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ELEMENTARY ALGEBRA
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ELEMENTAKY ALGEBRA
USE OF SCHOOLS AND COLLEGES
BY
CHARLES SMITH, M.A.
AUTHOR OF "a treatise ON ALGEBRA," "AN ELEMENTARY
TREATISE ON CONIC SECTIONS," ETC.
f
BE VISED AND ADAPTED TO AMERICAN SCHOOLS
BY
IRVING STRINGHAM, Ph.D.
prof:
\
professor of MATHEMATICS AND SOMETIME DEAN
IN THE UNIVERSITY OF CALIFORNIA
^ THE
THE MACMILLAN COMPANY
LONDON: MACMILLAN & CO., Ltd.
1902
AU righta reserved
'\c>X
COPYEIGHT, 1894,
By MACMILLAN AND CO.
Copyright, 1900,
By the MACMILLAN COMPANY.
Set up and electrotyped June, 1894. Reprinted September, 1894;
February, August, 1895; January, September, 1897; January, Au-
gust, 1898; October, 1899.
New revised edition August, 1900 ; April, 1902.
NorfajDOlf ^regg
J. S. Gushing & Co. — Berwick & Smith
Norwood Mass. U.S.A.
PREFACE TO THE AMERICAN EDITION.
The transition from the traditional algebra of many of
our secondary schools to the reconstructed algebra of the
best American colleges is more abrupt than is necessary
or creditable. This lack of articulation between the
work of the schools and the colleges emphasizes the
need of a fuller and more thorough course in elementary
algebra than is furnished by the text-books now most
commonly used. It is with the hope of supplying this
new demand that an American edition of Charles Smith's
Elementary Algebra is published ; a work whose excel-
lencies, as represented in former editions, have been
recognized by able critics on both sides of the Atlantic.
In the rearrangement of the work and in its adaptation
to American schools many changes have been made, too
many to be noted in a short preface, and a considerable
amount of new subject-matter has been introduced. The
following are innovations of some importance :
Chapter I., consisting of a series of introductory
lessons, is wholly new, and Chapter XIII. is partly
new and partly transferred from Chapter XXVIII. of
the second edition. Horner's synthetic division is made
prominent in the chapter on division, an early introduc-
tion to quadratic equations finds its appropriate place in
the chapter on factoring, the binomial theorem for posi-
'29;^
VI • PREFACE.
tive integral exponents is demonstrated by elementary-
methods in the chapter on powers and roots, and the
chapter on surds has been enlarged by a short discussion
of complex numbers. Some new collections of examples
have been introduced, and several of the older lists have
been extended.
As thus reconstructed, the book constitutes a rounded
course in what may be called the newer elementary alge-
bra, and includes the subject-matter specified by nearly
all American colleges as the requirement for admission.
It will prove especially helpful to students preparing for
such colleges as are using Mr. Smith's Treatise on Algebra
for advanced work.
Those who are familiar with the second edition will
notice that the chapters on permutations and combina-
tions, on logarithms and exponentials, on the binomial
theorem for negative and fractional indices, on scales of
notation, and on cube root have been omitted.* . . .
I am indebted to my colleagues of the mathematical
department of the University of California for valuable
suggestions, but especially to Professor Haskell and
Dr. Hengstler for contributions to subject-matter and
for reading many of the proof-sheets.
Special thanks are due to Mr. Smith, for allowing free
scope for this revision.
IRVING STRINGHAM.
University of California,
May, 1894.
* Chapters on logarithms, exponentials, and inequalities have been
introduced into this second briefer edition; and in the second complete
edition new chapters on the theory of equations are now incorporated.
L S. — Paris, December, 1899.
PREFACE TO THE SECOND EDITION.
In response to many appeals from teachers, a series
of short chapters introductory to the theory of equations
— Chapters XL. to XL VIII. inchisive — are added to
this second edition. In the interest of a better classifica-
tion of subject-matter, the chapter on determinants and
the chapter on scales of notation, of the first edition,
have been interchanged and renumbered in the second;
but with this exception there has been no renumbering
of chapters or articles.
Necessarily, in a brief survey of the theory of equa-
tions, many important topics are omitted, and the sub-
jects that are included are discussed but meagrely. For
a more thorough and systematic reading of this impor-
tant theory, especially in its modern phases, the student
will have recourse to the larger treatises.
Irving Stringham.
Paris, September, 1899.
Til
CONTENTS.
•o*— ^ —
HAPTBB PAGE
I. Introductory Lessons 1
II. The Language of Algebra. Definitions 17
Positive and Negative Quantities 27
III. Addition 32
Subtraction 37
Brackets 43
IV. Multiplication 46
Product of Multinomial Expressions 56
Continued Products 67
V. Division 71
Division by Multinomial Expressions 75
Homer's Synthetic Division 78
Miscellaneous Examples I 88
VI. Simple Equations 92
VII. Problems 101
VIII. Simultaneous Equations of the First Degree 109
Elimination by Addition and Subtraction 110
Other Methods of Elimination 114
Equations with Three Unknown Quantities 119
IX. Problems 122
Miscellaneous Examples II 128
X. Factors 132
Factors of ax"^ -^ bx + c 142
Factors by Rearrangement of Terms 147
Quadratic Equations 151
iz
CONTENTS.
CHAPTER
XI.
XII.
XIII.
XIV.
XV.
XVI.
XVII.
XVIII.
XIX.
XX.
PAGE
Highest Common Factors 156
Lowest Common Multiples 168
Miscellaneous Theorems and Examples 174-
Mathematical Induction. . . < 174
Factor Theorem. Remainder Theorem 179
Fractions 186
Reduction to a Common Denominator 193
Addition of Fractions 194
Multiplication of Fractions 201
Division of Fractions 202
Reciprocal. Infinity 203
Two Important Theorems 208
Equations with Fractions 216
Simultaneous Equations 221
Miscellaneous Examples III 229
Quadratic Equations 234
General Properties 239
Irrational Equations 245
Relations between the CoeflBcients and the Roots 249
Special Forms 251
Equations with Given Roots 253
Equations of Higher Degree than the Second. . ; 260
Simultaneous Equations of the Second Degree 266
Problems 279
Miscellaneous Examples IV .... 285
Miscellaneous Equations 289
Involution 297
The Binomial Theorem 302
Evolution 309
Square Root 310
CONTENTS.
XI
CHAPTER
XXI.
XXII.
XXIII.
^XIV.
XXV.
XXVI.
XXVII.
XXVIII.
XXIX.
XXX.
XXXI.
XXXII.
XXXIII.
XXXIV.
XXXV.
XXXVI.
PAGE
Fractional and Negative Indices 319
^urds ...;... 328
Complex Quantities 337'
Ratio 344
Proportion 349
Variation 357
Miscellaneous Examples V 363
Arithmetical Progression 367
Geometrical Progression 380
Harmonical Progression 393
Other Simple Series 398
Miscellaneous Examples VI 403
Inequalities 409
Limits 417
Indeterminate Forms 421
Exponentials and Logarithms 426
Exponentiation 427
Loga):ithmic O^^eration 430
Natural Logarithms 436
Convergency and Divergency of Series . 447
Indeterminate Coefficients 465
Application to Integral Functions 469
Application to Partial Fractions 472
Application to Expansion of Functions 476
Application to Summation of Series 480
Permutations and Combinations 485
The Binomial Theorem for an Integral Index 496
The Binomial Theorem : any Index 507
Exponential and Logarithmic Series 514
Logarithmic Computation 521
xu
CONTENTS.
CHAPTER
XXXVII.
XXXVIII.
XXXIX.
XL.
XLI.
XLII.
XLIII.
XLIV.
XLV.
XLVI.
XL VII.
XLVIII.
PAQK
Continued. Fractions 630
Scales of Notation J 548
Miscellaneous Examples VII ,/ 654
Determinants , 561
Rational Factors and Higher Equations 585
To Factor x^ + y^ -{- z^ - S xyzJ 585
To Factor the Cubic 586
To Factor the Biquadratic 588
To Factor ax^ + 2 hxy^ -{- by^ + 2 gx + 2fy + c. . 590
Simultaneous Equations 594
Horner's Method 598
Roots and Coefficients (60&:
Symmetric Functions 608 .
Graphical Methods 613
Binomial Equations 617
Derivatives 624
Maxima and Minima 631
Successive Derivatives 633
Taylor's Theorem 636
Continuity 638
Rolle's Theorem 639
Transformation 641
Reciprocal Equations 645
Criteria for Real Roots 650
Descartes' Rule of Signs 653
Sturm's Theorem 658
Miscellaneous Examples VIII 666
\ D B A ^ V
ALGEBRA.
-»o»{oo-
CHAPTER I.
Introductory Lessons.
1. Signs and Symbols in Algebra. In algebra a letter is
frequently used to denote a number. When so used in
a connected series of operations, a given problem for
example, the letter must be regarded as standing for the
same number in all the operations of the series. Several
distinct letters may be used to denote several distinct
numbers in the same problem.
In these Introductory Lessons the signs +, — , X, -^,
and =, have the usual significations given to them in
arithmetic. Their formal definitions are given in Arts.
12, 13, 14, 15, and 21, of the next chapter.
When letters are used to represent numbers, the sign
of multiplication is frequently omitted. Thus ab means
a xb.
2. The Method of Algebra. The statement of an equality,
made by placing the sign = between two numbers, or two
sets of numbers, indicating that the two numbers, or two
sets of numbers, are equal to one another, is called an
equation.
% TNTPvODUCroilY LESSONS.
Thus 2 a; = 5 and a; + 2 = a + 3
are equations. Problems stated in the form of equations
are said to be stated algebraically.
In this introductory chapter some of the uses of alge-
bra will be explained by stating and solving, with the
help of equations, certain classes of problems already
familiar to the learner who has studied arithmetic.
In applying the method of algebra, the first step is to
translate the ordinary language in which the problem is
expressed into a concise symbolical statement, and the
form used for this purpose is the equation. The solution
of this equation then solves the problem itself. Algebra
has no one universal rule that will put every problem
expressed in ordinary language into the algebraic form,
but rather it devises ways and means for solving the
various classes of equations that are produced by dif-
ferent practical problems.
Our first concern is to state the given problem in the
form of an equation, and the way to do this must be
sought in the language of the problem itself. A few
simple examples will best explain how.
Ex. 1. If % 10 were added to twice the money I have, the result
would be $ 150. How much money have I ?
The specifications of this problem may be expressed more con-
cisely thus :
Twice the money I have^ added to f 10, equals $ 150, or again,
using the signs + and = in the ordinary arithmetical sense,
Twice the money I have + $ 10 = $ 150.
The problem is here stated in the form of an equation, which as
yet, however, is not purely algebraic.
INTRODUCTORY LESSONS. 3
This analysis will not be in any way disturbed by the
replacement of 10 and 150 by any other numbers we
choose to insert, and we may analyze with equal facility
the following problem, in which a and h stand for any
numbers, although we impose, for the present, the con-
dition that h shall be greater than a.
Ex. 2. If a dollars were added to twice the money I have, the
result would be h dollars. How much money have I ?
Stated in the briefer form, this problem says ;
Twice the number of dollars I have + ^ a = $ 6.
Ex. 3. The sum of two numbers is 50, and their difference is 20.
What are the numbers ?
The first condition of this problem asserts that
The larger number + the smaller number = 50 ;
and since we obviously get the larger number by adding their dif-
ference to the smaller, the second condition asserts that
20 + the smaller number = the larger number.
Hence, replacing ' the larger number ' in the former of these two
statements by its equal ' 20 + «^e smaller number^' we obtain
20 + the smaller number + the smaller number = 50,
or, in briefer form,
20 + twice the smaller number^ 50.
From this statement the smaller number is at once seen to be 15.
Observe that here, as in Ex. 1, the numbers 20 and 50
may be replaced by letters representing any other num-
bers, without in any way affecting the analysis of the
problem, or the form of its solution.
These examples point to the fact (they do not prove
it) that no matter what the relation of the quantities
involved in a determinate problem may be, it can be
expressed in the form of an equation.
4 INTRODUCTORY LESSONS.
But although, the abbreviated statements thus far pro-
posed are all more concise than the corresponding state-
ments in ordinary language, they can be reduced to still
more convenient form. Thus, in Ex. 1, if we briefly
indicate ^ the number of dollars I have ' by soine symbol
or letter, as $ x, the equation
Twice the number of dollars I have -f- $ 10 = $ 150
becomes 2x$a; + $10 = $ 150,
and is now -more strictly algebraic. Any symbol, for
example ($?), may here be used to stand for ^the num-
ber of dollars I have,^ but it is the universal custom to use
a letter for this purpose, and obviously it may be any
letter.
It must be noticed that, in the algebraic statement of
a problem, x is an abstract number, which may turn out
to be 5, or 11, or |, or in fact any number, and that,
therefore, x dollars means x times one dollar, or x x {1
dollar). Similarly, x inches means a? x (1 inch), etc. In
other words, x dollars is a product, the first factor of
which is the abstract number x, the second being the
concrete unit one dollar, in terms of which all the quan-
tities involved in the problem must be expressed. The
unit beinp uhen well understood, we drop it out of the
equation, /hich in the case of Ex. 1 then becomes
2a; + 10 = 150,
a pure algebraic equation. The purely algebraical state-
ment of Ex. 2 is, similarly,
2x-\-a = b,
and that of Ex. 3 is 20 + 2 a; = 50,
in which x stands for ' smaller number/
INTRODUCTORY LESSONS. O
It is important that the beginner clearly understand
what these algebraic equations mean. They can be
easily translated backwards into the ordinary language
in which the problems were originally expressed. An
equation is thus simply a sentence in algebraic language
expressing the conditions of the problem. And besides
its conciseness of statement, giving at a glance the rela-
tions between the quantities involved in it, the theory of
algebra will eventually show that out of it may be
evolved, by simple processes, the value of the impor-
tant quantity ic, the unknown quantity, which, when
found, answers the question propounded by the problem.
State the following problems algebraically :
Ex. 4. Wliat number .is that whose half added to 73 will pro-
duce 85 ?
Ex. 6. What number is that whose nth part added to a will
produce b ?
Ex. 6. A boy had a apples. He gave a certain number of
them to his companions, and then had b times as many as he had
given away. How many did he give away ?
Let the learner now read the axioms of Art. 93.
3. Problems involving the Four Fundamental Processes of
Arithmetic. r •
Ex. 1. If the sum of x and 2 x and 3 a; be 42, whlft number does
X stand for ?
The algebraic statement of the problem is
x-{-2x + 3x = 42;
and this is the same as 6x = 42 ;
for, obviously, any number added to twice itself and three times
itself is six times the number. Hence, since 6 times x.is 42,
X = 7.
6 INTRODUCTORY LESSONS.
Ex. 2. If 5 ic be subtracted from the sum of 3 x, 4 x, and x + 6,
the result is 35. What number does x stand for ?
The algebraic statement of the problem is
Sx-{-4:X + x + 6 — 5x = S6.
In order to find what x stands for, we first combine the four
terms containing x into one term, which, by the rules of addition,
subtraction, and multiplication in arithmetic, obviously is 3 x ;
hence 3 x + 5 = 35.
From both sides of this equation subtract 5 ; then
3x = 30,
whence cc = 10.
Ex. 3. The sum of two numbers is 50, and their difference is 20,
What are the numbers ?
This is the third problem of Art. 2. It was there stated alge-
braically ; we now solve it in the purely algebraic form. Let x
represent the smaller number ; then 20 + x is the larger, and, by
the conditions of the problem,
20 + a: + « = 50.
We now subtract 20 from both sides of the equation, and obtain
2 X = 30 ;
whence x = 15, the smaller number,
and 20 + X = 35, the larger number,
Ex. 4. The larger of two numbers is 3 times the smaller, and
their difference is 92. What are the numbers ?
Let X represent the smaller number ; then 3 x is the larger, and
by the second condition of the problem,
3 X - X 1= 92.
But three times any number less the number itself is obviouslj
twice the number. Hence
2x = 92,
and X = 46.
INTRODUCTORY LESSONS. 7
Ex. 5. A, B, and C together have $75 ; A has twice as much as
B, and B has $ 5 less than C. How much has each ?
Let X represent B's share ; then 2 x is A's share, and x + 5 is
C's share, and the algebraic statement of the problem is,
a; + 2x + x + 5 = 75.
From this equation are obtained, by the processes already explained
in the preceding examples,
a; = 17|, 2ic = 35, x + 5 = 22^.
Ex. 6. An indicated sum, as (n + ^n — 1), is frequently placed
in a parenthesis in order to indicate that an operation is to be
performed upon the number which the sum represents. Thus, if
n = 6,
w + I w - 1 = 8,
and 2 X (n + ^ w - 1) = 2 X 8 = 16.
This result will again be the same if each term of the
sum be multiplied by 2 before the substitution of 6 for
n is made. Thus, whatever number n may represent, we
shall find it to be always true that
2x(n + iM-l) = 2xnH-2xin-2xl,
and we verify this result in the present instance by put-
ting 6 in place of n, thus obtaining
2x6 + 2x1x6-2 = 16.
It is of extreme importance to observe that we have
here, not a proof of a principle or formula, but a verifica-
tion of it in a particular instance-
Ex. 7. If n stand for 9, what numbers are represented by
2n, 5m, n + 3, w - 5, 2w + 3, 3n-2, 3(n + 1), 3(w - 4),
3 n + 2 w - 4 w, K^ - 3), 3(^ w - 1)?
8 INTRODUCTORY LESSONS.
Ex. 8. John is 4 years older than William, and 4 times John's
age is 5 times William's. What are their respective ages ?
Let X represent John's age ; then ic — 4 is William's age, and
the algebraic statement of the problem is
5(a; — 4) = 4 a;.
Here the parentheses enclosing a; — 4 indicate that both x and 4
are to be multiplied by 5. Performing this multiplication, we
obtain
5 X - 20 = 4 a^ ;
and from this equation, by adding 20 and subtracting 4 a;, perform-
ing the operations on both sides of the equation, we find
X = 20, John's age,
and then aj — 4 = 16, William's age.
EXAMPLES I.
1. Divide 15 into two parts whose difference is 7.
2. The sum of two numbers is a, and their difference is 6.
What are the numbers ?
3. A and B together own 140 acres of land, and B owns 3 times
as much as A. How many acres has each ?
4. A man having $92 spent a part of it, and then had 3 times
as much as he had spent. How much did he spend ?
5. A farmer sold 15 bushels of barley at a certain price, and
25 bushels of wheat at twice the price of the barley, and he received
$14 more for the wheat than for the barley. What were the
prices per bushel of the wheat and of the barley ?
6. Twenty-four coins, consisting of dimes and half-dollars,
amount to % 6. How many coins of each kind are there ?
7. A boy gave one-third of all the apples he had and one-third
of an apple more to his sister, and then had one apple left. How
many apples did he give to his sister ? !
INTRODUCTORY LESSONS. 9
8. What number is that whose double exceeds its half by 27 ?
9. Verify the following equations by putting some number (any
number) in place ot n:
(1) |w + n = fw,
(2) 3 x(n+l)+l = 3w + 4,
(3) 3 X (n - 4) + 2 = 3 « - 10,
(4) 5n-2(w- l)=5?i-2n + 2
= 3/1 + 2,
(6) J(n- l)+2 = ^w-i + 2
(6) |n-2(2-^n) = Jw-4 + n
= |n-4. .
10. A can do a piece of work in 10 days, A and B together in 6
days. In how many days can B do it alone ?
; 11.) The age of a father is now 7 times that of his son, but in 3
years it will be only 5 times that of his son. How old are each ?
12. At what time between 2 and 3 o'clock are the hour and
minute hands of a watch together ?
4. Problems involving the Eule of Three.
Ex. 1. What number has to 12 the same ratio as 57 to 9 ?
Let X represent the number. Then if the ratio of 57 to 9 be
denoted by 57 : 9, the conditions of the problem require that
X : 12 = 57 : 9 ;
or this statement may be made in the equivalent fractional form
12 9*
Multiplying both sides of this equation by 12, we obtain
10 INTRODUCTORY LESSONS.
Ex. 2. How long will it take to fill a cistern of 165 gallons by
a pipe that fills one of 120 gallons in 8 minutes ?
If X represent the number of minutes it takes to fill the larger
cistern, then by the rule of three (proportion)
a; ^165
8~120'
and by multiplication of both sides of this equation by 8,
„ 8 X 165
120
= 11.
Ex. 3. A train goes at uniform speed one-third of a mile in
20 seconds. What is its speed per hour?
Let X represent the number of miles per hour which denotes the
speed of the train. Then, by the rule of three, or proportion, the
algebraic statement of the problem is
3600" 20'
since there are 3600 seconds in an hour. From this equation, by
the proper multiplications and reductions, we find x to be
- 3600 ^g()_
3 x20
Ex. 4. If it require 15 men to build a house in 92 days, how
many men must be employed in order to build it in 60 days ?
This problem may be stated in the form of a proportion, or as
follows. It requires 15 x 92 days' work for the building of the
house ; hence, if x stand for the number of men that can build it;
in 60 days,
X X 60 = 15 X 92.
Therefore
« = 152192 = 23.
60
INTRODUCTORY LESSONS. 11
Ex, 5. If it require a men to build a house in 6 days, how
many men must be employed in order to build it in c days ?
It requires a x 6 days' work for the building of the house ;
hence, if x stand for the number of men that can build it in
c days,
xY.c = ay.h.
Therefore
„ a X 6
EXAMPLES II.
1. Divide $20 into two parts whose ratio to one another is
the same as the ratio of 3 to 7.
2. If 5 tons of hay cost $84, how many tons can be bought for
$270?
3. If the speed of a train be 45 miles an hour, how long will it
take it to go 360 miles ?
4. How long will it take to fill a cistern containing a gallons
by a pipe that fills a cistern containing 6 gallons in c minutes ?
5. A train goes at uniform speed a miles in h minutes. What
is its speed per hour ?
6. It requires a men to build a house in h days. How long
will it take c men to build it ?
7. In Ex. 6 put 15 for a, 75 for &, and 20 for c, and work
out the consequent numerical result.
8. A's age is now twice B's age ; but in five years their ages
will be in the ratio of 7 to 4. What is the age of each ?
9. If the wages of 6 men for a days be $60, what will be the
wages of 13 men for the same time and at the same rate ?
10. If 7 men earn $ 105 in 6 days, how many men, working for
the same rate of wages, will earn $ 157.50 in 7 days ?
12 INTRODUCTORY LESSONS.
11. If 10 men can reap in 3 days a field whose length is 1200
feet, and breadth 800 feet, what is the breadth of a field whose
length is 1000 feet which 12 men can reap in 4 days ?
5. Problems in Percentage and Interest.
Ex. 1. Of a herd of cows, 280 are Jerseys, and these are 35%
of the entire herd. How many cows are there in the herd ?
Let X stand for the number of cows in the herd. Then ^-^-^ x x
is the number of Jerseys. Hence, the algebraic statement of the
problem is
tVo X oj = 280,
and the value of ic, derived from this equation, is
x = M2i^ = 800.
35
Ex. 2. A town lost 7% of its population, and then had 6045
inhabitants. What was its population before the loss ?
Let 05 be the original, unknown number of inhabitants. Then
Y^^ X X was the loss, and x — j^ x x was the number of inhabi-
tants remaining after the loss. Hence
X- ^\^Y.x- 6045.
Now X less seven hundredths of x is ninety-three hundredths of x ;
therefore tV^ x x = 6045,
, ^ 100 X 6045 r,.r.r.
and X — — = 6500.
93
Ex. 3. The amount due on a note for $800 at 5% simple in-
terest was % 900. For what length of time was interest reckoned ?
The interest for one year is yf^ x 800 dollars, or $40; hence
the interest for x years is 40 x cc dollars, and the amount due after
X years is 800 + 40 x cc dollars, and this, by the conditions of the
problem, is $900. The equation expressing the conditions there-
fore is
800 + 40 a; = 900.
INTEODUCTORY LESSONS. 13
Subtract 800, and divide by 40, performing these operations on
both sides of the equation ; then
900-800_oi
40 ^
Thus the length of time for which interest was reckoned is 2^
years.
Ex.4. What principal amounts to $1456 in 2 years at 6%
simple interest ?
If X be the principal, ■^%^ x a; is the interest for one year, twice
this is the interest for two years, and x 4- (2 x ^^-^ x x) is the
amount. Hence
V 100 ;
the parenthesis being here used to indicate that the entire product
within it is to be added to x. (See Exs. 6, 7, 8, Art. 3.)
But y\^ of X added to x is |^§ of x. Therefore
IH X X = 1456,
and ^^100x1456^^3^
112
The required principal is $1300.
Ex. 6. The amount due, after Z\ years, on a note for $1500
bearing simple interest, was $1788.75. 'What was the rate of
interest ?
Let X stand for the rate of interest ; then 1500 x x is the interest
for one year, 3| x 1500 x x, or 5250 x x, is the interest for 3^
years, and 1500 + (5250 x x) is the amount due. Hence
1500 + (5250 X x) = 1788.75.
Subtract 1500, and divide by 5250, performing these operations on
both sides of the equation ; then
^^1788.75-1500^ Qgg^
5250
Thus the rate of interest was 5^ per cent.
14 INTRODUCTOKY LESSONS.
EXAMPLES III.
1. What number increased by | of 25 % of itself equals 315 ?
2. A city gained 13 % in population, and then had 80,456 inhabi-
tants. What was its population before the gain ?
3. The population of a city increased from 31,000 to 33,945
in ten years. What was the average per cent of increase per
annum ?
4. The annual rent of a house is $240, and this is 8% of its
value. What is its value ?
5. What sum bearing interest at 5i% will yield an annual
income of f 1500 ?
6. In what time will any sum of money double itself at 8%
simple interest ?
7. The amount due, after a years, on a note for h dollars bear-
ing simple interest, was c dollars. What was the rate of interest ?
6. Problems involving the Use of Negative Numbers. In
the second problem of Art. 2, whose statement in the
algebraic form was found to be
2 a; + a = 6,
and whose solution is therefore
a; = 1(6 -a),
we said that h must be greater than a. Now while this
restriction is necessary in arithmetic, the method of
algebra obtains a solution for all values of a and h. In
accordance with the enlarged algebraic view, the problem
under consideration may be stated as follows :
Ex. 1. I have no money ; but balance my accounts by setting
off debits against credits, then double the thus ascertained value
INTRODUCTORY LESSONS. 15
of my possessions, and add a dollars to it, and the result will be h
dollars. Can I pay my debts with the money due me, and how
much money, if any, shall I have after all accounts are settled ?
If the excess of credits over debits be x dollars, then by the con-
ditions of the problem 2 x + a dollars is equal to h dollars, or, in
the purely algebraic form,
2x + a = h.
From this equation we obtain, as previously,
Thus, when all accounts are settled, I shall have \ {h — a) dollars.
Let us see what will happen under the three different suppositions
that may here be made.
(i.) If the credits exceed the debits (in amount), I shall be
able to pay my creditors with the money due me and have some-
thing left. This something is \ (h — a) dollars, and obviously h
must be greater than a.
(ii.) If the credits be just equal to the debits, they cancel each
other in the final settlement of accounts, and I shall have just
nothing left. This nothing is again \ (fi — a) dollars, and in order
that I (6 — a) may be zero, h must be equal to a.
(iii.) But if the debits exceed the credits, I shall fail to cancel
all my indebtedness with the money due me, and shall have some
debts remaining and no money to pay them with. I no longer
have possessions, but debts. The amount of my indebtedness is
exactly \ (a — 6), and the condition for this state of affairs is
that a shall be greater than b.
Now algebra takes account of this condition by the simple device
of writing the minus sign before the positive number ^ (a — 6) ,
and calling the result negative ; thus
ic = 1 (6 — a) = — ^ (a — 6),
a negative number, representing indebtedness, if a be greater than
b. The number J (Jb — a) is therefore the true answer to the ques-
tion propounded by the problem, under all circumstances.
Consider a concrete illustration :
16 INTRODUCTORY LESSONS.
Ex. 2. Restate the problem with 55 in place of a, and 5 in place
of b. The solution then is
a; = 1 (5 _ 55) = _ 1 (55 - 5) = -25 ;
and the interpretation of this negative result is that the debits
exceed the credits by $ 25. Thus, if credits = $ 100, then debits
= I 125 ; or if credits = $ 5000, then debits = $ 5025, etc.
We may verify the conclusion here reached by performing the
operations indicated in the problem upon — 25. We first double
— 25 and obtain — 50, then to this negative number — 50 we add
the positive number 55 and obtain 5. These are some of the ordi-
nary processes of algebra whose meaning will be more fully ex-
plained in the next two chapters.
Ex. 3. A steam launch, whose speed in still water is a feet per
minute, goes up an estuary a distance of b feet in c minutes. Show
that the tide is running at the rate of (6 h- c) — a feet per minute, —
inward if 6 -=- c be greater than a, outward if 6 -r- c be less than a.
7. The foregoing problems have brought to view some
of the purposes of algebra and have shown how, by its
larger methods, some of the inconvenient limitations to
which arithmetic is necessarily subject may be removed.
Other like problems, to be considered in subsequent
chapters, will still further attest the superiority of these
methods. But since many of them cannot be success-
fully attempted until the fundamental operations of
algebra are learned, our next task must be a study of the
fundamental operations of algebra.
LANGUAGE OF ALGEBRA. DEFINITIONS. 17
CHAPTER II.
The Language of Algebra. Definitions.
8. Algebra is that branch of mathematics which treats
of the relations of numbers as expressed in equations and
in what are known as algebraic forms, or expressions.
These algebraic equations and expressions constitute
the language of algebra.
9. The Operations of Algebra* are conveniently grouped
in four main divisions, each of which comprises two con-
trasted kinds of operation, called respectively a direct
process and an inverse process. These eight processes,
thus classified, are :
Direct Processes : Inverse Processes :
I. 1. Addition, 2. Subtraction.
II. 3. Multiplication, 4. Division.
III. 5. Involution, 6. Evolution.
^IV. 7. Exponentiation, 8. Logarithmic Operation.
The direct process having been defined, its inverse is
described as that operation which annuls, or undoes, the
direct process.
* Another point of view enables us to regard all the operations of
algebra as forms of the addition process. But the investigations neces-
sary to establish this conclusion are not appropriate in a merely ele-
mentary treatise.
B
18 LANGUAGE OF ALGEBRA. DEFINITIONS.
A full explanation of the character of these processes,
as fundamental parts of algebraic science, forms a neces-
sary introduction to the problems that algebra under-
takes to solve. They will be discussed in the order here
indicated, the first four in Chapters III., IV., and V., the
others somewhat later, as occasion for their use arises.
10. The Symbols of Algebra. For the convenient rep-
resentation of these operations an algebraic symbolism
has been devised; and, as has been exemplified in the
problems of the Introductory Lessons, the signs used in
arithmetic are adopted as a part of it. »"'But for the
adequate representation of all its operations, algebra
requires an enlarged symbolism of which arithmetic
needs only a part.
The symbols of elementary algebra are of five kinds :
(i.) Symbols of Quantity, usually letters of the alpha-
bet, employed to represent numbers.
(ii.) Symbols of Operation, usually called the signs of
algebra, employed to indicate processes to be applied to
the symbols of number. ( + , — , ~, x, ^, % -y/, log.)
(iii.) Symbols of Relation, by means of which compari-
sons of numbers are expressed. ( = , =, >, <.)
(iv.) Abbreviations of frequently recurring words or
phrases. (.*., •••, •••.)^
(v.) Symbols of Aggregation. (Parentheses, braces,
square brackets, the vinculum.)
SYMBOLS OF QUANTITY.
11. In arithmetic, numbers are represented by figures,
each of which has one, and only one, meaning.
LANGUAGE OF ALGEBRA. DEFINITIONS. 19
^ In algebra, numbers are represented either by figures
or by the letters of the alphabet.
In an arithmetical product, such as 3 x 5, it is obvious that the
factors 3 and 5 may be interchanged, and this law of commutation,
as it is called, is easily proved to obtain for any two numbers.
But the statement of the law in the form 3 x 5 = 5 x 3 is the state-
ment of a particular case only. Now algebra states it in the form
a X 6 = 6 X a, in which a and 6 represent any two numbers what-
ever. /The gain in conciseness and generality of statement by the
use of letters' is thus apparent.
The word quantity is used in algebra as synonymous
with number.
SYMBOLS OF OPERATION.
12. The sign -f-, which is read ^plus,^ is placed before
a number, or a number represented by a letter, to indicate
that it is to be added to what has gone before.
Thus 6 + 3 means that 3 is to be added to 6 ; 6 + 3 + 2 means
that 3 is to be added to G and then 2 added to the result : so also
a+ b means that the number which is represented by b is to be
added to the number which is represented by a ; or, expressed
more briefly, it means that b is to be added to a.
13. The sign — , which is read ^ minus,' is placed
before a number to indicate that it is to be subtracted
from what has gone before.
Thus 6 — 3 means that 3 is to be subtracted from 6, a ~ b means
that b is to be subtracted from a, and a — b + c means that b is to
be subtracted from a and then c added to the result.
It should be noticed that in a series of additions and
subtractions the order of the operations is from left to right.
The meaning of the sign ~ is explained in Art. 46.
20 LANGUAGE OF ALGEBRA. DEFINITIONS.
14. The sign of multiplication is x, which is read
^multipHed by' or ^into.'
Thus 6x3 means that 6 is to be multiplied by 3, a x 6 means
that a is to be multiplied by &, and a x 6 x c means that a is to be
multiplied by h and then the result multiplied by c.
The sign of multiplication is generally omitted between
two letters, or between a number and a letter, and the
letters are simply placed side by side. Sometimes the x
is replaced by a point.
Thus ah OT a-h means the same as a x 6, and 2 ahc or 2a'b'C
the same as 2 x a x 6 x c.
15. The sign of division is -i-, which is read 'divided
by' or 'by.'
Thus 6-7-3 means that 6 is to be divided by 3, a -4- 6 means that
a is to be divided by b, and a -^ b -^ c means that a is to be divided
by 6 and then the result divided by c ; also a -^ b x c means that a
is to be divided by b and then the result multiplied by c.
The operation of division is often indicated by placing
the dividend over the divisor with a line between them,
or by separating the dividend from the divisor by an
oblique line called the solidus ; thus either - or a/b is
frequently used instead of a -r-b.
It should be noticed that in a series of multiplications
and divisions the order of the operations is from left to
right.
16. When two or more numbers are multiplied together,
the result is called the continued product, or simply the
product ; and each number is called a factor of the product.
LANGUAGE OF ALGEBEA. DEFINITIONS. 21
17. When the factors of a product are considered as
divided into two sets, each is called the coefficient, that is
the co-factor, of the other.
Thus in 3 ahx, 3 is the coefficient of ahx ; also 3 a is the coeffi-
cient of 6aj, and 3 ah is the coefficient of x.
When one of the factors of a product is a number
expressed in figures, it is called the numerical coefficient
of the other part of the product.
EXAMPLES IV.
Calculate the values of
1. 7 + 6 + 4. 3. il + 7-12-6. 6. 5 -- 3 x 4.
2. 6 - 3 + 4. 4. 7 X 6 X 4. 6. 11 x 7 -- 12 ^ 6.
If a = 1, 6 = 2, c = 3, and (Z = 4, find the numerical values of
7. c-6. 12. 13a-6& + 7c-6d.
8. d-a. 13. 18ft -3c-4d + 9a.
9. 7a -36. 14. 20a6-3cd.
10. 106 -6c. 15. 4<?a-26c.
11. 6a - 2 6 + 6 c - 4 d 16. a6c + bed -\- cda -{■ dab.
If a = 6, 6 = 2, c = 5, and d = 0, find the values of
17. 3ac + 2 6c + ca. 19. a x c ^ 6. 21. 2c -^ a -r- 6.
18. 7aKZ + 96c- 3ca. 20. a ^ c x 6. 22. 3ac-^6.
23. What are the coefficients of x in 3a;, 4 6a;, 5 bcx, and 16 a6ca; ?
24. What are the coefficients of xy in 4xy, 6axy, 7 abxy, and
19 abcxy ? What are the numerical coefficients ?
' 18. When a product consists of the same factor re-
peated any number of times, it is called a power of that
22 LANGUAGE OF ALGEBRA. DEFINITIONS.
factor. Thus aa is called the second power of a, aaa is
called the third power of a, aaaa is called the fourth power
of a, and so on. Sometimes a is called the^rs^ poiver of a.
Special names are also given to aa and aaa ; they are
called respectively the square and the cube of a.
19. Instead of writing aa, aaa, etc., a more convenient
notation is adopted as follows : a^ is used instead of aa,
a^ is used instead of aaa, and a" is used instead of aaa •••,
the factor a being taken n times ; the small figure, or
letter, placed above and to the right of a, showing the
number of times the factor a is to be taken. So also a^b^
is written instead of aaabb, and siniilarly in other, cases.
The small figure, or letter, placed above a symbol to
indicate the number of times that symbol is to be taken
as a factor, is called the index or the exponent. Thus, a"
means that the factor a is to be taken n times, or that
the nth. power of a is to be taken, and n is called the
index.
When the factor a is only taken once, we do not write
it a^, but simply a.
20. The quantity which when squared is equal to any
number a is called the square root of a, and is represented
by the symbol -^a, or more often by -^a : thus 2 is ■y/4:,
for 22 is 4.
The quantity which when cubed is equal to any num-
ber a, is called the cube root of a, and is represented by
the symbol -^a : thus 3 = ^27, for 3^ = 27.
/ In general, the quantity which when raised to the nth
power, where n is any whole number, is equal to any
LANGUAGE OF ALGEBRA. DEFINITIONS. 23
number a, is called the nth. root of a, and is represented
by the symbol -y/a.
/ The sign ^ was originally the initial letter of the word
radix. It is often called the radical sign.
/ A root which cannot be obtained exactly is called a
surd, or an irrational quantity : thus -^7 and ^4 are surds.
The approximate value of a surd, for example of -y/7,
can be found, to any degree of accuracy which may be
desired, by the ordinary arithmetical process ; but we are
not required to find these approximate values in algebra :
for us -^7 is simply that quantity which when squared
will become 7.
SYMBOLS OF RELATION.
21. The sign =, which is read ^equals,' or 'is equal to/
is placed between two quantities to indicate that they are
equal to one another.
Thus 6 4- 7 = 12, which is read five plus seven equals twelve.
The sign > indicates that the number which precedes
the sign is greater than that which follows it.
Thus a>b means that a is greater than b.
The sign < indicates that the number which precedes
the sign is less than that which follows it.
Thus « < 6 means that a is less than b.
The meaning of the sign = is explained in Art. 90.
SYMBOLS OF ABBREVIATION.
22. The sign •. • is written for the word because or since.
The sign .-. is written for the word therefore or hence.
24 LANGUAGE OF ALGEBRA. DEFINITIONS.
■ The sign of continuation, .••, is used for the words and
so on, or and so on to.
Thus 1, 2, 3, 4, ••• means 1, 2, 3, 4, and so on, and xi, x<i^ aJs, ••• Xn
means X\^ X2, Xs^ and so on to x„.
ALGEBRAIC EXPRESSIONS.
23. A collection of algebraic symbols, that is of letters,
figures, and signs, is called an algebraical expression.
^ The parts of an algebraical expression which are con-
nected by the signs -f- or — are called the terms.
Thus 2a — 3bx-\-5cy^ is an algebraical expression
containing the three terms 2 a, — 3 bx, and -f 5 cy^.
24. ''When two terms contain the same letters, every
one of which is raised to the same power in both, they
are called Hke terms.
Thus 3 abH^ and 7 ab^x^ are like terms ; hut although 3 a^bx^ and
7 a^b^x^ contain the same letters, they are not like terms, for all the
letters are not raised to the same power.
25y A monomial expression is one which contains only
one term, and a multinomial expression is one which con-
tains more than one term.
Thus 5 ab^cx is a monomial expression, and a ■{■ b is a multino-
mial expression.
An expression which consists of two terms is often
called a binomial expression, and an expression which con-
sists of three terms is called a trinomial expression.
Monomial and multinomial expressions are sometimes
called respectively simple and compound expressions.
LANGUAGE OF ALGEBRA. DEFINITIONS. 25
SYMBOLS OF AGGREGATION.
26. The parentheses ( ), the square brackets [ ], and
the braces \ \ , are used to indicate that all operations
denoted by signs within the enclosure are to be consum-
mated before any operation without it is performed.
Thus (a + 6) c means that b is to be added to a and that the
result is to be multiplied by c ; again, (a + by means that b is to
be added to a, and that the cube of the result is to be formed ; also
(a -\- 2b)(c — Sd) means that 2 6 is to be added to a and that
3 d is to be taken from c, and that the first of these results is then
to be multiplied by the second.
A line called a vinculum is often drawn over the expres-
sion which is to be treated as a whole : thus a — b — c is
equivalent to a — {b — c)j and Va -f 6 is equivalent to
When no vinculum or bracket is used, a radical sign refers only
to the number or letter which immediately follows it : thus ^^2 a
means that the square root of 2 is to be multiplied by a, whereas
\/2a means the square root of 2 a ; also y/a + x means that x is to
be added to the square root of a, but Va + x means the square
root of the sum of a and x.
' The line between the numerator and denominator of
a fraction acts as a vinculum, for ' is the same as
' 12
Note. — It should be carefully noticed that every term of an
algebraical expression must be added or subtracted as a whole,
as if it were enclosed in brackets. Thus, in the expression
a-hbc — d^e-\-f, b must be multiplied by c before addition, and
d must be divided by e before subtraction, just as if the expression
were written a + (be) — (<? -^■ e) + /.
26 LANGUAGE OF ALGEBRA. DEFINITIONS.
Ex. Find the values of
(i.) 2a-6c + cd-66 -a + ?-?, (ii.) (a + 6)^(26 - 3c)2,
a
(iii.) a + ftc + c«, (iv.) ^/{7 a^^-(h + c)^ +#},
when a = 4, & = 3, c = 1, c? = 0. The results are :
(i.) 2a-hc + cd-Qh-^a + —
a
= 2x4-3x1 + 1 X 0-6x3 --4 + ?-^
4
= 8 - 3 + 0 - I + 1- = 1.
(ii.) (a + &)3(2 6 - 3c)2 = (4 + 3)3(2 X 3 - 3 x 1)2 = 73 X 32
= 343 X 9 = 3087.
(iii.) a^ + ftc _^ c« = 43 4- 31 + 1* = 64^+ 3 + 1 = 68.
(iv.) ^{7a3 + (& + c)3 + #} = ^{7 X 43 + 43 + 03}
= ^{7 X 64 + 64 + 0} = ^(448 + 64) = ^512 = 8.
EXAMPLES V.
1. Write down the values of 2*, 33, 43, 4*, ^64, ^64, ^16, ^125,
^625, and ^32.
Ifa = 2, 6 = 3, c = 4, and <Z = 5, find the numerical values of
2. a2 + 62. 4. 6a2-2 62. e. a2&2 + 02^2.
3. c2 + c22. 5. 4 62 _ c2 + 5 <Z2. 7. a62c2 _ ^25^.
Ifa = 2,6 = 3, c = 4, and (2 = 5, find the numericar values of
8.—-^ + ^- ^. \hc + \ca + lah. 10. 10 c3# _ ig ^252.
5 3 27
11. Ia262c3_ia62c2^. 12. ab^_g^b^.
^ ^ 16 20
If a = 5, 6 = 3, c = 1, d — 0, find the numerical values of
• 13. (2a + 56)(36-6c). 14. (a + 26)(c + 2d).
15. (3a -46)3- 2(36 - 6c)2 + 2(a(2 + 6c)2.
16. 4a3 + 463 + 4c3- 3(6 + c)(c + a)(a + 6).
LANGUAGE OF ALGEBRA. DEFINITIONS. 27
17. 5(a + c)3(6-c)2- J3(a-2d)3(& + 3c)2.
18. Show that x^ — 5 x + 6 is equal to zero, if x = 2 or if x = 3.
19. Show that 2 x^ - 11 x^ + 17 x — 6 is equal to zero, if x = 2,
or if x = 3, or if X = 2'
If a = 5, 6 = 4, c = ^, find the numerical values of
20. Va2 - &2. 21. V6a. 22. V26c + 3a. 23. y/bc + a^.
24. ^(2a2 + 62_8c2). 25. ^(4a2 _ ^52 _,. jc2 - 1).
26. ^(a+b)^(iSab + 2bc).
27. Find the numerical value of
(a + 6)2 (X + y)2 - 4 (ax + 6y) (bx + ay),
if X = 5, ?/ = 8, a = 6, and 6 = 4.
28. Find the value of vs(s — rt)(s — 6)(« — c), when <z = 9,
6 = 12, c = 15, and s = 18.
29. Find the value of J^ {7, s when a = 9, 6 = 12,
\(s-6)(s-c)
c = 15, and 2s =a + 6 + c.
30. Find, when a=:8, 6 = 5, c = 3, the numerical value of
V{2 62c2 + 2 c2a2 + 2 a262 _ a* _ 54 _ c*}.
31. Verify that a^ — b^ and (a + 6)(a — 6) are equal to one
another (i.) when a = 6, 6 = 3 ; (ii.) when a = 9, 6 = 4 ; and (iii.)
when a = 12, 6 = 7.
32. Verify that the expressions
a^ - 63, (a - 6) (a^ + ab + 62), (a -6)8 + 3 ab (a-b),
and (a +6)3-3 a6(a + 6) - 2 63
are all equal to one another (i.) when a = 3, 6 = 2 ; (ii.) when
a = 6, 6 = 3 ; and (iii.) when a = 5, 6 = 2.
POSITIVE AND NEGATIVE QUANTITIES.
27. All concrete quantities must be measured by the
number of times each contains some unit of its own kind.
28 LANGUAGE OF ALGEBRA. DEFINITIONS.
Now a sum of money may be either a receipt or a pay-
ment, it may be either a gain or a Zoss; motion along a
straight line may be in either of two opposite directions ;
a period of time may be either before or after some
particular epoch; and so in very many other cases.
Thus many concrete magnitudes are capable of existing
in two diametrically opposite states.
28. Whatever kind of quantity we are considering,
4-4 will stand for what increases that quantity by 4
units, and —4 will stand for whatever decreases the
quantity by 4 units.
If we are calculating the amount of a man's property
(estimated in dollars), 4- 4 will stand for what increases
his property by $ 4, that is + 4 will stand for $ 4 that
he possesses, or that is owing to him ; so also — 4 will
stand for whatever decreases his property by ^4, that
is — 4 will stand for $ 4 that he owes. If, on the other
hand, we are calculating the amount of a man's debts,
4- 4 will stand for whatever increases his debts, that is
4- 4 will now stand for a debt of $ 4 ; so also — 4 will
now stand for whatever decreases his debts, that is — 4
will stand for f 4 that he has, or that is owing to him.
If we are considering the amount of a man's gains,
4- 4 will stand for what increases his total gain, that is
4- 4 will stand for a gain of 4 ; so also — 4 will stand
for what decreases his total gain, that is — 4 will stand
for a loss of 4. If, however, we are calculating the
amount of a man's losses, 4- 4 will stand for a loss of 4,
and — 4 will stand for a gain of 4.
Again, if the magnitude to be increased or diminished
is the distance from any particular point, measured in
LANGUAGE OF ALGEBRA. DEFINITIONS. 29
any particular direction, +4 will stand for a distance
of 4 units in that direction, and —4 will stand for a
distance of 4 units in the opposite direction.
29. From the above examples it will be seen that the
signs + and — will serve to distinguish between magni-
tudes of opposite kinds. Thus whatever + 4 may rep-
resent, — 4 will represent an equal magnitude, but of the
opposite kind. /The signs + and — are therefore used
in algebra with two entirely different meanings. ' In
addition to their original meanings as signs of the opera-
tions of addition and subtraction respectively, they are
also used as marks of distinction between magnitudes of
opposite kinds./
30. A quantity to which the sign -f is prefixed is
called a positive quantity, and a quantity to which the
sign — is prefixed is called a negative quantity. ^
The signs + and — are called respectively the positive
and negative signs.
31. The signs + and — are often called signs of affection
when they are used to indicate a quality of the quantities
before whose symbols they are placed.
The sign +, as a sign of affection, is frequently
omitted; and when neither the -f nor the — sign is
prefixed to a term, tjie + sign is to be understood.
32. The range of positive and negative algebraic
numbers is obviously double that of the numbers which
belong to arithmetic. Thus, the series of integers in
arithmetic is
0 +1 4-2 +3 +4 ....
30 LANGUAGE OF ALGEBRA. DEFINITIONS.
while the corresponding algebraic series is
... _4 -3 -2 -1 0 +1 +2 +3 +4 ...
In comparing the terms of this double algebraic series,
we adopt the convention that they are here arranged in
ths order of magnitude, so that
-3<-2<-l<0<l<2<3,
and in general, a and h being positive numbers,
— a < — 5. if a > &. [See Art. 47.]
33. ^The magnitude of a quantity considered inde-
pendently of its quality, or of its sign, is called its
absolute magnitude, or its absolute value.
Thus a rise of 4 feet and a fall of 4 feet are equal in absolute
magnitude ; so also + 4 and ~ 4 are equal in absolute magnitude,
whatever the unit may be.
34. Note. — Although there are many signs used in algebra,
the name sign is often used to denote the two signs + and — ex-
clusively. When the sign of a quantity is spoken of, it means the
+ or — sign which is prefixed to it ; and when we are directed to
change the signs of an expression, it means that we are to change
the -|- or — before every term into — and -l- , respectively.
EXAMPLES VI.
Calculate the values of
1. 5 _ 3 _ 4. 2. 5 X 3 - 19. 3. 21-^ 3 - 9.
If a = 1, & = 2, and c = 3, find the values of
4. a - 2 6. 6. 6c - 11 a. 8. 6 -4- a - c.
5. ab — c. t. ab -be. 9, a ^ b — c.
LANGUAGE OF ALGEBRA. DEFINITIONS. 31
10. The reading of a thermometer was + 40°, and its mercury
column then fell 53°. What was its reading after the change ?
11. If « = 2, 5 = 3, and c = — 10, which is the greater, axh
or c ?
12. Can the result of subtracting one positive number from
another ever be greater than the minuend ?
32 ADDITION. SUBTRACTION. BRACKETS.
CHAPTER III.
Addition. Subtraction. Brackets,
addition.
35. The process of finding the result when two or
more quantities are taken together is called addition, and
the result is called the sum.
Since a positive quantity produces an increase, and a
negative quantity produces a decrease, to add a positive
quantity we must add its absolute magnitude, and to add
a negative quantity we must subtract its absolute mag-
nitude.
Thus, when we add + 4 to + 6, we get + 6 -f- 4 = + 10 ;
and when we add — 4 to -|- 10, we get + 10 — 4 = + 6.
So also, when we add 4- 6 to a, we get a + 6 ; and when
we add — 6 to a, we get a — b.
Hence a + (+6) = a + 6 and a-f (— 6) = a — b.
We therefore have the following rule for the addition
of any term : To add any term, affix it to the expression
to which it is to be added, with its sign unchanged.
Ex. A "boy played two games ; in the first game he won 6
points, and in the second he won — 4 points (that is he lost 4).
How many did he win altogether ?
The total gain in the two games together is what is meant in
algebra by the sum of the gains.
To obtain the total gain, we must add — 4 to 6, and this opera-
tion is indicated by 6 + (— 4), which by the above is 6 — 4 = 2.
ADDITION. SUBTRACTION. BRACKETS. 3d
When numerical values are given to a and to b, the
numerical values of a + 6 and a — b can be found ; but
until we know what numbers a and b represent, we can-
not take any further step, and the process is considered
to be algebraically complete.
36. It should be noticed that when b is greater than
a, the arithmetical operation denoted by a — 5 is im-
possible ; for we cannot take any number from a smaller
number.
Thus, if a = 3 and b = 5, a — b will be 3 — 5, and we
cannot take 5 from 3. But to subtract 5 is the same as
to subtract 3 and 2 in succession, so that 3 — 5 = 3 — 3
— 2 = — 2 : we then consider that — 2 is 2 which is to be
subtracted from some other algebraical expression, or
that — 2 is two units of the kind opposite to that repre-
sented by 2. And if — 2 is a Jinal result, the latter is
the only view that can be taken.
In some particular cases the quantities may be such
that a negative result is without meaning ; for instance,
if we have to find the population of a town from certain
given conditions ; in this case the occurrence of a nega-
tive result would show that the given conditions could
not be satisfied, but so also in this case would the occur-
rence of a fractional result.
EXAMPLES VII.
Find the sum of
1. 4 and - 3. 6. 3 and - 11. 9. 2 a and - 3 6.
2. 3 and - 2. 6.-3 and - 9. 10. - 3 a and-2 b.
3. 6and-3. 7. 6, - 2, and 7. 11. 6a, -6 6, and -2c.
4. 7 and -8. 8. - 3, - 2, and 5. 12. -3 a, -4 6, and 7 c.
c
34 ADDITION. SUBTRACTION. BRACKETS.
37. It follows at once from the nature of addition
that ^,the sum of two or more algebraical quantities,
whether positive or negative, is the same in tohatever
order the quantities may he taken. This is known as the
Law of Commutation in addition, j
For example, to find how much a man is worth, we can take
into account the different items of property ;der)is "being con-
sidered as negative) in any order we please.
/ It also follows that to add any algebraical expression
as a whole gives the same result as to aCfd its terms sepa-
rately. This is known as the Law of Association in
addition, But to add any term, we have only to write
it down, with its sign unchanged, after the expression
to which it is to be added.
We have therefore the following
Eule. To add two or more algebraical expressions,
write down all the terms in succession ivith their signs
unchanged.
For example, the sum of a + & and c — d is a + b-\-c — d; also
the sum of a — 6 + c and — d + e — fisa — b + c — d-^e—f.
38. If some of the terms which are to be added are
^like' terms, we can and must collect all such terms
together before the process of addition is considered to
be complete. Of this we have three cases, as follows :
I. The sum of ' like ' terms which have the same sign
is a 'like' term which has the same sign, and whose
coefficient is the arithmetical sum of their numerical
coefficients.
For example, to add 2 a and 5 a in succession gives the same
result, whatever a may be, as to add 7 a ; that is, + 2 a + 5 a = + 7 a.
ADDITION. SUBTRACTION. BRACKETS. 35
Also to subtract 2 ab and 5 ab in succession gives the same result
as to subtract 7 ab ; that is, —2ab — ^ab=—7 ab.
II. The sum of two 4ike' terms whose signs ara
different is a * like ^ term whose coefficient is the arith-
metical difference of their numerical coefficients and
whose sign is that of the greater.
For example, +6a — 3a=+2a + 3a — 3a=+2a;
also -\-3ab - dab=-\-3ab-Sab -2ab = -2ab.
III. If there are several 'like' terms some of which
are positive and some negative, the positive terms can be
collected into one sum by I., as also the negative terms :
the final sum is then obtained by II.
Thus any number of 'like' terms can be reduced to
one term.
Ex. 1. Add 2 a + 3 6 and a -5b.
The sum is2a + 36 + a — 56; or, since the terras can be taken
in any order, the sum is 2a + a + 36 — 56 = 3a — 2 6, by col-
lecting the like terms.
Ex. 2. Find the sum of 6 a2 _ 6 a6 + 462, 2 62 - a6 - a^, and
6 a6 - 9 62 - 4 a^.
The sum is
6a2 - 6 a6 4- 4 62 + 2 62 - a6 - a2 4 6a6 - 9 62 - 4a2
= 6 a2 - a2 - 4 a2 - 6 a6 - a6 + 5 a6 + 4 62 + 2 62 - 9 6-2.
The terms 6 a% — a^^ and — 4 a2 can be combined mentally ;
and we have a"^. Similarly we have — 2 a6 and — 3 62.
Thus the required sum is a2 — 2 a6 — 3 62.
39. It is* best for beginners to place the different sets
of 'like' terms in vertical columns; so that the last
example would be put down in the following form, the
36 ADDITION. SUBTRACTION. BRACKETS.
-f sign being put before the terms 26^ and 5ab, which
have no signs : and then the sets of ' like ' terms can be
combined mentally.
6a2_6a& + 462
- a"^ - a& + 2 62
-4a2^_5a5_952
^2 - 2 a?) - 3 62
EXAMPLES VIII.
Simplify the following by combining ' like ' terms :
1. 2a+6 + 3c + 26 + 3a + 2c + 5a + c.
2. 6a-36 + 2c-4a-3c + 26.
3. 5a2_3a4-6-4a2 + 6a + 3.
4. 7 a3 _ 4 a + 9 - 3 a2 + 2 a + 7 - 3 a3 _ 16.
5. 5 a3 - 4 a26 + 3a52 _ 5 «3 _ 4 ^25 _ 3 53.
6. 3a2 + 6a6-462-2a2_4a6 + 362-a2-2a6 + 62.
Add together
7. a + 6 and a — b. 9. ^a + ^6 and - ^ a + ^ 6>
8. 2x — y and 2 x + y. 10. ^ a + | 6 and f 6 - J a.
11. a^ — a and a^ _}_ q;.
12. a + a2 + 4a3and2a3_a2-4a.
13. m2 + mn + n^ and w2 — mn — w2.
14. 3p2 _|. 5^g _ 6 g2 and 5 ^2 _ 4^^ _ 3^2.
15. 3 a2 - 2 a6 + 6'2 and a2 _ 2 ^6 - | 6^.
16. 2 a + 6 - 3 c, 2 6 + c - 3 «, and 2 c + a - 3 6.
17. 4 a - 3 6 - c, 4 6 — 3 c - a, and 4 c - 3 a — 6.
18. 4 a2 - 3 a6 + 62, 4 a6 - 3 62 + a^, and 4 62 - 3 a2 4. ab.
19. a — ^6 + ^c, 6-^c + ^«, and c - J a + ^ 6.
20. 4X-22/ + 1, -3x + 2-y, anda; + 32/-3.
21. -4a- 6 + 2, 2 + 8a- 56, and -a -46- 2.
^ Q^ --^DITION. SUBTRACTION. BRACKETS. 37
1 / \ ■ ■ 1 .
22. y?^1 x'^y - 2 x«/2, x'^y - 3 xy"^ - y^, and Sxy^-2y^- x^.
23. a3 + 4&3_5c3 + 3a&c, b^ + '^c^ - 3a^ + Qabc, and
c3 + 4 cjs _ 5 53 _ 9 Qjftc.
24. 5 a3 _ 2 a25 4. 9 a&2 + 17 &3^ _ 2 a^ + 5 a^ft - 4 a62 - 12 63,
&3 _ 4 a62 - 5 a26 - a3, and 2 a^ft _ 2 a3 - 6 63 _ a62.
25. I a^ - f a25 + 3 53^ ^^s _ 2 ^52 _ 3 ^,3^ and i a% -lah^-\ 68.
26. 3x3-4x + 5, 2x2 -6x+ 7, 6x3-2x2 -2x, and
3 + 8x-4x3.
27. x3 - 3ax2 + 5 a2x - a3, 2x3 + 4ax2 - 6a2x, 6ax2 _ 3a2x+ a*,
and - 2 x3 + 4 a2x - 5 a3.
28. 3x2 + 2/2 - 3y2 - 02, 2xy - Zy"^ + ^yz, and
- 4 x2 - 2 xy + y2 ^ ^2.
. 29. Show that, if x = 6 + 2c-3a, y = c + 2a-3 6, and
s = a + 26 — 3c; then will x-\- y -{■ z = 0.
30. Show that, if a = 5x — 3y — 2^?, 6 = 5?/ — 3^ — 2 x, and
c = bz — Zx — 2y', then will a + 6 + c = 0.
SUBTRACTION.
40. Since subtraction is the inverse of addition [Art. 9],
an addition and a subtraction of the same quantity pro-
duces no effect. Hence a is the same as a-f 6 — 6. Now
if we take + h from a + h — h^ what is left is a — h. So
that if we take away + h from a, the remainder is a — 6 ;
that is
a — (+6) = a — 6.
Again, if we take away — h from a + & — 6, what is
left is a 4- 6. So that if we take away — h from a, the
remainder is a + 6 ; that is,
a — ( — 6) = a + 6.
38 ADDITION. SUBTRACTION. BRACKETS.
Thus to give a numerical example,
+ 10-(+4) = + 10-4 = 6,
and + 6-(-4) = + 6 + 4 = 10.
We therefore have the following rule for the sub-
traction of any term : to subtract any term, affix it to the
expression from which it is to be subtracted, but with its
sign changed.
EXAMPLES IX.
Subtract
1. 3 from — 4.
3.-6 from 4. 5.-6 from a.
2.-4 from 3.
4. a from — 6. 6. — a from — 6,
Show that :
7. -4-(+3) = -
-7. 10. -b-(+a) = -b-a.
8. 3 -(-4)= 7.
11. -b-i-a) = -b + a.
9. 4 -(-6)= 10.
12. _(_6)_(_a)=6 + a.
41. Since subtraction au^ addition are inverse opera-
tions, and since we know that to add any algebraical
expression as a whole gives the same result as to add its
terms separately, it follows that to subtract an alge-
braical expression as a whole is the same as to subtract
the terms in succession. (Law of Association.)
We have therefore the following
Kule. To subtract any algebraical expression from any
other, write down its terms in succession with their signs
changed, after that other.
Thus, if2a-f-& — 4c be subtracted from 3 a — 4 6 -f c, the
result is
3a-4&-fc-2a-&+4c=3a-2a-46-6 + c-f4c=a-5 6 + 5c.
ADDITION. SUBTRACTION. BRACKETS. 6\)
The law of commutation for subtraction is the same
as that for addition, and has been enunciated in Art. 37 ;
but in applying it, the signs, as well as the letters to
which they are prefixed, must be interchanged. Thus,
+ a — 6 = —l)-^a
is an immediate consequence of
+ a + (-&) = + (-&) + a.
42. The expression which is to be subtracted is often
placed under that from which it is to be taken, * like '
terms being for convenience placed under one another .
and the signs of the lower line are changed mentally
before combining ' like ' terms.
Thus the example considered in Art. 41 would be written as
follows :
3a-46+ c
2a+ 6-4c
a — 66 + 6c
The terms of the result being obtained by combining mentally
3 a and —2 a, — 4 6 and — 6, and c and -1- 4 c.
As another example, if we have to subtract
Sa^-4a%-\-2 ab'^ - b^ from 4a3 + a^b - ab\
the process is written as follows :
4 a8 + a^b - ab"^
3a8-4a26 + 2a62_68
a* + 6 a26 - 3 ab'^ + b^
The terms of the result being obtained by combining 4 a^ and
3a3, a^b and + ^ffib, - ab^ and - 2a6^ 0 and + b^
43. We have hitherto supposed that the letters used
to represent quantities were restricted to positive values ;
40 ADDITION. SUBTRACTION. BRACKETS.
it would, however, be very inconvenient to retain this
restriction. In what follows therefore it must always be
understood, unless the contrary is expressly stated, that
each letter may have any value positive or negative.
Since any letter may stand for either a positive or a
negative quantity, a term preceded by the sign + is not
necessarily a positive quantity in reality ; such terms
are, however, still to be called positive terms, because they
are so in appearance ; and the terms preceded by the sign
— are similarly called negative terms.
44. We must now carefully examine whether terms
can be added and subtracted without knowing whether the
letters really represent positive or negative quantities.
Now we have seen in Arts. 35 and 40, that when h is
really positive,
4-(+5) = + 6, +(_6) = _6, _(4-6) = _6,
and — ( — 6) = + 6 ;
and we have to see whether the same laws hold good
although h may really be negative.
If h be really negative and equal to — c, where c is posi-
tive, then 4-5 = + ( — c) = — c, and — 6 = — (— c) = + c,
since c is positive. Hence, putting — c for + h, and -|- c
for — 6, the laws expressed above will be true, although
b is negative, provided
+ (-c) = -c, 4-(4-c) = + c, _(_c)= + c,
and — ( + c) = — c,
are true for all positive values of c, and this we know is
the case.
ADDITION. SUBTRACTION. BRACKETS. 41
Hence terms are added or subtracted in precisely the
same way whether the letters really stand for positive or for
negative quantities.
45. Zero, whose symbol is 0, may now be defined as a
difference, in the form
a — a = 0.
When it appears in any algebraic expression, as a term
in the form 0, or as a pair of terms in the form a — a,
or as a series of terms reducible to either of these forms,
it may obviously be erased without in any way affecting
the numerical value of the whole expression.
46. Def. The algebraical difference between any two
quantities a and h is the result obtained by subtracting
the second from the^rs^
For example, tlie algebraical difference of 5 and 4 is 5 — 4 = 1,
and the algebraical difference of 4 and 5is4 — 5 = — 1.
The algebraical difference between tvo quantities de-
pends upon the order in which they are given, and may
therefore not be the same as the arithmetical difference,
which is the result obtained by subtracting the less from
the greater.
The symbol a ~ 6 is used to denote the arithmetical
difference of a and h.
47. Def. One quantity a is said to be greater than
another quantity 6,' when the algebraical difference, a — b,
is positive.
From this definition it is easy to see that in the series 1, 2, 3,
4, etc., each number is greater than the one before it ; and that,
42 ADDITION. SUBTRACTION. BRACKETS.
in the series — 1, — 2, — 3, — 4, etc., each number is less than the
one before it. Thus 7, 6, 1, 0, — 5, - 7 are in descending order of
magnitude. [See Art. 32.]
EXAMPLES X.
Subtract
1. a — b from a + 6. ^. ^x — ly from ^x — ^y.
2. 2a -3& from 3a -26. 6. 3x - 4x2 from 4x - 3x2.
3. 2 X + ?/ from 2x-y. 6. x^ - 2 from 1 - 2 x".
7. 4a-26 + 3c from 2c+4&-3a.
8. 4 a2 - 2 a6 + 3 &2 from 2b'^ + 4.ab - Sa^
9. 3 x2 - 4 X + 2 from x2 + 6 x - 7.
10. 3x3-2x2 + 5 from 3x3-2x4- 5.
11. a — ^b — ^c from b — ^c — ^a.
12. ^x2 - ixy + |?/2 from 1 2/2- ix?/ + ^x2.
Find the difference between
13. a + 26 and a -26. 14. 3a - 76 and 7a - 36.
15. a2 + a6 + 62 and a2 - a6 + 62.
16. x2 - 3 xy and 3 x2 - 4 xy.
17. 3x3 + 5x2?/ + 4 jcy2 and 4x^y + 6 xy^ + 7 y^
18. 2 x8 - 7 x2?/ + 9 x?/2 - 4 ?/3 and 4 x^ - xSy + 9 x?/2 - 4 1/«.
19. What must be added to 2 a — 3 6 in order that the sum may
be 4 a - 6 6 ?
20. What must be added to a2 + 62 in order that the sum may
be 2 a2 - a6 ?
21. What must be added to 5 a6 — 2 6c + 3 ca in order that the
sum may be 7 a6 + 2 ca ?
22. What must be added to a2 + 3 62 + 2 c2 that the sum may
be62-3a2?
ADDITION. SUBTRACTION. BRACKETS. 43
23. Subtract from 3 a - 4 6 the sum of2a + 7 6, -4a-6&,
and 6 a — 5b.
24. Subtract from 3 x^ - 2 a; + 7 the sum of x^ - x + 9, 2x^
-f 7 X - 6, and 3 a;2 - 4 X - 5.
26. Subtract the sum of a'^-4:ah-^ b\ ab-ib'^-Sa^, and
62 _ 4 ^2 _ 3 a6 from 2 a^ -^ 2 b^ + 2 ab.
BRACKETS.
48. To indicate that any algebraical expression is to
be added as a whole, it is put between brackets (paren-
theses, braces) with the + sign prefixed. But, as we
have seen, to add any expression, we have only to write
down the terms in succession, with their signs unchanged.
Hence, when a bracket is preceded by a + sign, the
brackets may be omitted. Thus
4-(2a-6-fc) = 2a-& + c.
Conversely, any number of terms in an expression may
be enclosed in brackets with the sign + placed before
the bracket.
For example, we may write
a-26 + c + 2d--3e+/
in the form
a4-(-26 + c) + (2(«-3e+/),
or a-26 + (c-f2d)-|-(-3e-f/).
When the sign of the first term within the brackets is
-h, it is generally omitted for shortness, as in the preced-
ing example.
49. To indicate that any algebraical expression is to
be subtracted as a whole, it is put between brackets, and
44 ADDITION.' SUBTEACTION. BRACKETS.
the — sign prefixed. But, as we have seen, to subtract
any expression, we have only to write down the terms in
succession with all their signs changed.
Hence, when a bracket is preceded by the — sign, the
brackets may be omitted, provided that the signs of all
the terms within the brackets are changed.
Thus -(2 a — 6 + c) = -2a + 6-c.
Conversely, any number of terms in an expression may
be enclosed in brackets with the sign — placed before
the bracket, provided that we change the signs of all the
terms which are placed in the brackets.
Thus a-2b-{-c + 2d — 3e-\'f
may be written in the form
a-{2b-c)-(-2d + 3e -/) .
50. Sometimes one enclosure is put within another:
in this case the different pairs of brackets must be of
different shapes to prevent confusion.
Thus a— [6 + Jc — (d4-e)5] ; which means that we
are to add to b the whole quantity within the braces \ \,
and then subtract the result from a; and to find the
quantity within the braces JJ? ^® must add d and e, and
then subtract the sum from c.
When there are several pairs of brackets, they may be
removed one at a time by the rules of Arts. 48 and 49 ;
and it is best for beginners to remove at every stage the
innermost bracket.
Thus a - [6 + |c - (d + e) }] = a - [6 + {c - d -ej]
= a— [6 4-c — d — e] = a — 6 — c + d + e.
ADDITION. SUBTRACTION. BRACKETS. 45
EXAMPLES XI.
Simplify the following expressions by removing the brackets
and collecting like terms :
1. (a + b)-(a-b). 6. a -[a - {a -(2a - a)}].
2. a-b-(a-\-b). 7. 1 -[2 - {3 -(4 - 5)}].
3. a -(& + c) + (6 -c-a). 8. a + 6-[a-6+{a + 6-(a-6)}].
4. Sx-(y-2x) + (z-{-y-5x).9. 5 -[4 + {5 -(4 + S^^l)}].
6. x-{y -(z - x)}. 10. x-[y-{z-{x-y - «)}].
11. 3x-{2y + 50 - 3x + y}.
12. {2x-(i^y-Sz -h 7)] -li + {X -(Sy -\-2z + 5)]].
13. [2 a -{36+ (4c --36 + 2 a)}].
14. x-(y - z)-\-{2z- Sy-6x].
15. a - 2 6 - {3 a - (6 - c) - 5 c}.
16. a-[36 + {3c-((f-6)+a}-2a].
17. 3a -[26 — {4c -12a- (46 - 8c)} - (6 6 - 12c)].
18. {2x-(Sy-7z) + (Sx-2y + 9z)}
- {(y - 50)-(3x ~y-2z) + Sz],
19. a2-(3a6-462)-(2a2-3a6 + 662)
-{5 62-(3a6-7a2-62)}.
20. (m2 - w2) - {3 WW - (5 n2 - m^)}
+ [n2 - {3 mw - (5 m^ - 6 n2)} + 8 mn].
46 MULTIPLICATION.
CHAPTER IV.
Multiplication.
51. In arithmetic, multiplication is first defined to be
the taking one number as many times as there are units
in another. Thus, to multiply 5 by 4, we take as many
fives as there are units in four. As soon, however, as
fractional numbers are considered, it is found necessary
to modify somewhat the meaning of multiplication, for
by the original definition we can only multiply by whole
numbers. The following is therefore taken as the defi-
nition of multiplication :
Def. To multiply one number by a second we do to the
first what is done to unity to obtain the second.
Thus 4isl-f-l + l + l;
.-. 5 X 4 is 5 -f 5 + 5 + 5.
Again, to multiply 4 by f, we must do to -f-. what is
done to unity to obtain J; that is, we must divide ^ into
four equal parts and take three of those parts. Each of
5
the parts into which 4 is to be divided will be -, and
5x3 '^X^
by taking three of these we get - — -•
So also (-5)x4=-5+(-5)-f-(-5)-i-(-5)
= _5_5_5-5
= -20.
MULTIPLICATION. 47
With the above definition multiplication by a negative
quantity presents no difficulty.
For example, to multiply 4 by — 5. Since to^subtract
5 by one subtraction is the same as to subtract five units
successively,
... 4x (-5) = -4 -4 -4- 4- 4
= -20.
Again, to multiply — 5 by — 4. Since
-4 = -l-l-l-l;
.-. (-5)x(-4) = -(-5)-(-5)-(-5)-(-5)
= + 54-5 + 5 + 5 [Art. 40.]
= + 20.
We can proceed in a similar manner for any other
numbers, whether integral or fractional, positive or nega-
tive. Hence we have the following laws :
ax 6 = + a6 . . . (i.)
(—a) X 6 = — a6 . . . (ii.)
ax{—b) = —ah . . . (iii.)
(-a)x{-b) = -\-ab . . . (iv.)
The rule by which we determine the signs of the prod-
ucts is called the Law of Signs; this law is sometimes
enunciated briefly as follows : like signs give +, and
unlike signs — .
52. The factors of a product may be taken in any order.
It is proved in arithmetic that when one number,
whether integral or fractional, is multiplied by a second.
48 MULTIPLICATION.
the result is the same as when the second is multiplied
by the first.
The proof is as follows :
First, when the numbers are integers, a and h suppose,
write down a series of rows of dots, putting a dots in each
row, and take h rows, writing the dots under one another
as in the following arrangement :
a m a row
h rows.
Then, counting by rows, the whole number of the dots
is a repeated h times ; that is, a xb. Also, counting by
columns, the whole number of the dots is b repeated a
times ; that is, b x a. Hence, when a and b are integer s,
a xb = b xa.
Next, when the numbers are fractional, for example,
5x3
j- and J, we prove as in Art. 51 that ^ x f = = — j- And,
5x3 3x5
by the above proof for integers, = ; hence
7x44x7
T X ^ — 4 X y.
Hence ab = ba for all positive values of a and b ; and
being true for any positive values of a and b, it must be
true for all values, whether positive or negative; for,
from the preceding article, the absolute value of the
MULTIPLICATION. 49
product is independent of the signs, and the sign of the
product is independent of the order of the factors.
Hence for all values of a and h we have
ab = ba . . . . . . . (i.)
This is the Law of Oommutation in multiplication.
If in the above arrangement of dots we put c in the
place of each of the dots, the whole number of c's will be
ab ; also the number of c's in the first row will be a, and
this is repeated b times. Hence, when a and b are
integers, c repeated ab times gives the same result as c
repeated a times and this repeated b times. So that to
multiply by any two whole numbers in succession gives
the same result as to multiply at once by their product,
and the proposition can, as before, be then proved to be
true without restriction to whole numbers or to positive
values.
Thus, for all values of a, 6, and c, we have
a xb X c = ax (be) (ii.)
This is the Law of Association in multiplication.
From (i.) and (ii.) it follows that the factors of a
product may be taken in any order without altering the
result, however many factors there may be.
63. Since the factors of a product may be taken in
any order, we are able to simplify many products.
For example :
3a x4a = 3x4xaxa = 12a2,
(-3a)x(-46) = 4- 3a X 46=+3x 4 X a X & = 12a6,
(a6)2 =:ahxah = axaxhy.h = a^b^,
(^ a)2 = V2 a X V2 « = V2 X V2 X a X a = 2 a2.
50 MULTIPLICATION.
Although the order of the factors in a product is
indifferent, a factor expressed in figures is always put
first, and the letters are usually arranged in alphabetical
order.
54. By definition, c? — aa, a^ = aaa, a^ = aaaa, etc.
Hence a^ X a^ = aa x aaa = a^ = a^+^ ; ,
also . a^ X a^ = aaa x aaaaa = a^ = a^+* ;
and a x a'* = a x aaaa — a^ = a^+'*.
In the above examples we see that the index of the
product of two powers of the same letter is equal to the
sum of the indices of the factors. We can prove in the
following manner that the above is true whenever the
indices are positive integers :
Since, by definition,
a"* = aaaa • • • to m factors,
and a" = aaaa • • • to n factors ;
.-. a*" X a" = (aaa • • • to m factors) x (aaa • • • to n factors)
= aaa • • • to m + w factors
= a'"+'*, by definition.
Thus, for any positive values of m and n
a"^ X a'' = a"'+^
This result is called the Index Law.
55. Product of Monomial Expressions. The results arrived
at in the preceding articles will enable us to find the
product of any monomial expressions. These results
are :
(i.) The sign of the product of two quantities is
MULTIPLICATION. 51
4- when the factors are both positive or both negative ;
and the sign of the product is — when one factor is
positive and the other negative.
(ii.) The factors of a product may be taken in any
order.
(iii.) The index of the product of any two powers
of the same quantity is the sum of the indices of the
factors.
Ex. 1. Multiply 3 a262 by 6 a^fts.
3 a262 x6 ^253 = 3 X 6 X a2 X a2 X ?)2 X 68, from (ii.)
= 18 a2 + 252 + 3=18 a*65, from (iii.)
Ex. 2. Multiply - 3 a26 by - 6 ah^.
(-3 a2&) X (- 5 a&5) = + 3 a26 x 5 ah^, from (i.)
= 3x5xa2xax6x6®, from (ii.)
= 16 a2 + ifti + 6 = 16 a^h^, from (iii.)
Ex. 3. Multiply 2 a^h^d^ by - 6 a^h.
The whole work of finding the product of two monomial expres-
sions can be performed mentally and the result written down at
once. First put down the sign of the result; then multiply the
numerical coefficients to find the numerical coefficient of the
product ; then take each letter whiclf occurs to a power whose
index is the sum of the indices of that letter in the factors. Thus
the product of
2 a'^h^c^ and - 6 a^h is - 10 a'h^c^.
Ex. 4. Find the cube of 2 a^h.
The cube is 2 a2?, x 2 a26 x 2 a^h = 8 a^b\
52 MULTIPLICATION.
EXAMPLES XII.
Multiply
1. 3 a by 6 a. 10. 2 a by -4 &.
2. 5 a2 by 7 a. 11. 3 & by - 4 a.
3. 2 a^ by 5 a^. 12. a^ by - a.
4. rt6bya2&3. 13. _6a36by4a6.
6. 3a2&by2a62. 14. -2ah^\>y -1 aW.
6. 4 a63 by 7 a462. 15. _ 3 a^^c^ by 6 aftV.
7. 3a2&c3by6a6. 16. - 3 a&2c by 2 a&3c2.
8. 5 ahH^ by 3 aZ)^. 17. _ 2 a;?/* by - 5 a:V-
9. 2 aWc by a6c. 18. 2 ax^y^z by - 5 a2icy4;23.
19. 6 aWcH^y^z by - 12 a&3c4a;2?/32;5.
20. Find the values of (- a)2, (- of, and (- a)*.
21. Find the values of (- ^2)2, (- x2)3, and (- x'^y.
22. Find the values of (- a6)2, (- ahf, and (- aby.
23. Find the values of {a^h^Y, {a%^y, and (a2&3)4.
24. Find the values of (- 2 a%^y, (- 2 a354)3^ and -(2 a^ft*)*.
25. Find the squares of 3 aV^o^, — 2 a3?)c4, and — 4 a263c5.
26. Find the cubes of a^^ — «*, ah, and — a25.
27. Find the cubes of 2 a62, _ 3 a3&2^ _ 4 0,55^ and - 7 a255c4.
28. Find the values of (- a)2 x (- &)3, ( - 2 a)3 x (a2)2,
(- aW-Y X (- a2&)3, and (a262)3 x (-a63)4.
