >/^T). 17  ± (« + ^" = T). 18 ± {(« + /;)(,/ fc) ^"j.
9 19 . rtrt 4 ^0 , M .
19  1S + 11 1  2 ° 714'' 21 ' '•
1 3. 26 (3a 2 6 s ).
22. ^ + E J. 23. iTTT *•
5 5 a z + b
IX. a. Pages 88—90.
1. 35x 2 + 13.r12 = 0. 2. MiH.r' J + (u 2 wr) .c  mn = 0.
3. (;> 2 fl 2 ).r 2 + 4pga p* + q = Q. 4. x a  14.r + 29 = 0.
5. .r 2 + 10.r + 13 = 0. 6. x + 2px+p8<j = 0.
7. .r 2 + G.r + 34 = 0. 8. .r + 2(ix + a + b = 0.
9. x" + a 2  ,ab + b = 0. 10. 0.r ! + 11. c  19r + = 0.
11. 2ax s + {4a)x 2 2ax = 0. 12. .r J  8.r 2 + 17.r  4 = 0.
10 a  6
14. 3,0. 15. 2, . 16. — >.
fr 2  2ac 6c« (3ac  6") &*(&*4ac)
18, ~^— • 19  o7 • 20  ~aV •
21. 7. 22. 15. 23. 0. 24. x*  2 (p*  2q) x +p* (p*  l</)=0.
26. 1 ...,. (2) v . .  L . 27. «&= 1 + n)*ac.
28. aW  (6>  2ae) (a a + c) .c + (/, 2  9ac) 8 = < ».
29. x hinix{mn) = 0.
IX.
b.
Pages
92, 9a
1.
•J and
(it "
2.
4ff) (P a 
2
</)
5. b.r
(2) y
•_\/., f a = 0.
'4yA/ + 2 r
11.
1
6.
t
H. Ji. A. 34
530 HIGHER ALGEBRA.
IX. c. Page 96.
1. 2. 2. ±7. 5. (ln'l'nf=(lm'l'm)(mn'm'n).
7. (aa'  bb'f + 4 (ha' + Jib) (Jib' + Ji'a) = 0.
10. \bb'  2ac'  2a' c) 2 = (b 2  4ac) (b' 2  4a'c') ; which reduces to
(ac'  a'c) 2 = (ab'  a'b) (be'  b'c).
X. a. Pages 101, 102.
1.
1
4'
1
"2
5.
3,
2".
1
25
9.
9'
4
2.
± ^' ±L
3.
*l
4.
4 1
9' 4 
6.
1, 2 2 \
7.
25
27 ' 147 *
8.
9 4
13' 13
10.
1  1
' 32'
11.
2, 0.
12.
±1.
14.
±3.
15.
0.
16.
1,450.
13.  4.
17. 9, 7, 1± N / :I 24 18 2  ~ 4 > 1±*/71. 19 3 > "g' — X" ~
20. 4, J, y^». 2 i. 2, 8, 3 ±3x /5. 22. 3, §, 2 ^°.
23.
e 1 Si^/148
°'3' 3 *
24.
7 14 7± N
'' 3 ' 6
/37
•
25.
1 5± N / 201
A 2' 4
26.
K 7 8±V 415
'3' 6 '
27.
1, 3.
28.
5 1
5 '2*
29.
1 9  1 *
30.
a a
a, g. g
31.
2 ?
A 2*
32.
4 *?
' 3 '
33.
0, 5.
34.
6 ' "2*
35.
3 ±x /5
' 2 *
36.
o 1 ~W
d > 3'
^35
37.
WC 1 ^
38.
2 * 5 1
39.
3a, 4a.
40.
2a
41.
0, 1, 3.
42.
1±V!7
2

w
2
•
2
43.
3 2
2' 3*
44.
3, 1.
45.
±1. •
46.
13.
47.
4.
48.
o 63a
U ' 65 *
49.
' + ^) 2 4*
50.
±5.
51.
l±V75
D ' " *' 2
52.
1 li^31
3' 6
ANSWERS. 531
X. b. PAGES L06, 107.
8 15 „ 8 97
1. *=5, p y=4, — . 2. x = 2, _■ ,, .7,  vy
3. *=1,  — ; y = l, ^. 4. .i= ±o, ±3 ; y = ±3, ±5.
5. .r = 8, 2; y = 2, 8. 6. .r = 45, 5; y = 5, ]:,.
7. x=9, 4; y=4, 9. 8. x = ±2, ±3; 7/= ±1, ±2.
9. .r= ±2, ±3; ?/= ±3, ±4. 10. x=±5, ±3; y=±3, ±J.
11. s=±2, ±1; y=±l, ±3.
12. ^±^,±^^=0,^6^/1.
13. ar=5, 3, 4± v /^ 7 ; !/ = 3 > 5 > 4=f>/97'
14. a?=4, 2, ±JS /~15 + l; y=2, 4, ±„/~15l.
15. .r = 4, 2, d= v r ll + l; y=2, 4, ± J 111.
4 1
16. .t'=r> ~;y = 20, 5. 17. as=2, 1; y=l, 2.
o o
18. s=6, 4; y=10, 15. 19. x = 729, 343; y = 343, 729.
20. a; =16, 1; y=l, 16. 21. x = 9, 4; ?/=4, ( J.
5 2
22. # = 5; y =±4. 23. x = l, ~ ; y = 2, .
24. .r = 9, 1; y = l, 9. 25. <r= i25; y = ±9.
26. *=6, 2,4,3; y = l, 3, , 2.
27. x=±5, ±4, ±, ±2; y=±5, ±4, ±10, ±8.
; 107 , 48
28. a.^ — ;^!,.
. 1 + ^143 1± 3^/^143
29. a:=  6, — 5L '> V ~ 3 > 4 — — •
30. .r = 0, 9, 3; y=0, 3, 9. 31. .c = 0, 1, ^ ; y=0, 2, ^.
32. *=5,,0; y=3, ^' ~f ' 33> * = 2, •Vl, 2; T/2, 2/4, 6.
s* *=i.V^;y=2,3^/i.
35. #=±3, ±^18; y=±3, =f v /18.
36. .r=?/=±2.
37. x=o, > /a , ^ /rt • „ =0 ^* £^
tr a (26 a) _//' ((Sa
fe • ' b ; y ~u ' ~T
34—2
532 HIGHER ALGEBRA.
40. x = 0, ±« x /7, ift^/13, ±3a, ±«; y = 0, ^bs/7, ±6^/13, T &, T 36.
2a 2 „ a
41. #= ± 1, ± . ; ?/ = ± 2a, = . .
,716a 4 a 8 1 V 16 * 4 " 2  1
X. c. Pages 109, 110.
1. x = ±3; ?/ = ±5; z— ±4. 2. # = 5; ?/ =  1; 2 = 7.
3. af=5, 1; y = l f — 5; «=2. 4. # = 8, 3; ?/ = 3; 2 = 3,  8.
. 2±Vi5i _ . 2=7151 11
5. .x = 4, 3, ^ ; y = S, 4, 1 ; 2 = 2,  — .
6. x = ±3; ?/ = t2; 2= i5. 7. jr= ±5; y= ±1; «== ±1.
8. # = 8, 8; y = 5, 5; 2 = 3, 3. 9. .c = 3; ?/ = 4; 2 = ; w = .
a o
10. *=1; ?/ = 2; 2 = 3. 11. .r = 5, 7; y = 3, 5; 2 = 6, 8.
11
3 '
12. .t=1, 2; ?/ = 7, 3; 2 = 3,
13. # = 4, — ; */ = 6, —; 2 = 2, 6. 14. .r = a, 0, 0; y = 0, a, 0; 2 = 0, 0, a.
^^3' " 3 ' a 
16. «=a, 2a, ^ a; y = ±a, a,  ^ a;
2 = 2a, 4a, (l± v /15)
a.
X. d. Page 113.
1. z = 29, 21, 13, 5; y = 2, 5, 8, 11.
2. a?=l, 3, 5, 7, 9; 2/ = 24, 19, 14, 9, 4.
3. ar20, 8; y = l, 8. 4. a?=9, 20, 31; y = 27, 14, 1.
5. # = 30, 5; ?/=9, 32. 6. .t = 50, 3; y = 3, 44.
7. x=7p5, 2; y=5p4, 1. 8. s=l$p2, 11; y=6pl, 5.
9. .t = 21^9, 12; y = 8p5, 3. 10. £ = 17/), 17; ?/ = 13^, 13.
11. x = 19pW, 3; ?/ = 23^19, 4. 12. x = llp 74, 3; y = 30p25, 5.
13. 11 horses, 15 cows. 14. 101. 15. 56, 25 or 16, 65.