If a = 2, & = — 3, c = — 1, (? ,= 0, find the numerical values of
29. 6a6. 35. 3 a262c3 + 5 52^4.
30. 4a&c. 36. 2 a&2 + 3 5^2 + 4 C(Z2.
31. 8a262c2. 37. (a + 6)(c+(?).
32. ab^U^ca. 38. (a - &)2(c - d)2.
33. a3 4. 53 + c3 + (Z3. 39. (a2 + jc) (62 + cd;>),
34. a252 _|. ^,2^2 _|. c2(22. 40. (a - 6)3(c - a)^.
MULTIPLICATION. 63
56. We now proceed to the multiplication of multino-
mial expressions.
We first observe that any multinomial expression can
be put in the form
a-\-b + c-{- etc.,
where a, 6, c, etc., may be any quantities positive or nega-
tive.
For example, the expression ^x^y — ^ xy"^ — 7 xyz, which
by Art. 35 is the same as ^ary -{-{— ^xy^)-{-{—l xyz)^
takes the required form if we put a for 3a^y, h for
— ^xy"^, and o for — 7 xyz.
It therefore follows that in order to prove any theorem
to be true for any algebraical expression, it is only neces-
sary to prove it for the expression a -)- 6 -f- c -f etc., where
a, b, c, etc., are supposed to have any positive or negative
values.
57. Product of a Multinomial Expression and a Monomial.
Consider the product {a 4- b)c, where a, 6, and c have any
values whatever.
If c be a positive integer^ and a and h have any values
whatever; then
{a-\-b)c= (a-f 6) -f- (a-f 6) + {a-{- b) -\ repeated c times,
— a-{-b-\-a-\-b-\-a-\-b-\
= a4-a+a-f ••• repeated c times
-\-b-\-b-\-b-\- ...repeated c times,
=.ac-{-bc.
Hence, when c is a positive integer, we have
(a-h b)c = ac -f be.
54 MULTIPLICATION.
Since division is the inverse of multiplication, it fol-
lows that when d is a positive integer,
(a + b)-i-d = a-i-d-{-b-^d.
And since the operations of multiplication and division
to be performed on a + 6 can be performed on a and b
separately, this must also be the case for the complex
operation denoted by a fraction —
Thus (a -{-b)c = ac-\-bc for all positive values of c;
and being true for any positive value of c, it must also
be true for any negative value. For, if
(a + b)c = ac-{- be,
then — (a + b)c = — ac — be;
and therefore (a + 6) (— c) = a(— c) + ^(— c).
Hence, for all values of a, b, and c, we have
{a-]-b)c = ac -\- be.
58. Since (a-{-b)c = ac-\-bc for all values of a, b, and
c, it will be true when x-\-y is put in place of a.
Hence \ {x -\- y) + blc = {x -\- y)c -\- be = xc + ye + be;
.-. {x-\-y + b)c = xc-\-yc-\-bc.
And similarly
{x-\-y-\-z-{-p-] )c = xc-{-yc-\-zc-\-pc-\ ,
however many terms there may be in the expression
x-{-y-\-z-\-p-\-'".
Thus the product of any multinomial expressio7i by a '^
monomial is the sum obtained by multiplying the separate
terms of the multinomial expression by the monomial
MULTIPLICATION. 56
This result is called the Distributive Law in multi-
plication.
Ex. 1. Multiply a2 + a^hy a.
The result is a'^ x a + a^ x a = a^ + a^.
Ex. 2. Multiply a — 6 by c.
The result is a x c + {~ b)x c = ac — be,
Ex. 3. Multiply - 3x2 by a; - 1.
-3x2x(x-l) = (x-l)x(-3x2) = -3x8 + 3«2.
EXAMPLES XIII.
Multiply
1. a + 6 by 3. 8. 2 a^ - 3 a - 4 by - 3 a^.
2. 2 a -ft by 4. 9. 2 a^ - 3 a6 + 2 62 by a262.
3. 3 a - 4 & by 6. 10. be -\- ca -\- ab by abc.
4. a2 + a by a. 11. 2x^ - 3x2 + 5x - 4by - 5x2.
5. a2-abya8. 12. 4 - 3x2 + 3x8 - 4x*by- 6x»
6. a^ + lhySa^ 18. -6a6 by 3a2-2a6 + 7 &2.
7. 4a2_6a + lby a*. 14. - 6a86*by2a8-3a26-662.
Simplify
15. 2(a-6)-f 4(a + 6). 17. c^a -^ b)- c(a- b).
16. K^-2c)-|(c-2 6). 18. 7a(6-c)-26(a-c).
19. a262(c2 _ (f2) + c2d2(a2 - 6*2) + 62c2((f2 _ a2).
20. 2{3a6-4a(c-2 6)}.
21. 3a -2[6- {2c -6a -2(6 -2c)} -3(6 -2c)].
22. 4 aS - [(2 63 - 3c8) - 6 6c(6 + c) + 3 c(c2 + 2 62)].
23. 7 ac - 2 {2 c(a- 3 6)- 3(5 c-^^6)a}.
56 MULTIPLICATION.
59. Product of Two Multinomial Expressions. We have
now to consider the most general case of multiplica-
tion ; namely, the multiplication of any two multinomial
expressions.
We have to find '
(a + & + c+-..) x(x4-2/ + 2J---);
and, from Art. 56, this includes all possible cases.
Put Jf for x-^y-^z-i ; then, by the last article,
we have
(a + & + c 4- "')M=: aM+ hM-{- cM+ ...
= Ma-\-Mb-\- Mc + ... [Art. 52.]
= (x-^y + z-{-'")a + {x + y-\-z-{- ...)6
+ (a;-f-2/ + 2;+...)c-f---
= ax -\- ay -{- az -{- \- bx -{- by + bz -\
-{- ex -\- cy + cz -\ 1
Hence {a + b -\- c-\- "')(x -\-y -\-z-{- -")
= ax.-\- ay -\- az -\ \-bx-{-by-{-bz-\ \- ex -\- cy -{- cz -]
Thus the product of any two multinomial expressions is
the sum of the products obtained by multiplying every term
of the multiplicand by every term of the multiplier.
For example,
(a + 6) (c + ^) = etc + 6c + ad -f- bd.
Also
(2a+56)(3aH-26)=2ax3a + 56x3a + 2ax26 + 5&x26
= 6a^-fl5a6 + 4a6 + 10 52=6a2+19a6 + 1062.
MTJLTIPLICATION. 57
Again, to find (a — 6)x(c — d); we must first write
this in the form J a-f (— 6) j Jc + (— d) j, and we then
have for the product
ac + ( - 6)c + a( - d) + ( - 6) ( - d),
which by Art. 51 is equal to
ac — bc— ad-{- bd.
Note. In the rule given above for the multiplication of two
algebraical expressions it must be borne in mind that the terms
include the prefixed signs.
60. The following are important examples :
(i.) (a 4- by = (a + 6) (a -f- 6) = aa -h 6a + a6 + &6;
.-. (a + by = a'-^2ab-^b\
Plence, the square of the sum of any two quantities is
equal to the sum of their squares plus twice their product.
(ii. ) (a-by = (a-b) {cl^-th)=^ + ( - b) a + at - b')
-^{-b)(-b) = a^-ab-ab+ b')
.'. (a-by=a^-2ab-{-b\
Hence, the square of the difference of any two quantities
is equal to the sum of their squares minus twice their
product.
(iii.) (a + b){a-b) = aa + ba + a{—b) + b(- b)
= a^ -{- ab — ab — b^]
.'. {a-\-b){a-b) = a--b'.
Hence, the product of the sum and difference of any two
quantities is equal to the difference of their squares.
68 MULTIPLICATIOK.
61. It is usual to exhibit the process of multiplication
in the following form :
a2 + 2 a& - 62
a^ + 2a^h- o?h''
-2a^h-4.a^h^ + 2ab^
o?W + 2aW-h'
a* _4a262 + 4a63_54
or by a rectangular arrangement as follows :
-2ah
-2o?h-4.o?h^ + 2ah^
4- a262 + 2a63-6*
a* _4a262 4.4a53_54
The multiplier is placed under the multiplicand,
arranged horizontally, or to the left arranged vertically.
The successive terms of the multiplicand, namely a?,
+ 2 ah, and — Ir^, are multiplied by a?, the first term on
the left, or at the top, of the multiplier ; and the products
a^, + 2 a^h, and — a^h^, which are thus obtained, are put
in a horizontal row. The terms of the multiplicand are
then multiplied by — 2 ah, the second term of the multi-
plier, and the products thus obtained are put in another
horizontal row, the terms being so placed that ^like'
terms are under one another. The terms of the multi-
plicand are then multiplied by h^, the last term of the
multiplier, and the products thus obtained are put in a
third horizontal row, ^like' terms being again placed
under one another. The final result is then obtained by
MULTIPLICATION.
59
adding the rows of partial products ; and this final sum
can be readily written down, since the different sets of
' like ' terms are in vertical columns.
62. The following are additional examples of multipli-
cations arranged as in the preceding article :
a + 6 a — b a-{-b
a + 6 a — b a — b
a^ + ab
a^ — ab a^ + ab
-\-ab +b^
-ab +b^ -ab-
-b'
a^ + 2ab+ b^
a^-2ab-\-b^ a^
-W
a +^2b
a + 64-c
a -V2&
a
a- -h a6 + a^
a^-]--y/2ba
+ b
+ ab -{-b^ +
be
-^2ba-2b^
+ c
-hac +
bc + c"
2b'-
a^'\-2ab^2ac^-b^^-2bc-\-<?
63. While these arrangements are helpful to the be-
ginner, the practised worker in algebra writes out the
product of two simple multinomials without going
through this somewhat long and formal process. Let
the results in some of the examples of the following
list be written out in this way. Thus, in order to write
down the product of ic + 1 and ic^ -f- a; -|- 1, observe that
there can be only four kinds of terms ; namely, terms in
a^, a;^, x, and a term independent of x. The only partial
product containing a^ is xy,£-\ two partial products,
xy^x and 1 x a^, contain oi? ; two, x x 1 and 1 x a;, con-
tain X ; and one, 1 x 1, is independent of x. Hence the
product is
(a; + 1) («2 + a; + 1) = ar^ + 2 ar^ -f 2 a; + 1.
60 MULTIPLICATION.
EXAMPLES XIV.
Multiply
1. x + 2yhj x-2y. 11. 2 ?/ + 56 by 3?/ - 4 6.
2. a-Sbhj a + Sb. 12. 3 m2 - 1 by Sm^ + 1.
Z. 2x + Syhy Sx-2y. U. 2m^ + 6n^hy2 m^ - 5 ii^.
4. 5 a + 4 6 by a - 6. 14. a + ^ 6 by a - ^ 6.
5. x+7byx + 6. 15. 2a + ift by 3a + i6.
6. X - 7 by X - 6. 16. i a - 1 6 by i a - i 6.
7. ir + 7 by a; - 6. 17. x2 + x + 1 by cc - 1.
8. a + 9 by a - 5. 18. x2 - x + 1 by cc + 1.
9. 2 X - 4 by 2 X + 6. 19. a^ + a6 + 6^ by a - 6.
IQ. 3x-7by2x-l. 20. a^ - a6 + ft^ by a + 6.
21. 4 a'^ + 6 a6 + 9 62 by 2 a - 3 6.
22. 16^2 + 20m + 25 g2 by 4^) - 5 g.
23. x3 - 3 ax2 + 2 a^x by x.-\- 3 a.
24. a3 - 4 a26 + 6 a62 by a2 + 4 a6.
25. x3 - 3 x2 + 2 X + 1 by x2 + 3 x 4- 2.
26. x^ + x2 - 2 X + 1 by x2 - X + 2.
27. x2 + x?/ + ?/2 by x2 - x?/ + ?/2.
28. a* + a262 + 6* by a* - a26^ + 6*.
29. 2 x^ - 3 x2?/ + 2 x?/2 + 2/3 by x2 + 3 x?/ + 2y^,
30. x3 - 4 x2?/ + 6 x?/2 - 3 ?/3 by 3 x2 - 4 x?/ + 53/2.
64. If in any expression consisting of several terms
which contain different powers of the same letter, the
term which contains the highest power of that letter be
put first on the left, the term which contains the next
highest power be put next, and so on ; the terms, if any,
MULTIPLICATION. 61
which do not contain the letter being put last ; then the
whole expression is said to be arranged according to
descending powers of that letter. Thus, the expression
a^ 4- a^h + aW + W, is arranged according to descending
powers of the letter a. In like manner we say that the
above expression is arranged according to ascending powers
of the letter b.
65. Although it is not necessary to arrange the terms
either of the multiplicand or of the multiplier in any-
particular order, it will be found convenient to arrange
both expressions either according to descending or accord-
ing to ascending powers of the same letter : some trouble-
in the arrangement of the different sets of like terms of
the product in vertical columns will thus be avoided.
Hence, before beginning to find the product of two ex-
pressions, it is often desirable to rearrange the terms.
66. A term which is the product of n letters is said
to be of n dimensions, or of the nth degree. Numerical
factors are not to be counted in reckoning the number of
dimensions. Thus abc is of three dimensions, or of the
third degree ; and 5 a^b\ that is 5 aabbc, is of five dimen-
sions, or of the fifth degree.
67. The degree of an expression is the degree of that term
which is of highest dimensions.
• In estimating the degree of a term, or of an expression,
we sometimes take into account only a particular letter,
or particular letters : thus we say that ax^ + 6a; + c is of
the second degree in x, or is a quadratic expression in x ; also
that aoc^y + bxy^ is of the second degree in x, and of the
third degree in x and y.
62 MULTIPLICATION.
When all the terms of an expression are of the same
dimensions, the expression is said to be homogeneous.
Thus aj^ + 3a-6 — 56^ is a homogeneous expression, every
term being of three dimensions ; and ax^ + hcxy -\- dy^ is
a homogeneous expression in x and y.
EXAMPLES XV.
Arrange the following expressions according to descending
powers of a :
1. a^ + 6^-2 ab. 3. a^ + S^ + a% + ab^.
2. 2-4a2 + 5a-6a8. 4. 5^2 _ 4 _ 6a3 _ 2a.
5. aS + &3 + c8 - 3 abc.
6. a^ + 53 _|_ c3 -(- a^b + ab^ + a^c + ac2 + 62c + bc\
7. What are the degrees of the above expressions, and which
are homogeneous expressions ?
Bracket together the different powers of x in each of the fol-
lowing expressions :
8. x"^ -\- ax + bx + ab. 9. x"^ — ax — bx — ab.
10. x^ + ax"^ + bx^ + ca;2 -i- hex + cax + abx + abc.
11. asc^ + 6aj2 + ca;2 -f bcx + caaj + a&a; + abc.
12. a:.2(|/-;2)+?/2(;2-a:)+;3:2(^_y).
13. Simplify (Jb + c - a)x + (c + a - &)a: -\-{a + h — c)x.
14. Simplify (& — c)x+ (c — «)cc + (a — 6)a;.
15. Simplify {(6 - a)x + (c - d)?/} + {(a + 6)a; + (c + <?)«/}. '
68. Product of Homogeneous Expressions. The product of
any two homogeneous expressions must be homogeneous ;
for the different terms of the product are obtained by
multiplying any term of the multiplicand by any term
MTJLTIPLICATION. 63
of the multiplier, and the number of dimensions in the
product of any two monomial quantities is clearly the
sum of the number of dimensions in the separate quan-
tities ; hence if all the terms of the multiplicand are of
the same degree, as also all the terms of the multi-
plier, it follows that all the terms of the product are of
the same degree.
When two expressions which are to be multiplied are
homogeneous, that fact should in all cases be noticed by
a student ; and if the product obtained is not homo-
geneous, it is at once seen that there is an error.
69. We now return to the three important cases of
multiplication considered in Art. 60, namely,
{a-\-by = a\+2ah-\-b^ . . . (i.)
(a-by = a'-2ab-\-b^ . . . (ii.)
(a + 6)(a-6) = a2-62 ^ ^ (^^^ )
Def. A general result expressed by means of symbols
is called a formula.
Since the laws from which the above formulae were
deduced were proved to be true for all algebraical quan-
tities whatever, we may substitute for a and for b any
other algebraical quantities, or algebraical expressions,
and the results will still hold good.
We give some examples of results obtained by substi-
tution in the above formulae.
In the first place, let us put —b in the place, of b
in (i.) ; we then have
5a-f-(-6)J2 = a2^2a(-6)-f(-6)2-j
64 MULTIPLICATION.
that is, (a-by = a^- ^ab + b\
Thus (ii.) is readily seen to be included in (i.)
Now put 6 + c in the place of b in (i.) ; we then have
f a + (& + c) p = a^ + 2 a(6 + c) + (6 + c) 2 ;
.-. \a + b + cl^ = a'-j-2ab + 2ac + b^ + 2bc-^(^.
Thus {a+b-{-c\'=a^-\-b'+c'-^2ab-^2ac+2bc . (iv.)
Now put — c for c in (iv.), and we have
ja + 6 + (-c)P=a2 + 62 + (_c)2 + 2a5 + 2a(-c)
+ 26(-c);
... (^a + b-cy = a^ + b' + c' + 2ab-2ac-2 be.
Again put 6 + c in the place of b in (iii.) ; we then
have
la-\-{b-{-c)]la-{b-\-c)\==a'-{b + cy.
The following are additional examples of products
which can be written down at once :
(a2 + 52) («2 _ ^2) ^ (^2)2 _ (52)2 = ^4 _ ^4^
(a- 6 + c)(a + 5 -c) = {a-{b-c)}{a-h(ib - c)}
= ^2 - (6 - c)2 = a2 _ (52 _ 2 5c + c^) = ^2 _ 52 _|. 2 5c - c^
(a2 + a5 + 52) (a2 - ab + 52) = (a2 + 52 + ab) (^2 + 52 - a5)
= (a2 + 52)2 _ ^252 .^ a4 + 2 ^252 + 54 _ ^252 = ^4 4. ^252 + 54..
70. Square of Any Multinomial Expression. We have found
in the preceding article, and also in Art. 62, the square
of the sum of three algebraical quantities ; and we can
by the same methods obtain the square of the sum of
more than three quantities. The square of the sum
of any number of algebraical quantities may also be
found in the following manner.
MULTIPLICATION. 65
Suppose we wish to find
We know that the product of any two algebraical
expressions is equal to the sum of the partial products
obtained by multiplying every term of the multiplicand
by every term of the multiplier. If we multiply the
term a of the multiplicand by the term a of the multi-
plier, we obtain the term a^ of the product : we similarly
obtain the terms b^, c^, etc. We can multiply any term,
say b, of the multiplicand by any different term, say d,
of the* multiplier; and we thus obtain the term bd of the
product. But we also obtain the term bd of the product
by multiplying the term d of the multiplicand by the
term b of the multiplier, and we can obtain the term
bd in no other way, so that every such term as bd, in
which the letters are different, occurs twice in the
product. The required product is therefore the sum of
the squares of all the quantities a, b, c, d, etc., together
with twice the product of every pair.
Thus, the square of the sum of any mimber of algebraical
quantities is eq^ial to the sum of their squares together with
twice the product of every pair.
Ex. 1. Find the square oi a + b -\- c. The squares of the sepa-
rate terms are a^, 6"^, c'K The products of the different pairs of
terms" are ab, ac, and be.
Hence (a + 6 + c)2 = a^ + 62 + c2 + 2 a6 + 2 ac + 2 be.
Ex. 3. Find the square of a + 2b — Sc.
The required square
= a2 + (2 6)2^(-3c)2 + 2a(2 6)-|-2a(-3c)-i-2(2 6)(-3c)
= a2-f 4&2-f 9c2+4a6-6ac- 12 6c.
66 MULTIPLICATION.
Ex. 3. Find (« _ & + c - dy.
(a - & + c - (?)2 = a2 ^_(_ 5)2 + c2 + (- <^)2 + 2 a(- &) + 2 ac
+ 2a(-d)+2(-&)c + 2(-6)(-d)+2c(-d)
= a2 4- 62 4. c2 + d2 - 2 a6 + 2 «c - 2 ad - 2 6c ^- 2 &(Z - 2 ccZ.
After some practice the intermediate steps can be omitted and
the final result written down at once. To ensure taking twice the
product of every pair, it is best to take twice the product of each
term and of every term YflaXah. follows it.
EXAMPLES XVI.
Write down the squares of the following expressions :
1. 2 a + &. b. a^-bah. 9. a - b - c.
2. 4a + 36. 6. 2a2-3a6. 10. 2a + 25-c.
3. Sa-b. 7. -Sxy + 2y^. 11. 4a + 26-3c.
4. 5a -66. 8. 4^2-7^2. 12. 2a -56 -3c.
13. ic2+ic + l. 19. 3a + 26- 4c + d:.
14. a;2-a: + l. 20. 5a + 6 - 4c - 3c?.
15. x'^-xy + 2/2. 21. x^ + x^ + X + 1.
■ 16. x^ + x^y^ + 2/4. 22. x^-x^-\-x-l.
17. a-b-c + d. 23. 2x^ - x^y + xy^ - Sy^
18. 2a-26-3c + 3d!. 24. 2x^ - x'^y + 2xy^ - yK
Multiply
25. X - y + z hj X - y - z.
26. x-2y-h4:zhjx-2y — 4z.
27. Sx-y -6zhj Sx + y -bz.
28. -x + 2y-Sz\)yx-\-2y-3z.
29. x'^ -\- xy + y2 by x'^ — xy -{■ y^.
30. 3x2 - icy + 2y2 by 3x2 + 0;?/ + 2?/2.
MULTIPLICATION.
67
31. x^-x-\-7\)y x^-x-l.
32. 2 x2 - 3 X + 7 by 2 x2 + 3x + 7.
33. a + 6 + c + rf by a + 6-c-d
34. 2a-36 + 2c-4dby2a-36-2c + 4(2.
35. 2a + 36 + c-2fZby2a-3&-c-2d.
36. a - 3 6 - 4 c + <i by a + 3 6 - 4 c - d.
71. Continued Products. The continued product of
several algebraical expressions is obtained by finding
;he product of any two of the expressions, and then
.ultiplyiug this product by the third expression.
Ex. 1. Find the continued product (x + a) (x 4- b) (x + c).
The process is written as under ;
x + a
x + b
x^+ax
bx-\-ab
sc^ + (a + b)x + ab
x + c
a^ + (a + 6)x2 + a&x
cx2 + c(a + b)x
+ abc
x8 + (o + b'+ c)x2 + (a6 -\- ac+ bc)x + abc
If a -b = c, we have (x + a)^ = x^ + 3 ax^ + 'Sa^x + «».
The above result is arranged in a way which is frequently re-
quired, namely according to powers of x, and all the terms which
contain the same powers of x are collected together.
Ex. 2. Find the continued product of x^ + a^, x + a, and x — a.
The factors can be taken in any order ; hence the required prod-
uct
= (x - a) (x -f a) (X- -f a"-) = (x2 - a^) (x'-^ + a^) = x* - a*.
68 MULTIPLICATION.
Ex. 3. Find (x - ay(x + ay.
Since factors can be taken in any order,
(x - a)%x + a)2 =(x - a)(x +a)(x -a)(x + a)
={(x-a)(x + a)}2
= (a;2 - a^y = x^-2 aH'^ + a*.
72. Powers of a Binomial. We shall in a succeeding
chapter show how to write down any power of a binomial
expression, by a formula which is called the Binomial
Theorem. The square and the cube of a binomial should
however be learnt at this stage. These are given by
and (a ± 6)^ = a« ± ^ a% -\- 3 aV ± b\
EXAMPLES XVII.
Multiply
1. 3 a2 + a6 - 62 by a2 _ 2 a6 - 3 b'\
2. x^-xy -3 ?/2 hy Sx^ + xy - y^.
3. 6x^ -4:X^y + S xy^ hy x'^ + 4:xy+6 y^.
4. x^ -7 x'^y + Sxy"^ by a^* + 7 x^y - 3 x'^y^.
5. 3a;3_7x2 + 5ic-3by 2cc3 + 7ic2_5a; + 4.
6. a^ + Sa^-7a+Q\iySa^-a'^-h6a-4.
7. -|ic2 - ^xy + ?/2 by iic2 - 1 a;?/ - y2.
8. ^x2-fa:?/ + 12?/2by 12a;2 + |i«y-|y2.
9. a2 + 62 + c2 - 6c - ca - a& by a + & + c.
10. 4 a2 + 9 62 + c2 - 3 6c - 2 ca - 6 a6 by 2 a + 3 6 + c.
11. a^ -{-b^ + c^ + be -\- ca - ab hy a + b - c.
12. 9 a2 + 62 + 9 c2 + 3 6c - 9 ca + 3 a6 by 3 a - 6 + 3 c.
MULTIPLICATION. 69
13. x'^ + xy + fp- by -f -xy ^ x^.
14. a2 - 2 a6 + 4 6-2 by 4 62 + 2 a6 + a\
15. ax 4- a2x2 + aH^ by a^^-ax + 1.
16. (k2_|. 1^ ic+ 1, anda;-!.
17. a2 ^_ /c2^ 05 _^ X, and « — x.
18. a;2_|.4y2^ a; + 2y, anda;-2y.
19. 9 «2 4- 25 2/2, 3tc + 6 y, and 3 X - 5 y.
20. x* 4- y*, a;'^ + y^, x + y, and x - y.
21. (a2 + 62)2^ (« 4. 6)2^ and (a - 6)2.
22. (x2 + X + 1)2 and (x2 - x + 1)2.
23. x2 - xy + 2/2, x^ -{-xy + y^, and x* -x2y2 4- y*.
24. a* - a262 4- 54^ ^2 4- a& + b% and a2 - a6 + 62.
25. x2 — ax + a2, x2 4- ax + a2, a + x, and a — x.
26. X + 2/ 4- 2;, - X + y + 5r, X — y + 5r, and x + y - sr.
Find the following cubes :
27. (2 a + 3 6)8. 29. (3 a - 2 6)8. 31. (a - 6 +c)8.
28. (2a -3 6)8. 30. (^a + b + cy. 82. (a - 6 - c)*.
33. Show that
(i.) (2x + l)2 + (x- 1)2 = 4x24.(0:4. 1)24.1.
(ii.) (2x + 1)2 4- (X + 2)2 =(x - 2)2 4- 4x(x + 3)4-1.
(iii.) (x2 4- X + 1)2 + (x2 - X + 1)2 = 2 (x4 + 3 x2 4- 1).
34. Show that
a8 - 68 =(a - 6)(a2 4- ab +62) = (a - 6)3 4- 3a6 (a - 6)
= (a + 6)3 - 3 a6(a + 6) - 2 68.
35. Show that
(X + y)(x + z)-\-(y + z)(y + x) + (2r + x)(^ + y)-(x + y 4- z^
= yz + zx-\- xy.
70
MULTIPLICATION.
36. Show that
(6 - cy+ic - ay + (a - by - 3(6 - c)(c - a)(a -b) = 0.
37. Simplify
ix-]-y + zy-(-x + y + zy-\-(ix-y + zy-(x + y- zy.
38. Simplify
{x^y^z) (a;2 + 1/2 + ^2) _ yz{y + 2;) - zx{z -^x)- xy(x + y).
39. Show that
(6 -f c) (c + a) (a + 6) + a6c = (a + & + c) (he + ca + a&).
40. Show that, iix=2y+bz\ then will x^ = 8 y^ + 125 z^ 4- 30 xi/0.
41. Show that, iix=h + c-2a, y=c-\-a-2b, and z = a+b-2c;
then x2 4- 2/2 + ^2 _(_ 2 ?/0 + 2 ^x + 2 a;?/ = 0.
42. Show that aH6^+c3+3(6+c)(c+«)(a+6) = (a+& + c)3.
43. Show that a2(5 + c)2+62(c+a)2+c2(a+&)2+2a&c(a + & + c)
= 2(6c + ca + aby.
44. Show that (a + 6)3 + 3 c(a + 6)2 + 3 c2(a + 6) + c^
= (6 + cy + 3a(6 + c)2 + 3a2(6 + c)-ha^.
45. Show that
{(a + 6)^2 - (a2 + 62)x + a^ + ^sj ^(^a - 6)0^2 _ (^2 _^i^x + a^- b^
= (a2 - 62)x* - 2(a3 - 63)x3 + 3(a* - b^)x^ - 2Qa^ - b^)x + a^- b^
8BB
DIVISION. 71
CHAPTER V.
Division.
73. In multiplication we have two factors given, and
we have to find their product. In division we have the
product and one factor given, and we have to determine
the other factor, so that to divide a by 6 is to find a
quantity c such that b x c = a.
74. Since division is the inverse operation to that of
multiplication [Art. 9], and successive multiplications
can be performed in any order [Art. 52], it follows that
successive divisions can be performed in any order.
Thus a-^b^c = a-^c-^b.
It also follows, from Art. 52, that to 'divide by any
quantities in succession gives the same result as to divide
at once by their product.
Thus a^b ^c = a-i- (&c).
75. Not only may a succession of divisions be per-
formed in any order, but a succession of divisions and
multiplications may be performed in any order.
For example, axb-i-G = a-T-cxb.
For a = a -f- c X c ;
.*. a xb = a-7-c X c xb
= a-^cx&xc; [by Art. 52.]
72 DIVISION.
therefore, dividing each, by c, we have
ax6-^c = a-J-cx5.
Hence we get the same result if we divide the product of
two quantities by a third, as if we divide one of the quanti-
ties by the third and then multiply by the other.
76. The operation of division is often indicated by-
placing the dividend over the divisor with a line between
them, or by separating the dividend from the divisor by
an oblique line called the solidus : thus - or a/b means
When a-i-b is written in the fractional form -, a is
6
called the numerator, and b the denominator.
By putting a = l in ax6-^c = a-i-cx&,
we have lx6-^c = l-^cx6,
that is b -r-c = - xb = b X-,
c c
so that to divide by any quantity c gives the same result as
to multiply by —
c
77. Since a^ x a^ = a^, and a^ xa^= a^^, we have con-
versely a^ -i-a^—a^ = a^~% and a^^ -r-a^= a^°"^ = a', and
similarly in other cases.
Hence, if one power of a letter be divided by another
power of the same letter, the index of the quotient is equal
to the difference of the indices of the dividend and the
divisor.
DIVISION. 73
78. We have proved, in Art. 51, that a x{—b) = — ah;
.'. (— ab)-i-(—b) = a, Sind {—ab)-i- a = — b.
We have also proved that
(—a) X {—b) = -\-ab, and {-^ a) x {-{- b) = + ab;
. .-. (+a6)-^(— a) = — 6, and (f a6)^(+a) = + 5.
Hence, if the signs of the dividend and divisor are
alike, the sign of the quotient is + ; and if the signs of
the dividend and divisor are unlike, the sign of the
quotient is — ; we therefore have the same Law of Signs
in division as in multiplication.
79. Division of Monomial Expressions. The results arrived
at in the preceding articles will enable us to divide any-
one monomial expression by another. These results
are:
(i.) The sign of the quotient is + when the signs of
the dividend and divisor are alike, and the sign of the
quotient is — when the signs of the dividend and divisor
are different. (Law of signs.)
(ii.) The operations of multiplication and division
may be performed in any order. (Laws of commutation
and association.)
(iii.) When one power of any letter is divided by
any smaller power of the same letter, the index of the
quotient is equal to the difference of the indices of the
dividend and the divisor. (The index law for division.)
Ex. 1. Divide 18 a*b^ by 6 a^&s.
18 a*66 ^ 6 a263 = is ^6 x a^ ^ a^ x b^ -^ b^, from (ii.)
= 3 a*-265-3, from (iii.)
= 3a2&2.
74 DIVISION.
Ex. 2. Divide 15 a%^ by - 5 ahK
15a366^(-5a65) = -l5H-5xa3-ax66H-6^ from (i.) and (ii.)
= -3a3-i56-5^ from (iii.)
Ex. 3. Divide - 5 a'h^d^ by - 7 a%'^(^.
The whole work of dividing one monomial expression by another
can be performed mentally and the result written down at once.
First put down the sign of the result ; then divide the numerical
coefficient of the dividend by that of the divisor ; then take each
letter which occurs in the dividend to a power whose index is the
difference of the indices of that letter in the dividend and the
divisor.
Thus (- 5 a^&V) - (-7 a^&V) = + f a^h^, since c* -- c* = 1.
If, in the above example, we had used the rule for finding the
index of the quotient of c* ^ c*, we should have been led to the at
present meaningless result & : but c* -r- c* is obviously 1.
80. Division of a Multinomial Expression by a Monomial.
When a multinomial expression is divided by a monomial,
the quotient is equal to the sum of the quotients obtained
by dividing its separate terms by that monomial.
Thus {a-\-b t\- '") -i-x=: a-i- x-\-b -i-x-{- "'
To prove this, multiply by x ; then
(a + b-\ )^x xx = a-\-b-] — ;
also {a-i-x-{-b^x-\ — ) xx = a-irXXx + b-v-xxx-\ —
= « + & + ••• [Art. 57.]
Hence (a-\-b-\ )-i-x = a-^x-{-b-i-x-\
Ex. 1. Divide a^ + 2a'^ by a.
The result is a^ ^ a -h 2 a"^ -^ a = a'^ + 2a.
DIVISION. 15
Ex. 2. Divide a^x^ — 3 ax by ax.
The result is a^x^ -=- ax + ( — 3 ax) -4- ax = ax — 3.
Ex. 3. Divide 12 x^ - 5 ax2 - 2 a^x by 3 x.
The result is
12x3--3x + (-5ax2) ^ 3x +(- 2a2x) --3x = 4x2 - f ax - fa^.
EXAMPLES XVIII.
Divide
1. 10 a by - 5 a. 10. 25 a^b^c^ by - 5 a^bC*.
2. - 10 6 by 2 6 . 11. - 27 a^ftHc* by - 6 a^bK
8. -2xby-3x. 12. o6cd by -a&.
4. a2 by - a. 13. a^¥c*(f by - 2 a62c#.
6. 8 a6 by - 2 b. 14. 8 a^x^y'^z by 6 ax^y*.
6. - 4 a6 by 3 a. 15. - 3 a^b^cx'^y^ by - 2 a&cx=V^-
7. 12 a268 by - a&2. 16. 3 x^ - 5 ax by x.
8. -6 x2y8 by - 4 xy. 17. by* -Qy^hj - y^.
9. -3xVl3y2xy2. 18. 4a6 - Sa^ + 2a* by a*.
X 19. 12 a8 + 9 a* - 6 a5 by - 3 a2.
N 20. 15 a*66 _ 7 a^b'' + 9 a264 by - 3 a'^b\
V 21. 5(x - 1)2 - 3 a(x - 1) by (x - 1).
22. 12 a6(x -ay -6 a*(x - a)^ + 3 a^(x - ay by a^(x - ay.
81. Division by a Multinomial Expression. We have nov^
to consider the most general case of division, namely,
the division of one multinomial expression by another.
The object of division is to find by what algebraical
expression the divisor must be multiplied to produce
the dividend.
76 DIVISION.
Consider, for example, the division of
Sa^-\-Sa'b-\-4.ab^-\-b^hj2a + b.
The dividend and the divisor must first be both
arranged according to descending (or ascending) powers
of some common letter, and the terms of the quotient
will be found separately in the same order. In the
present case the arrangement is according to descending
powers of the letter a. Now the term of highest degree
in a in the dividend, namely, 8 a^, must be the product
of the terms of highest degree in a in the divisor and
the quotient ; hence the Jii^st term of the quotient must
be 8 a^ -7- 2 a = 4 al Now multiply the divisor by 4 a^
and subtract from the dividend: we then have the re-
mainder 4 a^6 + 4 ab"^ + b^. This remainder must be the
product of the divisor by the other terms of the quotient ;
and hence the first term of the remainder, namely 4 a^b,
must be the product of the first term of the divisor and
the next term of the quotient. Hence the second term
of the quotient is 4 a^6 -^ 2 a = 2 ab. Now multiply the
divisor by 2ab and subtract the product from the re-
mainder : we then get the second remainder 2 ab^ + 61
The third term of the quotient is similarly 2 a6^ -^ 2 a = b\
Multiply the divisor by b^ and subtract the product from
2 ab^ -\- b^, and there is no remainder. Since there is no
remainder after the last subtraction, the dividend must
be equal to the sum of the different quantities which
have been subtracted from it; but we have subtracted
m succession the divisor multiplied by Aa^j by 2ab,
and by b^; we have therefore subtracted altogether the
product of the divisor and 4 a^ + 2 a6 + b\ Hence the
required quotient is 4 a'^ + 2 a6 + b\
DIVISION.
77
The process is written in the following form :
2a + 6
8a^
-\-Sa'b-{-4.ab'-\-b'
-h^a'b
4:a'b-{-4:ab'-{-b^
4:a'b-{-2ab^
2ab'' + W
2ab^-{-b^
4a24-2a6 + 62
From the above example it will be seen that in order
to divide one multinomial expression by another we
proceed as follows :
(1) Arrange both dividend and divisor according to
ascending or both according to descending poivers of some
common letter.
(2) Divide the first term of the dividend by the first term
of the divisor : this will give the first term of the quotient.
(3) Multiply the divisor by the first term of the quotie^it
found above, and subtract the product from the dividend.
(4) Now treat the remainder as a new dividend, and go
on repeating the process.
82. The following are additional examples :
Ex. 1. Divide 3 x^ - 4 x2 + 2x - 1 by 1 - a;.
The order of the terms in the divisor must first be changed
to - X + 1.
3x8-3a;g | -Zx^ + x-l
- x2 + 2 X - 1
- X2+X
x-1
x-1
78
DIVISION.
Ex. 2. Divide a* -a%-\-2 a^"^ - ab^ + b^ by a"^ + b\
a^ -a% + 2 a%^ - ab^ + &*
a2+&2
«'-^ - a6 + 62
-a36 +
a262 -
- ab^ + M
-a%
-a63
+
a%^
+ 6*
+
a'^b^
+ 6*
Ex. 3. Divide a* + ^^252 + ^4 by ^2 _ ^^ ^ 52.
a* + a262
a^ - a% + a262
+ 6^
a2 _ a6 + 62
a2 + a6 + 62
+ a3?,
+ 6*
+ a^6
- a262 _f. ^53
+ a262 -
- a63 + 64
+ a262 _
- ab^ + 6*
In the last example the terms of the dividend were placed apart
in order that ' like ' terms might be placed under one another
without altering the order of the terms according to descending
powers of a. The subtractions can, however, be easily performed
without putting ' like ' terms under one another ; but the arrange-
ment of the terms according to descending (or ascending) powers
of the chosen letter should never be departed from.
83. Horner's Synthetic Division. In the foregoing process
of division there is a redundancy of work which may be
avoided by a different arrangement of the successive
steps. Fixing upon the idea of division as the process
that undoes multiplication [Art. 9], let us first multiply
together the divisor and quotient of Ex. 3 of the last
article, obtaining the dividend as their product and
arranging the work in rectangular form as follows:
DIVISION.
79^
D
— ab
a^^ab 4- b^
a^ + a% + a'b'
-a'b-a'b^-abK . .
-h oJ'b^ -^ ab^ +b''
a^ 4- 0 + CL-b^ + 0 + &'
= Q
=Pi
= P2
= P3
= P.
For brevity's sake the several multinomials of the
diagram will be referred to by means of the attached
letters D, Q, pi, p^, Psy P-
In order to retrace the steps here indicated and recover
the quotient Q from P and D (the dividend and (fi visor),
regarded as known or given, we have merely to subtract
from P the sum of the two partial products p^ and p^ and
divide the result by a? ; for P was obtained by adding
together p^, Ps, and a^Q. Or better, instead of subtracting
P2 + Pi from Pj we may change all the signs in pg and p^,
add together —p2, —Ps, and P, and then divide the
result by a^, as before. Now we can obtain the succes-
sive terms of —p2 and —p^ by multiplying the negatives
of the last two terms of D by the successive terms of Q
as fast as found; and these are found, by addition, as
fast as the columns in the above arrangement, proceed-
ing from left to right, are completed by the successive
multiplications here indicated.
In order to exhibit the process in a convenient form,
we modify the arrangement in the multiplication process
by interchanging P and Q, suppressing pi, and changing
all the signs in the lines in which p2 and p^ stand. Thus,
a'
+ ab
-^-a^b + a'W^aW . . . .
. -a'b'^ab^-b' . .
. =P
• =-P2
• =-Ps
a^-\-ab +b^i 0 0 . .
. =Q
80 DIVISION.
The successive steps of the work are as follows :
(1) The dividend and the divisor, with all its signs
changed except that of the first term, are written down
as indicated, arranged according to descending powers
of a, the zero coefficients of o? and a being retained.
(2) The first term a^ of the quotient is found as the
quotient of a^ by o?.
(3) The oblique column ~ _ 21,2 is got by multi-
plying + ah and — V^, the last two terms of the divisor
(with signs changed), by a^, the first term of the
quotient.
(4) The second term ah of the quotient is the sum
of the terms in the second column, divided by o?.
(5) The terms +a^6-, —aW of the second oblique
column are the products of ah, the second term of the
quotient, by + ah and — 61
(6) The third terra 6^ of the quotient is the sum of
the terms in the third column, divided by a^.
(7) The terms ah^, —W of the third oblique column
are the products of 6^, the third term of the quotient, by
+ ah and — 6^.
(8) The terms of the last two columns destroy one
another.
In this process, since the quantities sought are merely
the coefficients of the different powers of a in the
quotient, these powers of a may be omitted until the
coefficients are found, and be then inserted in their
proper places in the quotient. The work then appears
in the following abbreviated form :
DIVISION.
81
1
-62
1 + 0 + 6' + 0+6*
+ 6 + 62 + 6«
1 + 6 + 62; 0 0
Quotient : a^ + a6 + 6^.
The method here described is known as Horner's syn.
thetic division.
84. Other examples of the method :
Ex. 1. Divide ic*-2a;2+8a5-3bya;2 4-2a;-l.
8-3
1
-2
+ 1
1 + 0-2
-2 + 4
+ 1
6
2 + 3
1-2 + 3; 0 0
Quotient : x^ - 2 x + 3.
Ex. 2. Divide 4x^ - G x* + i x^ - 11 x'^ -{■ 1 hy 2 x^ - Sx - 1.
2x2
f3x
+ 1
4xS-6x4 + 4x3-llx2 + 0x+l
+ 6x* + 0 +9x2-3x
+ 2x8+0 +3x-l
- Quotient : 2 x^, + 0 x^ + 3 x
0
0
Observe that the terms of the quotient are here obtained by
dividing the sums of the terms in the respective columns by 2 x^.
85. If the division be exact, zeros will appear at the right
of the quotient, in the diagram of the synthetic method, and
their number tvill be one less than the number of terms in
the divisor.
For, when the dividend, divisor, and quotient are made
complete by inserting all the zero-terms (powers of the
leading letter with zero-coefficients), the number of terms
82 DIVISION.
in each is one greater than the highest exponent of its
leading letter, while the highest exponent of the dividend
is the sum of the highest exponents of the divisor and
the quotient. Hence the number of terms in the dividend
is one less than the number of terms in the divisor and
quotient combined. But the last row of our diagram has
the same number of terms as the dividend, and therefore
Number of zeros = (number of terms in dividend)
— (number of terms in quotient)
= (number of terms in divisor) — 1.
Should the zeros not make their appearance, as re-
quired in exact division, the algebraic expression which
takes their place is called a remainder. [See Art. 89.]
It is well to separate this remainder from the quotient
by a distinctive mark, a line, or a semicolon.
86. Note. The foregoing exposition of the division process in
its two principal forms (Arts. 81-84) has employed only particular
examples for the purpose, and is, therefore, not in the form of
strict algebraic proof. But in many such cases a formal proof is
not deemed necessary. A general principle can sometimes be
recognized through its application to a particular example.
87. The following formulae are very important, and
should be remembered ; they can easily be verified.
(flj^ + 2 ax -\- a^) ^ {x + a) = X -{- a,
{pi? — 2ax-\-a'^)^{x — a) = x — a,
{x^ — a^) -i- {x — a) = X -{- a,
{o? — o?) -i- {x — a) = x^ -\- ax -\- o?,
(ar^ 4- a^) -=r- {x -\- a) — o? — ax ■\- a^,
{x"^ — a'') -=-(« — a) =^-\- ax^ + a^x + a^,
{x^ — a^) -^ {x -{- a) = V? — ax^ -\- a^x — a^.
DIVISIOIT. 88
EXAMPLES XIX.
Divide
1. ic2-5a; + 6l)y x-2. 11. x^ - ^y^ hy x- ^y^.
2. a;2 + 5a;-24by x + 8. -^2. ^aj^ - ^^a^ by Jx^ - ^a^.
3. x2-23x+132by x-11. 13. 3x2 + 8x + 4 by 6x + 12.
4. a;2 - 4 X - 77 by X + 7. 14. x^ - ^V a; - 1 by 4 x + 3.
5. 3x2-4x-4byx-2. 15. a^ - b^ by a - b.
6. 3x2-llx+10by3x-6. 16. 8a8 + 27 &» by 2a + 36.
7. a2 ^ 62 by a - b. 17. a^x^ + 6^ by ^x + by.
B: x2 - 9 ?/2 by X - 3 j/. 18. 8 a6x6-125 f hy 2 a^x^-by'^.
9. x2 - 16 y2 by X + 4 y. 19. x^ - 8x - 3 by 3 - x.
10. 9x2-64y*by3x + 8y2. 20. 2x8 - 6x2 + 4 by 2 - x.
21. 2x*-9x8 + 10x2- 6x + 6by 3-x.
22. X* + 2 x2 - X + 2 by 1 -- X + x2.
88. We now give examples which require greater care
in the arrangement of the terms.
Ex. 1. Divide 2a2-2 62_3c2_55c-5ca-3a6 by a-26-3c.
Where, as in tliis example, more than two letters are involved,
we must not merely arrange the terms according to powers of a,
but h also is given the precedence over c. The terms are there^
fore arranged as under.
2a2_
2a2-
-3a6-
-4a6-
6ac -
Qac
-262-
-6^c-
-3c2 a-
2a
-26-3c
-f-6-l-c
a6 +
ab
ac -
-2 62-
-262-
-66c-
-36c
-3c2
+
ac
ac
-
-2 6c-
-2 6c-
-3c2
-3c2
84
DIVISION.
Ex. 2. Divide a^ + b^ + c^ - Sabchy a + b + c.
First arrange the dividend according to powers of a, and give
6 precedence over c througliout.
a* - 3 abc + 6^ + c^
« + 6 + c
a^ - ab - ac + 62 - 6c + c^
- a26-
- a25_
-a2c
-a62
- 3 a6c + 63 + c3
— a6c
- a2c + «62 -
-a2c
2 a6c + 63 + c8
a6c — ac2
+ a62-
+ a62
«6c + «c2
+ 63 + c3
+ 68 + 62c
—
a6c + ac^
abc
- 62c + C8
- 62c - 6c2
ac2 + 6c2 + c3
ac2 + bc"^ + c3
The work of this example is much shorter by the synthetic
method. Arrange the terms according to descending powers of a,
and retain only their coefficients and the term not containing a.
Thus,
-(b + c)
1 0 - 3 6c
-C6 + c) + (6 + c)2
+ (63 + c3)
-(6 + c)(62 + c2-6c)
l_(6 + c) + (62+ c2-6c); 0
Hence the quotient is
a^-(b + c)a + 62 + c2 - 6c = a2 + 62 + c2 - 6c - ca - ab.
Ex. 3. Divide a'^(b - c) + 62(c - a)+ c\a -b)hY a-b.
Arrange the terms according to descending powers of a, and
retain only their coefficients and the term not containing a. Thus,
1
+ 6
(5 _ c) - (62 - c2) + (62c - 6c2)
4- (6 -c)6+(6c2-62c)
(6-c) + (c2-6c); 0
Hence the quotient is (6 - c)a + (c - 6)c.
DIVISION. 85
89. In the process of division as described in Art. 81,
the dividend and divisor being arranged according to de-
scending powers of a, we subtract from the dividend, at
the first subtraction, an expression such that the term in
it which contains the highest power of a is the same as
the term in the dividend which contains the highest
power of a ; hence it follows that the highest power of a
contained in the remainder after the first subtraction will
be less than the highest power of a contained in the divi-
dend. Similarly, the highest power of a contained in
every remainder will be less than in the preceding re-
mainder; and hence by proceeding far enough, we must
come to a stage where there is no remainder, or else
where there is a remainder the highest power of a in
which is less than the highest power of a in the divisor,
and in this latter case the division cannot be exactly
performed.
It is convenient to extend the definition of division to
the following : To divide Aby B is to find an algebraical
expression C such that B x C ^s either equal to A, or differs
from A by an expression which is of lower degree, in some
particular letter, than the divisor B.
For example, if we divide x'^ + S ax -\- a^ hy x + a, we have
X \x^ + Sax + a^
— a \ —ax — 2a'^
X +2a; -a\
Thus
(x2 + 3 aa; + a2) -J. (x + a) = « + 2 a, with remainder - «'. [Art. 84.]
We may express the above in tJie following form :
a;2 + 3ax + a"-2=(x-|-2a)(x + a)-a^.
86
DIVISION.
We have also, by arranging the dividend and the divisor
differently,
- a;a - 2 a;2
a + 2 X ; — aj2^
and the remainder is now
Hence, when there is a remainder, a change in the order of the
dividend and the divisor leads to a result of different form. This
is, however, what might be expected considering that in the first
case we find what the divisor must be multiplied by, in order to
agree with the dividend so far as the terms which contain x are
concerned ; and in the second case we find what the divisor must
be multiplied by, in order to agree with the dividend so far as the
terms which contain a are concerned.
When, therefore, we have to divide one expression by another,
both expressions being arranged in the same way, it must be
understood that this arrangement is to be adhered to.
Divide
1.
10.
U.
EXAMPLES XX.
a;4 + ic2 + 1 by ic2 + X + 1.
1 + X* + x8 by 1 - x2 + X*.
a;4 ^_ 4 a;2 4- 16 by x2 + 2 X + 4.
16x4 + 36x2 + 81 by 4 x2 - 6x + 9.
x* + 2 x2 + X + 2 by x2 + X + 1.
2 ic* - x3 + 2 x2 + 1 by 1 - X + x2.
x5-4x3 + 2x-4by 2-x.
x6 _ 3 x2 - 52 by 2 - X.
I_x + 5x3-3x4by l-2x + 3x2.
a;6 _|. 2 x5 - 4 x* - 2 x3 + 12 x2 - 2 X - 1 by x2 + 2 X -1.
l-4x8 + 3x*by (l-x)2.
DIVISION. 87
12. l-6a^ + 4a^by (x-iy.
13. 35 + 4x-16a;2 + i9a:3-6x*by 7 + 6x-Sx^.
14. 3a^ - 12x* + 17 x3 - 19x2 + 1 by (1 - x)3.
15. x5 - 5a;2 + 5 X - 1 by 1 + x2 - 2x.
16. 16x6 -97x*- 84x2 + 77x + 8by 4x2+ 11 x + 1.
17. 1 + 2 x5 + x6 + 2 x^ by 1 + X + x2.
18. xio + x5 + 1 by x2 + X + 1.
19. 6 x6 - 7 x^y + x3?/2 + 20 x2y3 - 22 xy* + 8 y5 by 2 x2 - 3 xy + 4 y\
20. 8?/5 _ 22xy^ + 20x2^/3 + x^ - 7x4i/ + 6x5 by 4y2 _ Zxy + 2x2.
21. x2 — 2/2 _ a:^; + 2/2 by X - y.
22. x2 - j/2 _|. 2 X2 - 2 ?/0 by X - 2/.
23. a2 - 62 + ac - 6c by a + 6 + c.
24. x2 - 2/2 - ;s2 _|_ 2 2/2; by 2/ - 2; - X.
26. rt2 _ 4 52 _ e2 _ 4 6c by 2 6 + c - a.
26. a2 + 4 62 + 9c2 - 12 6c - 6 ca + 4 a6 by 3 c - 2 6 - a.
27. 2 a2 - 2 62 - 2 c2 - 4 6c + 3 ca + 3 a6 by a + 2 6 + 2 c.
28. 3 a2 - 16 62 - 4 c2 + 20 6c - 11 ca + 8 a6 by a + 4 6 - 4c.
29. x3 + ?/8 _ ;53 ^ 3 ;;c?/s by X + 2/ - 0.
30. 8x8 - 2/^ + «3 + 6x2/0 by 2/ - 2 - 2x.
31. 27a3_863 + c8+ 18a6cby 3a-26 + c.
32. a^ + 3a26 + 3a62 + 6^ + c^ by a + 6 + c.
33. a-* + 6* + c* - 2 62c2 - 2 c2a2 - 2^262 by a + 6 + c.
34. (x+l)3+2/3by X+2/+I. 35. {x^-yzy+8fz^hyx^-^yz.
36. (x2 - 2 yzy -27 y^z^ by x2 - 6 yz.
37. 9 a2 + 6 a6 + 62 - 4 c2 - 4 c<Z - d2 by 3 a + 6 - 2 c - (i.
38. x3+(4a6-62)x-(a-2 6)(a2 + 3 62) byx-a + 26.
88 MISCELLANEOUS EXAMPLES I.
39. a3(6 - c) + h^{c - a)+ c^{a - h) by
a\h - c) + 62(c - a) + c%a - &).
40. a^(h - c)+ 6*(c - a)+ c\a - h) by
a2(6 - c) + 62(c _ a) + c2(a - h).
MISCELLANEOUS EXAMPLES I.
A. 1. Find the value of {y - zf + (s - xy + (x - yY when
ic = — 1, ?/ = 0, and z — \.
2. Add together 3 a2 _ 2 ca - 2 a6, 2 62 + 3 5c + 3 6a, and
c2 - 2 ac - 2 6c.
3. Show that
a3 4. 53 4. c8 _ 3 a6c - a(a'^ - be) - b(b^ - ca) - cif' - ab) = 0.
4. Multiply 2x2-^a; + |by^x + 3,
and ioc;^ + ixy + y^hy \x^ — ^xy + y^.
6. Simplify {x + 1) (x + 2) (x + 3) - (x - 1) (x - 2) (x - 3).
6. Divide 2 - 12x5 + lOx^ by 1 - 2 x + x2.
B. 1. Find the value of c2 + 6(c + a), and of \/2 be — a, when
a = II, 6 = 3, and e = 6.
2. Subtract 5 a* - 3 a^ft 4. 4 ^252 from 5 6^ - 3 6% + 4 62a2.
3. Simplify x -(I - 1 - x), 3 x - 7 - 4x,
and 6 — 2a — {c — a— (6 — a + c)}.
4. Multiply 9 x2 — 1 by x2 + I, and a + 6 + cbya + 6 — c.
6. Show that
2(a2 + 62 + c2 - 6c - ca - a6) = (6 - c)2 + (c - a)2 + (a - 6)2.
6. Divide x^ - 3 x2 + 3 x + ?/3 - 1 by x + ?/ - 1.
C. 1. Find the numerical value of
Va2 + 2 6c , \/62 -{-ea, Vc2 + a6
1 1 }
a 6 c
when a = 4, 6 = 3, and c = — 2.
MISCELLANEOUS EXAMPLES I. 89
2. Add together 3a'^b-6 ab^+7 b^ and 2 a^ _ i a^ft + 5 a62 _ 4 53,
3. Multiply ax2 — X + a by ax2 + X + a, and find the square of
2 a + 6 - 3 c.
4. Simplify (x+yy-(x-\-y)(x - y)- {x(2y-x)-y(2x-y)}.
5. The product of two algebraical expressions is
x° + x^y + x*2/2 _ x^y^ + y^,
and one of them is x^ + xy + y^ ; what is the other ?
6. Prove
(i.) a(6 -c)+ 5(c-a) + c(a- 6)=0,
(ii.) a2(6-c) + &2(c-a) + c2(a-6) + (6-c)(c-a)(a-6) = 0.
D. 1. If a = 1, 6=2, and c= -3, find the value of 6a + 36+4c,
andof aS + b^-^c^-Sabc.
2. Simplify 3{x - 2(y - z)} -[^y + {2y -(z - x)}].
3. What must be added to {a + b + c)^ that the sum may be
(a-b-cy?
4. Multiply a^ + 6^ by a — 6, and divide the result by a + 6.
6. Divide x^ + ^^ by x 4- 2, and from the result write down the
quotient when (x + yy + z^ is divided hy x -^ y + z.
6. Prove
(i-) (X + y)ix -y) + (y+ z)(y - z) + {z + x)iz - x) = 0,
(ii.) {x-yy+{y-zY-\-{z-xy
= 2(x - y)(x - z)+2(y -z)(y- x)+ 2(2 - x)(z - y).
E. 1. Find the value of
2 V(a2 + 62),^ ^(a2 + 52 - c^ + 6(f^)
d — C + b — a
when a = 3, 6 = 4, c = 6, and d = 6.
2. Simplify 4 x - (y - x) - 3 {2 ?/ - 3(x + y)} and
2x-3y-4(x-2y)+5{3x-2(x- y)},
and multiply the two results together.
90 MISCELLANEOUS EXAMPLES I.
3. Show that {a + by = a^ + b^ + S ab(a + h), and verify the
result when a = 1 and b = — 2.
4. Find the coefficient of x^ in the product of x^ — (^a— b)x — ab
and x2 + (a + b)x + ab.
6. Divide the difference of (2 a + 3 by and (3 a + 2 by by
a + 6 ; divide also the sum of (2 a — 3 6)3 and (3 a — 2 &)3 by a — &.
6. Show that
(a + b + c - d)(a + b - c -\- d) + (a - b + c + d)(- a + b -h c -\- d)
= 4(a6 + cd).
F. 1. Find the value of
(g + 6) (c + <?)-(& + c) (d + g)
g6 + 6c — cd — (?g
when g = 3, 6 = 4, c = 5, and d = — 4:.
2. Find the value of (a — x)(a — ^y){a — ^z).
3. If x^ + 7x + c is exactly divisible by x + 4, what is the
value of c ?
4. Divide x^ — y^ — z'^ +2yz by a; + ?/ — 0, and find the co-
efficient of X in the quotient obtained by dividing 8 x* + xy^ — y^
by x-^y.
5. Show that
(g2+ 62) (c2+ c?2) = (ac + 6(^)2+ (gd! - 6c)2= (ac - bdy-i- (ad + 6c)2.
6. Show that
(6 - c)2 + (a - 6) (g - c) = (c - a)2 + (6 - c) (6 - g)
= (g - 6)2 + (c-g)(c- 6).
G. 1. Simplify 5 x - 3[2 a; + 9 ?/ - 2{3 a: - 4(?/ - x)}].
2. Subtract 6 g2 - 4 g6 + 7 62 from 2 g2 - 4 g6 + 2 62, and from
the remainder subtract 4 g2 _ 6 g6 -}- 5 62.
3. Multiply g2 -1- 62 + 1 - g6 - g - 6 by g + 6 + 1.
4. Divide g^ -)- 6^ - 8 c^ + 3 g26 + 3 g62 by g + 6 - 2 c.
MISCELLANEOUS EXAMPLES I. 91
5. The product of two algebraical expressions is x^ — 64 x, and
one of the expressions is x^ - 2 x + 4 ; what is the other ?
6. Prove that
(1 + X+ X2)(l _ X + X2)(l - X2 + X*)(l _ X* + X8) = 1 + X8 + Xl6,
and that
a3(6-c) + 63(c_a) + c3(a-5) + (6-c)(c-a)(a-6)(a + ft + c) = 0.
92 SIMPLE EQUATIONS.
CHAPTER VI.
Simple Equations.
90. A statement of the equality of two algebraical
expressions is called an equation, and the two equal ex-
pressions are called the members or sides of the equation.
When the statement of equality is true for all values
of the letters involved, the equation is sometimes called
an identical equation. An identical equation is, however,
generally called an identity, and the name equation is
reserved for those cases in which the equality is only
true for certain particular values of the letters involved.
Thus a + « = 2 a, and (a + hy = a^ -\-2ah + b^, which are true
for all values of a and 6, are identities ; and 5 a + 2 = 12, which
is only true when a is 2, is an equation.
In an identity the sign = is frequently used instead of
the sign = . Thus, a + a = 2a.
Note. — Eor the sake of distinction, a quantity which is sup-
posed to be known, but which is not expressed by any particular
arithmetical number, is represented by one of the first letters of
the alphabet, a, b, c, etc., and a quantity which is unknown, and
which is to be found, is represented by one of the last letters of the
alphabet, x, y, or z.
91. To solve an equation is to find the value, or values,
of the unknown quantity for which the equation is true ;
and these values of the unknown quantity are said to
\
\ SIMPLE EQUATIONS. 93
satisfy the equation, and are called the roots of the
equation.
Def. In more precise terms we therefore define a root
of an equation to be any quantity which, when substi
tuted for the unknown quantity, reduces the equation
to an identity.
This definition supplies us with a convenient test for a
solution. Thus 2 is a root of
because when 2 is substituted for x, the equation is
reduced to the identity
8-8 + 2 = 2.
Ex. 1. Show that 1 is a root of each of the following equations :
(i.) x^-x = 0, (iv.) 2x8 + x = 3,
(ii.) rK2-3ic + 2 = 0, (v.) aic^ - ax + x = 1,
(iii.) x8-l = 0, (vi.) x^-lr^O.
Ex. 2. Show that o is a root of each of the following equations :
(i.) x2_fl(2 = o, (iii.) x«-a» = 0,
(ii.) ax — a^ + x = a^ (iv.) x* — ax^ + x* _ ^x,
(v.) x2 - (a + 9)x + a6 = 0.
92. An equation which contains only one unknown
quantity, x suppose, is said to be of the first degree when
X occurs only in the first power ; it is said to be of the
second degree when a^ is the highest power of x which
occurs, and so on.
Equations of the first degree are, however, generally
called simple equations, and equations of the second degree
are generally called quadratic equations.
94 SIMPLE EQUATIONS.
In the present chapter we shall only consider simple
equations which contain one unknown quantity.
93. In the solution of equations frequent use is made
of the following axioms :
(i.) If we add to equals the same quantity, or equal
quantities, the sums will be equal.
Thus, ii a = b, then a + c = b + c.
(ii.) If we take from equals the same quantity, or
equal quantities, the differences will be equal.
Thus, if a = 6, then a — c = b — c.
(iii.) If we multiply equals by the same quantity, or
by equal quantities, the products will be equal.
Thus, if a = 6, then ac = be.
(iv.) If we divide equals by the same quantity not
zero, or by equal quantities, the quotients will be equal.
Thus, if a = 6, then a ^ c = b -i- c. But although 2x0 = 3x0,
we cannot divide by 0 and say that 2 = 3.
94. Let a -f- b = c —d.
Add — 6 to both sides, then the equality still holds
good by Axiom (ii.) >
.-. a-{-b — b = c — d — bf
that is, a = c — d — b.
Thus the term b has been cancelled from one side of
the equation, and it appears on the other side with its
sign changed from + to — .
SIMPLE EQUATIONS. 95
Again, add d to both sides of the last equation j then,
by Axiom (i.),
a-{-d = c — d — b-{-d'y
.'. a -\- d = c — b.
Thus the term d has been cancelled from one side of
the equation, and it appears on the other side with its
sign changed from — to -|-.
We can proceed in a similar manner in any other
case ; hence any term may be moved from one side of an
equation to the other, provided its sign is changed.
When terms are changed from one side- of an equation
to the other side, they are said to be transposed.
95. We may change the signs of all the terms of an
equation; for, by Axiom (iii.), we do not destroy the
equality by multiplying both sides, and therefore every
term, by —1; and this multiplication will change the
sign of every term.
96. The method of solving simple equations will be
seen from the following examples :
Ex. 1. Solve 8x + 7 = 4a;+27.
Transpose the terms 4 x and 7, then
8a; -4a: = 27 -7.
Combine like terms, then
4x = 20.
Divide both sides by 4, the coefl&cient of x ; then
x = 6.
96 SIMPLE EQUATIONS.
Ex. 2. Solve
6a;-7:
= 7a;-15.
By transposition, we have
6X — 1X:
= - 15 + 7.
Combining like terms,
, we
get
-2X:
= -8.
Divide both sides by -
-2,
the coefficient of x ; then
•
X:
= 4.
Ex. 3. Solve
| + -
4^2
Multiply every term by 4, the L. C. M. of the denominators, then
the fractions will be got rid of, and we shall have
2 X + 8 = ic + 10.
Transposing, 2aj — x=10 — 8;
,:x = 2.
Ex. 4. Solve ^-t^ _ 1 (a; _ 1) z= 1.
Multiply by 12 to get rid of fractions ; then
\H^ + i)-¥-(^-i)=i2,
that is, 3 (« + 1) - 4 (a; - 1) = 12 ;
.-. 3ic + 3-4x + 4 = 12.
Transposing, 3aj — 4a; = 12 — 3— 4;
r. — x = 6,
or, by changing the signs, x = — 5.
Ex. 5. Solve ax -}- b^ = bx -[- a^.
By transposition, ax— bx = a^ — b^,
that is, (a - &) X = a2 - bK
Divide by a — 6, the coefficient of x ; then
X = (a2 - 62) -^ (a - 6) = a + 6.
SIMPLE EQUATIONS. 97
Note. — The student should test his results by seeing that the
values obtained really satisfy the given equations, that is, reduce
them to identities.
For example, if we put 5 for x in Ex. 1, we have
8x5 + 7 = 4x5 + 27,
that is, 40 + 7 = 20 + 27,
which is clearly true.
Again, if we put a + 6 for x in Ex. 5, we have
a (a + 6)+ 62 = 6 (a + 6)+ a\
that is, a^j^ ah + b'^ = ab + b^-\- a%
which is an identity. [See Art. 91.]
97. From the above it will be seen that the different
steps in the process of solving a simple equation are as
follows. First, clear the equation of fractions, and per-
form all the algebraical operations which are indicated.
Then transpose all the terms into which the unknown
quantity enters to one side of the equation, and all the
other terms to the other side. Next combine all the
terms which contain the unknown quantity into one
term, and divide by the coefficient of the unknown quan-
tity ; this gives the required root.
Note. — In order to acquire a habit of clear and accurate thought,
the way in which each new equation is derived from the preceding
should always be indicated.
A beginner should be cautioned against the common mistake of
putting a meaningless sign of equality at the commencement of a
line.
98. The following are additional examples of simple
equations.
98 SIMPLE EQUATIONS.
Ex. 1. Solve (x - 1) (a; - 2) + 5 = (x + l)^.
Removing the brackets, we have
a;2 -3x4-2 + 5 = a;2 + 2x4-1.
Transposing, we have
x2-3x-x2-2x = l-2-5;
.-. -5x= -6.
Divide by — 5 ; then ^ = f •
Ex. 2. Solve 3 (x - 1) - {3 X - (2 - x)} = 5.
Removing the brackets, we have
3x-3-3x + 2-x = 5;
.-. 3x-3x-x = 5 + 3-2;
.♦. -X =6,
or ^ X = — 6.
Ex. 3. Solve3x2-l=(3x + 2)(x-5).
Removing the brackets, we have
3x2-1 = 3x2- 13x- 10.
Transposing, we have
3x2-3x2 + 13x= -10 + 1.
Hence . 13x = — 9 ;
.-. x=-tV
Ex. 4. Solve a(x - a) + 6(x - &) = 2 ah.
Removing the brackets, we have
ax - a2 _|_ 5x - &2 = 2 ah.
Transposing, we have
ax + &x = a2 4. 52 _|. 2 ah,
that is, x(a + 6) = (a + hY ;
.-. a;=(a + &)2-(a + &)=a+ 6.
SIMPLE EQUATIONS.
99
EXAMPLES XXI
Solve the following equations :
1. 3a; + 4 = ic + 10.
2. x + 7 = 4a; + 4.
3. 5ic-12=6x-8.
4. 7a; + 19 = 5x + 7.
5. 3(x-2)=2(x-3).
14. x +
6. 5(x + 2)=3(x+3)-fl.
7. a;-(4-2x)=7(x-l).
8. 5(4-3x)=7(3-4x).
9. 2(x-3)=5(x+l) + 2x-l.
10. 4(l-x)+3(2 + x)=13.
11. 2(x - 2) + 3(x - 3) + 4(x - 4) - 20 = 0.
12. 2(x-l)-3(x-2)+4(x-3)+2=0.
13. 5x + 6(x+ l)-7(x + 2)-8(x+ 3) = ©.
2
10.
16. ^ + ^^
2 2
3.
18. K2-a;)-K5^ + 21) = x + 3.
19. ^(x-2)+KaJ-3)+Ka^-4)=10.
^
22. ^+ 1 I a? + 2 , g;H-3
23. 2x-[3-{4x + (x-l)}-5]=8.
24. l-2{x-3(l + x)} = 0.
25. (x+l)(x + 2) = (x-2)(x-4).
26. (x-l)(x-2) = (x-3)(x-4).
27. 2x2=(x + l)2 + (a; + 3)2.
28. 3x2=(x+l)2-f(x + 2)2 + (x + 3)2.
100 SIMPLE EQUATIONS.
29. {x-2){x -b) + (x- 3)(x- 4)= 2(x - 4)(a; - 5).
30. (x - 1)2 + ^(x - 3)2 = b{x + 5)2.
31. 5(x + l)2 + 7(a; + 3)2 = 12(x + 2)2.
32. (a;-l)(x-4)=2a; + (x-2)(a;-3).
33. (X - 1)3 + (X - 2)3 + (x- 3)3 = 3(x - l)(x - 2)(x - 3).
34. ^+i_2^:zi+ii^0. 35. ^±i_21_4^^
2 5 * 8 2
36. .5x + 37.5 = 5.25 a; -1. 37. .25x + 4 - .375x= .2x - 9.
38. .15 a; + 1.2 - .875 X + .375 = .0625 a;.
39. 1.2x-K.18ic-.05)=.4x+8.9.
40. a{x- a)=h(x-h). 41. 2(x - a)+ 3(x - 2a) = 2a.
42. \(x+ a + h)-\-l{x+ a- b) = b.
43. (a + b)x + (a - b)x = a^. 44. (a + 6)x + (& - a)x = 62,
45. i(a + x)+ i(2« + x)+ K3a + x)= 3 a.
46. ^ + ^ = a2^_52. 47. (a+6x)(6+ax)=a6(x2-l).
6 a
48. (a2 + x) (62 + x) = (a6 + x)2.
49. a(x + a)+ 6(6- x) = 2 a6.
60. (x + a + 6)2+ (X + a - 6)2 = 2 x2.
51. (x-a)(x-6) + (a + 6)2=(x + a)(x+ 6)7)
62. (x + a + 6 + c) (x + a - 6 - c) =
(x - a — 6 + c) (x — a + 6 — c).
53. ax(x + a) + 6x(x + 6) = (a + 6) (x + a) (x + 6).
64. (X - a)3 +(x - 6)3 + (x - c)3 = 3(x - a)(x ~ 6)(x - c).
PKOBLEMS. 101
CHAPTER VII.
Problems.
99. With some of the general methods of algebra
now at our command, we return to the subject introduced
in the first chapter ; namely, the solution of problems.
In order to solve a problem, the relations between the
known and unknown quantities must be expressed by-
means of algebraical symbols : we thus obtain equations,
the roots of which are the required values of the unknown
quantities.
100. In the present chapter we shall only consider
problems in which there is one unknown quantity, and
in which the relation between the known and the un-
known quantities is expressed algebraically by means of
a simple equation.
Of such problems the following are examples :
Ex. 1. A has $ 20, and B has $ 3.75. How much must A give
to B in order that he may have just four times as much as B ?
Let X be the number of dollars that A gives to B.
Then A will have 20 — x dollars, and B will have 3.75 + x dol-
lars. But A now has four times as much as B. Hence we have
the equation
20-a; = 4(3.75-hx).
^That is 20- x = lb + ix.
Transposing — x — 4x = 15 — 20,
that is — 5 X = — 5.
102 PROBLEMS.
Divide by — 5, then x = l.
Thus A must give one dollar to B.
Note. — It should be remembered that x must always stand for a
number. It is also to be noticed that in any problem all concrete
quantities of the same kind must be expressed in terms of the
same unit ; for example, in the above all sums of money were
expressed as dollars.
Ex. 2. A man has 12 coins, some of which are half-dollars and
the rest dimes, and the coins are worth 4 dollars altogether. How
many are there of each kind ?
Let X be the number of half-dollars ; then 12 — ic will be the
number of dimes. The half-dollars are worth ^ x dollars, and the
dimes are worth xV(12 — x) dollars. Hence, since the coins are
worth 4 dollars altogether, we have the equation
4.
Ix + jW2-x)
Therefore
lx-j\x
at is
t\^
.: IK = 28 - 4
7.
Hence there are 7 half-dollars and 5 dimes.
Ex. 3. A father is six times as old as his son, and in four years
he will be four times as old. How old is each?
Let the son be x years old. Then the father must he 6x years
old. After four years the son will be x -f 4, and the father will be
6 oj -I- 4 years old. Hence by the question
6x-{-4: = 4(X-\-4:)
that is
6a; + 4 = 4x4- 16.
Hence
6x-4x = 16-4,
that is
2 X = 12 ;
.-. x = 6.
Hence the son is 6 years old, and the father is 36 years old.
PROBLEMS. 103
Ex. 4. A can do a piece of work in 12 hours which B can do
in 4 hours. A begins the work, but after a time B takes his place,
and the whole work is finished in 6 hours from the beginning.
How long did A work ?
Let X — the number of hours that A worked.
Then 6 — a; = the number of hours that B worked.
Since A can do the whole work in 12 hours, the part done by A
in 1 hour is ^^.
Therefore the part done by A altogether is — •
Since B can do the whole work in 4 hours, the part done by B
in 1 hour is \.
Therefore the part done by B altogether is |(6 — x).
But A and B together do the whole of the work. Hence we
have the equation
Multiply by 12, then
a; + 3(6-x)=12,
that is a;+ 18 -3ic = 12.
Transposing — 2 x = — 6 ;
.-. a; = 3.
Hence A worked for 3 hours.
Ex. 6. Find the time between 3 and 4 at which the hands of
a clock are together.
Suppose that the hands are together at x minutes after 3 o'clock.
At 3 o'clock the hour-hand is 15 minute-spaces in front of the
minute-hand, and after x minutes they are together. Hence
while the minute-hand moves through x minute-spaces the hour-
hand will move through x — Vo such spaces. But the minute-
hand moves twelve times as fast as the hour-hand, and therefore
in any time the minute-hand passes over twelve times as many
minute-spaces as the hour-hand.
104 PROBLEMS.
Hence jc = 12(ic - 15),
that is cc = 12 a; - 180 ;
.-. llx = 180,
or x= Vt- = 16/t.
Thus the time required is IQfj minutes past 3 o'clock.
EXAMPLES XXII.
1. Find two numbers whose sum is 200, and whose difference
is 2. ' ;' .
2. r^id two numbers whose sum is 66, and whose difference
is 20. {S
3. Divide 25 into two parts whose difference is 5. / ^
4. Divide 100 into two parts whose difference is 45. ^^ -><
5. What number is that to which if you add 40 the sum will
be three times the original number ? "a^ t?
6. What number is that from which if you take away 14 the
remainder is one-third of the original number ? v f
7. The difference of two numbers is 15, and one number is
four times the other. Find the numbers. •
8. Find two numbers whose difference is 10, and one of which
is three times the other.
9. The sum of two numbers is 38, and one of them exceeds
twice the other by 2. What are the numbers ? ' >.
10. Find two numbers the sum of which is 31, and which are
such that one of them is less by 2 than half the other.
11. Find a number whose fourth part exceeds its fifth part by 2.
12. Find a number whose third part exceeds its seventh part
by 80.
13. Find a number which when multiplied by 4 is as much
above 35 as it was originally below it.
PROBLEMS. 105
14. Find a number which when multiplied by 8 exceeds 27 as
much as 27 exceeds the original number.
15. Four times the difference between the fourth and fifth parts
of a certain number exceeds by 4 the difference between the third
and seventh parts. What is the number ?
16. Fifty times the difference between the seventh and eighth
parts of a certain number exceeds half the number by 44. What
is the number ?
17. Divide 100 into two parts such that three times one of the
parts plus five times the other is 410.
18. Divide 100 into two parts such that twice one part is equal
to three times the other.
19. The difference of two numbers is 20, and one-half of one
of the numbers is equal to one-fifth of the other. Find them.
20. The sum of two numbers is 36, and their difference is half
the greater. Find them.
21. A has $ 100, and B has §20 ; how much must A give B in
order that B may have half as much as A ?
22. A and B play for a stake of 5s. If A loses, he will have as
much as B, but if A wins, he will have three times as much as B.
How much has each ?
23. A and B have $ 50 between them. A wins from B as much
as he had originally, and he then has three times as much as B.
What had A at first ?
24. A, B, and C have a certain sum between them. A has one-
half of the whole, B has one-third of the whole, and C has $ 50.
liow much have A and B ?
25. A and B together have $75, and A has $5 more than B ;
how much has each ?
26. A has $ 5 more than B, B has .f 20 more than C, and they
have 1 360 between them. How much has each ?
106 PROBLEMS.
27. A has ^ 15 more than B, B has $ 5 less than C, and they
have $65 between them. How much has each ?
28. A has $ 5 less than B, C has as much as A and B together,
and they have 3 100 between them. How much has each ?
29. Divide $150 among 10 men, 20 women, and 40 children,
giving to each man $3.75 more than to each child, and to each
woman as much as to two children.
30. Divide $ 15 among 3 men, 5 women, and 20 children, giving
to each man one dollar more than to each woman, and to each
child half as much as to each woman.
31. A man of 40 has a son 10 years old ; in how many years
will the father be three times as old as the son ?
32. One man is 70 and another is 50 years of age ; when was the
first twice as old as the second ?
33. A father's age is three times that of his son, and in 10 years
it will be twice as great ; how old are they ?
34. In 5 years a father will be just four times as old as his
daughter, and in 10 years he will be just three times as old ; how
old is he now ?
35. A sum of money is divided between three persons — A, B,
and C — in such a way that A and B have $ 60 between them, A
and C have $65, and B and C have $ 75. How much has each ?
36. Four persons — A, B, C, D — are possessed of certain sums
of money, such that A and B together have $ 49, A and C together
have $ 51, B and C together have $ 53, and A and D together have
$ 47. How much has each ?
37. What is the price of beef if a reduction of 20 per «ent in
the price would enable a purchaser to obtain 6 lbs. more for five
dollars ?
38. How much are eggs a score, when a rise of 20 per cent in
the price would make a difference of 80 in the number which could
be bought for five dollars ?
PROBLEMS. 107
39. A man leaves one-half of his property to his wife, one-third
to a son, and the remainder (which is $ 2000) to a daughter. How
much did he leave altogether ?
40. A man left his property to be divided among his three
children in such a way that the share of the eldest was to be twice
that of the second, and the share of the second twice that of the
youngest. It was found that the eldest received $ 750 more than
the youngest. How much did each receive ?
41. A purse contains 30 coins which altogether amount to
$ 136,50. A certain number of the coins are dimes, one-fifth of
that number are half-eagles, and the rest are eagles. Find the
number of each.
42. A purse contains 36 coins which altogether amount to
$ 52.80. A certain number of the coins are eagles, there are three
times as many half-eagles, and the rest are dimes. Find the num-
ber of each.
43. Find the time between 6 o'clock and 6 o'clock when the
hands of a watch are together.
44. Find at what times between 9 and 10 o'clock the hands of a
watch are at right angles to one another.
45. There are two numbers one of which exceeds the other by
3, while its square exceeds the square of the other by 99. Find
the numbers.
46. In a mixture of spirits and water half of the whole plus
25 gallons was spirit ; and a third of the whole minus 5 gallons
was water ; how many gallons were there of each ?
47. A garrison of 1000 men having provisions for 60 days was
reinforced after 10 days, and from that time the provisions only
lasted 20 days. Find the number in the reinforcement.
48. A laborer was engaged for 36 days, upon the condition
that he should receive $ 3.25 for every day he worked, but should
pay $0.75 for every day he was idle. At the end of the time he
received $ 29. How many days did he work ?
108 PROBLEMS.
49. A sum of money is divided among three persons. The
first receives $ 10 more than a third of the whole sum ; the second
receives $ 15 more than a half of what remains ; and the third
receives what is over, which is $ 70. Find the original sum.
50. A purse of sovereigns is divided amongst three persons, the
first receiving half of them and one more, the second half of the
remainder and one more, and the third 6. Find the number of
sovereigns the purse contained.
51. From a sum of money $ 20 more than its half was taken
away ; from the remainder $ 30 more than its third part ; and
from what then remained f 40 more than its fifth part ; after which
there was nothing left. What was the sum ?
52. At a cricket match the caterer provided dinner for 24
persons, and fixed the price so as to gain 121 per cent upon his
outlay. Three of the cricketers were absent. The remaining 21
paid the fixed price for their dinner, and the caterer lost 25
cents. What was the charge for dinner ?
53. A can do a piece of work in 30 days, which B can do in
20 days. A begins the work, but after a time B takes his place,
and the whole work is finished in 25 days from the beginning.
How long did A work ?
54. A can do a piece of work in 20 days, which B can do in
30 days. A begins the work, but after a time B takes his place
and finishes it ; and B worked for 10 days longer than A. How
long did A work ?
55. How many men are there in a regiment which can be drawn
up in two hollow squares with the men 3 and 5 deep respectively,
if the one square will just fit within the other, and the same num-
ber of men be in each square ?
SIMULTANEOUS EQUATIONS. 109
CHAPTER VIII.
Simultaneous Equations of the First Degree.
101. A single equation which contains two unknown
quantities can be satisfied by an indefinite number of
pairs of values of the unknown quantities. Consider,
for example, the equation a; — 3?/ = 24. It is clear that
the equation is satisfied by the values y = 0 and re = 24;
or 2/ = 1 and a; = 27 ; or y = 2 and x = 30. In fact, we
may suppose y to be equal to any quantity k, provided
that X is taken equal to 24 + 3 k.
If there are two equations containing two unknown
quantities, each equation taken by itself can, as we have
just seen, be satisfied in an indefinite number of ways ;
but this is not the case when both equations are to be
satisfied by the same values of the unknown quantities.
Def. Two or more equations which are satisfied by the
same values of the unknown quantities contained in
them, are called simultaneous equations.
102. The degree of an equation which contains the two
unknown quantities x and y, is the degree of that term
which is of the highest dimensions in x and y. [Art. 67.]
Similarly the degree of an equation which contains
the unknown quantities «, y, and z is the degree of that
term which is of the highest dimensions in x, y, and z.
110 SIMULTANEOUS EQUATIONS.
Thus 4a; + 5?/ = 20 is an equation of the Jirst degree;
and Soc^ — 2xy = 7 is an equation of the second degree.
The equations ax^ -|- 6?/^ + c = 0 and x-\- y = xy are also
of the second degree.
103. In the present chapter we shall show how to
solve simultaneous equations of the first degree.
It will be seen that any two simultaneous equations
containing two unknown quantities can be solved by de-
ducing from the given equations a third equation from
which one of the unknown quantities is absent ; and
this will give the value of that unknown quantity which
is retained.
104. Elimination by Addition and Subtraction. We will
show how to solve two simultaneous equations of the
first degree by taking as an example the following pair
of equations :
3a; + 52/ = 22,
and 7a; — 42/ = 20.
Multiply each member of the first equation by 7, and
each member of the second equation by 3 ; the equations
then become
21a; + 352/ = 154,
and 21a; -122/= 00.
The coefficients of x are now the same in the two equa-
tions ; hence if we subtract each member of the second
equation from the corresponding member of the first, we
shall obtain an equation from which x is absent : the
equation will be
472/ = 94;
whence 2/ = 2-
SIMULTANEOUS EQUATIONS. Ill
If we put this value of y in the first of the given
equations, we have 3 a; -1-10 = 22; hence 3 a; = 22 — 10,
or ic=4. Observe that this value of x could equally well
be found fi m the second equation.
We may, if we please, find the value of x before finding
the value of y. To do this, multiply the first equation
by 4, and the second by 5 ; the equations then become
12 a; -h 20 2/ = 88,
and 35a; -202/ = 100.
The coefficients of y in the two equations are now
equal in magnitude but opposite in sign ; hence if we add
each member of the second equation to the correspond-
ing member of the first, we shall obtain an equation
from which y is absent ; the equation will be
47 a; =188;
whence a; = 4.
We .can now find the value of y by putting a; = 4 in
the first of the given equations.
It will be seen from the above example that in order
to solve twc simultaneous equations of the first degree
we can proceed as follows : Multiply the given equa-
tions by numbers such that in the resulting equations
the coefficients of one of the unknown quantities may be
equal in magnitude ; then by addition or subtraction we
shall obtain a simple equation which contains only one
of the unknown quantities, and which can be solved as
in the preceding chapter.
The unknown quantity which has been got rid of by
the above process is said to have been eliminated, and the
112 SIMULTANEOUS EQUATIONS.
particular method above explained is known as elimina-
tion by addition and subtraction.
It is generally best to eliminate that unknown quan-
tity which has the smaller coefficients in the two equar
tions.
105. The following are additional examples :
Ex.1. Solve 3x + 22/= 13, 7ic + 3y = 27.
Multiply the equations by 3 and 2 respectively ; we then have
9a; + 6?/ = 39,
and 14 jc -f- 6 ?/ = 54.
By subtraction we eliminate y, and have
-6 a; = -15;
.-. x = 3.
Put this value of x in the first of the given equations ; then
9+2y = 13;
.-. y = 2.
Thus X = 3, y = 2 are the values required.
Ex.2. Solve2x~32/ + 14 = 0, -4x-|- 52/ = 26.
We have 2 x - 3 ?/ = - 14,
and -4x + 5?/ = 26.
Multiply the first equation by 2 ; then
4x-62/ = -28.
From the last two equations we have by addition
or 2/ = 2.
SIMULTANEOUS EQUATIONS. 113
We then have from the first equation,
2x-6=-14,
that is 2a: = -14 + 6=-8;
.-. x = -i.
Hence the values required are
X = — 4, y = 2.
Ex. 3. Solve ^-^ = 9, 6x-^ = 29.
3 4 4
Clear of fractions by multiplying the equations by 12 and 4
respectively ; then
20a; -3y = 108,
24a;-7y = 116.
Multiply the first equation by 7 and the second by 3 ; then
140x-21y = 756,
72 a; - 21 2/ = 348.
By subtraction, we have
68 a; = 408;
.-. x = 6.
Then 120 -3?/ = 108,
whence «/ = 4.
EXAMPLES XXIIL
Solve the equations :
1. 7a; + 4 2/ = 1, 4. 8a;-21y = 5, 7. 19a; + 85y=350,
9x + 4y = 3. 6x+14y = -26. 17a;+119y=442.
2. 3x + 5?/ = 19, 5. 34x-152/=4, 8. 8x-lly = 0,
6x-4y = 7. 51x + 25?/ = 101. 25x-17y=139.
3. x-lly = l, 6. 39x-15y = 93, 9. 3x-lly = 0,
llly-9x = 99. 65x-|-17y=113. 19x-19y = 8.
114 SIMULTANEOUS EQUATIONS.
10. ^+1=1, 12. f+32/ + 14 = 0, 14. ^±^ + 4?/ = 2,
43 5^^^ 11 2
11. f-^ = l, 13. ^+52/=:-4, 15.2^+^+^6=2,
36 2 5 57
^-^ = ^. l(4-5a. = 4 2x-5y a;+7^-.
5 10 2 5 * 3 4'
16. 4x-K2/-3)=5a:-3, 18. ^^-^-^ = 0,
2,+ i(2x-5) = 21^^1I. 2x-5_ll-2y^Q^
6 5 7
"• |-J(y-2)-j(x-3) = 0, 19. «:=-?_2^ = 0,
106. Other Methods of Elimination. Instead of proceeding
as in Art. 104, we may solve two simultaneous equations
in any one of the following ways :
Ex. Solve the equations 3 a; — 5 y = 2, and 5 a; — 2 y = 16.
(i.) From the first equation we have ^x = 5?/ + 2, and there-
fore a; = 1(5 y 4- 2). Now substitute this value of x in the second
equation, and we get
|(5y + 2)-2^/ = 16,
which is a simple equation containing only one of the unknown
quantities ; the solution of which gives y = 2. Then, since y = 2,
we have
x = KS^ + 2)=K10 + 2)=4.
This method is known as elimination by substitution.
(ii.) We may also proceed as follows: From the first of the
given equations we have S x = 6 y -{- 2, or a; = i (5 ?/ + 2). Also
from the second equation we have 5 x = 2 ?/ + 16, or x = ^ (2 y + 16) .
SIMULTANEOUS EQUATIONS. 115
Hence, by equating the two values of a;, we get ♦
i(5y + 2)=i(2y+16).
The last equation gives y — 2, and then, as before, we 'find that
This method is known as elimination by comparison.
(iii.) Or, we may proceed as follows : Multiply the .first of the
given equations by A:, an arbitrary multiplier, and to the resulting
equation add the second of the given equations. Thus, adding
together
3 A;x - 5 A:?/ = 2 ^^ and 6 X - 2 y = 16,
we obtain (3 A; + 5) x - (5 A; + 2) y = 2 A; 4- 16.
This equation is true for all values of k. We may therefore give
to k two values in succession that will make, first the coefficient of
x, and then the coefficient of y, vanish. Thus, taking A: = — |, we
have - ( - 2j5 -I- 2) y = - ^ + 16,
which determines y, and taking A; = — f , we obtain
(-| + 6)a;=- I +16,
which determines x. This method is known as elimination by
undetermined (or arbitrary) multipliers.
No one method of elimination is preferable for all cases. The
practised worker in algebra chooses that form which is best suited
to each example as it arises.
107. The following are examples of common types of
simultaneous equations :
Ex. 1. Solve 57 a; + 25 2/ = 3772, and 25 x + 57 y = 1148.
We have 67 x + 25 ?/ = 3772,
25x + 57y = 1148.
In this example, the direct application of the method of Art. 104
would lead to troublesome arithmetic ; but, since the coefficients
of X and y are simply interchanged in the two given equations, we
should, by addition or subtraction, obtain equations in which the
coefficients of x and y would be equal. Hence we proceed thus :
116 SIMULTANEOUS EQUATIONS.
By addition, S2x-\-82y = 4920 ;
.♦. cc + ?/ = 4920 -- 82 = 60.
By subtraction, S2x-S2y = 2624 ;
.-. X - ?/ = 2624 - 32 = 82.
And from the equations x + y = QO, x — y = S2, we obtain at
once X = 71, and y = — 11.
Ex.2. Solve (X - l)(iy - 2)-(x - 2) (y - 1)= -2,
(X -^2)(y + 2)-(x- 2) (y - 2)= 32.
On removing the brackets, the first equation becomes
xy-y-2x + 2-xy + 2y + x-2=-2,
or —x + y = —2.. (i.)
The second equation becomes
xy + 2x-\-2y + 4:-xy + 2x-}-2y-4: = S2,
or 4a: + 4?/ = 32 . . (ii.)
Dividing (ii.) by 4, we have x-{- y = S . . (iii.)
From (i.) and (iii.) we have by addition
2 y = 6, and therefore y = S.
From (i.) and (iii.) we have by subtraction
2 ic = 10, and therefore x = 5.
Ex. 3. Solve ^ + - = 8, ^ + ^ = 13.
X y X y
These equations can be solved as two simultaneous equations
containing the unknown quantities - and - •
X y
Multiply the equations by 5 and 3 respectively ; then we have
X y
and 1^ + 1^ = 39.
X y
SIMULTANEOUS EQUATIONS. 117
Hence we have by subtraction
2
- = 1, and therefore y = 2.
y
We then have from the first equation
X 2
that is ' § = 8-2=6;
X
.'. 3 = 6a;, or x = ^.
Thus X = ^, y = 2 are the values required.
Ex. 4. Solve ax-{- by = 2 o.b, bx — ay = b^ — a^.
Multiply the first equation by a, and the second equation by b ;
we then have
a'^x -h aby = 2 a^b,
and 6% — aby = b^ — a%.
Hence by addition a^x + 6% = a^b + b^,
that is (a^ + b'^)x = b{a^ + 62);
.'. x = b.
Put this value for x in the first of the given equations ; then
ab ■\-by = 2ab\
.'. by = ab;
.-. y = a.
Thus X = b, y = a B,re the required values.
Ex.5. Solve6«H-3y -13 = 3a;-2y + 7 = 2a; + 6y-4.
Since 6a; + 3y-13 = 3a;-2?/ + 7,
we have 2x + 5y = 20 (i.)
Also, since Bx-2y-^7 = 2x-^6y-4,
we have x-Sy= -11 (ii.)
The solution of (i.) and (ii.) by the method of Art. 104 gives
r = 5 and y — 2.
118 SIMULTANEOUS EQUATION^.
y EXAMPLES XXIV.
Solve the equations :
1. 17 X + 23 ?/ = 86, 4. 29.x + 85?/= 31,
23 cc + 17 ?/ = 74. 13ic-43y = 95.
2. 15ic+19y = 18, 5. x+.2?/ = .3,
19a: + 15?/ = 50. 1.7 a; + .01 y = .345.
3. 51a;- 14?/ = 287, 6. .5a; + ?/ = 2.75,
14 a; - 51 2/ =- 157. 3.4 x + -02?/ = 1.76.
7. 3a;-4?/ + 2 = 5a;-6?/-2 = 7x + 22/ + 4.
8. 4a;-6?/-3 = 7a; + 2?/-4=-2x + 3?/ + 24.
9. 3x + Z?/-2 = ll?/- — = 20.
2i 5
10. 5a; + ^-l = 3?/+--2 = 4.
5 3
11 7 + a;^9 + y^ll + x4-y ^g ^-±1 = l^-±-^ = ^-+
3 5 7 ' '234
13. (a;+l)(?/ + 5) = (a;+5)(?/ + l),
a;2/ + a;-f y=(x + 2)(?/ + 2).
14. a;2/-(a;-l)(?/-l) = 6(?/-l), a;-?/ = l.
15. ? _ § ^ 2, 19.
X y
18 + 8 = 10.
a; ?/
J6. - + - = 7, 20.
a; ?/
§_?=ii.
a; 2/
17. § + ^ = 3, 21.
X y
6_2^j
a; y
18. ^-^ = 9, 22.
a; y
7_2
I-? = 6.
.+?=
y
3^-? =
y
7
2'
.26
3*
2x_§ =
y
:3,
8x + l*i.
+ 6 =
1-3.
= 8,
^^
= 6.
a;
= 57,
^ + 22/
= 7.
SIMULTANEOUS EQUATIONS. 119
23. x-^ + 7 = Sx-^-
y y
-ll = 7x + 21. 24. ^ + y = 2a,
x-y = 26.
26. ax + 6y = (a + 6)2,
32.
ax - 6y = a2 + 62,
ax— by = a^ - b^.
x + y = 2a.
26. ax + by = a^ + b%
33.
a2x + 62?/ = c2,
bx + ay = 2 ab.
a^x + b^y = c3.
27. X- y = a-b,
ax-by = 2a^-2 b\
34.
ax + 6?/ = 1,
6x + ay = 1.
28. ax + 6y = a2 - 62,
6x + ay = a2 - 62.
35.
X y
29. b'h^-ah) = (i,
bx-\- ay = a + b.
X y
30. x + y = a + 6, 36. (a + 6)x +(a + c)y = a + 6,
ax-by=:b^- a\ (a + c) x + (a + 6) y = a + c.
31. bx-ay = b^, 87. (a + 6)x -(a - 6)y = 3a6,
ax-by = a'. (a + 6) y - (a - 6) x = a6.
108. Simultaneous Equations with Three Unknown Quantities.
Three simultaneous equations of the first degree, con-
taining the three unknown quantities x, y, and z, can be
solved in the following manner :
Multiply the first and second of the given equations
by such quantities that, in the resulting equations, the
coefficients of one of the unknown quantities, z suppose,
may be equal ; then by addition or subtraction we elimi-
nate z.
Then take the first and third, or the second and third
of the given equations, and eliminate 2; in a similar man-
ner. We thus obtain two simultaneous equations con-
taining the two unknown quantities x and y j and these
can be solved as in Art. 104.
120 SIMULTANEOUS EQUATIONS.
Ex. 1. Solve the equations
5x — 42/ + 40 = 9.
Multiply the first equation by 2 ; then
4x + 8?/ + 2;2=14.
Subtract the second equation ; then
x + Qy = Q (1.)
Now multiply the first equation by 4 ; then
8 ic + 16 ?/ + 4 0 = 28.
Subtract the third equation ; then
3x + 20«/ = 19 (ii.)
From (i.) and (ii.) we obtain ic = 3, y = I. Substitute these
values in the first of the given equations ; then
6 + 2 + 0 = 7; therefore z= -\.
Thus a; = 3, 2/ = 1 , and z = —\ are the values required.
Ex. 2. Solve the equations
^ y 2
_1_
X
5,
1+1.
Z X
_1_
y
:3,
1+1.
X y
z
:1.
Add the first two equations ; ■
then
2 1
- = 8, whence z = -'
z 4
Add the first and third equations ; then
2 1
- = 6, whence y = -•
y 3
Add the second and third equations ; then
- = 4, whence x = —
a? 2
SIMULTANEOUS EQUATIONS. 121
EXAMPLES XXV.
1. y-\- z = U, 10. x + y-\-z = l,
z-{-x=is, ^-^y + iz = h
x + y = 2^. 2 4
8.. + . = 2a, f + '-f-t'-
z-\-x = 2b,
x + y = 2c. 11. ax + by =1,
8. x + y + z = l, by + cz = h
2x + Sy + z = ^, cz-^ax=l.
^x-\-9y-{-z = l6. ^2. cy-hbz = bc,
4. 5 X 4- 3 y + 7 0 = 2, az+ cx = ca,
2 x - 4 y + 9 ^ = 7, bx-\-ay = ab.
Sx + 2y + Gz = S.
4 13. x + y-\-- = S,
5x + y+3« = 5, 1
2x-3; + 4. = 20. 2a: + 3y +
x-{-2y-Sz = 6, 3x-2y + -=ll.
2x + 4i/-72 = 9, ^
x4-2w + 32; = 4, a6 z
2x + Sy + 4:Z = Q, 2x + §i[ + £ = 2,
3x + 4y + 5iS = 8. a b z
8. 3x + 2y + 5. = 21, 3x_2j,^3c^,,^
2x-3?/ + 4^ = ll, a 0 z
x-\-Sy + 7z = 20. 15. a; _ ay + a2^ = qjB^
9. lOx-2y + 40 = 5, x-by + b^z = b^
Sx+by-Sz = 7, x-cy + c''z = cK
x + Sy-2z = 2.
122 PROBLEMS.
CHAPTER IX.
Problems.
109. We shall now give examples of problems which
involve more than one unknown quantity, and in which
the relations between the known and unknown quantities
are expressed algebraically by means of equations of the
first degree.
Many of the problems given in Chapter VII. really con-
tain two unknown quantities, but the given relations are
in those cases of so simple a nature that it is easy to find
an equation giving one of the unknown quantities in
terms of the known quantities, and when one of the
unknown quantities is found, the other is immediately
determined.
Ex. 1. Find two numbers such that the greater exceeds twice
the less by three, and that twice the greater exceeds the less by 27.
Let X and y be the numbers, of which x is the greater.
Then we have by the conditions of the problem
a; - 2 y = 3,
and 2x— y = 27.
Multiply the first equation by 2 ; then
2x-4:y = 6.
Now subtract the members of this last equation from the corre-
sponding members of the second equation, and we have
Sy = 21,OTy = 7.
PROBLEMS. 123
Then from the first equation
x = S-\-2y = S + Uz=n,
Thus the numbers are 17 and 7.
Ex. 2. A number of two digits is equal to seven times the sum
of its digits, and the digit in the ten's place is greater by four than
the digit in the unit's place. What is the number ?
Let X be the digit in the ten's place, and y be the digit in the
unit's place.
Then the number is equal to 10 a; + y, for the x represents so
many tens ; also the sum of the digits is x + y.
Hence we have 10 a; + y = 7(a; 4- y)j
that is 10x + 2/ = 7aj + 7y;
.♦. 3 a; = 6 y, or X = 2 y.
We have also x = y + 4.
Hence 2 y = y + 4, or y = 4,
and therefore x = 8.
Thus the required number is 84.
Ex. 3. Find the fraction which is equal to \ when its numerator
is increased by unity, and is equal to ^ when its denominator is
increased by unity.
Let X = the numerator of the fraction, and y = the denominator.
Then we have by the conditions of the problem
^+1 = 1, and ^ -1
y 2' y + 1 3
Multiply the first equation by y ; then
.+i=|.
Multiply the second equation by y + 1 ; then
a; = i(y + l).
124 PEOBLEMS.
Subtract the corresponding members of the last two equations ;
and we have
from which we jfind that y = S. Then, since y is 8,
2
Thus the fraction is f .
Ex. 4. A man and a boy can do in 15 days a piece of work
which would be done in 2 days by 7 men and 9 boys. How long
would it take one man to do it ?
Let X = the number of days in which one man would do
the whole ;
and let y = the number of days in which one boy would do
the whole.
Then a man does -th of the whole in a day ; and a boy does
X
- th of the whole in a day.
y
Now by the conditions of the problem a man and a boy together
do jijth of the whole in a day.
Hence we have - + - = —
X y 16
We have also, since 7 men and 9 boys do half the work in a day,
1 + 5 = 1.
X y 2
Multiply the first equation by 9 and subtract the second ; then
2_Z=1_1
X X 15 2'
9 1 1 f1
thatis ±=:±-,oT^ = ^;
X 10 X 20
.-. x = 20.
Thus one man would do the work in 20 days.
PROBLEMS. 125
EXAMPLES XXVL
1. A and B have $50 between them, but if A were to lose half
his money, and B | of his, they would then have only $"20. How
much has each ?
2. A number of two digits has its digits reversed by the addi-
tion of 9. Show that the digits differ by unity.
3. A man bought 8 cows and 50 sheep for $ 1125. He sold the
cows at a profit of 20 per cent, and the sheep at a profit of 10 per
cent, and received in all $ 1287.50. What was the cost of each
cow and of each sheep ?
4. Twenty-eight tons of goods are to be carried in carts and
wagons, and it is found that this will require 15 carts and 12
wagons, or else 24 carts and 8 wagons. How much can each cart
and each wagon carry.
6. A and B can perform a certain task in 30 days, working
together. After 12 days, however, B is called off, and A finished
it by himself 24 days after. How long would each take to do the
work alone ?
6. If the numerator of a certain fraction be increased by 1 and
its denominator diminished by 1, its value will be 1. If the nu-
merator be increased by the denominator and the denominator be
diminished by the numerator, its value will be 4. Find the
fraction.
7. Find the fraction such that if you quadruple the numerator
and add 3 to the denominator the fraction is doubled, but if you
add 2 to the numerator and quadruple the denominator the frac-
tion is halved.
8. The first edition of a book had 600 pages, and was divided
into two parts. In the second edition one quarter of the second
part was omitted and 30 pages added to the first. The change
made the two parts of the same length. What were they in the
first edition ?
126 . PEOBLEMS.
9. If A were to receive $ 10 from B, he would then have twice
as much as B would have left ; but if B were to receive $ 10 from
A, B would have three times as much as A would have left. How
much has each ?
10. A farmer sold 30 bushels of wheat and 50 bushels of barley
for $93.75. He also sold at the same prices 50 bushels of wheat
and 30 bushels of barley for $96.25. What was the price of the
wheat per bushel ?
11. A rectangle is of the same area as another which is 6 yards
longer and 4 yards narrower ; it is also of the same area as a third,
which is 8 yards longer and 5 yards narrower. What is its area ?
12. A and B can together do a piece of work in 15 days. After
working together for 6 days, A went away, and B finished it by
himself 24 days after. In what time would A alone do the whole ?
13. An income of $ 120 a year is derived from a sum of money
invested partly in 3| per cent stofek and partly in 4 per cent stock.
If the stock be sold out when the 3 1 per cents are at 108 and the
4 per cents at 120, the capital realized is $ 3672. How much stock
of each kind was there ?
14. A number of two digits is equal to seven times the sum of
its digits ; show that one digit must be twice the other.
16. Find all the numbers of two digits, each of which is equal
to four times the sum of the digits.
16. $ 1000 is divided between A, B, C, and D. B gets half as
much as A ; the excess of C's share over D's share is equal to
one-third of A's share, and if B's share were increased by $ 100,
he would have as much as both C and D. Find how much
each gets.
17. A number has two digits, of which the second is double
the first ; and, if the digits be reversed, the new number exceeds
the original number by 36 ; find the number.
PROBLEMS. 127
18. A certain number consists of two digits, and another num-
ber is formed from it by reading it backwards. If the sum of the
two numbers is 99 and the difference is 45, lind the digits. .
19. In a certain proper fraction the difference between the
numerator and the denominator is 12, and if each be increased
by 5, the fraction becomes equal to f . Find it.
20. The wages of 10 men and 8 boys for a day amount to $ 31 ;
and four men receive $ 5.50 more than six boys. How much does
each boy receive ?
21. A farmer has two farms, for each of which he pays a rent of
$ 7.60 an acre, and his total rent is $ 3375. If the rent of one
farm were reduced by $ 1.25 an acre, and that of the other by
$ 2.50 an acre, his rent would be $ 2500. What is the acreage of
each of the farms ?
22. A man has one pound's worth of silver in half-crowns,
shillings, and sixpences ; and he has in all 20 coins. If he changed
the sixpences for pennies, and the shillings for sixpences, he would
have 73 coins. How many coins of each kind has he ?
23. The price of a passenger's ticket on a French railway is
proportional to the distance he travels ; he is allowed 25 kilo-
grammes of luggage free, but on every kilogramme beyond this
amount he is charged a sum proportional to the distance he goes.
If a journey of 200 miles with 50 kilos, of luggage cost 25 francs,
and a journey of 150 miles with 35 kilos, cost 1C| francs, what
will a journey of 100 miles with 100 kilos, cost ?
24. A has twice as many dimes as dollars ; B, who has 60 cents
more than A, has twice as many dollars as dimes ; together they
have one more dime than they have dollars. How much has each ?
25. Of the candidates in a certain examination, one-quarter fail.
The number of marks required for passing is less by 2 than the
average marks obtained by all the candidates, less by 11 than the
average marks of those who pass, and equal to double the average
marks of those who fail. How many marks are required for
passing ?
128 MISCELLANEOUS EXAMPLES II.
MISCELLANEOUS EXAMPLES IL
A. 1. Find the value of
ab + 2hc-Zcd gg + &^ - c3
6+c + d a2 + 62_c2'
when a. = 1 , & = 2, c = — 3, and cZ = 0.
2. Subtract 2a-3 («-&-«) from 2 6 - 3 (& - « - 6).
3. Show that
(a^+3)(y+3)-3(a;+l)(2/+l)+3(a^-l)(y-l)-(a:-3)(2/-3)=:0,
and that
(x+2)(^/+2)-4(.'c+l)(y+l)+6a;2/-4(x-l)(2/-l)
+ (x-2)(i/-2)=0.
4. Divide x^ - 5x* + 7 x^ - x^ - 4a; + 2 by x^ - 3a:2 + 3a; _ i.
5. Solve the equations :
,. . a;a; + la; + 2a; + 3^^
^■^2345
(ii.) f + | = 5, 2a;+|-17 = 0.
3 5 3
(iii.) aaj — &y = a'^ — 62 ^
6x-a?/=62_a2i .
6. A number of marbles were divided among thi*ee boys^so that
the first boy had 10 less than half of the whole, the second had
10 more than a quarter of what was left, and the third had 20.
How many marbles were there ?
B. 1. Find the numerical value of
(a - 6)2 + (6 - c)2 +(c - a)2 +(6 -c){c- a)(a - 6),
when a = 1, 6 = 2, and c = ^.
2. From the sum of (2 a - 6)2 and (a - 2 6)2 take the square
of 2(a-6).
MISCELLANEOUS EXAMPLES II. 129
3. Show that
n^ = n(n- l)(n - 2)+ 3w(n - 1)+ n,
and n4=n(n-l)(w-2)(w-3)+6w(n-l)(w-2) + 7 7i(n^l) + n.
4. Find the algebraical expression which, when divided by
x2 — 2 a; + 1 , gives a quotient x^ + 2 x + 1 and a remainder x + 1.
5. Solve
(i.) 3(x + 3)2 + 5(x + 5)2 = 8(x + 8)2.
(ii.) ^(x + y)-K^-y)=S \
cm.) 1 + 4^1, Z_8^1.
X y X y 6
6. A man paid a bill of £100 with sovereigns and crowns,
using in all 130 coins. How many coins of each sort did he use ?
C. 1. Simplify a -[a -^ b - {a -{- b ■}- c - a -\- b + c + d}'].
2. Find the sum, the difference, and the product of a + 6 and
a _b
2 2
3. Show that
ab - 3(a - 1)(6 - 1) + 3(a - 2)(6 - 2) -(a - 3) (6 - 3) = 0,
and that
ab - 4(a - 1)(6 - 1)+ 6(a - 2)(6 - 2)
- 4(a -3)(b- 3)4-(a - 4)(6 - 4)= 0.
4. Divide a^ + 6^ + c^ - 3 a6c by a + 6 + c, and from the result write
down, without division, the quotient when Sx^ -\-Sy^ + ^ — 12 xyz
is divided hy 2x-\- 2y -\- z.
5. Solve the equations :
(i.) x = K5?/ + 2), 2/ = K^-l)-
(ii.) ax -\- by = a^, bx + ay = b^.
6. A and B each shoot 30 arrows at a target. B makes twice
as many hits as A, and A makes three times as many misses as B.
Find the number of hits and misses of each.
X
130 MISCELLANEOUS EXAMPLES II.
D. 1. Find the value of
abc(ab + bc + cd -i- da) -=- (a + 6) (a + c) (a + d),
when a = 1, & = 3, c = — 5, and d = 0.
2. Simplify (a + 6)2 - [2 a^ - {(a - 6) (a + 2 &) - 6(a - 6)}].
3. Find the continued product of x + a, x + b, and x + c ; and
from the result write down the continued product of a — x, a — y,
and a — z.
4. Divide 76a%^c^ - Iba^b^c^ by ba'^bc^, and
a2 _ 2 62 _ 6 c2 4- a6 - ac + 7 6c by a - 6 + 2 c.
5. Solve the equations :
(i.) ^(ix-l)-iix-2)+l(x-S) = 0.
Cii ^ ^~^ = y -^ = a; + y
^ '^ 2 3 9
(iii.) § + ^=22, l-i = 20.
X y X y
6. A square grass plot would contain 69 square feet less if each
side were one foot shorter. How many square feet does it contain ?
E. 1. Simplify
b^ + {b(a — c)+ ac] — a{b — a - c] + b(c - b r- a) — ab.
2. From what must the sum of 3 a^ + 2 a6 — ac, 3 6^ -j. 2 6o — a6,
and 3 c2 + 2 ca — 6c be taken in order that the remainder may be
a2 + 62 + c2 ?
3. Show that (x + y)2 + (y + zy + (0 + xy
+ 2(a; + y)<ix+ z)+ 2(y + zXv + x)+2{z + x){z + y)
= 4(x + y + zy.
4. Divide ^x^ + -\^ x"^ - ^x + ^ hj ^x + 2, and
acx^ + {ad - bc)x^ — (ac + bd)x + be by ax — 6.
MISCKLLANEOUS EXAMPLES n. 131
5. Solve the equations :
(i.) (a;-l)(x + 2) = a;2 + 3.
+ .2?/ = 1.8J
(ii.) 2x-\- Ay z= 1.2
bx
(iii.) (a-6)x = (a+ 6)y)
x-\-y = 2a /
6. Find two numbers such that twice their difference is greater
by unity than the smaller number, and is less by two than the
larger number.
F. 1. Find the value of ^(x -\- y ■\- zy - (t? -\- y^ + z^) , when
jc = 3, ?/ = — 5, and z = 1.
2. Find the continued product of a* + x*, a^ + x^, a + x, and
a — x.
3. Show that (1 - a)(l - 6) + a(l - 6) + & = 1,
and that (1 - «) (l _ 6) (1 _ c) + a(l - b) (1 - c) + 6(1 - c) + c = 1.
4. Divide 4y* - 9y^ + 6y - I hy 2y^ -^ Zy - 1.
Divide also a^ + 2 62 - 3 c2 + &c + 2 ac + 3 a6 by a + 6 - c.
5. Solve the equations :
(i.) (a + x)(6 + x) = (c + x)(d + x).
(ii.) Sx~7y = 4, 7x-9y = 2.
6. If 12 lbs. of tea and 3 lbs. of coffee cost ^8.76, and 12 lbs.
of coffee and 3 lbs. of tea cost $6.24, what is the price per lb.
of tea and of coffee ?
132 FACTORS.
CHAPTER X.
Factoes.
110. Definitions. An algebraical expression which does
not contain any letter in the denominator of any term is
called an integral expression.
Thus a + h and \ a^ _ i ^2 are integral expressions.
An expression is said to be integral with respect to any
particular letter when that letter does not occur in the
denominator of any term.
Thus — 4 is integral with respect to x.
a a + h
An expression is said to be rational when none of its
terms contain square or other roots.
111. In the present chapter we shall show how factors
of rational and integral algebraical expressions can be
found in certain simple cases.
In arithmetic we mean by the factors of a number its
integral divisors only; and similarly, by t\\Q factors of an
algebraical expression, we mean the rational and integral
expressions which exactly divide it.
112. Monomial Factors. When any letter, or group of
letters, is common to all the terms of an expression, each
term, and therefore the whole expression, is divisible by
that letter, or group of letters.
FACTORS. 133
Thus ab + ac = a(^b + c),
a'^b + a62 = ab(a + 6),
and 3 ay^y + 6 axY = 3 0x2^/ (x + 2 y2) .
Monomial factors, if there are any, can be seen by-
inspection, and the whole expression can be at once writ-
ten as the product of the monomial factor and its co-factor,
as in the above examples. In what follows therefore we
need only consider expressions which have been freed
from monomial factors.
EXAMPLES XXVII.
Find the factors of the f ollowmg expressions :
1. x^ + X. 7. ic* -bx^y-\- 20x2^/2.
2. a2 - ab. 8. a^ - a*x + a2ic2.
3. ab - be. 9. 6 ax* - 5 a^x^ + 20 a2x3.
4. 2 ax -I- ^ x2. 10. 3 a^x*y^ - ^ oh^yK
5. 4x8-3x2. 11. Ila25c2- ia&2c3.
6. a8-3a26. 12. p'^q^r^ - 7 j^q^.
113. factors found by comparing with Known Identities.
Sometimes an algebraical expression is of the same forin
as some known result of multiplication : in this case its
factors can be written down. We proceed to apply this
principle in the case of the most important forms of
algebraical expressions.
114. We know that
and a^ - 2a6 + 62 = (a - bf.
134 - FACTORS.
Hence, when a trinomial expression consists of the sum
of the squares of any two quantities plus tlvice the product
of the quantities, it is equal to the square of their sum.
And when a trinomial expression consists of the sum of
the squares of any two quantities minus tivice their product,
it is equal to the square of their difference.
Hence it is easy to recognize when a trinomial is a
complete square ; for two of the terms must be squares,
and the remaf~ang term must be twice the product of the
quantities v^ ares are the other terms.
The two tt 3h are squares must both have the
same sign.
Thus a2 + 6 a& + 9 b'^, that is a"^ + (3 &)2 + 2 a (3 &), consists of
the squares of a and 3 b together with twice the product of a and
3 b, and hence
a2 + 6a6 + 962= (a + 3&)2.
Again,
^ab - a^ - 4:b'^ = - (a"^ + 4b^ - i ab) = - {a"^ + (2 6)2 - 2 a (2 6)}
=:-(a-2&)2.
As other examples, we have
lea* 4- 8a-^ + 1 =(4a2)2 + 2 (4^2) + i = (4a2 + 1)2,
ics ^ 4 x^y^ + 4 icy* = a; (x2 - 4 xy^ + 4 y*)}
= x{x^ - 2 a; (2^/2)+ (2?/2)2} = x (x -2y'^)\
and (a + 6)2 - 2c (a + &)+ c2 = {(a + b)-cf.
Note. — We may consider that a^ — 2ab-\-b'^ is equal to
(b - a)2 instead of (a - 6)2, for by the Law of Signs (& - a)\
that is {-{a - b)f, is equal to (a - 6)2. In fact, when we find
that an expression is equal to the product of two factors, we may
equally well consider it is as the product of the same two factors
with all the signs changed.
I
FACTORS. 135
EXAMPLES XXVIII.
Find the factors of the following expressions :
1. 4x2 + 4a;+l. 9. 3a2 + 6a6 + 362.
2. 9x^-Qx+ 1. 10. 5a4 _ lOa^b + 5 b^
3. l-8x2 + 16x*. 11. a»-6a%-{-9ab^.
4. 4a2_i2a6 + 962. 12. 3 a^ - 30 a*68 + 75 a^fe^
5. 9 a* + 24 a262 + 16 6*. 13. 4^22^2 .a4_4y4.
6. x2 + xy + iy2. 14. 8x2 ^,.^^^-4.
7. 4 a2x2 + 4 a6xy + &2y2. 15. 4 ^i/S Ja'J4 x22/2 + x^y.
8. 25a*x^-S0a'^b^xy + 9b*y^. 16. x22/2 + ^^y + ^ xy8.
17. (a + 6)2 + 4c(a + 6)+4c2.
18. (X2 + 2/2)2 _ 2 (X2 + y') 02 + ;j4.
19. 4x2y2 + 4 (a + b) xy + (a + b)\
20. 9 (a + 6)2 _ 6 c (a + 6) + c2.
115. From the formula
a'-b' = (a + b){a-b),
we see that the difference of the squares of any two quan-
tities is equal to the product of the sum and the difference
of the quantities. ^
Thus a2 - 4 &2, or a^ - (2 6)2, is equal to (a + 2 6) (a - 2 &).
Also 9x8 - 4xy2, that is x{(3x)2 - (2 yY); is equal to
x(3x + 2i/)(3x-22/).
As other examples, we have :
8 axy - 18 a^x^y^ = 2 axy (4 - 9 a^xY') = 2 axy {22 - (3 axyy]
= 2 axy (2 - 3 axy) (2 + 3 axy),
a* - 6* = (a2 + &2) (^2 _ 52) ^ (aj2 + ^,2) (a + 5) (« _ 5)^
and 792 - 712 = (■yg +71) (-79 _ 71) ^ 150 x 8 = 1200.
136 FACTORS.
We may deal with the squares of multinomial expres-
sions in precisely the same way as with the squares of
monomial expressions.
Thus (a + by - c^ = {{a + 6) + c] {{a + 6) - c}
= (a + 6 + c)(a + &-c),
ai + a252 + 54 = (^2 + 52)2 _ («5)2 ^ («2 + 52 _^ «5) (^2 + 52 _ <25)^
and (2a- 6 + 2c)2-(a + 46 + c)2
= {(2 a - & + 2 c) + (a -f- 4 6 + c) } {(2 a - 6 + 2 c) - (a + 4 6 4- c)}
= (3 a + 3 6 + 3 c)(a - 5 6 + c).
EXAMPLES XXIX.
Find the factors of the following expressions :
1.
a2 - 9.
19.
X* - 2/4.
2.
16 - 62.
20.
X* - 16 yK
3.
25 a2 _ 62.
21.
81 a* - 16 6*.
4.
a;2 - 9 y2.
22.
625 a* -256x4.
5.
16x2 -9 2/2.
23.
x*^/* - a*64.
6.
64 a2 _ 49 62.
24.
a464 _ 81 c4#.
7.
4a2-81&2.
25.
81 xV - 1.
8.
x2-9y*.
26.
16 a464c4 _ 1.
9.
36 a* -49 62.
27.
16-81xV,
10.
4a262_9c2.
28.
X8 - ^8.
11.
9«2x2_49622/2.
29.
a8 _ 68^8.
12.
49a262c2_36x2i/2«2.
30.
aio - a2.
13.
4x2/2-9x3.
31.
{a + 6)2 - c2.
14.
8 a62 _ 18 aK
32.
(a + 6)2-4c2.
15.
3 «& - 108 aK
33.
4 (X + 2/)2 - 1.
16.
7 a^ - 28 a9.
34.
9 (X - y)2 - 4.
17.
8 x^y - 32 xy^.
35.
(x + 2/)2-(x-2/)2.
18.
7a6c2-7a868.
36.
(2a + 6)2-(26 + a)2.
FACTORS. 137
37. x^-(x-yy. 41. (a2 + 52)2 _ 4 <2262.
38. a2 _ (2 6 _ a)2. 42. (a + 5 + c)2 - (a - 6 - c)2.
39. 4(a + &)2-(a-6)2. 43. (3a +6_ 2c)2-(<i+3 6-c)2.
40. 9 (a; + 2/)2 - 4 (X - y)2. 44. (a - 2 6 + 3c)2 -(a - c)2.
46. (3 x2 + a; - 2)2 _(a;2 _ a; _ 2)2.
116. From the formulae
/■
a^ + a^ = (a; + a) (a^ — aa; + a^),
and x^-a^ = {x-a){a?-\-ax + a-), [Art. 87.]
we see that the sum of the cubes of any two quantities is
divisible by the sum of the quantities ; and that the dif-
ference of the cubes of any two quantities is divisible by
the difference of the quantities.
Thus 8 a3 4. 27 68, that is (2 «)» + (3 by, is divisible by 2 a + 3 6,
and the quotient is (2 aY - (2 a) (3 h) + (3 6)2, that is
4 a2 _ c a& + 9 h\
Also 27 x^ - 8, that is (3a;)» - 2^, is equal to
(3ic-2){(3:?)2 + (2.3x)+22} = (3x-2)(9x2 + 6x4-4).
As other examples,
asjs _ 1 c8 =(a6 _ ^c)(a262 + ^ a&c + ic2),
and a6 _ 56 _ (fljs + 68) (^^a _ ^,8)
. =(a + 6)(a2 -ab + b^)(a - b)(ia^ + ab + 62).
The cubes of multinomial expressions may be dealt
with in precisely the same way.
Thus (a + 6)3 - c3 = {(a + 6) - c}{(a + 6)2 + (a + 6) c + c2},
B,nd (x-2yy-(y-2xy
= (x - 2y - y - 2x){{x - 2yy -\-(x -2y)(y - 2x)-^(y - 2xr-}
= (3a;-3y)(3a;2-3a:y + 3!/2).
188 FACTORS.
EXAMPLES XXX.
Find factors of each of the following expressions :
1. a3_8 63. 11. S a^b + 24: ab\
2. 8a^ + 63. 12. 40 a^bc - 5 fe^c*.
3. 8a3_i25ic8. 13. a^ - 64.
4. a3- 125x6. 14. 64 a'- -729 66.
5. 4a3 + 32 63. 15. x^^ - a^¥.
6. 27x3-i?/3. 16. (x-i-2yy-y^
7. x3y3 + . ^1^0^353. 17. (x + 2yy + Cy + 2xy,
8. 8 a666 + x^. 18. (2 y - x)^ - (2 x - i/)3.
9. 2xy^-l xK 19. (x - 3 ?/)3 - (y - 3 x)^.
10. 9a*62-ia6^ 20. (2?/ - x)^ + (2x - y)3.
117. By multiplication we have
(a? + a) (» + 6) = a^ + (a + 6) ic + ab.
Hence conversely, if an expression of the form
a^ -\- px -\- q he the product of the two factors x + a and
x-\-b, the given expression must be the same as
a^ + (a 4- 6) a; + a6 ;
we must therefore have p = a + b and q = ab. Hence
a and b are such that their sum is p, and their product q.
For example, to find the factors of x^ + 7 x + 12. The factors
will be X + a and x + 6, where ab = 12 and a 4- 6 = 7. Hence we
must find two numbers whose product is 12 and whose sum is 7.
Pairs of numbers whose product is 12 are 12 and 1, 6 and 2, and
4 and 3 ; and the sum of the last pair is 7. Hence
x2 + 7 X + 12 = (x + 4) (x + 3).
Again, to find the factors of x^ — 7 x + 10, we have to find two
numbers whose product is 10, and whose sum is — 7. Since the
product is + 10, the two numbers must both be positive or both
negative ; and since the sum is — 7, they must both be negative.
FACTORS. 139
The pairs of negative numbers whose product is 10 are — 10 and
— 1, and — 5 and — 2 ; and the sum of the last pair is — 7. Hence
a;2_7x+10=(x-5)(a;-2).
Again, to find the factors of x^ + 3 x — 18, we have to find two
numbers whose product is — 18, and whose sum is 3. Since the
product is — 18, one of the numbers is positive and the other
negative. The pairs of numbers whose product is — 18 are — 18
and 1,-9 and 2,-6 and 3,-3 and 6,-2 and 9, and — 1 and
18 ; and of these pairs the sum of 6 and — 3 is 3. Hence
x2 + 3 X - 18 = (x + 6) (x - 3).
EXAMPLES XXXI.
Find the factors of each of the following expressions :
1. x2 + 4x + 3. 13. x2 + 6x-14.
2. x2-4x + 3. 14. X2-X-132.
8. x2-6x + 8. 16. x2 + 18x + 72.
4. x2-8x + 15. 16. x2-5x-84.
6. x2 - 11 X + 18. 17. x2 - 25 X + 150.
6. x2 + 9x4-20. 18. x2 + 5x-150.
7. x2 + 2x - 3. 19. x2 + 11 X - 180.
8. x^ + ix-5. 20. X2-X-156.
9. a:2 + x-6. 21. x2-31x + 240.
10. x2 - X - 6. 22. x2 - 17 X - 200.
11. x2 4-2x-35. 23. x2-34x + 288.
12. x2-3x-10. 24. x2-35x-200.
118. By multiplication we have
{ax + b) {ex -\-d) = ac3? -}- {ad -{■hc)x-\- bd.
Hence conversely, if an expression of the form
px^ -{-qx-{-r be the product of the two factors ax-\- b
and ex -f d, the given expression must be the same as
140 FACTOES.
acx^ -f {ad-\- bc)x + bd-, we must therefore have ac =p,
hd = r, and ad-{-hc = q. We can in simple cases find by
trial the values of a, b, c, d which satisfy these relations.
For example, to find the factors of 3x^ — 16x+ 6. The Sx^
can only be given by the multiplication otSx and .r. The 5 could
only be given by the multiplication of 5 and 1, or — 5 and — 1 ;
and, since the middle term is negative, the latter pair must be
taken. We have now only a choice between (3 oj — 5) (ic — 1) and
(3x — l)(a: — 5), and it is at once seen that the latter must be
taken in order that the coefficient of x in the product may be — 16.
Thus the factors required are 3 ic — 1 and cc — 5.
Again, to find the factors of 5 x^ + 32 a; — 21, the 6x^ can only
be given by the multiplication of 5 x and x. The end term — 21
can be the product of —21 and 1,-7 and 3, —3 and 7, or —1 and
21. The possible pairs of factors, so far as the two end terms
are concerned, are therefore (5x =F 21)(x ± 1), (5x T 7)(x i 3),
(5 X T 3) (x ± 7), and (5 x T 1) (ic ± 21). It will be found that of
these pairs the one which will give 32 x for the middle term is
(5x-3)(x+7).
119. The factors of oc^ — 5 xy -\- 4:y^ can be found in
the same way as the factors of cc^ — 5 ic + 4. For we
must find two quantities whose product is 4 y^, and whose
sum is — 5y: these are — 4 2/ and — y. Hence
a^ — 5 xy + 4:y^ = (x — 4:y) (x — y).
So also the factors of 3x^ — 16 xy -\- 5y^ can be found in
the same way as the factors of 3 o^ — 16 a; + 5 ; and the
factors of either can be written down as soon as the fac-
tors of the other are known.
For example, if we know that 5 x^ + 32 x - 21 = (5 x - 3) (x + 7),
we must have 6 x^ 4- 32 xj/ — 21 2/2 = (5 x - 3 y) (x + 7 y).
FACTORS. 141
EXAMPLES XXXII.
Find the factors of each of the following expressions :
1. 3a;2-10a; + 3. 20. 7x2 + 123x-54.
2. 3a;2-17x+10. 21. 24x2-30x-75.
3. 2x2+ 11 x + 12. 22. a:2 + 4x?/ + 3t/2.
4. 2x2 + 3x - 2. 23. x* _ 6xy + 8?/2.
5. 3x2 + 7x- 6. 24. x2 - llxy + I82/2.
6. 4x2 + x-3. 25. x2 + 5xi/-14y2.
7. 6x2-38x + 21. 26. x2 - 25 xy + 150 y2.
8. 3x2 + llx-20. 27. x2-35xy-2002/2.
9. 7 x2 - 33x - 54. 28. 3x2 - 17 xy + 10 y\
10. 6x2-38x + 48. 29. 7x2 - 33xy - 541/2
11. 7x2 + 75x-108. 30. 24 x2 - 70 xy - 75 y2.
12. 9x2 + 130x-75. 81. x*- 13x2 + 36.
13. 4 x2 + 21 X - 18. 32. X* - 25 x2y2 + 144 y*.
14. 4x2 + 4x-15. 33. 36 x* - 97 x2y2 + 36 y*.
15. 6x2+55x-50. 34. x3-3x2-18x.
16. 10 x2 + 3 X - 1. 86. x^y - x2y2 _ 2 xy*.
17. 132x2 + x-1. 86. 15x4?/-4xV-4x2y8.
18. 4x2-5x+l. 37. 75x2/3 -130x2?/4_ 9 a;8y5.
19. 12x2 + 50x-50.
120. It is clear that the method of finding by trial the
factors of an expression of the form pa? + ga; + r, where
p, q, r are known numbers, would be very tedious if
there were many pairs of numbers whose product was p,
and msiiij pairs whose product was r, for there would
then be very many pairs of factors which would agree
with the given expression so far as the end terms
were concerned, and out of these the single pair which
would give the correct middle term would have to be
142 FACTORS.
sought. It would, for example, be almost impossible to
find the factors of 2310 x" - 2419 x — 9009 in this way.
Again, we not unfrequently meet with such an expres-
sion as a;^ + 6 a; + 7 which cannot be written as the prod-
uct of two factors altogether rational, and in such a case
it would be impracticable to try to guess the factors.
We therefore need some method of. finding the fac-
tors of a quadratic expression which is applicable to all
cases. This method we proceed to investigate.
121. We first note that since a^ -f 2 aa; -f a^ is a perfect
square, namely {x-\-aY, it follows that in order to co^^
plete the square of which x^ and 2aa; are the first two terms,
we must add the square of a ; that is, we must add the
square of half of the coefficient of x.
For example, x^ + 6 a; is made a perfect square by the addition
of (1)=^ ; and the square \sofi + Qx + 9, which is (x + 3)2.
So also, Qci'^ -\- hx is made a perfect square by the addition of
(1)2 ; and the square is x^ + 5 aj + (-\^), which is (x + f )2.
And x2 — 7 X is made a perfect square by the addition of (- |)2,
that is of \^ ; and the square is x^ — 7 x + Y, which is (x — |)2.
Whatever a may be, x^ -|- ax is made a perfect square by the
addition of ( - i , f or
• /
1212. We will now find the factors of ax^ -{-hx-\-c by
a method which is applicable in all cases.
The problem before us is to find two factors which are
to be rational and integral with respect to x, and are there-
fore of the first degree in a;, but which are not necessarily
FACTORS. 143
rational and integral with respect to arithmetical num-
bers or any other letters which may be involved.
We first apply the method to some examples. «
Ex. 1. Find the factors of
a;2 + 6 X + 8.
As we have seen in the last article, x"^ + 6x is made a perfect
square by the addition of 3'^ ; hence, adding and subtracting 3^, we
have
x^-\-6x + S = x^ + ex + 9-9 + S
= (x + 3)2-1.
The last form of the expression is the difference of two squares,
and the factors are therefore x + 3 + 1 and x + 3 — 1, that is x + 4
and X + 2. Hence the factors of x^ + 6 x + 8 are x + 4 and x + 2.
Ex. 2. Find the factors of x^ - 3 x - 28.
x2 — 3 X is made a perfect square by the addition of (— |)2, that
is of I ; hence, adding and subtracting |, we have
a;2 - 3x - 28 = x2 - 3x + I - f - 28
= (X-3)2_121
= (a;_|)2_(Y)2.
Now the factors of the last expression are x — f + -^ and
a^ - f - ¥•
Thus the required factors are x + 4 and x — 7.
Ex. 3. To find the factors of x2 + 6 x + 7.
x2 4- 6 X is made a perfect square by the addition of 3^ ; hence
we write
x2 + 6x+7 = x2 4-6x + 9-9 + 7
= (x + 3)2 - 2
= (x + 3)2-(v2)2.
Now the factors of the last expression, and therefore of
x2 + 6 X + 7, are X + 3 + V2 and x + 3 - ^2.
144 FACTORS.
Ex. 4. To find the factors of 3 ^2 - 10 a; + 3.
3x2 - lOx + 3 = 3 (x2 - -V*x + 1).
Now x2— ^-x is made a perfect square by the addition of
(- f)2, that is of -%^- ; hence
x2 _ -ijO X + 1 = x2 - -i/x 4. 2^ _ 2^5 4. 1
The factors of the last expression are x — f + f and sc — | — f ,
that is X — 1^ and x — 3 ; hence
3x2 - lOx + 3 = 3 (x - i)(x - 3).
From the above examples it will be seen that the
method of finding the factors of a quadratic expression
consists in changing the expression into an equivalent
one which is the difference of two squares.
The factors of ax^ -\-bx-{-c are therefore found in the
following manner :
ax^-{-bx-\-c = a[a:^ -\--x-\- -]'
\ a aj
Now a^H — X is made a perfect square, namely fx-\ ) ,
• ^ /" bY b^ . ^ ^^-^
by the addition of ( — ] = — ^- And, by adding and sub-
tracting — - to the expression within brackets, we have
4a^
\^ a a J ( a \2aJ Aa^ a)
FACTORS. 145
The factors of the expression in square brackets are,
by Art. 115,
2 a \\4a2 a) 2a \\4a^ aj
Hence ax^ -\-hx + c
This formula solves the most important case of factor-
ing in elementary algebra, and in form represents a class
of cases, also important, in which the problem is: to
resolve an algebraic expression of higher degree in x, or
in X and y, into a product of factors of the first degree
in the same letters. In this problem the coefficients of
X and of y in the factors may, and frequently do, involve
irrational quantities (surds).
The principle of factoring, thus enunciated, as the stu-
dent will have occasion to observe in the sequel, underlies
all elementary methods for solving algebraic equations.
123. Instead of Avorking out every example from the
beginning, we might use the formula found in the last
article, and we should only have to substitute for a, h,
and c their values in the particular case we are consid-
ering.
Ex. 1. Find the factors of x^ + 7 a: -}- 12.
Here a = 1, & = 7, c = 12.
Hence x2 + 7 x -H 12
= {a^+i + V(¥-12)}{x + |-V(¥-12)}
= (x + 4)(x + 3).
K
146 FACTORS.
Ex. 2. Find the factors of 5 a;2 + 32 xy - 21 y'^.
Here a = b, b = S2y, c = -21y^.
=V{
266f±my^') ^19y
25 / 5 *
C 5 5Jl 5 6J
= 6(ix + 7y)(x-^-^\
= (x + 7y)(i6x-Sy).
124. If in the formula of Art. 122 we put 2 for a and
4 for 6, we obtain the factors of the expression a^+2a;4-4 ;
these factors are x + 1 + ^(1 — 4) and a; -f 1 — -^(1 — 4);
that is, a; + l4-V(— 3) and x-\-l—^{ — 3).
Now all squares, whether of positive or of negative
quantities, are positive. Hence it is impossible to find
any real number whose square is —3. Such an expression
as V— 3 is, however, often used in algebra; and the
meaning to be given to V— 3 is simply that expressed
by V^^ X V^^ = - 3.
The square root of a negative quantity is said to be
imaginary ; and a; + 1 + V— 3 and « + 1 — V — 3 are said
to be imaginary factors of x^ + 2 a; + 4,
By referring to the formula of Art. 122, it will be seen
that the factors of ax^ -i-bx-{- c are imaginary whenever
• — : is negative.
4a^ a
TACTORS. 14T
125. Factors found by Eearrangement and Grouping of
Terms. The factors of many expressions can be found
by a suitable rearrangement and grouping of the terms.
No general rule can be given, but in many cases it will be
sufficient to arrange the expression according to powers of
one of the contained letters, and the factors will then often
become obvious, particularly if the given expression only
contains one power of that letter.
Ex. 1. Find the rational factors oix^ -Zx"^ -\- x -Z.
x8-3a;2 + x-3 = x\x - 3) + (x - 3) = (x2 + i)(a; - 3).
Ex. 2. Find the factors of ax -\- by -{■ hx -\- ay.
Arrange in powers of x ; then we have
x{a+h)-\-hy + ay = x{a + h)^-y{a+h) = (x-\-y){a-\- b).
Ex. 3. Find the rational factors of aot^ — x -\- a — I.
Arrange in powers *of a ; then we have a (x^ + 1) — (aj + 1), and it
is now obvious that ic + 1 is a factor.
Ex. 4. Find the factors of the first degree of
a\b - c)+ fe2 (« - «)+ c2 (a - 6).
Arrange in powers of a ; then we have
a2 (6 - c) - a (&2 _ c2) + ftc (6 - c).
It is now obvious that 6 — c is a factor, and the given expression
= (6 - c){a2 - a (& + c)+ bc}=(b - c)(a - 6)(a - c).
Ex. 5. Find the factors of the first degree of
a2 _ 3 52 _ c2 _ 2 a6 + 4 be.
This is a quadratic expression in a, or in b, or in c. We therefore
proceed as in Art. 122. The given expression
= a2 - 2 a6 - 3 62 - c2 + 4 6c.
148 FACTORS.
The terms which contain a, namely a^ — 2 ab, will be made a
complete square by the addition of b^. We therefore write the
expression in the form
a2 - 2 a6+ 62 _ 4 62 _ c2 + 4 6c = (a - 6)2 _ 4 62 _ c2 + 4 6c
= (a - 6)2 -(2 6 - c)2 ={(a - 6)-(2 6 - c:)}{(a - 6) + (2 6 - c)}
= (a-36 + c)(« + 6-c).
126. The expression a^ -\-b^-\- & — 3abc is of frequent
occurrence and its factors should be known. It is easy
to verify that
= (a + 6 + c) (a^ + 62 _|_ c2 _ 5c - ca - ab).
The expression x'*^ + x^ + 1 is also of frequent occur-
rence. Writing it in the form (aj^ + 1)2 — x^, we see that
a;^ + 1 4- 07 and x^ -^1 —x are factors ; and we do not pro-
ceed any further, for both x"^ -\-x-\-l and,a^ — a; + 1 have
imaginary factors of the first degree.
The following miscellaneous examples will afford suit-
able practice in the methods above described.
EXAMPLES XXXIII.
Express in rational factors :
1. 16x4-625y*. 7. (a + 6)2-(c-d)2.
2. 81 a* -16 6*. 8. (a + 6)2 - 4 (c - ^)2.
3. iK5-81a;. 9. (a + 6)2 - (a + c)2.
4. 1 + 27 «». 10. {X + yy - (x- yy.
5. 27 + 8a;3. 11. (a + 6 + c)2 - 4c2.
6. 27x* + 8a;. 12. (a + 6 - 3c)2 - 9c2.
13. (a2 + ab + 62)2 _ (^^2 _ 52)2.
14. (a2 + a& + 62)2 -(a2-a6 + 62)2. .
15. (x3 + 3x)2-(3x2 + l)2.
FACTORS. 149
16. (a;2 4- 5 ccy + y2)2 _ (^^2 - xy + y'^y.
17. (a + b + c + d)^-(a-b + c-dy.
18. (3a + 26 + c)2-(a + 2& + 3c)2.
19. (2x-\-Sy)^-\-(3x + 2yy. 26. 72 (x2 - 1) - 17 x.
20. (a + 36)3 -(3a + 6)3. 27. x4-5a;2 + 4.
21. ic2 _ 3 a;y _ 1 y2. 28. ic* - 13a;2?/2 + 36y*.
22. x"^ + la + ^\xy + y^- 29. a6 (x2 + y2)+ a-?/(a2 + 62).
23. a;3 + -y-x2 + x. 30. a6(a;2_?/2)+a;?/(a2-62).
24. 6x'^-5xy-6y^. 31. (a + 6)2 - 5c (a + 6)+ 6c2.
25. X* - f x^y - x22/2. 32, (x + y)2 _ 7 « (x + ?/) + 10 z-
83. (a + 6)2 - 8 (a + 6)(c + d!)+ 15 (c + (Z)2.
34. tc(a; + 2)-y(y + 2).
35. a: (a; + 4) -?/(?/ + 4). 46. a262 - a2 - 6^ + 1.
36. x3 - 5 x2 + a; — 5. 47. ac -\- bd - ad - be.
37. x3 + ic2 _ 4 a; _ 4. 48. ac2 + bd^ - ad^ - bc\
38. 2 a;3 _ 3 a;2 _ 2 X + 3. 49. x^ - y'^ -\- xz - yz.
39. 5x3-x2-5x+l. 60. a2 - 6^ - (a - 6)2.
40. ax8 + X + a + 1. 51. a* + a262 - 62c2 - c*.
41. ax3 + 6x + a + 6. 62. a^ - b'^ + be - ca.
42. x3 + 6x2 _ (j2x _ a25. 53. a"^ _ a - d^ + c.
43. 6x3 _|. ax2 -}- 6x + a. 64. 1 + 6x - (a2 + a6) x2.
44. ax2 + 6y2 + (a + 6) xy. 65. 1 - a6x3 + (6 - a2) x2.
45. a262 + a2 + 62 + 1. 66. a'^c'^ -{- acd -^ abc + bd,
67. a2x + a6x -\- ac -{- b'^y + aby + be.
68. a2 - 62 + c2 - d^ - 2 (ac - 6d).
69. 4 a262 - (a2 + 62 - c2)2.
60. (a2 - 62 + c2 - (|2)2_ (2 ac- 2 6(^)2.
61. X4 - 23x2 + 1. 63. X* - 11 x22/2 + y4.
62. X* - 7 x22/2 + y4. 64. x4- 3x2+1.
65. (x2 + 4x)2 - 2 (x2 + 4x)- 15.
66. (x2 + 7 X + 6) (x2 + 7 X + 12) - 280.
150 FACTORS.
127. Zero Pactors. Since zero was defined [Art. 45] as
a difference, in the form a — a, a zero factor in any prod-
uct may be replaced by a — a. If, then, F have a finite
value, the product
0 X F= (a - a) F= aF- aF
is itself zero, and it is obvious that a product will not
be zero unless one of its factors is zero. Hence, exclu-
sive of infinite factors,*
In order that a product may he zero, it is necessary and
sufficient that one of its factors be zero.
128. Quadratic Equations. In particular, the theorem
of Art. 127 says : the product {x — 2){x — 4) is zero, if
a; — 2 is zero, or if aj — 4 is zero ; and, in order that this
product may be zero, one of these factors must be zero.
The roots of the equation
(a; - 2) (a; - 4) = 0
are therefore 2 and 4. [Definition of a root, Art. 91.]
Again, the product
(a; -3) (a;- 4) (a; -5)
vanishes if a; — 3 = 0, or if a; — 4 = 0, or if aj — 5 = 0,
* The combination of zero and infinite factors in a product introduces
what are known as indeterminate forms. Thus, for the value x = a,
the expression (x'^ — a^)x 1/ (x — a) assumes the form 0x1/0, or
0 X 00 , where 1 / 0 is said to be infinite and is represented by the symbol
00. But this product is not necessarily zero, when x = a; for, suppress-
ing the common factor x — a from x^ — a^, and x — a, we have
(x2 - a2) X 1 / (x - a) = ^^~^^ = X + a ,
x — a
and X + a = 2 a when x = a. [See Treatise on Algebra, Art. 217.]
FACTORS. 161
and in no other case ; therefore 3, 4, and 5 are the roots
of the equation
{x - 3) (a; - 4) (cc - 5) = 0.
Applied to the general quadratic expression ax^ -\-hx +c,
this principle of vanishing products asserts that
ax'^ -\-hx-\-Cy
or its equivalent [Art. 122]
«{-f.W(&-9}{-^-V{£-;)}
becomes zero, if and only if one of the factors here writ-
ten is zero, and that therefore the values of x that satisfy
the equation
aa^ -\-hx-\-c = 0
are derived, one of them from the equation
x +
2^^Vfe~a)"^'
the other from the equation
Thus the problem of solving any equation of the
second degree is reduced to that of solving two equations
of the first degree. This subject will be taken up again
in the chapter on quadratic equations.
The general principle, of which the above examples
are particular cases, asserts that the equation
{x -a){x- h) {x - c) = 0
is equivalent to the system of alternative equations
a; — a == 0, or x — h = 0, or a; — c = 0, etc.
Note. — Observe the distinction we make between alternative
equations and simultaneous equations [Art. 101] .
152 FACTORS.
129. From the examples considered in the last article
it will be apparent that the solution of an equation of any
degree can he written down at once provided the equation is
given in the form of a product of factors of the first degree
equated to zero.
The following are examples of such equations :
Ex.1. Solve (x-l)(x+l) = 0.
The equation is satisfied if ic — 1 = 0, or if x + 1 = 0, and in
no other case.
Hence we must have
x-l=0,OTx-]-l = 0;
that is, a; = 1, or X = — 1.
Thus the roots of the equation are 1 and — 1.
Ex.2. Solve x(x+l){x + 2)=0.
The equation is satisfied if ic = 0, or if a; + 1 = 0, or if a; + 2 = 0,
and in no other case.
Hence we must have
x = 0, ora; + l = 0, orx + 2 = 0;
that is, a: = 0, or oj = — 1, or a; = — 2.
Thus the roots of the equation are 0,-1, and — 2.
Ex.3. Solve x(2a;-l)(2a; + 3) = 0.
The equation is satisfied if cc = 0, or if 2 a; — 1 = 0, or if
2 X + 3 = 0, and in no other case.
Hence we must have
x = 0, or2x-l = 0, or2x + 3 = 0;
that is, X = 0, or X = |, or x = - f .
130. Since all the terms of any equation can be trans-
posed to one side, an equation can always be written with
FACTORS. 153
all its terms on one side of the sign of equality, and zero
on the other side.
It therefore follows from the last article that tte prob-
lem of solving an equation of any degree is the same as the
problem of finding the factors of an expression of the same
degree.
Hence the process of solving any equation which in-
volves only one unknown quantity is as follows :
First write the equation with all its terms on one side of
the sign of eqvMity, and zero on the other side ; then resolve
the whole expression into factors {of the first degree in the
unknown quantity), and the values obtained by equating
each of these factors separately to zero will be the required
roots.
In the following examples the resolution into factors
can be performed by inspection.
Ex. 1. Solve the equation x^ — 3 x = 0.
Since x^ — 3 x = x (x — 3),
we have x (x — 3) = 0.
Whence x = 0, or x — 3 = 0 ;
the roots required are therefore 0 and 3.
Ex. 2. Solve the equation x^ — 9 = 0.
Since x2 - 9 = (x - 3) (x + 3),
we have (x - 3) (x + 3) = 0.
Whence x-3 = 0, orx + 3 = 0;
the roots required are therefore 3 and — 3.
Ex. 3. Solve the equation x^ — 2 = 0.
Since x2 - 2 = (x - v2) {oc + y/2),
we have (x - ^2) (x + ^2) = 0.
Hence x — y^ = 0, or x + v^ = 0.
Thus y/2 and — ^2 are the required roots of the equation.
154
FACTORS.
Ex. 4. Solve the equation x^ - 4x = 0.
Since x^ - 4x = x (x^ - 4) = x (x - 2)(x + 2)^
- we have x(x - 2)(x -{- 2) = 0.
Hence x = 0, or a: - 2 = 0, ot x + 2 =0.
Thus 0, 2, and — 2 are the roots of the equation. ■
Ex. 5. Solve 9x^ = ix.
Transposing, we have
9x^-4:x = 0;
that is, x(9x^ - i) = 0,
or x(Sx-2)(Sx-{-2) = 0.
Hence a: = 0, or .3 a; - 2 = 0, or 3 x + 2 = 0 ;
that is, X = 0, or a; = I, or X = - |.
Ex. 6. Solve x2 + 6 = 5 X.
Transposing, we have
x2-5x + 6=0;
that is, (X - 2) (x - 3) = 0.
Hence x - 2 = 0, or x - 3 = 0.
Thus 2 and 3 are the required roots.
EXAMPLES XXXIV.
Solve the equations
1. (x-l)(x + 2) = 0. 7. x(x-2)(x + 3) = 0.
2. (x-3)(x-4)=0. 8. x(x + 2)(x-4) = 0.
3. (x + l)(x + 2) = 0. 9. x(x-3)(x + 4)=0.
4. (2x + l)(2x-l)=0. 10. x(2x-l)(3x + 4) = 0.
5. (3x-l)(3x + l) = 0. 11. x(5x-2)(6x-7) = 0.
6. (2x-5)(x-4)=0. 12. x(x-3)(3x+7) = 0.
13. (x-l)(x-2)(x-3)(x-5) = 0.
14. (x-2)(x-l)(x+l)(x + 2) = 0.
15. (3x-l)(4x+ l)(5x-2)(2x + 7) = 0.
16. (2x-3)(3x-4)(4x-6)(5x + 6)=0.
FACTOKS. 155
17. x^-x = 0. 25. 3x2 = 6a;.
18. x^-2x = Q. 26. 5ic2 = -6x.
19. x2 + 3x = 0. 27. x2 = 6x.
20. 2 x2 - 3 X = 0. 28. ax2 = 6x.
21. 2x2-5x = 0. 29. 6x2 = x-2 + 5.
22. 3x2 + x = 0. 30. 5(x2 + 5)=3(x2 + 25).
23. x2 = 6x. 31. 5(x2 + 4)=4(x2 + 9).
24. 2x2 = x. 32. 2(x2 + 7)=7(x2 + 2).
33. 5(x2 + 3)-(x-5)(x + 5)=76.
34. 7(x2-l)-(x + 3)(x-3) = 56.
35. 3x2 + (5x + 2)2 = 20x + 32. 43. x2-6x-84 = 0.
36. 17 + 3 X = |(x + 3)2 _ 28. 44. x2 - x - 156 = 0.
37. x2-5x + 6 = 0. 45. x2+5x- 150 = 0.
38. x2-7x+12 = 0. 46. x2 + 2x = 3.
39. x2-12x + 20 = 0. 47. x2 + 4x = 45.
40. x2 - 9x + 20 = 0. 48. x2 - 3 x = 10.
41. x2 - 11 X + 28 = 0. 49. x3 + 11 x2 - 180 x = 0.
42. x2-25x + 150 = 0. 60. x8-x2 = 132x.
156 HIGHEST COMMON FACTORS.
CHAPTER XI.
Highest Common Factors.
131. A Common Factor of two or more algebraic expres-
sions is an expression which will exactly divide each of
them.
Thus a is a common factor of ab and ac.
The Highest Common Factor of two or more algebraic
expressions is the expression of highest dimensions which
will exactly divide each of them.
Thus a^ is the highest common factor of a^b and a^c.
Instead of Highest Common Factor it is usual to write
H.C.F.
We proceed to show how to find the H. C. F. of given
expressions.
132. H. C. F. of Monomial Expressions. The highest com-
mon factor of two or more monomial expressions can be
seen by inspection.
Take, for example, a^'^c and a%^(^,
The first expression is clearly divisible by a, or by a^, or by a^,
but by no higher power of a ; and the second expression is divisi-
ble by a, or by a'^, but by no higher power. Hence a^ will divide
both expressions, and it is the highest power of a which will divide
them both. Also b'^, but no higher power of b, will divide both
expressions ; and c, but no power of c above the first, will divide
both expressions.
HIGHEST COMMON FACTORS. 167
Hence the H.C.F. of a^h^-c and a^ftV is a^y^c.
Again, to find the H. C. F. of ahHH"^ and aH^d^.
The highest power of a which divides both expressions is a ; 6
will not divide both expressions ; the highest power of c which
divides both is c^ ; and the highest power of d which divides both
is (?*. Hence the H. C. F. is acH\
Also a^hd^^ a%HH^ and a^hd^d^ are all divisible by a^, by 6, and
by c* ; and therefore the H. C. F. of the three expressions is a%c^.
From the above examples it will be seen that the
H.C.F. of two or more monomial expressions is obtained
by taking each letter which is common to all the expres-
sions to the lowest power in vjhich it occurs.
If the expressions have numerical coefficients, the
G. C. M. of these can be found by arithmetic, and pre-
fixed as a coefficient to the algebraical H. C. F.
EXAMPLES XXXV.
Find the H. C. F. of
1. a862anda268. 7. 9 a^ftSa^y^ and 8 xV-
2. a6c2 and a25c8. g. f a^ft^cs and 2 ft^c*.
8. 9 a68 and 6 a%. 9. 42 axy"^^ and 77 hhf^.
4. 4 x^y and 10 xy^. 10. ah^, a^bc, and abc^.
6. 2i a%^x^ a.nd eO a'^b*3fi. 11. Sx'2yz^,l5xy^z'^,Sindl0x'^y^z'^.
6. a'^b^x^ and 3 b^x. 12. ab^c^x\ a^bc^x^, and a^b'^cx^.
133. H. 0. F. of Multinomial Expressions whose factors are
known. When the factors of two or more multinomial
expressions are known, their H. C. F. can be at once
written down. The H. C. F. will be the product obtained
by taking each factor which is common to all the expres-
sions to the lowest power in which it occurs.
158 HIGHEST COMMON FACTORS.
Consider, for example, the expressions (x — ay(x+ b)^ and
(x — ay{x + by. It is clear that both expressions are divisible by
(x — ay, but by no higher power of (x— a). Also both expressions
are divisible by (x + by, but by no higher power of x + b.
Hence the H. C. F. is (x - ay(x + bj^
Again, a^b (a - b)'^ {a + by and ab^ (a - by (a + by are both
divisible by a, by 6, by (« — by, and by (a + by. Hence the
H. C. F. is ab (a - by (a + by.
In the following examples the factors can be seen by-
inspection, and hence the H. C. F. can be written down.
Ex. 1. Find the H. C. F. of a%^ - a%^ and a^&a + a%^.
Since a^b"^ - a^b^ = a^b^ (a^ - 6^) :^ 0^2^,2 (^^ _i)^(^a + b),
and a453 4. ^354 ^ ^3^3 (^ + 6),
we see that the H. C. F. is a'^b'^ (a + b).
Ex. 2. Find the H. C. F. of
«3 + 3 a'^b + 2 a&2 and a* + 4 a^ft + 3 aS^a.
a^ + Sa'^b + 2 ab'^ = a (a^ + 3 a& + 2 6-2) = a (a + &) (a + 2 6),
and
a* + 4 a35 + 3 ^2^2 = cfi ((^2 + 4^5 + 3 52) =a'^(^a+b){a + ^b)\
hence the H. C. F. is a (a -f b).
Ex. 3. Find the H. C. F. of 3 a^ + 2 a2 _ « and 5 a* + 3 a^ _ 2 a2,
3 a3 + 2 a2 - a := a (3 a2 + 2 a - 1) = a (3 a - 1) (a + 1),
and 5 a* + 3 a3 _ 2 a2 ^ ^2 (5 ^2 + 3 q^ _ 2) = ^2 (5 ^^ _ 2) (a + 1) ;
hence the H. C .F. is a (a + !)•
EXAMPLES XXXVI.
Find the H. C. F. of
1. x2 (x - ay and x^ {x - ay. 2. a2 _ 52 and (a + by.
3. 3 a6 (a + 6)2 and 2 (a - &) (a + &)3.
4. 64^6 (5 _i_ c)2 and ft^cS (& + cy.
6. ^353 ^_ ab^ and a^ _ <x254. 7. (^-2^3 + 2 a3x2 and aH'^ - 4 ^4x2.
6. a26 + 3 63 and a^ - 9 a'^b^. 8. a''x2 - 4 a+x^ and a6x2 _ 16 a'^oi^.
HIGHEST COMMON FACTORS.
159
9. x2 + 3 X + 2 and x^ + 6 x + 8.
10. x3 + 3 x2?/ + 2 X2/2 and X* + 6 o(?y + 8 xV-
11. 3x2-4x+ 1 and4x2-5x+ 1.
12. 3 a2 - 4 a6 + 62 and 4 a* - 5 a^h + a262.
v/ 134. Although we cannot, in general, find the factors
'oi a multinomial expression of higher degree than the
second, we can always find the highest common factor of
any two expressions by the following process, which will
be seen to be analogous to that used in arithmetic to find
the greatest common measure of two numbers.
Eule. Arrange the two expressions in descending powers
of some common letter, and divide the expression ivhich is
of the highest degree in the common letter by the other : if
both expressions are of the same degree, it is immaterial
which is used as the divisor. Take the remainder, if any,
after the first division for a new divisor, and the former
divisor as dividend; and continue the process until there
is no remainder. The last divisor will be the H. C. F. of
the two given expressions.
For example, to find the H. C. F. ofx^ + x2 - 2 and x^ + 2 x2 - 3,
the process is as follows :
First dividend,
Second dividend,
Third dividend,
Thus the H. C. F. is x - 1.
x8 +
X3-
a;8 4. 2 x2 - 3
X8 -1- X2 - 2
x2-2
X
X2-1
X +1
x^-\-x ^2
X2 -1
• •
X2-1
X -1
X2-X
X +1
X -1
X -1
X* + x2 — 2 First divisor
Second divisor
Third divisor
160
HIGHEST COMMOIT FACTORS.