16. To pay 3 guineas and receive 21 halfcrowns.
17. 1147 ; an infinite number of the form 1147 + 39 x 56p.
18. To pay 17 florins and receive 3 halfcrowns.
19. 37,99; 77,59; 117,19.
20. 28 rams, 1 pig, 11 oxen; or 13 rams, 14 pigs, 13 oxen.
21. 3 sovereigns, 11 halfcrowns, 13 shillings.
ANSWERS.
533
XI. a. Tacks 122—124.
1. 12.
4. 6720.
8. 6.
12. 1440.
2. 221.
5. 15.
9. 120.
13. 6375G00.
3. 40320, 0375600, 10626, 11628.
6. 40320; 720. 7. 15, 860.
16. 1140, 231. 17. 144.
20. 56. 21. 360000.
24. 21000. 25. yJ^p.
29. 2903040.
28. 9466.
33. 1956.
10. 720.
14. 360, 144.
18. 224, 896.
22. 2052000.
26. 2520.
30. 25920.
11. 10626, 1771.
15. 230300.
19. 848.
23. 3(19600.
27. 5700.
32. 41.
34. 7.
XL b. Pages 131, 132.
1. (1) 1663200. (2) 129729000. (3) 3326400.
3. 151351200. 4. 360. 5. 72.
7. n r . 8. 531441. 9. V n .
11. 1260.
15. 4095.
19. 127.
12. 3374.
16. 57760000.
20. 315. 21.
13. 455.
14.
2. 4084080.
6. 125.
10. 30.
yunrwr '
17. 1023. 18. 720; 3628800.
inn
22. 64; 325. 23. 42.
24. (1) *Ji£l>i<£zi> + l; (2)
{\m) n \ii '
' P(pl){p e <
6
6
3>(gl) (p2) _ g(gl)(g2) , !
6 6
27. 113; 2190. 28. 2454.
hi. 26. (p + l)*l.
29. 6666600. 30. 5199960.
XIII. a. Pages 142, 143.
1. ««  15a; 4 + 90a; 3  270.r 2 + 405a;  243.
2. 81a: 4 + 216a; 3 // + 216.r// 2 + 96.r?/ 3 + 16// 4 .
3. 32.c 5  80x*y + 80.r 3 y°  40.ry 3 + lO.r// 4  if.
4. 1  18rt 2 + 135a 4  540« G + 1215a 8  1458a 10 + 729a M .
5. a; 10 + 5a; 9 + 10a; 8 + 10a; 7 + 5a; 6 + a; 5 .
6. 1  7.iv/ + 2 la; 2 */ 2 ~ 35a; 3 */ 3 + 35a; V  tlxhf + 7«V  x7 'f'
81a; 8
7. 1648a; 2 + 54ar l 27.c 6 +
16
8. 729o«  972a 5 + 540a 4  160a 3 + ^  ^ + ~
Ix 21x 2 35a; 3 35a; 4 21x n 7x G x 7
9 1+ T + ~T~ ~8 _ + l<T + ~32~ + 64 + 128"
534
HIGHER ALGEBRA.
64a; 6 32a; 4 20a
10 * 729~~2~r + ^~
135 _ 243 729
+ 4a; 2 8a; 4 + 64a; 6 '
1 a 7ft 2 7ft 3 35ft 4 n _ _ , . _ _
11. ^T^ + ytv + t? +^T +— 5~ + 7ft 5 + 7ft 6 + 4ft 7 + ft 8 .
2ob lb lb 4 8
, 10 45 120 210 252 210 120 45
12. 1 — + Z 5 + —  r + — ^ +
a'
X'
X
x*
14.
a; 1
x v
13.  35750a: 10 .
130
16. J~(5a;) 3 (8?/) 7 . 17.
19.
;273_
10500
a: 3
 112640a 9 .
40ft 7 6 3 .
20.
70x 6 y
10
22. 2x (16a; 4  20a; 2 ft 2 + 5ft 4 ).
24. 2 (3 65  363a; + 63a; 2 a; 3 ).
25. 252.
27. 110565ft 4 .
30.
33.
189ft 17 21
8
'"I6 a
In
19
28. 84ft 3 & 6 .
7
18*
31.
i(»r)!4(H + r)'
x
:8
15.
18.
21.
23.
26.
29.
32.
,.io *
10
iC 9 x
 312a; 2 .
1120
81 MK
2x 4 + 24a; 2 + 8.
140^2.
^a; 14
1365, 1365.
18564.
34. (  1)»
Bn
»
2/i*
XIII. b. Pages 147, 148.
1. The 9 th . 2. The 12 th . 3. The 6 th . 4. The 10 th and 11*.
5. The3 rd = 6. 6. The 4 th and 5 th =Jrj> 9. x = 2, ?/ = 3, n=5.
10. 1 + 8.r + 20a; 2 + 8a; 3  26a; 4  8.r 5 + 20a; 6  8a; 7 + x\
11. 27a; 6  54«a; 5 + 117ft 2 a; 4  116« 3 .c 3 + 117ft 4 a; 2  54ft 5 a; + 27ft 6 .
12.
n
r1 nr+1
x ri a nr+l.
14. 14.
I2n+1
13. ( l) p  ? , „.,  z*>*"+\
p + 1 2n p
15. 2r = n.
XIV. a. Page 155.
_ 2 3 2 __8_
3> 5* 25* 125
5. 1  x  a; 2   a; 3 .
o
7. laj+ga^g^ 3 
9  1+X+ 6U
x
3 3 „ 1
2. i + 2 a + _a;__a:.
4. l2a; 2 + 3a; 4 4a; 6 .
14
6. l + a; + 2a; 2 + — a; 3 .
o
8. 1 — a; + ;r a;  — — ar .
10. l2ft + ft
«r
ANSWERS. 535
U. 4(l + ... + ..). !4. i(l + * + §*« + «»).
2 a i \ a 2 a 2 </••/ lb 250
1040 ._ 10//
18.  lr a« 19. 2  43rt ,. 20. (r + 1)**
(r + l)(r+2)(r + 3) 1 . 3 . 5 ... (2r3)
21 x 2 3 a?. 22. (1) _ r
23 ( lr _, 11.8 . 5.2 .1 .4,..(3r14)
»' I *J 3 r r *•
10719
24. 1848.1 13 . 25. i^z<\
XIV. b. Pages 161, 162.
1.3.5.7...(2rl) (>• + !) ( r + 2)(r + 3)(r + 4 )
*• [ x > ^y • • p •
3 ^1.2.5... (3r4)^, 4> 2.5.8. ..(8rl)
t
3' r
x r .
x r .
5 /_nrfe±llt±2) r r fi 3.5.7... (2r + l)
*•{*) ^_ *~. 6. 
b r r + 1
2 .1.4...(3r5) ^ 1.3.5... (2,1)
9< 3'lr V>' U " [ } \r
2.5.8... (3rl) r (n + l)(2n + l)... (rl.» + l) .r r
13. The 3**. 14. The 5 th . 15. The 13 th . 16. The 7 ,h .
17. The 4 th and 5* h . 18. The 3 r<1 . 19. 989949.
20. 999333. 21. 10 00999. 22. G 99927. 23. 19842.
24.
100133.
25.
•00795.
26. 50009G.
27. l 6 .
28.
IK>
29.
5x
1 "8"'
1 5
30. ^x.
31  ^MO*'
32.
1 71
3 3G0*'
35. 1
4# + 13:r 2 . 36.
29 297 .
2+ T* + 32^
XIV. c. Patqeb 1G7— 1G9.
1. 197. 2. 142. 3. (1)" 1 .
4. (l)»(n»+2n+2). 6. v/8=(l) .
53G HIGHER ALGEBRA.
/ 2\' v / 1V H 2n
14. Deduced from (1  a; 3 )  (1  #)3 = 3:r  3.r 2 . 16. (1) 45. (2) 6561.
18. (1) Equate coefficients of x r in (1 + x) n (1 + re) 1 = (1 + x) 9 ^ 1 .
(2) Equate absolute terms in {l + x) n f 1 +  ) =z 2 (l + o;) n  2 .
20. Series on the left + (  1)» q n 2 = coefficient of x n in (1  .t 2 )* .
1 2w
21. 2 2 " 1  J
2 ' In ire
[Use (c +c 1 + c 2 + .. .c J 2  2 ( Co c, + c lCs + ...) = c 2 + Cl 2 + c 2 2 +. . .c n 2 ].
XV. Pages 173, 174.
1.
 12600.
2.
168
5.
9.
6.
8085.
9.
10.
10.
3
2*
13.
59
16"
14.
1.
17.
l2a 2 + 4a; 3
+ OX*
 20.r 5 .
3.