The work can be arranged more concisely as follows :
a:3+ 3^2 _ 2
x^ + x^ -2
x^-x
x^-1
x'^-x
jc2 4- X - 2
X2 -1
X -1
X -1
a; -1
Note. — It is only multinomial factors which are to be found by
the above rule ; the H. C. F. of any monomial factors of the given
expressions must be found by inspection.
For example, to find the H. C. F. of
a2^4 _^ Qj2^3 _ 2 «2 ^ and abx^ + 2 abx*
Since ^2.y.i _l «2^3
and
- 3 abx^.
a'x' + a^x"" — 2aH — aH (x^ + a^^ - 2),
abx^ + 2 «&x* - 3 a6x2 _ ^^5^2 (3^3 + 2 a;2 _ 3),
it is clear that the H. C. F. of the monomial factors is ax. We have
already found that the H. C. F. of x^ + a;2 _ 2 and x^ + 2 x2 - 3
is X — 1.
Hence the whole H. C. F. is ax (x — 1) .
' )^35. The rule given in Art. 134 for finding the H. C. F.
of two expressions which, have no monomial factors, may
be proved as follows.
Let A and B stand for the two expressions, which are
supposed to be arranged according to descending powers
of some common letter, and let A be of not higher dimen-
sions than B in the common letter.
Let B be divided by A^ and let the quotient be Q, and
It the remainder, so that the first stage of the process of
finding the H. C. F. of A and B is as under,
A)B{q
AQ
B
HIGHEST COMMON FACTORS. 161
Then we have B = AQ + R (i.),
and E = B-AQ (ii.).
Now an expression is exactly divisible by any other if
each of its terms is so divisible; hence, from (i,), B is
divisible by any common factor of A and R.
Thns any common factor of A and Ii is also a common
factor of A and B.
Again, any factor of both B and A will be a factor of
B — AQ', that is, from (ii.), will be a factor of M.
Thus any common factor of A and B is also a common
factor of A and R.
It follows therefore that the common factors of A and
B are exactly the same as the common factors of A and R.
Hence the H. C. F. of A and B is the H. C. F. of A and R.
If now we divide A by R, and the remainder is S, the
H. C. F. of S and R will similarly be the same as the
H. C. F. of A and R, and therefore will be the H. C. F.
required. And so on ; so that the H. C. F. of any divisor
and the corresponding dividend is the H. C. F. required.
But if at any stage there is no remainder, the divisor
must be a factor of the corresponding dividend, and that
divisor is clearly the H. C. F. of itself and the corre-
sponding dividend. It must therefore be the H. C. F.
required.
It should be remarked that by the nature of division
the remainders are successively of lower and lower dimen-
sions ; and hence, iHiless the division leaves no remainder
at some stage, we must at last come to a remainder which
does not contain the common letter, in which case the
given expressions have no H. C. F. containing that letter.
We have already remarked that the process we are
162
HIGHEST COMMON FACTORS.
considering is only to be used to find the H. C. F. of the
multinomial factors of the given expressions ; and, bear-
ing this in mind, it is clear that any of the expressions
which occur may be divided or multiplied by any mono-
mial expression without destroying the validity of the
process ; for the multinomial factors will not be altered
by such division or multiplication.
Ex.1. Find the H. C. F. of
x^-\-4:X^ -Sx + 24: and x^ -x^ + 8x-8.
Neither expression has any monomial factors; we therefore
proceed as in Art. 134, performing the work by the synthetic
method.
X3
X*-
-x3+0 +8a:-8
-4x2
-
-4x3 + 20x2
+ 8x
+ 8 x2 - 40 X
-24
- 24 X + 120
x-5; 28x2 -56x + 112
The remainder 23 x2 - 56 x -f 112 = 28 (x2 - 2 x + 4) ; and since
the numerical factor 28 is clearly not a factor of the given expres-
sions, we reject it, and continue the process with x2 — 2 x + 4 as
the new divisor :
+ 2x
-4
x3 + 4x2- 8X-1-24
+ 2x2 + 12x
- 4x-24
x + 6; 0
Thus the H. C. F. is x2 - 2 x + 4.
0
Ex. 2. Find the H.C.F. of x^-4a^x+lba^ and x*+a2x2+25a*.
X3
+ 4a2x
-15a2
x* + 0 + a2^2 + 0
+ 4 «2x2
— 15 a%
+ 25a4
x; 5a2x2_ I5a3x + 25a4
Now 5 a2x2 - 15 a^x + 25 a* = 5 a2(x2 - 3 ax + 5 a^).
HIGHEST COMMON FACTORS.
163
We reject the monomial factor 5a^, and continue the process
with x2 - 3 ax + 5 a- as the new divisor :
x3 + 0 - 4 a2x + 15 a3
+ 3ax2+9a2x
- 5 a2x - 15 a3
x^
+ 3ax
-5a2
x + 3a; 0 0
Hence x^ — 3 ax + 5 a^ is the required H. C. F.
Ex. 3. Find the H. C.F.otx^-y^ and x' - y"^.
x' — y"^ |x^ — y^
xV|x2
Before using the remainder as a divisor, we reject the monomial
factor y^ ; we then have x^ _ yi.
x6-y5
x2-!/2
X6 - x82/2
X8 + X«/2
X«y2
xV
_y6
-Xt
xy* - y5
We reject the monomial factor y* of the remainder.
xy
x + y
xy
xy_
t.
Hence x — y is the required H. C. F.
Ex. 4. Find the H. C. F. of 2x2-5x + 2 and x3+4x2-4x-16.
To avoid the inconvenience of fractions, we multiply
x8 + 4x2-4x-16
by 2. We have shown in the preceding article that we may do
this, as no additional common factors can be thereby introduced.
It will be seen that we multiply a second time in the course of the
work. The following arrangement will be found convenient :
164
HIGHEST COMMON FACTORS.
First dividend x^ + 4:X^- 4 x - 16
Multiply by . . 2 First divisor.
X
2x3 + 8^2- 8a: -32
2 a:3 - 5 x2 + 2 x
2x2- 5a: + 2
2x2 -4a:
13x'^- lOx-32
Multiply by . . 2
-x + 2
-a:+2
13
26a:2_20a:-64
26x2 -65 a: + 26
0 0
2 a:
Divide by . .
Second divisor .
45i45x-90
X- 2
Thus a: - 2 is the H. C. F.
136. The H. C. F. of more than two expressions is
sometimes required, when the factors cannot be deter-
mined by inspection.
Now it is clear that any factor which is common to
each of t'hree or more expressions, must be a factor of
the H. C. F. of any two of them. Hence we first find the
H. C. F. of two of the given expressions, and then find
the H. C. F. of this result and of the third expression,
and so on.
Ex. Find the H.C.F. of
x3 + x2-x-l, a:3 + 3a:2
X
3, and a:3 + x2-2.
The H. C. F. of the first two expressions is a:2 - 1. The H. C.F,
of a:2 — 1 and the third of the given expressions is x — 1. Hence
X — 1 is the H. C. F. required.
137. If the numerical coefficients of the first terms of
the given expressions are large, it is often desirable to
arrange the terms of both expressions in the reverse
order before applying the rule for finding their H. C. F.
HIGHEST COMMON FACTORS. 165
For example, in order to find the H, C. F. of
7 x* + 2x3 - a:-2 + 8x + 1 and 9x* - 2 a;3 + 3 a;2 -f 6 X + 1
it is best to arrange the expressions in the form
1 + 8x-x2 + 2x3 + Tx^andl +6x + 3x2 -2x3 + 9x*.
138. The highest common factor of two expressions is
sometimes, but very inappropriately, called their greatest
common measure [G. C. M.].
The inappropriateness of the term greatest common
measure can be seen by an example. If a^ is a factor of
two expressions, so also is a, and a^ is of higher dimen-
sions than a ; but, as a may represent any number what-
ever, a^ is not necessarily greater than a ; in fact, if a is
positive and less than unity, d^ is less than a.
It should also be noticed that if we give particular
numerical values to the letters involved in any two ex-
pressions, and in their H. C. F., the numerical value of
the H. C. F. is by no means necessarily the G. C. M. of the
values of the expressions. This may not be the case even
when the given expressions are integral for the particular
values chosen. For example, the H.C.F. of 2a^-f-15a;-f 13
and 6 a;^ -t- 17 a; + 11 will be found to be a; -f 1 ; but if we
suppose X to be \^ the numerical values of both expres-
sions, and therefore their G. C. M., will be 21, whereas
the numerical value of the H. C. F. will be f .
EXAMPLES XXXVII.
Find the H.C.F. of
1. x2 - 6 X -I- 4 and x^ - 5 x^ + 4.
2. x2 — 5 xy -|- 4 y2 and x* — 5 x^y + 4 xy^.
3. 2x2 -6x + 2 and 4x3+ 12x2- X- 3.
166 HIGHEST COMMON FACTORS.
4. 2 x2 - 5 a;?/ + 2 y2 and 4^:^ + 12 x^^y - xy^ - S yK
5. a;* + 3 x2 - 10 and a:4 - 3 0^2 + 2.
6. x^ + 3x*y - 10 x2?/2 and x^-Sx^y + 2 y^.
7. 2 a2 - 5 a + 2 and 2 a3 - 3a2 _ 8a + 12.
8. 2 62 _ 5 5 + 2 and 12 63 _ 8 6-2 _ 3 5 _^ 2.
9. ic2 - 3x + 2 and x^ -Sx-j- 2.
10. xY - 3 a;2^2 + 2 and x^y^ - S x^y^ + 2.
11. x^-Sa^x-2 a^ and x^ - aic2 _ 4 a^.
12. 2 a3 + 3 a26 - 6^ and 4 a^ + ^62 _ ^,3.
13. a^ + 68 and a* + a262 + 54.
14. 8 a3 + 1 and 16 a* + 4 a2 + 1.
15. x3 + 2 ic2y - a;^/2 _ 2 ?/3 and x^ - 2 a:2y - xy'^ + 2 y^.
16. 2x3 + 5x2 + X - 8 and 3x3 - 4x2 4- 9x - 8.
17. 3x3 + x2 + X - 2 and 2x3 - x2 - x - 3.
18. x3 - 4x + 15 and -X* + x2 + 25.
19. 3 x2 - 38x + 119 and x3 - 19x2 + 119x - 245.
20. 3x3 - 3x2y + X2/2 - 2/3 and 4x2?/ -6xy^ + y^.
21. 12x2- 15x?/ + 32/2 and 6 x3 - 6 x2?/ + 2 x?/2 -2y^.
22. 2 x2 - 14 X + 20 and 4x (x2 + 5) - 25 (x + l)(x - 1).
23. 16 X* + 4 x2 + 1 and 8 x* - 16 x3 + x - 2.
24. a2_4a;2+ 12X-9 and a2 + 2a- ' 8x-3.
25. 21 - 7 X + 3x2 - x3 and 35 + 19 x2
26. 2x* +9x3 + 14x + 3 and 2 + 9x + 14x3 + 3x*.
27. x3 + 3 x2 - X - 3 and x* + 4 x3 - 12 x - 9.
28. 3x* + 5x3-7x2 + 2x4-2 and 2x* + 3x3 -2x2 + 12x + 5.
29. y3 _ 2?/2 + 3 2/ _ 6 and 2/4 - 2/3 _ 2/2 _ 2 y.
HIGHEST COMMON FACTORS. 167
30. y* - 3 yz^ + 20 z^ and 5y^ -Sy^z-^ 64 z^.
31. 2ic3_iix2 4- llx+4 and 2x4-3x3 + 7x2- 12x-4.
32. 2x4 + 4x3 + 3x2-2x-2 and 3x* + 6x3 + 7x2 + 2a + 2.
33. x* - x3 + 2 x2 - X - 1 and 2 X* - 2 x2 + X - 1.
34. 2x*-6x3 + 3x2-3x+ 1 and x^ - 3x^ + x« - 4x2 + 12x-4.
86. x8 + (to - 3) x2 - m (2m + 3) X + 6 m^
and x8 + (5TO-3)x2 + 3m(2m- 5)x-18to2.
86. mn (x^ + y^) -\-xy (m^ + ri^) and mn (x^ + y^) +xy (m^y + w^x).
^•c*
168 LOWEST COMMON MULTIPLES.
CHAPTER XII.
Lowest Common Multiples.
139. A Common Multiple of two or more algebraical
expressions is an expression whicli is exactly divisible
by each of them.
The Lowest Common Multiple of two or more algebraical
expressions is the expression of lowest dimensions which
is exactly divisible by each of them.
Instead of lowest common multiple it is usual to write
L. C. M.
We proceed to show how to find the L. C. M. of given
expressions.
140. We first consider monomial expressions, the
factors of which can be seen by inspection.
Take, for example, a%'^c and a%H^. The highest power of a
which occurs in either expression is a^ ; hence any common multi-
ple of the given expressions must contain a^ as a factor. Any-
common multiple must also contain h^ as a factor, and it must
also contain c* as a factor. Any common multiple must therefore
contain a^^C^ as a factor ; and hence the common multiple of
lowest dimensions must be a^6^c*.
Again, to find the L. C. M. of a%c^, a%^cH and a'^hcM'^, the
highest power of a in the" given expressions is a* ; hence any com-
mon multiple of them must contain a* as a factor. Any common
multiple must also contain 6^^ ^s^ ^^(i ^-2 ^g factors. Any common
multiple must therefore contain d^hH^d'^ as a factor; hence the
L. C. M. required is a^c^hH^.
LOWEST COMMON MULTIPLES. 169
From the above examples it will be seen that the
L. C. M. of two or more simple expressions is obtained
by taking every letter which occurs in the different^ expres-
sions to the highest power which it has in any one of them.
If the expressions have numerical coefficients, the
L. C. M. of these can be found by arithmetic, and pre-
fixed as a coefficient to the algebraical L. C. M.
EXAMPLES XXXVIII.
Find the L. C. M. of
1. a362anda268. 7. 9 aSftSa^y and 8 a^.
2. a6c2 and a'^b(^. 8. | a^b^c^ and 2 ¥<fi.
3. 9 ab^ and 6 a^b. 9. 42 axy^z^ and 77 6^-
4. 4 y^y and 10 xy^. 10. ab^, a%c, and abd^.
5. 24a363a4 and ma%^7^, 11. Zx'^yz^, \bxy^z% and \(ix^yH\
6. a%H^ and 3 b^x. 12. aft^c^x*, a^bd^T?, and a^'^cx^.
141. When the factors of two or more multinomial
expressions are known, their L. C. M. can be at once
written down.
The L. C. M. will be the product obtained by taking
every factor which occurs in the different exjwessions to the
highest power which it has in any one of them.
Consider, for example, the expressions (x — a)(x — by (x — c)^
and (x — ay (x — 6) (x — c).
It is clear that any common multiple must contain (x — a)* as a
factor ; it must also contain (x — by and (x — cy as factors. Any
common multiple must therefore contain (x — a)* (x — by (x — cy
as a factor; and hence the common multiple of lowest dimensions
must be (x — a)* (x — by (x — cy.
170 LOWEST COMMON MULTIPLES.
Ex. 1. Find the L. C. M. of a^V^ - a%^ and a^h^ + a%^.
Since a^h"^ - a^h^ = a^b'^ («-&)(«+&),
and a*63 + ^3^4 ^ ^353 (a + 6) ,
we see that the L. C. M. is a^b^ (a -&)(« + &).
Ex. 2. Find the L.C.M. of a^+Sa%-\-2ah'^2ind a^+^a^b-{-Sa^b\
a3 + 3a26 + 2a62 = «(« + &)(« + 2 6),
and aj4 _^ 4(^35 4. 3 ^2^,2 ^ q^2((^ 4. 5)^^ + 3 5) .
hence the L. C. M. is a'^(a + 6) (a + 2 6) (a + 3 &) .
We leave the L. C. M. in the above form, as it is generally most
convenient to have it expressed in factors.
Ex. 3. Find the L. C. M. of ^2 + 2a: + 1, ^2 - 2 x - 3, and
x2 + 4x4-3.
a;2 + 2a:+l=(a: + l)2,
a;2_2x-3=(x+l)(x-3),
and x^ + ^x + S=(x+l){x + S);
hence the L. C. M. is (x + l)%x - 3)(x + 3).
EXAMPLES XXXIX. _
Find the L. C. M. of
1. {a — x)(a — 2x) and (a — 2ic)(a — 3x).
2. ax^(a — x)(a — 2x) and a%(a — 2 x) (a — 3 x).
3. a2 - 62 and (a + b)^.
4. 6 «6(a + 6)2 and 4 a2(Qj2 _ 52).
5. x2 + 3x + 2 and x2 + 5x + 4.
6. x2 + 3 xy + 2 ?/2 and x2 + 5 xy + 4 y^.
7. x2 - 4 X + 3 and x2 - 5 x + 6.
8. 3 x2 - 4 xy2 + ?/4 and 6 x2 - 5 x?/2 + y*.
9. (a + 6)2, (a - 6)2, and a^ - b^.
10. (x + 2 ?/)2, (x - 2 ?/)2, and x'^-iy^.
11. x2 + 7x + 12, x2 + 6x + 8, and x2 + 5x + 6.
12. x2 - 7 xy + 12 2/2, x2 - 6xy + 8 y2, and x2 - 5xy + 62^2.
LOWEST COMMON MULTIPLES. 171
142. When the factors of the expressions whose
L. C. M. is required cannot be seen by inspection, we
must use the rule for finding the H. C. F. given in the
preceding chapter.
Thus, to find the L. C. M. of a^ + ^2 - 2 and x^-\-2x'^-3.
The H, C.F. of the given expressions is x— 1, and we find by
division that
a;3 + x^ — 2 =(x - l)(a;2 + 2x + 2),
and x^-h2x^-S=(x- l)(a;2 + 3a: + 3).
Then, since x'^ -\-2x-\-2 and x2 + 3 x + 3 have no common factors,
the L. C. M. required is
'{X - l)(x2 -f 2a; + 2)(x2 + 3x + 3).
143. Let A and B stand' for any two algebraical
expressions, and let H stand for their H. C. F., and L
for their L. C. M.
Let a and h be the quotients when A and B respec-
tively are divided by /f; so that
A = H y^a,
and B=H xb.
Since H is the highest common factor of A and B, a and
b can have no common factors. Hence the L. C. M. of A
and B must he Hx a xb. Thus
L = H-a'b (i.).
It follows from (i.) that
L=.Hax^ = Ax% .... (ii.),
and also that
LxH=HaxHb = AxB . . . (iii.).
172 LOWEST COMMON MULTIPLES.
From (ii.) we see that the L. C. M. of any two algebrai-
cal expressions is found by dividing one of the expressions
by their H. C. F., and multiplying the quotient by the other
expression.
From (iii.) we see that the product of any two expres-
sions is equal to the product of their H. C. F. and L. C. M. -
144. To find the L. C. M. of more than two expres-
sions, whose factors cannot be determined by inspection,
we first find the L. C. M. of two of the given expressions,
and find the L. C. M. of this result and of the third
expression, and so on.
EXAMPLES XL.
Find the L. C. M. of
1. 3a26c, 27a^b^c:^, and 6a6%. 3. ia^ 6a% and Sab^.
2. a^b% b% and aV. 4. 2a2, 6ab% and 4ta^b^.
5. x^(x - 2/)2, y^{x + ?/)2, and xy(x^ - y^).
6. «» + a% a(a-b), and a'^ - b^.
7. xy'^ — y^, x^y^ + xy'^, and x'^y — y.
8. 2axy(x - y), Sax:\x^ - y2), and iy^^x + yy.
9. 6(x2-9), 9(x + S), 15(aj-4), and 10(a;2 - x - 12).
10. 4 a + 4 6, 6 a2 _ 24 b'^, and a'^-Sab-\-2 62.
11. 4 a63 _i_ 4 1)3^^ 0^-252 _ 52(^2^ and 8 a^bd - 8 abcP.
12. ^2 - lOic + 24, x2 _ 8 jc + 12, and x'^-6x + 8.
13. x2 - 9ic - 10, a;2 - 7a; - 30, and x^ + 4a; + 3.
14. 2ic2-8, 3ic2-9x + 6, and 6x^ + 18x+ 12.
15. x2-3ic + 2, 2x2 -5C- 6, and 3x2-2x-l.
16. 6 x2 + X - 2, 21 x2 + 17x + 2, and 14x2 - 5x - 1.
17. 3x2 - 10 x?/ + 3 y^, 3 x2 -ixy + y^, and x2 - 4 xy + 3 y^.
18. x8 + 2x2-3x and 2x^ + 5a;2-3x.
LOWEST COMMON MTJLTIPLES. 173
19. xV _ 9y2^ r^2y _y^_Qy^ Siud d^ + x^ - 6 X.
20. x^-a^, 3i^-{- a^, and a^ + a'^x^ + a*.
21. x2 - 1, ic3 ^_ a;2 _f. a; ^ 1^ and a^ - x2 + a; - 1.
22. 9a^ - x - 2 and Sic^ - 10x2 - 7x - 4.
23. x8 - ax2 - a^x + a^ and x^ + ax2 - a^x - a^,
24. x3 + x2-4x-4 and x34-6x2 + llx + 6.
26. X* - x3 + 8 X - 8 and x3 + 4x2 - 8 X + 24.
26. a^-{-6a^ + na-hQ and a^ + lOa^ + 29a + 20.
174 MISCELLANEOUS THEOREMS.
CHAPTER XIII.
Miscellaneous Theorems and Examples.
145. Mathematical Induction. A- method of proof, ordi-
narily called mathematical inductioyi, is frequently em-
ployed in the demonstration of mathematical propositions.
The method is best explained through its application to
simple examples.
Ex. 1. The following equations are evidently true :
1 + 3 = 4 = 22, ^
1 + 3 + 5 = 9 = 32, '
1 + 3 + 5 + 7^=16 = 42,
1 + 3 + 5 + 7 + 9 = 25 = 52; 5
and they at once suggest that perhaps the sum of the first n odd
numbers is equal to n2. Let us assume provisionally that it is.
This hypothesis algebraically expressed is
1 + 3 + 5 + ... + 2w-l = w2,
in which n is an integer and equal to the number of terms in the
sum.
To both sides of this equation add 2 w + 1. The result of the
addition is
1 + 3 + 5 + ... +2w-l + 2w + l = n2 + 2/1 + 1
= {n + 1)2,
and shows that if the sum of the first n odd numbers is n2, then
the sum of the first w + 1 odd numbers is (7i + 1)2.
But the formula above written is obviously true when n = 1 and
when n = 2 ; hence it is true when n = 3. And being true when
MISCELLANEOUS THEOREMS. 175
n = 3, it is true when w = 4, and so on indefinitely. It is there-
fore true when n is any integer whatever.
Thus the proposition that the sum of the first n odd integers is
equal to ri^ is proved.
Ex. 2. Let it be proposed to find the sum of the first n natural
numbers. We begin by writing down the following equations,
which are evidently true :
1 + 2 = 3 = J. 2 (2+1),
1 + 2 + 3 = 6 =^.3(3 + 1),
1 + 2 + 3 + 4 = 10 = ^.4(4 + 1),
1 + 2 + 3 + 4 + 5 = 15 = ^.5 (5+1).
They at once suggest that perhaps the sum of the first n integers
is ^ n (n + 1). We accordingly assume, provisionally, that
l+2 + 3 + 4 + ... + n = ^n(w + l).
Adding n+ 1 to both sides of this hypothetical equation, we obtain
1 + 2 + 3 + ... + n + n + l = |w2 + ^n + « + l
= i(,i + l)(n + 2),
a result which shows that if \n{n -\- V) is the sum of the first
n integers, then ^(w + l)(7i + 2) is the sum of the first n + 1
integers.
But the formula is obviously true when n = 1 ; hence it is true
when 11 = 2. And being true when n = 2, it is true when n= 3,
and so on indefinitely. It is therefore true, whatever integer n
may represent.
These examples may suffice to explain a method of
proof that will be made use of, as occasion requires, in
subsequent pages of this book. If its logical rigor be
not at first entirely evident to the student, a careful
study of its applications in a few examples that require
its use should remove the doubt.
Ex. 3. Prove that the sum of the first n even integers is n(n + 1).
176 MISCELLANEOUS THEOREMS.
Ex. 4. Prove that the sum of the squares of the first n integers
is^n(n + l)(2w+l).
Ex. 6. Prove that
I.2 + 2.3 + 3.4+... + w(w + l)=|n(w + l)(7i + 2).
146. The term induction does not correctly describe the
above method of proof. In the natural sciences the true
method of induction is used for the purpose of inferring
the universality of natural laws from particular mani-
festations of their truth in natural phenomena. It does
not seek to attain absolute certainty, but achieves its
object as soon as it establishes a high degree of proba-
bility.
The so-called method of mathematical induction, how-
ever, is as logically rigorous as any other process in mathe-
matics. It is more correctly described as reasoning by
progressive transformations, in which the uniqueness of
an algebraic form persists, — monomorphic transforma-
tions, or monomorphosis.
147. Theorem. The expression ic" — a** is divisible by
X — a for all positive integral values of n.
It is known that x — a,x^ — a^, and a^ — a^ are all divisi-
ble by a; — a.
We have x'' — a'* = x'^ — ax'^~'^ -f ax^~^ — a"
= x""-^ (x — a)-i-a (x""-^ — a'*-^).
Now, if x — a divides cc*""^ — a**"^, it will also divide
a;n-i (^x — a)-\-a {x""-^ — a""-^),
that is, it will divide x^ — a\ (The if of this sentence
is important.)
MISCELLANEOUS THEOREMS. 177
Hence, if x — a divides a;"~^ — a"~^, it will also divide
ic" — a".
But we know that x — a divides oi? — a^; it will there-
fore also divide a;* — a*. And, since x — a divides x^ — a^,
it will also divide ar^ — a\ And so on indefinitely.
Hence a;** — a" is divisible by x — a, when n is any posi-
tive integer [Art. 145].
Since x" + a" = af* — a" + 2 a", it follows that when
x^-\- a*" is divided by a; — a the remainder is 2a''y so that
a;" -f- a" is never divisible by a; — a.
If we change a into —a, x—a becomes a;— (—a) =x+a;
also a;** — a" becomes af* — ( — a)" ; and x" — ( — a)" is
a;« -f a" or «" — a** according as w is odd or even.
Hence, when n is odd,
x^ + a** is divisible by a; + a,
and when n is even,
a;« — a" is divisible by a; + a.
Thus, when n is a positive integer,
x — a divides a;" — a" always,
x — a . . . . af -f a" never,
x-\-a .... af — a** when n is even,
and x-{-a .... a;" + a" ... . odd.
We have in the above shown that the four cases are
all included in the first : we leave it as an exercise for
the student to prove each case separately.
The above results may be written so as to show the
quotients : thus
178 MISCELLANEOUS THEOREMS.
— - x''-'^ + x^'-^a + x^'-^a' H h a"-\
X— a
x""-^ — x^~^ a + x^'-^a^ ± a~-^
x"" ± a ._i „_2
x-}- a
the upper or lower signs being taken on each, side of the
second formula according as n is odd or even.
EXAMPLES XLI.
1. Find the factors of x^ — x^ — x -{■ 1.
2. Find the factors of c(fi - x^ - x"^ + 1.
3. Find the factors of I — x — x^ + x^.
4. Find the factors of 1 — x^ — x^ + x^^.
5. Find the factors of x^ - y"^ - 2ax - 2by -{■ a^ - h\
6. Find the factors oi x^ - y'^ -Zx- y ^2.
7. Find the factors of a-.2 - 2 xy - 8 ?/2 _ 2 x + 20 ?/ - 8.
8. Find the factors of a^ - 62 + c2 _ /2 _ 2 ac - 2 hf.
9. Write down the quotient in each of the following divisions :
(i.) (x5 - y^) - (X - y), (ii.) (x^ + f) - (a; + y),
(iii.) {x^ -y'^^-^ix^ y).
10. Show that, if n is any positive integer, 7" — 1 is divisible by
6 ; show also that SS^^+i + 1 is divisible by 36. .
11. Show without actual division that
(3x2 - 2 a; + 1)3 -(2a;2 + 3x - 5)8
is divisible by x2 — 5 x + 6.
12. Write down the result of dividing (2 a + 3 6)8 + (3 a +2 6)8
by 5 a + 5 6.
13. Write down the result of dividing
(2 a + 4 6 - 4 c)3 + (a - 6 + 7 c)8 by a + 6 + c.
14. Show that (1 — x) 2 is a factor of 1 — x — x° + x^.
MISCELLANEOUS THEOREMS. 179
15. Show that (1 — xy is a factor of 1 — x — x" + x'*+\ n being
any positive integer.
16. Show that (x - 1)2 is a factor of wx"+i - (n + l)x" + 1, and
also of x'' — wx + w — 1, w being any positive integer.
148. Factor Theorem. If any rational and integral ex-
pression which contains x vanish when f is put for x, then
will x—fhe a factor of the expression.
Let the expression, arranged according to powers of
Xj be
ax''-\- hx""-^ + caf"-^ -\
Then, by supposition,
Hence aa;" + dx""^ + cx""'^ -\
= aaf» + &af*"^ + ca;'*-^ h (a/" + ft/''-^ + c/— ^ + ..•)
= a (x» -/") + h (a;"-i —f~^) +c (af'-^ —f-^) + ...
But, by the last article, a;"-/**, a;'-^-/^-!, a;"-2_yn-2^
etc., are all divisible by a; — /
Hence also ax"" + hx""'^ + cx""^ + ••• is divisible by x—f.
149. Eemainder Theorem. If any expression which con-
tains X he divided by x—f the remainder is equal to the
result obtained by putting f in the place of x in the
expression.
Divide the expression ax'* 4- 6af~^ + cx"~^ -\ by a; — /,
continuing the process until the remainder, if there be
any remainder, does not contain x\ and let Q be the
quotient, and R the remainder.
Then, by the nature of division,
asf" + 6af-i + csf'-^ + ... = Q(a; -/) + R,
and the two members of this equation are identical.
180 MISCELLANEOUS THEOREMS.
Now since R does not contain x, no change will be
made in R by changing the value of x\ put, then, x=f,
and we have the required result,
ar + ¥"-' + cp-' + - = Q{f-f) + R = R.
This is known as the remainder theorem. It includes
the factor theorem of Art. 148 as a particular case, or
corollary; for, a/'* + ^/^^-^ + c/"-^ + . . . = 0 when R=0,
and if i? = 0, the expression ax"" -\- hx^~^ + cx""'^ + ••• is
exactly divisible by x—f.
Ex. 1. Find the remainder when x^ — 4a;2 + 2 a; + 1 is divided
by a; - 3.
The remainder is 33 - 4 • 32 + 2 • 3 + 1 = - 2.
Ex. 2. Show that x - 2 is a factor of «* - 3 a;2 + 2 a; - 8.
The condition is 2* — 3- 22 + 2- 2-8 = 0; and this condition
is satisfied.
Ex. 3. Determine whether aj^ — 3 aj + 2 and x^ — 13 a: + 12 have
a common factor.
The factors of a;^ — 3 a; + 2 are obviously a; — 1 and a: — 2 ; and
of these a; — 1 does, and x — 2 does not, satisfy the above condition
of being a factor of a;^ — 13 x + 12.
Ex. 4. Show that any rational and integral expression in x is
divisible by x — 1 if the sum of the coefficients of the different
powers of x is zero.
For if the sum of the coefficients is zero, the expression will
vanish when we put x = 1, and therefore x — 1 is a factor.
150. Symmetry. An expression is said to be symmetri-
cal with respect to any two letters when it is unaltered
by an interchange of the two letters.
Thus a + 6 and a^ + 2>^ are symmetrical with respect
to a and 6, for neither expression would be altered by
changing a into h and h into a.
MISCELLANEOUS THEOREMS. 181
Also a + b -{-c and a^ -{-b^ -\-c^ — Sabc are symmetrical
with, respect to any two of the three letters a, b, c.
The only expression of the first degree which is
symmetrical with respect to the three letters a, b, c
is pa-\-pb+pc, where p is some numerical coefficient.
These examples suggest the following definition :
Def. An expression is (completely) symmetrical with
respect to a specified set of letters when it is symmetri-
cal with respect to every pair of them.
Ex. The following expressions are symmetrical with respect to
the letters contained in them :
a2 + 62 + c2 - 6c - ca - ab,
(6-c)4+(c-a)* + (a-6)*,
bed + cda + dab + abc.
151. A partial symmetry may exist in expressions
containing more than two letters. Thus, although
ab^ -i-bc^ -\- ca^ is not completely symmetrical, but in
fact changes its form when any two of its letters are
interchanged, yet it remains unaltered when a is changed
to b, b to c, and c to a. Such a series of changes is
called a cyclic displacement, and the symmetry, repre-
sented in such an expression as ab^ -\-bc^-\- ca^, is called
cycle-symmetry.
Def. An expression is cyclo-symmetrical with respect
to the letters a, b, c, cZ, • • • Z if it remains unaltered when
a is changed to 6, 6 to c, c to d, etc., and I to a.
Ex. The following expressions are cyclo-symmetrical with
respect to the letters contained in them:
a\b - c) + 62(c - a) + c2(a - 6),
(6-c)3 + (c-a)8+(a-6)8,
(6 - c - l)(c - a - l)(a - 6 - 1).
182 MiSCELLAi^EOtJS THEOREMS.
152. The arrangement of the terms in a symmetrical
expression is of some importance. Consider, for example,
the arrangement of the expression bc-\- ca-\- ah. The
term which does not contain the letter a is put first, and
the other terms can be obtained in succession by chang-
ing a into h, b into c, and c into a. In the expression
a\b — c) -f 6^(c — a) -f c^^a — b) the same arrangement is
observed ; for by a cyclic displacement of the letters in
a^{b — c) we obtain b'\c — a), and another cyclic change
will give c^{a — b).
By reason of this law of the derivation of terms
from one typical term, symmetrical expressions are fre-
quently not written out in full, but indicated by placing
the Greek letter S before the typical term. Thus, when
two letters are. involved,
^a'b = a'b + b%
or, if three letters are involved,
^a'b = a'b -\- ah -f b'^c + b'^a -f c^a + (?b.
Similarly, when three letters are involved,
2a.2 + %bc = a^ + 52 _|_ c^ _l_ 5c -f ca + ab.
153. We now proceed to consider some examples
which will illustrate the theorems proved in the pre-
ceding articles.
Ex. 1. Show that, if w be a positive integer, x" — 2/** is divisible
by X - y.
If X = 2/, then
ajn _ yn _ gjn _ /jjn — Q^
Hence the proposition is an immediate consequence of the factor
theorem, which is itself a corollary of the remainder theorem.
MISCELLANEOUS THEOREMS. 183
• Ex. 2. Show that (& - cy + (c - ay + (a - by
is divisible by (& — c)(c — a)(a — 6).
If we put 6 = c in the expression
(6 - cy + (c - ay + (a - by,
the result is (c — a)^ + (a — c)^
which is zero.
Hence (6 — c) is a factor, and we can prove in a similar manner
that c — a and a — b are factors.
Ex. 3. Show that
(a -{■ b -h cy -(b + c - ay -(c + a -by-ia + b - cy = 24 abc.
If we put a = 0 in the expression
(a + & + c)8 - (& + c - ay -(c + a-by-(a + b- cy (i.),
it is easy to see that the result will vanish for all values of b and c.
Hence a is a factor of (i.).
So also b and c are factors of (i.).
Now (i.) is an expression of the third degree ; it can therefore
only have three factors. Hence it is either equal to abc, or is
equal to abc multiplied by some number.
Thus
(a + b + cy-(b-^c - ay -(c-^ a - by-(a + b - cy =Labc(u.'),
where L is some number which is always the same whatever a, 6,
and c may be.
We can find the value of L by giving particular values to a, 6,
and c.
Thus, let a = 6 = c = 1. Then (ii.) becomes
38 _ 18 _ 18 _ 18 ^ X.
.-. X = 24.
Ex. 4. Find the factors of
a8 (6 - c) + b^ (c-a)+ c« (a - 6).
If we put 6 = c in the expression
a3 (6 - c) + 68 {c-a)+ c* (a - 6),
the result is <fi (c — a)-{- c^ (a — c),
whinh is clearly zero.
184 MISCELLANEOUS THEOREMS.
Hence 6 — c is a factor of the given expression ; and we can
prove in a similar manner that c — a and a — h are also factors.
Now the given expression is of the fourth degree ; hence,
besides the three factors we have found, there must be one other
factor of the first degree, and as this factor must be symmetrical
in a, 6, c, it must be a + ft -f c.
Hence the given expression must be equal to
i (& - c) (c - a) (a - 6) (a + & + c),
where i is a number.
We can find L by giving particular values to a, 6, and c ; or, by
comparing the coefficients of a^, we have at once
h -c = -L{h-c).
Hence L =—\.
Thus S 53 (c _ a) = _ (6 - c) (c - a)(a - &) (a + 6 + c).
154. The following is an important identity :
d^^W^c^-^ahc = (a + & + c) (a^ -^-y -\-c^ - he - ca- ah).
It should be noticed that
a?j^W J{. c" -he - ca-ah=\\{h-cy+ {c-ay+ {a-hy\.
Since a-\-h-\-c is a factor of a^ + 6^ + c^ — 3 ahc, it
follows that a-^ 4- &' + c' - 3a6c = 0 if a + 6 + c = 0.
Hence o? -\-W + (?=3 ahc for all values of a, h, and c,
provided only that a 4- & + c = 0.
That is, the sum of the euhes of any three quantities is
equal to three times their product, provided that the sum of
the three quantities is zero.
For example, the letters involved being a, 6, c,
2 (& - c)3 = 8 (& - 6) (c -«)(«-&),
2 (ft + c - 2 a)3 = 3 (6 + c - 2 a) (c + a - 2 6) (a + 6 - 2 c) ;
also, since {x - a)(h - c)-{-{x - h) (c - a) + (x - c) (a — 6) = 0,
we have
2 (a; - a)3 (6 _ c)8 = 3 (x - a)(x - 6)(ic - c)(h-c){G -a){a - 6).
MISCELLANEOUS THEOKEMS. 185
EXAMPLES XLII.
Let the student employ the 2 notation for the purpose of abbre-
viating such of the following expressions as admit of its use.
1. Show that a2 (& _ c) + &2 (^c- a)+ c^ (a - ft)
= 6c (6 - c) + ca (c — a) + ab (a —6) = - (6 - c) (c - a) (a - 6).
2. Show that a^ (6 - c)+ &* (c - a)+ c* (a - 6)
= -(6 - c)(c - a)(a - &)(a2 + 52 _|_ c2 + &c + ca + a&).
8. Show that (6 - c)^ + (c - ay + (a - 6)5
= 5 (6 - c)(c - a)(a - 6)(a2 + ^a _^. c2 _ ^c - ca - a6).
4. Show that
(a + 6 + c)* -(6 + cy - (c + ay - (a + 6)* + a* + 6* + c*
= 12 a6c (a + 6 + c).
6. Show that a (6 - c)3 + 6 (c - ay -{- c (a - by
= {b- c){c- a)(a-b){a + b + c).
6. Show that 62c2 (b-c)+ c^a'^ {c-a)-{- a^b'^ (a - 6)
= — (6 — c) (c — a) (a — 6) (6c + ca+ ab) .
7. Show that a» (62 - c2) + 68 (c2 - a2) + c^ (a2 - 62)
= — (6 — c) (c — a) (a — 6) (6c + ca + ab).
8. Show that a* (62 - c2) + 6* (c2 - a2) + c^ (a2 - 62)
^-(6 -\-c){c + a)(a-\- 6)(6 - c)(c - a)(a - 6).
9. Find the factors of 6c (62 - c2) + ca (c2 - a2) + a6 (a2 - 62).
10. Find the factors of a (6 + c - a)2 + 6(c+a-6)2+c(a + 6-c)2
+ (6 + c - a) (c + a - 6) (a + 6 - c).
11. Find the factors of a2 (6 + c - a) + 62 (c+a - 6) + c2 (a+6-c)
- (6 + c - a) (c + a - 6) (a + 6 - c).
12. Find the factors of a (6 + c) (62 + c2 - a2)
+ 6 (c + a)(c2 + a2 - 62)+ c (a + 6) (a2 + 62 - c2).
18. Find the factors of
(y - «)^ + (2 - x)^ -{-{x - y)^ - k{y -z){z- x){x - y).
14. Find the factors of
a6(62 _ c2)3 + 56(c2 _ ^2)8 4. c6(a2 _ 62)8.
186
FRACTIONS.
CHAPTER XIV.
Fractions.
155. To obtain the arithmetical fraction ^, we must
divide the unit into 7 equal parts and take 5 of those
parts. So also to obtain the fraction -, where a and b
h
are positive integers, we must divide the unit into b equal
parts and take a of those parts.
156. The numerator and the denominator of a frac-
tion, defined as in Art. 155, must both be positive inte-
gers : we cannot, for example, with that definition, have
3
such a fraction as —^ ; for to say that the unit is to be
divided into — f equal parts, and that | of such parts
are to be taken, is without meaning.
We must therefore suppose the letters in - to be re-
ft
stricted to positive integral values, or we must alter the
definition of - ; and as we cannot restrict the values of
b
the letters, we must entirely dispense with the fractional
form, or make some modification in its meaning.
Now, with the definition of Art. 155, to multiply the
fraction - by 6, we must take each of the a parts b
FRACTIONS. 187
times ; we thus get ab parts, the parts being such that
each b of them make up a unit, and therefore the whole
ab parts will make up a units. Thus
^Xb==a ...... (i.).
Dividing each side by b, we have
g=«^6 («•)•
Now we may define the fraction - as that quantity which
when multiplied by b becomes a; for, as we have just
seen, this new definition agrees with that of Art. 155,
whenever the definition of Art. 155 has meaning ; and by
taking this new definition we do away with the necessity
of ascribing only positive integral values to the letters.
We may similarly dejirie the fraction - as the quotient
ob'tained by dividing a by b.
Hence, instead of the definition of Art. 155, which is
inapplicable to an algebraical fraction, we have either of
the following equivalent definitions.
Def. I. The algebraical fraction -, where a and b are
b
supposed to have any values whatever, is that quantity
which, when multiplied by b, becomes equal to a.
Def. II. The algebraical fraction - is the quotient
obtained by dividing a by b.
The fractional form — has already been used with the
b
188 FRACTIONS.
meaning a-^b, and henceforth the notation a/b will also
be frequently employed to denote a fraction.
157. We now proceed to consider the properties of
algebraical fractions ; and we shall find that algebraic:il
fractions are added, subtracted, multiplied, divided, aiid
simplified, precisely in the same way as arithmetical
fractions. It will be assumed throughout the discussion
that the quantities involved are all finite and different
from zero.
158. The value of a fractmi is not altered by multiply-
ing its numerator and denominator by the same quantity.
We have to prove that
a_am
b bm'
for all values of a, &, and m.
Let a; = -. Then xxb = -xb.
b b
But - X 6 = a, by definition.
b '
.: xb = a', and therefore xbm = am.
Divide by bm, and we have x = am -j- bm ;
. r , • a am
that IS, - =
b bm
Thus the value of a fraction is not altered by multiplying
its numerator and denominator by the same quantity.
159. Since, by the last article, the value of a fraction
is not altered by multiplying both the numerator and the
denominator by the same quantity, it follows conversely
FRACTIONS. 189
that the value of a fraction is not altered by dividing
both the numerator and the denominator by the same
quantity.
Hence a fraction may be simplified by the rejection of
any factor which is common to its numerator and denom-
inator. For example, the fraction a^y/a^y takes the sim-
pler form aya^, when the factor y, which is common to
its numerator and denominator, is rejected.
When the numerator and denominator of a fraction
have no common factors, the fraction is said to be in its
lowest terms.
160. Eeduction of Tractions to their Lowest Terms. To
reduce a fraction to its lowest terms, we must divide
its numerator and denominator by their H. C. F. ; for we
thus obtain an equivalent fraction whose numerator and
denominator have no common factors.
Ex. 1. Reduce ^ to its lowest terms.
6 a'^xy
The H. C. F. of the numerator and denominator is 3 axy ; and
Zax^y _ 3 ax^y -f- 3 axy _ _x_
6 a^xy 6 a'hcy -h 3 axy ~ 2a
Ex. 2. Reduce ^^ to its lowest terms.
a^b^
The H. C. F. of the numerator and denominator is a^ft* ; and
a^b* a^b* ^ a^b^ a^b
Ex. 3. Reduce ^ «^&^^y^ to its lowest terms.
3 ab^x^y'^
The H. C. F. of the numerator and denominator is ab'^xy^ ; and
2 aWxy^ _ 2 a^b'^xy^ ^ ab^xy^ _ 2jcfiy^
3 ab^x^y^ 3 abH^y^ -h ab'^xy^ 3 bx^ '
190 FRACTIONS.
161. Instead of reducing a fraction to its lowest terms
by dividing the numerator and denominator by their
H. C. F., we may divide by any common factor, and
repeat the process until the fraction is reduced to its
lowest terms.
Thus a^6^c3 ^ ^253^3 ^ giffi ^ (jfl
a^b*c^ 6*0* 6c* be
The above process may be written down more compactly as
follows : a^
a^d^c* be
be
EXAMPLES XLIII.
Reduce the following fractions to their simplest forms :
- o^ - 15 aWcPx^
ab"^' ' 26a'^¥c^x^'
2 ^^ g 125 a6%3#
xY ' 150a46W
2 a^be^ g 3 a^x^yz-^
4 a^b^c 5 a¥xy^z
^ 6 a^b^c* jQ '5 a%^&xy^
4a765c4* • 7b*c^xy^ '
6 ^!^. 11 14 ab'^c^xy'^z^
x'52/2^2' • 21 a^b^cx^y^z
g 12a;Vgio j2 3 a^bc^x^y'^z
16x^y^z^' ' ^a^b^cx^y^z^'
162. When the numerator and denominator of a frac-
tion are multinomial expressions whose factors can be
seen by inspection, write the numerator and denominator
as the product of factors of the lowest possible dimen-
sions; the factors which are common to the numerator
3.
FRACTIONS. 191
and denominator will tlien be obvious, and can be
removed.
, ^ a^ — ax
Ex. 1. Simplify ^TT^'
a^ — ax _ aja — x) _ a
Ex. 2. Simplify
a2 - x2 {^^ x){a + x) a + x
x^ — X-
x*-l
x*-! (x»'-l)(x2 + i3 a;2^1*
a;2 _ 7 X + 10
Ex.3. Simplify ^,_,^^g
a;2 - 7 X + 10 ^ (a; - 5)(x - 2; ^ X - 5
x2_5x + 6 (x-3)(x — 2) x-3'
Ex.4. Simplify |5^.
x'' — ax _ x(x — q)
a2 _ x2 (o - x)(a + «)*
Now X — a = — (a — x) ; hence, dividing the numerator and
denominator by a — x, we have the equivalent fraction,
o + X
If we divide the numerator and denominator by x — a, we have
X
-(a + x)*
By the Law of Signs in Division ~ ^ and are both
^ -a + x -{a + x)
equal to ; — . and this latter is the form in which the result is
a + x'
usually left.
Note. — It should be remarked that the value of a fraction is not
altered by changing the signs of all the terms in the numerator
and also of all the terms in the denominator ; for this is equivalent
to multiplying both numerator and denominator by — 1.
192 FRACTIONS.
EXAMPLES XLIV.
Simplify
^ 2ab g 4x- 16 ^„ 6ax- 15 g^
18.
3^ u_^^^^ jj_ j^iZlil. 19.
21.
a2 + ab
a^V
10.
(k2 - x^y'^
a?-- ah
a2 + ah
11.
ic2-f aa;
a;2-a2
12.
(X - 1)2
X2-1
13.
X2 - x2?/2
(a: + xy)2
14.
x2+2x
15
x2-4
x* + a:2
x^-l
16.
x2 - 9 X + 20
x2 + 6x-
55
1-92/2 4-20?/*
1+6^/2-
55^4
x2 - 8 x?/
+ 72/2
x2 — 3 x?/ ■
-282/2
1 _ 8 a2?,2 + 7 05454
X2-
-16
2x3
-4x*
X2.
-4x4
X —
■2
4-
X2
a ~
-3
9-
a2
a2x^
'-X*
x*
-a*
X4-
-^2
a2-
-ax2
X5-
-a2x3
x*
-a*
^2x2
2/2 - x^y^
x-^ - y a;^
3 x2 - 12 ffx
48a2_3a;2'
a2 - 2 ax 4- x2
a2 _ x2
rt4 + 2 «2/,2 4. 54
-1
X3-1
X4-1
23.
24.
X6- 1
x2 - 5 X + 6
x2 - 7 X + 12'
1 - 5 a + 6 a2
30.
27. -^ - -^^y -^ > y . 3j
x*2/4 - a* 1 - 7 a + 12 a2
25 a;2-9x + 20 gg (g^ _ a;3) (a + x)
(q3-68)(a2_(^5 4.52)
{a^-h'^){a + h)
(a^-h^)(a-b) _
1 - 3 a262 _ 28 a464 " (^3 _ 63)(^a4 _ ^4)*
163. When the factors of the mimerator and denom-
inator of a fraction cannot be found by inspection, their
H. C. F. can be fonnd by the rule given in Chapter XI. ;
and the fraction will be reduced to its simplest form by
dividing the numerator and denominator by their H. C. F.
FRACTIONS. 193
Thus to simply ^-^^x + 10^
The H. C. F. will be found to be x^ - 5 x + 2, and
a:8 _ 23x + 10 =(x2 - 5x + 2)(x + 5),
5x3 - 23 x'^ + 4 =(x2 - 5x + 2)(5x + 2).
Hence the given fraction is equal to
x + 5
5x + 2*
EXAMPLES XLV.
Simplify
J x8-3x + 2 g 2x8 + 3x2 + 4x-3
2x8-3x2 + 1* ' 6x8 + x2-l
2 3x2-8x+5 ^ x8-3x-2
x8-4x2 + 5x-2" ■ x* + 2x3 + 2x2 + 2x+ 1*
g x8 + 3x2-20 g X* - x8 - X + 1
x* - x2 - 12 ' ' X* - 2x8 - x2 - 2 X + l'
^ x8 + 6x2+ llx + 6 g x* + 2x8-3x2-7x-2
x8 + 6x2 + 6x ' * 2x* + x8-6x2-6x-l*
g 2x8 + ax2 + 4«2a;-7a8 ^^ x* -20x2 - 15x + 4
* x8-7ax2 + 8a2x-2a8' ' x* + 9x3 + 19x2 -9x - 20*
164. Eeduction of Fractions to a Oommon Denominator.
Since the value of a fraction is unaltered by multiplying
its numerator and denominator by the same quantity
[Art. 158], any number of fractions can be reduced to
equivalent fractions, all of which have the same denomi-
nator.
The process is precisely the same as in arithmetic, and
is as follows.
First, find the L. C. M. of all the denominators ; then
divide the L. C. M. by the denominator of one of the
194 FRACTIONS.
fractions, and multiply the numerator and denominator
of that fraction by the quotient ; and deal in a similar
manner with all the other fractions : we thus obtain frac-
tions equal to the given fractions, all of which have the
L. C. M. for denominator.
For example, to reduce
X y
, and
a%{x + ay ab^ix-a)' ah {x^ - a^)
to a common denominator.
The L. C. M. of the denominators is a^js (-3,2 _ ^2-)_ Dividing the
L. C. M. by (jfih (aj -fa), ah"^ (x — a), and ah {od^ — a"^), we have the
quotients & (x— a), a{x-\- a), and ah respectively. Hence
XX h{x — a) _ hx(x — a)
and
a'^h (x + a) a% (x + a)xh(x-a) a^h"- (x^ - a'^y
y _ y X a{x + a) _ ay {x + a)
ah^ {X - a) ah^ (^x - a) x a^x + a) a%'^ (a;2 - a^)'
z z X ah ahz
ah (x2 - a2) ah (x^ — a^) x ah a^h'^ {x? - a^)
Note. — In the above process it is not absolutely necessary to
take the lowest common multiple of the denominators ; any com-
mon multiple will do equally well, but by using the L. C. M. there
is a saving of labor.
165. Addition of Fractions. The sum (or difference) of
two fractions which have the same denominator is a
fraction whose numerator is the sum (or difference) of
their numerators, and which has the common denomi-
nator. This follows from Art. 80.
When two fractions have not the same denominator,
they must first be reduced, by the method explained in
the last article, to equivalent fractions which have the
FRACTIONS. 195
same denominator : their sum or difference will tlien be
found by taking the sum or difference of their numera-
tors, retaining the common denominator.
Thus « + ^ = ^+-^, -*
XX X
a_b_ a — b
XX X
q,b_ a X y b x x _ay ,bx_ ay -f bx
X y XX y y X X xy xy xy
and a + ^=:«l^ + ^ = «|ML&.
y y y y
When more than tWo fractions are to be added, or
when there are several fractions, some of which are to
be added and the others subtracted, the process is pre-
cisely the same. The fractions are first reduced to a
common denominator, and then the numerators of the
reduced fractions are added or subtracted as may be
required.
Thus qj^b_c_^a + b-c^
^ ^ JO tJu
and ^ + ^ _ ^ = a X yx b x xz _ c x
X y z XX yz y x xz z x xy
_ayz . bxz _ cxy _ ayz + bxz — cxy
xyz xyz xyz xyz
Note. — It may be necessary to remind the student that when
there is no sign between a fraction and a letter or number, tho
sign of multiplication is understood. Thus, 2- means 2 x - and
b b
not 2 -f - as in the arithmetical form 2i.
196 FRACTIONS.
Ex. 1. Find the value of — h ^
X — a X + a
The L. C. M. of the denominators is (x — a)(x + a); and
1,1_ x + a X — a
x — a x-\- a {x — a){x + a) {x + a){x — a)
_a; + a-\- x — a _ 2a;
Ex. 2. Find the value of — ^— + —^ —
a — xx^ — a^
Beginners should always see that the denominators of the frac-
tions which are to be added are all arranged according to descend-
ing powers, or all according to ascending powers, of some particular
letter. This is not the case in the present example ; hut
ax _ ax _ — ax
X'i — a^~- (a-2 _ a;2) ~ «2 _ x^"
We then have
a \ ~ ^^ — ^(^ + ^^1 I — (i^
_ a{a + x)— ax
~ a-2 - ic2
Ex. 3. Simplify -J— + —1— + — ^ + - ^
V
1-X l-\-X 1 + X2 1+X*
It is sometimes best not to add all the fractions at once ; this
is particularly the case when the denominators are not all of the
same dimensions.
1 1 ^(l + a;) + (l-a;)^ 2 .
1 - X 1+ X (1 - X) (1 + x) 1 - x2 '
then
2(1 +x2)+2(l-x2)_ 4
1-X2 l-\-X^ (1-X2)(1 + X2) 1-X*'
and finally
4 4 ^ 4(l + x*)+4(l-xO ^ 8
1 _ a4 "•" 1 + x* (1 - X*) (1 4- X*) 1 - x8*
FRACTIONS. 197
Ex. 4. Find the value of — — ^ + ^
x-1 x+l x-2 x+2
Here again it is best not to reduce all the fractions to a common
denominator at once : the work is simplified by proceeding as
under.
ce+l-(a;-l)
x~l x + 1 (x-l)(x+l) x^-l'
1 1_ ^ x + 2-(x-2) ^ 4
x~2 x + 2 (x-2)(x + 2) a;2-4'
and 2 4 ^2(a;2-4)+4(x2-l)^ 6a;2-12
a;2_i aj2-4 (x2-l)(x2-4) a:* -6x2 + 4
Ex. 6. Find the value of ^
x2-Sx-\-6 a;2_7a. + i2'
1
x^-bx + Q (a;-2)(x-3)
and 1 1 *
a;2_7x + 12 (x-3)(x-4)'
hence the L. C. M. of the denominator is (x — 2)(x — 3)(« — 4).
Hence we have
x-4 x-2
(«-2)(ic-3)(ic-4) (x-2)(x-3)(x-4)
(X - 4) - (x - 2) -2
(x-2)(x-3)(x-4) (x-2)(x-3)(x-4)
2
(x-2)(x-3)(x-4)*
Ex. 6. Find the value of
(o-6)(a-c) (6-a)(6-c) (c-a)(c-6)
In examples of this kind it is best for beginners to arrange all
the factors in the denominators of the fractions so that a precedes
h or c, and that b precedes c. We therefore change b — a into
— (^a — b), c — a into — (a — c), and c — b into — (& — c).
198
FRACTIONS.
Thpn
1 1 -1
(6-a)(6-c) -(a-6)(&-c) (a-&)(&-c)'
and
1 1 1
(c - a)(c - b) {-(a - c)}{-(6 - c)} (a - c)(6 -c)
Hence we have to simplify
1^-1.1
(a — 6) (a — c) (a -:-&)(& — c) (a — c) (6 — c)
The L. C. M. is (a - 6)(a - c)(6 - c) ; and
1 ^ b-c
{a — b)(^a — c) (a — b){a — c)(b — cy
-1 ^ -(a-c)
(a-6)(&-c) (a-6)(6-c)(a-c)'
and 1 ^ - ^
(a — c) (6 — c) (rt — 6) (a — c) (6 — c)
Hence we have
b — c — (a — c)-\-a — b_ b — c — a + c+a — b 0_
{a — b){a — c)(b — c) (a— 6)(a— c)(6 — c) {a — b){a—c)(b — c)
EXAMPLES XL VI.
Reduce to their lowest common denominator :
1. A, ±, 7 g 3 4
3a;' 5a;' 30a; ' 2a;-2' 3a;-3
7.
1 1 1
2 ax' 6 6x' 8 ex
^ a b c «
6c ca ab
. be ea ab «
a 6 c
5. -i ^— 10.
5
3
6X + 12'
1
8a; +16
3 5
a;+l' 2
6
x + 2' x2-l
2 4
5a;- 5'
a
3x + 3' a;2-l
X a2
a; + l' 2 a; + 2 ' x - a' a - a;' a;2 - a^' a^ - x^
11.
12.
13.
14.
FRACTIONS. 199
2a b 3a2 6 6^
a-b' 2b -2a 4(a^ - b^y 6(62 -a^)*
1 2x 3x2
a;+ l' (x+l)2' {x + iy
1 1 1
(x - a) (a; - &) (6 - x) (c - x) (x — c) (x — a)
1 1 1
ia-b)(ia-cy (6_c)(6-a)' (c-a)(c-6)'
Reduce to one term :
IK „ 1 , 2 2Q 6a-56 4a-76
iU.
" ^-^a + l-
16.
a + x+ ^' .
a -X
17.
x-2y
18.
a — b . a — 6
a^b ab^
19.
a-bb a-Sb
21 Q^ ~ '^^ I 3q — ft
22. 2^-111-3^ + ^.
3 4
23. 5a;-2y_3y-2x^
6 4
24. ? - ^~^ + ^ ~ ^
4 3 6
25 2x-3y x + 2y 3x-2y
3 4 6
Simplify
26.
a
-+ ^
a - i
& 6-a
27.
X — a a — X
28.
X
X2-
1 ^
a2 ' a2 -
x2
QQ
1
2
80.
1
2 -X
4
4-x2
81.
1
3 + x
-A
82.
1
x-3
1
x-2
Ait
1
1
1 — X 1 — x2 X — 4yiX — 5y
!00
FRACTIONS.
34.
1 1 .n 2a .
2 6 a2 + 62
' Sx-2y bx-2y a + ft '
a-h a'-h'^
^iS
1-
1-
^x \-x ^j a
^ 1
-X \-\-x a-\
a{a - 1)
36.
a+26a-26 ^^ 2
a-26a + 2& x-2
1 x + 6
x + 2 x2 + 4
37.
1 1 2 43 1
1 x + 3
X -
- 2 ' (x - 2)2 "" x-1 2{x + 1) 2(x2 + 1)
38.
a -
a 1 2 a6 44 ^ ~ ^
-2& (a-2 6)2 ' 1 -3x
3 + x l-16x
l + 3x 9x2-1
39.
x + y x-y ^ 4.y'^ 45 « +
X — y X + y x2 — y2 a — ic
a , 2a2
a + X a2 + x2
46.
a 1 a 1 2a2 ^ 4 a*
a-x a + x a2 + x2 a* + x*
47.
1 11 1 ,
x+1 x+2 x+3 x+4
48.
\ ?+ \- 49. 1
x-lxx+1 a
2 + 1 •
a+1 a+2
50.
1 3 3 1
x-3 x-1 x+1 x+3
51.
1 3 1 3 1
a a+1 a+2 a+3
52.
1 4 1 6 4,1.
X— 2 x-l'x x+1 x+2
53.
1 4,6 ^ 1 ^
a a+1 a+2 a+3 a+4
54.
1 2 1
x2-5x + 6 x2-4x + 3 x2-3x
+ 2
55.
. 1 . .- . .^ _ .
1
x2 + 5 ax + 6 a2 x^ + 4 ax + 3 a^ x2 + 3 ax + 2 a2
FRACTIONS. 201
gg x-l 2(x - 2) x-S
(x-2)(x-S) (x-S)(x-l) (x-l)(x-2)
X(X - 1) 1 - X2 iC(x + 1)
gg 2g x — a 2
(ic-2a)2 a;2-5ax + 6a2 x-3a*
59 3 + 2g 2 -3a 16a -q^
2 -a 2 + a a2-4
gQ a; + gy x - ay . 2^^ -^ «^y^
x — ay X -{• ay x^ — a^y^
61 a^^-2a; a;4-3 4x gg 11 l+2a;-g;2
68.
X2-1 X+ll-X 1-XX+l 1
_J_ + _«i^^ L_.
x+3y x2-9y2 3y-x
a;^-(y-g)g y2_(^^a;)2 g2-(x-y)g
(X + y)2 - Z-2 '^ (y + Z)^ _ x2 "^ (;2 + x)2 - y^'
g2_(6 + c)2 62-(c + a)2 c2-(a-6)2
(a + 6)2 - C2''' (6 - c)2 - a2"^ (c - a)2 - ft**
166. Multiplication of Fractions. We have now to show
how to multiply any two algebraical fractions.
Let the fractions be - and -•
b d
Let aj = ^X^.
0 a
Multiply by 6 X d ; then
xxb X d = ^x-x bxd
b d
= ^x6x^xd, by Art. 52.
202 FRACTIONS.
But -xb = a, and - x d = c ;
0 a
.', xbd = ac.
Divide by hd ; then x = —
hd
That is, «x-^ = ^.
h d hd
Thus the product of any two algebraical fractions is
another fraction whose numerator is the product of their
numerators, and whose denominator is the product of their
denominators.
The product of any number of fractions is found by
the same rule.
For example :
^x-x- — — x- = — •
h d f~hd f hdf
Again, a c d = ^
0 d a hda o
And a^ + 1 .. a; + 2 ^ a; 4- 3 ^ (a; + l)Ca; + 2){x + 3) ^ ^
a + 2 X + 3 X + 1 (ic + 2) (x + 3) (x + 1)
167. Division of Fractions. Let - and - be any two f rac-
a c b d ^
tions, and let a; = - -r- -•
b d
mi. caeca
Then a; x - = -^- x - = - ;
d b d d b
c d ad
,-. a; X - X - = - X —
d c b c
FRACTIONS. 203
But X X- X-=x\
d c
a d
.'. X= -X —
b c
Thus to divide by any fraction is the same as to multiply
by its reciprocal. [See definition of reciprocal, Art. 168.]
Ex. 1. Divide - by ?•
a^ a
a^ ' a a2 ^ a
Ex. 2. Divide - by ab.
b
g . q^—Qy 1 -.1,
b ' ~ b ab~b^
Ex. 3. Divide ^-=1^ by ^i:^'-
x3 + a^ x-\-a
X — a , x^ — a^_x — ax + a
X* 4- a^ x + a x^ + a^ x^ — a'
1
(x^ -ax+ a2) (x2 + ax+ a^) (x* + a'^x^ + a*)
168. When two quantities are such that their product
is 1, each is said to be the reciprocal of the other. Thus
- is the reciprocal of -, and - is the reciprocal of a.
a b a
A quantity is small or large according as its reciprocal
is large or small. For example,
10000 ^j-m^Thi^TJi^ ^y
but 10000 > 1000 > 100 > 10 > 1.
204 FRACTIONS.
If then a be very small, 1/a is very large, and if a be
very large, 1/a is very small.
When a becomes zero, the fraction a/h assumes the
form 0/5 and, by the proposition of Art. 127, is itself
zero. But when h is zero, the fraction assumes the form
a/0 and, regarded as a quotient, has no meaning ; for
division by zero is an impossible operation. It becomes
merely symbolic.
Yet it frequently makes its appearance in algebraic
operations, and since it may be regarded as having arisen
by virtue of a decrease of the denominator from a finite
quantity to zero, that is, by an increase of the fraction
itself beyond measurable limits, it is called an infinite
quantity, or infinity, and for convenience the special sym-
bol 00 is used to represent it.
But this nomenclature must be regarded as conven-
tional, and the symbol oo must be used in algebraic
operations with extreme caution. A complete study of
its legitimate use requires more extended discussion than
is at present appropriate. [See Treatise on Algebra,
Art. 217.]
3. -H-
4.
EXAMPLES XLVii.
duce
the following fractions to their simplest f(
2a,
3c '
4a
5.
h d ac
5a2
6 6c
X ^^'
10 ca
6.
«' X ^' X c2
be ca ab
«
c
7.
b^b^c^
c
d
a c ' a
2a2
6c
3a&
8.
a\b^ . c2
b^ c^' a^
FRACTIONS.
9. ?«x2»xi«.
be ca ab
5 62 10 62a;
10. 2x» 3,3 5^
3yz bzx 2xy
,„. 5a26c . 5a26c2
3 62c2a ■ 3a62c»
205
13. By what must -^ be multiplied, that the product may be — ?
5 x^y 5 X
14. By what must ^-^ be divided, that the quotient may be — ?
Express in their simplest forms :
16. ^-iJLx-^+iL. 21. ^^x^^x^^.
x^ -\-zy xy — y^ x — 2 x-S a:-4
16.
g + 6 ^ q6 - 6^ gg x+1 ^ x+2 x+3
a8-a26 a6 + a2 ' (a;+2)2 (x+3)2 (x+l)2*
j^ x2 + 2 X ^ z2 _ 3 a; ^^ x2-3x + 2^^x2-7x+12
a;2_9 x2-4 x2-5x+6 x2-5x+4
18 x2-y2 ^x-2y. ^^ x2-l ^^ x2-25
* x2-4y2 x-^y * x2+3x-10 x2-3x-4
19. ^ + 3a;2 ^ x + 3 g^ a^ - a^ ^ x + 2a_
X + 4 * x2 4- 4 X * x2 — 4 a2 x - a
20 <^ + ^^ ^ q6 + 4 62 gg g^ - x» ^ (g - x)2.
g2 + 6 g6 ■ g8 + 5 g26' * g* + x^ ' g2 - x2 *
87. g + ^ X ^^ ~ ^^ ' g* - X*
(g - x)2 g2 + a;2 (a + x)8
(g-6)8 g*-6* " g2 + 62 *
^ (g - 6)2 - c2 ^ 62- (c - g)2
80.
31.
32.
(g-c)2-62'^c2-(g-6)2
x^-(y + zy , y^-(x + zy
x-^-Cy-zy • yi-{x-z)^'
ofi + y^ X '^~y ^ a;* - x2y2 + y*
X^ — y^ X -\- y X* + x2?/2 -j. y4
206
FRACTIONS.
169. We will now give some examples of more com-
plex fractional expressions.
d ad
'be
Ex. 1. Simplify ^ /--
hi d
a / c_a d
Ex.2: Simplify (I +iy(^ + l).
\b 1/ [a I b I a b b + a b
Ex.3. Simplify (^-^ -^^^Wf «±^+ «II^V
\a — X a + xj/ \a — X a + xj
a + x a — X _ (a + x) (a + x) — (a — X) {a — x) _ 4: ax
a — x a + x~ (a — x)(^a + x) ~ a'^ — a;^'
a2-x2
a-\-xa-x_(a + x)(a + x) + (a — x)(a- x) _ 2a^ + 2x^
2L11Q. — -|- ; — 7- r-7 ; r ^_— — -^-—
a — X a-\- X (a — ic) (a + a;)
Hence the given fraction is equal to
4 ax / 2 a2 + 2 a:2
4: ax /2a;
a^-x^l a-
Ex. 4. Simplify
x + 2
X
x + 1 (x + 2) ic - (x + 1)
X
a;(a;2 + a;-l)
~ X- ^'^ + ^^ ~ ^(^^ + X - 1) - (x2 + 2x)
X2 + X - 1
_ X(X2 + X - 1) _ X2 + X - 1.
x8 - 3» x2 - 3 '
FRACTIONS. 207
a — b , b — c
T^ E o- ^■e^ I + ab 1 + bc
Ex. 6. Simplify ; t—tt r—
1 (a-b)(b-c)
(l + a6)(l + &c)
a-b b- c _ (a - 6) (1 + be) + (6 - c) (1 + ab)
l + ab l + bc (1 + ab) (1 -^ be) ^
_a — b -{- abc — b'^c + b — c -{■ ab^ — abc
il + ab)(l + bc)
_a-c + ab'^-cb^
(l + a6)(l + 6c)'
^ (a-b)(b-c) ^(l + a6)(l + &c)-(a-6)(6-c)
(1 + a6)(l -f be) (1 + a6)(l + be)
_ 1 4- a6 + 6c + abH - ab -\- b^ -\- ae - be
(l + ab){l + bc)
_ 1 + gc + fe^ + ab^e
(l + ab)il + bc)'
Hence the given fraction is equal to
a - e + ab^ - eb"^ . 1 + ac + 6^ + ab^c
(l + a6)(l + 6c) ■ (l-hab)(l + be)
_ a-c + ab^-cb^ __ (g - c)(l -f- &^) _ a-c
1 + ac + 62 + acb^ (1 + ae) (1 + 62) i ^. ^c
EXAMPLES XLVIII.
Ig— 2x X — 2g/ \2g— x a — xi
2. [ ^ I ^~^] ' f ^ 1-^] ■
*\l + x x/ll+x xJ
4. 2 ^ ^
Simplify
1_1 i_y i_?
X ^ X y
208 FRACTIONS.
X y 2 1 4- a; 1 + a;^
1+? 1+r 1+1 10. 1 + ^^ i + "l
^y ^x x^y 1 + x^ 1 + x«
1 + x3 1 + x*
^4.2^ + 2 ? + ?'-2 ,, 1 1
X y X 11.
6. ^__ri_4-2 •" "• ^ T
x + y x-2/ x+
3. 1
a+& _^6 »
a + 26 a 12. x3- ^
a + 6 & ^ 1 -X
a a+26 ^ , 1
X + 1 13.
« r^- .+ 1
o — a
x + 2 _ 4x + 5 j^ 1
2x + 3 5x + 6 * ^2 x« + 1 '
2x + 3 _ 3x+4' ^ _j_ 1
3x + 4 4x+5 x-1
170. The following theorems are of importance.
Theorem I. If the frdctions a/h, c/d, e//, etc., he all equal
to one another, then will each he equal to the fraction
pa -{• qc -\- re -\
ph-\-qd + rf-\-'"
Let each of the equal fractions be equal to x.
Then, since a/h = x, c/d = x, e/f= x, etc.,
.'. a=bx, c = dx, e=fx, etc.;
.', pa =phx, qc = qdx, re = rfx, etc.
FRACTIONS. 209
Hence, by addition,
pa-^qc + re -\ =pbx + qdx + rfx -\
= (pb-i-qd-\-rf+'")x',
pa -\- qc -{• re -i- ' • ' __ _ a _
As a particular case, each of the equal fractions a/&, c/d, c//,
etc.,isequalto" + ^ + ^ + ''
6 + d+/+-
Theorem II. If the denominators of the fractions a/h,
c/dj e/f etc.f be all positive, then will the fraction
— be greater than the least, and less than the
a+ c + e4-
b-i-d+f +
greatest, of the fractions a/b, c/d, e/f, etc.
Let a/b be the greatest of the fractions, and let
a/b = X.
Then c/d < x, e/f< x, etc.
Hence, as b, d,f"- are all positive, we have
a=bx, c<dx, e<fx, etc.
Hence, by addition,
a-\-c + e-{-'" <bx + dx-\-fx-{- ... < (6 + ^+/+ ...)a;;
a -\-c + e -\-
b + d-\-f+
<x.
a I c I 6 I , , ,
Hence —^ — — — — — is less than the greatest of the
b-hd+f-h'"
fractions ; and it can be similarly proved to be greater
than the least of the fractions.
210 FRACTIONS.
Ex. 1. Show that, if ^ = ^, then will ^i^ = ^-±-^.
b d a—bc—d
Let ? = X ; then - = x. Hence a = bx, and c = dx.
0 d
Hence
Also
a + b _
a-b
_bx+b _
bx-b
_x + \
x-1
c-\-d _
_dx + d _
.x + 1,
dx — d X — \
And, since ^-i-? and ^-+-^ are both equal to ^ "*" , they must
a—b c—d x—1
be equal to one another.
Ex. 2. Show that, if ^ = ^, then will ^!-+_^
b d " ■ ^'
Let| =
= x;
then ^ = X. Hence a = 6x, and c = dx.
d
Hence
Also
a2 + 62 _ 52a;2 + 62 _ 62(a;2 + 1) _ ^2
C2 + ^2 ^2^.2 + ^2 ^(a;2 + 1) d2
a2 _ 62 _ 62x2 - 62 _ 62(x2 - 1) _ 62
c^-d^ d^x^-d^ d2(x2-l) d2
Hence
a2 + 62 _ a2 _ 62
C2 + (?2 c2 - d2
Ex. 3.
Show that, if ^^ + ^^ = «^ + ^^ = ^^ + «2^,
in will
bcx _ cay _ a60
— al + bm + en al — bm + en al -h bm — en
By Theorem I., each of the given equal fractions is equal to
— a(cy + bz) + b(az + ex) + e(bx + ay) _ 2 6cx
— al + bm + en — al + bm -\- en
And similarly each of the fractions can be proved to be equal
to 2cay ^^ ^^ 2abz
al — bm + en al + bm — en
FRACTIONS.
211
Simplify
24 a66c8x2*
EXAMPLES XLIX.
4.
x'^-Qxy -ifby^
«* — 6 x^y + 4 xy2
a^-lOx^y- lla;y2
«2y _8a;y2_9y8 '
6.
7.
9.
(a - 6) (a* - h^)
(q2_62)(a6_7)6)
(a + h)\a - 6)2 '
ac8 + 3x2-x-3
x* + 4x8- 12x-9*
3a^+5a^-7a;2 + 2a; + 2
2x*+3ic8-2a;2 + 12a; + 5'
IQ 2a:* + 9x8y + 14a;y8 + 3y*
3ic* + 14ic8y + 9a;y8 + 22/**
11.
2 a^ - 11 a;2y -f 11 a;y2 + 4 ys
12. 2^-±-i^-3^^
3 5
2x*-3x8y + 7x2
12 xy^ — 4 y*
a
13.
14.
16.
16.
^x-2y ^y-2x
4 6
4-a;
1
X2 -16
3 .
S-x a;2
a;
9
X - 2 y (2 y - x)2
17
18.
19.
20.
21.
+
4ab
a -4b (4 6-a)2
_1 1_ + J^ L_.
a+4 a + 6 a + 6 a + T*
+
a+3 a+6 a+9
x3
?/"
x2-9y2
X2_v2
x + 3y
y-^.
x2_4y2 2y-x
22. |/^l + lV-J--_A_UI/'i-iV-i-V
Ha 6/ ab {a-byi ' \\a b) ab )
23. ( ^ I Mxf^-^ ^V
Vx-2 x-8/ V3x-8 x + 2y
(x— y)*— xy(x— y)2— 2x2y2
(x-y)(x8-y3)+2x2y2
25.
I 2x + l
1 X2 + X + 1
212 FRACTIONS.
«« 1 . 1 3
27.
6a-26 3a + 26 6a + 2b
1 1.6
8a + 26 2&-8« 16a2-62
a6 + &6 ^ a-& ^ gi - ^252 4, 54
a6 _ 56 ^2 + 62 • a;4 + ^262 + 6**
m2+ w2
n m2 — n2
11 m^ — n^
n m
/l + ^ + _^i_\|l ?i_l
I a + 6 (a+6)2J I (a + 6)2J
I (a + 6)3i I a + 6i
31. x + '</ x-y 4 y2
' X — y x + y x^ — y^
32
33.
I a(a-36)i I 26-ai
xyiy^ — a;2) x2 + ic?/ xy — y^
34. a+6 ^ a2-62 6-a. 35 ^-^- ^^
a+6 6 25 62-a2 a -56 a-3 6 a+3 6 9 62-a2
36. ^ "^^^ + 1 ^-^
2(x-l) x2-7a;+10 2cc2-9a;+18
37. ^ 2 x-6 33 3 2 -.-7
a;_3 a;_4 (x-2)(x-5) x-4 x-S (x-6)(x-3)
39. (x-y 1 lx?i+^.
I x-y J
40. fa^ + y i 1 x^'-y'.
I X + 7fJ
FRACTIONS. 213
"• l^ 1 + x^ l-x2 /''2X + 1
a;-2 ^ x-4: 1
42. ?-2
X
a;_2 ^ x-4 ^
X — 5 X — 4
43. .6_^__7_^^ 1
x2 + 2x-8 X2 4-X-12 x2-5x + 6
44 rxM v«^f'^' + ^' ^'-y'U ^ • l^+y ^-y]
^ ^1x2-2/2 a;2+«/2J ^2 - Xy + J/^ ' \ jg-y x-^y (
V 2a6 ja8 + 68 ■ a2 - a6 -f 62*
47. I 1 + 1^ U/_^±iL_^y_l.
1 3x2- 14 xy+ 15 2/2 3x2-2 XI/-5 2/2/ lx-3y x+Syi
48 ^ ~ ^ I c — g , g — 6 , (6 — c)(c — a)Ca — 6)
* a + x 6 + x c 4- a; (a + x)(6 + x)(c + x)
I.\.a + ft/ a + 6 i lU-fc/ u-b I
61. 16 21 16 21
17 x-9 x + 4 x + 7
11 7 11 7
x-6 x-7 x+2 x+4
x + 7x + 8 x + 9 x+10
x+1 x+2 x+3 x+4
214 FRACTIONS.
{a + h-cY-d^ (b + c-ay-d^ (c + g-by-d'^
(a + by - (c + dy "^ (6 + cy - (a + dy "^ {c+ay - ib + dy'
55. Simplify ^ 1.1
axy aQii^ a) {y - a) y{x - a){y - a)
56. Simplify ^+-^ + ^±^ + ^-±-^
xix-y){x-z) y{y - z){y ~x) z{z-x){z-y)
ab
57. Find the value of - — — when x =
b -2x a + b
58. Find the value of
a + b + 2c a + b + 2d^
a + b -2g a + b -2d
when a+b = -i^-
c + d
59. If - = -, prove
b d
(i-)
(ii.)
a
-b
c-d
n
a -
-2b
c~2d
a2
c2
(c + dy
(i
(a
+ 6)2
f
X
. y _
z
gg _ 52 ^ C^ - ^2
a2 + 62 c2 + d2*
jpa2 4- qah + rb^ _ pc"^ + qcd + rd^ _ ,:
Za2 + ma& + nbf^ Id^^ mod + wd2
go. If -^^L- = —y— = _£_ ; then will x + y + z = (i.
a — b b — c c — a
Gl. If y+^ __gjL^^_Ml_ . then will each fraction = ^
ay+bz az + bx ax+by
62. Prove that, if a, b, c be unequal, and
ay+bz az + bx ax+by a+b
b — c c — a a — b
X y z
then will x + y + z = 0 and ax + by + cz = 0.