3360.
7.
30.
11.
1.
211
15  8
4.
1260a 2 fcV.
8.
1905.
12.
4
81'
16.
1ia,
7 2
— X 1 ,
8
18. w(l^x* + 3x* + ^x«~x? + ^xA.
XVI. a. Pages 178, 179.
1. 8,6. 2. 2,1. 3. ^, i. 4. 4, .
_4_4 2 157.42
3' 5* b  5' 2' ~2* 7 3' " ' "3' 3'
2 3
8. 61oga + 91ogb. 9. loga + logb.
4 1 2 1
10. loga + logfc. 11. glogalog&.
7 1
12. —log a log b. 13. 2^°8 fl  14  51ogc. 16. log 3.
logc . 19 . 51 °^
log a  log b ' 2 log a + 3 log & *
log a + log b 41ogm logw
^ 21ogcloga + log6* ' X ~ \oga' J log 6 '
1 1 log (a  &)
22. loga; = g(a + 36),log?7 = g(a26). 24. log(fl + b y
answers. 537
XVI. b. Pages 185, 186.
1. 4,1,2,2,1,1,1.
2. 8821259, 28821259, 38821259, 58821259, 68821259.
3. 5, 2, 4, 1.
4. Second decimal place ; units' place ; fifth decimal place.
5. 18061800. 6. 19242793. 7. 11072100. 8. 2*0969100.
9. 11583626. 10. 6690067. 11. 3597271. 12. 0503520.
13. f5052973. 14. 44092388. 15. 1948445. 16. 1915631.
17. 11998692. 18. 10039238. 19. 9076226. 20. 178141516.
21. 9. 23. 301. 24. 346. 25. 429. 26. 1206. 27. 14200.
28. 4562. 29. ._£«». fr  lo S 2
log 3 log 2' J log 3 log 2'
31og321og2 log3 , .,«*,
32. ££««, ?, l0g J = 5614.
2 log 7 log 2
XVII. Pages 195—197.
1. log, 2. 2. log, 3 log, 2. 6. 0020000000666670.
9. e*~cy~. 10. 8450980; 10413927; 11139434. In Art. 225 put
7i = 50 in (2) ; ?* = 10 in (1); and ?i = 1000 in (1) respectively.
12 . ( . lr ..r±i>. 13 . < 1 » r "' 3r + 2r ^.
r r
„ L (2.r) 2 (2x) 4 (2x) r )
14 . 2 ji + ^. + i_X + ... + i_^ + ...{.
/>» /)»4 /y»6 /y»*7* ,*•
H" 1 " 1L~ 11 } E 1^+iog.a*).
24. 69314718; 109861229; 160943792; a= log, (l  M =105360516;
b =  log, ( 1  A^  040821995 ; c = log, f 1 + iA = 012422520.
XVIII. a. Page 202.
1. £1146. 14s. 10J. 2. £720. 3. 142 years.
4. £6768. 7s. 10hd. 5. 96 years. 8. £496. 19*. 4frf.
9. A little less than 7 years. 10. £119. 16s. 4^7.
XVIII. b. Pack 207.
1. 6 percent. 2. £3137. 2s. 2U. 3. £110.
4. 3 per cent. 5. 28J years. 6. £1275. 7. £920. 2s.
8. £6755. 13s. 9. £183. 18s. 10. 3} per cent. 11. £616. 9s. l£d
13. £1308. 12s. 4 U. 15. £4200.
538 HIGHER ALGEBRA.
XIX. a. Pages 213, 214.
8. a 3 + 2& 3 is the greater. 12. x s > or < # 2 + x + 2, according as x > or < 2.
14. The greatest value of x is 1. 15. 4 ; 8.
22. 4 4 . 5 5 ; when x = 3. 23. 9, when #=1.
XIX. b. Pages 218, 219.
, n 3 3 .5 5 . /3 3 /2
XX.
Page
228.
1.
10 9
7 '' 4*
2.
9;
1
9'
3.
1 5
2 ; 3*
4.
15 
8 ;6 '
5.
1;
*
6.
0; 30.
3
7 ' "2
8.
log a  log b.
9.
2.
10.
me ma . 11.
1
2V«*
12.
1
3*
13.
1.
14.
x/2a
aj'd + l'
15. N /a.
16.
0.
17.
3
2*
18.
2
e a .
XXI. a. Pages 241, 242.
1. Convergent. 2. Convergent. 3. Convergent.
4. #<1, or# = l, convergent; x>\> divergent.
5. Same result as Ex. 4. 6. Convergent. 7. Divergent.
8. #<1, convergent; x>l, or x = l, divergent.
9. Divergent except when p > 2.
10. #<1, or x=l, convergent; x>l, divergent.
11. If x < 1, convergent ; x > 1, or x = 1, divergent.
12. SameresultasEx.il. 13. Divergent, except when ^>1.
14. x<l, or x — 1, convergent; .t>1, divergent.
15. Convergent. 16. Divergent.
17. (1) Divergent. (2) Convergent.
18. (1) Divergent. (2) Convergent.
XXI. b. Page 252.
1. x<l, or x=l, convergent; x>l, divergent.
2. Same result as Ex. 1. 3. Same result as Ex. 1.
11 1
4. x <  , or x =  , convergent ; x> , divergent.
e e e
5. x<e, convergent; x>e, or#=e, divergent.
ANSWERS. 5:5!)
5. x<l, convergent; x >1, or x1, divergent. 7. Divergent,
1 1 1 ,.
8. x< y convergent; x>  , or x — , divergent.
e e e
9. x<l, convergent; x>l, divergent. If x = l and if 7~a/3 is positive,
convergent ; if y a  (3 is negative, 01 zero, divergent.
10. .r<l, convergent; .r>l, or x = l, divergent. The results hold for all
values of q, positive or negative.
11. a negative, or zero, convergent; a positive, divergent.
XXII. a. Page 256.
l. n(4n«l). 2. jn(n+l)(n+2)(n+3).
3. in(n+l)(n+2)(3n+5). 4. ?t 2 (2n 2 l).
5. in(w + l)(2/i + l)(3» 2 + 3»il). 6. p 3 ^ 2 .
7. & 3 = 27a 2 <Z, c*=27acP. 8. ad=bf, 4a%0»=8ay.
13. flic + 2/^ft  a/ 2  6jy 3  c/i 2 = 0.
XXII. b. Page 260.
1. l + Zx + 4x 2 + 7x s . 2. l7xx 2 4Zx :i .
1 1 3 „ 1 , 3 5 11 , 21 .
3.  + l x x  +  X 3. 4.  +  X+s  X 2 +mX .,.
5. 1  a X + a (a + 1) x*  (a 3 + 2a 2  1) X 3 .
6. a = l, b = 2. 7. « = 1, &=l, c = 2.
9. The next term is + 00000000000003.
11.
a n
(1  a) (1  a 2 ) (1  a 3 ) (1  a") *
XXIII. Pages 265, 266.
4 5 7 5 4
!• i «r^ — ^ n • *• « i. — ■} ?; • 3.
l3.c 12.T* ' 3.r5 4x+H' l2x 1x'
2 3 4 ,11 8
4. zr +  n ^. 5. 1+
Xl x 2 a: 3' .r 5 (x  1) 5(2.r + 3)
1 1 3
.t  1 a: + 2 " (x + 2) 2 '
17 11 17
7. x2 +
8.
16(s + l) 4(.c+l) 2 1(5 (x 3)*
41.r + 3 15 3.r
X*+l x + 5' ' .r 2 + 2.r5 .r3"
5 7 13
10. , TTl7— 7T,+
(*!)« (.l•l) ;,T (.rlf T Il ,
540 HIGHER ALGEBRA.
1 1 3 3
Xl X + l {x + l) 2 {X + 1)3 {X+l)*'
12 ' BiiWsoW ±Pnr4n*.
13 _il 4 • 1 flll( 1 >'V
ld • 3(1*) 3(2 + *)' 3\ + V~* J '
1 + 3(* + 5) 3(x + 2)' 1 ' 3V5' +1 2^+V
15. L.  1_  ,4 5 {1+ ( l) r  1 2 r +2}^.
l# 1 + a; 12* l v ' '
"■ 8(I^j3(I^) + (r^y !{9r + 8+(l)'2~}V.