63. Prove that, if 5Li| = J-tS. = ^iL«^,
a - 6 2(6 - c) 3(c - a)
then 8a + 9& + 6c = 0.
FRACTIONS. ^15
64. Prove that, if ^^ - ^V = ^x-az^ ^^^^^ ^^^^ .^ ^^^^ ^^ ay-bx^ -
b — c c—a a — b
65.
Prove that,
if
a
b
c
2y+ 2z-
-Sx
2z
'. + 2x-
-Sy
2x
+ 2y-
-Sz
en
X
_
y
z
a + 2b + 2c b + 2c + 2a c + 2a + 26
66. Prove that, if
x — a _y — b _z — c
P Q ~ r *
and x+y + z=ia + b + Cj
then x = a, y = b, and z = c.
g^ jf ^y±cz^cz±ax^qx±by^2^^
a b c
prove that cix ^ by ^ cz ^^^
b + c — a c-\-a — b a + b — c
and therefore x=bc(b + c — d)^
y = ca{c + a — b)y
M = ab (^a •{■ b — c).
216 EQUATIONS WITH FRACTIONS.
CHAPTER XV.
Equations with Fractions.
171. In the present chapter we shall give examples of
equations which contain fractional expressions.
Ex.1. Solve ^^_3^-1^43-5:«.
^ ^^ ^^ %
We may multiply both sides of the equation by 120, the L. C. M.
of the denominators of the fractions, without destroying the
equality ; we thus get rid of fractions, and have
24(x - 1) - 15 (3x - 1) = 20 (43 - 5x) ;
... 24a; - 24 - 45x + 15 = 860 - lOOx.
Transposing, we have
24x - 45x + 100a; = 860 + 24 - 15 ;
.-. 79 a; = 869.
TT 869 n 1
Hence a; = ^^^ = 11.
79
Ex. 2. Solve — ^ +
a;-3 x + 3 a;+5
The L. C. M. of the denominators is (x — 3)(a; + 3)(x + 5).
Multiplying both sides of the equation by the L. C. M.,* we get
rid of fractions, and have
(a; + 3)(a; 4- 5)+ 2 (a; - 3)(x + 5)= 3 (a; - 3)(a; + 3);
.-. a;2 + 8x + 15 + 2a;2 + 4x-30 = 3a;2-27.
* See Art. 181.
EQUATIONS WITH FRACTIONS. 217
By transposition, we have
a;2 + 2x2 - 3x2 + 8x + 4a; = - 27 - 15 + 30 ;
.-. 12x = -12,
or x = — l.
Ex.3. Solve 1 . 1 _ 1 . i
x+ 1 x+7 x+S x+5
In this case it is best not to multiply at once by the L. C. M. of
the denominators ; the work is simplified by proceeding as under.
Wehave — ^ + — ^^ ^ = 0;
x+1 x-^S x+7 x+5
x + 3-(x+l)x-f5-(x + 7)_
•• (x+l)(x + 3) (x + 5)(x + 7) -"'
that is = 0 ;
(x+l)(x + 3) (x + 5)(x + 7)
, .-. (x+l)(x + 3) = (x + 5)(x + 7),
that is x2 + 4x + 3 = x2+ 12X + 35;
.-. 4x-12x = 35-3;
.-. -8x = 32,
.-. x = 32/(-8) = -4.
Ex. 4. Solve the equation
X — 1 , x+5 _ x + 1 x + 3
x+1 x+7~x+3 x+5
x+5_j_ 2 x+l_i 2
Since -
X
-1
+ 1
:1-
and
wehave
1--
2
x+l'x + 7 x + 7'x + 3 x + 3'
x_+3_ j__2_
x + 5 x + 5'
x + 1 x+7 x + 3 x + 6*
2 2^2 2
x+1 x+7 x+3 x+5
which is the same equation as that in Ex. 3.
218 EQUATIONS WITH FRACTIONS.
Ex. 5. Solve the equation
4x+5 , a;+ 5_2x + 5 a;'^ - 10
x-\-l x + 4 x + 2 x + 3
By division, we have
4a; + 5_^ J 1 a; + 5 _ .^ 1
a;+l jc4-lic + 4 x + 4
ic + 2 a; + 2 x + 3 x + 3
Hence 4 + -1-+ 1 +-l- = 2 + -l-- fx-3--^V^
x + 1 x + 4 x + 2 V x + SJ
■ 1 + 1-1 ■ 1
x+1 x+4 x+2 x+3
. _J 1_^_J. 1_.
**x+l x + 2 x + 3 x + 4'
(X + 2) - (X + 1) ^ (X + 4) - (X + 3)
(x + l)(x + 2) (x + 4)(x + 3) '
that is
(x + l)(x + 2) (x + 4)(x + 3)
.-. (x+l)(x + 2) = (x + 4)(x + 3),
that is x2 + 3x + 2 = x2 + 7 X + 12 ;
.'. x2 + 3x-x2-7x=12-2;
.-. -4x = 10;
10 5
2
EXAMPLES L.
1 a? — 2 X — 3_x— 7 „ X x — 2_x o^ — 3
6 4~10' *6 5~5 4*
o3x — 1 oa^-l_2 — X ^ x-2 gX-l,. ^x-2
11 6 10 3 5 6
EQUATIONS WITH FRACTIONS. 219
g l-3a; 3x + l^ 2 ^ S^ 5_J ]_ ^ ^
2 2 l-3x' * 2x + 2 4ic + 3 4x
g 3-4a; 5-8x_l-a; ^^ 2x- 5_2x-7
6 12 1 +x ' 3x-7 3x-5*
„ Sx-1 o«+l_o 3-6x „„ 6x-2 3x+r
7. — z- — -—z — ad. - =
4 X — 3 8
8 2x 3^ — 1 — 3 1— 4x rto
6 X + 1 10
1.1 2
8x
-5
4x+8
X-
-9
2x-
-5
3x
-7
6x-
-4
3-
2x
2x-
-7^
10.
4x + 6 6x + 4 2x + 3 x-5 4-x
1 I 2^2. gg a;+l x-3^8
3x + 9 5x+l a; + 3 'x-1 x4-3 x
J J ^^ 3 _ 4 gg x + 2 a;-2_ 8
4x+12 4x+l 2x+6* 'x-2 x + 2~x+l'
12. _A_ + _5_ = _J_. 27. ^^-^±2 + 10 = 0.
2x + 3 4x + 6 6x + 8 x+3 x-4 x
13. _i^ ^ = 3. 28. -^-3^^+ ^ =0.
x+lx-2 x + 2 x-2 x+1
14. ^ ^ = 2. 29. 3^^ + 2^+1 = 5.
X + 6X + 6 x + 1 x-1
15. _6x x_^5^ 3Q 5^^:i2_2X^^3^
X-7X-6 x+2 x+3
16. ^^ i^+2=0. 31. -l^ = -i L..
X + 3X+7 x3-lx-4x + 4
17^ 1 I 2_3 o«12x,2_2
18.
x + 4 x + 6 x + 5 x3-9 x + 3 x-3
_3 2_^^_. 33 4 1^1
x+1 x + 2 x + 3 *x2-l x+1 1-x
19. ^-i §-J_ = -i_. 34. ^ • 1 - 2
2x + 4 2x + 2 x + 6 x2-9 x + 3 3
220 EQUATIONS WITH FRACTIONS.
OK 2a; + l 8 2a^-l
2a^-l 4x2-1 l + 2aj
36.
3x + 5 5 _8 + 3cc
3x-l 1-9x2 l + 3x*
?7
1 3 _ 5
x-5 2x-6 (x-3)(x-6)
38
1 3 _ 7x+l
2x-4 x-5 (x-2)(x-6)
39.
^ + 1^ + 4 =0.
2-3x 6-6x 10 + x
40.
« 1 ^ 1 -0.
3x-l 3-7x 5 + x
41.
1111
x+5 x+6 x+6 x+8
42.
1111
x+7 x+1 x-\-l x + 3
43.
1 + 1 _ 1 + 1
x + 2 x+10 x + 4 x + 8
44.
1111
x-2 x-6 x-4 x-8
45.
1 + 1 = ^ + 1.
x-5 x+2 x-4 x+1
46.
X x-9_x+lx-8
X — 2 x-7 X — 1 X — 6
47. 2x+l ^ 2x + 9^2x + 3 2x + 7_
*x+l x + 5 x + 2 x + 4
2x-3 2x-4 2x-7 2x-8
48.
2x-4 2x-5 2x-8 2x-9
49 »^ + 7 x + 9 _ x + 6 ^ x + 10
x+6 x+7 x+4'x+8'
EQUATIONS WITH FKACTIONS. 221
gQ 16 a; - 13 40 x - 43 ^ 32 a; - 30 20 x - 24
' 4x-3 8x-9 8x-7 4x-5*
51 a;-7 x-8^2x-7 2x-ll
x-5 x-6~2x-5 2x-9*
7 11 7 11
53.
x-9 x-4 x+2 x+3
16 _ 21 ^ 16 _ 21
x-17 x-9 x + 4 x + 7*
^ x-\-a j a^ + ^_2.
* X— 6 X — a
55 ^ + <^ _ X — Z) _ 2(g + 6)
' x—a x+b X
-fl ax . 6x ^ , 7,
66. 1 = a + 0.
a + x 6 + X
-- q + c , 6 + c _ q + 6 + 2c
' x + 26 x + 2a x + a + 6 *
68 ''^~ ^ _ ^ — <"' — 2(a — 6)
x — a X — b X — a — b
172. Simultaneous Equations. Any pair of equations of
the first degree in two unknown quantities can be reduced
to the form
ax-{-by + c = Of
a'X'{-b'y-{-c'=0,
and any formula that completely solves them solves also
every pair of equations of this class. We shall now
derive such a formula, applying for this purpose the
method of elimination by undetermined multipliers.
[Art. 106.]
To solve the simultaneous equations
ax + &?/ + c = 0,
a'x + b'y + c' = 0.
222 EQUATIONS WITH FRACTIONS.
Add the first equation to k times the second, thus :
(a + ka')x + (6 + kh')y + c + A:c' = 0, . . . (i.)
and make h + kh' = Q,
for the purpose of eliminating y. Then
k = -^,
b'
and, substituting this value of k in equation (i.), we obtain
(«-^«')
x-\.c--'C' = 0
h'
whence x =( — c + — • c' j /( a 1 ' ^'\
Again, for the purpose of eUminating x, make
^ a^ka' = 0.
Then lc = -^^
a
and, substituting this value of k in equation (i.), we obtain
\ a' I a'
whence y = (-c + ^^- c<\/ lb -^'h'\'
When simplified, these fractional expressions for the values of
X and y have the form
x = (bc> -b'c)/{ab' -a'b),
y =(ca' —c'd)/{ab' — a'b).
Observe that either one of them may be derived from the other
by interchanging a with b and a' with b'. Thus the principle of
symmetry here finds an application to elimination ; and it fre-
quently happens that by taking advantage of this principle, one-half
or two-thirds of the work of elimination by the ordinary process is
avoided.
EQUATIONS WITH FRACTIONS. 223
173. A convenient device in elimination, sometimes
called the rule of cross-multiplioation, is suggested by the
cyclo-symmetry observable in the equations
y = (ca'-c'a)/(ab'-a'b)y
when they are written in the form
y
6c' — b'c ab' — a'6' ca' — c'a ab' — a'b'
or more concisely, in the form
a; _ y _ 1
be' — b'c ca' — c'a ab' — a'b
For the purpose of making the symmetry of the
expressions more apparent, c is changed to cz and c' to
c'z, in the two equations to be operated upon, which then
become
a>x -{■ by -\- cz = 0,
a'x 4- b'y + c'z = 0 ;
and the process of elimination (in any of its forms)
gives, not now the values of x and 2/,,but of the fractions
x/z and y/z, in the form already written above, namely,
a? _ y _ 2
be' — b'c ca' — c'a ab' — a'b
in which z has now taken the place of 1 in the third
numerator.
The rule of elimination by cross-multiplication may
now be stated as follows : Write three fractions with x,
y, and z for their respective numerators. Under x write
the difference of cross-products be' — b'c, formed from
224
EQUATIONS WITH FRACTIONS.
the coefficients of the terms that do not contain a;, and
under y and z respectively write the similar differences
ca' — c^a and ab' — a'b, observing the cyclic order in the
displacement of the letters. By this simple rule the
values of x/z and y/z are written down by inspection.
These cross-product differences are sometimes exhibited
in the form of square matrices, which are called determi-
nants, thus :
b
c
c
a
a
b
b'
c'
>
c'
a'
J
a'
b'
and when so written they suggest at once the rule of
cross-multiplication. For example,
^ - =bc'- b'c,
and the other two similar identities are derived from
this one by a cyclic displacement of the letters.
The method here used is applicable to the solution of
simultaneous equations with any number of unknown
quantities. [See Treatise on Algebra, Arts. 145, 432.]
Ex. 1. Write, by the rule of cross-multiplication, the values of
x/z and y/z which satisfy the simultaneous equations
Sx + 6y-\-2z^0,
2x + 1y + 4z = 0.
Solution
or
X
y
=
z
20
-
14
4-
12
21-
lo'
X _
6~
y
-8
=
z
11
)
••
X _
z
6
y
z
= -
8^
"ll*
EQUATIONS WITH FRACTIONS. 225
Ex. 2. Write, by the rule of cross-multiplication, the values of
x/z and y/z which satisfy the simultaneous equations
2a; + 72/-42! = 0.
Solution :
_ y -
or
20-14 4 + 12 21 + 10
? = ^ = — •
6 16 31 '
'*' z~Z\ z~Vl
Ex. 3. Determine the relations that must exist between the
coefficients of ax^ + ftx + c and a'x^ + 6'x + c', in order that the
equations
ax2 + 6x + c = 0,
a'x2 + 6'x + c' = 0,
may have a common root.
By a common root is meant a value of x which satisfies both
equations. Hence, if we assign to x, for the moment, the particular
value in question, the two equations become simultaneous in x
and x^.
Then, by the rule of cross-multiplication, we have
6c' — 6'c (M^ — da ah' — a'b '
and, placing the product of the first and third of these fractions
equal to the square of the second, we obtain
(be' - b'c) {ab' - a'b) {ca' - c'a)'^
whence, after dividing by x^ and inverting,
(be' - b'c)(ab' - a'b) = (ca' - c'aY,
which is the relation sought.
226 EQUATIONS WITH FRACTIONS.
174. When a set of equations can be conveniently
written in terms of a series of equal fractions, Theorem
I. of page 208 provides still another useful method of
elimination.
Ex. 1. Solve the simultaneous equations
y + z _ z + x _ x + y _ j^
a b c '
By Theorem I., page 208, we have
— y - z + z + x + x + y _ 2x _ j^
— a + 6 + c — a + b + c
.'. x = lk(-a + b + c);
and by a cyclic displacement of the letters
y = lkia-b + c),
z = ^k{a + b - c).
Ex. 2. Solve the simultaneous equations
a b c
By Theorem I., page 208, each of these fractions is equal to
a-\- b + c
and the original equations become
a b c ^,
Putting ^(a + & + c) in plac*"" c k in the results of Ex. 1, we
have
a: = i(a + 6 + c) (- a + 6 + c),
2^= i(a + 6 + c)(a-6 + c),
0= i(a + 6 + c)(a + 6-c). •
I
EQUATIONS WITH FRACTIONS. 227
Ex. 3. Solve the simultaneous equations
y-\-Sz_z + x_Sx + 5y_ ^
4 ~ 5 ~ 2
By Theorem I., page 208, we have
6y + l^z — 16z — lBx — Sx — by_ ^
20-75-2 '
.-. -18x = -57, x = f|;
y + Zz-Sz-Sx + Bz + 6y_^
4-15 + 2
.-. 6y = -9, y = -i;
6y + 15g + 3g + 3a;-3a;-6y_ ^
20 + 15-2
.-. 180 = 33, z = ^.
EXAMPLES LI.
Solve the following simultaneous equations, obtaining one set of
values of the unknown quantities in each case.
- 1 1 X ?/ 2 2
^ a h a h a^ y^
2.
X y_-^, __y a; _ 1
a + b a — b ab a -{• b a— b ab
3.
a b
:1, f + ^ =
6 a
1.
4.
X
x-\-\
-.-It'
1
x + l
6.
X
■-^-1,
.: -\-y = a
+ &.
x-a y
6. ? + I^ = l, ^ + i^ = l.
a b c a' b' c'
7. _ + - = c, — + - = c'-
X y x y
228 EQUATIONS WITH FRACTIONS.
X Sy xy
X y ' y z ' g^ic
10. 1 + 1 = 7, 1 + 1 = 9, 1 + 1 = 12.
y z z X X y
11. '^ = y = ^ = (x + y + zy-
a 0 c
12 3y-^-ig-3a;_a;+y_2
2 3 5'
a 6 c
J, ^ cy + &^ _ az + ca; _ 6a; + ay _ i. *
Z m n
15. Pind two sets of values for aj, ?/, 0, that satisfy the equations
X _y _z _ 1
a & c x + y + z
16. Find three sets of values for x, y, z, that satisfy the equa-
tions
^ = y=^ = xyz.
a b G
17. Find a value of c that will make the equations
2a;2-a: + c=:0, 4a;2+4a: + l = 0
have a common root, and then determine the common root.
18. Determine two values of a, either of which will make the
equations
a;-2 + 2 a; - 3= 0, aa;2 - 11 X + 10 = 0
have a common root, and find the common root in each case.
* See Ex. 3, of Art. 170.
MISCELLANEOUS EXAMPLES III. 229
MISCELLANEOUS EXAMPLES III.
A. 1. What must be added to x^ — 3 x^y + 3 xy^ that the sum
may be
Sx^y-Sxy^ + y^?
2. Show that
x(y -i- zy-^ y (z + xy + z (x + yy - 4 xyz = (y+z)(z + x) (x + y).
3. Find the factors of (l) x^ - y^, (ii.) a^ -b^ -\- 2a -2b, and
(iii.) x^-Sx-2S.
4. Find the H. C. F. of x* + 9a; - 20 and 5a;* + 9x^ - 64.
Simplify ^^ + ^^-^Q .
6. Solve
(ii.) Sx-4y-\-2z=5x-6y-2=7x-\-2y-\-i.
(iii.) -JL-4. ^ =^L±A.
x + b x + a X
6. If A were to give $ 10 to B, he would then have three times
as much as B ; but if B were to give $ 5 to A, A would have four
times as much as B. How much has each ?
B. 1. Find the value of
a (a + 6) (a + 6+ c) — a (a - 6) (a — 6 — c),
when a = 5, 6 = 3, and c = — 6 ; also when
a = — 3, 6 = — 2, and c = 4.
2. Show that a^ + 3(a - 2 6)2 = S(a-by + (a-S by, and that
8. Find the factors of (i.) x^ -4:X^ + ix, (ii.) 2 a;2 _ 6 -^ + 2, and
(iii.) asS - a;2 + a; - 1.
230 MISCELLANEOUS EXAMPLES III.
lift
4. Simplify -— + ---^--^-^, and show that
\ y-x /\ y -X J \ y -X J
y -X J \ y
6. Solve
X
= 1
x—1 X — 2 X— 3 x.y_a
ha b _
6. Find the fraction which becomes ^ when 2 is added to its
numerator, and ^ when 2 is taken from its denominator.
C. 1. Prove that
(a;- 1)2(2/2 + i)_(a;2 + l)(2/ _ 1)2 = 2(x - y){xy - 1),
and that
{X + y) (a; + 2) - x2 = (y + z) (2/ + x) - 2/2 = (0 + x) (^ + y) - ^2.
2. Prove that the product of any two numbers is equal to a
quarter of the difference of the square of their sum and the square
of their difference.
3. Find the factors of (i.) 1 + 18 cc - 63 a;2, (ii.) 3 a:^?/ - 24 y*,
and (iii.) (x^ ^ 2x)^ -{Bx^ + iy.
4. Simplify ^^-^Q^ + ^\
and 1 ^-4 ' ^-^
(x-3)(a;-2) (a;-l)(x-3) (x-l)(a-2)
6. Solve
^' \ x — S x — l I x + 1 x — S_f^
00-^ i~'^"6 8~-^'
(ii.) ax + by = c, a2ic + b^y = c^.
6. A and B have $ 100 between them ; but if A were to lose one-
half of his money, and B were to lose one-third of his, they would
then have only $ 55 between them. How much has each ?
MISCELLANEOUS EXAMPLES III. 231
D. 1. Find the value of Va^ -{- b^ + c^ -{a - b - c)2, when
a = 3, 6 = — 3, and c = 4.
2. Show that
(x-5)(x-4)-3(sc-2)(a;-l)+3(x+l)(a;+2)-(x+4)(a;+5)=0,
and that
1 + n + Jn(n + 1)+ in(« + l)(n + 2) = Kw + l)(n + 2)(n + 3).
3. Find the factors of (i.) x^ + x^y - 6 xy% (ii. ) x^ + a5c2 - a^a; - a',
and (iii.) xV _ a;2 - y2 _|. i.
4. Simplify
a ^ ^^ ~ ^ I 4 ax , g + a;
a + X a^ _ a;2 ^ - x
^ ^ x-4 x-5 x-6
6. Solve
(i.) 3x+i_2(6-x) = ^^^-^^.
(ii.) 3y + ? + 6 = 0, y-]-- = S.
X X
6. A hare is pursued by a greyhound, and is 60 of her own leaps
before him. The hare takes 3 leaps in the time that the greyhound
takes 2 ; but the greyhound goes as far in 3 leaps as the hare does
in 7. In how many leaps will the greyhound catch the hare ?
E. 1. Show that
(m + n)(^m + n — 1) = w(m — 1)+ 2mn + n(n — 1),
and that
(m + n) (m + n — 1) (m + n — 2) = m(m — 1) (m — 2)
+ 3wi(m- l)n + 3mn{n - 1)+ n{n - l)(n-2).
2. Divide x* — y* by x — y, and from the result icrite down the
quotient when (x + yy — IQz^ is divided hy x + y ~2z.
232 MISCELLANEOUS EXAMPLES in.
3. Find the factors of
5 x2 + 24 oj - 6, and of a^ - 2 ahc - ah"^ - m'^.
4. Find the L. C. M. of 8 a^ _ is aV^, 8 a^ + 8 a^d - 6a62, and
4 a2 - 8 a6 + 3 6-2.
6. Show that
(i.)l4--^+ ^^ - ^'
X — a {x — a){x — h) {x — a){x — h)
(ii.) !+_«-+ "^
X — a {x — a){x — h) {x — a){x — 6) (x — c)
^ ^
{x — a)(x — b) (x — c)
6. A debtor is just able to pay his creditors 25 cents on the
dollar ; but if his assets had been five times as great, and his debts
two-thirds of what they really were, he would have had a balance
in his favor of $ 700. How much did he owe ?
F. 1. Find the value of
{a~{b- c)}2 + {6 -(c - a)}2 + {c-(a- b)f
when a = — 1, 6 = — 3, and c = — 5.
2. Divide x^ - 2 a*x* + a^ by x^ + ax^ + a^x + a^.
3. Find the L. C. M. of x^ - 1, (x - 1)2, (x + 1)2, x» - 1, and
xS+1.
4. Add together » > and — : and show
2x + 2/ 2x-y y2_4a;2'
that
(x -a)(y- a) (x-b)(y-b) (x - c) (y - c) ^
(a - 6)(a - c) "^ (6 - c)(6 - a) "^ (c - a)(c - 6)
5. Solve (i.) -^—+ ^ ^
x + a x + 6 x + a + 6
(ii.) 5x + 22/ + 3^ = 13j
Sx + 7y-z= 2 ^«
X — 2y + 2= sJ
MISCELLANEOUS EXAMPLES III. 233
6. When the arable land of a farm was let at $ 7.50, and the
pasture at $10 an acre, the total rent of a farm was $2750.
When the rent of the pasture was reduced by $ 1.25 an acre, and
that of the arable land by $2.50 an acre, the whole rent was
$ 1937.50. What was the total acreage of the farm ?
G. 1. Solve the simultaneous equations
X — a . y — h _ ^
c a + c — b '
x-\- a — c . y — a_
a + c a — b a + c
2. Solve the simultaneous equations
yJ:S-a=^±^-b='^^±y-c = x + y + z.
a b c
3. Solve the simultaneous equations
^/y + c/g _ c/z + a/x _ a/x + b/y _a . b . c ^
a b c X y z '
4. A man rides one-half of the distance from A to B at the rate
of a miles an hour, and the other half at the rate of b miles an hour.
If he had travelled at the uniform rate of c miles an hour, he would
have ridden the whole distance in the same time. Prove that
2 = 1+1.
cab
234 QUADRATIC EQUATIONS.
CHAPTER XVI.
Quadratic Equations.
175. Let it be required to state the following problem
in the form of an equation.
If the speed of a certain train were increased by 4
miles per hour, the time it now requires to make a trip
of 120 miles would be decreased by one hour. What is
the train's speed in miles per hour ?
Let X = the speed of the train in miles per hour. Then
120/ic = the number of hours the train now requires
for the trip of 120 miles ;
120/(a;-|-4) = the number of hours it would require
for the same trip, were its speed increased as indicated,
and by the conditions of the problem this is one hour
less than 120/a; hours ;
120 120
X-i-4:
■M.
Multiplying both sides of this equation byic(a;+4),
we have
120 a; + 480 = 120^ -{-3^ + 4:X,
whence ic^ + 4 ic = 480,
The problem of finding the speed of the train is thus
reduced to that of solving an equation which contains
both the first and second powers of its unknown quantity.
QUADRATIC EQUATIONS. 235
Some equations of this class were considered in Arts.
128, 129. We now proceed to a more complete analysis
of them.
176. A quadratic equation is an equation which contains
the unknown quantity to the second, but to no higher
power.' Thus x^ = 4:, 3a?'+4a;=7, and ax^ + 5rc + c = 0
are quadratic equations.
When all the terms are transposed to one side, a quad-
ratic equation must be of the form
ax^ -{- hx -\- c = 0,
where a,' b, c are supposed to represent known quantities.
It was shown in Art. 122 how the factors of the
expression on the left side of this equation can be found
by chayiging the expression into an equivalent one which is
the difference of two squares.
We first apply the method to some examples ; but the student
will profit by reading Art. 122 again before considering them.
Ex. 1. Solve x^ + 12 X + 35 = 0.
Since x^ + 12 a; is made a perfect square by the addition of the
square of half the coefficient of x, that is the square of 6, we add
and subtract 62. The equation then becomes
x2 + 12 X + 62 - 62 + 35 = 0,
that is (x + 6)2 - 1 = 0 ;
... ((x+6)+l}{(x4-6)-l} = 0,
that is (x + 7) (x 4- 6) = 0.
Hence x + 7 = 0, or x + 5 = 0.
Thus X = — 7, or X = — 5.
Ex. 2. Solve 3 x2 = 10 X - 3.
By transposition we have
3x2-10x + 3 = 0.
236 QUADRATIC EQUATIONS.
We must first divide by 3 in order to make the coefficient of y?
unity ; the equation then becomes
Half the coefficient of a; is — f , and therefore x^ — J^O- x will be
made a perfect square by the addition of (— |)2 ; and adding and
subtracting (— 1)^, that is ^-^ the equation becomes
... (X- 1)2 -1^ = 0,
that is (x_ 1)2 -(4)2^0;
... (a;_5 + |||^_5_||^0;
... (x-i)(x-3)=0. . .,
Hence x — | = 0, or oj — 3 = 0.
Thus ic = ^, or a; = 3.
Note. — In many cases the factors can be written down at once,
as in Arts. 117 and 118, without completing the square. The
student should always see if this can be done : he will thus save
himself a great deal of unnecessary work.
Ex. 3. Solve 4 ic - x2 = 2.
By transposition, we have
4x-ic2_2 = 0.
Change all the signs so as to make the coefficient of x'^ unity ;
then we have
cc2- 4ic + 2 = 0.
Add and subtract 4, the square of half the coefficient of x. Then
x2-4x + 4-4 + 2=0;
.-. (x- 2)2 -2 = 0;
that is (ic - 2)2 - ( ^2)2 = 0 ;
... (x-2+V2)(x-2-V2) = 0.
Hence x - 2 + ^2 = 0, or x - 2 - ^2 = 0.
Thus X = 2 - v/2, or X = 2 + V2.
QUADRATIC EQUATIONS. 287
EXAMPLES LIT.
Solve the following equations :
1. a;^-4x = 45. 22. 17aj2 + 8 = 70a;.
2. x^-6x = 91. 23. 51x2 4-19z = 10.
3. (a;-3)2 = x + 3. 24. 110x2 - 21a; + 1 = 0.
4. (x-4y = x-2. 25. 21x2 -13 a; = 20.
6. (x-l)(x-2) =20. 26. 21x2 + 23 x = 20.
6. (x + l)(x + 3) = 2(x + 3). 27. 6x2 + 6 = 13x.
7. 4x + 3 = x(x + 2). 28. 6x2 = 5x+l.
.8. 4x-3 = x(2-x). 29. 9x2 - 63x + 68 = 0.
9. (x+l)8=(x-l)8 + 26. 30. 16x2 + 3= 16x.
10. (x-l)8 = (x + l)8-56. 31. 29x2 -41x- 138 = 0.
11. 9x2 + 6x + l=0. 32. 29x2 + 11 x = 138.
12. 9x2 + 16 = 24x. 33. x2 - 16 = 216 - lOx.
13. x2+150 = 53x. 34. (x + 2)2 = 4(x - 1)2.
14. x2 = 2 X + 99. 86. (x + 6)2 = 16(x - 6)2.
16. 3x2 + 3 = 10x. 86. (x + 8)2 = 9x2.
16. 3x2 + 11 X = 20. 87. (x - 7)2 = 49(x + 2)2.
17. 4x2+21x = 18. 88. x2 - 3ax + 2a2 = 0.
18. 6x2 + 65x = 50. 89. x2 + 3a2 = 4ax.
19. 24x2 = 30x+75. 40. x2 + 2a6 = 62 + 2ax.
20. x2 - 200 = 35x. 41. 4 x2 + 4 ax = 62 _ a^
21. 19x2-39x + 2 = 0. „ 42. x2 + 2ax = 62 + 2a6.
177. To solve the equation
ax^ -\-bx-{- c = 0.
Divide by a, the coefficient of ar'; the equation then
becomes
238 QUADRATIC EQUATIONS.
Now add and subtract the square of half the coefficient
of X, that is the square of - — Then we have
Z a
a \2aJ \2aJ a
The first three terms are now a perfect square, namely,
Hence we have (x -\ ) — ( — - — - )= 0,
V 2aJ \4:a^ a)
Hence [Art. 115]
We therefore have
2a 2a 2a
b -^{W-^ac) _~h-^{W-4.a6)
or flJ = — pr pr — ^ •
2a 2a Za -
It follows from the above that a quadratic equation has
two and only two roots.
QUADRATIC EQUATIONS. 239
178. Instead of working out every example from the
beginning, we may use the general formula found in
Art. 177, and substitute for a, 6, and c their values in the
particular equation we are considering.
Thus, the roots of the equation ax^ + bx + c = 0 being
2a 2a
we find the roots of 3 ic^ - 10 x + 3 = 0 by putting 3 for o, - 10
for 6, and 3 for c in the above formula.
Hence the roots of 3 aj^ — 10 x + 3 = 0 are
-10 V(100-36)
6 6 '
that is \° ± I ; and hence the roots are 3 and \.
179. General Properties. In discussing the quadratic
equation aa^ -\- bx -\- c = 0, it is assumed, unless a contra-
dictory hypothesis be made, that the coefficients a, b, c
are finite and different from zero ; and the general prop-
erties of this equation are enunciated under this limi-
tation.
In Art. 177 we found that the equation ax^ -f 5x -f c = 0
had two roots, namely,
b v(&^-4ac) b V(6^-4ac)
2a^ 2a 2a 2a
(i.) If b^ — Aac be positive, \^(b^ — 4ac) is the square
root of a positive quantity and has a numerical value,
either rational in the form of an exact square root, or
irrational in the form of an indicated square root. The
quantity -y/(6* — 4ac) is here said to be real. The roots
of the quadratic equation are then also real, and they
are obviously different from one another. Also, unless
240 QUADRATIC EQUATIONS.
6^ — 4ac be positive, V(^^ — 4ac) has no numerical value
and is not real. Hence :
In order that the roots of ax^ + 6a; + c =0 may he real, it
is necessary and sufficient that 6^ > 4 ac.
From the formula of Art. 177 it is clear that the roots
of the equation aa? -\- hx -\- c = 0 are irrational unless
6^ — 4 ac is a perfect square.
It is also clear that if either of the roots of a quad-
ratic equation be irrational, they are both irrational.
Let the student here call to mind the meaning of the
expression irrational quantity. It is a quantity whose
exact value cannot be expressed either as an integer, or
as a fraction whose numerator and denominator are
integers. In algebra it is usually expressed symbolically,
as in the case under consideration by means of the sign
V- [See Art. 20.]
(ii.) If 6^ — 4ac = 0, both roots reduce to — h/2a, and
are thus equal to one another. In this case we do not
say that the equation has only one root, but that it has
two equal roots. Also, the roots are unequal unless
6^ — 4ac = 0. Hence:
In order that the roots of ao? -\-hx-{-c = 0 may he equal
to one another, it is necessary and sufficient that 6^ — 4 ac = 0.
When 6^ — 4 ac = 0, the expression ax^ -[-hx-\-c is a
perfect square so far as x is concerned; for ax^-\-hx-{- c,
that is, a(a^ + -x-{--\ will then be a(x-\ ), since
^, ^ \ a aj \ 2aJ
4a^ a
(iii.) When &^ — 4ac is negative, y(6^ — 4ac) is the
indicated square root of a negative quantity, and has no
<^
QUADRATIC EQUATIONS. 241
numerical value. It is therefore excluded from the
category of so-called real quantities, and is called an
imaginary quantity. [Art. 228.] Hence :
In order that the roots of ax^ + 6a; + c = 0 may he imagi-
nary, it is necessary and sufficient that h^<,^ac. '
It should be noticed that if either of the roots of a
.quadratic equation be imaginary, they are both imaginary.
180. The above criteria of the character of the roots
of a quadratic equation are more concisely stated thus :
(1) If 6^ — 4ac be positive, the roots are real and
unequal.
(2) If 6^ — 4ac = 0, the roots are equal.
(3) If 6^ — 4ac be negative, the roots are imaginary*
and unequal.
(4) If 6^ — 4ac be a perfect square, the roots are
rational and unequal.
(5) If 6* — 4 cic be not a perfect square, the roots are
irrational and unequal.
181. By multiplying both sides of an equation by the
same quantity, we do not destroy the equality, and
therefore the new equation is satisfied by all the values
which satisfy the original equation; if, however, we
multiply by any integral expression which contains the
unknown quantity, the new equation will be satisfied not
only by the values which satisfy the original equation,
* Complex quantity is a more general and more accurate name for
expressions of the form in-\- V^n. The^term imaginary is then ap-
plied only to expressions of the form V— n. [See Chapter XXII.]
Q
242 QUADRATIC EQUATIONS.
but also by any value which, makes the expression by
which we multiply vanish.
For example, the equation x^ = 9 is satisfied by the values
X = 3, or x = — 3. If we multiply both sides of the equation by
aj — 2, we have the new equation
and this new equation is satisfied not only by x = 3, and by
X = — 3, but also by X = 2,
Thus additional roots are introduced whenever both
sides of an integral equation are multiplied by any inte-
gral expression which contains the unknown quantity.
When an equation contains fractions in whose de-
nominators the unknown quantity occurs, the equation
may, however, be multiplied by the L. C. M. of the
denominators without introducing any additional roots ;
for we cannot divide both sides of the resulting equation
by any one of the factors of the L. C. M. without re-intro-
ducing fractions.
The student must, however, be careful to multiply by
the lowest common multiple of the denominators, for
otherwise the resulting equations will have roots which
are not roots of the original equation.
We may, for example, multiply both sides of the equation
2x 10 ^ 7
x-1 x2-l x-t-1
by x2 — 1, the L. C. M. of the denominators of the fractions ; we
thus obtain
2x(x + l)-10 = 7(x-l),
and this new equation is not satisfied by either of the values
obtained by equating to zero the factor by which we have multi-
QUADRATIC EQUATIONS. 243
plied, and hence no additional roots have been introduced by the
multiplication.
If, however, we multiply the original equation by (x-l)(x2 — 1),
we shall have
2x(x+l)(x-l)-10(x-l)=7(x-l)2, .
and one of the roots of this equation, namely x = 1, is not a root of
the original equation.
Note. — The student should always remember that when both
sides of an equation are divided by any factor ichich contains the
unknown quantity^ the resulting equation will not give all the
roots of the original equation ; to find the remaining roots we must
equate to zero the factor by which we have divided.
For example, if we divide both sides of the equation
(x2 - 4)(x + 2) = (x2 - 4)(2x + 1)
by x2 — 4, we obtain the equation
x + 2 = 2x-l,
from which we obtain x = 3. But x = 3 is not the only solution of
the original equation ; to obtain the other solutions we must equate
x2 _ 4 to zero.
EXAMPLES LIII.
Solve the following equations :
X. ^.H
g x+lx + 2_29
• X + 2 X + 1 10
- -i=f-
7 1 2 _3
* x-2 x + 2 5
2 a; 3
8 ' 1 ^ -^
x+4'4-x 3
. x+5 4 .
' 4 X + &
_3,
2*
9. \4- ^=3.
x-2 x+1
6. \+==-V
X- 1 X
13
6*
10 5 10 _ 2
'x + 1 x + 10 3x-3
244 QUADRATIC EQUATIONS.
11.
i+ 1 + 1 =0.
2 3 + X 2 + 3x
17.
2x-l 13_3x + 5
2x+l 11 3x-5
12.
1+ ^ + ^ -0.
3^j» + 3 x + n
18.
2x-2 3-3x_ 5
2x-3 3x-2 8x-12
13.
.^x-1 2^+1 = 5.
x+l x-1
19.
3x 4 4 _o^
x-2 x+3 2-x
14.
3 4 _ 15
x-1 x-3 x+3
20.
5x 6 19 _Q
x-3 x+2 3-x
16.
4x + 3_6x-13 7x-46
9 18 5x + 3
21.
x-lx+l_ 5x
x+l x-1 X2-1
16.
2x-3_x-5 5X-16
4 12 ' x-1
22.
1 + 1 = ^•
x2-3x x2 + 4x 2x2
A
23. ^
x2-3x 9
J
■X2"
13
16 X
«1 1 I
5
J_
7X _Q
/^- Sx-eU
-
x^'
72(x + 2)
25.
1 17 1
27.
1+ 1 = ^ + ^ •
X x+4 x+l x+2
x2-l 1-x 8 x+l
26.
1 3^1
28.
X x+l_x-2 x-1
x+l x+2 X— 1 X
a.2__4 2-x '3x + 6
09 ^ a:-
3
_{_
X x+3_2
* x-3 X
X
+ 3x3
80.^ + ^ + ^-
X+l X —
6.
4
X-
+ 4 x-5
X-
+ 2 x-3
31.^-3 X-
J
_x
-6 x-7
X — 4x — 5x — 7x — 8
32. x2 + 2 a2 := 3 ax. 35. a(x2 + 1) = x(a2 + i).
33. 9 x2 - 6 ax = a2 - &2. 36. a;2 - 2(a - 6)x + &2 = 2 ab.
34. a(x2 - 1) + x(a2 - 1) = 0. 37. x2 + 2(6 - c)x + c2 = 2 be.
38. (b - c)x2 + (c - a)x + (a - 6) = 0.
QUADRATIC EQUATIONS. 245
39. (a + 6)x2 + c«-a-6-c=0. 41. fecx^ + (62 + c2)x + 6c = 0.
40. a6a;2-(a2 + 62)a; + a6 = 0. 42. (a'^ - b^) {x^ - 1) = 4: abx.
43. (62 - o2)(a:2 + 1) = 2(a2 + 62)a;.
44. x + - = a-\-^
a X
49. tl^ + -=- + ,
a X a 0
45. _^ + _^ = 4. 50. 1_- = 1 + 1 + 1.
a+x^a-x x+a+b x a b
46 _l_ + _i_ = l + l. 61. __L_ = 1-1_1.
■ x-a x-b a b x-a-b x a b
47. _«L. + _^ = « + ^. 52. J-4-l-=-i-+-^
x-a x-b b a x+a x+6 c+a c+6
48. ^i- + ^^ = a + 6. 53. -£-+-^=-^+-^.
x-6 X — a x+a ic+6 c-\-a c-{-u
g. x + q ■ x + b _x — a . x-b
' x — a x — b x + a x + b
65. -J^ + -^= ^-^ + ^-^^ ♦
x + a x+6 x+a-c x+6+c
182. Irrational Equations. An irrational equation is
one in which, square or other roots of expressions' con-
taining the unknown quantity occur.
In order to rationalize an equation it is first written
with one of the irrational terms standing by itself on
one side of the sign of equality : both sides are then raised
to the lowest power necessary to rationalize the isolated
term ; and the process is repeated as often as may be
necessary. J
Ex. 1. Solve the equation v^x2 _ 9)+x = 9.
By transposition, we have
y'(x2-9) = 9-a;.
246 QUADRATIC EQUATIONS.
Square both sides ; then
x2 - 9 = (9 - a;)2 ;
.-. 18x = 90; .-. x = 6.
Ex. 2. Solve the equation
V(2x+8)-2 V(a^ + 6)=2.
Squaring both sides, we have \
2x + 8 + 4(a: + 5)-4v(2a:+8)V(a; + 6) = 4,
that is 6a; + 24-4v(2x + 8)V(a; + 5) =0;
.-. 3a: + 12 = 2V(2a; + 8)V(a;+5).
Square both sides of the last equation, and we have
9x2 + 72x + 144 = 4(2x + 8)(x + 6);
.-. x2-16 = 0.
Hence x = 4, or x = —4.
Note. — If we put 4 for x in the original equation, we should
have ^16 _ 2^9 = 2, that is 4 - 6 = 2,
which is not true.
Again, if we put — 4 for x, we should have
V0-2V1=2,
which is not true.
Thus neither of the values we found for x appears to satisfy the
equation.
The failure is due to our having supposed that V(2x + 8) and
■y/{x + 5) were necessarily positive, whereas every algebraical
quantity has two square roots, one positive and the other negative,
and the square-root symbol does not necessarily represent one
only to the exclusion of the other.
Bearing this in mind we shall find that x = 4 does satisfy the
equation, if the ambiguous signs be inserted ; for the condition is
±y/16- 2(±V9) = 2, that is ±4 -(±6)= 2,
which is true when the lower signs are taken.
QUADRATIC EQUATIONS. 247
If an irrational equation is to be solved on the supposition that
the radicals it contains are to be necessarily positive {or necessarily
negative), the equation will be accompanied by a definite statement
to that effect.
Ex. 3. Solve the equation
V(2x + 7) +^(3x - 18) +y/(7x + 1) = 0.
Since the sum of three positive numbers cannot be zero, this is
not a possible equation unless one or more of the square roots
involved in it be negative.
By transposition, we hkve
V(2 X + 7) -Jv(3 X - 18) = - V(7 a; + 1).
Square both members I then I
2x + 7 + 2V(2a; + 7) V(3a: - 181+ 3a; - 18 = 7x + 1.
By transposition and division by 2, wb have
V(2a: + 7)V(3x-18)=a; + 6.
Square both members ; then
(2a; + 7)(3x-18)=(a; + 6)2;
.-. 6a;2-27a;-162 = 0.
Hence a; = 9, or a; = — y-.
By applying the test for a solution [Art. 91] it will be found
that X = 9 is a solution when ^^(7 x + 1) is taken negatively, and
that X = — 18/5 is a solution when y/{Sx — 18) is taken negatively,
the two remaining square roots being taken positively in each case.
The process of rationalizing equations introduces factors that
contain the unknown quantity. Hence, in solving irrational equa-
tions it is important to apply the test for a solution. [See Art. 181.]
Ex. 4. Solve the equation
-\-^(x^-9)-(x-9)=0,
wherein + y/(x'^ — 9) indicates that only the positive square root is
to be taken.
Multiplying by +y/(x'^ — 9)-{-(x — 9), we have
a;2_9_x2-f 18x- 81 =0;
whence 18 x = 90, and x = 5.
248 QUADRATIC EQUATIONS*
But 5 is not a root of
+ V(^2_9)_(a._9)^0,
and it is a root of the equation
by whose left member the original equation was multiplied. It
may be shown that there is no finite number that will satisfy the
equation
+ V(^"^-9)-(a^-9)=0.
EXAMPLES LIV.
Solve the following equations, taking account of both positive
and negative square roots of the quantities under the radical sign.
Apply the test for a solution in each case.
1. V(^-3) = 3. • 14. y/{9 + ^x)=2x-S.
2. V(»^-7) = 4. 15. Sx=^-{-y/(30x-n).
3. y/(iSx+l)=4. 16. 2 a:- 5 V^- 3.
4. 5v'(x + 7)=:4V(3a;-2). 17. x -\- 3 + ^(x + S) = 6.
6. 3y/(x + Z)=2y/(Sx + 6). 18. 2x + 1 = v'C^ a; + 3).
6. ^(x + 5)=2y/(Sx + 4). 19. 7a: = V(3a;-ll)+33.
7. 3V(aJ + 7)=5v(3x-2). 20. y/(x+ 10) + y/(^x + 1)=1.
8. y/(x + 2) = x. 21. ^x + y/{bx+l)=6.
9. y/{x-\-20) = x. 22. ^x + y/{6x+l) = 2.
10. x + y/(x + l)=6. 23. V(2^ + 9)-V(« + 4)=l.
11. x-y/(x + 2) = 4t. 24. V(7a:+l)-V(3^+10) = l.
12. x-2 + S^(x-2) = 0. 25. V(2a;+ll)-V(2a;-5)=2.
13. x-6 + 2^(x-b)=0. 26. V(4a; + l)-V(a^ + 3) = 2.
27. V(8a^ + 5)-2V(2a: + l)=l.
28. V(a^ + 4) + v'(^+20)-2V(a; + ll) = 0.
29. V(4a;+l)-V(a^ + 3) = >/(»5-2).
QUADRATIC EQUATIONS. 249
80. V(2a; + 4) + v(3x + 7) = V(12a; + 9).
31. V(6aJ + l) + v2(l-x) = V(7a; + 6).
32. 2V(2x+3)-V(5x + l) = V2(x-l).
33. V(x - l)(x - 2) + V(x - 3)(x - 4)=y2.
34. v(a;+l)+_^_^ = 2.
86. y/(ix+l) + ^x ^
V(l+^)
36.. y/iS + X)-\-y/Xz=-^.
y/X
38. V(«-^) + V(^ -a;) = \/(« + ^-2x).
39. y/iax + 62) + ^(6x + a2) = a _ 5.
40. V(« + a;) + V(& + «) = \/(« + ft + 2x).
41. y/(a-x) + y/ib-x) = y/{2a-{-2b').
183. Eelations between the roots and the coefficients of a
quadratic equation.
We have seen [Art. 122] that cc^ -\-px-\-q can always
be expressed in the form {x — a) {x — p).
We have also seen [Art. 128] that if
a^ + px -}- q = {x - a) (x - /3),
then a and /8 are the roots of the equation
a^ -{- px -{- q = 0.
But, if
x"^ -\-px -{-q = {x — a){x — I3) = x^ — {a-^ P)x-{- aft,
we have « -f- /3 = —p and ajS = q.
250 QUADRATIC EQUATIONS.
Hence in the equation
Qi? +pa; + g = 0,
fhe sum of the roots is —p and the product of the roots is q.
The equation aa:^+bx-\-c=0 becomes, on dividing by a,
a a
Hence, from the above, if a and p be the roots of
aa^ + bx -\- e = Oj
b c
we have a-\- 3 = and a3 = -•
a a
The above relations between the roots of a quadratic
equation and the coefficients of the different powers of
the unknown quantity are of great importance.
Analogous relations hold good for equations of the
third and of higher degrees. [See Treatise on Algebra,
Art. 129.]
These relations may also be deduced as follows :
We have found in Art. 177 that the roots of
ax^ + 6a; + c = 0
b ^(b'-4.ac)
are — o— ± o -'
2a 2a
Call these roots a and p respectively ;
then « = __+Vi-^- ^
^ 2a 2a
QUADRATIC EQUATIONS. 251
By addition and by multiplication we then have
a
Again, the quotient of a-{- (3hj ap is
a+^^1 1__6J
a/3 a'^ (3 c'
and observing that
. /I 1\2 2 1 1
we find a^4-/3^ = ^^_2_c = ^!^^,
a^ a or
184. Special Forms. If one or more of its coefficients
vanish, the quadratic equation and the formula for its
roots become simplified.
(i.) If c = 0, the equation reduces to
X {ax -f 6) = 0,
the roots of which are 0 and — b/a.
(ii.) If c = 0 and also 6 = 0, the equation becomes
ax^ = 0, both roots of which are zero.
(iii.) If 6 = 0, the equation reduces to
aa^ + c = 0,
252 QUADRATIC EQUATIONS.
the roots of which are -\-^—c/a and —^—c/a. The
roots of a quadratic equation are therefore equal and
opposite when the coefficient of x is zero.
(iv.) If a, 6, and c are all zero, the equation is clearly
satisfied for all values of x.
(v.) The case in which either a = 0, or a = 0 and
6 = 0 (c not zero), is best studied through the general
expressions for the roots of asi? -\-hx-\- c = 0, but in the
changed form
2c 2c
That these expressions are respectively equal to
_5 4-^(62_4^c) -6-V(62-4ac)
2a ' 2a
follows from the identity
5 _ 5 4- ^(62 - 4ac) IS - 6 - V(^' - 4ac) | = 4ac.
If now a = 0, one root becomes — c/6, and the other
2c/0. And if a = 0 and 6 = 0, both roots become 2c/0.
A fraction, such as 2 c/0, whose numerator is a finite
quantity and whose denominator is zero, was called, in
Art. 168, an infinite quantity, or infinity, and was denoted
by the symbol oo. Thus 2c/0 = oo.
In accordance with this definition and with the results
here obtained, when it becomes necessary to interpret
the equations
0.a;2 + 6a; + c=0,
0.a^ + 0.a; + c = 0,
QUADRATIC EQUATIONS. 263
we say that the former has one infinite root, the latter
two infinite roots. Such equations present themselves
in investigations in analytic geometry.
Ex. 1. Solve the equation
Vax + 1 + Vx + 4 = 3.
Squaring both members, we have
(a + l)x + 5 + 2\/(ax+ l)(x + 4) = 9;
then transposing the rational terms to the right side and squaring
again, we obtain
4(ax + 1) (X + 4) = 16 - 8(a 4- l)x + (a + 1)^x2,
whence, by the proper reductions and transpositions,
(a - 1)2x2 - 12(2 a + l)x = 0.
Hence, one root is x = 0, and for this one both square roots
must be taken positively. A second root is
12(2 g+l)
and it requires \/x + 4 to be taken negatively. ^
If a = 1, the coefficient of x2 is zero, and the second root is
infinite.
Ex. 2. Solve the equation
y/(ax)-{-y/(x + l) = y/b,
and determine for what value of b one root is zero, and for what
value of a one root is infinite.
185. Equations with given Koots. Although we cannot
in all cases find the roots of a given equation, it is very
eas3^ to solve the converse problem, namely, the problem
of finding an equation which has given roots.
For example, to find the equation whose roots are 4 and 5.
We are to find an equation which is satisfied when x = 4, or
254 QUADEATIC EQUATIONS.
when X = 5 ; that is, when x — 4 = 0, or when x — 5 = 0 ; and in
no other case. The equation required must be
(a;_4)(x-5)=0;
for this is an equation which is a true statement when x — 4 = 0, or
when aj — 5 = 0, and in no other case.
If we get rid of the brackets by multiplying out, the equation
becomes
x2 - 9 X + 20 = 0 *
Again, to find the equation whose roots are 2, 3, and — 4.
We have to find an equation which is satisfied when x — 2 = 0,
or when x — 3 = 0, or when x + 4 = 0, and in no other case. The
equation must therefore be
(a;_2)(x-3)(x + 4) = 0,
that is x3 - x2 - 14 X + 24 = 0.
Similarly, the equation whose roots are 0,-1, and — ^ is
x(x+l)(a;+D=0.
that is x8 + fx2+ix = 0.
Ex. 1. Find the equation whose roots are the squares of the
roots of the equation x^ + 5 x — 7 = 0.
Let a, /3 be the roots of the given equation ; the«i a^, ^ will be
the roots of the required equation. Hence the required equation is
(x-a2)(x-/32) = 0,
that is x2 - (a2 + /32)x + a2/32 = 0 (i.)
* The equation a;2 — 9 x + 20 = 0 is certainly an equation with the
proposed roots ; but to prove that it is the only equation with the pro-
posed and with no other roots we must assume that every equation has
a root.
If the equation x^ + 7x^ — 2=^0, for example, had no roots, then
(x — 4) (x — 5) (x5 + 7 x2 — 2) = 0 would be an equation vnth the pro-
posed roots and with no others.
The proof of the proposition that every equation has a root is too
difficult for an elementary book.
QUADRATIC EQUATIONS. 255
We have therefore to find a^ + ^ and a^p^. Now, by Art. 183,
we have
a+/3=-5,
and a)8 = — 7.
Hence a2 + /32 =(a + i3)2 -2a/3 =(- 5)2 - 2(- 7)
= 25 4- 14 = 39 ;
also a2/32 = 49.
Substituting in (i.), we have the required equation, namely
a:2_39x + 49 = 0.
Note. — "We might have obtained the required equation by
finding the roots of the given equation, and squaring them ; but
it is best to use the relations found in Art. 183.
Ex. 2. If a, /3 are the roots of the equation
ax^ -h bx -\- c = 0,
find the equation whose roots are - and ^.
/3 «
The required equation is
(-|){.-!)-.
(i*i}
that is x2-p + tl x + l=0.
Now - + § = ?1±^.,
/S a a/3 '
and, by Art. 183, we have
a/3 = ^anda2 + /32 = ^_2c
a a^ a
... tt^ + /32^62 ^
* * a/3 ac '
Hence the required equation is
«2-(^_2^a; + l = 0,
or, multiplying by ac,
acx'^ - (62 _ 2 ac)x + ac = 0.
256 QUADRATIC EQUATIONS.
Ex. 3. Show that no mimerical value of x will make aj^— 4x+ 5
zero ; and find its least value.
Since a:2 - 4 x + 5 = (a; - 2)2 + 1,
and {x — 2)'^ is always positive, \-\-{x — 2)2 is always greater than
1, except when a; — 2 = 0, and then it is equal to 1.
Thus x2 — 4 a; + 5 can never be zero, and its least value is 1.
EXAMPLES LV.
Write down the equations whose roots are
8. 0 and - 4.
9. 5,-3 and 0.
10. V2 and - V2.
11. ^5 and - y'S.
12. 0, ^3 and - v3.
13. 2 - V3 and 2 + ^3.
14. 5-^7 and 5 + ^7.
15. a —y/h and a + ^6.
16. Write down the product of the roots of each of the follow-
ing equations :
(i.) cc2 + 4x+ 1=0, (iii.) 3a:2 + 5x-7 = 0,
(ii.) ic2 + 7 X - 2 = 0, (iv.) 5x2 - x - 1 = 0, ■
(v.) 9ax2 + 36x + 4c = 0.
17. Write down the sum of the roots of each of the following
equations :
(i.) x2-4a2 = o, (iii.) x2-5x + 3 = 0,
(ii.) x2 + 3x - 5 = 0, (iv.) 2 x2 - x - 7 = 0,
(v.) 6ax2 + 77>x + 8 = 0.
18. Show that, if a, ^ be the roots of ax2 + 6x + c = 0, then
1^1= V(&2-4ac)
o~/3 c
1.
2 and - 2.
2.
3 and - 4.
3.
- 3 and - 2.
4.
\ and — |.
5.
\ and - 1 .
6.
- i and - \.
7.
0 and 3.
^
QUADRATIC EQUATIONS. 257
19. Show that, if a, /3 be the roots of the equation ax^+bx+c=0,
then
(i.) aB + ^3 = ifcftf, (i,) 1^ + i 3«6c_^s,
(iii.) ^+^' = ^1=11^^ + 2.
20. Find the sum of the squares of the roots. of the equation
x2 + 4 a; + 2 = 0.
21. Find the sum of the squares of the roots of the equation
x2 + ipx + p2 _ 0,
22. Show that the sum of the squares of the roots of the equa-
tion x2 + ax + & = 0 is the same as the sum of the squares of the
roots of the equation
x2 + 3ax + b +*4a2 = 0.
23. For what value of a are the roots of the equation
equal to one another ?
24. For what values of a are the roots of the equation
4x2'+(l + a)x+ 1 =0
equal to one another ?
26. Find the value of a in order that one of the roots of
100 x2 + 60 X + a = 0 may be double the other.
26. Show that one of the roots of x^ -{- px -\- q = 0 ia double the
other, if 9^ = 2p2,
27. Show that, if a, /3 be the roots- of 2 x^ - 5 x + 3 = 0, the
equations whose roots are - and ^ is6x2— 13x + 6 = 0.
28. Show that, if a, /3 be the roots of 2x2 — 15x + 4 = 0, the
equations whose roots are - and ^ is 8 x^ — 209 x + 8 = 0.
/3 a '
258 QUADRATIC EQUATIONS.
29. Form the equation whose roots are a + j8 and - + -, where
a ^
a, jS are the roots of x2 - 11 ic + 22 = 0.
30. If a, j3 be the roots of a;^ ^ 7^; -|- 9 = 0, find the equation
whose roots are " "*" ^ and " "^ ^.
31. Show that the roots of x^ -(p"^ -2q)x + q- = 0 are the.
squares of the roots oi x'^ -\- px + q = 0.
32. Find the condition that the roots of a^x^ 4- 6% + c^ = 0 may-
be the squares of the roots of ax^ -\- bx -{- c = 0.
83. Show that, if a, /3 are the roots of px^ + gx + r = 0, the equa-
tion whose roots are a + j3 and _f^£_ is pqx^ + {pr -{■ q^)x + qr = 0.
a + /3
34. Show that, if a, /3 are, the roots of ax^ + 6ic + c = 0, the
equation whose roots are a/3 and — is acx^ — (a^ -\- c^)x + ac = 0;
a/3 1
show also that the equation whose roots are a + /? and ^, a is
a&a;2 + (a^ + &2)a; _,. ^5 _ 0.
35. Show that, if the difference between the roots oix^+ax+b=0
be equal to the difference between the roots of x"^ -]- px -\- q = 0, then
a2_46=;)2_4g.
36. Show that, if a, /3 be the roots of x"^ + px + q = 0, the equa-
tion whose roots are (a + /3)2 and (a — j8)2 is
a;2 _ 2(^2 _ 2 g)a; +^2(p2 _ 4 g) = 0.
37. If a, /3 be the roots of x^ -{- px + q = 0, show that the
equation whose roots are a + - and /S + - is
/3 a
gx2 +p(l + g)a; + (1 + g)2= 0.
38. Show that, if a, /S be the roots of ax^ -j- bx -{■ c = 0, the
roots of the equation (2 b^ + ac)x^ + 3 abx + a^ = 0 are
and
a + 2j8 jS + 2a
QUADRATIC EQUATIONS. 259
39. Find the relation that must exist between k and the coeffi-
cients of the equation ox^ + 6x + c = 0, in order that one root may
be k times the other.
40. Determine the relation that must exist between a and 6, in
order that one of the roots of the equation
y/{ax) + y/{x + a) = ^{x + h)
may be zero, the value of a that will make one of its roots infinite,
the value of a that will make both of its roots infinite, and the
relation between a and h that will make its two roots equal.
260 EQUATIONS OF HIGHER DEGREE.
CHAPTER XVII.
Equations of Higher Degree than the Second.
186. It is beyond the range of this book to show how
to solve equations of higher degree than the second, when
the equations are in their most general forms; we give
however some easy examples of such equations.
Ex. 1. Solve x* - 6 cc2 4- 8 = 0.
Here the equation contains only two powers of the unknown
quantity, one of which is the square of the other ; we therefore
proceed as in Art. 176.
We have ic*-6a;2 + 9-9+8 = 0,
that is (x2 _ 3)2 _ 1 = 0.
Hence {(^2 - 3) - 1} {(x^ _ 3) + 1} = 0 ;
. •• a;2 — 4 = 0, giving x = ± 2,
or x2 — 2 = 0, giving x = ± ^2.
Thus there are four roots, ±2, ±^2.
Ex. 2. Solve (x2 + a;)2 + 4(x2 + x) - 12 = 0.
In this, as in the former example, the unknown quantity occurs
in two terms, one of which is the square of the other. In all such
cases we can bring all the terms over to the left side of the sign of
equality, and then resolve the expression on the left into two fac-
tors as in the preceding chapter.
In the present instance the factors can be seen by inspection.
For ^2 _|. 4^ _ 12 =(^ + 6)(^ - 2);
therefore the equation may be written
{(x2 + x)+6}{(x2 + x)-2} = 0.
EQUATIONS OF HIGHER DEGREE. 261
Hence a;2 + x + 6 = 0, or x^ + a; - 2 = 0.
The roots of a;2 ^ x + 6 = 0 are - i ± ^^-
The roots of x^ + a; - 2 = 0 are 1 and — 2.
Hence the given equation has the four roots
1, -2, -\±\yr^^.
Ex.3. Solve -^ + ^±2 = 37.
X + 2 x2 6
Here one of the terms in which x occurs is the reciprocal of the
/J.2
other. If we put y = , we find y from the quadratic equation
y 6
or, multiplying by 6 y,
6 ?/2 _ 37 y -(- 6 = 0,
that is (6 y - 1) (y - 6) = 0.
Hence y = |, or y = 6.
Thus ^_=1, or-^=6.
a; + 2 6 x + 2
In the first case, we have 6 x^ — x — 2 = 0, the roots of which
are \ and — \.
In the second case, we have
a;2 _ 6 X - 12 = 0,
the roots of which are 3 ± V^l-
Hence the given equation has the four roots,
I, -I, 3±V21.
Ex. 4. Solve 4x2-6x + 3v(2x2-3x+ 7) = 30.
Equations of this form, in which the ratio of the coefficients
of x2 and x in the terms under the radical is equal to the ratio of
the coefficients of x2 and x in the terms outside the radical, are not
uncommon.
262 EQUATIONS OF HIGHER DEGREE.
. To solve the equation, put
V(2x2-3x+7)=y,
then 2 x2 - 3 a; + 7 = y2,
2 a;2 - 3 a; = 2/2 _ 7^
and 4x2-6ic = 2 2/2-14.
Hence, from the given equation, we have
22/2 -14 + 32/ = 30;
that is 2 2/2 + 3 2/ - 44 = 0,
that is (2/-4)(2 2/+ 11)=0;
hence 2/ = 4, or y = — y^ ;
. •. 2/2 = 16, or 2/2 = if i.
Since 2/2 = 2 a;2 _ 3 a; _|_ 7^ we have
I. 2x2-3x + 7 = 16,
that is 2 x2 - 3 x - 9 = 0,
that is (2 X + 3) (x - 3) = 0 ;
. % X = 3, or X = — f .
II. 2x2-3x+7 = 4J-,
that is 2 x2 - 3 X - -9^ = 0,
the roots of which are - ± :s/}^.
4 4
187. It has been shown [Art. 148] that if a rational
and integral expression containing x vanishes when any
particular value a is given to a?, then a; — a is a factor of
that expression.
For example, x^ — 7 x + 6 vanishes when 2 is put for x ; there-
fore, by the above theorem, x — 2 is a factor, and by division we
find that
x3 - 7 X + 6 = (X - 2) (x2 + 2 X - 3) .
Again, x^ — 4x2 + 2x+l vanishes when x = 1 ; therefore x — 1
is a factor, and by division we find that
ic3 _ 4x2 + 2x + l = (x - l)(x2 - 3x - 1).
EQUATIONS OF HIGHER DEGREE. 263
188. It follows from the theorem enunciated in the
last article that when one root of an equation is known,
the degree of the equation can be lowered.
By the application of this principle, the roots of the
important equation oc^ — 1 = 0 can be found.
Ex. 1. Solve the equation x^ — 1 = 0.
Since a^ - 1 ={x - l^^x"^ + x + 1),
we have (x - 1) (x^ + x + 1) = 0.
Hence x = l;
or else x^ + x + I = 0,
the roots of which are — a ± \/(~ !)•
Hence there are three roots of the equation a^ = 1, one
being real and the other two imaginary. Thus there are
three quantities whose cubes are equal to 1 ; that is, there
are three cube roots of 1, which are
1. -i + V(-f) and _^-V(-f).
[See Treatise on Algebra, Art. 139.]
Ex. 2. One root of the equation x^— 7x + 6 = 0 is 2; find the
other roots.
Since x^ — 7 x + 6 vanishes when x = 2, x — 2 must he a factor,
and by division we find that
x8 - 7 X + 6 = (X - 2) (x2 + 2 X - 3) .
Hence (x - 2)(x2 + 2x - 3) = 0.
Hence the other roots of the equation are those given by
x2 + 2x-3 = 0,
that is (x + 3)(x-l) = 0.
Thus the cubic equation x^ — 7x + 6 = 0 has the three roots 1,
2, and - 3.
264 EQUATIONS OF HIGHER DEGREE.
Ex. 3. For what values of x will
0-4-25x2 + 30a; -9,
and X* - 8 ic3 + 19 a;2 - 14 x + 3
both vanish ?
If both expressions vanish for the s^me value of x, say cc = a,
they will both ha^^e a; — a as a factor. Now we can find the com-
mon factor of any two expressions by the ordinary process of
finding their H. C. F.
In the present case, the H. C. F. will be found to be
x2 - 6 X + 3 ;
and since ic2 — 6 x + 3 is a factor of both expressions, and is their
highest common factor, both expressions will vanish for the
values of x given by
x2-5cc + 3 = 0,
and for no other values. Thus the values required are the roots of
a;2 - 5 X + 3 = 0,
and these roots are
5 ± V13,
EXAMPLES LVI.
Find the roots of the following equations :
c4 _ 2 x2 - 8 = 0.
a^
4. X*- 7x2 -18=0. 10. (x2-x)2-8(x2-x) + 12=0.
5 x2-l ^ 1 _i^Q 11- (x2+x)2-22(x2+x) + 40zz0.
9 ^'^ 12. (x2 + x)(x2 + x+l)=42.
6. x2 + l^ = 29. 13. _^ + ^+l = 2.
^^ X + 1 X2
7. x2 + l=a2 + l. 14. ^_+?i±I = ^.
•^ ^a;2 a2 x2+l^ x 2
EQUATIONS OF HIGHER DEGREE. 265
16. _^+£^ = lI.
x-1 x^ 4
16. (x2 + x+l)(a;2 + x-f)+l = 0.
17. (xHa;+l)(x2+x+2) = 12. 2Q ^. , ^ , ^ ^ 42
18. (.+ 1)V4(. + 1) = 12,
a;2 + x
21. 2x2-4 a;-Va;2-2x-3=9.
19 x2+2 x2+4x+l_5
• X24-4X+1 X24-2 ~2 22. x2+ Vx2+3x+7=23-3x.
23. 2x2 4-6x = l-V(aj2 + 3a; + i).
24. x2 + v(4x2 + 24x)=24-6x.
25. 2(2x-3)(x-4)-V(2a;2-lla;+15)=60.
26. x2 + (X - 2) (X - 3) + V2 x2 - 5 X + 6 = 6.
27. (x+6)(x-2)-36=V(x + 4)(x-l).
Solve the following cubic equations having given one of the roots
of each :
28. x8-2x+ 1 =0, [x= 1].
29. x8-6x + 4=:0, [x = l].
80. x8-49x + 120 = 0, [x = 3].
81. x8 - 3x2 - 7 X + 21 = 0, [X = 3].
82. x8-2x2-7x-4 = 0, [x = -l].
88. 5x8 - 16 x2 + 3 X + 14 = 0, [x = 2].
34. For what values of x will x* + 2 x2 + 9 and x* - 4 x + 16
both v&,nish?
35. The equations
2x8 + 21x2 + 34x- 106 = 0,
2x3- x2-76x- 105 = 0
have one root in common ; find it.
Solve the following equations :
36. x8-21x + 20 = 0. 39. x8 - 3x2 - 60x - 100 = 0.
37. x8+2x2-9x-18=0. 40. x3-19x + 30 = 0.
38. x8-x2-22x+40=0. 41. 2x8 + 7x2 - 5x - 4 = 0.
42. 4x8 +4x2- 9x- 9 = 0.
266 SIMULTANEOUS EQUATIONS.
CHAPTER XVIII.
Simultaneous Equations of the Second
Degree.
189. We now proceed to consider simultaneous equa-
tions, one at least of wMcli is of the second degree.
We first take the case of two equations which contain
two unknown quantities, one equation being of the first
degree and the other of the second degree.
Example. Solve the equations
From the first equation we have
X = 6 — 2y.
Substituting this value of x in the second equation, we have
(5_22/)2 + 22/2 = 9;
.-. 6y^-20y + 16 = 0;
.-. 32/2- 102/ + 8 = 0;
that is (3 y - 4) (?/ - 2) = 0.
Hence ?/ = |, or y = 2.
If 2/ = t, x = 6-2y = S-^ = i;
and if 2/ = 2, a; = 5 — 2?/ = 5 — 4 = 1.
Thus we have x = |, y = § ', or x = 1, y = 2.
[The result should not be written in the ambiguous form a: = |,
orl; 2/ = I, or2.J
SIMULTANEOUS EQUATIONS. 267
From the above example it will be seen that two
equations, one of which is of the first degree and the
other of the second degree, can be solved in the following
manner:
From the equation of the first degree find the value of
one of the unknown quantities in terms of the other unknown
quantity and the known quantities, and substitute this value
in the equation of the second degree : one of the unknown
quantities is thus eliminated, and a quadratic equation is
obtained, the roots of which are the values of the unknown
quantity which is retained,
Ex. 1. Solve the equations
Sx + Ay = 5, 2x^ -xy + y^ = 22.
From the
first equation, x = ~ ^- Hence, substituting in
the second, we have
K^T-r-^^)-^--'
which reduces to
63 y2_95y_ 148 = 0,
that is
(y+l)(53y-148) = 0.
Hence
— - = f-
If
v = -i,. = 411 = 3.
And, if
148 6-4y 109
^53' 3 53
Ex. 2. Solve xy-hx = 25, 2xy-3y = 2S.
Multiply the first equation by 2, and subtract the second ; then
2 X + 3 2/ = 50 - 28 = 22.
268 SIMULTANEOUS EQUATIONS.
H«nce y = ^^:
and substituting in the first equation, we have
f22-2x\
+ a; = 25 ;
.-. 2x2 -25a: + 75 = 0,
whence we have x = 5 or x = Y-, the corresponding values of y
being 4 and |, respectively.
Ex.3. Solve
2«2_3aj-42/ = 47,
Multiply the second equation by 2 ; then
6x^-\-Sx + iy = ns\
Adding the last equation to the first, we ha^e-
8x2 + 5a; = 225,
whence x = 5 or x = — -*/, The values of y are then found by sub-
stituting in either of the given equations.
EXAMPLES LVII.
Solve the following equations :
1. X + y = 6, 6. X + 2/ = 15,
x2 — j/2 = 24. 4:X — 4:y = xy.
2. X — y = 10, 7. 2 X — 2/ = 5,
x2 + y2 = 58. x + Sy = 2xy.
3. 3x+3y = 10, 8. 3x-31 = 5y,
xy=l. x2 + 5x2/ + 25 = 2/2.
4. 2x + 3y = 0, 9. 3x + 2y = 5,
4x2 + 9x2/ + 92/2 = 72. x^ - 4xy + 5y^ = 2.
5. 2x-52/ = 0, 10. 2x-72/ = 25,
x2- 32/2= 13. * 5x2 + 4x2/ + 32/2 = 23.
2 + 3 = 6.
X y
12.
a; + 2y = 7,
? + « = 5.
X y
13.
a: + y = 5,
1 + 1 = 1
X y Q
14.
x-y = l.
x_y _5
y X 6*
15.
2x-y = 4,
^4-^ = 1.
X y
SIMULTANEOUS EQUATIONS. 269
31. x + y = 2, 16. 7x-3y + 7 = 0,
^-1^ = 6.
X y
17. xy + x=15j
xy — x = 8.
18. xy + 2x=5f
2xy-y = Z.
19. a;2-y = 29,
a;2 + a: + y = 49.
20. a;2 + 3a;-2y = 4,
2a;2-6x + 3y + 2 = 0.
190. It should be remarked that the methods we have
thus far explained do not enable us to solve any two
equations which are both of the second degree ; for the
elimination of one of the unknown quantities will fre-
quently lead to an equation of higher degree than the
second, from which the remaining unknown quantity-
would have to be found ; and we have not j^et learned
how to solve an equation of higher degree than the
second, except in very special cases.
For example, if we have the equations x^ + a; + y = 3 and
x2 + 2/2 = 5. We have from the first equation y = 3 — x — a;^ ; and,
by substituting this value of y in the second equation, we get
X2 + (3 - X - X2)2 = 5,
that is a4 + 2x8-4x2-6x + 4 = 0;
and this equation of the fourth degree cannot be solved by any
methods which are within the range of this book.
270 SIMULTANEOUS EQUATIONS.
191. We can always solve two equations of the second
degree when all the terms which contain the unknown
quantities are of the second degree. The method will be
seen from the following example.
Ex. 1. Solve the equations
x?/ + 4 2/2 = 8.
Divide the members of the first equation by the corresponding
members of the second equation ; we then have
a;2 + 3 a;y _ 28 _ 7^
o-y _l_ 4 y2 ~ 8 ~ 2'
Hence 2(x2 + 3 x^) = 7(a:!/ + 4 y2) ;
.-. 2^2 -xy- 28 2/2 = 0,
that is (2x + 7?/)(ic-4 2/) = 0.
Hence x = 4 y, or x = — | ?/.
I. If X = 4 2/, we have from the second equation
42/2,+ 42/2 = 8; .-. y=±l.
And therefore x = 4 «/ = ± 4.
II. If X = — 12/, we have from the second equation,
-12/2 + 4^2 = 8;
.-. y = ±4.
And therefore x = — 1 2/ = =F 14.
Thus there are four sets of values, namely, x=4, 2/=l; x = — 4,
2/ = — 1 ; X = 14, 2/ = — 4 ; and x = — 14, 2/ = 4.
We have in the above written down the factors of
2 x2 - X2/ - 28 2/2
by inspection : when this cannot be done, the factors can be found
by the method of Art. 122.
SIMULTANEOUS EQUATIONS. 271
Ex. 2. Solve x2 - 3x2/ = 0, 5x2 + 3^2 = 48.
The first equation is x(x — 3 y) = 0.
Hence x = 0, or x = 3 y.
If X = 0, the second equation gives Sy"^ = iS; .*. 2/ = ± 4.
If X = By, the second equation gives 45 ^24.3^2—48 j . •. y— _j- 1^
and then x = ± 3.
192. Any pair of equations which are homogeneous, so
far as the terms which contain the unknown quantities
are concerned, can be solved by the method adopted in
the last article. Sometimes, however, the equations can
be solved by special methods.
For example, to solve
x2 + y2 _ 74^
2 «y = 70.
By addition, we have
x2 + 2 xy + 2/2 = 144,
that is (x + y)2 - 122 = 0 ;
.-. (x + 2/-12)(x + 2/ + 12) = 0.
Hence x + y = l2 (1.),
x + y=-12 . . (ii.).
Again, from the given equations, we have by subtraction,
aj2 - 2 xy + 1/2 = 4^
that is (X - t/)2 - 22 = 0 ;
.'. (x-2/-2)(x-2/4-2) = 0.
Hence x — y = 2 Oii-)>
x-y = -2 (iv.).
From (i.) and (iii.), we have x = 7, y = S.
From (i.) and (iv.), we have x = 5, y = 7.
From (ii.) and (iii.), we have x = —5, y — —7.
And, from (ii.) and (iv.), we have x = — 7, y = — 5.
272 SIMULTANEOUS EQUATIONS.
Thus there are four pairs of values, two of which are given by
aJ = ± 7, ?/ = ± 5, and the other two hy x = ± 5, y =±7, it being
understood that in both cases the upper signs are to be taken
together, and the lower signs are to be taken together.
Again, taking the equations considered in Art. 191, namely,
x^ + Sxy = 28,
«!/ + 4 ?/2 = 8.
We have by addition x^ -\- 4:xy -\- iy^ = 36,
that is (x + 2 y)2 - 62 = 0 ;
.-. x + 2y = 6 (1.),
or x + 2y = -6 (ii.).
We can now complete the solution as in Art. 189, by taking
(i.) and (ii.) with either of the given equations.
EXAMPLES LVIII.
1. a;2_2xy = 0, 9. x^-Sy'^ = lS,
4aj2 + 9i/2 = 225.. Sx^-y^ = n.
2. 2x2-3xy = 0, 10. Sxy + x^ = 10,
y2 -I- 5xy = 34. 5xy - 2 x2 = 2.
8. x2 + 3x2/ = 45, 11. x2 + 3x2/ = 54,
y2_x2/ = 4. xy + 4^2 = 115.
4. x2_x?/ = 63, 12. x(x + y) = 40,
y^ + Xy = 22. y(^^_y^^Q^
5. x2 + a;?/ = 24, 13. x^ - 3xy + 2y^ = S,
2 2/2 + 3xy = 32. 2x^ + y^ = 6.
6. x2-3xy = 10, 14. a;2-2xy + 5 = 0,
42/2-xi/ = -l. (a;_y)2_4 = 0.
7. x2 + x?/-2y2 = _44, 15. x^ + 5y'^ = S^,
xy + 3 2/2 = 80. 3x2 + 17 xy - y2 = - 84.
8. x2 + 3x2/ = 7, 16. x2-7x2/-9y2 = 9,
y^ + xy = 6. x2+ 6x2/ + 11 2/2 = 5.
SIMULTANEOUS EQUATIONS. 273
17.
4a;2-3a;y = 10,
y^-xy = 6.
23.
3 2 3
18.
x^ +3xy = 40,
2- + ? = 0.
4y^+xy = 9.
19.
x^ + xy + y-i = 7,
6x^-2xy + y'^ = 6.
24.
2 5_6
« 2/ 6'
20.
2 x2 - 2 a;?/ + 3 2/2 = 18,
3x2-2^/2 = 19.
21.
jc+2/ + l=0,
22.
a: 2/ 12
a; + y + 3 = 0,
1 + 1-1 = 0.
X y 6
25.
26.
(a; + l)(j/ + l)=10,
a;2/ = 3.
(x + 3)(2/ + l)=4,
xy + 1 = 0.
193. The following examples will show how to deal
with some other cases of simultaneous equations.