"• 4lI^) + 4(l 1 ^p ;4M ( 12 + llr ) a: '
18  Tfx + (TT^2T3i' (V (►+•£?)*
31 — 3a; 1 3 —
19 ' 2la^I) + 2(rT^) ;? ' eVen '2 {{  1)2  3}a;r;r0dd ' h 1 + {  1)2 }XT '
2 3 2
(lsc) d (lx) 2 1sc
j ^ r+2 *"  + 2 c^" 2 )
* j (a  b) (a  c) + (6  c) (6^o) + (ca)(cb) 1 X '
5 2 1 2 ( 5r + 9)
22 ' (2^p2^ + (I^ )2 + r^; f+» W *
23 11) 1 ! * * 1
f9\ _JL J_l_ 1 1
(1a) 2 \l + a n x l + aP+tx 1 + x 1 + ax) '
1 1 ( x x* x n+l
25 <
.r(la•)(l.'c 2 )• * (l.r) 2 (l3 l^ i_ x n+i
XXIV. Page 272.
*' (TT^ 25 (4r+1) ^ 2  l + x2x* > { 1 +( 1 ) r2r ^ r 
3  12* + Use 2 1
5 ' l=^flE?r®5^ + tr + 1 )" r  6 ' 3 + 2nl;(3«l) + 2nl.
7. (2. 31 3. 21)^1 2(13^") 3(12^*)
' 13* 12* *
v ; ' l4x 1Sx
ANSWERS. 541
, l.r» l3».r" l2' l .r"
11. ?/ n 3» n _ 1 + 3» u _oj/ n _3=:0; M»4« n _ 1 +6M w _ a 4ti to _a+«»_4=0.
12. S n =5 aD S, where 2 = smn to infinity beginning with (n + l) th term.
This may easily be shewn to agree with the result in Art. 325.
13. (2n+ 1)3+ (2**+i+l).
XXV. a. Pages 277, 278.
2 13 15 28 323 074
L 1' 6' 7 ' 13' 150' 313*
12 7 9 43 95 JU3
2 ' 2' 5' f7' 22' 105' 232' 1497'
3 10 13 36 85 121 1174
1 ' 3 ' T ' 11 ' 26 ' 37 ' "359 '
1 1 1 1 1 1117
*• + 2 + 2 + 2+ 1 + 1 + 2 + 2 ' 12 "
1 1 1 1 1 157
* D + 4+ 3+ 2+ 1+ 3 ; "30" '
JL _L JL JL J_ JL *• ^
3+ 3+ 3+ 3+ 3+ 3+3' 109"
_1_JLJ^JL_! Ill 11
7 ' 3T 5+ IT 1+ 3+~ 2+ 1+ 5 ; 35"
J__l_ J_ J_ _1_1. _7 J_J^AJl_ X ?5i
2+ 1+ 2+ 2+ 1+ 3 ; 19" 1 + 7+ 5+ 6+ 1+ 3 5 223'
6.
10.
11111111 G3
3+ 3+ 3+ 6+ 1+ 2+ 1+ 10' 208*
J__l_l 259 1 7 8 39 47
n ' 4 + 3+ o+ 3' GO* i' 29' 33' 161' 194*
16. n  1 + 7 r — , jt = ; and the first three convergents are
(n + l) + (nl)+ 7i + l
//  1 n 2 ?j 3  n 2 + n  1
~T~' n+~l' ~n*~ "'
XXV. b. Packs 281—283.
1 j 1 151
and
4.
(203) 2 * "2(1250)** " 115'
11 11 a*+3a+3
a+ (« + !)+ (a +2) + a + 3' a a + 3a + la + 2 '
542 HIGHER ALGEBRA.
XXVI. Pages 290, 291.
1. a = 7m + 100, 2/ = 775f + 109; a=100, y = 109.
2. x = 519t73, ?/ = 455t64; a=446, y = S91.
3. x = 3934 + 320, ?/ = 436t + 355; x=320, ?/ = 355.
5 4
4. Four. 5. Seven. 6. , .
^3117 \ L ^ 1_
12' 8*' 12' 8 ; ° r 8' 12 ; 8' 12*
8. £6. 13s. 9. x = 9, y = S, z = S. 10. x = 5, y = 6, 2 = 7.
11. x = 4, y=2, z = l. 12. as=2, ?/ = 9, s=7.
13. <c = 3, 7, 2, 6, 1; y = ll, 4, 8, 1, 5; 2 = 1, 1, 2, 2, 3.
14. aj=l, 3, 2; y = 5, 1, 3; z = 2, 4, 3.
15. 280« + 93. 16. 181,412.
17. Denary 248, Septenary 503, Nonary 305.
18. a=ll, 10,9, 8, 6,4, 3; 6 = 66, 30, 18, 12, 6,3, 2. v
19. The 107 th and 104 th divisions, reckoning from either end.
20. 50, 41, 35 times, excluding the first time.
21. 425. 22. 899. 23. 1829 and 1363.
XXVII. a. Pages 294, 295.
1 1 26 „ 1 2889
1+2+ ••"15* n 4+"" , 1292"
" 1 1 485 1 1 99
3 ' J + 2T4T' ; 198 4  2 + IT4+  ; 35"
11 3970 J. J_ 1 1 J_ 119
5  d + 3+"6+" ; 1197 ' 6 ' + 1+ 1+ 1+ 1+ 6+" ; 33
1111 116
7. 3f
1+ 2+ 1+ 6+ ""' 31
_L J_ JL J_ JL J_ . 197
+ 1+ 2+ 4+ 2+ 1+ 8+ "'•' 42 *
_1_ J_ 1351 JL_ J_ JL 1_ 198
9  d + 2+ 6+ '•• ; "390"' 1+ 1+ 1+ 10+  ; 35 *
111111 161
11. 6
12. 12 +
1
13.
1+ 2+ 2+ 2+ 1+ 12+ ' 21
11111111 253
1+ 1+ 1+ 5+ 1+ 1+ 1+ 24+ ' 20 '
111111 12
4+ 1+ 1+ 2+ 1+ 1+ 8+ "" 55*
_1_ Jj_ J_ _1 1_ . 47_ _1 1_ > 5291
' 5+ 1+ 2+ 1+ 10+ '"•' 270' 10+ 2+ '"' 4830*
1111111111 280
ie .
±D ' 1+ 3+ 1+ 16+ 1+ 3+ 2+ 3+ 1+ 16+ ■'•' 351*
ANSWERS "»4.*>
4030 1(577 1 I 1
9 ' 401 * 20  483 * 21, 2 + 2+2 + ""
( 1111 111
22. 4 + r— —  — —  ... 23. 1 + —
1+ 1+ 1+ 4+ ' 2 + 3+ 1+ "
„„ a 1 ! 11111
24 ' 4+ 3+3T ; rT2+8+3+3+ • 25 ' ^
26. Positive root of x + 3« 3 = 0. 27. Positive root of 3x 2  lOx 4 = 0.
28. 4^/2. 30. .
a
1. a +
XXVII. b. Pages 301, 302.
1 1 1 8a 4 + 8a 2 + l
2a + 2a+ 2a+ ""' 8a :, + 4a
J. 1_ 1 1 8a 2 8a + l
2 * * + 2T 2(al)T 2+ 2(al)+ * ,; 8a4
, 1 1 1 1 2a 1
3. al + .
4. 1 + 
6. a +
1+ 2(al)+ 1+ 2(al)+ "■' 2a
1111 8a 2 + 8a + l
2a+ 2+ 2a + 2+ "' ' 8a 2 + 4a '
1111 2a 2 6 2 + 4a& + l
6. alt
7.
b+ 2a + b + 2a + ' 2a&*+26 '
1111 2a/il
1+ 2(nl) + 1+ 2(al)+ "" ' " 2/i
432a 5 + 180a 3 + 15a
141a 4 + 36a 2 +1
XXVIII. Page 311.
l. s=7 or 1, 2/ = 4; s=7 or 5, y=6. 2. # = 2, y = l.
3. x=3, y=l, 11; *=7, y=9, 19j x = 10, y = 18, 22.
4. x=2, 3, 6, 11; y=12, 7, 4,3. 5. x = 3, 2; y=l, 4.
6. x = 79, 27, 17, 13, 11, 9; y=157, 61, 29, 19, 13, 3.
7. x = 15, ?/ = 4. 8. x = 170, y = 39.
9. x=32, y=5. 10. x = lG4, y = 21. 11. x=4, y = l.
12. 2.r = (2+ x /3) n +(2 v /3)»; 2 V '3 . y = {2 + s /3) n  (2  v /3)»; /t being any
integer.
13. 2x = (2 + v /5) n +(2^/5)' 1 ; 2^/5 . ^ = (2 + ^5)" (2  v /5)»; n being uny
even positive integer.
14. 2x = (4 + v /17) n +(4 v /17) n ; 2 V /17. ?/ = (4 + x /17)»  (4  v /17)»; n being
any odd positive integer.
The form of the answers to 15 — 17, 19, 20 will vary according to the
mode of factorising the two sides of the equation.