Ex.1. Solve x-y = 2 (i.),
x3- 2/8 = 386 (ii.).
From (i.), we have x = y + 2.
Substitute this value in (ii.), and we have
(y + 2)8 - 2/8 = 386,
that is 6 2/2 + 12 2/ + 8 - 386 = 0 ;
. ♦. 2/2 4- 2 2/ - 63 = 0.
Hence y = 7,ory = — 9.
If2/ = 7, x = 2/ + 2 = 9.
If 2/ = - 9, x = j/+2 = -7
Thus X = 9, 2/ = 7, or X = — 7, y = — 9,
Ex. 2. Solve x-3y = 2,
x2 - 9 2/2 = 8.
274 SIMULTANEOUS EQUATIONS.
Divide the members of the second equation by the correspond-
ing members of the first ; then
X + 3y =4;
and from x + 3 y = 4 and aj — 3 ?/ = 2,
we have x = 3, y — \.
Ex. 3. Solve
' We may consider
107, Ex. 3.
^and
X
1
y
as the unknown qna,n titles, as in Art.
Square both members of
0
:.) ; then
l + ~ + l = 49 (111.).
a;2 xy y'^ ^ ^
From (ii.) and (iii.), we have
— = 49-25 = 24 (iv.).
xy
From (11.) and (iv.), we have
1^A + 1 = 25-24 = 1,
x^ xy y^
that Is
\x y)
.•.1-1=±1. . . \ . . (v.).
X y
The values of - and - are at once found from (1.) and (v.).
X y
Ex.4. Solve x2 - x?/ + ?/2 = 61 (1.),
X4 + ic2y2 + y4 = 1281 (ii.).
Divide the members of (11.) by the corresponding members of
(1.) ; we then have
x'^ + xy -\-y'^ = 2l (Hi.).
SIMULTANEOUS EQUATIONS. 275
From (i. ) and (iii. ) we have, by subtraction,
2xy = - 40,
or xy = -20 ..... . (iv.).
From (i.) and (iv.), we have
a;'-^- 2x^ + 2/2 = 81;
.-. a:-y = +9 (v.),
or x-2/ = -9 (vi.).
From (1.) and (iv.), we have also
a;2 + 2a;y + 2/2 = 1;
.-. x4-2/ = l (vii.),
or x-\-y = —\ (viii.) .
Then combining either of the equations (v.) and (vi.) with
either of the equations (vii.) and (viii.) we get the four pairs of
values x = ±5, y=T4; x = ±4, y = q=6.
Ex.6. Solve x2_2xy = 3y (i.),
2x2-9y2 = 9y (ii.).
Multiply both sia'cc of (i.) by 3 ; then
3x2-6xy = 9j,- . ^. (iii.).
Hence 2x2 - 9y2 = 3x2 -65// , ^
.-. x2-6xy + 9y2 = o, ^^^
that is (x - 3 y)2 = 0 ;
.-. x = 3y.
Substitute in (11.) ; then
2(3z/)2-92/2 = 9y,
or 9 y2 _ 9 y = 0 ;
.♦. y(y-i)=o.
Hence y = 0, or y = 1.
If2/ = 0, x = 3y = 0.
Ify = l, x = 3«/ = 3.
Thus X = 0, J/ = 0 ;
or else x = 3, y = 1.
276 SIMULTANEOUS EQUATIONS.
Ex. 6. Solve X - ?/ = 2, x^-y^ = 242.
Put x = z-\-l, then y = z — 1.
Hence
x^-y^=(z + iy-(iz-iy
= z^+6 z^ + 10 z^+lO z'^-\-6 z+1- (z^-b z^+lO z^-10 z^+5 z-1)
=:lOz* + 20z^ + 2.
Hence 10 z^ + 20 z"^ -\- 2 = 242 ;
.'. z* + 2z'2 = 2^',
Hence z = ±2, or z = ± V- 6.
If « = ± 2, X = 3 or - 1, and ?/ = 1 or — 3.
It z=± ^/^^, X = 1 ± V^^, and y=-l i.V^^.
Ex. 7. Solve x?/ + x0 = 27 (i.),
yz + yx = S2 (ii.),
ZX + zy = S5 (iii')'
From (i.) and (ii.), we have by addition
2xy + xz + yz-bQ .■>'•;. . . . (iv.).
From (iv.) and (iii.), we have by subtraction
2 xy = 24 ;
.: xy = 12 (v.).
Hence, from (L), x^ = 15 (vi.).
And, from (ii.), yz = 20 (vii.)-
From (v.) and (vi.), we have by multiplication
x'^fz = 180 (viii.).
From (vii.) and (viii.), we have by division
x2 = 180/20 = 9 ;
.-. x = ±3.
Hence,, from (v.) , «/ = 12/ ( ± 3) = ± 4.
And, from (vi.), ^ = 15/(± 3) = ± 5.
Thus x = ±3, y = ±4, z = ±5.
All the upper signs being taken together, and all the lower signs
being taken together.
SIMULTANEOUS EQUATIONS. 277
EXAMPLES LIX.
1. x-y = S,
x3 _ ?/3 = 279.
2. x-y = 2,
x3 _ ?/3 = 98.
3. x + y = l,
x3 + 1/3 = 91.
4. x-\-y = 1,
a^ + yi = 61.
6. x^ + xy + y^ = lS,
Xi + xY + y4 = 91.
6. x^ — xy + y^ = 9,
x* + x2?/2 + y4 = 243.
7. xy(x-\-y) = 2^0,
x^ + y^ = 280.
8. xy(x- y)= 12,
x3 - w3 = 63.
19. 2x2-x2/ + 2/2 = 2y,
«ll2 2x2 + 4xi/ = 5y.
X y
1 . 1
13.
x-^y = l.
x2y2 + 13 x?/ + 12 = 0.
14.
X + 2/ = 5,
4x?/ = 12 -xV- ^
15.
x2 - xy + 2/ = - 6,
y^-xy-x = 12.
16.
x2 + x?/ - y = 9,
2/2 + X2/ - X = - 3.
17.
x4-i = 3,
18.
^_l1 18
y 7
'-i=^-
20. x3 4- 1 = 9 y,
;„ x2 + y'2 = 20. x2 + x = 6|/.
21. 2x3 + 5 1/8 = 115,
10. ? + l = l, 3x3 + 7y8^186.
X 1/
2 3 1 _ f. 22. x22/ + a;?/2 = 180,
^■^^"^^2-^- x2y2 = 400.
3 i_ 23. a^ + x2y2 + yt = 91,
"• ^-^2-^' (x2-Xl/ + l/2)(x-l/)2 = 28.
A_JL4.2__3^ 24. a;24-3x2/+2/2+4(x+y)=13,
x2 xy 2/2 * •3x2-xi/ + 3?/2+2(x+2/)=9.
12. 1-A = 3, 26. X2/ + ? = ^
4 1/2 y 3
x2 xy 4i/2 ^^x 6
278
SIMULTANEOUS EQUATIONS.
X xy S
y 3
28. a(x-a)= b(y-b),
xy = ax + by.
29. X = y = a
y X ^
yz = 4, zx = 9, xy = 16.
27. ax = by,
(x - a)(y -b)= ab. 30.
31. x'^yz = 6, y'^zx = 12, ;s2a.y ^ ig.
32. a;(x+?/ + 2!) = 8, ?/(x + y +2)= 12, 2;(sc + ?/ + 0)= 6.
33. x(y + z)=e, y(z-j-x)=12, z(ix + y)= 10.
34. (_x + y)(ix + z)=2, (y + z)(y + x)=S, {z -\- x) (z + y) = 6.
35. y^; = a^, zx = b^, xy = c^.
36.
y±z_z-hx _x
= 2 xyx.
a b c
37. (l/ + 5?)(a; + y + ;2)=6 + c,
(^z -\- x)(x + y + z)= c + a,
(^ + y) (aj + w 4- is;) = a + &.
PROBLEMS. 279
CHAPTER XIX.
Problems.
194. We now give examples of problems in whicli the
relations between the known and unknown quantities are
expressed algebraically by means of quadratic equations.
In the solution of problems it often happens that by
solving the equations which are the algebraical state-
mrnts of the relations between the magnitudes of the
known and unknown quantities, we obtain results which
do not all satisfy the conditions of the problem.
The reason of this is that the roots of the equation
are the numbers, whether positive or negative, integral or
frattional, which satisfy it; but in the problem itself
there may be restrictions, expressed or implied, on the
numbers, and these restrictions cannot be retained in the
equation. For example, in a problem which refers to
a number of men, it is clear that this number must be
integral, but this condition cannot be expressed in the
equations.
Thus there are three steps in the solution of a prob-
lem. We first find the equations which are the algebrai-
cal expressions of the relations between the magnitudes
of the known and unknown quantities ; we then find the
values of the unknown quantities which satisfy these
equations ; and finally we examine whether any or
all of the values we have found violate any conditions
/^ OF THE
I UNIVERSITY
280 PROBLEMS.
which are expressed or implied in the problem, but which
are not contained in the equations.
The following are examples of problems which lead to
quadratic equations.
Ex. 1. How many children are there in a family, when eleven
times the number is greater by live than twice the square of the
number ?
Let X be the number of children ; then we have
11 X = 2x2 + 5;
.-. 2x2-11x4-5 = 0,
that is (2x-l)(x-6) = 0.
Hence x = 5, or x = |.
Thus there are 5 children, the value | being inadmissible.
Ex. 2. Eleven times the number of yards in the length of a
rod is greater by five than twice the square of the number of
yards. How long is the rod ?
This leads to the same equation as before, only in this case we
cannot reject the fractional result. Thus the rod may be five
yards long, or it may be half a yard long.
Ex. 3. A number of two digits is equal to twice the product
of the digits, and the digit in the ten's place is less by 3 than the
digit in the unit's place. What is the number ?
Let X be the digit in the ten's place ; then x + 3 will be the digit
in the unit's place. The number is therefore equal to
10 X + (x + 3).
Hence, by the question,
10x + (x + 3)=2x(x + 3) ;
.-. 2x2- 5x- 3 = 0,
that is (x - 3) (2 X + 1) = 0.
Hence x = 3, or x = — |.
PROBLEMS. 281
Now the digits of a number must be positive integers ; hence the
second value is inadmissible.
Therefore the digits are 3 and 6, and the required number is 36.
Ex. 4. The square of the number of dollars a man possesses
is greater by 1000 than thirty times the number. HoW much is
the man worth ?
Let X be the number of dollars the man is worth ; then, by
the conditions of the problem
a;2 = 30x4-1000;
.-. x2-30x- 1000 = 0,
that is (x - 50) {x + 20) = 0.
Hence a; = 50, or x = — 20.
Both of these values are admissible, provided a debt is considered
as a negative possession.
Hence the man may have $ 50, or he may owe $ 20.
Ex. 5. The sum of a certain number and its square root is 42 :
what is the number ?
Let X be the number ; then
aj+ ^x = 42,
that is y/x = 42 — X.
Square both sides ; then, after transposition, we have
x2_85x+ 1764 = 0,
the roots of which are 36 and 49.
The value 49 will not however satisfy the condition of the
problem, if by a square root of a number is meant only the arith-
metical square root.
Ex. 6. The sum of the ages of a father and his son is 100
years ; also one-tenth of the product of their ages, in years,
exceeds the father's age by 180. How old are they ?
Let the father be x years old ; then the son will be 100 — x
years old. Hence by the conditions of the problem,
282 PROBLEMS.
■^\x(100-x') = x+ 180;
.-. x2-90x + 1800 = 0,
that is (x - 60) (x - 30) = 0.
Hence x — 60, ov x = 30.
The second value is inadmissible, although it is a positive
integer, for it would make the son older than his father.
Hence the father must be 60, and the son 40 years old.
EXAMPLES LX.
1. Find two numbers one of which is three times the other and
whose product is 243.
2. Find two numbers whose sum is 18 and whose product is 77.
3. Find 'two numbers whose difference is 20, and the sum of
whose squares is 650.
4. Divide 25 into two parts whose product is 156.
5. Divide 80 into two parts the sum of the squares of which is
3208.
6. A certain number is subtracted from 36, and the same
number is also subtracted from 30 ; and the product of the re-
mainders is 891. What is the number ?
7. A rectangular court is ten yards longer than it is broad ; its
area is 1131 square yards. What is its length and breadth ?
8. The product of the sum and difference of a number and its
reciprocal is 3| : find the number.
9. The number of tennis balls which can be bought for a pound
is equal to the number of shillings in the cost of 125 of them.
How many can be bought for a pound ?
10. The number of eggs which can be bought for 25 cents is
equal to twice the number of cents which 8 eggs cost. How
many eggs can be bought for 25 cents ?
PROBLEMS. 283
11. A cask contains a certain number of gallons of water, and
another cask contains half as many gallons of wine ; six gallons
are drawn from each, and what is drawn from the one cask is put
into the other. If the mixture in each cask be now of the same
strength, find the amount of water and wine.
12. The cost of an entertainment was $ 20, which was to have
been divided equally among the party, but four of them leave
without paying, and the rest have each to pay 25 cents extra in con-
sequence. Of how many persons did the party consist ?
13. A man buys a certain number of articles for § 5, and makes
$3.82 by selling all but two at 4 cents each more than they cost.
How many did he buy ?
14. A man bought a certain number of railway-shares for
$9375; he sold all but 15 of them for $10,450, gaining $20 per
share on their cost price; how many shares did he buy ?
16. A crew can row a certain course up stream in 8f minutes,
and if there were no stream they could row it in 7 minutes less
than it takes them to drift down the stream ; how long would they
take to row down with the stream ?
16. A boat's crew can row 8 miles an hour in still water. What
is the speed of a river's ciirrent if it take them 2 hours and 40
minutes to row 8 miles up and 8 miles down ?
17. Two trains run without stopping over the same 36 miles of
rail. One of them travels 15 miles an hour faster than the other,
and accomplishes the distance in 12 minutes less. Find the speed
of the two trains.
18. A person having 7 miles to walk increases his speed one
mile an hour after the first mile, and is half an hour less on the
road than he would have been had he not altered his rate. How
long did he take ?
19. A and B together can do a piece of work in a certain time.
If they each did one half of the work separately, A would have
to work one day less, and B two days more than before. Find the
time in which A and B together do the work.
284 PROBLEMS.
20. The price of photographs is raised 50 cents per dozen ; and,
in consequence, six less than before are sold for $ 5. What was
the original price ?
21. What are eggs a dozen when two more for 30 cents would
lower the price 2 cents a dozen ?
22. A woman spends 75 cents in eggs ; if she had bought a
dozen less for the same money, they would have cost her 3| cents a
dozen more. How many did she buy ?
23. The price of one kind of sugar per pound is 2 cents more
than that of a second kind, and 10 pounds less of the first kind
can be got for $2.40 than of the second. Find the price of each
per pound.
24. One-half of the number of cents which a dozen apples cost
is greater by 2 than twice the number of apples which can be
bought for 30 cents. How many can be bought for $ 2.50 ?
25. Divide $ 3620 among A, B and C so that B shall receive $ 20
less than A, and C as many times B's share as there are dollars
in A's share.
26. Find two fractions whose sum is |, and whose difference is
equal to their product.
27. Two men start at the same time to meet each other from
towns which are 25 miles apart. One takes 18 minutes longer
than the other to walk a mile, and they meet in 5 hours. How
fast does each walk ?
28. The men in a regiment can be arranged in a column twice
as deep as it is broad. If the number be diminished by 206, the
men can be arranged in a hollow square three deep, having the
same number of men in each outer side of the square as there
were in the depth of the column. How many men were there at
first in the regiment ?
29. The area of a certain rectangle is equal to the area of a
square whose side is six inches shorter than one of the sides of the
rectangle. K the breadth of the rectangle be increased by one inch
MISCELLANEOUS EXAMPLES IV. 285
and its length diminished by two inches, its area would be unal-
tered. Find the lengths of its sides.
30. The diagonal and the longer side of a rectangle are together
five times the shorter side, and the longer side exceeds the shorter
by 35 yards. What is the area of the rectangle ?
31. If the greatest side of a rectangle be diminished by 3 yards
and the less by 1 yard its area would be halved ; and if the greater
be increased by 9 yards and the less diminished by 2 yards its area
would be unaltered. Find the sides.
32. Two trains A and B leave P for Q at the same time as two
trains C and D leave Q for P. A passes C 120 miles from P, and
D*140 miles from P. B passes C 126 miles from Q, and D half
way between P and Q : find the distance from P to Q.
MISCELLANEOUS EXAMPLES IV.
A. 1. Simplify 2x - [Sx - 9y - {2x - Sy -(^x + 6 y)}].
2. Multiply a2_|. 25 62+402+6 a6-2ac+ 10 6c by a-56+2c.
3. Divideic8+(4a6-62)a;_(a_2 6)(a2+3 62) by jc-a+2 6.
4. Find the factors of (i.) (2 x + y - 5?)2 - (a; + 2 y + 4 z)^.
(ii.) x'^y^ -x^ -y^+1, and (iii.) x'^y^z^ - x^z - y'^z + 1.
(ii.) ^Lul 2 ^-2 ' ^-^
(x-2)(x-3) (x-3)(x-l) (x-l)(x-2)
6. Solve the equations :
X — a x — b X — c
(ii.) 4x2 -25x- 21 = 0.
7. Show that x2 — 5 x + 7 can never be less than |.
8. The difference of the cubes of two consecutive integer num-
bers is 919 : find the numbers.
286 MISCELLANEOUS EXAMPLES IV.
B. 1. Simplify {x + 3)3 - 3(a; + 2)3 + Z{x + 1)3 - x\
2. Show that
{x - ay + (y - 6)2 + (a2 + 52 _ Y)^,^ ^y2_y^
= (xa + by - 1)2 + (6a; - ayy.
3. Divide x^ -2 a^x^ -\- a^ by x^-2ax-\- a^.
4. FindtheL. CM. of
8a;3 + 27, 16x4 + 36x2 + 81, and 6x2 -5x- 6.
6. Simplify
2 1 x + 3
.(i.)
1 X + 1 X2 + 1
Cii >) {(q + 6)(a + 6 + c)+ c^Ka + hy - c^}
^ '^ {(a + 6)3 - c3}(a + 6 + c)
6. Solve the equations :
(i.) (6-x)(l + 2x)+3x(x + 6) = (x + l)2-x.
(ii.) 5x2 + 7x = 160.
(iii.) x2 + xy = 10, y"^ -xy = 3.
7. Show that the sum of the squares of the roots of the equation
x2-5x + 2 = 0 is 21.
8. At a concert $ 6 was received for reserved seats, and the same
sum for unreserved seats. A reserved seat cost 50 cents more than
an unreserved seat, but 6 more tickets for unreserved than for
reserved seats were sold. How many tickets were sold altogether ?
C. 1. Simplify 12a-3{6-2(a-36)-2a}.
2. Multiply a3 + 2 a26 - a62 + 2 63 by a3 _ 2 a% -ah'^-2 63.
3. Divide 24 x* - 10 x^y - 8 x2?/2 + 10 xy^ - 4 2/* by 2 ?/2-xy-4x2.
4. Find the factors of 9 x2 + 9 x + 2,
and of 4(a6 - cdy - {a?- + 62 -&- d^y.
MISCELLANEOUS EXAMPLES IV. 287
6. SimpUfy __l + xy_^
1 <y - ^)
l + xy
and show that ^-::i^ + ^^1^ + ^^^ + ^^ - ^^^" - ^^^ - ^^ = 0.
a 6 c a6c
6. Solve the equations :
3 6 6
(ii.) ic + 1/ = 2 a, x2 + ?/2 = 2 a2.
7. Find the least value of x^ -\-Qx-\- 12, and the greatest value
of 6ic-x2-4.
8. A and B have 45 coins between them, which are all dollars
and dimes. A has four times as many dollars as dimes, and B has
just as many dollars as dimes ; also A has $ 9.50 more than B.
How much money has each ?
D. 1. Show that
(6 + c)2 _ a2 + (c + ay - h"^ +{a + by - c"^ = {a + h + cy.
2. Arrange (1 + xy + 2(1 —x-^-x"^) according to ascending
powers of X.
3. Show that the difference between the squares of any two con-
secutive numbers is one more than double the smaller number.
4. Show that if - -|- - + i = 0, a* + 62 + c2 =(« + ft + cy.
a b c
6. Find the L. C. M. of
a;2 - 5a: - 14, x2 - 4a; - 21, and x^-Sx"^- 25x - 21.
For what value of x will all three expressions vanish ?
6. Solve the equations
(i.) 4_3^5^6^3^10,
X y X y
X a b X— a — b
(ill.) x + y = 3, 3 x2 - 11 y2 =: 1.
288 MISCELLANEOUS EXAMPLES IV.
7. If oji, CC2 are the roots of ax^ + bx + c = 0, prove that
Xi,X2_ b^ — 2 ac
X2 Xi ac
8. Out of a cask containing 60 gallons of alcohol a certain
quantity is drawn off and replaced by water. Of the mixture a
second quantity, 14 gallons more than the first, is drawn off and
replaced by water. The cask then contains as much water as
alcohol. How much was drawn off the first time ?
E. 1. Find the numerical value of
2. Show that (a + by - (a^ - b^y = 4ab(a + by, and that
2(a - 6)(a - c)+ 2(& - c)(6 - a)+2(c - a)(c - b)
= {b~ cf + (c - ay + (a - 5)2.
3. Divide (« + 2 6 - 3 c + d)2 - (2 a-f 6 + 3 c - df)2 by a + 6.
4. Find the L. C.U.oiQx'^ - bax -Q a^ and 4 a;^ _ 2 ax2 - 9 a^.
6. Simplify (1^ + ^)^ +
\x yly'i-x'^
6. Solve the equations :
X — b X — a
(ii.) {X - l)(x - 2) + (a; - 2)(x - 3) + (« - 3)(x.- 1)= 11.
(iii.)x + l = fl, y+^ = \
?/ 10 X 3
7. For what values of x are - + - and - + - equal to one an-
a X a c
other ?
8. A tricyclist rode 180 miles at a uniform rate. If he had
ridden 3 miles an hour slower than he did, it would have taken
him 3 hours longer. How many miles an hour did he ride ?
Xy — y2 a;2 4. y'l
MISCELLANEOUS EQUATIONS. 289
F. 1. Add together S a^ - A a'^b -\- 2 ah\ Sa'^b-i ab^ + 2 63,
3a6-2_4 63 + 2a3, and 3 63 _ 4 a3 4. 2 a26.
2. Show that (a^ + ft^ + c^) (x^ + y2 + z^) - (ao; + 6y + czy
= {ay - 6x)2 + (6z - cyy + (ex - azy.
3. Divide 2 + 11 x + 11 a;2 + x^ - x* by x2 + 3x + 2^
4. Find the H. C. F. of x3-x2-2x+2 and x*-3x3+2x2+x-l.
What value of x will make both expressions vanish ?
6. Simplify (i.) ^ ■ 1.3
(ii.)
x(x-2) x2-6x + 6 x(3-x)
1
1-
1- 1
1-x
6. Solve the equations :
(i^ a;4-3 6-x^ 7,
^•^5 10 10
(ii.) ^±1 + EZL| = 2, ax + 6y = a2 + 62.
a + 6 a— 6
(iii.)^±i + ^±2=:2.^ + 3
x-1 x-2 x-3
7. Find the difference of the squares of the roots of the equation
x2 - 7 X + 9 = 0.
8. What number exceeds its square root by 166 ?
MISCELLANEOUS EQUATIONS.
1. (x - 1) (x - 4) + (X - 3) (X - 5) = 2(x - 3)2.
2. i(x-2)-K^-4)+K^-5)=0.
8.
X -
1
-3 x-5
X
1
-2
1
x-4
4.
15:
T
r + 17y =
79,
17 X
+ 16y-
:81,
290 MISCELLANEOUS EQUATIONS.
5. A man drives to a certain place at the rate of 8 miles an
hour. Returning by a road 3 miles longer at the rate of 9 miles
an hour, he takes 7| minutes longer than in going. How long is
each road ?
6. i(x-2)-^^^+5?-=^=0.
7. S(x + l)(a; - 3) + (x + 7)(3x - 13) = (2x - 5)(3x - 3).
g 2a; + 9 3a;-2_5a; + 14
' x + 2 x + 3 x + 4: '
9. ? + y = 82, ^ + y = 8S.
7 8 ' 8 7
10. A pound of tea and 5 pounds of sugar cost 90 cents, but if
the tea were to rise 20 per cent and the sugar 25 per cent in price,
they would cost f 1.10. What was the price of the tea ?
11. ^(6x + 3)+3^(7a; + 6)+TV(9a^ + 2)=2x+l.
12. ? + ^ = «_:^.
abba
13 a; + 3 x — 6 __x i- 4: x — 5
' x+1 x-4 x+2 x-S
14. X + 2/ = 2, (a + b)x +(b-a)y = 2 b.
15. A poultry keeper obtained 100 eggs more in May than in
April, and the daily average in May was 3 more than in April.
How many did he have in May ?
16. 3a; - 3[4x - 2(2x - 5)]= 9 - 2[3x - b{x - 6)].
17. ^-6 = 2, 4 + 12 = 3.
X y X y
18. 15x2 + 34^ + 15 = 0.
19. a? + 2 a;-3_.^^
* X — 4 » — 6
MISCELLANEOUS EQUATIONS. 291
20. A father's age is three times that of his youngest son, and
in 7 years he will be twice as old as his oldest son ; also the oldest
son is 5 years older than the youngest. What are their present
21. K^ + l)(x + 3)= i(x + 5)(x+2)+lix-l)(x^ 4).
22. (6 + c)(x-a)-(^c + a) (x - 6) = (a + b) (x - c).
23. 17x- 19y = 4, 27x-29y =24.
24 2x-3 3x+8_.
25. Two passengers, who have altogether 360 lb. of luggage, are
charged 80 cts. for excess above the weight allowed free ; if one
passenger had taken the same weight of luggage, he would have been
charged $ 1.30. What weight is allowed free ?
26 ^^- 1 J. 5x4-1 _ 9;x + 1 _ 1 -X
5 "^ 6 8 3 *
27. ^-±l=2x-y-2, ^y-^ = y-x + 5.
3 2
28. (a + b)x + (a - b)y = a^ - 6^
{a - b)x -\- la + b)y = a^ + bK
29. _i_+_§_=3.
3-x 8-x
30. A father's age is equal to the united ages of his five children,
and five years ago his age was double their united ages. How old
is the father ?
31 2x x + 3 _ 3x+ 19
'x-lx+1 x+6 •
32. l{x-2)=l(il-y), 26x + 32/ + 4=0.
33. 1 17 1
x2-l 1-a; 8 x + 1
34. y/x + Vbx + l=2.
292 MISCELLANEOUS EQUATIONS.
35. Two cyclists started at the same time, and one rode from
B to A and the other from A to B along the same road. They
reached their destinations 3 hours and 1 hour 20 minutes, re-
spectively, after they passed each other. When did they meet ?
39. \/4 X + 4 - Vx-\-S = Vx - 4.
40. A certain sum of money was divided between A, B, and C.
A had one-ninth of the shares of B and C together, B had one-
third of the shares of C and A together, and C had 6 dollars more
than A and B together. How much had each ?
"• i(-!)-(^-T)--«-
42. ^±^-^=^ = 1.
x — S X — 4
43. x-\-2=y/{4 + xVS-x}.
44. 3 X - 2 y = 12, 9 x2 - 4 2/2 = 576.
45. A man about to invest in the 2f per cents observed that, if
the price of the stock had been 7| less, he would have got f per
cent more interest for his money. What was the price of the
stock ?
j,^ 4x^ + 4x2 + 8x4-l_2x2 + 2x+l.
x + 1
5) =11, 3,V(55y-12) = ^-37.
2 x2 + 2 X + 3
47.
tV(5^ + 6)-^V(112/
48.
X X— 1
49.
V« + Vx + 4 = -^'
MISCELLANEOUS EQUATIONS. 293
50. A fast train left Cambridge for London at the same time as
a slow train left London for Cambridge. They arrived at their
destinations three-quarters of an hour and one hour and twenty-
minutes respectively after they passed one another. Find how
long each took for the journey.
x-1 SVx-l 3J
23
lO(x-l)
52. 6x + 2y -l=Sx-y + U = x-^19y + 6.
53. V(ic2-a2)=2- ^
y/ix^-a^)
54. 3x2 - 2 a;?/ + 4 1/2 = 36, 4x'^-y^ = 7.
55. Find three consecutive integers such that the. sum of the
cubes of the greatest and least exceeds twice the cube of the
middle number by 42.
66 ^--^ ^ _2a;-l 9} - x
3 6 ~ 3 "•" 21 *
57 a;-3 x-2 2 . 3
2 3 x-S x-2
58. ax - &?/ = a2 - 62 _ 2 ab, bx -\- ay = 2 ab -\- a^ - b\
59. a;2 4- a;?/ + y2 _ g^ ^^ 4. y^^yi + y* = 243.
60. A man bought a certain number of articles, of which he sold
half at 5 per cent profit, one-third at 10 per cent profit, and the
rest at 30 cents each. His total profit was at the rate of If per
cent. What was the cost of each article ?
61. (a-f b)(x-\-a-b) + {a- 6)(x - a - 6)+ 2a(a; - 2c)=0.
62 3a;-2 2a;-3^2
2x-b 3x-2 3*
63. 2x2 -3V(«2 + 2x-M4)-f4x- 49 = 0.
64. 2x2+ 3x2/ = 8, y'^-2xy = 29.
294 MISCELLANEOUS EQUATIONS.
65. A man bought a certain number of sheep for $ 290. Having
lost five of them, he sold one quarter of the remainder for $ 63,
making a profit of 6 per cent on the sale of these. How many
did he buy ?
66. 1 :. + 5 _ 3
6a; -1 1-25x2 1 + 5x
67. (a'+ h)x + hy-ax^-{a+ b)y = a^ + &».
68. ^(x-a) + y/(x-b): ^ ' ^
y/(x-a) ^(x-b)
69. x"^-^ y^ = xy + 7, x^ — y^ = xy - 1.
70. A number of articles were bought for $ 25, and sold again
at an advance of 12 1 cents in the price of each ; and at the advanced
price .1 25 was received for the sale of all but 10 of the articles.
How many were there ?
71. \iSx + 1)~ ^(2x - S)= j'^{6x - 1)- i(7 X -3).
72. --- — ---.
c x~ a' c
73. ^(x+l) + V2^=^(i6x + l).
74. x^ -xy-\-Sy = ll, y^ - xy -Sx + 1 =0.
75. The value of 185 coins consisting of dollars, dimes, and
half-dimes amounts to $30.50. If there were twice as many
dimes, half as many half-dimes, and three times as many dollars,
the total value of the coins would be $ 72. How many coins are
there of each kind ?
-g 3x+l l-5x_ic-fl 2-9«
5 3 3
77.
x+1 a;2 4. 1 _ 29
x^ + lz + 1 10
78.
(x -a)(y -b) = ab, bx = ay.
79.
x2 = 2y-f24, ?/2 = 2x + 24.
MISCELLANEOUS EQUATIONS. 295
80. 1120 square feet of paper will just cover the four walls of a
room which is 8 feet longer than it is wide ; but, if the room were
4 feet higher, the same quantity of paper would just cover the two
smaller and one of the larger walls. What are the dimensions of
the room ?
81. Ka^_2)-^ + 5^ = 0.
82. (a-xy+{b-xy=ia-\-b-2x)^.
83. (x-6)2+(y-6)2 = 2(a;y-40), x = y + l.
84. 4.xy = 96-x^y% x + y = 6.
85. A merchant gained as many eagles on the sale of a certain
quantity of coal as there were half-dollars in the cost price of a
ton, or half-dimes in the retail price of a cwt. How many tons
did he sell ?
86 2x-3 2x-4_2x-6 2a;-7
* 2ic-4 2a;-6~2a;-7 2a;-8*
87. 2xCa-b-2x) + ia-l)(b-l)=0.
88. y/(6x+l) + V2{l-x)=zy/7x-\-6.
89. xy(x + y) = 30, x^-\-y» = - 91.
90. A rectangular enclosure is half an acre in area, and its
perimeter is 198 yards. Find the length of its sides.
91 g; + 3 3a;-|-7_2a; + 1
2a;+l 2a;-2 x-2
92. x-\-2y -\-Zz = i, x + Sy + 2 = iz, x -\- 2z -\- S z= iy.
93. ^ia-x) + -^(x-b) = ^(ia-b).
94. {x - 3)2 +(2/ - 3)2 = 34, xy - 3(x + y)= 6.
95. Two trains start at the same instant, the one from B to' A,
the other from A to B, the distance between A and B being 100
miles. The trains meet in 1 hour 15 minutes, and one train gets
to its destination 1 hour 20 minutes before the other. Find the
rates of the trains.
296 MISCELLANEOUS EQUATIONS.
gg ax - hy _{2 a + h)x - (^a + 2 b)y _ ^^ ^^
3 7
97. 13x2+ 12 = 80 ic.
98. (a: + l)(x + 2)(x + 4)(x + 6) = 4.
99. x^iy + l)+y^{x + l)= 109, xy = 12.
100. A starts to walk from P to Q at 10 a.m., B starts from Q to
walk to P at 10.24 a.m. They meet 6 miles from Q. B stops 1
hour at P, and A stops 2 hrs. 54 min. at Q, and returning they
meet midway between P and Q at 6.54 p.m. Find the distance
from P to Q.
101. y^ + z'^ = z^ + x'^ = x^-^y^ = axyz.
102. ax(y -\- z)= hy{z -{■ x') = cz(x + y)— xyz.
103. x'^ ^- 2yz = y"^ -\- 2. zx = z"^ -^ 2xy = 12.
104. Resolve x^ -\- y^ -\- z^ — Z xyz into the three factors
X -\- y -\- z, X + cjy + a>%, X + <a^y + uz,
where w = — ^ + ^V^,
and determine three alternative relations between a, 6, and c, one
of which must be satisfied in order that the equations
x^ + y^ -^ z^ — S xyz — 0,
ax+ by + cz = 0,
a b c
may be simultaneous in x, y, and z.
POWERS AND ROOTS. 297
CHAPTER XX.
Powers and Roots.
196. The process by wMcli the powers of quantities
are obtained is called involution ; and the inverse process,
by which the roots of quantities are obtained, is called
evolution. [Art. 9.]
INVOLUTION.
196. When m and n are any positive integers, we have
by definition
a*" = aaaa. . . to m factors,
and a" = aaaa... to n factors ;
. • . a^xa*' = (aaaa. . . to m factors) x (aaaa. . . to n factors)
= aaaa... to m + w factors
= a"*+**, by definition.
Hence when m and n are any positive integers,
^m _|_ ^n _ ^m+n
Thus, the index of the product of any two powers of the
same quantity is the sum of the indices of the factors.
This result is called the Index Law.
From the Index Law we have
a"^ xa"" xa^ = »"*+" X a" = a'"+"+^,
and so on, however many factors there may be.
298 POWERS AND ROOTS.
Hence a"* x a** x a/ • • • = a'^+^+p+-.
Thus, the index of the product of any number of powers
of the same quantity is the sum of the indices of the factors.
197. We have
ttxaxaxa...tom factors
a'^/a'' =
axaxax a... to n factors
Now, if m is greater than n, the n factors of the denom-
inator can be cancelled with n of the factors of the
numerator : we then have m — n factors left in the
numerator.
Thus, when m is greater than w,
If, however, n is greater than m, the m factors of the
numerator can be cancelled with m of the factors of
the denominator : we then have n — m factors left in
the denominator.
Thus, when m is less than n,
198. To find (a'")" when m and n are positive integers.
By definition
(oC^Y = a*" x a*" x a"* X ••• to 71 factors
Hence («'")'* = a"*^
POWERS AND ROOTS. 299
Thus, to raise any power of a quantity to any other
power, its original index must he multiplied by the index
of the power to which it is to be raised.
199. To find (ab)^.
{ab)'^= ab X ab X ab... to m factors, by definition,
= {aaa. . . to m factors) x (bbb... to m factors) [Art. 52.]
= a"* X 6"", by defijiition.
Hence (ab)"* = a'"6'*.
Similarly
(abc)"^ = abc x abc x abc... to m factors,
= (aaa... to m factors) x (bbb... to m factors)
X (ccc... to m factors)
= a"* xb"^ X C*.
Hence (abc)"^ = a'*6"*c*,
and so on, however many factors there may be in the ex-
pression whose mth power is required.
Thus, the mth power of a product is the product of the
rath powers of its factors.
200. The most general monomial expression is of the
form a'b^c'...
To find (a'b"c'...y.
(a'b*C...y = {a''y{b«y{c'y... [Art. 199.]
= a'^b^'^c'"*... [Art. 198.]
Thus, any power of a monomial expression is obtained by
taking each of its factors to a power whose index is the
product of its original index and the index of the power to
which the whole expression is to be raised.
300 POWERS AND ROOTS.
201. The following is an important case.
To find '^
-) =-x-x---«tom factors, by definition,
oj 0 0 0
a X a X a... to m factors p . , 1 fi7 1
b X b X b... to m factors
^.
b) ~b^'
202. It should be noticed that all powers of a positive
quantity are positive, and that successive powers of a
negative quantity are alternately positive and negative.
This follows at once from the Law of Signs ; for we have
(_a)2 = (-a)(-a) = + a^
(-ay = {-ay{-a) = i-\-a'){-a) = -a';
(-ay = {-ay{-a) = {-a')(-a) = + a'',
and so on.
Thus ( _ a) 2" = -f a^% and ( - a) 2"+i = - a^^+i.
From the above it is clear that all even powers, whether
of positive or of negative quantities, are positive, and that
all odd powers of any quantity have the same sign as the
original quantity.
203. We have already proved the following cases of
the involution of binomial expressions.
POWERS AND ROOTS. 301
and (a + by = a'-hSa'b + 3ab' + ¥.
If we multiply again hj a-\-b, we shall have
(a + by = a* + 4a36 + Ba^ft^ + 4.ab^ + 6*.
By multiplying the last result by a + 6 we should obtain
(a-^by, and by continuing the process we could obtain
any required power of a + 6 ; but to find in this way any
high power, for instance to find (a + b)^, would clearly
be very laborious.
We shall shortly prove a theorem, called the Binomial
Theorem, which will enable us to write down at once any
power of a binomial expression.
The above formulae are identities, and are true for all
values of a and b ; hence we can write down the squares
and the cubes of any binomial expressions.
Thus (a* - ¥y =(a4)2+ 2(a4)(- 6*) + (- b^y
= a8-2a*6* + 6S
and (5 a2 - 3 62)3 = (5 ^2)8 + 3(5 o2)2( _ 3 52)
+ 3(5a2)(-3 62)2^.(_3ft2)8
= 125 a» - 225 a*b^ + 135 a^b* - 27 6«.
Also (a + 6 + cy={a + (6 + c)}«
= a8 + 3a2(6 + c)+ 3a(6 + cy+{b + cy.
204. An important case of involution is considered in
Art. 70, where the square of a multimonial expression is
obtained.
302 POWERS AND EOOTS.
EXAMPLES LXI.
Write down the value of each of the following :
1.
(a3)5. -
21.
(«3> 63)8.
2.
(X5)3.
22.
(2 a2- 3 62)3.
3.
(-a2)8.
23.
(3a2_2 62)8.
4.
(- a3)2.
24.
(a2 + 62 + c2)2.
6.
i-2a^y.
25.
(a3-2 63 + 3c8)2.
6.
C-sa^y.
26.
(02 _ 4 ^,2 _ 3 c2)2.
7.
i-ab^y.
27.
(x2_3x-6)2.
8.
{a^h^y.
28.
(3^2 -X- 5)2.
9.
i-ab^y.
29.
(2x2 + 5x-l)2.
10.
i-Sa'^b^cy.
30.
(3x2 -6x- 6)2.
11.
C-ab'^c^y.
31.
(1+X + X'^ + X8)2.
12.
(-5a263c4)8.
32.
(X3 - X2 + X - 1)2.
13.
(-Y-
33.
(X3 + x2 - 2 X - 2)2.
Wl
34.
(a + 26 + 3c + 4d)2.
14.
16.
[ bH^j
35.
36.
37.
(2a-6 + c-2c?)2.
(x2 + X + 1)8.
(X2 - X + 2)8.
16.
(2 a* + 3 63)2.
38.
(3x2-6x + l)8.
17.
18.
(a^ - 2 6^)2.
39.
l-*lh\{'-w
19.
20.
(-a2 + 2a6)2.
(a2 + 62)8.
40.
'-^^4-
THE BINOMIAL THEOREM.
205. The numerical coefficients on the right side of
the formulae of Art. 203 may be so constructed as to
exhibit the law of their formation. Thus," written in the
POWERS AND ROOTS. 303
order of their occurrence in the formulae, they and their
reconstructed equivalents are
1, 2, 1 = 1, |, |l|, v|P^
1, 3, S, 1-1, -, j-^, ^-^-^,
1 A fi i 1-1 f 4-3 4-3.2 4-3.2.1
1, 4, b, 4, 1-1, ^, ^ ^, ^-^Tg, 1.2.3.4-
If this law of their formation obtains for higher
powers of (a 4-6), we shall be able to write out, by a
very simple rule, the entire series of terms in the expan-
sion of (a + by, wherein n is any positive integer. Thus,
if the law holds for all positive integral values of n,
{a + by = a' -h 5a^6 + ^a'^' + 1^4^«'&'
and in general
(a + by = a" + na^-'b + ^(^--^)a"-^> + ...
J. * ^
1.2.3...r --
Here r is some integer less than w, and
n(n-l)... (n-r + l)^n-.y
1.2.3...r '
which is called the general term, may be made to assume
in succession the form and value of every term in the
series, by giving to r the successive values
0, 1, 2, 3, 4, ...n-1, n.
304 POWERS AND ROOTS.
Thus, if r = 5, this general term becomes
n{n-l)(n-2)(n-S)(n-4:) ,,
1.2.3.4.5
We proceed to prove that the above formula is true
for all positive integral values of n. The proof here
given is by mathematical induction (monomorphic trans-
formation) explained in Art. 145.
For brevity's sake we write the formula thus :
(a + 6)" = Coa'* -f Cia'-^d + C2a''-"-b^ H
in which the letters c have the respective values
-, 71(71 — 1)
Co = l, Ci=n, c2 = -5^— y^, •.•
V
r _y^(^-l) ••• (n-r+1) _.
"'" 1.2.3...r ' ^'''~
We assume provisionally that the formula is true
when the index is n. Can we justify this assumption ?
Multiplying both sides of it by (a + 6), we obtain
(a + 6)"+^ = Coa"+^ + (co + c{) a^^h + (ci + Cg) a'^-^t^
+ (C2 + C3) a"-263 + ^. + (c,_i 4- c,) a"-'-+i6''+ ...
But Co + Ci = n + 1,
c, + c. = . + -i^ = ^^(^,
' ' 1.2 1.2.3
_yi(n — l)(ii + l)
1.2.3
POWERS AND ROOTS. 305
and in general
_n(n—iy*'(n—r+2) n(n — 1) »•• (n— r+ 1)
Cr-i-\-Or- i.2.3...^_i + 1.2.3...r
_n(n — l) '" (n —r -\-2)(r -{-n — r -\'l)
"■ 1.2.3-..r
_ (yi + l)n(n — 1) ... (n + 1 — ^ + 1)
1.2.3...r
Substituting these results for Cq, Cq -\- Ci, Cj + Cg, etc., in
the formula, we have
(a + by+^ = a"+i + (w + l)a"6 + ^^"^ ^^^a"-^6^ + ...
1 * ^
4. (n + l)n(n-l)... (n + l-r + 1) ^^i_, ,
"^ 1.2.3...r
+ ••• +6"+'.
The series thus arranged is exactly the same in form
as that for (a + 6)", w + 1 having now taken the place of
n. Hence, if the formula be true for any positive inte-
gral value of n, it is true when n is replaced by the next
higher integer.
But it is obviously true when w = 1 ; hence it is true
when n = 2. And, being true when n = 2, it is true when
n = 3, and so on indefinitely. Therefore it is true for
every positive integral value of n. [Art. 145.]
The proposition thus established is known as the
Binomial Theorem for positive integral exponents. It is
one of the most important theorems of algebra, and will
be discussed more fully in a subsequent chapter to be
devoted to this special topic.
u
306 POWERS AND ROOTS.
The special numerical coefficients which have been
designated above by Cq, q, C2, c^, ••• c^, ••• c„ are called the
binomial coefBcients. They are frequently used in other
algebraic formulae and should be committed to memory.
Ex. 1. If in the binomial formula we write w = 3, we have
(a + 6)3 = a3 + § ^25 ^hlab^ + b^
= a3-j. 3^25 + 3^52+ 53.
Ex. 2. If a = 2, 6 = ?/2^ and w = 3, we have
(2 + y)3 = 23+1-22.2/2 + ^.22/4 + 2^
= 8 + 12^2 + 62/4+2/6.
Ex. 3. If a = -, 6 = -, and w = 4, we have
2 y
/i + iV=l + i.^ i + ili 1 1 I ^--^-^ i A + ±
V2 y) 24 l*23'2/'^1.2*22'2/2 1.2."'-^* ^^ ^
•2.1.1.1
3*2*2/^2/*
l + l.i + §.l+2.1 + i.
16 2 2/2 2/^ 2/3 y*
206. The formula for the expansion of (a — &)" is
obtained by writing — 6 in place of h in the previously
written formula
(a + by = a'* + Cia"-i6 + i^a'^-^h'^ + Cga^-^d' H h 6",
the even terms of which become negative by this change,
because odd powers of a negative quantity are negative.
Thus,
(a - 6) " = a" - c^al'-^h + c^a^'-^h'^ - c^a^^-^b^ + • • • ± &",
and the last term has the positive or the negative sign
according as n is odd or even.
POWERS AND ROOTS. 307
Ex. 1. If a = x, h = 2y, and w = 4, the result of substitution
in the formula for the expansion of (a — &)" is
(x - 2 2/)4 = X* - 4 x^(2 y) + — x2(2 yy
1 * ii
= X* - 8 a;8y _|. 24 x2«/2 _ 32 a;y8 + 16 y*.
Ex. 2. If a = 2 X, 5 = 3 2/2, and n = 5, the result of substitu-
tion in the formula for the expansion of (a — 6)" is
(2 X - 3 y2)6 = (2 xy - 6(2 x)*(3 y^) + ^ (2 x)8(3 y2)2
-(3 1/2)6
= 32 x5 - 240 X42/2 + 720 x^j/* - 1080 x^ + 810 xy8
-243yw
207. General Term. By inspection of the formula, it is
at once obvious that the number of any term in the
expansion of (a ± 6)" is one greater than the suffix of c
(or number of factors in the denominator of c) in that
term. Thus c^a'^'^h^ is the fourth term. Hence, in gen-
eral, the (r 4- l)th term is
n(n-l)(n-2)...(n-r-H)
"^ 1.2.3...r '
and by means of this formula any isolated term may be
written down.
Observe that r is at once the exponent of h, the
number of factors in the denominator and the number of
factors in the numerator, and that the sum of the expo-
nents of a and h is n.
308 POWERS AND ROOTS.
Ex. 1. The eleventh term in the expansion of (1 + x^y^ is
15.14.13.12.11.10.9.8.7.6
1.2.3.4.5.6.7.8.9. 10
(a:2)io = 3003x20.
Ex. 2. The sixth term in the expansion of (x — ^y'^y^ is
_ 11. 10. 9. 8. 7 231
1.2.3.4.5 ^^^^ 16 ^
The sign of this term is negative because even terms in the ex-
pansion of (a — 6)" are negative.
EXAMPLES LXII.
Write out the following expansions :
1. (x + ay. 3. (Sx-2yy. 6. (2x2-1)6.
2. (l-x2)5. 4. (2a + 3«2)4. 6. (y-xy.
7. Find the third term of (a - 3 &)io.
8. Find the fifth term of (2x - x^y^.
9. Find the sixth term of (2 a - 1)8.
10. Find the seventh term of (1 — xy^.
11. Find the eighteenth term of (1 + x)^.
12. Find the twenty-first term of (1 - x)^.
13. Expand (a + ^/by + (a - ^/by.
14. Expand (a -f ^by + (a - ^/by.
15. Expand (a + y/by -f (a - y/by.
16. Find the middle term of (1 + xy.
17. Find the middle term of (1 -f xy^.
18. Find the middle term of (2 x - 3 yy.
19. Show that in the expansion of (1 + a;)'»+'', the coefficients
of x"* and x" are equal.
20. Expand (^ + -) *
21. Expand ^^x + ij-i^x-^)*-
POWERS AND ROOTS. 309
EVOLUTION.
We know that there are two square roots of a?,
namely ± a ; we also know [Art. 188] that there are
three cube roots of o?, of which a is one, and the other
two are imaginary.
There is therefore an important difference between
powers and roots ; for there is only one nth power ^ but
there is more than one nth root, of a given expression.
209. An expression which when raised to the ?ith
power, where n is any positive integer, becomes equal to
a given expression, is called an nth root of the given
expression.
We have shown in Art. 199 that the mth power of a
product is the product of the mth powers of its factors ;
hence, conversely, the mth root of a product is the
product of the mth roots of its factors.*
Thus ^ahc — ^a ^h ^c,
and Vah = ^a ^h.
Again, we have shown in Art. 200 that the nth power
of a monomial expression is obtained by multiplying the
index of each of its factors by n.
It follows conversely that if we divide the index of
each factor of a given expression by n, we shall obtain
an nth root of the expression. For by raising to the nth
* It should be noticed that the proof, in Art. 52, that the factors
of a product may be taken in any order, only holds good when those
factors represent integral or fractional numbers, and does not enable
us to assert that VaxVh = Vhx Va, when Va or Vh is really a surd.
For a proof that surds obey the Fundamental Laws of Algebra see
Treatise on Algebra, Art. 162.
310 POWERS AND ROOTS.
power the result obtained by such division of the indices,
we must clearly get the original expression.
Thus one value of -y/a* is a^^ one value of Vc^h^^ is aH>H^^ and
one value of Va^b^ is aPb^.
When the square root of an expression which is not a
perfect square, or the cube root of an expression which
is not a perfect cube, is required, the operation cannot
be performed. We can, for example, only write the
square root of a as ^a, and the cube root of a^ as -^/a^,
and similarly in other cases.
SQUARE ROOT.
210. We now proceed to consider the square root of
multimonial expressions.
In Art. 114 we have shown how to write down the
square root of any trinomial expression which is a com-
plete square.
Having arranged the expression according to ascending
or descending powers of some letter, the square root of
the whole expression is then found by taking the square
roots of the extreme terms with the same or with diffei-
ent signs according as the sign of the middle term is
positive or negative.
Thus, to find the square root of
4a«-12a463^96«.
The square roots of the extreme terms are
±2a^and ±36«.
Hence, the middle term being negative, the required
square root is ± (2 a* — 3 b^) .
POWEES AND EOOTS. 311
Note. — In future only one of the square roots of an expression
will be given, namely that one for which the sign of the first term
is positive : to find the other root all the signs must be changed.
As other examples
V(49 aW + 28 a5 + 4) = 7 a5 + 2
V(l + 6xy3 + ^^xV)= 1 + f a^y^,
and V{«^ + 2 a(6 + c) + (6 + c)2}= a -\- b -\- c.
211. When an expression which contains only two
different powers of a particular letter is arranged accord-
ing to ascending or descending powers of the letter, it
will only contain three terms. For example, the ex-
pression
a^-^b^ + c^ + 2bc-\-2ca-\-2ab,
which contains no other power of a but a^ and a, when
arranged according to powers of a, is
a2 + 2a (6 + c) -f (6^ + c^ + 26c).
Thus any expression which only contains two different
powers of a particular letter can be written as a trinomial
expression ; and since we can write down the square root
of any trimonial expression which is a complete square,
it follows that the square root of any expression which
is a complete square can be written down by inspection,
provided that the expression only contains two different
powers of some particular letter.
For example, to find the square root of
a2 + 62 + c2 -h 2 6c + 2 ac + 2 a6.
Arranging the expression according to the descending powers of
a, we have a'^ + 2 a{h + c) + (62 -f 2 6c + c2),
that is, a2 + 2 a(& -1- c) + (6 -H c)2,
which ifl {a -h (6 + c)f.
312 POWERS AND ROOTS.
Thus V(«^ + 62 + c2 + 2 6c + 2 ca + 2 a6) = (a + 6 + c).
Also, to find the square root of
(c* + 4?/4 4- 9^4 + 4icV _ 6a;2^2 _ I2y%2.
Arrange the expression according to descending powers of x ; we
then have
x^ + 2x^(2y^ - 3^2) + 4^4 + 9^4 _ I2y^z\
that is, X* + 2 a;2(2 «/2 - 3 z^) + (2 ?/2 _ 3 ^2)2,
which is {^2 + (2 r - 3 2:2)}2.
Thus
V(X* + 4 2/4 + 9 5!4 + 4 a:2?/2 _ 6 ^202 _ 12 y2^2)=,(a;2 _|. 2 2/2 _ 3 02).
Again, to find the square root of
a2 + 2 a6x + (62 + 2 ac)x2 + 2 bcx^ + c2aj*.
Arrange the expression according to powers of a ; we then have
a2 + 2 a(6a: + cx2) + 62^2 + 2 bcx^ + c^x:^,
that is, a2 + 2 a(6x + cx2) + (6x + cx2)2,
which is ^ {a + (hx + cx2)}2.
Hence the required square root is (a ■\-hx + cx^) .
And to find the square root of
x6_2x5 + 3x4 + 2 x\y - 1) + x2(l - 2 y) + 2 xy + y^.
The expression only contains y'^ and y ; we therefore arrange it
according to powers ofy; we then have
?/2 + 2 y(x^ - x2 + x)+ x6 - 2x5 + 3 x* - 2x3 + x2.
Now if the expression is a complete square at all, the last of the
three terms must be the square of half the coefficient of y ; and it
is easy to verify that
x6 - 2 x^ + 3 X* - 2 x3 + x2 is (x« - x2 -f x)2.
Thus the given expression is
?/2 + 2 y{x^ - X2 + X) + (X3 - X2 + X)2.
The required square root is therefore
y + x^ — x2 + X,
POWERS AND ROOTS. 313
From the above it will be seen that however many
terms there may be in an expression which is a perfect
square, the square root can be written down by inspection,
provided only that the expression contains only two difs
ferent powers of some particular letter.
EXAMPLES LXIII.
Write down the square roots of the following expressions.
1. 9 a;2 _ 30 xy + 25 2/2. 5. x^ -6x^y^ + 9y\
2. 25 a;4- 30x22/2 + 9 y4. 6. 9x12-6x62/3 + 2/6.
3. 4 X* - 12 x22/2 + 9 2/*. 7. ^ x^ - ^ x^ + i J/«.
4. 4x10-12x62/8 + 92/6 8. 1 x^ - I x^i/' + i.
9. 25 X82/6 _ 40 a268x42/8 + 16 a'^b\
10. ^+8x22/3 4. i6a22/6.
a2
11. 9^-24x22/'« + 16^^
a2 52
12. ^_42^+49aio.
x6 x8
13. a2+462 + 9c2+ I26c + 6ca + 4a6.
14. 4 a2 + 62 + 9 c2 + 6 6c - 12 ca - 4 ab.
16. 4 a* + 6* + c* - 2 62c2 _ 4 c2a2 + 4 a^b^.
16. 25 a* + 9 6* + 4 c* + 12 62c2 + 20 c2a2 - 30 a^b"^.
212. In order to show how to find the square root of
any algebraical expression, we will take an expression
and form its square, and then show how to reverse the
process.
314 POWERS AND ROOTS.
Consider, for example,
x' + 2xy + 3f (i.),
whose square is
x^ + 4.:ty-\-10o^y^-{-12xf-[-^y^ . . . (ii.),
both expressions being arranged according- to descending
powers of x.
We may write the square of x^ -{■ 2 xy -\- ^ y^ in either
of the following forms :
[o? -^ {2xy + ^f)Y = x' ^-2x\2xy ^^y'')
j^{2xy^Sfy (iii.),
{{:>? + 2xy)-\- ^f\' ^{0^ + 2xyy + 2{a? ^-2xy)^f
•\-{^fY (iv.).
Now it is clear that the first term of (ii.) is the square
of the first term of (i.). Hence the first term of the
root of (ii.) is found by taking the square root of its
first term.
Again, we see from (iii.) that when we have sub-
tracted x^ (the square of the first term of the root), the
term in the remainder which contains the highest power
of a; is 2 ic^ X 2 xy, which is twice the product of the first
and second terms of the root.
Hence, after subtracting from (ii.) the square of the
first term of the root, the second term is obtained by
dividing the first term of the remainder by twice the first
term of the root.
Again, we see from (iv.) that when we have subtracted
{x^ ■\-2xyy, that is, the square of the part of the root
already found, the term in the remainder which contains
POWERS AND ROOTS. 315
the highest power of a; is 2 cc^ X 3/, which is twice the
product of the first and third terms of the root.
Hence, after subtracting from (ii.) the square of that
part of the root already found, the next term of the root
is obtained by dividing the first term of the remainder by
twice the first term of the root.
If we now subtract the square of x^ -\-2xy -^-^y^ from
the given expression, there will be no remainder; and
hence Qi?-{-2xy -\-^y^ \^ the required root.
We will now consider the most general case.
Suppose we have to find the square root of {A + BY,
where A stands for any number of terms of the root,
and B for the rest ) the terms in A and B being arranged
according to descending {or ascending) powers of some
letter, so that every term in A is of higher {or lower)
dimensions than any term in B.
Also suppose that the terms in A are known, and that
we have to find the terms in B.
Subtracting A^ from {A -f B)^, we have the remainder
{2A + B)B.
Now from the mode of arrangement it follows that the
term of the highest (or lowest) degree in the remainder
is twice the product of the first term in A and the^rs^
term in B.
Hence, to obtain the next term of the required root,
that is, to obtain the highest (or lowest) term of B, we
subtract from the whole expression the square of that part
of the root which is already fonnd, and divide the first
term of the remainder by twice the first term of the root.
The first term of the root is clearly the square root of
the first term of the given expression; and when we
316 POWERS AND ROOTS.
have found the first term of the root we can find each of
the other terms in succession by the above process.
For ex'ample, to find the square root of
xi ^4x^y -^ 10 xV + 12 xy^ + 9 y^.
x* + 4 x3y+ 10 x'^y^+12 xy^-^9y^ (pfi-\-2xy+ 3 y^
(X2)2 = ^^ *~
(x2 + 2 xyy = x* + 4 x^y + 4 x^y^
(x2 + 2xy + 3?/2)2 = x^-{-4.x^y+ lOxV + 12x^3 ^
The given expression must first be arranged according to ascend-
ing or descending powers of some letter.
We then take the square root of the first term of the given
expression : we thus obtain x2, the first term of the required root.
Now subtract the square of x2 from the given expression, and
divide the first term of the remainder by 2 x^ : we thus obtain 2 x.v,
the second term of the root.
Now subtract the square of x2 + 2 xy, which is
X* + 4 x3?/ + 4 x2?/2,
from the given expression, and divide the first term of the
remainder, namely 6 x^y"^^ by 2 x2 : we thus obtain 3 !/2, the third
term of the root.
Subtract the square of x2 + 2'xy + 3?/2 from the given expression,
and there is no remainder.
Hence x^ -\- 2 xy -\- S y^ is the required square root.
The squares of x2, x2 -f 2 xy, etc. , are placed under the given
expression, like terms being placed in the same column: the
remainder left after taking away any square is then obvious by
inspection.
213. Instead of finding each square independently,
some labour can be saved by making use of the previous
square. Thus the process of finding the square root of
an algebraical expression is generally written as follows,
the same example being taken as before :
POWERS AND ROOTS. 317
a^ + 4 x^y + 10x2y2 + 12 icy3 + 9 y4 (x2 + 2 xy + 3 y2
2x2 + 2ic?/) 4x^y
4 ic^y + 4 x2y2
2x2 + 4xy + 3?/2) 6x22/2
6x2?/2+ 12x?/3 + 92/*
The first term of the root is x2. Having subtracted the square of
x2, namely x*, the first term of the remainder is 4 x^y.
Now double the part of the root already found and divide the
first term of the remainder by the result ; we thus obtain the
next term of the root, namely 2xy. Add this term of the root to
2 x2, placing the sum in the ordinary position for the divisor. Now
multiply the sum by 2xy and subtract the product from the
remainder 4x^y -\- '■■, we then have the remainder Qx^y^+---.
Repeat this process as often as may be necessary.
[It is of importance to notice that by the above process we
have at the end of the second stage subtracted altogether -
(x2 + 2xy)2,
since (x2 + 2 xy)^ = (x2)2 + (2 x2 + 2 xy) x 2 xy.
Similarly at the end of every stage it will be seen that what is sub-
tracted on the whole is the square of that part of the root which
has been found up to that stage.]
EXAMPLES LXIV.
Find the square roots of the following expressions :
1. x* + 2x8 + 3x2 + 2x+ 1.
2. 4x4 -8x8 + 4x+ 1.
8. 9x*- 36x8 + 72x4- 36.
4. 1-xy - Y- x2y2 4. 2 x3y8 + 4 x*y*.
6. 4x* + 4x8- ^x + tV
6. x4-2x3 + |x2- 1x + tV-
318 POWERS AND ROOTS.
7. x* + 2 x"'?/ + SxV + 2 X2/3 + ?/4.
9. l + 4x+ 10x2+ 12x3 + 9ic*.
10. 4x*-4x3 + 3x2-x + ^.
11. (1 + 2x2)2 _ 4x(l -x)(l+2x).
12. x'''-4x5 + 6x*-8x'^ + 9x2-4x + 4.
13. 9x8- 12x5 + 22x4 + x2+ 12X + 4.
14. x6 - 22 x* + 34 x3 + 121 x2 - 374x + 289.
15. a2_aja;+ ix2 + 8a-4x+ 16.
16. x8 + 2x7+ x6-4x5- 12 X* - 8x^ + 4x2 + 16x+ 16.
17. 16x2 -96x + 216-^ + ^-
X X2
18. x6-6x* + 15x2-20 + lf--^+l.
* X2 X* X^
19. X* + 2 x3(y + 0) + x2(?/2 + 2.2 + 4 2/0) + 2 x?/0(?/ + z) +y^z^.
20. 2x2(?/ + 0)2 + 2l/2(2; + x)2 + 2 02(ic + 2/)2 + 4a;y2(X + 2/+0).
FBACTIONAL AND NEGATIVE INDICES. 319
CHAPTER XXL
Fractional and Negative Indices,
214. We have hitherto supposed that an index was
always a positive integer ; and this is necessarily the case
so long as we retain the definition of Art. 19 ; for, with
that definition, such an expression as a^, for example,
has no meaning whatever.
It is however found convenient to extend the meaning
of a", so as to include fractional and negative values of n.
Now it is essential that algebraical symbols should
always obey the same laws whatever their values may be ;
we therefore do not begin by assigning any particular
meaning to a", when n is not a positive integer, but we
first impose the restriction that the meaning of a" must
in all cases be such that the fundamental Index Law,
namely,
a"^ xa''^^ a'"+'',
shall always he true ; and it will be found that the above
restriction is of itself sufficient to define the meaning of a"
in all cases, so that there is no further freedom of choice.
For example, to find a meaning of a^ consistent with the Index
Law.
We must have
a^ X a^ = a~ ^ = a^ = a.
Thus a* must be such that its square is a, that is a* must
mean y/a.
320 FRACTIOISrAL AND NEGATIVE INDICES.
Again, to find a meaning of a~i consistent with the Index
Law.
We must have a^ x a~^ = a^~^ = a^ ;
a
Thus a-i must mean —
a
We now proceed to consider the most general cases.
215. To find the meaning of a", where n is any posi-
tive integer.
By the index law
111
a^ X a"" X a"" X • • • to n factors
1
Hence a" must be such that its nth power is a, that is
1
a" = ^a,
216. To find the meaning of a", where m and w are
any positive integers.
By the index law
m m m
a" X a** X a" X • • • to n factors
™ + ^ + ^ 4- ... to n terms ^
Thus a" is equal to the nth root of a"^, that is
FRACTIONAL AND NEGATIVE INDICES.
We have also
111
a" X a** X a" X • • • to m factors
321
= a
'- + '- + '- +
~ n n
a".
Hence a** may be considered as the mth power of a", and
by Art. 215,
1
a" is -l^a
m
Thus we may consider that a** is the nth root of the mth
power of a, or that it is the mth power of the nth root of a ;
which we express by
a" = V («") = ( V")"
Note. — It should be remarked that it is not strictly true that
^(a"*)={;y(a)}"' unless by the wth root of a quantity is meant
only the arithmetical root. For example y/{a^) has two values,
namely ± a^, whereas ( ^ay has only the value + a^.
Examples
(i.) 8^ = (^8)2 = 22 = 4, (ii.) 4^ = {/48 = ^64 = 8,
(iu.) 3^=^35=^43.
217. To find the meaning of a^.
By the index law
a''xa'^= a°+'"
a™;
.-. a** = a^a"* = 1-
Thus, whatever a may he, a^ = 1.
322 FRACTIONAL AND NEGATIVE INDICES.
218. To find the meaning of a~"*, where m has any
positive value.
By the index law
and, by the preceding article, a*' = 1 ; -a
.'. a"* X a-"*= 1,
so that a-"* = — , and a"" = —^
Thus we can transfer any factor from the numerator to
the denominator, or from the denominator to the numerator,
of a fraction, provided we change the sign of its index.
For example 2-2 = — = 1,
- = a6 1 = — -»
and ^ = a^b'^x-^y-^ =
219. In the preceding articles we have found that
in order that the fundamental Index Law, namely
«»* X a" = a'"+**, may always be obeyed, a"* must have a
definite meaning when n has any given positive or nega-
tive value. It can be proved ^'^ that, with the meanings
thus obtained,
a'" X a~ = a'"+", (a"*)" = a"*", and (aby = a'*6*»,
are true for all values of m and n.
It therefore follows that we may deal with quantities
containing fractional or negative indices in precisely the
same way as if the indices were positive integers.
* See Treatise on Algebra, Art. 168.
FRACTIONAL AND NEGATIVE INDICES. 323
For example {ah^-y = a^^'^ft^''^ = ah.
a^ xa^ = a^ ^ = a* = «/a.
1 1 _2 '+i-2
a^ X a^ y- a ^ = a- ^ 3 — ^o = i.
a*
a-2/a-* = a-2-(-'*) = a^.
^(a86-8c*) = ^(a3) . ^(6-3) • ^(c*)= a^&"M =ah-^c\
EXAMPLES LXV.
Find the numerical values of
1. si 3. 16-^. 6. (^\)-i 7. (27)-i 9. i^:^^)-^.
2. rl 4. (^V)'^- 6- (fl)"^- 8- (100)-^. 10. (t?^)-!
Simplify
11. a^ X a"i 14. a x a"i 17. (a^ft"^)^.
12. a^ X J. 15. a^6"^ x a~h\ 18. (a^6'^)«.
13. a~^ X a. 16. (a^6)2. 19. (a~h^)^.
20. (a*6"^)l ^''^- (^'-0^ X (^-^)' X C^'"')*"-
i 28. ^-^^.
21. {(a-2)2}t.
22. {(a"bT^.
23. {(a"^)8ri ^^
80.
(xyzy+v+'
24. a^ X a -f X a^^. x^+'j/^+^z^+y
25. a' X a"^ x (a2)-H«T^)*. 31. (a^^C^h^ ^ (aM&^.
26. a:P+« x x^-« h- x'^p. {b^)^a^ (p^y
324 FRACTEONAL AND NEGATIVE INDICES.
a2y3
33. X^P+<1 X XP-"^ X (a;2)9-2'- ^ CC^-8r.
r P-?W ff^ir-kpj- r-p'\q p-g q-r r-p
Z^. \xP^ )\x^' f\x^ I -^x "- P « .
|-r P2i:\?Z!!l'!:^ p-g 9-r r-j>
35. Llic '^ M J ' XX '• !> "^ « .
Express with fractional indices and simplify
36. > X > X >. 4o_ ^^, ^ ^^, ^ ^-i ^ ^1
-37. ^(a^y)x^(a2/2). ^^^ ^a^^{aV(«-^^)}-
38. ^(a2^5)xv(a3x). ^ ^^^ ^(a26i0c5)^^a66,3).
39. ^a^ x ^a^ x a"^ -- a^i". 43. V(«^^"^)^ ^(a-*&^).
44. ^(a^feM) x 6"^ X (cV)"^.
Express the following quantities with radical signs and positive
indices.
45. J - a-\ 47. a^6-i - ah'^.
46. a-46"^. 48. a-3&"^ + 3-2a3V^.
220. In the following examples the foregoing princi-
ples are applied, but the results of the operations of
addition, etc., which have to be performed on the indices
are given without any of the intermediate steps.
Ex. 1. Multiply a^ + 1 + a~^ by a^ - 1 -I- a^
a^ + 1 +a'^
a^ - 1 + a~^
a^ -^a^
1 _i
a^ -\- a 3
+ 1 + a"^.
FRACTIONAL AND NEGATIVE INDICES.
325
Ex. 2. Divide a+b + c -Sa^b^c^ by a^ + b^ + c^.
We perform the work by the synthetic method [Art. 83.].
f
(6^ + c^)
0 -Sahh^ +(6 + c)
a3 (6^ + c3) + a^(b^ + c^ + 2 b^c^) -(b + c)
aJ(^b^ + c^) + 6^ + c3 - dM
0.
Ex. 3. Find the square root of
•t 4- rf.
ic^ - 4 X » + 2 x^ + 4 a; - 4 X 5 + jc
4
^ - 4 x^ + 2 x^ + 4 X - 4 x^ + x^ (x^ - 2 x^ + a;^
(X^)2 = X^
(x^ - 2x^)2 = x^ - 4x^ + 4x
(x^ -2 x^+x^)2 = x^ - 4 x^ + 2 x^ + 4 X - 4 x^ + x^.
EXAMPLES LXVI.
Multiply
1. x^ + y^ by x^ - y^.
^_.A
4. x^ + y3 by x^ - yi
S. x^ + y^ by x^ - y^.
6. x^ + x"^ + 1 by x^ - 1.
3. xf + 1 by x^ - 1. 6. x^ - x^^y^ + y^ by x^ + y^.
7. a^ + a^ft^ + b^ by a^ - b\
8. a^ + a^&^ + 6^ by a^ - ah^ + h^
5 3 1 _1 3 1
9. x^ — X* + x^ — X ^ by x^ 4- x^.
10. x^ - x^ + x^ - X by x^ + ic^.
n _n ,
11. x« + x2 + 1 by x-« + x 2 4-1.
12. x^ + x"y" + y2n by a;-2n _^ x-"y~" 4- y~^.
13. a^ + 6^ 4- c^ - bh^ - c^a^ - ah^ by a^ 4- 6^ 4- ci
326 FRACTIONAL AND NEGATIVE INDICES.
14. ^J- tV a&^ + tV «^& - 2V ^^ by I «"^ + i ^^'
15. 81 x^ - 27 x^y^ + 9 x^y 3 _Sx^y + y^ by 3 x^ + yi
Divide :
16. a-b by a^ - bK 19. x^ + y^ by a;^ + yi
17. x^-y^hyx^-yK 20. a; - 243 ?/ 3 by x^ - 3 2^^.
3n 3n n n
18. x'^ - y^ hj x^ - y^, 21. ?/*4-&V+&* ty y-6^y^+6.
22. x^y~^ + 2 + x"%^ by x'fy"^ - 1 + x~^y^.
23. a^ + a^b^ - ah^ - ab + ah^ + b^ by a^ + b\
1 _4 2 _2
24. a3 - 2 4- a 3 by a3 - a 3.
25. (a; - x-i) - 2(x^ - x"^) + 2(x^ - x"^) by x^ - x^.
26. x^^/""^^ + y~^'x~^ by x^y"^ + y^x~^.
27. Simplify (aft-^cS)^ x (a^&^c-i)^ x (a-^bc^)^.
28. Simplify (a - 4 b^ (a + 2 a^&^ + 4 63) (a - 2 a^6^ + 4 5^).
29. Show that 1 - a^ + b^-a-^-b-^ ^ (g _ a-i)(6 _ 5-1).
2
30. Show that -^ —- p— + -^ — = 2 + xl
x^ - 1 x^ + 1 x^ - 1 x^ + 1
31. Multiply 4x2 -5x-4- 7x-i + 6x-2 by 3x-4 + 2x-i,
and divide the product by 3 x — 10 + 10 x~i — 4 x~^.
32. Simplify — ^— + ^ — + — ^ + ^
1-x^ l + x^ 1+x^ 1+^
33. Multiply
(a + 6) ^ + (a2 _ 62)i + (oj _ 5)f by (^ + 6)i _ (« _ 6)i
FRACTIONAL AND NEGATIVE INDICES. 327
34. Write down the square roots of
(i.) a;^ + 2 x^ + 1, (ii.) 4x^ - 4:X^y^ + y^,
(iii.) ax^ - 2 aV + a^x.
35. Find the square root of
4a;2a-2 - 12xa-i + 25 - 24a;-ia + 16x-^aK
36. rind the square root of
25 a;2y-2 + ^ y^x-"^ - 20 xy-i - 2 yx-^ + 9.
37. Find the square root of
x^ — 2 a~^x'^' + 2 a'^a;' + oT^x^ — 2 a^x^ + a^.
88. Find the square root of
x^ - 4x^^ 4- 4a; + 2a;^ - 4a;* + xi
Solve the following equations :
89. a;^-2x^ + l=0. 41. a;^ - 26 a;^ - 27 = 0.
40. a; - 8a;^ + 12 = 0. 42. x^ + Sx~^ = 4.
48. 4a;^-3a;"^ = 4.
328 SURDS. COMPLEX QUANTITIES.
CHAPTER XXII.
Surds. Complex Quantities.
221. Definitions. A surd is a root of an arithmetical
number which can only be fouDd approximately.
Thus ^2 and ^4 are surds.
An algebraical expression such as -^/a is also often
called a surd, although a may have such a^value that ^a
is not in reality a surd.
Surds are said to be of the same order when the same
root is required to be taken. Thus ^2 and 6^ are called
surds of the second order, or quadratic surds ; also ^4
and 5 are surds of the third order, or cubic surds;
and -^5 is a surd of the nth. order.
Two surds are said to be similar when they can be
reduced so as to have the same irrational factors.
Thus 2,y/2 and 6^2 are similar surds.
The rules for operations with surds follow at once
from the principles already established.
Note. — It should be remarked that when a root symbol is
placed before an arithmetical number it denotes only the arith-
metical root, but when a root symbol is placed before an algebraical
expression it denotes any one of the roots. Thus -^a has two
values, but y/2 is only supposed to denote the arithmetical root,
unless it is actually written ± y/2.
SURDS. COMPLEX QUANTITIES. 329
222. Any rational quantity can be written in the form
of a surd.
For example
and 2 = ^2 _ ^23 = «/2«.
Again -^2 = ^l^2'\ = ^2'.
Also, since -^a x ^b = -Vab, we have
3V2 = V3' X V2 = V(3' X 2) = V18,
2^5 = ^2^ X -^5 = ^(2^ X 5) = ^40,
and a^yft = ^a« X V^ = ^^^•
Conversely, we have
V18 ^ V(9 X 2) = V^ X V2 = 3 V2,
^135 + ^40 = ^(3^ X 5) + ^(2« X 5)
= 3^5 + 2^5 = 5^5,
and -^a^ 4- Va6^ = ay'ct 4- 6^a = (a + 6) y'a.
t. -4ny ^i«o surds can be reduced to surds of the same
order.
This follows at once from the fact that -y/a = "^'a*".
For example, to reduce ^2 and ^5 to surds of the
same order.
We have
-2/2 = ^1^2^ = ^2«, and ^5 = ^l^5'\ = ^5^;
thus the equivalent surds of the same order are ^8
and ^25.
Again, to reduce ->/a and ^b to equivalent surds of
the same order.
330 SURDS. COMPLEX QUANTITIES.
We have
^a = ^{^a^^l ='Va^ and ^b = Vf V^"i ="76".
Hence the required surds are "4ya'" and ''^b''.
Ex. 1. Reduce ^3 and ^5 to surds of the same order.
The L. C. M. of 4 and 6 is 12. Hence we have
^3 = ^{^33} =1^27, and ^5 = ^{^62} =1^25.
Thus ^^27 and 1^5 are the required surds.
Ex. 2. Which is the greater, ^14 or ^6 ? We must reduce the
surds to equivalent surds of the same order.
Thus ' ^14=^142=^196,
and ^6 = ^63 = ^216.
It is now obvious that y/6 is greater than ^14.
224. The product of two surds of the same order can
be written down at once from the formula
^a x^b = Vab.
When surds are of different orders their product
cannot be simplified until they are reduced to equivalent
surds of the same order.
Ex. 1. Multiply V5 by V20.
We have ^6 x V20 = VC^ X 20) = VlOO = 10.
Ex. 2. Multiply y/2 by ^3.
V2 X ^3 = ^3 X ^32 = ^(28 X 32) = 4/72.
Ex. 3. Divide ^2 by V^-
^2/ V6 - ^27 ^6-3 = ^(22/63) = ^3^j.
SUKUS. COMPLEX QUANTITIES. 331
Ex. 4. Multiply 4 ^3 + 4^2 by 2 V3 - 2 ^2.
The process is as under :
4V3 + 4V2
2V3-2V2
8x3 + 8^/6
- 8y/6 - 8 X 2
24 -16 =8.
EXAMPLES LXVII.
Simplify
1. V27 + >/48. 15. ^i2xV24.
2. V50 + V98. 16. 2Vfx3v^
3. V45 + 2V126. ^^ 3^
4. 2V180-V406. ^ ^
6. 2V28-V63. "• V12XV27XV75.
6. 5V2-08-3V3-26. "' ^^« >< ^« >< ^•
7. 3 3.5 + ^625. ^- ^12x^75x^30.
8. 3^72 - 2 </243. ^l. </6 x ^12 x ^18.
9. 4^448-15^7. 22. VlO x ^00.
10. V512 - V^O - y/9S. 23. ^4 X ^.
11. 3 V12 - V27 + 2 V75. 24. y/m-^y/50.
12. V147-2V27-V3. 26. V63^Vn2.
13. 5 Vi - \/8 + -^- 26. V20 X y/9Q - ^30.
■v/18
14. y/15 X V60. 27. ^147 -f- ^35 x ^736.
28. Which is the greater, ^3 or ^(5^)?
29. Arrange in order of magnitude ^^50, ^^344, and ^402.
30. Multiply V6 - V3 hy y/Q + ^3.
31. Multiply 2 V5 + 3 ^3 by 3 V5 - 4 V3-
32. Multiply ^2 + ^3 + V^ by 2 v2 + 3 V3 + v'B.
33. Multiply 1 + ^3 + ^6 by 1 + ^3 - ^5.
332 SUKDS. COMPLEX QUANTITIES.
34. Find the square of ^2 + V^ + V^-
35. Find the continued product of ^2 + y'3 + ^5, —^-\-^Z
+ V5, V2 - V3 + V5, and V2 + V^ - V^-
225. When surds occur in the denominators of frac-
tions they can be got rid of, and the denominators are
then said to have been rationalized.
The following examples will sufficiently illustrate the
process :
3 ^ 3xV5 ^3 .^
3V2^3V2x V7^3 ,^^
2+ V5^(24-V5)(V5 + l)^2V5 + 5 + 2+V5
V5-1 (V5-1)(V54-1) 5-1
= i(7 + 3V5),
a-^b^ (a--y/b) (a - ^b) ^a^-2a^b-\-b
It is important to notice that a ± ^b is made rational
by multiplying by a if ^b ; also that ^a ± ^b is made
rational by multiplying by -y/a T ^h.
When the denominator of a fraction is rationalized its
numerical value can be more easily found.
226. The following is an important proposition.
If a-\- ^b = « + ^^, where a and a are rational, and
^b and -^(S are irrational; then will a= a, and b = p.
For we have (a — a)+ -^b = ^fi.
SUEDS. COMPLEX QUANTITIES. 333
Square both, sides ; then, after transformation, we have
2(a -a)-y/b = l3- b- {a - a)'.
Hence, unless the coefficient of ^h is zero, we must have
an irrational and a rational quantity equal to one another,
and this is impossible.
The coefficient of ^h in the last equation must there-
fore be zero ; hence a = a. And when a = a we have
from the given relation, ^h = ^p, ox b = ft.
Hence, if the sum {or difference) of a rational quantity
and a quadratic surd be equal to the sum {or difference) of
another rational quantity and a quadratic surd, the two
rational quantities must be equal to one another, as also
the two irrational quantitie,s.
Note. — It should be noticed that when a -f ^/b = a + ^^/S we can
only conclude that a = a and 6 = /3 provided that y/b and ^/^ are
really irrational. We cannot, for example, from the relation
S + y/i = 2 + y/9, conclude that 3 = 2 and 4 = 9.
227. The square root of a binomial expression which
is the sum of a rational quantity and a quadratic surd
can sometimes be found in a simple form. The- process
is as uncjer.
To find V(^ + V^)' where ^b is a surd.
Let ^{a-\-^b)=^a-\--^/ft (i.)
Square both sides ; then
Now, since -^/b is a surd, we can equate the rational and
irrational parts [Art. 226] ; hence
«=«+^l (ii.)
6 = 4a/J J
334 SURDS. COMPLEX QUANTITIES.
Hence [Art. 183] a and ^ are roots of the equation *
a;2 _ aaj + - = 0 ;
and these roots are ;
Thus v(^+v^)=V{-'"^f~-j;.,,^^^ ^'
It is clear, that unless -yjio? — h) is rational, the right
side of (iii.) is much more complicated than the left.
Thus the above process is of no utility unless a^ — & is a
square number ; and as this condition will not often be
satisfied, the process has no great practical utility.
From (ii.) we see that we have to find two numbers
whose sum is a and whose product is - ; and if two rational
numbers satisfy these conditions, they can generally be
found at once by inspection.
Ex. 1. Find y/{\b + 2 V^B).
Let V(15 + 2 V56) = V« + V^-
Square both sides ; then
15 + 2 V56 = a + ;8 + 2 ^afi.
Equating the rational and irrational terms, we have
o + )8 = 15,
o3 = 56.
The numbers which satisfy these relations are obviously 7 and 8.
Hence yj(l^ + 2 y/bQ) = V7 + V^-
-Tf-^ SURDS. COMPLEX QUANTITIES.
^^t --^x. 2. Find ^(6-^/35).
\ I Let V(6 - V35) = y/a- y/$.
...Square both sides ; then
6 - V35 = a + iS - 2 y/a$.
:ce, equating the rational and irrational terms, we have
a+)8 = 6,
335
Jy mspection, or by solving the equations for a and jS, we find that
EXAMPLES LXVIII.
Rationalize the denominators of
1
3
V7
8.
3^2
2v3
5 2
1+V2
2.
2
V5
4.
V6.
V5
6. V2-1
v^ + 1
7
2V5
V6 + V3
11 1
1+V2+V3
8.
15 + 14^3
15-2V3
12. 3
2 + V3 + V5
9.
V6 + 3 ^3
2 V6 - V3
13 ^ 1 ^
V»/2 - 1 ' ^ + 1
10.
V6 - 3v/12
2V6+ y/l2
14. ^ 1 ^
^ - 1 \S/9 + 1
15.
Simplify
1
^ 1
(2-V3)2 ' (2+V3)2
16.
Simplify
1
/■ct
, 1 .
/0\« ' /-O 1 *o\«
(2-V3)« (2+V3)»
17. Simplify (3 + V2)(5-v^)
(3-v2)(6 + V2)
336 SURDS. COMPLEX QUANTITIES.
18. Simplify 2V15-3V5 + 2V2^
V15+V2
19. Simplify (7-2V5)(5 + V7)(31 + 13V5)^
(6-2 V7)(3+V5)(11 + 4V7)
21. Show that ,^1^0"^^^ = 2 + V2 + V3 + V6.
22. Simplify
1 , 1
+ ^ :: -^
1+V2 + V3 -1+V2+V3 1-V2+V3 1+V2-V3
24. Show that
26. Find the value of '^ — : — to three places of decimals.
V2-1
27. Find the value of "^ + - — ^ to three places of decimals.
2, — -y/O 2 + -yjL
Find the square roots of
28. 6 + V20. 32. 101 - 28 ^13.
29. 16 + 6V7. 33. 117 + 36 VlO.
30. 12-6V3. 34. 280 + 56V21.
31. 28-5V12. 35. 4^ + 2 V2.
Simplify
36. 3V6-V2+V(7+2V10). 37. 6-4^3 + ^(16-8^3).
SUKDS. COMPLEX QUANTITIES. 337
Find the square roots of
38. 11 + 2(1+V5)(1+V7)- 40. 2a + 2v(a2-cc2).
39. 2x + 2V(x^-l)- 41- 3a;-l+2V(2x2+x-6).
Simplify
42. ^ ^ ^ , • 43
V(16 + 6V7) V(15 + 2V56)
44. V(7 + 2V10) + V(7-2V10).
45 VC3 + 2V2)-V2.
V2+V(3-2V2)
2 + a/3 2-V3
V2 + V(2 + V3) V2 - V(2 - V3)'
46.
47. Show that V^ + V^^ = 3 • 632 . ..
48. Find the value of x^ — 4 x 4- 5 when x = 2 + y/b.
49. Show that, if x2 = a; + 1, then x^ = 2x + 1 and x^ = 5x + 3.
50. Show that, if
x2 = 3 X + 6, then x« = 14 x + 16 and x* = 67 x + 70.
COMPLEX QUANTITIES.
228. In the chapters on quadratic equations, and on
equations of higher degrees, we encountered expressions
which involve the square roots of negative quantities.
Such square roots present themselves in the form ^—c,
in which c is a positive number. This is called a pure
imaginary as distinguished from the mixed binomial form
a-{-y/—c, which is called a complex quantity.
Such expressions do not enter largely into the applica-
tions of elementary algebraic principles, and a detailed
study of them would displace other matters of more
immediate importance to the student. But some account
338 SURDS. COMPLEX QUANTITIES.
of their behaviour in the operations of algebra should be
given. The following discussion consists, in the main,
of explanations, without formal demonstration, of the
principal laws of operation with complex quantities.
229. In admitting imaginary forms to the category of
algebraic quantity, it becomes necessary to enquire
whether the algebraic processes may be applied to them
without limitation. And we find in fact at the outset^
that if we attempt to apply the i-ule that the sign ^ is
distributive over the factors in a product, and assert that
V— cxV — c = V(— c) x{—c) — c,
this is at once in conflict with the other rule that
which asserts that to square an indicated square root
has merely the effect of removing the radical sign.
Driven to a choice of interpretations, we reject the
former, adopt the latter, and define outright that
{y/—cy = -e,
and accept the other consequent interpretations to which
this definition leads.
230. If now we assume the commutative and associa-
tive laws in multiplication [Art. 52], and write
ax-\/ —c = ^ —c xa,
(a X V— c) X 6 = a X (V— c x 6),
we have
Vc X V— 1 X Vc X V— 1 = Vc X \/c X V— 1 X V^^,
SURDS. COMPLEX QUANTITIES. 339
whence ( Vc X V— 1)^ = — c ;
and comparing this with
we see that it is legitimate to write
V— c = VcxV^.
The form Vc x V— 1, which is the one universally
employed, is usually written in the abbreviated form hi,
in which i now takes the place of ^—1, and b of yc.
The symbol ■\J—1, or its equivalent i, is called the imag-
inary unit.
231. With the interpretation oi ■^J—l we have now
adopted, we have
^« = _1, V = -i, ?•«=!, etc.,
so that i, — 1, — i, 1, are the only values that the inte-
gral powers of V~^ ^^^ assume.
232. For the imaginary unit in its combinations with
real quantities we have assumed the commutative and
associative laws in multiplication, namely
(axi)xb = ax{i X b),
and it is equally essential that the like laws in addition
a + i=i-\-ay
(a + i) + b = a-h{i-\-b),
and the law of distribution
i x(a-{-b) = i xa-{-i xb
340 SURDS. COMPLEX QUANTITIES,
shall also obtain. With these laws and the interpreta-
tion of ^— 1 definitely agreed to, we can perform the
fundamental algebraic operations upon complex quanti-
ties. [See Treatise on Algebra^ pp. 221, 222 ; also
Chrystal's Algebra, Vol. I., Chapter XI.]
The following examples illustrate the application of
the rules of addition, subtraction, multiplication, and
division.
Ex. 1. The sum of a + hi and c + di is
{a-^U)^(c-\-di) = ia + c) + (h-\-d)i.
Ex. 2. The difference of a + hi and c -{■ di is
{a + hi)-{c + di) = {a-c)-^{h-d)i.
Ex. 3. The product of « + hi into c + di is
(a 4- hi) (c + di) =ac -{- adi + hci + hdi^
— {ac — hd) + {ad -f hc)i.
Ex. 4. The quotient of a + hi by c + di is
a + hi _{a -\- hi) (c — di)
c + di (c + di) (c — di)
_ac — adi + hci — bdi^
c2 — cdi + dci — d'H'^
_ (ac + hd) + (6c — acg)i
C2 + d2
Ex. 5. The square of a + hi is
(a + hi)-^ = a'^-b^-\-2 abi.
Ex. 6. The reciprocal of a + hi is
1 _a — hi
a + hi~ a^ + b"^
From these examples it is at once evident that sums,
differences, products, and quotients of complex quantities
are themselves complex quantities. It may likewise be
SURDS. COMPLEX QUANTITIES. 341
shown, that from the other algebraic operations, when
applied to complex quantities, only other complex quan-
tities (or possibly reals and imaginaries*) can result.
[See Stringham's Uniplanar Algebra, Chapter III.]
233. When two complex quantities have their real
parts the same and their imaginary parts different only
in sign, they are said to be conjugate to one another. For
example, —a + hi is conjugate to — a — hi.
Both the sum and the ^woduct of two conjugate complex
quantities are real.
For, let the expressions be a + hi and a — hi, both a
and h being real. Their sum is
a-\-hi-\- a — hi = 2 a,
and their product is
(a + hi) (a - hi) = a^ - ahi + hai - hH^ = a^ -f- h^.
Conversely, if the sum and the product of two complex
quantities he hoth real, the complex quantities must he con-
jugate.
For, let the expressions be a + hi and c -f- di, a, h, c,
and d being real. Their sum is
a -\- hi -\- c -\- di = a -\- c -{- {h ■\- d) i,
which cannot be real unless 6 -f d = 0 ; and their prod-
uct is
(a + hi) (c + di) = ac + adi -f- hci + hdi^
= {ac — hd) + (ad -f hc)i,
which cannot be real unless ad-{-hc = 0.
* Reals and imaginaries may be regarded as particular cases of
complex quantities, the former being complex quantities with zero
imaginary parts, the latter complex quantities with zero real parts.
342 SURDS. COMPLEX QUANTITIES.
But if b -\- d— 0 and ad-\-bc = 0, then b{c — a) =0,
whence, either c = a, or 6 = 0. Then :
(1) If b = 0, from b -\-d = 0 it follows that d is also
zero and the original expressions are real.
(2) If c = a {b not zero), it follows, because d = — b,
that c -\-di= a — bi,
which is conjugate to a + bi. q.e.d.
234. When an expression is written in the form a -j- bi,
it is understood that a and b are both real.
If a-{-bi = 0, then a = 0. and 6 = 0.
For^ if a + bi = 0, then a-= — bi.
But a real quantity cannot be equal to an imaginary
one unless both are zero.
.-. a = 6 = 0.
This proposition is a lemma to the following.
Two complex quantities cannot be equal to one another
unless their real and their imaginary parts are respectively
equal
For, if a 4- 6i = c + di,
then a — c = {d — b)i'y
whence, as previously proved,
a — c = 0, d — b = 0,
that is a = c and d=b. q.e.d.
EXAMPLES LXIX.
Simplify each of the following expressions by reducing it either
to a form having no imaginary part, or to a form having no real
part ; or if it have both real and imaginary parts, reduce it to the
typical form a + bi.
SUBDS. COMPLEX QUANTITIES. 343
1- h *■ r^*- ■>■
3i 1 -I
m'
^ 5. %±^- 8. (l±i\*.
1+i 2-5i V V'2 /
3. i±i. 6. ii:^tO-'. 9 /^-^y.
l-i 2i-l ' \ I /
10. (-3 + 20(3 + 20. ^^ 1 + t l-t
11. (5i + 2)(15 + 60. 1-2? l + 2i
12. a + JV3)^. 15 2-3i ^ 2 + 3i
13. (i-^V3)«. ' ^ + ^*" ^-2*''
16. Write the equation whose roots are 1, i, — i, and — 1.
17. Write the equation whose roots are 1, \ + \iy/Z, and
18. li ta=-\-\-\ 1 ^3, prove that 1 + w + «2 = 0.
344 RATIO. PROPORTION. VARIATION.
CHAPTER XXIII.
Ratio. Proportion. Variation.
235. Definitions. The relative magnitude of two quan-
tities, measured by the number of times which the one
contains the other, is called their ratio.
Concrete quantities of different kinds can have no
ratio to one another : we cannot, for example, compare
with respect to magnitude miles and tons, or shillings
and weeks.
The ratio of a to 5 is expressed by the notation a : 6 ;
and a is called the first term, and b the second term of the
ratio.
Sometimes the first and second terms of a ratio are
called respectively the antecedent and the consequent.
It is clear that a ratio is greater than, equal to, oi-
less than unity, according as its first term is greater than,
equal to, or less than the second.
A ratio which is greater than unity is sometimes called
a ratio of greater inequality, and a ratio which is less than
unity is similarly called a ratio of less inequahty.
236. Magnitudes must always be expressed by means
of numbers ; and the number of times which one number
is contained in another is found by dividing the one by
the other. Hence
a : 6 is equal to —
b
Thus ratios can be expressed as fractions.
RATIO. PROPORTION. VARIATION. 345
237. A fraction is unaltered in value by multiplying
its numerator and denominator by the same number.
[Art. 158.]
Hence also a ratio is unaltered in value by multiplying
each of its terms by the same number.
Thus the ratios
2:3, 6:9 and 2 m : 3 m
are all equal to one another.
Again, the ratios 4:5, 7:9 and 11 : 16 are equal respectively to
36 : 46, 36 : 46 and 33 : 45.
Hence the ratios 4:6, 7:9 and 1 1 : 16 are in descending order
of magnitude.
238. A ratio is altered in value when the same quan-
tity is added to each of its terms.
For example, by adding 1, 10 and 100 to each of the terms of
the ratio 4 : 5, we obtain respectively the ratios
6:6, 14 : 16 and 104 : 105 ;
and these new ratios are different from the given ratio and from
each other.
Since
we see that by adding the same quantity to each of the terms of
the ratios 4 : 5, a new ratio is obtained which becomes more nearly
equal to unity as the quantity added becomes greater.
This is a particular case of the following general proposition.
239. A)iy ratio is made more nearly equal to unity by
adding the same positive quantity to each of its terms.
By adding x to each term of the ratio a : b, the ratio
{a -{- x) : {b -{- x) is obtained. We have to show that
(a + a)/(6 + x) is more nearly equal to 1 than is a/b.
PROPORTION
VARIATION.
t-'-
a —
b
b
a-\-x
_1_
a-
-b
346 RATIO
Now
also ^_ — ,
b -j- X b -i-x
and it is clear tliat the absolute value of (a — b)/(b + x)
is less than that of (a — b)/b, for the numerators are the
same and the denominator of the first is greater than
that of the second : this proves the proposition.
Now when a is greater than b, a/b is greater than 1,
and so also is {a-\-x)/{b-{- x)-, hence, as (a -\- x) / (b -\- x)
is more nearly equal to unity than a/b is, it follows that
{a -\- x) / {b -{- x) is less than a/b.
Thus a ratio which is greater than unity is diminished
by addi7ig the same positive quantity to each of its terms.
If, however, a is less than b, a/b is less than 1, and
so also is (a + x)/{b + x); but (a + x)/{b + x) is
more nearly equal to unity than a/b' is, and therefore
{a -\-x)/(b -^ x) is greater than a/b.
Thus a ratio which is less than unity is increased by
adding the same positive quantity to each of its terms.
Also, when x is very great, the fraction {a — b)/(b-\-x)
is very small ; and we can make (a — b)/(b + x), which
is the difference between {a -}- x) / {b -\- x) and 1, as small
as we please by taking x sufficiently great. This is ex-
pressed by saying that the limit of (a-\-x)/(b-\-x) when x
is very great, is unity.
240, The following definitions are sometimes required :
The ratio of the product of the first terms of any
number of ratios to the product of their second terms
is called the ratio compounded of the given ratios.
RATIO. PROPORTION. VARIATION. 347
Thus ac : bd is called the ratio compounded of the
ratios a : b and c: d.
The ratio a^ : b^ is called the duplicate ratio of a:b.
The ratio a^ : b^ is called the triplicate ratio of a:b.
The ratio -y/a : -^b is called the sub-duphcata ratio of
a : b.
241. Incommensurable Numbers. The ratio of two quan-
tities cannot always be expressed by the ratio of two
whole numbers ; for example, the ratio of a diagonal to a
side of a square cannot be so expressed, for this ratio is
^2, and we cannot find any fraction which is exactly
equal to ■y/2.
When the ratio of two quantities cannot be exactly
expressed by the ratio of two whole numbers, they are
said to be incommensurable.
Although the ratio of two incommensurable numbers
cannot be found exactly, the ratio can be found to any
degree of approximation which may be desired ; and the
diiferent theorems which are proved with respect to ratios
of commensurable numbers can be proved to be true also
for the ratios of incommensurable numbers.
EXAMPLES LXX.
1. Arrange the ratios 5 : 6, 7 : 8, 41 : 48, and 31 : 36 inglescend-
ing order of magnitude.
2. For what value of x will the ratio 3 + a; : 4 + x be equal to
5:6?
3. For what value of x will the ratio \b -\- x : 11 + x be equal
to ^?
4. What must be added to each of the terms of 3 : 4 to make
the ratio equal to 25 : 32 ?
348 RATIO. PROPORTION. VARIATION.
5. Find two numbers whose ratio to one another is 5 : 6, and
whose sum is 121.
6. Two numbers are in the ratio 3 to 8, and the sum of their
squares is 3577 : find them.
7. What is the ratio of x to y, if ^J^"^^^ = 2 ?
Zx-y
8. If 4 x2 -f ?/2 — 4 xy^ find the ratio of x to y.
9. Find x : y, having given x^ + Qy^ = 5 xy.
10. A certain ratio will be equal to 2 : 3 if 2 be added to each
of its terms, and it will be equal to 1 : 2 if 1 be subtracted from
each of its terms : find the ratio.
11. Find two numbers such that their sum, their difference, and
the sum of their squares are as 7 : 1 : 76.
12. What is the least integer which must be added to the terms
of the ratio 9 : 23 in order to make it greater than the ratio 7:11?
13. Write down the ratio compounded of the ratios 2 : 3 and
15 : 16 ; also the ratio compounded of the ratios 5 : 6 and 18 : 25.
14. Write down two quantities which are in the duplicate ratio
of 2 ic : 3 ?/.
15. Find x in order that a; + 1 : a; + 4 may be the duplicate
ratio of 3:5.
16. The ages of two persons are as 3 : 4 and thirty years ago
they were as 1 : 3, what are their present ages ?
17. Show that, if from each term of a ratio the inverse of the
other be taken, the ratio of the differences will be equal to the
original ratio.
18. Show that, if a and x be positive and a > x, then a^ — x^:
a2 + x^ will be greater than a — x : a + x.
RATIO. PROPORTION. VARIATION. 349
PROPORTION.
242. Definition. Four quantities are said to be propor-
tional when the ratio cf the first to the second is equal to
the ratio of the third to the fourth.
Thus a, b, c, d are proportional, if
a:b = c:d.
This is sometimes expressed by the notation
a:b::c:dy
which is read " a is to 6 as c is to d."
The first and fourth, of four quantities in proportion,
are sometimes called the extremes, and the second and
third of the quantities are called the means.
243. If the four quantities a, b, c, d are proportional,
we have by definition
b^d
Multiply each of these equals by bd ; then
ad = be.
Thus the product of the extremes is equal to the product
of the means.
Conversely, if ad = be then a, b, c, d will be propor-
tional.
For, if ad = be,
then qd^bc, . a^e^
bd bd b d
that is a:b = e: d.
350 RATIO. PROPORTION. VARIATION.
Thus, if a:b = c:d, then ad = bc;
and conversely if
ad = be, then a:b = c:d.
244. It follows from the last article that the four rela-
tions
a:b = c:d,
a:c=b:dj
b:a = d:Cf
and b: d=:a:c,
are all true, provided that ad = bc.
Hence the above four proportions are all true, when
any one of them is true.
245. If a:b = c:d,
then will a-\-b: a — b = c-\-d:c — d.
For a-{-b:a — b = c-\-d:c — d,
if {a-^b)(c — d) = {a — b){c-\-d),
that is, if ac-\-bc — ad — bd = aG—bc-\- ad — bd,
or, if be = ad.
But this condition is satisfied, since a : b — c : d.
The above proposition has already been proved in Art.
170, Ex. 1.
246. Definitions. Quantities are said to be in continued
proportion when the ratios of the first to the second, of
the second to the third, of the third to the fourth, etc.,
are all equal.
RATIO. PROPORTION. VARIATION. 351
Thus a, b, c, d, etc., are in continued proportion if
a:6 = 6:c = c:d, etc. ,
that is, if ^ = ^ = 2 = etc.
bed
If a:b = b :c, then h is called the mean proportional
between a and c ; also c is called the third proportional to
a and b.
247. If a, b, c be in continued proportion, we have
a_b
b~c'
. • . b"^ =ac, OT b= -\/ac.
Thus the mean proportional between two given quantities
is the square root of their product.
Also ^x^ = ^x^
b 0 c b
that is ^' = -.
Hence a : c — a^ : b^.
Thus, if three quantities are in continued proportion the
ratio of the first to the third is the duplicate ratio of the
first to the second.
248. It is often very convenient to represent a ratio
by a single letter, as in Art. 170. The following are
additional examples :
Ex. 1. If a'.b = c:d,
then a2 + a5 : c2 + cd = &2 _ 2 a6 : (22 - 2 cd.
Let - = x; then also - = x.
b d
352 RATIO. PROPORTION. VARIATION.
Hence a = bx, and c = dx.
(jfi + ah ^ 6^x2 + h-^x ^ h'^jpfi + x) ^ 6^
' ' c^ + cd d^x'^ 4- d^x d\x:^ + x) d^'
Also 6'^ - 2 q^ ^ 62 _ 2 6% ^ 62(i_2x) ^ 6^^
d^-2c(^ #_2(i2a; (i2(^i_2x) d^'
Hence ai±^& ^ ^!zil^,
that is a2 + a& : c2 4- cd = &2 _ 2 a& : ^2 _ 2 c(?.
Ex. 2. If a: b = c:d = e:f^ show that
a3 + c3 + e3 : 63 + ^3 ^^-3 ^ ace : 6d/.
Let a/6 = x ; then c/d = x and e/f = x.
Hence a = bx, c = dx, and e=fx;
a^ -{- c^ -\- e^ _ b^x^ + d^x^ + f^x^ _ „3
•*• 684.^34.^-3- 534. ^3 4.^3
And — = ^^'^^'f^ - ic8
bdf bdf
Hence «i±ci±^ ^ ace,
63 + #+/3 6d/
that is a^ + c8 + e8 : 68 4. (^3 +y3 _ ^qq . ^^f^
Ex. 3. Show that, if
x:26 + 2c-a = t/:2c + 2a-6 = 0:2a + 26-c,
then will a:2?/4-20 — x = 6:22! + 2ic — y = c:2x + 2y-2;,
We have —
26 + 2c-a 2c + 2a-6 2a + 2.6-c
Put A for each of these equal fractions ; then
ic = A(26 + 2c-a), 2/ = A(2c + 2a-6),
;s = \(2a + 26 - c).
Hence 22/4-22! — a; = 9 aA, and similarly
20 + 2x — ?/ = 9 6a and 2x+2y — 0 = 9cA.
a 6 c
Whence
2y + 20-x 2« + 2x-i/ 2x + 2y-;2
EATIO. PROPORTION. VARIATION. 353
249. The definition of proportion given in Euclid is
as follows: Four quantities are proportional, when if
any equimultiples be taken of the first and the third, and
also any equimultiples of the second and the fourth, the
multiple of the third is always 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.
If the four quantities a, b, c, d satisfy the algebraical
test of proportionality, we have
b~d''
therefore, for all values of m and n,
ma _mc
nb nd
Hence
mc>= or <nd, according as ma > = or < w6.
Thus a, b, c, d satisfy also Euclid's test of proportionality.
Next, suppose that a, b, c, d satisfy Euclid's definition
of proportion.
If a and b are commensurable, so that a:b = m:n,
where m and n are whole numbers ; then
a
b^
m
na =
= mb.
But, Euclid's definition of the proportion a:b=c:d
asserts that
nc > = or < md, according as wa > = or < mb.
z
354 RATIO. PROPORTION. VARIATION.
Hence, as na = mb, we must have
nc — md ;
c _m _a
' d n b
Thus a, b, c, d satisfy the algebraical definition.
If a and b are incommensurable we cannot find two
whole numbers m and n such that a : b = m : n. But, if
we take any multiple na of a, this must lie between two
consecutive multiples, say mb and (m + 1)6, of b, so that
na > mb and na < (m + 1)6.
Hence, by the definition,
nc > md and nc < (m -f l)c? ;
.*. — >— and -< — ^!^ —
d n d n
Hence both a/b and c/d lie between
™ and '?^+l.
n n
Thus the difference between a/b and c/d is less than 1/n ;
and as this is the case however great n may be, a/b
must be equal to c/d.
EXAMPLES LXXI.
1. Show that, if a:b::c:d, then
(i.) ac'.bd-.id^icP.
(ii.) abicd: :a^:c^.
(iii.) a2:c2::a2_52.c2_<22.
3. Show that, if a:b = c:d, then
2a + 3c:3a + 2c = 26 + 3d:36+2(!.
RATIO. PROPORTION. VARIATION. 355
3. If a : 6 :: a + c : 6 + cZ, then
c:d::c -{- a:d + b.
4. If a : b = c : d, then
la + mb:pa + qb = lc-\- md:pc + qd.
5. Show that, if 3a-56:3c-5d = 5a + 36:5c-p3d,
then a:b = c:d.
6. Find a mean proportional to a^b and ab^.
7. Find a mean proportional to (a -\- by and (a — by.
8. Find a third proportional to a and a^ ; also to
(a - by and a^ - 62.
9. If a:b::c:d, then will ab + cd he a mean proportional
between
a2 + c2 and b"^ + d^.
10. Show that, if a:b::c:d, then
(i.) a: a + c: :a-\- b: a + b -{■ c + d.
(ii.) «2 + 0^6 4. 52 . <j2 _ 055 ^ ^2 . . c2 + cd + d-^ : c2 - cdJ + d2.
(iii.) a + 6 : c + d : : Va^ + 6-^ : VcM^.
(iv.) Va2 + 62 : Vc2 + (^^ : : v^a^ + 6* : v^cS + d*.
(v.) a2c _|. ac2 : 62d + 6^2 : : (a + c)* : (6 + d)».
(vi.) \/a»» + 6" : Vc" + d" : : >/a'- - 6'' : \/c^ - (?•.
11. If ? = ^ = ?, then will ^L±y = yjtl = L±J^, and also
a b c a+6 6+cc+a
(a2 + ^,2 4. c2)(x2 4. y2 4. 2-2) = (^X + 6^ + C«)2.
12. Show that, if
bz — cy _cx_—_az _ ay — bx
a b c
then will ? = 2^ = 1
a b c
356 RATIO. PROPORTION. VARIATION.
13. Show that, if (^Y + (^)' = 2^'
then will a:h = c:d.
14. Show that, if
lx{ny — mz), mijilz — nx) and nz{mx — ly)
be equal and not zero, then will
mn + nl-{- Im = 0 and 2/0 + 0X + ojy = 0.
15. Show that, if
X V _ y _ z
h -{■ c — a c + a — b a + b — c
then will (b — c)x + (c — a)y + (« — b)z = 0.
16. Show that, if a:b = c:d = e:f, then
a^ + a^c + ace : b^ + b^d + bdf: : ace + ac^ + c^ : 6(?/+ 6(^2 ^_ ^
17. If a : a; : : 6 : 2/ : : c : 0, prove that
a* + a--^62 + 6* : a;4 + ic2?/2 + y* : : 6* + 62c2 _|. c* : yi + ?/2^2 + ^.
18. Show that, if a : b = pa — qc : pb — qd, then will
c : d = pa + qc : pb + qd.
19. Show that, if a:b — c:d, then
bade
20. If 4 a - & : 4 a + 6 : : 1 : 2, find the value of
7a + 36:7a-3 6.
21. If icy + 3 : a;0 + 1 = 2/2 + 3 : 2/0 + 1,
show that x = y or ?/ = 3 0.
22. Show that, if a + 6 — c:c + d + a = a — c:2(f,
then b : a — c = a -\- c — d -.2 d.
23. Show that, if x - z:y — z — x^-.y'^,
then X + 2:1/ + 2r = a;2 + 2 xy : 2/2 + 2 a;2/.
RATIO. PROPORTION. VARIATION. 357
24. Show that, if a(y-z)+ h(z - x) + c{x -y) = 0,
then y — z:b — c = z — x:c — a = x — y. a — b.
25. Show that, if a:b: :c:d,
then will a^ + b^ + c^ + d^ : (a + by + (c + d)^ : : (a + g)2^
+ (6 + rf)2:(a + 6 + c + (i)2.
VARIATION.
250. When any substance is sold at a fixed price per
pound, the cost of any amount of it is so related to its
weight that when the weight is doubled the cost is also
doubled, when the weight is halved the cost is also
halved, and so on, the ratio of any two values of the cost
being equal to the ratio of the corresponding Aveights.
When two quantities are related in this way they are
said to vary as one another.
Definition. One magnitude is said to vary as another
when the two are so related that the ratio of any two
values of the one is equal to the ratio of the correspond-
ing values of the other.
Thus, if tti, tta, be any two measures of one of the
quantities, and b^, h^j be the corresponding measures of
the other, we have
-^ = -1 ; and therefore -^ = ^•
Hence the measures of corresponding values of the two
magnitudes are always in the same ratio.
251. The symbol oc is used for the words varies as:
thus ^ oc jB is read " A varies as B."
358 RATIO. PROPORTION. VARIATION.
If a X 6, the ratio a: b is constant ; and if we put m
for this constant ratio, we have
-=m; .'. a = mb.
b
To find the constant m in any case it is only necessary
to know one set of corresponding values of a and b.
For example, if a x 6, and a is 10 when b is 2, we have
a = mb; .•. 10 = w x 2 ; .'. m = 6.
Hence a = 5 &.
252. Definitions. One quantity is said to vary inversely
as another when the first varies as the reciprocal of the
second.
Thus a varies inversely as 6, if ax-, that is if a = m x -, or
I, b b
ab = m.
One quantity is said to vary as two others jointly, when
the first varies as the product of the two others.
Thus a varies as b and c jointly if a x &c, that is if a = mbc.
One quantity is said to vary directly as a second and
inversely as a third, when the ratio of the first to the
product of the second and the reciprocal of the third is
constant.
Thus a is said to vary directly as b and inversely as c, if
a:b X- is constant, that is if a = m-, where w is a constant,
c c
In all the different cases of variation defined above,
the constant will be determined when any one set of
corresponding values is given.
RATIO. PROPORTION. VARIATION. 359
For example, if a varies as 6 and inversely as c, and a is 6 when
6 is 2 and c is 9, we have
a = W-, and therefore 6 = w--
c 9
H .nee w = 27, and therefore a = 27 -•
c
253. Theorem. If a depends only on h and c, and if
a varies as b when c is constant, and varies as c when h is
constant ; then when both h and c vary, a will vary as he.
Let a, b, c; a', b\ c and a", 6', c' be three sets of corre-
sponding values.
Then, since c is the same in the first and second
ah ...
a=b' ('•)
Also, since 6' is the same in the second and third cases,
we have
a' c
(ii.)
Hence, from (i.) and (ii.),
a^ a' _ be
a'^^'~b^''
that IS — 77 = -r-:»
a" b'c'
which proves the theorem.
The following are examples of the above proposition :
The cost [O] of a quantity of meat varies as the price per
pound [P] if the weight [WT] is constant, and the cost varies as
the weight if the price per pound is constant ; hence, by the*
proposition, when both the weight and the price per pound change,
the cost varies as the product of the weight and the price.
360 RATIO. PROPORTION. VARIATION.
Thus, if C cc P, when W is constant,
and C cc W, when P is constant ;
then C cc PW, when both P and TT change.
Again the area [^] of a triangle varies as the base [J5] when
the height [iZ] is constant; the area also varies as the height
when the base is constant ; hence, when both the base and the
height change, the area will vary as the base and height jointly.
Thus, if Ace B, when H is constant ;
and if Ace H, when B is constant ;
then A qc BH, when both B and R change.
Again, the pressure [P] of a gas varies as the density [i>]
when the absolute temperature [ T ] is constant ; the pressure also
varies as the absolute temperature when the density is constant ;
hence, when both density and temperature change, the pressure
will vary as the product of the density and temperature.
Thus PccDT.
Ex. 1. If ^ oc B, and if also Ace C; then will B ce C.
For, since ^ oc P, we have A = mB, where m is some constant.
And, since Ace C, we have A = nC, where n is some constant.
Hence B = —C, where — is some constant; and therefore
poca
Ex. 2. If O oc WP, then will fFoc ^.
P
For, since C oc WP, we have C = m - WP, where m is some
constant.
1 r' 1 c
Hence W=— —, where — is some constant ; therefore Wee —
m P m P
Ex. 3. The pressure of a gas varies jointly as its density and its
absolute temperature ; also when the density is 1 and the tempera-
ture 300, the pressure is 15. What is the pressure when the
density is 3 and the temperature is 320 ?
RATIO. PROPORTION. VARIATION. 361
Since
Fee TD,
we have
P = mTD,
where m is some constant.
Also,
by the conditions of the problem,
15 =
wi X 300 X 1 ;
.'. m =
A.
.-. P =
^V ™.
Hence when D is 3 and T is 320, we have
P =
jV X 320 X 3
-
48.
EXAMPLES LXXII.
1. A varies as B, and ^ is 5 when jB is 3 ; what is A when
Bis 5?
2. W varies inversely as P, and TF is 4 when P is 15 ; what
is W when P is 12 ?
3. U xccy and y cc z, then will xz qc y"^.
4. If x^ X y and z"^ « y, then will xz qc y.
6. If a; cc - and y cc -■, then will xcc z.
y " z
6. A varies as B and O jointly ; also ^ = 4 when P = 2 and
0 = 6: find the value of A when P = 2 and 0 = 9.
7. A varies as P and inversely as C; also A = 2 when P = 3
and 0 = 4: find the value of P when ^ = 6 and 0 = 3.
8. The area of a circle varies as the square of its radius, and
the area of a circle whose radius is 10 feet is 314.159 square feet.
What is the area of a circle whose radius is 12 feet ?
9. The volume of a sphere varies as the cube of its radius, and
the volume of a sphere whose radius is 1 foot is 4.188 cubic feet.
What is the volume of a sphere whose radius is 3 feet ?
362 RATIO. PROPORTION. VARIATION.
10. The velocity of a falling body varies as the time during
which it has fallen from rest, and the velocity at the end of two
seconds is 64. What is the velocity at the end of five seconds ?
11. The distance through which a heavy body falls from rest
varies as the square of the time it falls, and a bjdy falls through
144 feet in three seconds. How far does it fall in two seconds ?
12. Given that the area of a circle varies as the square of its
radius, show that a circle of 5 feet radii:is is equal to the sum of a
circle of 3 feet radius and another of 4 feet radius.
13. The volume of a gas varies as the absolute temperature and
inversely as the pressure. Also when the pressure is 15 and the
temperature 280 the volume is 1 cubic foot ; what is the volume
when the pressure is 20 and the temperature 300 ?
14. If d?' — 62 varies as c^, and if c = 2 when a — h and 6 = 3:
find the equation between a, 6, and c, and show that 6 is a mean
proportional between a — 2 c and a + 2 c.
15. The volume of a right circular cone varies jointly as its
height and the square of the radius of its base ; and the volume of
a cone 7 feet high with a base whose radius is 3 feet is 66 cubic
feet. Find the volume of a cone 9 feet high with a base whose
radius is 14 feet.
16. The volume of a sphere varies as the cube of its radius ; if
three spheres of radii 6, 8, and 10 inches respectively be melted and
formed into a single sphere, find its radius.
17. The distance of the offing at sea varies as the square root of
the height of the eye above the sea level, and the distance is 3 miles
when the height is 6 feet ; find the distance when the height is 50
yards.
MISCELLANEOUS EXAMPLES V. 363
MISCELLANEOUS EXAMPLES V.
A. 1. If — = — = - = 1, find the numerical value of
3 4 2
(a - 6)2 + (6 - c)2 + (c - a)2.
2. Divide
9 a263 _ I2a46 + 3 65 + 2 0^62 + 4 a^ _ n «54 by 3 63 + 4 a^ - 2 a62.
3. Find the H. C. F. of 1 - x - x^ + x^ and 1 - ic^ _ a^ _ ^7.
4. Simplify a^-a^b-^-l + b-^.
^ ^ a + a6-i + 1 + 6-1
6. A person bought 15 ducks and 12 geese for $26.25 ; and the
prices were such that 2 more ducks could be bought for $4.50, than
geese for $5. What was the price of a duck ?
6. Show that the difference of the roots of the equation
x^ — px -\- q = 0 is equal to the difference of the roots of the equa-
tion x2 - Spx + 2i)2 + g = 0.
7. What is the least integer which must be added to the terms
of the ratio 3 : 4 to make it greater than the ratio 19 : 21 ?
8. Show that, if a + 6, 6 + c, c + a are in continued proportion,
then b + c : c + a : : c — a : a — b.
B. 1. Find the value of
(Xy + Vl - X2V1 - 2/2) ^ (xVl - !/2 _ yy/l _ a;2),
when X = I, y = |.
2. Multiply x2 + (a - l)x + a + 1 by (a - l)x - a2 - a - 1.
3. Find the factors of
(i.) x8- 13x2^4- 42 xy2,
(ii.) (a + 2 6 + 3 c)2 - 4(a + 6 - c)2,
and (iii.) x2 — 4 xy + 4 2/2 _ 9.
4. Simplify ^ + -^^- + ^^^ ♦
^ ^ x-2y x-\-2y iy^-x^
364 MISCELLANEOUS EXAMPLES V.
6. Solve the equations :
,j. 10 3 ^ 10
^ X x-\-2 x + 1
(ii.) x + y4-v'(^ + y)=12l
iK2 - 2/2 _ 21 j '
6. If a; + - = 1 and ?/ + - = 1, prove that z + - = 1. and that
y " z "^ X
xyz + l= 0.
7. Find the square root of
9 a;6 - 12 x'^y'^ + 30 x^y^ + 4 x^y^ - 20 xy^ + 25 ?/«.
8. Find two numbers such that their sum, their difference, and
the sum of their squares are as 3 : 1 : 15.
9. It a:b : : c:d, show that
la+mb:lc+ md: : Va^ + b'^d : Vac"^ + #.
C. 1. Subtract a - 2(6 - c) from 3 {& + 2(a - c) - 5(a - 6)}.
2. Multiply a + ?) + - + — bya-6 + ^-^•
a 6 * a b
3. Multiply ^'^"^^^ by 27 (a - &), and divide
9(a2 _ ^2)
9a2_i6 62 Sa-4tb
a+b ^ a^-b^'
4. If a = y + 0, b = z -{- X, c = jc + y, then
a2 + 62 _|. c2 _ 6c — ca — a6 = a;2 _i_ ^2 ^. ;j;2 _ 2/;2! _ 5!X — a;?/.
5. Find the square root of
aJ + 4 ay^ + 10 a^y^ + 12 a^y^ + 9 y^.
6. If x = 2 + V2, prove that (x -l)(x-2) = x. .
7. ^ and 5 start simultaneously from two towns to meet each
other ; A travels two miles an hour slower than B, and they meet
in seven hours ; if B had travelled one mile an hour faster than he
did and A at only half his previous pace, they would have met in
nine hours. Find the distance between the towns.
MISCELLANEOUS EXAMPLES V. 865
8. If a(y + z)= b(z + x)= c(x+ y) ; then
y — z _ z — X _ x — y
a{h — c) h(c-a) c(a — b)
D. 1. Show that («i^ V - ( ^+^Y = 4 aKa^ + b^).
\ a — b J \a -\- b J
2. Divide
a4 - 2 6x8 - ((j2 _ 52)ic2 + 2 a^bx - a'^b^ by x^ - (a + b)x + ab.
8. Find the L. CM. of
x2-7x+ 12, 3x2 -6x- 9 and 2x8- 6x2 -8x.
X2; 2:y + z^
5. Solve the equations :
5x + 7_2x-J^3 14^
^ ^ 2 3
(ii.) y2_a;y = 161
x2 + a;y = 14 i
6. The fore-wheel of a carriage makes 64 revolutions more than
the hind-wheel in travelling one mile, and the sum of the revolu-
tions made by each in a journey of 10 miles is 7040 : find the cir-
cumference of each wheel.
7. Divide 2 x^ - 17 xy^ + 17 x^y - 15 y^ by 2 x^ - 15 y^.
8. Prove that if the ratios a: a,b : ^^ and c : y are all equal,
then each is equal to the ratio a-|-64-c:o-|-/34-7.
E. 1. Simplify the expression
2{x-2a -|(x- a)+ a] - l{(x - a)-(a - b) + y -b}.
2. Divide 2 a2x2 - 2(3 6 - 4c) (& - c) 2/2 -|- a6xy by ax+2(6-c)y.
3. Pind the factors of the following expressions :
(i.) 4a26*-16a*62, (ii.) x2-26x-87, and (iii.) 'Sx^-xy-lOyK
866 MISCELLANEOUS EXAMPLES V.
4. Show that
1 + 1 + 1 =f_!_ + _L- + -J_Y.
(y - z)'^ iz-x)^ (x-yY \y - z z-x x-y)
5. A man bought a number of articles for $ 108. If he had got
five more for the same money each would have cost 24 cents less
than it did ; how many did he buy ?
6. Multiply y/(x - a) + y/a — y/x'by V(^ - «) - V« + ^x.
7. Show that the mean proportional between x"^ and y^
is xy
xy
8. li a:h'.:c:d, show that
a2 + 62 : c2 + d:2 . . (a + 5)2 . (c + ay,
F. 1. Multiply a — h + c — d\iYa-\-h — c-\-d.
2. Divide a^ + 63 + c^ + 3(6 + c) (c + a) (a + 6) by a + 6 + c.
3. Find the H. C. F. of x^ - 19 x + 30 and 5 x^ - 19 a;2 + 36. For
what values of x will both expressions vanish ?
4. Show that /a^&y_/a:^Y^8a6(a2 + 62).
6. Solve the equations :
^'^ 9x + 6 12x4-8 2*
(ii.) 4x + 62/= 3|
4x2 4-9x?/ + 9y2=^ 11 j *
_2
and x^y^ x ( — W {x"*y"^}.
7. Find the square root of
a;i2 _ 6x10 + 13x8 - 14x6 + lOx* - 4x2 + i.
8. If a : 6 : : c : d, show that aH^ : 63^2 : : a^ + c^ : ft^ + d^.
ABITHMETICAL PROGRESSION. 367
CHAPTER XXIV.
Arithmetical Progression.
254. Def. A series of quantities is said to be in Arith-
metical Progression when the difference between any term
and the preceding one is the same throughout the series.
Thus a, h, €, d, etc., are in Arithmetical Progression
(A.P.) if b — a=zc — b = d—c, etc.
The difference between each term of an AP. and the
preceding term is called the common difference.
The following are examples of arithmetical progressions :
1,
3,
6,
7, etc.,
2,
6,
10,
14, etc., ^
9,
8,
7,
6, etc., -^
and 3, -
-1,
-5,
-9, etc.
In the first series the
common
difference is 2, in the second
series it is 4, in the third it is
— 1, and in the last it is —4.
255. If the first term of an arithmetical progression
be a, and the common difference d, then
the 2d term will he a + d,
the 3d term will be a 4- 2 c^,
the 4th term will he a -\- 3d,
and so on, the coefficient of d being always less by unity
than the number giving the position of the term in the
series.
368 ARITHMETICAL PROGRESSION.
Thus the nth term is
a-\-{n — l)d.
We can therefore write down any term of an A.P.
when the first term and the common difference are given.
For example, in the A.P. whose first term is 5 and whose com-
mon difference is 3,
the 9th term is 5 + (9 - 1)3 = 29,
and the 27th term is 5 + (27 - 1)3 = 83.
256. When any two terms of an A.P. are given, we
can find the first term and the common difference, and
therefore any other term of the series.
Suppose, for example, that the 10th term of an A.P. is 25 and
the 15th term is 5.
Let the first term he a and the common difference d.
Then the 10th term will be a + 9 d and the 16th term will be
a + 14 d.
Hence a + 9 d = 25,
and a + 14 d = 5.
By subtraction 5d = — 20;
.-. d = -4.
Then a =i 25 - 9d = 25 - 9(- 4)= 6L
Thus the series is 61, 57, 53, etc.
Ex. The 12th term of an A.P. is 15 and the 19th term is 36 ;
find the 30th term.
Let the first term be a and the common difference d.
Then the 12th term will he a + lid and the 19th term will be
a-f 18d.
Hence, by the conditions of the problem,
a + 11 ^ = 15,
a + 18d = 36.
ARITHMETICAL PROGRESSION. 369
By subtraction 7d = 21;
.-. d = S.
Then a = 15 - 11 c2 = 15 - 33 = - 18.
Hence the 30th term = a + 29 (Z = - 18 + 29 x 3 = 69.
257. When three quantities are in arithmetical pro-
gression, the middle one is called the Arithmetic Mean of
the other two.
Thus, if a, b, c are in A. P., 6 is the arithmetic mean of
a and c.
By the definition of arithmetical progression, we have
b —a = c — b',
Thus the arithmetic mean of any two quantities is half
their sum.
258. When any number of quantities are in arith-
metical progression, all the intermediate terms may be
called arithmetic means of the two extreme terms.
Between any two given quantities any number of
arithmetic means may be inserted.
For example, to msert four arithmetic means between 10 and 25.
We have to find an A.P. with four terms between 10 and 25, so
that 10 is the first and 25 is the sixth term of an A.P.
Let d be the common difference.
Then the 6th term is 10 -f- 5 d ;
.-. 10-f-5d = 26;
.-. d = S.
Thus the series is
10, 13, 16, 19, 22, 25;
and the required arithmetic means between 10 and 25 are
13, 16, 19, 22.
2a
370 ARITHMETICAL PROGRESSION.
We now consider the most general case, namely to
insert n arithmetic means between a and b.
We have to find an A. P. with n terms between a and
&, so that a is the first and h is the (n + 2)th term of
the A.P.
Let d be the common difference.
Then the {n + 2)th term is a + (n + l)c^;
.'. {n-\-l)d=b — af
.-. d = ^-^^.
n-\-l
Thus the series is
, b —a , nb — a , 6 — a,
n-\-l n-\-l 71 + 1
and the arithmetic means required are
, b — a . (yb —a , r.b — a , b — a
n-t-l w + l n + 1 n-\-l
that is,
7ia-{-b (n — l)a + 2b (n — 2)a-^3b ^^^ a -hub
n -j- 1 n + 1 n + 1 ' ' n + l
EXAMPLES LXXIII.
1. Find the 30th terms of each of the following arithmetical
progressions :
(i.) 3, 5, 7, etc. (ii.) 1, 5, 9, etc. (iii.) 12, 9, 6, etc.
(iv.) i, A, I, etc. (v.) a -{- b, a, a — b, etc.
2. Find the last term of each of the following series :
(i.) 3, 6, 9, etc., to 24 terms. (iv.) 14, 46, 78, etc., to 12 terms,
(ii.) 5, 9, 13, etc., to 30 terms, (v.) 6, 8f, 11^, etc., to 14 terms.
(iii.) 6, 5, 4, etc., to 10 terms. (vi.) i, -^, -|, etc, to 25 terms.
AEITHMETICAL PEOGKESSION. 371
3. The 10th term of an A.P. is 6 and the 6th term is 10. Find
the first term.
4. The 12th term of an A.P. is 15 and the 20th term is 25.
Find the common difference.
5. The 7th term of au A.P. is 5 and the 12th term is 30. Find
the common difference.
6. The 3d term of an A.P. is 40 and the 13th term is 25. Find
the first term.
7. What is the 10th term of the A.P. whose first term is 7 and
whose third term is 13 ?
8. What is the 12th term of the A.P. whose first term is 20 and
whose 6th term is 10 ?
9. The 3d term of an A.P. is 10 and the 14th term is 54. Find
the 20th term.
10. The 7th term of an A.P. is 5 and the 6th term is 7. What
is the 12th term ?
11. Which term of the series 5, 8, 11, etc., is 65 ?
12. Which term of the series |, |, |, etc., is 18 ?
13. Which term of the series 9, 13, 17, etc., is 229 ?
14. Which term of the A.P. 16a -8 6, 15a -7 6, 14a -66,
etc., is 8a?
16. Write down the arithmetic mean of (i.) 7 and 13 ; (ii.) 9
and —9, and (iii.) a + 6 and a — h.
16. Insert 6 arithmetic means between 8 and 29.
17. Insert 8 arithmetic means between 50 and 80.
18. Insert 7 arithmetic means betweeen 269 and 295.
19. Insert 15 arithmetic means between 67 and 43.
20. Insert 25 arithmetic means between 84 and 40f .
21. Insert 10 arithmetic means between 5 a — 6 6 and 5 6 — 6 a.
372 ARITHMETICAL PROGRESSION.
22. Insert 8 arithmetic means between a — 6b and b — 5a.
23. K a, b, c, d are in A.P., show that a + d = b + c.
24. The sum of the 1st and 4th terms of an A. P. is 19, anw the
sum of the 3d and 6th terms is 31. What is the first term ?
25. The sum of the 2d and 5th terms of an A. P. is 32, and the
sum of the 3d and 8th terms is 48. What is the first term ?
26. The sum of the 3d and 4th terms of an A.P. is 187, and
the sum of the 7th and 8th terms is 147. What is the second term ?
27. The sum of the 2d and 20th terms of an A.P. is 2, and the
sum of the 9th and 15th terms is 8. What is the sum of the 6th
and 7th terms ?
28. Show that, if the same quantity be added to every term of
an A.P., the sums will be in A.P.
29. Show that, if every term of an A.P. be multiplied by the
same quantity, the products will be in A.P.
30. Show that, if every alternate term of an A.P. be taken
away, the remaining terms will be in A.P.
31. Show that, if between every two consecutive terms of an
A.P. their arithmetic mean be inserted, the whole will form another
arithmetical progression.
32. Show that, if four quantities are in A.P., the product of
the 1st and 4th is always less than the product of the 2d and 3d.
259. To find the sum of any number of terms of an
arithmetical progression.
Let a be the first term and d the common difference.
Let n be the number of the terms whose sum is required,
and let I be the last term.
ARITHMETICAL PROGRESSION. 373
Then, since I is the 7ith term, we have
l=za + {n — l)d.
Hence, if ;S^ be the required sum,
,S = a + (a + d) + (a + 2 d) + ... + (Z - 2 d) + (Z - d) + ?.
Now write the series in the reverse order ; then
^ = Z + (Z - d) + (^ - 2 d) + ... + (a -h 2 d) + (a + d) 4- a.
Hence, by addition of corresponding terms, we have
2>S = (a4-0 + (a4- I) -\- {a -\- 1) -\ to w terms
=zn{a-Ji-l)'^
.-. ^ = ^(« + 0 (i-)
But a + Z = a4-Sa + (w — l)d| = 2a + (n — l)d;
.-. S = '^\2a + {n-l)d\ (ii.)
The formulae (i.) and (ii.) are both important and
should be remembered.
It should be remarked that the formula (ii.) gives the
value of any one of the four quantities a, d, n, and S
when the other three are known.
Ex. 1. Find the siun of the first 20 terms of the series
6 + 8 + 11 + etc.
Here a = 6, d = 3, n = 20;
.-. ^ = |{2a+(n-l)d}
= 670.
374 ARITHMETICAL PROGRESSION.
Ex. 2. Show that the sum of any number of consecutive odd
numbers, beginning with unity, is a square number.
The series of odd numbers is
1 + 3 + 5 + 7 + ...
Here a = 1 and d = 2; hence the sum of n terms is given by
S = ^{2a + {n - l)d}
■ =|{2+(n-l)2}
Thus the sum of n consecutive odd numbers beginning with
unity is n^. [Compare Art. 146.]
Ex. 3. The sum of 20 terms of an arithmetical progression is
410, and the first term is 30. What is the common difference ?
We have ' /S = ^{2a +(ri - 1)(^},
where 8 - 410, a = 30, and n = 20.
Hence 410 = ?^{60 + 19d}.
From which we find that d = — 1.
Ex. 4. How many terms of the series 11 + 12 + 13 + etc. must
be taken in order that the sum may be 410 ?
We have S = '^{2a +(n - l)d},
where a = 11, d = 1, and aS' = 410.
Hence 410 = ^{22 +(« - 1)};
2
.-. w2 + 21 w- 820 = 0,
that is (w-20)(w + 41) = 0;
.'. n = 20, or n = — 41.
ARITHMETICAL PROGRESSION. 375
When a, S, and d are given, n is to be found by solving a quad-
ratic equation, and one of the roots of the equation is generally
inapplicable, for any value of n which is not a positive integer is
without meaning.
In the present case, — 41 is to be rejected : thus the only value
of n is 20.
Ex. 6. How many terms of the series 24 + 21 + 18 + etc. must
be taken in order that the sum may be 105 ?
We have
S = ^{2a+in-l)d},
where
a = 24, d = - 3, and S= 105.
Hence
105 = ^{48 + (n-l)(-3)};
.-. 210 = n{48-3n + 3};
... w2_i7^ + 70 = 0.
Hence
n = 7, or n = 10.
Since both values of n are positive integers, they are both
answers to the question. Whenever the sum of an A. P. is the
same for two values of n, the additional terms for the greater value
have a zero sum ; in the present case these additional terms are 3,
0, -3.
EXAMPLES LXXIV.
Find the sum of the following series :
1. 2 + 4 + 6 + .•• to 20 terms.
2. 15 + 14 J + 14 + ... to 16 terms.
3. 1+2^ + 3^ ... to 12 terms.
4. _ 5- 1 + 3 + ... to20terms.
6. i + i + i H to 7 terms.
6. i - § - Y - ••• to 61 terms.
7. 10 + ^^-1- ^/+ ••• to 7 terms.
376 ARITHMETICAL PROGRESSION.
8. 1 + 1 + f -f •.. to 15 terms.
9. 3i + 2^ + 1| + ••• to n terms.
10. 1^ + IH- + Hj + — ton terms.
11. 8 + 7^ + 6f + •••• to 19 terms.
12. 4J + 4^ + 3j9^ + •.. to 31 terms.
13. 5 + 6 . 2 + 7 • 4 H to 21 terms.
14. — = 1- a/2 + — h ••• to 7 terms.
V2 + 1 ^ ^V2-l
15. ^Jui + Vl=1 + ?Lzi3 + ... to n terms. -
n n 71
16. w-f 1 +(2w + 3) + (3w + 5)+ ••• to n terms.
17. (a + 6)2+ (a2 + 62) + (a - 6)2+ ... to n terms.
18. The 3d term of an A.P. is 15, and the 20th term is 23^ ;
find the sum of the first 20 terms.
19. The 5th term of an A.P. is 37, and the 13th term is 81 ; find
the sum of the first 24 terms.
20. Find the sum of 20 consecutive odd numbers of which the
least is 25.
21. Find the sum of 40 consecutive odd numbers of which the
greatest is 99.
22. Insert 29 arithmetic means between 5 and 50, and find
their sum.
23. Insert 40 arithmetic means between 10 and 100, and find
their sum.
24. There are 27 terms of an A. P., of which 3 is the first and
107 is the last. Find the middle term and the sum of all the terms.
25. There are 71 terms of an A. P., of which the first is 7 and
the last is 1015. Find the middle term and the sum of all the
terms.
ARITHMETICAL PROGRESSION. 377
26. There are 19 terms in an A. P., of which the first is 8 and
the last — 4. Find the sum of the series.
27. Find the sum of 20 terras of the series 3, 5, 7, etc., beginning
at the 7th.
28. Find the sum of 35 terms of the series 6, 9, 12, etc., begin-
ning at the 5th.
29. The sum of 10 terms of an A. P., whose first term is 2, is
155. What is the common difference ?
30. The sum of 25 terms of an A. P., whose first term is 6, is
25. What is the common difference ?
31. The sum of 10 terms of an A.P. is 100, and the 6th term
is 11. What is the first term ?
32. The sum of 28 terms of an A.P. is 133, and the 5th term
is 0. What is the common difference ? '
33. The sum of 10 consecutive terms of the series 3, 8, 13, etc.,
is 705. Which is the first of them ?
34. The sum of 25 successive terms of the series 5, 8, 11, •••, is
1025. Which is the first of them ?
35. How many terms of the series |, 1, |, must be taken in
order that the sum may be zero ?
36. How many terms of the series 15, 12, 9, •••, must be taken
in order that the sum may be 45 ?
1 q C
37. How many terms of the series 1 — , 1 — , 1—-, •••,
a a a
must be taken in order that the sum may be —6a?
38. How many terms of the series — 8 — 7 — 6 — ••• will
amount to 42 ?
39. The first term of a certain A.P. is 1, the last is 99, and the
sum of all the terms is 450. How many terms are there ?
40. The last term of an A.P. of 20 terms is 62, and the sum is
670. What is the common difference ?
378 ARITHMETICAL PROGRESSION.
41. The ninth term of an A. P. is 136, and the sum of the first
19 terms is 2527. Find the sum of the first 40 terms.
42. Find the sum of 15 terms of an A. P. of which the eighth
is 6.
43. Find the sum of 35 terms of an A.P. of which the eighteenth
term is 15.
44. Find the sum of 2 n + 1 terms of an A.P. whose {ti + l)th
term is 100.
45. Show that, if any odd number of quantities are in A.P., the
first, the middle, and the last are also in A.P.
46. Show that, if unity be added to the sum of any number of
terms of the series 8, 16, 24, etc., the result will be the square of
an odd number.
• 47. Find the sum of all the odd numbers between 100 and 200.
48. Find the sum of all the even numbers which are between
101 and 999.
49. Find the sum of all the numbers between 100 and 500 which
are divisible by 3.
60. A man saved each year $ 10 more than he did in the pre-
ceding year, and he saved $ 20 the first year. In how many years
will his savings amount to $ 1700 ?
61. The sum of 10 terms of an arithmetical series is 145, and
the sum of its fourth and ninth terms is five times the third term.
Determine the series.
62. In an A.P. consisting of 2 n + 1 terms, show that the sum
of the odd terms is to the sum of the even terms a,s n + 1 : n.
63. In an A.P. consisting of 21 terms, the sum of the three last
terms is 117, and the sum of the three middle terms is 90. Find
the series.
64. The sum of n terms of the series 87, 85, 83, etc., is equal to
the sum of n terms of the series 3, 5, 7, etc. Find n.
ARITHMETICAL PROGRESSION. 379
' 55. The sum of n terms of the A. P. 43, 45, 47, etc., is equal to
the sum of 2 w terms of the A.P, 45, 43, 41, etc. Find n.
56. Divide 80 into four parts which are in A.P., and which are
such that the product of the first and fourth is two-thirds of the
product of the second and third.
57. Find four numbers in A.P. such that the sum of their squares
shall be 120, and that the product of the first and last shall be less
than the product of the other two by 8.
58. If -, i, - be in A.P., prove that (s - a)2, (s - 6)2,
a h c
(s — c)2 are also in A.P., where 2s = a + 6 + c.
59. Show that, if a, 6, c be in A.P., then will a\h + c),
62 (c + a), c2(a + h) also be in A.P.
380 GEOMETRICAL PROGRESSION.
CHAPTER XXV.
Geometrical Progression.
260. Def. A series of quantities is said to be in G-eo-
metrical Progression when the ratio of any term to the pre-
ceding one is the same throughout the series.
Thus a, b, c, d, etc., are in Geometrical Progression
(G.P.) if - = - = -, etc.
a b c
The ratio of each term of a geometrical progression to
the preceding term is called the common ratio.
The following are examples of geometrical progressions :
1, 3, 9, 27, etc.,
4, 2, 1, J, etc.,
and I, -1, f, -f, etc.
In the first series the common ratio is 3, in the second series it
is ^, and in the last it is — |.
261. If the first term of a G.P. be a, and the cominon
ratio r, then
the 2d term will be ar,
the 3d term will be ar^,
the 4th term will be ar^,
and so on, the index of r being always less by unity than
the number giving the position of the term in the series.
Thus the nth term is ar"~^
GEOMETRICAL PROGRESSION. 881
We can therefore write down any term of a G.P. when
the first term and the common ratio are given.
For example, in the G.P. whose first term is 5, and whose
common ratio is 3,
the 4th term is 5 x S^-i = 5 x 33,
and the 10th term is 5 x 3io-i = 5 x 3^.
262. When any two terms of a G.P. are given, we can
find the first term and the common ratio: the series will
thus be completely determined.
Suppose, for example, that the 5th term of a G. P. is | and
the 7th term \\.
Let the first term be a, and the common ratio r.
Then the 6th and the 7th terms are ar^ and ar^ respectively.
Hence ar* = |
and a»< = II ;
.*. by division, »^ = I ;
.*. r = ±f.
Then, since a x |f = f , a = |.
Thus the series is f , ± 3, 2, ± f , etc.
Ex. The 4th term of a G.P. is 189 and the 6th term is 1701 ;
find the 8th term.
Let the first term be a, and the common ratio r ] then the 4th
term will be ar* and the 6th term will be ai*.
Hence, by the conditions of the problem,
ar* = 189 and ar^ = 1701.
By division, r^ = 9; .-. r = ± 3.
Then a = 189/ ( ± 3)8 = ± 7.
Hence the 8th term = ar^ = ±7(^±sy = 15309.
263. When three quantities are in geometrical progres-
sion, the middle one is called the geometric mean of the
other two.
382 GEOMETRICAL PROGRESSION.
Thus, if a, b, c are in G.P., b is the geometric mean of
a and c. By the definition of geometric progression, we
have
a 6'
.•. b^ = ac;
.-. b = ± Vac.
Thus the geometric mean of any two quantities is the
square root of their product.
It should be noticed that quantities which are in con-
tinued proportion [Art. 246], are in geometrical pro-
gression.
264. When any number of quantities are in geometrical
progression, all the intermediate terms may be called
geometric means of the two extreme terms.
Between two given quantities any number of geo-
metric means may be inserted.
For example, to insert three geometric means between 5 and 80.
We have to find a G.P. with three terms between 5 and 80, so
that 5 is the first term and 80 is the 5th term.
Let r be the common ratio ; then the 5th term is ar*.
Hence 5r* = 80; .-. r* = 16; .-. r = ±2.
Thus the series is 5, ± 10, 20, ± 40, 80.
The required means are therefore ± 10, 20, i 40.
We now consider the most general case, namely to
insert n geometric means between a and b.
We have to find a G.P. with n terms between a and 6,
so that a is the first and b is the {n + 2)th term of the
G.P.
is a?-'*+^;
GEOMETRICAL PROGRESSION. 383
Let r be tlie common ratio. Then the {n -\- 2)th term
r — — ,
Hence the required means are ar, ai^, ar^, •••, ar", where
n+lj
r
EXAMPLES LXXV.
1. Find the 6th terms of each of the following geometrical
progressions :
(i.) 9, 3, 1, etc. (ii.) 2, - 3, f, etc. (iii.) a^, ah, b"^, etc.
2. What is the 5th term of the G.P. whose first term is 3 and
whose 3d term is 4 ?
3. The 3d term of a G.P. is 1 and the 6th term is - ^. What
is the 10th term ?
4. The 4th term of a G.P. is .016 and the 7th term is .000128.
What is the first term ?
6. The 6th term of a G.P. is 156 and the 8th term is 7644.
What is the 7 th term ?
6. The 3d term of a G.P. is 2.25 and the 7th term is 11.390625.
What is the 5th term ?
7. The 3d term of a G.P. is 4 and the 6th term is - f What
is the 10th term ?
8. The 4th term of a G.P. is y^ and the 7th term is — j^-^.
What is the 6th term ?
9. Find the geometric mean (i.) of 4 and 9, (ii.) of 7 and 252,
and (iii.) of a^b and ab^.
384 GEOMETRICAL PROGRESSION.
10. Insert two geometric means between 1 and — 8 ; also insert
three geometric means between 12 and f , and four between |
and f|.
11. Insert five geometric means between 3 and .000192.
12. Insert four geometric means between a^b~^ and a~^¥.
13. The sum of the 3d and 4th terms of a G.P. is 40, and the
sum of the 6th and 7th teems is 2560. What is the first term ?
14. The sum of the 1st and 2d terms of a G.P. is 72, and the
sum of the 3d and 4th terms is 8. What is the first term ?
15. If a, b, c, d are in G.P., show that ad = be.
16. If all the terms of a G.P. be multiplied by the same quan-
tity, the products will be in G.P.
17. Show that the reciprocals of the terms of a geometrical pro-
gression are in G.P.
18. If every alternate term of a G.P. be taken away, the remain-
ing terms will be in G.P.
19. If between every two consecutive terms of a G.P. their geo-
metric mean be inserted, the whole will form another geometrical
progression.
20. Show that the product of any two terms of a geometrical
progression, which are respectively equally distant from the first
and the last terms, is equal to the product of the first and last
terms.
265. To find the sum of any number of terms of a
geometrical progression.
Let a be the first term, and r the common ratio. Let
n be the number of the terms whose sum is required,
and let I be the last term.
Then, since I is the nth- term, we have
GEOMETRICAL PROGRESSION. 385
Hence, if JS be the required sum,
S=za-\-ar + ar^-\ \- ar""-'^ + ar'*-^
Multiply by r ; then
Sr = ar + ar^ -{- ar^ -\ h ar"" ^ + ar^.
Hence, by subtraction,
S — Sr = a— ar"" ',
that is /S(l - r) = a(l - r»);
1 — r
The above formula may be written in another form ;
for, by substituting the value I = ar**~^, we have
e a — rl
Ex. 1. Find the sum of 10 terms of the series 1, 2, 4, etc.
Here a = 1, r = 2, n = 10 ;
o 1 — r*»
1-r
1 -2^0
1-2
= 210 - 1 = 1023.
Ex. 2. Find the sum of six terms of the series
2-3 + I -, etc.
— 3
Here a = 2, r= — , and n = 6 ;
\ — r
_gl-(-f)^-g !!
~ l-(-f) " 1 + f
2b
386 GEOMETRICAL PROGRESSION.
266. Sum to Infinity of a G-eometrical Progression. It
might at first be thought that, by taking a very great
number of terms of any geometrical progression, the
sum could be made as great as we please ; an example
will, however, show that this is not the case. Suppose
that we have a line two inches long, and divide it
into two equal parts and take away one; then bisect
the remainder and take away one of the parts ; and sup-
pose this process continued to any extent. Then the
lengths in inches of the successive parts taken will be 1,
1 1 \, etc. Now it is clear that the sum of all the parts
taken away can never exceed two inches, but, as the part
which is left over will diminish without limit as the
number of the operations is increased, the sum of the
parts taken away can be made to differ from two inches
by a length which, is smaller than any conceivable length;
hence, by taking a sufficiently large number of terms,
can be made to differ from 2 by an arbitrarily small
quantity.
From Art. 265 we have
S — <^(^ — ^") _ ct _ gr"
1 — r 1 — r 1 — r
Kow if r is Si proper fraction, whether positive or nega-
tive, the absolute value of r"" will decrease as w increases ;
moreover r" can be made as small as we please by suffi-
ciently increasing the value of n.
Hence, when r is numerically less than unity, the sum
of a very large number of terms of the series can be
made to differ from a/(l — r) by as small a quantity as
we please.
GEOMETRICAL PROGRESSION. 387
It is customary to express this fact in conventional
language by the statement : The sum to infinity of the
geometrical progression a -^ ar -{- ar^ + '•-, in which r is
numerically less than unity, is a/{l — r).
Ex. 1. Find the sum to infinity of the series
(i.) 1 + ^ + ^ + ^ + ....
(ii.) 9-6 + 4......
In (i.) a = l, r=J;
•••^=^^.=^^i-
In (ii.)
«=. .=-«=-!;
.: S= " = » =
9^27
!-»• l-(-f) f 6*
Ex. 2. Find the value of .234.
A. .234 = .23444. ..
= 2. + A + A + A + A+etc.
10 10'-2 103 ^ 10* ^106
xr 4,4, , . « ., 4/108 10^ 4 _ 4
^^^IP + ioi+-'^^^^"^'y = W,= 9- ^10^-960*
Hence .234 = ^^ + ^k + ^h = Uh-
Ex. 3. Find the geometrical progression whose sum to infinity
is 18 and whose second term is —8.
Let a be the first term and r be the common ratio.
Then the second term is ar, and the sum to infinity is a/(l — r).
Hence ar = — 8,
and a/(l-r)=18.
388 GEOMETRICAL PROGRESSION.
Hence, by division, we have
r(l _ r) = ^ ;
^ ^ 18 '
.♦. r2-r:=f ;
.-. r = -i, or r = f
If r = -l, a ==^ = 2i.
3 r
Thus the series is 24, — 8, f , etc.
The value r = | is inadmissible, for the sum of a geometrical
progression is not given by the formula
1 —r
except when r is numerically less than unity.
EXAMPLES LXXVI.
Find the sum of the following series :
1. 8 + 12 + 18 + 27 + •.. to n terms.
2. 12 + 9 + 6f + ••• to6 terms.
3. a + - + - + — to n terms.
r r^
4. li + \\ + l^V + ••• to 8 terms.
5. f + i + IH to 6 terms.
6. 12-9 + 61 to infinity.
7. 4 + 1.2 + .36 + .108 + ... to infinity.
8. 4 + .8 + .16+ ••• to infinity.
9. a2_^ 055 _^ 52 _^ ... to w terms.
10. 3 - 2 + I - ••• to 'infinity.
11. Find the sum of the successive powers of 2 from 2 up to
4096 inclusive.
12. Find, correct to four places of decimals, the sum to infinity
of the series 1 + — + - + ••..
■^2 2
GEOMETRICAL PROGRESSION. 389
13. Find the geometric mean of 4 x^ — 12 x4- 9 and 9 x^ + 12 x + 4.
14. Show that the product of any odd number of terms of a
G.P. is equal to the nth power of the middle term, n being the
number of the terms.
16. Find the G.P. whose sum to infinity is 4, and whose 2d
term is f .
16. Find the G.P. whose sum to infinity is 9, and whose 2d
term is — 4.
17. The sum of the first 10 terms of a certain G.P. is equal to
33 times the sum of the first 5 terms. What is the common ratio ?
18. If the sum of a geometrical series to infinity is n times the
first term, show that the common ratio is 1
n
19. If the common ratio of successive terms of a G.P. be posi-
tive and less than ^, show that each term is greater than the sum of
all that follow it.
20. The 6th teim of a G.P. is 8 times the 3d term, and the sum
of the first two terms is 24. Find the series.
21. Find the sum to infinity of the series
_«_+f_«_y+f_«_y+...
a + b \a-hbj \a + bj
22. Find the sum to infinity of the series
23. Show that, if
Si = a + ar + ar^ + ar* + — + ar^
and S2 = a — ar-\- ar^ — ar^ + h ar2»» ;
then S1S2 = a"^ + a^ -{■ ah^ + — + a^r^.
24. Sum the series VI + iV^ + f Vl + ••* ^o infinity.
25. The 6th term of a G.P. is jslj^ and the 2d term is ^j.
Find the sum to infinity.
390 GEOMETRICAL PROGRESSION.
26- Find three numbers in G.P. such that their sum is 14, and
the sum of their squares is 84.
27. The sum of the 1st, 2d, and 3d terms of a G.P. is 42, and
the 4th term exceeds the first by 126. What is the series ?
28. The sum of the 1st, 2d, and 3d, terms of a G.P. is to the
sum of the 3d, 4th, and 5th terms as 1:4, and the 5th term is 8.
What is the series ?
29. Show that, if a, 6, c, d be in G.P., then will a + ?), 6 + c,
c + d, and also a^ + h\ h"^ + c2, c^ + d^ be in G.P.
30. Show that, if a, 6, c be the jDth, gth, and rth terms respec-
tively of a G.P., then will a?-'-6'-i>cP-2 = 1.
31. Show that, if (ofi + 62) (52 + ^2) = {ah + 6c) 2, then will a, 6,
c be in G.P.
267. Sometimes we are not told the law which con-
nects successive terms of a series; but when a certain
number of the terms are given, the law can in simple
cases be at once determined.
Suppose, for example, we have the series
3, 9, 15, etc.
Here 9-3 = 6, and 15 - 9 = 6.
Thus, the series is an arithmetical progression^ whose common
difference is 6.
Again, in the series 3, 9, 27, etc.,
Since 9 — 3 = 6, and 27 — 9 = 18, the series is not an arith-
metical progression. We then see whether it satisfies the condi-
tions for being a geometrical progression, namely | = ^-^^ which is
the case. Thus the series is a geometrical progression^ whose
common ratio is 3.
GEOMETRICAL PROGRESSION. 391
EXAMPLES LXXVII.
Sum the following series :
1. 3, 2.7, 2.4, etc., to 21 terms.
2. 2, 18, 162, etc., to 7 terms.
3. 1, .2, .04, etc., to infinity.
4. a, 6, — , etc., to n terms.
a
6. .3, .03, .003, etc., to infinity.
6. 3 + 4.3 + 5.6 + ..• to 11 terms.
7. a + i«_±^ + 5^±2^+...tol9terms.
3 3
g. J _ ^ + ^ _ etc., to 8 terms.
9. 3J + 1^ + ^ + etc., to infinity.
10. 2J + ej + 15| + ••• to 5 terms.
11. Sum the following series to 6 terms, and when possible to
infinity :
(i.) 2^ + 6^ + 10 + ... (iii.)4-34-f-- (v.)i + i + f + -
(ii.) 9 + 6 + 1 + ... (iv.) I + i + I + ... (vi.) ^ - 1 + I _ ...
12. Show that, if an odd number of quantities are in geomet-
rical progression, the first, the middle, and the last of them are
also in geometrical progression.
13. The sum of the first 7 terms of an A.P. is 49, and the sum
of the next 8 terms is 176. What is the series ?
14. The arithmetic mean of the 1st and 3d terms of a geo-
metrical progression is 5 times the 2d term. Find the common
ratio.
16. If the 4th term in an arithmetical progression is the
geometric mean of the 2d and 7th terms, show that the 6th term
is the geometric mean of the 2d and 14th.
392 GEOMETRICAL PROGRESSION.
16. If P be the continued product of n terms of a geometrical
progression whose first term is a and last term I, show that
P2=(a?)n.
17. The continued product of three numbers in G.P. is 216, and
the sum of the products of them in pairs is 156. Find the numbers.
18. Divide 25 into five parts which are in A.P., and which are
such that the sum of the squares of the least and greatest of them
is one less than the sum of the squares of the other three.
19. Insert between 6 and 16 two numbers such that the first
three may be in A.P. and the last three in G.P.
20. If a, b, c be in geometrical progression, and x, y be the
arithmetic means between a, 6, and 6, c, respectively, prove that
? = ! + !, and2 = « + -».
h z y X y
HARMONICAL PROGRESSION. 393
CHAPTER XXVI.
Harmonical Progression. Other Simple Series.
268. A series of quantities is said to be in Hannonical
Progression when the difference between the first and the
second of any three consecutive terms is to the difference
between the second. and the third as the first is to the
third.
Thus a, h, c, dy etc., are in harmonical progression, if
a — 6:6 — c :: a'. Cy
b — c: c — d: : b : dj
and so on.
269. If a, 6, c are in
harmonical progression,
we
have
by definition
a —
6:6
— c::a:c.
Hence
c{a-
-6):
= a(6-c),
or
ca-
-be.
= ab — ac.
Divide by abc ;
then
1
_1
1 1
6
a
"c-6'
which shows that
1
a'
1 1
6' c
are in arithmetical progression.
Thus, if quantities are in harmonical progression, their
reciprocals are in arithmetical progression.
394 HARMONICAL PROGRESSION.
The above relation between quantities in harmonical
progression is much more frequently employed than that
given in the preceding article.
270. When three quantities are in harmonical pro-
gression, the middle one is called the harmonic mean of
the other two.
If a, b, c are in harmonical progression,
111
a b c
are in arithmetical progression ;
" b a c &'
b a c
a-\- c
Thus the harmonic mean of any two quantities is twice
their product divided by their sum.
271. If we put A, G, H for the arithmetic, the geo-
metric, and the harmonic means respectively of any two
quantities a and b, we have
A^Ua-\-b), G = Vab, and ^=-^-^^
Hence, AxH=\{a + b) x-?-^ = a&;
a -\- 0
.'. AH= 0\
Hence O is the geometric mean of A and H.
Thus the geometi^ic mean of any two quantities is also
the geometric mean of their arithmetic and harmonic means.
HARMONICAL PROGRESSION. 395
272. Many questions concerning quantities in har-
monical progression can be solved by considering the
arithmetical progression whose terms are the reciprocals
of the terms of the harmonical progression.
For example, to insert n harmonic means between
any two quantities a and b.
ft 1 -I
Insert n arithmetic means between - and — These,
1 b
by Art. 258, will be
l_^l/b-l/a^l_^,l/b-l/a^^^^
a n-\-l a n + 1
Thus 1, l + lA^»,l+2l/^^,...,l
a a w+1 a n+l 6
are in A. P.