544 HIGHER ALGEBRA.
15. x = 11 fi  3/t 2 , y = m?  2mn. 16. x =  m 2 + 2mn + n~; y = m 2  ?i 2 .
17. x=2mn, y = 5m?n*. 18. 53, 52; 19, 16; 13, 8; 11, 4.
19. mn\ 2mn; m 2 + w 2 . 20. m 2 n 2 j 2mn + n' 2 .
21. Hendriek, Anna ; Claas, Catriin; Cornelius, Geertruij.
XXIX. a. Pages 321, 322.
1
1. ^n(n + l)(n + 2)(ti + S). 2.  n (n+ 1) (n +2) (« + 3) (/t + 4).
3. — (3>i  2) (3n + 1) (3/i + 4) (3n + 7) + ^ = ~ {21 n* + 90/i 2 + 45u  50) .
71 11
4. ( n + l)(n + 6)(/i + 7). 5. (n + l)(n + 8)(n+9).
fc n , n 1
6. ; 1. 7.
n+1' ' 3;i+l' 3*
o 1 l l « 1 1
8. t^  tt^ =^^ ^r ; ,~. 9. ^T 
12 4(2/1 + 1) (2/7 + 3)' 12" ' 24 6(3/i + l) (3// + 4) ' 24*
5 2/1 + 5 5 11 2 1
'4 2 (7i + l) (?? + 2) 5 4* 6 h + 3 + (u + 3)(h + 4) ; 6*
12 ' SnT2 + a(n + l)(i» + 2) ; i' ^ j^+l><« + >H« + «>(* + «>'
1 n
14. ?i 2 (n 2 l). 15. j (»l)(w + l)(n + 2)(2n+l).
16. — (n + 1) (n + 2) (3/r + 36n 2 + 151n+ 240)  32.
1? (nl)«(n+l)(n+2) ig »(n+l)(n+2) n
6(2n+l) 3 n+1"
n(/t + 3) , 3 2 1 1
19, ~^~ + 2~^2~ (n+l)(n + Z)' 20  H + 1 7l+~r
XXIX. b. Pages 332, 333.
1. 3» 2 + 7i; ?i(7i + l) 2 . 2. 5/t 2 + 3/7;  n (n + 1) (5/7 + 7).
o
3. 7t 2 (/i + l); — «(n+l)(n+2)(3n+l).
4. 4ra a (n3); 77 (/7+1) (n 2  3u  2).
5. rc(»+l)(n+2)(n+4); ^n(n+l) (n+2)(n+3)(4»+21).
l + .r 2 la; + 6.x 2 2.r 3 2.r + .r 2
(l.r) 3 * 7  (la;)3 * 8  (I*; 3 '
_ 1aJ 1 + lLc + lla^+g 3 9
(1+.t) (lz) 4
12. gj. 13. 3.2» + /7 + 2; 6(2"l) + W ^ + 5 ^ .
ANSWERS.
545
14. n«(n+l) s ; ^(Sn»+2n*15n26). 15. S*i + n; 3 *+ n ~+" ]
16. 2»+>n s 2tt; 2»+ 2 4 t»(u+1)(2«+7).
17. 31 + 1 n (n +g) . 1 (3 „ +1 _ 8) + »(» + l)(. + 5) _ „
18.
1  x n nx 11
20.
22.
(1  a;) 2 1  x
1 1
24.
n+1 ' 2 n '
n (n + 1) (3k 3 + 27k 2 + 58n + 2)
15
k(k + 1)(9h 2 + 13k + 8)
12
26. 1
2»t+i
k + 2'
28. (nl)3» +1 +3.
n
30.
n + 1
2».
1 n + 1
32> 2""^T2
19.
1  x n >i.r n )i (n + 1) x n
(1  xf " (1*)" " 2(1 a?)
n  1 4 n + 1 2
21 ' 7TT2~^~ + 3
23.
n (k + 1) (12k 3 + 33n a + 37n + 8)
60
1 1
25.   
2 2'1.3.5.7 (2k + 1)'
27. (k 2 k + 4)2"4.
29 * 135 (2/t+l )
2 2.4.6 (2k + 2)"
31. J l J
33. 1
4 2(k+1)(k + 2) *3"'
k + 4 1
(K + l)(K+2) ' 2 n + 1 '
XXIX. c. Pages 338—340.
1.  (e x  e~ x )  x.
(e x  c~ x
5. (l + x)e x .
8. k(2«1).
11. lo g< 2£,
ie ix + ie~ ix ).
2.
4.
6 {p + q) r
II
9. 0.
12. 3(<3l).
1+ — log(lar).
1
(r2)rT
7. 1.
10. 4.
13. e*log(l + .r).
n 6 ?i 5 n 3
... ?l' 71° ?f 71* K
14 ' < X > 7 + 2 + 26 + 42'
n 8 ?i 7 7k b 7k 4 ir
12 24 + 12
(J)
15. lot'. 17. (1) n + 1.
x 2 \ 1 + k + n 2 J v ' k + 1
20. (1 + ' )2 loa+.T) 3 ^. 21. »(nfl)2»~».
4.r
22  W 3 I 1 + 2»+i + (!)«+'}• <) 5l2 + (_1) (n+l)(n+2)/'
H. H. A.
35
54G HIGHER ALGEBRA.
XXX. a. Pages 348, 349.
1. 3, G, 15, 42. 2. 1617, 180, 1859. 6. 18.
7. 23. 33. 8987.
:. b. Pages 356—358.
20. x = 139t + (jl, where t is an integer.
XXXI. a. Pages 367—369.
2. 1 h —  . 18. 1 ; it can be shewn that q n =l +j> n 
XXXII. a. Pages 376, 377.
5. 2 to 3.
2197
10.
16.
20825 '
11
2.
8
663"
3.
1
56*
4.
3
8*
6.
4
8.
43 to 34.
9.
36 : 30 : 25
270725 '
1.
7
952 to 71
n [n
5.
1)
14.
1
6"
15.
2
7 '
4165* (m + 7i) (m+n1)'
XXXII. b. Pages 383, 384.
1.
5
36'
2.
16 52
5525' 77*
4.
16
21*
*i
6.
72
289 "
7.
2197 . 2816
1 ' 20825' [ ' 4105'
8.
4651
7776"
209
9 * 343
10.
1
7*
11.
91 10
216* 13 ' 19'
14.
63
256*
1
15 ' 32*
16.
16 12
37' 37'
9
37*
22 13
17> 35' 35"
•
18. n 
 3 to 2.
19.
13 to 5.
45927
20 ' 50000 '
XXXII. c. Pages 389, 390
1.
2133
3125'
2.
5
16'
3.
4
9"
6.
17s. 2d.
o
7.
4
63*
8.
7
27'
™ . 1
4. Florins. 5. .
me 1
9. 11 to o. 10. b .
o
ANSWERS.
.j 47
11. A £5; B £11.
MX 250 ... 276
14. (1) 7770 ; (2) w
12.
20
27
15. 4a".
16.
13. I, 1 , shillings.
17. 31+ in.
2
XXXII. d. Pages 309, 400.
1.
"1
5'
2.
1
35"
12
3. 1? .
5.
2
I)"
6.
32
11
7 " 15'
n (j
i+
10.
1
8'
11.
40
41"
11
12 ' 50*
15.
£R.
17.
n1
n1
2 4
*■ L '5 ; a iV
8. 2s. 3d. 9.  .
o
13. £1. 14. (1) \ ; (2) 1
vm  1 ' mn  tu  1
18.
13
14
1. 7 to 5.
XXXn. e. Pages 405—408.
12393
1
126 '
6.
9.
14.
: 6 : yy : w : \q) '
13
28'
_1 1
168' 126*
20. One guinea.
28. 7 .
4
32. If &>, the chance is 13
10.
16.
22.
29.
343
1695'
25
216'
140
141'
1
4*
3.
7.
12500 '
16
21*
5.
275
504'
11. 11 to 5.
149
8. 6 ; each equal to ., .
13 A m B *5
13. il, 324 , 1>, 324 ,
17.
23.
30.
18.
33
2401"
n (n+1)
1265 . £ 5087
1286' 5144'
1000' 60*
shillings. 26. 15 to 1.
31.
If &<„, the chance is
(47'
( S W
XXXIII. a. Pages 419, 420, 421.
1. 7. 2. 0.
5. l+x 2 + y" + z".
3. 1.
6. xy.
10. 3.
20.
11. 3atoa 3 Z/ 3 c 3 = 0.
4. a be + 2fgh  af"  % 2  ch.
0. 8. 4a6c. 9. 0.
13. (1) x = a,orb; (2) .t = 4.
6 2 + c 2
ca
ab
c n ~ + a*
cb
ac
be
22. \ ! (\ 2 + a 2 + & 2 lc 2 )*.
35—2
548
HIGHER ALGEBRA.
26. The determinant is equal to
27.
u w' v' =0.
w v
v' v!
w
10
a?
a
1
X
1
2x
X 2
.
b 2
b
1
1
2y
y 2
c 2
c
1
1
2z
z 2
28.
u w'
v'
h
u
10'
v'
a
B
to' V
u'
w
' V
u'
b
v' u'
10
v'
u'
tv
c
a
b
c
XXXIII. b. Pages 427, 428.
1. 1. 2. ; add first and second rows, third and fourth rows.
3. (a + 3)(al) 3 . 4. a 2 + b 2 + c 2  2bc  2ca  2ab.
6. 6 ; from the first column subtract three times the third, from the second
subtract twice the third, and from the fourth subtract four times
the third.
_ _/, 1 1 1 1\
6. abcd(l +  + T +  + ).
\ a b c a)
7. (x + y + z) (y + z  x) (z + x  y) (x + y  z).
9. a'. 12. .„»»»»■«)
8. (axby+cz)''
k(kb)(kc) B
13  x =—, TVT (' &c 
a (a  b) (a  c)
(a — b) (a c)
; &c.
"• X {ab){ac)(ad)'
XXXIV. a. Pages 439, 440.
1. 102. 2. 3a + 6 = 27.
3. x 3 2x 2 + x + l; loz+11. 4. a = 3.
5. x 4 + 5x~ 5 + 18x~ 6 + 54:X 7 ; U7x i B56x 5 + 90x & + 4S2x 7 .
6. (bc){ca) {ab) (a + b + c).
7.  (b  c) (c  a) (a b)(b + c) (c + a) (a + b).
8. 2iabc. 9. (£> + c) (c + a) (a + b).
10. (&c)(ca)(a&)(a 2 + 6 2 + c 2 + &c + ca + aZ>).
11. 3a6c(& + c) (c + a) (a + b). 12. 12a6c (a + 6 + c).
13. 80a&c(a 2 + J 2 + c 2 ).
14. 3 (b  c) (c  a) (a  b) (as  a) (x b) (xc).
x
28.
30.
(x a) (x b) (xc)'
(p x)(q x)
(a + x) (b + x) (c + x) '
29. 2.
31. 1.
32. a + b + c + d.
5. 0.
28 (a 2 + bc)(b 2 + ca)(c 2 + ab).
XXXIV. b. Pages 442, 443.
7. A = ax+by + ay, B = bxay.
ANSWERS
549
XXXIV. c. Pages 449, 450.
1. x 3 + xy 2 + ay = 0.
4. y = a(xSa).
7. 6 4 e 4 + c 4 a 4 + a 4 6 4 = cfbcJ 2 .
9. a 4 4</c 3 + 36 4 = 0.
abed
2. z + rt = 0.
5. a*a*=l.
3. aB 3 +y»=o*
6. ««+y s =2a s .
8. ?/ loa;=fc 2 (a:+a)".
10. « 4 2a 2 & 2 & 4 + 2c 4 = 0.
11.
+
3=1. 12. 5a 2 6 3 = 6c 5 .
1 + d
14. a 3 + 6 3 + c 3 +abc = 0.
16. a° + b 2 + c 2 ±2abc = l.
l + a 1 + 6 1 + c 1+<Z
13. ab = l + c.
15. (a+6)*(a5)$=4c$.
17. abc = (i ab  c) 2 .
20. c 2 (o + b  1)  c {a+ b  1) (a  2a6 + & 2  a  b) + ab = 0.
1 1 1
18. « 2 4a&c + ac 3 + 46 3 &V = 0.
22
23.
x i +
+
(a b)cr+(a c) bq (b  c) ap + (b  a) cr (c  a) bq + (c  b) ap
1
bcqr + carp + abpq
ab'  a'b ac'  a'c ad'  a'd
ac'  a'c ad'  a'd + be'  b'c bd!  b'd
ad'  a'd bd'  b'd cd'  c'd
= 0.
XXXV. a. Pages 456, 457.
1. 6x 4 13x 3 12x 2 + 39.r18 = 0. 2. x* + 2xr i  lis* 12.r 3 + 3Gx 2 = 0.
3. x 6  ox*  8x 4 + 40x 3 + 1 G.r 2  80r = 0.
4. re 4  2 (a 2 + b 2 ) x 2 + {a 2  b 2 ) 2 . 5. 1,3, 5, 7.
1 l p
2' 2' ~°
_3 3 1
2' 4' 3"
113
V 2' 4*
6.
3 3 .
— 4
2* 2'
7.
9.
4' 3,4.
10.
12.
8 2 1
9 ' " 3 ' 2 '
13.
15.
4, 1,2, 5.
16.
18.
(D 92 ; 2 2i,r ;
(2)
2> 2 " 2?
r 2
20.
2g, 3/.
8. 6, 2, g .
/o 3 1
11 ±n/3, j,
2*
14. ,,W2.
16. f,,2,3.
17.
19. (1) 6 2J (2)
21. 2(/ 2 .
4 3 5
3' 2' 3'
<1
XXXV. b. Pages 460, 461.
2 ldb^3
L d ' ~3' ~2~ ~"
3. 1=^/2, l^V^ 7 !
5. 1, ±v/3, 1*2^/^1.
7. x 4 8x 2 + 3G = 0.
2. g, 3,2 ±N /3.
4. ±7^1, _2±VTT.
6. .t 4 2x 2 + 25 = 0.
8. ar»+ 1G = 0.
.)
50 HIGHER ALGEBRA.
9. s 4  10^+ 1=0. 10. re 4  10a; 3 19.r 2 + 480a; 1392 = 0.
11. ar»  6a; 3 + 18x 2  26a; + 21 = 0. 12. a; 8  1 6.r 6 + 88a; 4 +192.t 2 + 144 = 0.
13. One positive, one negative, two imaginary. [Compare Art. 554.]
15. One positive, one negative, at least four imaginary. [Compare Art. 554.]
16. Six. 17. (1) pq = r; (2) ph'=q*. 20. q 2 2pr.
21. pqr. 22. ^3. 23. pqSr,
24. prls. 25. p i 4p 2 q + 2q 2 + 4:pr4LS.
XXXV. c. Pages 470, 471.
1. a, 4 6a; 3 + 15a: 2 12a; + l. 2. a: 4  37a; 2  123a:  110.
3. 2a; 4 + 8a; 3 a; 2 8a; 20. 4. a; 4 24a; 2 l.
5. 16aa;7i (.r 6 + 7a; 4 /* 2 + 7a; 2 /t 4 + h 6 ) + 2bh (5.T 4 + 10.r 2 /r + 7i 4 ) + 2ch.
10. 2, 2,  1,  3. 11. 1, 1, 1, 3. 12. 3, 3, 3, 2, 2.
1 = ^/^3 1±J~Z 1 1 1
13.2, 2 , 2 14. ^ > 2 ' 2 '
15. 1, 1, 1, 1, 1, 2. 16. ± x /3, i^/3, 1 = ^/^1.
/3 l±J7 /3 l±,7r23
17. a, a, a, 6. 18. ^Wg'" 2 — ;± \/2' 4
19. 0, 1, ,  ; 0, 1, , . 20. n"j** = 4p»{n2)»*.
22. (1) 2; (2) 1. 27. 5. 28. 99,795.
XXXV. d. Pages 478, 479.
2. ?/ 4 5?/ 3 + Sy' 2 9y +27=0.
4. 3 ±2^/2, 2 = ^/3.
6 2,2l,L(l±Jl3).
1.
2/ 3
24y 2 + 9y
24 = 0.
3.
1,
1, 2,
1
"2
i
5.
1,
l=fc v /
2
3
>
3 = ^/5
2 '
7.
4,
*s
8.
6,
3,
2.
2'2' 2
in 1 1 X X
1U * 4' ' 2' ~5*
11. y*2y + l = 0. 12. ?/ 4 4?/ 2 + l = 0. 13. y 5 7y 3 + 12y 2 7y=0.
14. 2/ 6 " 6 0i/ 4  3 20r  717?/ 2  773?/  42 = 0.
15. t/ 3 ^ 2 +^^ = 0. 16. 2/ 5 + ll2/ 4 + 42 2 / 3 + 57y 2 132/GO = 0.
17. ?/ 3 8?/ 2 + 19j/15=0. 18. ?/ 4 + 3?/ 3 + 4y 2 + 37/ + l = 0.
19. ?/ 3 + 33?/ 2 + 12*/ + 8 = 0. 20. ri/ + kqi/ 2 +k* = 0.
21. y' i q 2 y 2 2qr 2 yr 4 = 0. 22. r\f  qif 1 = 0.
23. ?^ + </(lr)2/ 3 + (lr) 3 = 0. 24. ?/ 3 2^ 2 + 9 2 y + r 2 =0.
25. 2 / 3 + 3?^ 2 + (5 3 + 3r 2 )2/ + r 3 = 0.
26. ;V + 3?^ 2 +(3r 2 + </ 3 )?v/ + r(r 2 + 2 2 3 ) = 0. 28. ±1, ±2, 5
ANSWERS. 551
XXXV. e. Pages 488, 489.
1. 5, d*±£Lll. 2  10,5±7V^8, 3. 4, 2±5j~^S.
a
4. 6, 3±4 7^~3. 5. .iVJ?. 6. 11,11,7.
10. 4, 2, ld=7^1. U. * 1,4*^/6. 12. 1,2,2,3.
13. 1±72, 1±7^1. 14. 1, 3,2±75.
15. 2,2,, . 16. 1, 4 ±715,  3 ^.
17. 4, 4, 4, 3.
18. g»+8r*=0;?, I 3 ^ 5
22. 2±76, ±72, 2 ±72. 23. sV + 2s(l*) 2 ?/ 2 + r(ls) 3 j/ + (l4 4 = 0.
25. 2±73. 26. °— ^?.
28. .t 4  8.r 3 + 21a; 3  20a; + 5 = (x 2  5x + 5) (a; 2  3a; + 1) ; on putting x = 4  y,
the expressions a; 2  5a; + 5 and a;  3a; + 1 become y 2  3// + 1 and
y 2  5// + 5 respectively, so that we merely reproduce the original equation.
MISCELLANEOUS EXAMPLES. Pages 490—524.
2. 6, 8. 3. Eight.
4. (1) 1±75; 1±275.
(2) 35=1, y = S, z=  5 ; or x=  1, y=  3; 2 = 5.
a + 2b 1
6. (1) 1,  q , . (2) 3. 7. First term 1 ; common difference  .
8. ^3; pCp 8 ^); (p 2 q)(p°3q).
9. (oft + a^" 1 ). 10. ^. 13. A, 7 minutes; ZJ, 8 minutes.
14 . a 4 + &4 + c 4 = 6 2 c 2 + c 2 a 2 + a 2 62<
15. x = y 1 = : ; or = ,=£;
J a + b + c ca ab
where Jca (a + b + c 2  be ca ab) = d.
16. One mile per hour.
17. (1) {b+e)(e+a){a+b). (2) ^/g + ^ " ' 2 ~. 18. ^ ; 22G8.
13. (1) ^ 105
(2)x = y=±7^; a^6= (J+26) =± \/y+a6a«
22. 1«*5; nine. 23. i {(1 + 2 + 3+ .. . + /<) (l 2 + 2* + 3 2 + ... +n)\.
552 HIGHER ALGEBRA.
24. Wages 15s.; loaf 6d. 25. 6, 10, 14, 18.
29. x = 3fc, ?/ = 4fc, z = 5fc; where F = l, so that fc=l, w, or or. 30. 480.
31. Either 33 halfcrowns, 19 shillings, 8 fourpenny pieces;
or 37 halfcrowns, 6 shillings, 17 fourpenny pieces.
32. a = 6, ft = 7. 33. 40 minutes.
35. 1 + x + ^x 2  x i  — x\
37 . l^y3 ^ pr 1^/gj 1 [a ,.4_. r _5 (a;2 + a; + l) = 0.]
38. a = 8;^— . 40. The first term.
.r 5
l + 4ft 2 c 2 + 9c 2 a 2 + a 2 ft 2
41. 13, 9. 42. .,,.,, ■, •
' a 2  o + c
43. (1) 3,  2, — ^^—  . [Add a 2 + 4 to each side.]
(2) x = l, , 1, 0, 0;
ssl, , 0, 1, 0;
2=1, \, 0, 0,1. 17. 5780.
a
48. 150 persons changed their mind; at first the minority was 250, the
majority 350. 50. 936 men.
„. _ 2 m l  . ad be
51  C 1 ) 0, 7J rrt. (2) .
i* '2 m +l K ' abc+d
[Put (a  c)(6 d) = {(x  c)  (x  a)} {(x  d)  (x  ft)} ; then square.]
53. 6, 577. 55. m =  r — — , , »=, — ^St.
30 ^/a + ^/6 v/a + ^/6
58. (1)1. (2) ±4 ^[putting a; 2  16 = y 4 , we find y*16 4y(?/ 2  4) = 0.]
60. —, — — males; ^— — — females. 63. 0, a + b, — .
bc bc a+b
64. Common difference of the A. P. is ; common difference of the A.P.
n1
which is the reciprocal of the H.P. is => — . [The r th term is
ab (n  1)
a(nr) + b(rl) ... ,... , . aft(nl)
—  ' — ^  ; the in  r + l) th term is — . e — rr — rv 1
n1 ' a(nr) + b (r  1)
68. 19. 69. £78.
l±V3 1*^3
* U ' 2 ' 2 *
[(a + ft) 3 a 3 ft 3 = 3aft(a + ft), and (a ft) 3  a 3 + 6 3 = 3aft(a  ft).]
95
ANSWERS. 553
72. (1) .,*]«» *4U (2) *=* 2( ^? l0 S 2) ~*l189.
v ' log b 1  log 2
73. 7, 2. 74. 8 hours.
. *■* * !f * a a + 6 + c ... ,
79. (1) =*f==sO, or  . 2) .<c = y = 2 = l.
v ' a o c a6c
80. a = 3, 6 = 1. 81. [Put xa = u and y  b = v.] 82. ar = 3. 84. 126.
85. Sums invested were £7700 and £3500: the fortune of each was £1100.
86. 503 in scale seven. 91. 25 miles from London.
T _ 5 ! ItthE}. ?/ _ 3 3 25 + 10^ /5 !
' * 6 ' *» 29 >*> d ' 5 ' 2 y ' 96 " \/3' "21'
_„ r , .. , ,. . l + 4x 2(l2'\r») l(l)».r»
100. Generating function is z „ ., ; suru= ^ — = —  \ '
1x 2i 1  2.c 1 + x
n ih term= {2" + (  l) n } x n ~K
107. a + 6  c  d. 108. 12 persons, £14. 18s.
109. (1) x = a, y = b, z = c. (2) x = 3, or 1; f=l, or 3.
111. 1+ JL = i L ,— 1; x = 948, y = 492. 113. £12. 15*.
1+12+1+1+1+9 *
117. (1) x = a, y = b; x = a, y = 2a\ x = 2b, y = b.
(2) x = 3 or 1, y = 2, 2=1 or 3;
12 ° W l ~^TW
(2) a(J "' 1) + d {*»+" + **+>  (n + 1) 2 * 2 + (2*« + 2li  1) x  n 1 .
121.  . 122. (1) y or ^ .
(2) x = 0, y = 0, 2 = 0; x=±2, y=±l t 2= ±3.
13.c  23 lO.r1 t r + 4
3 (.r 2  3x  1) 3 (**+*+ 1) ' 2^+1 '
125. 1 = 1; scale of relation is 1  x  2x ; general term is { 2' 1 ~ 3 + ( l) n_1 } .r"'.
127. (1) .r=6, 2; y = 9, 3. (2) *=; y=y
128. (1)^". (2)^. 129. 12, 16; or 48, 4.
130. (1) x= ±7.
(2)  = I =  = ± JL where & 2 = 26 2 c 2 + 2c 2 a 2 + 2a 2 6 2  a 4  6 4  c 4 .
a b c 2abc
124
133. 11, r1. 134. 384sq.yds. 136. a= ±2, 6 = 3, c = ±2.
V2' y " v/2
138. £3. 2s. at the first sale and £2. 12*. at the second sale.
137. (1)*=±^, y=±^. (2) ±^; *^/^
554
HIGHER ALGEBRA.
139. (1) i«(«+l)(2« + l). (2) ~n(n + l)(n + 2)(Bn 2 + 6n + l).
(3) n(»+l)(4wl).
141. (1) x = l or y; ?/ = 3 or y.
(2) x, y, z may have the permutations of the values 3, 5, 7.
142. y 3 + <jy 2  g 2 y/  q' s  8r = 0.
x (x n  1) w
143. (1)
(xiy
xl'
(3) 2' l +i + u(/t + 7)2.
3 + Ua; 157a: 2
( ' 1 + 5x 50a; 2 8a; 3 '
144. 2 (6 3  d 3 ) = 3 (& 2  c 2 ) (b  a).
145. 2, 2, 2,
3*
11,13,15,17,19,21,23,
14,15,16,17,18,19,20,
miles,
146. A walks in successive days 1, 3, 5, 7, 9,
B walks 12, 13,
so that B overtakes A in 2 days and passes him on the third day ; A
subsequently gains on B and overtakes him on B's 9 th day.
^61
5 ~ *
147.
150.
148. (a + b + c), (a + wb + u) 2 c),  (a + (o' 2 b + wc).
n (a n — b n )
n th term is * — L a;" 1 ; Sum = A  B,
ab
, a(lna n x n ) a 2 x (1  a n  1 x n ~ i )
where A = — y — ' H *_
1  ax (1aa;) 2
sponding function of B.
151. qif  2p 2 y 2  5pqy  2p 3  q 2 = 0.
153. (1) 7, 7±8 y~ 8  ( 2 ) ^ ±8 > 4 «
and B denotes a corre
154. 3 days.
156. (1)  ,  ^ , 0± ^ 4 89 . [(12a;  1) (12a;  2) (12s  3) (12*  4) = 120.]
„. , ,„„ T92 11 2 "I
157. 22 years nearly. 161. 44 hours.
162. (1) x
_ 7±V217 .
=r= r* ; s=±i,
±2; y==F2, Tl;ar=y=±V3
(2) x = h (6 4 + c 4  a 2 6 2  a?e 2 ) , &c. , where 2/i; 2 (a 6 + b G + c 6  3a 2 6 2 c 2 ) = 1 .
[It is easy to shew that a 2 x + b 2 y + c 2 z = 0, and
a 2 y + b 2 z + c 2 x = a; 3 + y' i + z' i  Sxyz = a 2 z + b 2 x + c 2 y .]
163. 2 (a + b + c)x = (be + ca + ab) ± J (be + ca + ab) 2  4abc (a + b + c).
[Equation reduces to (a + b + c) x 2  (be + ca + ab) x + abc = 0.]
164. (1) ~n(n + l)(n + 2)(Sn + 13). (2) 2e5.
ANSWERS. 555
166 . { l )x J 1 + f* /2 a,y= 1 £. [Eliminate..]
(2) #, ?/, z are the permutations of the quantities 2,  ., , g
167. (*+y+*) 9 =3* 9 . 168. 2. 169. *»+*»+** Says.
170 He walks 3 miles, drives 7* miles, rides 10 miles per hour.
AB = 37h £C = 30, CA = 15 miles.
172. (1) a; = 13 or 10, i/ = 10 or 13.
d iab) c(.ft). b(dc) , a(dc)
M £3200. "6. rr + 3^ + (3,r)^r0.
177. 7> = ( ac * M ) fa ± / /l ) + ( 6c * ad } t? T '?) ''
q = (be T ad) (<?0 ±/>0  ( ac * bd ) if if T eh >'
178. . = 6, 5; £ > 2 '
18±*/^47 14*^/^74
y = 5, 6; g . 2
[Put a;  y = M and ay =0, then u 2 + 2y = 61 , u(61 +v) = 91.]
1 3=^5
182. 8987. 183. ?/ 3  ft»/ 2 aci/  c 2 = 0. 1, 8, j, 2 ■
I + n/^3 1n/£?
186. (1) x, ?/, s are the permutations of the quantities 1, g—  , 2
a(Z> 2 + c 2 ) _
187. Conservatives; English 286, Scotch 19, Irish 35, Welsh 11.
Liberals ; English 173, Scotch 41, Irish 68, Welsh 19.
191
, (1) 7, 9, 3. (2) 2±J3, 2±Jh
a ~ b . 07, n J. I."— . 201.
aft _ jm + n2
192. 2a w =.+6+yr; 2b n =a+bgr. 201. m _ x )n _x
n 4«+ 1 + 4(l) w + 1
202. 54, 26, 14±840 N /1. 204. ^, 4 „+! _(_!)»■
206.
3rw 3 + nm 2 g3Ti 3 207. 81 years nearly.
m 3 + nvi 2 q + n 8
209. 7 Poles, 14 Turks, 15 Greeks, 24 Germans, 20 Italians.
210 1_? _l+^iog(l + a;).
J1U * 2 4 2.c 6 v '
212. (1) jn(n+l)(n+2)(n+3); W "Ti+iyr ' (3) »•
213.
556 HIGHER ALGEBRA.
223. (l)^ = l (±15±V 33 )j , = 1 (±15W ^ );
or* = 4, 6, 4, 6;
y = 6, 4, 6, 4;
2 = 5, 5, 5, 5.
(2) a; ~ fl  V~ b _ gc
a(6c) 6( C rt)~c^rT6) X '
where (& c )( C a)(« & )x = a2 + 62 + c2 _ &c _ ca _^
226. 12calves, I5pigs, 20 sheep. 229. Lim f (^l)} =; convergent.
230. Scale of relation is 112., + 32. Xtenn =  1 {«« + 8— .
2*»i 23«i 5
^n = — s H
3 T 7 "21
11
231  2T3
232. ar = ± ^ ~ P + ? * N /a=* + 6« c a, & c .
233. a 3 + V + cS=aZ(b + c) + bZ(c + a) + c*(a + b).
235. (1) (l*)^=l+4o;+^.(n + l)3 a .n +(3B8+6lf _ 4) ^ I
 (3m 3 + 3n 2  3k + 1) «*M + »%»+«
(2) J 1
' 8 (n + l)2( 7i + 2 )3
236. 1 + a*x* + a *x* + fl «x" + ««*!« + a*V» + a u x * + „^ 28 + ^ + a20;r , 6<
237. 3hours51min. 240 . 2 or  ^ . 242 . _ 140 .
244. 3,4,5,6. 246. a*(c*Sd?)*=:(ab*+2&) (ab*£f.
247. 2, 6, 1, 3. 248. ± .
13
249. (1) 2*H2jU(n + l)(2,t + l).
(2) _^!L_ _ 2
(n + l)(n + 3) S'
< 3 > T^ + *££P he . U even , If*^ + 'Jiz*^)
when n is odd.
250. (l)* = 2/ = * = 0or. If however X > + y > + z * + yz+zx + xy = 0) then
* + 2/ + * = a, and the solution is indeterminate.
(2) —, V z
a(a + b + c) b(ab + c)^(^b^c)
1
flM (i ± s/l a + *> + cjja b + c) (a + b~^~c) '
253. &+B E1 C,HA, + B , + c. )<4 , B , + , )(Alf + £+ Xre
A = y/a (6  c), &c.
ANSWERS. 55?
256. (1) .>— 1, w, to ;
z=(a + b), (rtw + W), (i/w' + /yu).
(2) x = S, or 7  2 = 0, or 4
y = 7, or 3 ) m = 4, or 6
257. To at least 3r2 places. 258. Tea, 2s. Go".; Coffee, 1*. 6d.
262. 2g a 6pr+24». 263. 11 turkeys, 9 geese, 3 ducks.
266. (1) x, y, z have the permutations of the values
a, la(bl+sJb*2b3), \ a (b  1  JJfi^W^ 3).
(i j_ b f c
(2) as=f/=« = l; # = =— ; &c. 267. 0.
268. 16 Clergymen of average age 45 years ;
24 Doctors of average age 35 years ;
20 Lawyers of average age 30 years.
269. (a a 2  afl (a. 2 a 4  a 3 2 ) = (afy  a 2 2 ) 2 ;
or a a. 2 a i + 2a x a. 2 a 3  a a 3 2  afa A  a 2 3 = 0.
270. X= ± : — , &C. u = ±  — T &c 273. c~$.
Va 2 + 6 2 + c 2 V« 2 + ^ 2 + c 2
274. (1) fl?Viog(la;)2. (2) ^ jl # , w '^— , j ■
v '\ xj a1 ( (a + l)(a + 2)...(a + n))
275. (1) *=?, ?, 2;
?/=l, g, 1;
3 3
2 ' 4' 4'
(2) z=±4, 2/= ±5, t<=±2, w=±l.
5 Vi> ^ t2 v!' u= 4v3' u=± \^
276. a 2 + ft 2 + c 2 + d 2 + \. 277. p 1 8 + 3^0 Bp 3 .
279. ^, 6 birds; B, 4 birds. 281. 2.
287. a, 5a, 5a. 289. S 1=  ft"? ^'fl'fr"?? , &*
291. .4 worked 45 days ; i>, 24 days; C, 10 days.
294. (ft 2 + c 2  a 2 ) (a 2  ft 2 + c 2 ) (a 2 + ft  c 2 ).
300. Walked 3 miles, worked 4 hours a day ;
or walked 4 miles, worked 3 hours a day.
* =± 3
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L. B. x:at. No. ii
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WELLESLEY COLLEGE LIBRARY
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4?212
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Higher al gebra
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