That is
1 (nb + a) (n-l)b + 2a 1
a {n-\-l)ab' {n + l)ab ' '"' b
are in A.P.
Hence the reciprocals of these are in H.P.
Thus the required harmonic means are
(n-\-l)ab {n-{-l)ab ^^^
nb-^a ' (n-l)6 + 2a'
Note. — It is of importance to notice that no formula can be
found giving the sum of any number of terms in harmonical
progression.
273. The conditions that a, b, c may be in A.P., G.P.,
or H.P., respectively, may be written
a — b:b — c:.a:a,
a — b:b — c::a:by
a — b:b — c::a:c.
396 HARMONICAL PROGRESSION.
The first and third follow from the definitions of A.R
and H.P., and the second can be easily verified by mul-
tiplication.
Ex. 1. The second term of a harmonical progression is 2, and
the fourth term is 6 ; find the series.
The second term of the corresponding arithmetical progression
is ^, and the fourth term is ^. Hence, if a be the first term,, and
d the common difference, we have
a-\- d — \^ and a + Zd = \,
whence « = I, and d= — ^.
Thus, the arithmetical progression is
f> 5> 3' o' ®tc.
Hence the harmojiical progression is
f, 2, 3, 6, etc.
Ex. 2. Show that a + 6, 2 6, and 6 + c will be in H.P. if a, 6, c
are in G.P.
The three quantities a + 6, 2 5, and 6 + c will be in H.P. if
1.1 2
a + 6 6 + c 2h
or if
b(b + c)+ b(a + &) = (« + b)(b + c) ;
that is, if
62 4- 6c + a6 + 62 = a6 + 62 + ac 4- 6c,
or if
62 = ac ;
and this is the case when a, 6, c are in G.P.
EXAMPLES LXXVIII.
1. Show that if the terms of a harmonic progression be all
multiplied by the same quantity, the products will be in harmonical
progression.
2. Insert 5 harmonic means between 1 and 7.
3. Insert 4 harmonic means between f and f .
4. Show that, if the arithmetic mean of two quantities is 1,
their harmonic mean will be the square of their geometric mean.
HARMONICAL PROGRESSION. 397
5. Show that, if a^, 52^ ^^ are in arithmetic progression, 6 + c,
c + o, and a + b will be in harmonical progression.
6. Show that, if x, y, ^ be in H.P.,
(2/ + 0-x)2, (0 + x-2/)2, (x + y-zy
will be in A.P.
7. Show that, if x, y, 0 be in H.P., then will — ^, — ^»
—=- — , be also in H.P.
a^ + y
8. Show that, if x, y, z be in H.P., then will
X y z
y + z — x' z + x — y' x-\-y — z
be also in H.P.
9. If a, 6, c, d be in H.P., then will , ^ ,
c ^ <i V • TJX. 6 + c + d c + d+a
, and be in H.P.
d-h a-\- b a-\-b -\- c
10. If a, b, c be in H.P., then will 2a — b, b, and 2c — 6 be
in G.P.
11. If a, 6, c be in H.P., then will a, a — c, and a — 6 be in H.P.
12. Show that
(rt2 + 62) (<j2 + a6 + 62)^ ^4 4. ^252 + 54^ ^
and (a2 + b'^) {a^ - ab -\- 62)
are in H.P.
13. Show that, if x, ai, 02, y be in A.P., x, gi, 92, V in GJP.,
and X, ^1, ^, y in H.P., then will
giQi _ ai + q2.
14. If a, 6, c be in H.P., then will
b+ a _^ 6 + c _ 2_
6 — a 5 — c
15. If a, 6, c be in G.P., then will -^— , -, and — ?— be in A.P.
a + 6 6 6 + c
398 HARMONICAL PROGRESSION.
16. If a, b, c be in A. P., and 6, c, d in H.P., then will
a:b = c:d.
17. Show that, if
a c h — a h — c
then 6 = a + c, or else a, &, c are in harmonical progression.
18. If «, 6, c, d be in H.P., then will
3(6 - a){d -c) = (c- b){d - a).
19. Show that, if a, 6, c be in H.P., then
2 1^1
b b — a b — c
20. Three numbers are in A.P. The product of the extremes
is five times the mean, and the sum of the two largest is three times
the least. Pind the numbers.
21. Show that, if ±±A^ b, ^-±-^ are in A.P., then a, -, c
1 — ab 1 — be b
will be in H.P.
22. If a, 6, c be in A.P., and a^, b"^, c^ in H.P,, prove that
— -, b, c are in G.P., ov a = b = c.
23. Three numbers are in A.P,, and if 1 be added to the greatest,
they will be in G.P. ; show that the smallest is equal to the square
"iiZ the common difference.
24. Show that, if x, a, 6, y be in A.P., and x, w, v^ y in H.P.,
then av = bu = xy.
25. Show that, if ai, ^2, as, 04 be in H.P., then
aia2 + 0,2(23 + cisdA = 3 aiGi.
26. Show that, if a, 6, c be in A. P., and a, c, b be in G.P., then
will &, a, c be in H.P.
274. Other Simple Series. There are, besides the pro-
gressions, many other series the successive terms of which
are formed according to simple laws. The following are
examples of such series :
SIMPLE SERIES. 399
12 4. 22 + 32 +...+^2,
1 . 2 -f 2 . 3 H- 3 . 4 + ... + w(ri + 1),
and J_ + J_-f J_+...+ 1
1.22.33.4' w(n + l)
275. We proceed to find the sum of n terms of some
of these simple series.
The sum of n terms of a series is generally denoted
by Sn, and the sum of the series when continued without
limit by S^.
Ex. 1. Find the sum of n terms of the series
1. 2 + 2. 3 + 3. 4 + 4.5 + . ..
Here
>S„ =1 . 2 + 2 . 3 + 3 . 4 + ... + (n - l)n + n(n + 1).
Let
S'= 1 . 2 . 3 +
2.3-4 + 3-4-5 + ..- +(n-l)n(n + 1)+ n(M + l)(n +2).
Shifting each term one place to the right, we have
2 = 1.2-3 + 2.3-4 + ... +(n-2)(n-l)n+(n-l)n(n + l)
+ n(n+-l)(7i + 2).
Now subtract the last result from the preceding, taking each
term from the one above it ; then we have
0 = 1 - 2 - 3 + 3 - 2 . 3 + 3 . 3 - 4 + ... + 3(n - l)n + 3 n(n + 1)
-n(w + l)(n + 2).
Hence
3{1 - 2 + 2 - 3 + 3 - 4 + - + (w - l)ri + n(n+l)}= n(n+l)(w + 2) ;
.-. Sn = inin-\-l)in + 2).
Ex. 2. Find the sum of n terms of the series
1-2.3 + 2-3-4 + 3.4-5 + ....
Here
/§; = 1.2-3+2. 3-4+3-4- 5 + .- + (n-l)n(n+l) + n(n+l)(n+2).
400 SIMPLE SERIES.
Let
S = 1.2. 3. 4 + 2. 3. 4- 5 + 3. 4.5.6 + •••
+ (w - l)n(n + l)(n + 2)+ n(n + l)(w + 2)(w + 3).
Then
S= 1. 2. 3. 4 + 2. 3.4.5+. ••
+ (w - 2)(w - l)w(n + 1) + (n- l)w(w + l)(n + 2)
+ n(w + l) (n + 2)(/i + 3).
Hence, subtracting each term from the one above it,
0 = 1.2.3.4 + 4.2.3.4 + 4.3.4.5+...
+ 4(w-l)w(w + l)+4n(w+ l)(w + 2)
-w(?i + l)(w + 2)(n + 3);
.-. 4(1.2.3 + 2.3.4 + 3.4.5+... + n(w+ l)(w + 2)}
= n(w+ l)(7i + 2)(n + 3);
.-. ^„ = jn(w + l)(n + 2)(n + 3).
Note. — This series and the preceding are examples of an impor-
tant type of series in which (i.) each term contains the same
number of factors, (ii.) the factors of any term are in A.P., and
(iii.) the first factors of the successive terms form the same A.P.
as the successive factors of the first term.
The S series by means of which the required sum can always be
found is the series whose terms are formed according to the same
law but with an additional factor at the end.
Ex, 3. To find the sum of n terms of the series
12+22 + 32 + ..-.
Here /^„ = P + 22 + 32 + ... + w^.
Now n^ = n{n + 1) — n ;
.-. /S'„ = 1.2 + 2.3 + ... + w(w+l)-l-2 n.
But, by example 1,
1.2 + 2.3 + 3-4+ ... + n(w + l)=^w(w + l)(n + 2);
also 1 +2 + 3+ ... + w = ^w(w + 1).
Hence 8n = \n{n-\- V){n + 2)- ^n{n + \)
= in(w+ l)(2w+ 1).
SIMPLE SERIES. 401
Ex. 4. To find the sum of n terms of the series
13 + 23 + 33+... + n8.
Here Sn = l^ + 2^ + Z^ -h — + n\
Now 4 w3 = {w(w + 1)}2 — {(n — l)n}2, for all values of n.
Hence we have
4. 13 =(1.2)2,
4 . 28 = (2 . 3)2 _ (1 . 2)2,
4.33 =(3.4)2 -(2.3)2,
4(n - 1)8 == {(w -V)nf- {(n - 2) (w - \)f,
4 w3 = {n(n + 1)}2 - {(n - \)nf.
Hence, by addition,
4(18 + 23 + 38 + ... + w8} = {n{n + \)f,
.-. 13+28 + 38+... + w3 = I n2(w+ 1)2.
This result may be expressed differently ; for since
l + 2 + 3 + ... + n = Jw(w + 1),
we have 18 + 28 + 38 + ... + n8 = (1 + 2 + 3 + ... + n)2.
Thus the sum of the cubes of the first n integers is equal to the
square of their sum.
Ex. 6. Find the sum of n terms of the series
-JL + ^ + _JL+....
1.2 2.3 3.4
Here ;S^„ = J_ + -1- + J-+ ... +
1-2 2.3 3.4 («-l)n n(n + l)
Let 2 = 1 + 1 + 1 + 1 + ... + ! +
1234 n n+1
then S= 1 + 1 + 1 + .. . + _1_ + 1 +
123 n-lww+1
2o
402 SIMPLE SERIES.
Hence, by subtraction,
0 = 1- 111 -1 11
1-2 2-3 3-4 (n-l)»i w(n + 1) n + 1
Hence J_ +_!_....+ 1 = i_ 1
1-2 2-3 w(n + l) w + 1
When n is infinite, is zero ; hence the sum to infinity of
n + 1
the series 1 1 + •■• is 1.
1-2 2.3 3-4
Ex. 6. Find the sum of n terms of the series
1 +__! + 1^+....
1.2.3 2.3.4 3.4.5
Take a series S formed according to the same law but with one
factor less in each term, and proceed as before. The result is
2 11 -2 (n + l)(n + 2)/
2
Ex. 7. Find the sum of n terms of the series
l-\-2x + Sx^ + 4x^+ ....
Let /S' = 1 + 2 a; + 3 x2 + 4 ic3 + ... + nx^-'^ + (w + l)a?».
Then Sx = x + 2 x'^ + S x^ + "• ■{• wx" + (n + l)a?»+i.
Hence, by subtraction,
S(l-x)=l + x + x'^-h x^ -{-'•' +x^-(n + l)a;"+i.
But 1 + a; + a;2+ ... 4-cc« = ^!^
1 —X
Hence, S(l -x)= Lz^!!!^ _ (n + l)a;«+i ;
^ = 1^^;_(, + 1)^+^
ii-xy " 'i-x
MISCELLANEOUS EXAMPLES VI. 403
EXAMPLES LXXIX.
Find the sum of the foUowmg series to n terms, and when
possible to infinity.
1. 2.4 + 4.6 + 6.8 + .... 3. 1.4 + 4.7 + 7.10 + ....
2. 3.5 + 5.7 + 7.9 + .... 4. 2. 6 + 6. 8 + 8- U + -.
5. 1.3.5 + 3.5.7 + 5.7.9 + ....
6. 2 . 7 . 12 +7 . 12 . 17 + 12 . 17 . 22 + ....
7. -i- + -1-+ — + ... 8. — + -i-+-^ + ....
1.33.55.7 2. 66. 8 8. 11
9. _JL_ + _l_ + _i_+....
1.3.53.5.76.7.9
10. _L_ + _^- + __L_ + ....
2.5.8 6.8.11 8-11.14
11. \ + 1 + I +....
(a; + l)(x + 2) (x + 2)(a: + 3) (x + 3)(a; + 4)^
12. ^ + ^ ^ 1 + ....
(l+x)(l+2a;) (l+2a;)(l+3a;) (l+3a;)(l+4a;) ^
13 1 I 1 I 1 I ^ I
* 11+2 1+2+3 1+2+3+4
14. 12 + 32 + 52 + .... 15. 18 + 38 + 58 + ....
16. a(a + &) + (a + 6)(a + 2 6) + (a + 2 6)(a + 3r6)+....
MISCELLANEOUS EXAMPLES VL
A. 1. Simplify
(y+3)(2^-l)-3(y+l)(y2-9)+3(y-l)(y2_9)-(2,-3)(y2_i).
2. Prove that
(2y-x)»-(2x-y)» (2y-x)H(2x-y)» ^ ^^ ,_
3(y-a;) x+y " "
404 MISCELLANEOUS EXAMPLES VI.
3. Find the H.C.F. of Sx^ - ISx^ + 2Sx - 21 and 6x^-\-x^
-44x + 21.
For what value of x will both expressions vanish ?
4. Solve the equations :
(i.) Sx + --l = 12x + ^+U = ^-2x-U.
y y y
(ii.) x2 - (a + 6 + 2 c)a: + (« + 6 + c)c = 0.
5. A man buys 9 oxen and 20 sheep for £230 ; by selling the
oxen at a gain of 25 per cent, and the sheep at a loss of 20 per
cent, he gains £ 35 altogether. Find the price he gave for each.
6. If a and yS be the roots of the equation a;^ + 4 x + 3 = 0,
show that the equation whose roots are " "^ and ° "^ ^ is
a iS
3 a:2 - 16 a; + 16 = 0.
7. Multiply ah-^-2 ai6-^ + 4-8 a-^&^+16a-H by a^+2h^.
8. Rationalize the denominator of -^ — ^,
5-4V3
and simplify V25 + 4^34.
9. Sum the series 5 — 3| + 2i|— etc., to 8 terms, and find its
limit when continued to infinity.
10. Insert 20 arithmetic means between 100 and 300, and find
their sum.
B. 1. Simplify Z{x - 2{y - z)]-l'iy + 2{x - y - z}'].
2. Find the factors of
wnx2 + m'^xy + n'^xy + mny^^
and of x^ — x'^y + xy^ — y^.
3. If the sum of two numbers be equal to 4, show that their
difference is one-quarter of the difference of their squares.
4. Solve
(i.) -^-+-^ = 1.
x + a x + b
(ii.) V12 X - 3 + Va: + 2 + V7 X - 13 = 0.
MISCELLANEOUS EXAMPLES VI. 405
6. Show that x^ — 8 x + 22 can never be less than 6.
6. Find the square root of a^ + 2 a^h"^ - 2 a^c + ft* - 2 hH + c\
7. If a : 6 : : c : (?, show that
Ct2 + 62 . c2 + (J2 . .(oj + 5)2 .(c + d)2.
8. Sum the following series :
(i.) 5 - 1 - 7 - etc. to 20 terms.
(ii.) 2J + 1 + f + ••• to infinity.
9. Find the sum of all the numbers which are less than 1000,
and are divisible by 7.
10. A train travelling at the rate of 37 J miles an hour passes a
person walking on a road parallel to the railway in 6 seconds;
it also meets another person walking at the same rate as the other,
but in the opposite direction, and passes him in 4 seconds. Find
the length of the train.
C. 1. Find the value of V(y^ + ^) _ ^ - <y - ^),
V(« -\-y) x-z(y- X)
when x = 0, y = 1, z = li.
2. Divide 4x* - 9xV -|- I2xy^ - 4y* by 2x2 + 3xy -2y'^.
8. Show that the square of the sum of two consecutive numbers
is equal to four times their product increased by unity.
4. Find the L. C. M. of x^ - 6 ax + 9 a^, x^-ax-6a^ and
3x2-12a2.
6. Simplify _2_ + ,^_+ ^ ^
a-2 1 - a a+ 1 a + 2
6. If 40 minutes would be saved in a journey by increasing the
rate of the train by 5 miles an hour, and 1 hour would be lost by
diminishing it by the same amount, find the rate of the train and
the length of the journey.
7. Simplify
2 + ^8 + V2 - V27 - Vl2 + V76 - V(19 + ^ V2)-
406 MISCELLANEOUS EXAMPLES VI.
8. Prove that any ratio is made nearer to unity by adding the
same number to each of its terms. Also show that ^^ "^ ^^ is
. mh -\-nd
intermediate between - and -•
h d
9. Find the sum of the following series :
(i.) 21 + 15 + 9 + ... to 8 terms,
(ii.) 5 + 2 + .8 + ... to infinity.
10. The arithmetic mean of two numbers is 17, and their geo-
metric mean is 15. What are the numbers ?
D. 1. Simplify {x{x + a)- a(x - a)}{x(x -a)- a(a - x)].
2. Showthat (w + 1)*- w4=(2n + l)(2w2 + 2w+ 1),
and that (a + 6)* - a* - 6* = 4 ab(a + 6)2 - 2 a'^b'^.
3. Simplify |i_3a:-20|r^_8a:-42|,
4. Solve the equations :
(i.) §-^ = 21, ^-f Z = -ll.
X y X y
(ii.) 3x2 - 4a;?/ + 2 ?/2 = 33, x"^ - y"^ = 16.
5. A number consists of two digits, of which that in the unit's
place is the greater; the difference between the squares of the
digits is equal to the number, and if the digits were inverted the
number thus formed would be 7 times the sum of the digits. Find
the number.
6. Find the equation whose roots are the cubes of the roots of
the equation a;2 — 4 x + 2 = 0.
7. Find the square root of x^ - 2 x2 + 8 + x-2 - 8 x"* + 16 x'^.
8. If a : & : : c : c?, prove that ^^^-±^ = (^Lzi^V.
c^-\-d^ \c-dj
9. Sum the following series :
(ii.) 9-6 + 4 to infinity.
MISCELLANEOUS EXAMPLES VI. 407
10. The series of odd numbers is divided into groups as follows :
1 ; 3, 5 ; 7, 9, 11 ; 13, 15, 17, 19 ; and so on. Show that the sum
of the numbers in the nth group is n^.
E. 1. Simipmyx:2-2xy+y^-ix^-hxy-{-y^)-{x(-2y+x)+y^}.
2. Divide6a4+4&*-a5& + IS ab^ +2 a-b"^ by 2 a^ + 4^ b"^ - Sab.
3. Find the highest common divisor and the lowest common
multiple of 3 a* - 3 a^b + ab'^ - b^ and 4 a^ - 6 a6 + 6^,
4 ^ x-1
4. Simplify -
1 1
x—1 X
6. If a and /3 are roots of x^ -\- mx -\- n = 0, find a^/S + /S^a in
terms of m and n.
Test your result on the equation ic^ — 3 a; + 2 = 0.
6. Find the value of x for which Sx^ -{- 5 x -\- S has its least pos-
sible value, and show that the least value is ||.
7. Show that if a, fe, c, d be in continued proportion,
(a- by^a
[b - c) d
%. Find the square root of
4 x^ - 12 xy^ -7x^y + 24 x^ y^ + 16 y*.
9. Sum the series :
(i.) — 3 — 2 — 1 ••. to n terms,
(ii.) 1 — J + J^ ••• to infinity.
10. A train 72 yards long passed another train 60 yards long,
which was going in the opposite direction on a parallel line of rails,
in 4 seconds. Had the first-mentioned train been travelling at
twice its actual speed, the trains would have passed each other in
3 seconds. Find the number of miles per hour at which the trains
were travelling.
408 MISCELLANEOUS EXAMPLES VL
P. 1. Simplify
(x + y + zy-{- x + y + zy+ (x - y + zy- (x + y- z)\
2. Divide a^ - 3 a^fe + 3 a62 _ fts _ ^3 by a-b- c.
3. Prove that if x = a -\- d, y = b + d, z = c + d,
then x^ + y^ + z^ — yz — zx — xy = a^ + b^ + d^ — be — ca — ab.
4.
Simplify
1 2a
1
a-2x 4x^-
a'^ a
; + 2ic
5.
. Solve the equations :
(i.) 2x-^4:y-Sz
= 22
4x — 2y + bz
= 18
■•
5ic+ y-2z
= 14,
.... a; - 1 a; - 2 .
X —
5 X-
-6
'^x-2 ic-3 x-6 ic-7
(iii.) ?i±l + ^-I=^ = ^ 1
x2 + ?/2 ^ 90 J
6. Show that, if any integer be put for x in the expression
x6 _ 4x5 + 14x4 - 32x3 + 49x2 - 60x + 36,
the result will be a square number.
7. Keduce to the best form for numerical cal-
V(12-V140)
culations, and find its value to 4 places of decimals.
8. Show that the ratio a — x : a + x is greater or less than the
ratio a2 — x2 : a2 + x2, according as the ratio x : a is greater or less
than unity.
9. If _ + i varies inversely as x + y, then x^ + y^ varies as xy.
X y
10. If a, b, and c are the sums of n, 2 n, and 3 n terms respec-
tively of a geometric progression, then a2 + 62 = a(& + c).
INEQUALITIES. 409
CHAPTER XXVII. "^
Inequalities.
276. The statement p — q is positive is expressed alge-
braically hyp — q > 0, and the statement p — q is negative
is similarly expressed hj p — q < 0. With this notation
the definition of an algebraic inequality becomes
p>q,iip-q>Oj [Art. 47.]
and it immediately follows that
p<q,iip — q<0.
For example, since — 3— (— 5) = 2>0,
.-. - 3 > - 5, and since -9-(-6)=-4<0,
... _9<_6.
This definition necessarily involves the converse propo-
sitions that
p - g > 0, if p > 9,
^ _ g < 0, it p<q.
We shall call either of these two forms of inequalities
the reverse of the other.
The equational statement
p = q, it p-q = 0
is the transitional form through which either of these
inequalities passes into its reverse.
410 mEQUALITIES.
277. The simpler properties of inequalities are direct
consequences of the foregoing definition.
I. Every inequality can he expressed either in the form
p > 0, or in that of its reverse p < 0 ; — an obvious conse-
quence of the definition of an inequality.
II. If p^q and q > r, then p>r.
For, {p — q)-{-{q — r) =p — r > 0.
Cor. Ifp>q>r>S'-'>z, then p > z.
III. Any term may be transferred from one side of an
inequality to the other, provided its sign be changed.
For, ii p -\- r > q, then by definition {p -\- r) — q > 0,
that is, jp — (Q' — ^) > 0, and therefore p > q — r. Sim-
ilarly, if p > g V Sj it may be shown that p — s > q.
IV. The signs of all the terms of an inequality may' be
changed, provided the symbol of inequality be reversed.
For, if m-\-p — r'>n-{-q — s, then by III.,
— n — g4-s>— wi —p -f r,
that is, —m—p + r<—n — q-{-s.
278. The terms of an inequality may be combined with
the terms of an equation in accordance with the follow-
ing laws :
V. If p> q and r = s, then p±r'>q±s.
For, (p±r) — {q±s) = {p — q)±{r — s)=p-q=Si,
positive quantity. ^,^
VI. If p > q and r = s, then
pr > qs and p/r > q/s, if r and s be positive ;
pr < qs and p/r < q/s, if r and s be negative.
(^
INEQUALITIES. 411
For, since t=s, and p — g is positive by hypothesis,
.'. pr-gs, p/r-q/s = {p-q)r, {p-q)/r;
whence it follows that pr — qs and p/r — q/s are positive
when r and s are positive, but negative when r and s are
negative.
CoR. 1. If q be positive^ and p > qr, and r > s, then
p > qs.
Cor. 2. If the terms of an inequality be fractional, they
can be made integral by the proper multiplications.
Thus, from f > f we derive 5 x 5 > 4 x 6, and from > — ;
weobtain (-4)x(-6)<5 X 5. ^ ~^
279. The terms of two or more inequalities may be
associated with one another, but under certain important
limitations.
VII. If Pi>qi,P2> 72, Ps > ^3, ••' Pn > Qny
then ^i-i-p2+P3+--- +P„> 71 + 92 + 93+ •••+9n.
For, since {pi - qi),{p2 -q'i),"'{pn- Qn) are all posi-
tive and
{Pi +P2 + —Pn) - (7i + 72 + ••• +7«)
= (i>i-7i)+-+(P«-g«),
so is (pi+PaH hi>„)-(7i + 72H 1-7„) also posi-
tive, and therefore
Pi +P2 + — +i>n > 7l + 72 + ••• + 7«-
From Pi > qi andpa > 72 we cannot infer p^— i?2>7i— 72-
For example, 8 > 7 and 5 > 2, but 8 — 5 < 7 — 2.
VIII. If Pi>quP2> 72, Ps > 73, '•'Pn> qny
then PiPiPi'-'Pn > 7i7273--- 7n,
provided all the quantities be positive.
412 INEQUALITIES.
For, since p^ > qi and paPs '-'Pn is positive,
••• PlP2"'Pn>qiP2P3"'Pny
and since p2 > ^2 ^-nd QiPs-'-Pn is positive,
.-. PiP2"'Pn>qi<l2P3"'Pn,
and proceeding in this way until all the p's on the right
are replaced by the corresponding g's, we have finally
PlP2"'Pn>qiq2'"qn-
From pi > gi and ^2 > 9^2 we cannot infer Pi/p2 > Q'i/5'2-
For example, 8 > 6 and 4 > 2, but 8/4 < 6/2.
Cor. J/*!) and q be positive and n z= a positive integer^
and if p > g, then p" > q"".
IX. i/* j9 a?id g he positive, n= a positive integer, and
^\/n^ gi/n^ (jlenote the positive real n*^ roots of p, q ; and if
p>q, then p^^"" > q^^\
For, if this be not so, then p^^"" < or = g^/" ; but since
p^/" and g^/" are both real and positive, fromp^/" < or = g^/"
it follows, by the corollary of VIII, that (p^/")"<or
__ (gV»)»i^ whence p< or = g, which contradicts the
hypothesis p>q.
CoR. Ifp and q he positive and m,n = positive integers,
and if p> q, then p"*/** > g«/"; provided, if n he not 1,
only the positive real n'* roots he taken.
280. The following examples illustrate some of the
uses of the method of inequalities :
Ex. 1. The sum of any real positive quantity and its reciprocal
is greater than 2.
Let the quantity be a, and let a ^/^ denote its positive square root.
Then (aV2 - l/aV2)2 = «_2 + l/a>0;
.-. a-f l/a>2. [byllL]
INEQUALITIES. 413
Ex. 2. For what values of xisx^ — 4x + 3> — 1?
Completing the square for x, we have
a;2 _ 4x + 3 = (x - 2)2 - 4 + 3.
Hence, the given inequality is satisfied if
(a; _ 2)2 - 1 > - 1,
that is, for all real values of x except the value a; = 2.
Ex. 3. For what values of x is (2 x - l)/(x + 2) >, =, or < 1 ?
Let /=(2x-l)/(x + 2).
We have /> or < 1,
according as (2 x - 1) / (x + 2) - 1 > or < 0, [by III. ]
according as {(2 x - 1) - (x + 2)}/ (x + 2) > or < 0,
according as (x - 3) / (x + 2) > or < 0,
according as (x - 3) (x + 2) > or < 0 ; [by VI.]
.•:/>or<0,
according as x > or < 3, if x be positive,
or, according as — x > or < 2, if x be negative ;
and /= 1, if x = 3.
Ex. 4. The arithmetic mean of any two positive quantities is
greater than their geometric mean.
If the two quantities be a and 6, the arithmetic mean is \{a -\- b),
the geometric mean is VCaft), and we have to prove that
' K« + ^) > V(a&).
Since (a - 6)2 > 0,
.-. (a-6)2 + 4rt6>4a6, [by V.]
that is, (a + 6)2 > 4 a& ;
.-. a + 6>2v(a6), [by IX.]
and dividing by 2 completes the proof.
This example is a particular case of the following :
Ex. 6. If X, y, z, '•' the n real quantities,
(n-l)Sx2>2Sxy.
414 INEQUALITIES.
The student will have no difficulty in verifying the indentity,
by comparing the like terms of its two members. But S(x — ?/)2
is essentially positive, and hence,
(w- l)Sx2-2Sx«/>0,
unless oc = y = z = '•• = t, for which S(x — yy = 0.
.-. (ri-l)2x2>2Sx?/.
We have here used the S notation, as explained in Art. 152.
The student should note its conciseness and its great utility in
examples of this kind.
281. The theory of simultaneous inequalities, involv-
ing two or more unknown quantities, is most clearly pre-
sented with the aid of certain geometrical constructions,
which are not available to the student who is not familiar
with at least the elements of analytic geometry. But
the ordinary process of elimination by addition can be
applied to some of the simpler examples of this class.
It must be remembered that multiplications, with the
corresponding terms of two inequalities, can be per-
formed only with limitation, and subtractions and divis-
ions not at all. [Art. 279.]
Ex. 1. Resolve the simultaneous inequalities
2xf?/-6>0, -3ic + 2?/ + 6>0, x-y -]-l>0.
We apply the ordinary process of elimination by addition.
From the first and second inequalities we obtain
6x + Sy -1S>0
-6x + 4y + 12>0
Ty- 6>0, .-. 2/>f.
From the first and third
2x+y-e>0
x-y+l>0
3 X - 5 > 0, . •. X > |.
INEQUALITIES. 415
From the second and third
-3x + 2y + 6>0
2x-2y + 2>0
-x + 8>0, .-. x<S.
From the second and third
Sx-2y -6<0
. -3a; + 3y-3<0
y-9<0, .'. y<9.
Thus, the values of x and y are confined within the following
limits :
8>a;>|, 9>2/>f.
Observe that if x and y are to be limited to finite values, the
inequalities must have both positive and negative terms.
Ex. 2. Resolve the simultaneous inequalities
bx + ay — ab> 0, xy < 0.
Reverse the first inequality, changing its signs for this purpose,
and then add the two inequalities together. We obtain
xy — bx — ay -{- ab <, 0,
that is, (x — a) (y — 6) < 0.
.'. x>a, y <b, or x<a, y>b.
If a and b are both positive, since x and y are of opposite sense,
either x > a, y < 0, or y > 6, x < 0.
EXAMPLES LXXX.
Determine the range of values of x that satisfy each of the fol-
lowing inequalities :
1. (x-5)(x-2)>(a;-l)(a;-3).
2. a;2-4a;-i-3>, =, <0.
> Z. x^-Sx+ 10 >, =, <0.
4. (x + 5)/(x-3)> or < 1.
6. (2x-3)/(x+3)>, =, or <1.
6. (3x-4)/(2x-l)>, =, or <1.
416 INEQUALITIES.
7. Compare the values of -^(x + 1) + ^/{'X, — 1) and V(2^) for
real values of x and positive values of the square roots.
8. Resolve the simultaneous inequalities
a;+22/-4>0, -3x + 22/ + 6>0, x-2/>0.
9. Resolve the simultaneous inequalities
x-i/ + 3>0, 2a;-4?/ + 8<0, x+by-b<0.
10. Determine the values of x and y that make
(x?/ + l)/(x + ?/)>,=, or <1.
Discuss the cases oj + ?/ > 0 and x + y < 0.
. 11. If X, y, z be real, prove that
Sx8>, =, or <.Sxyz, according as Sx>, =, or <0.
Use the result of Ex. 5, Art. 280, and the identity of Art. 164.
[Chrystal, Algebra, Vol. II., p. 40.]
12. If X, y, z be real and not all equal, prove that
(Sx)8> 27x^2; and <9Sx8.
13. If a > 6, show that
{X + a)/^(ofi + a'^)>, =, or <(x + &)/ V(«^ + &^)»
according as x>, =, or <v'(a&).
14. Prove that, if a, 6, c, c? be all positive and
a/h = b/c = c/d,
then (a-d)>3(6-c).
16. Prove that, in a geometrical progression having an odd
number of positive terms,
(sww of all terms)'^ (number of all terms) x {middle term).
LIMITS. LNDETERjynNATB FORMS. 417
CHAPTER XXVIII.
Limits. Indeterminate Forms.
282. Functional Nomenclature. Any expression which
involves x in such a manner as to undergo a change
of value whenever x changes is called a function of x.
If two or more letters, as x, y, z, • • • are involved in it,
the expression may be regarded as a function of the sev-
eral letters x, y, z,"-.
These letters, through whose values or changes the
function is itself determined or changed in value, are
called the variables, and all other letters (including mere
numbers) are then considered, for the time being, as not
susceptible of change, and are called constants.
If the function does not contain, or can be so reduced
as not to contain, fractions whose denominators involve
^he variables, it is said to be integral ; but if such denom-
inators are an indispensable part of the function, it is
said to be fractional.
The term rational, or irrational, is applied to the function
according as its variables do, or do not, appear every-
where in it in the rational form.
The following examples will sufficiently illustrate this
functional nomenclature :
Ex. 1. (i.) axy + bx + cy -\- d is a rational-integral function of x
and y of the second degree.
2d
418 LIMITS. INDETEKMINATE FORMS.
(ii.) ax^-\- hxy + cy'^ is a rational integral homogeneous function
of X and y of the second degree.
(iii.) (ax + by -{- c) / (l + z) is a rational fractional function of
X, y, and z. It is, however, a rational integral function of x and y.
(iv.) axy + bVx + y is an irrational function of x and y.
(v. ) x^ + y^ + z^ -\- xyz is a rational symmetrical function of x,
y, and z. [Art. 150.]
Ex. 2. Describe the following functions :
(i.) ax^ + 6?/^ + Scxyz.
(ii.) ax + 6?/ + 3c(xy0)i/8.
(iii.) (ax + by + c) / Vr+~z.
(iv.) ax2 + 2 6x + c.
(v.) a(x^-\-if)/Vxy.
In all of the above examples the letters x, y, z ave regarded as
variables, a, 6, c, (2 as constants ; and, in general, constants are
denoted by the letters of the first half of the alphabet, variables by
the letters of the second half.
283. It is important to have the means of using the
ex^Tessions function of x, function ofx, y, etc., in mathe-
matical formulae, and for this purpose the abbreviations
f{x), f(x, y), etc., are employed.
Thus, such an expression as ax^ + bxy -\- cy^ -\- d is briefly repre-
sented by /(x, ?/) . And if
/(x, y) = ax^ + bxy + cy'^ + d,
and we assign x = 1, y = 2, then we write
/(I, 2)=a.l + 6.1.2 + c-22+d
= a + 26 + 4c + (^;
similarly f(l, 0)= a -h d.
LIMITS. INDETERMINATE FORMS. 419
284. A function of x is continuous, or varies continuously,
if its change, due to an arbitrarily small * change in x,
is also arbitrarily small.
Thus, f{x) is continuous, at the value x, if the difference
f(x + h) — f{x) may be made arbitrarily small by making h arbi-
trarily small.
A function is discontinuous, at any value of x for which
it does not fulfil the foregoing condition.
285. Definition of a Limit. If by making x approach
the fixed value a, in such a way that x — a becomes arbi-
trarily small, we make f{x) approach as near as we please
to the fixed value Z, then I is said to be the limit of f(x)
as X approaches a. We express this briefly by writing
It will be convenient to use the sign = as an abbre-
viation for the expression approaches as a limit.
The following examples will help to make the mean-
ing of this definition clear.
Ex. 1. What is the value of (x^ - a^)/(x - a) when x = a?
If a be substituted for x in (x^ — a^)/{x — a), the result is 0/0,
an expression which cannot be interpreted as an algebraic opera-
tion. There is therefore no proper answer to the question as
above formulated. But expressed in the form, what limit does
(x^ — a^) / (x — a) approach when x = a, the question can be
answered in the language of limits as follows :
Since (x'^ — a^) / {x — a) = {x -\- a){x — a) / (x — d)= x + a,
for all values of x not absolutely equal to a, we see at once that
* Smaller than a previously assigned quantity whose value we
may make as small as we please.
420 LIMITS. INDETERMINATE FORMS.
when X differs arbitrarily little from a, {^ — a*^) / (x — a) differs
arbitrarily little from 2 a ;
, limit a;2-a2 _ ^
hence, ^ ^ = z a.
Ex. 2. Find the sum to infinity of the series
By the ordinary process of division we have
^^1^ = X + a:2 + x3 + ... + x"-i,
I —X
X x^
that is, x + x2+ ... + a:'»-i =
1 — X 1 — X
and this formula is true for all finite values of x.
Putting X = I, we have
If the number of terms of the series be now increased beyond
finite range, l/2'*-i = 0, and therefore '
n = CO \^2 4 2«-v
This series was discussed in Art. 266, and its limit was there
determined from geometrical considerations.
286. If, when the variable approaches a limit, the
function exceeds every finite value however large, we
extend the language of limits to this case and say that
the function has the limit oo . This use of the term
limit must be understood to be exceptional.
For example,
limit x^ + X + 2 _ limit a; + 2 _ ^^
X=lx3-x2-X+l X-ly?- -\
The value 0 may obviously occur as a limit.
^ 1 limit x^-\ _ r.
For example, ^ ^ ^ -^^ - ^'
LIMITS. INDETERMINATE FORMS. 421
INDETERMINATE FORMS.
287. Certain combinations of 0 and oo produce what
are known as indeterminate forms. Such a form made
its appearance in Ex. 1 of Art. 285, where the substi-
tution of a for X in {x^ — o?) / {x — a) resulted in the
illusory fraction 0/0. The evaluation of indeterminate
forms is essentially an application of the method of
limits. The simpler forms are 0/0, oo/oo, 0 x co, oo — cc.
The following example^ will show how they may arise
and exemplify the method of their evaluation.
Ex. 1. (1 - x5)/(l - a;-2) becomes 0/0 when x = \. But
l-x5_l-t-a; + a;^ + a^ + a^ 1 -a;.
I_x2~ 1 + a; ' \-z'
limit \ — x^ _ 5
• • X = 1 1 _ x'-i ~ 2'
Ex. 2. (2 ic8 + 3 xS) / (3 x2 + 5 x*) becomes 0/0 when a; = 0. But
2x»-|-3x5_2a; + 3a^ a^.
3x2 + 6x4 3 + 6x8 ' x^'
. limit 2x8 + 3x5_0_Q
• * « = 0 3x2 + 5x4 3 •
Ex. 3. (a + &'*)/(c + 6") becomes qo/qo when 6 > 1 and n = 00.
If numerator and denominator be multiphed by 6-«, then
Umit g + ft"_ limit a&-» + 1 _ ^
n = Qo c + 6'»~w = Qo c^-** + 1 ~ *^
If?)<l, say b = l/k, where A;>1, then
(a + &") / (c + &") = (a + A:-") / (c + A:-«),
and ^^"^'^ « + ^'" = «.
n = Qo c + A;-« c
If 6 = 1, no inference can as yet be drawn.
422 LIMITS. INDETERMINATE FORMS.
Ex. 4. The function (x - 2)/(\/2x - 2) assumes the form 0/0
when X = 2, or the form co/oo when x = co.
In the first case we have
limit X — 2 _ limit y/x + v/2 _ „
a^ = 2 ^(2 x)-2~^ = 2"^2 ~*
In the second case
limit X — 2 _ limit 1 - 2/x _1_
a^ = '» V(2a^)-2~^ = °o V(2A)-2/x"0~**
Ex. 5. The function
•^^^> = (^+^)*^' + ^-2>
assumes the form oo x 0 when x = 1. But
1 _L « —^ + a — 1
X — 1 X — 1
and X2 + X-2 =:(x + 2)(x- 1).
. limit -, X limit , , i\/^ , oa q^
••• a.^i/(a;) = ^^l(a; + a-l)(x + 2)=3a.
Ex. 6. The function
/(x)=a/(l+x)-2&/(l-a;2)
assumes the form go — go when x = — 1. But
a/(l + a;)=a(l-a;)/(l-x2),
and therefore
/w^^^^ff^-
(i.) Now, in whatever manner x attains the value — 1, the numer-
ator of this fraction assumes the unique value 2 a — 2 &, and the
denominator becomes zero. But 1 — x:^ is and remains positive if
x2 = 1 through increasing values a little less than 1, or it is and
remains negative if x^ = 1 through decreasing values a little greater
than 1. In the former case /(x) = + 2(a — 6) x co, and if a > 6,
limit ^/ N
LIMITS. INDETERMINATE FORMS. 423
In the latter case/(x) = — 2(a — 6) x oo,
and if a > Z>,
The function has two branches, one for values of cc less than
— 1, the other for values of x greater than —1, and each branch'
has its own limit at the value a; = — 1.
(ii.) If & = a, the function becomes
/•/3.N^-a(l + a;)^ a . 1 + a;
^^ ^ l_x'-i x-\ 1 + x
As an indeterminate form it has been changed to 0/0, and its
limit, for ic = — 1, is obviously —\a.
288. The fundamental indeterminate form is 0/0, to
which all the others may be reduced. This property
results from the definition of infinity, as given in Art.
168, in accordance with which, if a and h be any finite
quantities, the symbol oo may be replaced ad libitum by
any such value as a/0 or 6/0.
Hence we may write
QO_a /b__a 0_0-a_0
a b 0-a — O'b 0
00 — oo := = = —
0 0 0 0
Ex. 1. The function a/(l -x)-2 a/ (I - x^) becomes oo - oo
when x = l. But
a 2o__ a(l -\- x)—2a
1 - ic 1 - x2 ~ 1 - x2 '
and in this form it becomes 0/0 when a; = 1.
424 LIMITS. INDETERMINATE FORMS.
Ex. 2. The function (1 - a'*)aV«, where a > 1, becomes 0 x oo
when n = + 0. But
and in this form it becomes 0/0 when w = + 0.
289. That any quantity whatever may appear as the
result in the evaluation of any of the indeterminate
forms thus far considered is at once evident from the
consideration that the one form 0/0 may be taken as the
general representative of the entire set of four forms,
and that, a being an arbitrary finite quantity,
a X 0 = 0, 0x0 = 0, and O/oo =0;
for, from these three equations we derive, by the proper
multiplications and divisions,
a = 0/0, 0 = 0/0, and ao = 0/0,
respectively. The multiplications and divisions here
performed are, of course, merely symbolic.
The important fact to be remembered, in this connec-
tion, is that the indeterminate form gives no information
concerning the critical value of the function that gives
rise to it.
EXAMPLES LXXXI.
Evaluate the limits of the following functions for the values of
X indicated, taking account of positive square roots only.
1. (x2 - X - 2) /(x2 + a; - 6) when x = 2.
2. (x3 -x^-x + l)/(x5 + x^ -5x + S) when x = 1.
3. (ic« - a") /{x — a) when x^a.
4. (a:3/2 _ «3/2) / (^1/2 _ ^1/2) when a; = a.
5. (V^c- l)/>/(a: - 1) whena; = 1.
L
LIMITS. INDETERMINATE FORMS. 425
6. {2 - ^(x + 3)}/(l - x) when x = 1.
7. -—^ ^ — wheii?; = l.
V(l-v'^) VCl-^)
8- V^-^-VC^-l) whenx = l. ^QtXu<Hv^-v'wv\.
(« - 1)2 ^a; - 1 -J
2 1
9. when x = 2.
2-x 2- ^'Ix
10. {2 - V(l + a;)}/{3 - V3^} when x = 3.
1 1 ax* + 6x2 + ex + <Z
11. — ' =!— when x = 00-
It^ + mx + w
1 o ax* + 6a; + c ,
12. — - — ■ — ^ — when x = oo.
Zx^ + wx2 + n
18. Solve the equation + ^{y? _ 9) - (x - 9) = 0. [See page
248.]
14. Solve the equation + VC^^ + 2 ax) - (x + a) = 0.
i26 EXPONENTIALS AND LOGARITHMS.
CHAPTER XXIX.
Exponentials and Logarithms.
290. The inverse operations of raising to powers and
extracting of roots, considered in Chapter XX., may be
regarded as presenting themselves through the medium
of the equation
in the two inverse problems :
(1) Given a Jcriown number x and a known integer k, to
find y. {Involution.)
(2) Given a known number y and a known integer k,
to find X. {Evolution.)
In these problems the exponent, as it appears either in
x^ or in y/*, is always to be regarded, either as an integer,
positive or negative, or as the reciprocal of such an inte-
ger ; the case where it is a fraction whose numerator is
not unity calling for the application of the processes of
involution and evolution in succession.
Thus, if the form 6^/^ presents itself, we first find 6^, or
216, and seek the fifth roots of 216 ; or, we may reverse
the order of this work and get first the fifth roots of 6
and then cube each one of them.
Involution and evolution, therefore, deal primarily
with such expressions as
2% 3^ 8^^ etc.,
and 2V3,. ^y\ lliA etc.
EXPONENTIALS AND LOGARITHMS. 427
291. If now the form of the expression d" be changed
to }f, for the purpose of indicating that the exponent
may become an unknown quantity (the variable), the
equation
gives rise to the two inverse problems :
(1) Given x and b, any known numbers, to find y.
(2) Given y and b, any known numbers, to find x. ^
The operation upon b which produces b' is called expo-
nentiation, b is called the base, and 6* (sometimes also
written exp^a;) is called the exponential of x with respect
to the base b.
The inverse operation of producing x, when b and y are
given, is called logarithmic operation, and is expressed sym-
bolically by the equation
x = nog y,
in which *log y is read logarithm of y with respect to the
base b.
It is an immediate consequence of this definition that
EXPONENTIATION.
292. The process of exponentiation seeks the value of
expressions of the form b', in which x and b are any
real numbers.f But in thus generalizing the combined
* A third case may also present itself ; namely, given x and y,to
find h. But this is identical with (1), for the equation h' = y may
be replaced by 6 = yV*.
t In the most general interpretation, exponentiation also allows
h and x, in the expression 6^, to have imaginary and complex values.
428 EXPONENTIALS AND LOGAEITHMS.
operations of involution and evolution, it so restricts (by
arbitrary definition) the interpretation of its results that
to an unique pair of values of x and h there corresponds
one and only one value of If, so that while 6^/*, regarded
as an example in evolution, has h roots, as the exponen-
tial of l/k to base h it has but one value. The expo-
nential will therefore be so defined as to embody this
restriction.
As already indicated, this restriction is arbitrary, but
in the chapter on exponential and logarithmic series, a
definite algebraic form will be assigned to 6'=, namely, a
series whose terms are integral powers of x having co-
efiicients that involve only h and certain numerical frac-
tions, and in this form the uniqueness of the value of Jf,
for a given value of x, will be apparent.
293. In defining If for our present purposes we suppose
that h is real, positive and greater than 1, and should x
be irrational, we replace it by one of the rational frac-
tions m/n, or (m + V)/n,^ which, by properly choosing
the integers m and n, may be made to differ from x by
an arbitrarily small quantity, that is, by a quantity that
is smaller than 1/w, where n is as large as we choose to
make it. [See Art. 249.] The values of h"" we have to
But the discussion of these more general cases does not fall within
the scope of this elementary treatise. For a fuller Interpretation
of exponentiation and logarithmic operation the student is referred
to Chrystal's^ Z^e&ra, Vol. II., Chapter XXIX., andtoStringham's
Uniplauar Algebra, Chapter III.
*It must not be inferred that these restrictions are necessary in
the general theory of exponentials. [See foot-note to Art. 292.]
EXPONENTIALS AND LOGARITHMS. 429
consider, are therefore of the form ft*^/", in which m and
n are integers, and we retain only the real positive n^^
root.
For negative values of x, say a; = — m/n, h' becomes
1/6'"/'*, namely, the reciprocal of an exponential with
positive exponent.
294. The Laws of Indices as applied to expressions of
the form 6*, under the limitations above imposed upon
the values of h and x, have been fully explained and
justified in Chapters XX., XXI., and when we require
their application to exponentials and logarithms we may,
therefore, write without hesitation,
6^6*6*..- = 6"+"+'+-,
h'/h" = ¥-^,
(p')y = {byy = h^,
{ahC"'y = a'h'<f"'.
295. If a series of numbers he in arithmetical progres-
sion, their exponentials, with respect to any base, are in
geometrical progression.
This proposition follows immediately from the equa-
tions
6- =6-, b'+^^b'b'',
Here d is the common difference of the arithmetical
progression
X, X -\-d, x-\-2d,"-x-{- nd,
and ¥ is the common ratio of the geometrical progression
b', b'b", h^b'^r-b^b'^-
430 EXPONENTIALS AND LOGARITHMS.
LOGARITHMIC OPERATION.
296. We have already defined the logarithm to be a
value of X obtained from the equation y. =]f^ when 6 and
y are given, and we denote this value of x by
JB=?=^l0g2/.
The properties of logarithms follow immediately from
the fundamental laws of indices. We proceed to deduce
them and to apply them to a few important examples.
297. Properties of Logarithms.
I. Since IP =\ for all finite values of 6, it follows
that 4og 1 = 0; that is :
The logarithm of 1 is 0, whatever the base may he.
II. Since h^ = h for all values of b, it follows that
*log & = 1 ; that is :
The logarithm of any number with respect to that num-
ber as base is always 1.
III. li^logx=p and *log?/ = g; then, by definition,
x=b^ and y z=b^-^
.'. xxy = bPxb^ = ¥'^^ ;
.-. 4og(a;2/)=j9 + Q' = *loga;4-nogy.
Similarly it can be proved that
^\og{xyZ'") = ^\ogx-\-^\ogy -\-^\ogz-\ .
That is : The logarithm of a product is the sum of the
logarithms of its factors.
EXPONENTIALS AND LOGARITHMS. 431
IV. If4ogic=p and ^logy = q; then, by definition,
x=b^ and y=¥',
.-. xjy= 6^/6« = 6P-*;
.\^\og{x/y)=p-q = nogx-'>\ogy. .
That is : The logarithm of a quotient is the algebraic
difference of the logarithms of the dividend and the divisor.
V. If ic = 6^ ; then x"" = bP"" for all values of m. Hence
*log af* = Hog bP^=p xm = Hog x x m.
That is : The logarithm of ayiy power of a member is
the product of the logarithm of the number by the index
of the power.
VI. Let *log x=p and *log x = q.
Then ' a; = a^, and x — ¥.
Hence a^ = 6', and therefore
a = 6'/^ and b^a^f"-,
and, by definition of a logarithm,
Hog a = q/p, and "log b = p/q ;
.-. Hoga X "log 6 = ? X - = 1.
P <1
Again, from q=p x Hog a,
we have *log x = Hog x x Hog a.
Hence : TJie logarithms of a series of numbers to the
base b are found by multii^lying the logarithms of those
numbers to the base a by the constant multiplier Hog a.
432 EXPONENTIALS AND LOGARITHMS.
VII. Since 6+* = + oo , and 6"* = 0, for all positive
finite values of h, \h > 1],
4og( + oo) = + oo, and 4ogO = — oo.
Hence : In every system of logarithms the logarithm of
-j-oo is +00, and the logarithm of 0 is -co.
298. The logarithms in any system whose base is real,
positive and greater than 1 [see Art. 293] have the fol-
lowing properties.
VIII. The logarithm of a negative number cannot he a
real number.
For, if b be positive, and a; be a real number, 6^ cannot
be negative ; that is, x cannot be the logarithm of a nega-
tive number.
IX. Numbers greater than 1 have positive logarithms^
numbers less than 1 have negative logarithms.
For, if b"" — u and 6 > 1, then, since for n = a positive
number 6" > 1, and 6~" < 1,
.-. x>, or <0, according as w >, or < 1 ;
that is, since x = *log u,
4og w >, or < 0, according as w >, or < 1.
Note. — In a system whose base is positive and less than 1,
numbers greater than 1 have negative logarithms, and numbers less
than 1 have positive logarithms. For, by VI., Art. 297,
^ Hog{l/a) nog a
and if a and b be both greater than 1, by the proposition just
established, ^loga is positive, and *logw is positive or negative,
according as u is greater or less than 1.
EXPONENTIALS AND LOGARITHMS. 433
299. If a series of numbers be in geometrical progression
their logarithms are in arithmetical progression.
This is the proposition of Art. 295, stated in terms of
logarithms instead of exponentials. Thus, the loga-
rithms of the several terms of the geometrical , progres-
sion
are 4og y, 4og y -f- 4og r, • • • 'log y -{-n 4og r.
300. The following examples will illustrate this part
of our subject.
Ex. 1. Find 21og8, nog 2, wioglOOO, and ^log ^100.
8 = 23, 2 = 8^, 1000 = 103, and ^100 = loi
Hence 21og8 = 3, 81og2 = ^, loiog 1000 = 3, and loiog ^100 = f.
Ex. 2. Having given .ioiog2 = .3010300 and i<51og3 = .4771213,
findioiog40, i^og 12, ioiogl5, and loiog ^880.
By the laws of logarithmic operation we have :
ioiog40 = loiog (4 X 10)= ioiog2^ + i^loglO
= 2 X .3010300 + 1 = 1.6020600.
loiog 12 = loiog (22 X 3) = 2 X ioiog2 + ioiog3
= 2 X .3010300 4- .4771213
= 1.0791813.
ioiogl5 = ioiog^i^^^= wioglO + ioiog3 - wiog2
= 1 + .4771213 - .3010300
. = 1.17G0913.
loiog ^880 = I X loiog (10 X 25 X 32)
= K^°log 10 + 5 X i'^log2 + 2 X wiog3)
= 1(1 + 1.5051500 + .9542426)
= 1.1531308.
2e
434 EXPONENTIALS AND LOGARITHMS.
Ex. 3. Solve the equation a'^^+<' = d.
Taking the logarithms of both sides of the equation, with respect
to base a, we have
6x + c = «log d.
. '-logd-c
• h
Ex. 4. Solve the equation
f(x) =(2^ . ai*-3 - 2* \ a^-2 - 4 • a^x-i + 4 == o.
The left member of the equation may be factored thus :
fix) = 2* .;a^-2 (a3x-i _ 1) _4(a3^-i - 1)
= (2«.a^-2-4)(a3x-i_i).
Hence 2* • a'-- -4 = 0, or a^x-i _ 1 _ q ;
and, therefore, by logarithmic operation, in any system,
xlog2+ (ic - 2) log a - log 22 = 0,
{X - 2)(log 2 + log a) = 0, a; = 2,
or (3 X — 1) log a = 0, x = \.
EXAMPLES LXXXII.
Find:
1. loiogl. 6. loiogV-Ol. 9. ^2iog2.
2. loiog 10000. 6. loiog^lOO. 10. ^logSv^.
3. loiog.Ol. 7. 2log32. 11. nog 3^3.
4. loiog.OOl. 8. nog 32. 12. ^log ^64.
13. Having given ^log 2 = .3010300 and i^log 3 = .4771213, find
loiogf, loiogOO, ioiog4500, and ^log ^25.
14. Given loiog 3 = .4771213 and ^log 5 = .6989700, find
i^log3.75, ioiogl.28 and wiog(33 x 'o^f2J).
15. Given ^log 5 = .6989700 and i^log 6 = .7781513, find ^ ,g 324,
loiog 1.458, and ^log. 00432.
16. Given I'^log 5 = .6989700 and loiog 7 = .8450980, find
ioiogl.25, ioiogl.28, and ^log (23 x 1^/^^).
EXPONENTIALS AND LOGARITHMS. 435
"^ 17. Given i^log 12 = 1.0791813 and ^log 18 = 1.2552726, find
io;og8 and ioiog9.
18. Given ^log 24 = 1.3802113 and wiog 36 = 1.5563026, find
loiog 72 and i^log ^432.
19. If a, 6, c be tlie bases of any three systems of logarithms,
p .ve that
51oga:=M«.alogX.
«log6
20. Prove that, in a series of logarithmic systems, whose bases
are respectively a, &, c, etc.,
"log a; ^ ^loga; _ ''logx _ ^^^^
«logy ^logy «logy
21. Given «log2 = 69315 and «log 5 = 1.60944, find ioiog80,
61og64, and 2log 125.
22. Given «log 5 = 1.60944 and 'log 3 = 1.09861, find 6log27,
81og25, and ^log 15.
23. If loiog 30.6614 = 1.48659 and ^log 3.1299 = .49553, what is
the relation between 30.6614 and 3.1299 ?
24. If loiog .3.0501 = .4813141, i^log 9.458055 = .9758019 and
I'^log 3.1009 = .4914878, what is the relation between 3.0501, 3.1009,
and 9.458055?
X 25. Given ioiog2 = .301030, find x from the equation 2* = 10.
26. Given ^log 2 = .301030 and ^log 5 = .698970, find x from
the equation 2=^-^ = 10*.
27. Solve the equation 2^ - 2*+^ -f 1 = 0.
28. Solve the equation 52* — 5*+i + 6 = 0, obtaining two values
of X in terms of the logarithms of 2, 3, and 5.
>•' 29. Solve the equation a^~^ - a^-^ — a* + 1 = 0.
30. Solve the equation a'^^ -\- b ■ a'' — c = 0.
^ 31. Solve the equation 2* • 3^'- 3* - 9 • 6* + 9 = 0.
32. "l^olve the equation 3*'-2*+2 =10, obtaining two values of x
in terms of the logarithm of 3 to base 10.
33. Solve the simultaneous equaii6'fi%
2x+2y - 10, 53y-i = 25*+y,
obtaining x and y in terms of the logarithm of 2 to base 10.
436 NATURAL LOGARITHMS.
CHAPTER XXX.
Natural Logarithms.
301. The limit which the ratio {^logy'—^logy)/{y^— y)
approaches when 2/' and y both approach the common
limit 1 is called the modulus of the system of logarithms
whose base is 6- Let y' = ry ; then this ratio becomes
4og?*z/— 4og?/ 4ogr
ry-y y{r — iy
and when y = l it is (4og r)/{r — 1).* Hence the modu-
lus may be defined as the limit which (*logr)/(r — 1)
approaches when r approaches 1 ; or, symbolically, if M
denote the modulus,
r = 1 r — 1
When the modulus is 1 the corresponding system of log-
arithms is called the natural or Napierian system,t and the
base of such a system is called natural or Napierian base.f
* This modulus is otherwise described as the rate of change
of ^logr at the instant when r = 1. [See Stringham's Uniplanar
Algebra, Art. 23.]
t The logarithms of the natural system must not be confounded
with the numbers of Napier's original tables, which were not cal-
culated with respect to a base, and are therefore different from the
so-called Napierian logarithms of modern tables.
NATURAL LOGARITHMS. 4B7
and is denoted by the letter e. We therefore have, by
definition,
limit "logr*^^
r= 1 r— 1
302. From this definition we are able to derive a for-
mula by means of which an approximate value of the
number e may be calculated, and this being done, the
modulus corresponding to any base b will be given by
the formula
limit 'log ^
M=
r=z 1 r — 1
= ' = 1F^'1°S«' [Art.297,VL]
that is, since limit (4og r)/(r — 1) = 1,
3/=*loge = l/'log6.
In place of *log we shall henceforth use the shorter
symbol In, made up of the initial letters of logarithm
and of natural or Napierian.
303. We first derive a formula which determines e as
a numerical limit.
For this purpose let
In r = n, that is, r = €",
and let u/{e^ — 1) = h, or e"li = h-\-u.
* This fraction is positive whether 1 be greater or less than 1,
for the logarithm of any number less than 1 is negative [Art. 298].
Hence r may approach 1 through values either a little smaller or
a little greater than 1.
438 NATURAL LOGARITHMS.
Then u = 0 when r = 1, and /i = 1 when u — 0, and we
have [Art. 301] ^
limit Inr limit u
But from eVi = h -{- u we obtain
. e'' = (l+i6//i)'^/";
and hence, since h = l when u=0,
limit
= limit ^ = 1.
e =
„ = o(i + «)■'"•
We have here supposed u to approach zero from the
positive side, but the supposition was not a necessary-
one. If we assign to u Si negative value, say —v, the
formula for the value of e becomes
limit
u = 0
e= Zai^-v)-'-.
304. Now it may be shown that in whatever way u is
made to approach zero, whether positively or negatively,
the limit of (1 + it)^/" is a finite number, and that when
u passes continuously towards zei'O, (1 + 2^)^/" diminishes,
and (1 — w)^/" increases, continuously towards a common
limit ; and that e is therefore a definite number.
For a formal proof of these statements, which is not
properly within the scope of this treatise, we refer the
student to Chrystal's Algebra, Vol. II., Chapter XXV.,
§ 13, and only present here a few numerical verifications.
By ordinary arithmetical computation we easily find :
(1 + 1)^ =2.25.
-i)-' =4.
(1 + 1)3 ^2.370.
-i)-3 =3.5.
(1+1)4 =2.44140625.
(1-
-^)-4 =3.1604903...
{1 + ^y =2.48832.
-1)-^ =3.0517578125.
a 4. _ij^)io = 2.5937424601.
-^y -10 = 2.867971990..
NATURAL LOGARITHMS. 439
Thus, as the exponents m (1 -j- ?*)^/" and (1 — r^)"^/"
are made numerically larger, the two series of numbers
produced by these formulae slowly approach one another,
those in the left column by increase, those in the right
column by decrease.
We can infer from this calculation that e is greater
than 2.5 and less than 2.9. None of the decimal figures,
however, in any of these numbers are as yet correct for the
value of e, and further extensions of this method become
rapidly tedious.* We therefore do not use the formula
(1 -h m)^/" for obtaining close approximations to the value
of e, but replace it by a rapidly converging series.
305. Such a series may be obtained by expanding
(1-j- l/z)" by the binomial theorem, on the supposition
that 2 is a positive integer and ultimately becomes infi-
nitely large. t In this investigation we shall use the
convenient symbol n ! to denote the product of the first
n natural numbers ; that is,
n ! = 1 • 2 • 3 • • • to 71 factors.
With this notation it is obvious that
nl= {n — 1)\ n = (?i — 2)!(n — l)n, etc.
By the binomial theorem we have, for integral values
of z [Art. 205],
♦ The value of e, correct to seven places of decimals, is 2.7182818,
and the value of (1 + .001)^ooo gives only two of these decimal
figures. In fact, the first eight figures in the computed value of
(1 + •001)«wo are 2.7169239.
t The demonstration is taken from Chrystal's Algebra, Vol. II.,
Chapter XXVIII., § 2.
440 NATURAL LOGARITHMS.
\ zj z 21 z^ 3! s^
, z(g-l)...(2;-n + l)l , ,1
n! 2** 2;*
that is,
(^ 2^ ^2! 3! ^
_^ (i-iA)-(i-»-iA) _^ _g
where
^ ^(l-lA)...(l-nA) (l-l/.)...(l-n+lA)
(71 + 1)! (n + 2)! ^ '
Let n have any /a;ec? integral value and let z be an
arbitrarily large integer, at least larger than n. The
series B^ will then terminate ; and each of its numera-
tors will be less than 1, for
1/z < 2/z <3/z<"' < n/z <"'< (z - l)/z < 1.
Hence
r» ^ 1 I 1 I 1 I 4- _
" (n -h 1) ! (n + 2) ! (n + 3) ! 2 !
<_^^ 1
(n + 1)! 1-1/(71 + 2)
If now in the second of the above series for (1 + l/z)'
we put 2J = 00, each of its numerators becomes 1, and we
have
i^""!! fi + -T= 1 + T + li + li +•••+-.+ -^^
z = Gc\ zJ 12! 31 ill
NATURAL LOGARITHMS.
441
where Bn satisfies the condition
1
Rn<
1
(n + l)!l-l/(n + 2)
Finally, if n be made to increase beyond finite range,
J?„ decreases towards the limit zero, the number of terms
in the series becomes indefinitely great, and we have
limit j^i ^ ly _ 1 + 1 ^ 1 _^ ... 1 ^ ... ^^ infinUum.
z=ccy zj 12! r!
306. This is the series by means of which the value
of e is calculated to as great a degree of accuracy as is
desired. The third term is .5, the fourth is .5/3, the
fifth is this last quotient divided by 4, and so on. Set-
ting down these results in succession as far as the term
1/10 ! we have
1 + 1 = 2.
1/2! =
.5
1/3! =
.1666667
1/4! =
416667
1/5! =
83333
1/6! =
13889
1/7! =
1984
1/8! =
248
1/9! =
27
1/10 ! =
3
.*. approximately, e = 2.7182818.
This result is an approximate value of the series
1
l + l + i+i^..,
^1^2!^3!^
+ j^+i2io
442 NATURAL LOGARITHMS.
with, the remainder Riq omitted. It will be useful to
observe how the neglect of this remainder, and of the
errors in the seventh decimal figures of the successive
terms, might affect the final result. Now, by the formula
fori2„,
i2io<j^<. 000000005,
and the neglect of this number could not affect the
seventh decimal figure ; moreover, the aggregate of the
errors in the several terms could not exceed 8 x .5, or 4,
in the seventh place, and this could not affect the fifth
decimal figure. Hence the above computation gives an
approximate value of e correct to at least five places of
decimals.
307. The following are important examples of limits
that involve exponential and logarithmic functions.
Ex. 1. Find the Hmit of (a' — l)/x when x = 0,
/yx 1 fixlna 1
Since = ^— • In a,
X xlna
... «»i'^«l^ = l„a. ■ [Art. 303.]
Ex. 2. Find the limit of (a* — a^)/(x — n) when a = n.
a* — a** „„ a*-" — 1
Since
x — n X — n
.^Tio^-i^=^-' CEX.1.]
Hmit^-^^^„j^^^
x = n
X
Ex. 3. Find the limit of a^/x when a; = oo, a > 1.
Whatever rational positive value x may have we can always
1,
NATURAL LOGARITHMS. 443
express it as the sum of a positive integer n and a positive finite
proper fraction /, and we may therefore write
9l. — ^^^^ — of ^ ^
X /+ n /+ n w'
in which / and af remain positive and finite for all positive values
of X. But, because/ is finite,
limit n _ limit 1
and therefore
limit g^_ f limit a^
in which w, always a positive integer, approaches oo by successive
integral steps.
Also, since by hypothesis a > 1, we may write
a = l + A, ^>0;
and hence, by the binomial theorem for a positive integral index,
a« = 1 + w/i + ^^^ ~ ^^ h^ + nx (other positive terms),
— = i+^4-— (n-l)+ other positive terms,
n n 2
and
limit a_" ^ limit ^ (^ _ j) + other positive terms = oo.
n = oo^ 71 = 002
. limit a^ _ „f ^ ^
a; = oo a;
Ex. 4. Find the limit of y "log y when
?/ = + 0, a>l.
Let y = a-" ; then
y «log y = — xa-* = — x/a*,
and a; = + 00 when y = + 0. But
limit
f-^) = -^=0. [Ex.3.]
limit
••y=+0y''^°S2^=^-
444 NATURAL LOGARITHMS.
Ex. 5. rind the limit of x* when x= + 0.
By Art. 291, a;* = 6=^1° =»;
. limit limit .m^.^o-i pEx 41
Ex. 6. Eind the limit of (x - 1)1"* when a: = 1.
We may write
(X - 1)1"=^ = {(X - l)=-l}(lnx)/(x-l).
But ^"^l (X - l)-i = 1, [Ex. 4.]
and ^^^^J(lnx)/(x-l)=l. [Art. 301.]
.-. ^^^i*^(x-l)ln-=ll = l.
308. Exponential Indeterminate Forms. In order to dis-
cover under what conditions the function u" will assume
an indeterminate form, as in Exs. 5 and 6, Art. 307, we
write
f^W, logf=vlogu,
the logarithms being taken in any system, and seek the
conditions that make vlogu indeterminate, and these
will also be the conditions that make /or u" indetermi-
nate. Since log 1 = 0, log co = oo, and log 0 = — oo, these
conditions obviously are
-y log 2^ = ( ± oo) X 0, or w" = 1 ±*,
vlogu =^ 0 x{+co), OT u" = co^,
vlogu = 0 x{—oo), OT u^ = 0^.
These are known as exponential indeterminate forms,
and the list is evidently exhaustive for forms that result
from combinations of values of u and v. But of course
NATURAL LOGARITHMS. 445
either u ot v may by itself assume one of the algebraic
forms 0/0, co/oo, 0 x co, oo — oo.
Ex. 1. The limit of (1 + x/n)^ when n = co (x = a finite quan-
tity) is
Ex. 2. Find the limit of xV* when x = oo.
Let X = l/z, then 2 = 0 when x = oo,
and
xi/« = (i/2)- = l/2*.
Ex. 3. Find the limit of (In x)'-^ when x = 1.
We proceed as in Ex. 6, Art. 307, and obtain
(lnx)=^-i = {(lnx)^°'}(^-i)/i°*,
i^^\Vlnx)-^ = li = l.
EXAMPLES LXXXIII.
Evaluate the limits of the following functions for the values ot x
indicated :
1. In(x3 — 1) — ln(x — 1) when x = 1.
2. In(x2 _ 1)+ In / 1+ — ^"j when x = 1.
3. In {2 - VCl + X)} - In {3 - V(3 x)} when « = 3.
4. (x2 - 1) ln(x - 1) when x = 1.
6. (1 + l/x2)* when x = co.
6. (1 + l/xy^ when x = oo.
7. x^/^' when x = 1.
446 NATURAL LOGARITHMS.
8. xi/(a=2-i) when x = l.
9. {(x3 - l)/(x^ + 1)}''-^ when x = l.
10. (x)^^ when x = 0.
11. (a-^ - 1)^ when x = 0. ,
12. (a=* - a-^)^ when x = 0.
13. Solve the equation x^''+^ — 2x2*' — a; + 2 = 0.
14. Solve the equation
a^ . x^+i — X ■ a^ - x'' -\- a'' = 0,
in which a is positive and greater than 1. (Take account of the
result of Ex. 3, Art. 307.)
CONVERGENCY AND DIVERGENCY. 447
CHAPTER XXXI.
CONVERGENCY AND DIVERGENCY OF SERIES.
309. A succession of quantities, formed in order ac-
cording to some definite law, if finite in number, consti-
tute a finite series ; but if their number exceed every finite
quantity, however great, they are said to form an infinite
series.
We have already seen [Art. 266] that, when r is
numerically less than unity, the sum of the n terms of
the geometrical progression a -f ar + ar^ -f • • • -f ar^~^ can
be made to differ from a/(l — r) by an arbitrarily small
quantity, by sufficiently increasing n. Many other series
have this property, — that the sum of the first n terms
approaches a finite limit when n is increased ad infinitum.
310. Definitions. When the sum of a finite number of
terms of an infinite series approaches a finite limit (or
zero), as the number of terms is increased ad infinitum,
the series is said to be convergent, and this limit is called
the sum, or limit of the series.
If the series have ± oo as its limit, it is said to be di-
vergent; and if it have neither a finite nor an infinite
limit, it is said to be neutral or indeterminate.*
* Series of this last class are also called divergent series in many
text-books of algebra.
448 CONVERGENCY AND DIVERGENCY.
Ex. 1. The series ^ + I + I -h •■■ ad inf. is convergent, and has
the limit 1. [Art. 285, Ex.2.]
Ex. 2. The series 1 + 2 + 3 + ••• acZ inf. has + oo for its limit,
and is therefore divergent.
Ex. 3. The series 1 — 1 + 1 — IH ad inf. has no definite
limit, but is 1 or 0 according as the number of terms is odd or even;
it is therefore indeterminate.
311. It is evident that an infinite series whose terms
are either all positive or all negative cannot be indeter-
minate, but must have either a finite or an infinite limit.
If each term of an infinite series be finite (not zero)
and the terms are either all positive, or all negative, the
series must be divergent. For, if each term be not less
than a the sum of n t^rms will be not less than na, and
na can be made larger than any finite quantity, by suffi-
ciently increasing n.
312. The successive terms of an infinite series will be
denoted by Wi, U2, ^3? ••• w„j •••? the sum of the first n
terms by 17^, and the limit of the series, if it be conver-
gent, in which case alone it has a limit, by U. Thus
C4 = ^i + W2H hw„,
and U =Ui + Wg + ^^3 + ••• <xc? inf.
We shall frequently omit the words ad inf., and denote
the limit of the entire series by Wi + 1^2 + ^3 + ••••
313. In order that the series Ui-\-U2-\- f-^n + ***
may be convergent, it is, by definition, necessary and
sufficient that each of the finite sums Un, C^+u C^«x2>
etc., shall approach the common limit U, as n increases.
Hence U^, U^+i, Un+2, etc., must differ from U, and
CONVERGENCY AND DIVERGENCY. 449
therefore from one another, by quantities that converge
to zero, as n increases without limit.
But Un+i- Un=U^^j
U- Un = W„+l + W„+2 + Wn+3 + ••••
Hence : In order that a series may he convergent^ the
(n + 1)^^ term, and also the sum of any number of terms
beginning with the {n-\-l)th^ must have zero as a limit
when n is increased ad infinitum.
For example, the series 1/2 + 1/3 + 1/4 + ••• + 1/n + ••• can-
not be convergent, although limit „=oo(l/w) = 0 ; for the sum of n
terms, beginning at the (n + l)th, is
1 +_!_,+ 1+...+ 1
n+1 n+2 n+3 2n
which is greater than (1/2 n) x n, that is, than 1/2.
314. Tests for Oonvergency. The following are the most
important criteria for the convergency of infinite series.
It is convenient to state them as applying only to series
whose terms are all positive (Criterion IV. and the cases
in which negative terms are chai-acteristic of the series
excepted). But they are also effective when applied to
series whose terms are all negative, and (provided the
signs of all negative terms be changed) to that class
of series, containing both positive and negative terms,
which remain convergent when all the negative terms
are made positive. For, if a series be convergent when
all its terms are positive, it is obviously convergent when
either all or some of its terms are negative.
27
450 CONVERGENCY AND DIVERGENCY.
For example, if m^ + m.^ + Wg + •••
is convergent, so ig — (Wj + M2 + % + *•*)»
and also w^ — Wj + % — W4 + •••.
The term series means always infinite series unless
specific statement to the contrary is made.
315. Test by the difference between the terms of the
given series and the corresponding terms of a convergent
series.
I. A series {whose terms are all positive) is convergent
if, after any particular term, each of its terms is less than
the corresponding term of a known convergent series.
Let there be two series
XJ=,S-\-u^-\-U2-\-u^-\-'", V^Vi + v., + v.^-\ ,
in which S is the sum of a finite number of the terms of
U; suppose V to be convergent, and that u^ < v^ for all
values of r. Then U—S is certainly less than F, that is,
U< F+ S,
and since Fand 8 are finite, so also is XJ) and CT must
have a limit, for all its terms are positive.
II. It needs only a slight modification of this reason-
ing (which the student can easily make) to prove that
an infinite series {having only positive terms) is divergent
if, after any particular term, each of its terms is greater
than the corresponding terms of a divergent series.
CONVERGENCY AND DIVERGENCY. 451
Ex. 1. Apply the test for convergency to the series
1 1.2 1.2.3 1.2.3.4
The terms of this series are, in the order of occurrence, less
than the corresponding terms of the geometrical progression
1+ 1 +^j_,+ 1 .
1 1-2 1.2.2 1.2.2.2
whose common ratio is ^, and which is therefore known to be
convergent. The given series must therefore also be convergent.
Ex. 2. Test for convergency the series
1 + 3 + ^+33 3* 3^ _
2!3!4!6!
After the fifth term each term of this series is less than the
corresponding term of the geometrical progression
S6 Rd 37
4 . 4 1 42 . 4 ! 48 . 4 I
whose common ratio is 3/4, The given series is therefore con-
vergent.
316. Test by the ratio of the terms of the given series
to the corresponding terms of a convergent series.
III. If the ratio of the corresponding terms of two series
(of positive terms) be always finite, the series mil both be
convergent or both divergent.
Let the series be respectively
?7=Wi + W2 + %H J V=v^-{-v^-\-Vz-\-"'.
Then, since all the terms are supposed to be positive,
^/Fmust be greater than the least and less than the
greatest of the fractions Ui/vi, U2/V2, u^/v^, ••• [Art. 170,
452 CONVERGENCY AND DIVERGENCY.
Theorem II.]. Hence C7/F is finite; and, therefore, JJ
is finite or infinite according as Fis finite or infinite.
Ex. 1. The two series
2-3 4-5 6-7 2 7i(2w+l) '
3 + 5 + 7 + -+2^r^+-'
are both convergent or both divergent. For, the ratio of the n*^
terms is (n + l)/2 w, or 1/2 + 1/w, which cannot be greater than
3/2 or less than 1/2 for values of n between 1 and co.
Ex. 2. Prove that the two series of Ex. 1 are divergent by com-
paring the terms of the second one with the terms of the divergent
series 1 + 1/2 + 1/3 + h l/n -\ .
317. Test by ratio of convergence.
IV. A series is convergent if, after any particular term,
the absolute value of the ratio of each term to the preceding
is always less than some fixed finite quantity which is itself
less than unity.
Let the terms be supposed to be all positive, and in
accordance with the conditions here assigned, beginning
with the (r + 1)*^ term, let
u^+\/u^ < k, w^+a/'^r+i < ^? u^+Ju^^2 < ^j etc., where A; < 1.
Then u^^^ < ku^, u^j^2 < ^^^r+i < ^^^r? ^r+s < ^^^rj 3,nd so on
ad infinitum. Hence
Ur + u,+i + Wr+2 + '•• < u,{l + A) + A^^ + -) [Art. 279.]
< uj{l - k), '.' k < 1. [Art. 266.]
Therefore, the limit of the series beginning with the
7*^ term is finite, and since the sum of any finite number
of terms is finite, the entire series must be convergent.
CONVERGENCY AND DIVEKGENCY. 453
This demonstration remains valid for series having
both positive and negative terms, if only u^ u^^ etc., be
replaced by abv u^, abv u^^^, etc., where abv is an abbre-
viation for "absolute value of"; for, if the series be
convergent when all its terms are positive, it is certainly
convergent when some of them are negative.
We shall hereafter frequently use the abbreviation abv
to denote absolute value of.
Def. The ratio u^+i/u^ is called the ratio of convergence
of the series Ui + u^-^- u^ + •••,
V. A series {of positive terms) is divergent if after any
particular term, the ratio of each term to the preceding is
either equal to or greater than unity.
If every term after the r^ be equal to u^ then
W,+l + Wr+2 -\ h y^r+n = flU^
or, if after the r*^ term the ratio of convergence be
greater than 1, then
W,+1>W„ Wr+2 > W,+i > W^, "'U,^^>U^
that is,
W,+l -f Wr+2 H h ^^r+n > nU^
In either case the sum of these n terms can be made
greater than any iinite quantity by sufficiently increasing
71. Hence this part of the series, and therefore the entire
series, is divergent.
Ex. 1. The ratio of convergence of the series
1+1+1+...+A+...
2 22 28 2"
is {(n + l)/2«+i}/(n/2«) = (n + l)/2 n, and this is less than f for
all values of n greater than 2. Hence the series is convergent.
454 CONVERGENCY AND DIVERGENCY.
Ex. 2. The ratio of convergence of the series
2 22 23 2* 2«
1.22.33.44.5 w(w+l)
is 2(n ^ l)/(w + 1), which is greater than 1 for all values of n
greater than 2. Hence the series is divergent.
Ex. 3. The ratio of convergence of the series
12 _|. 22x + 32x2 + ... +n2x«-i + —
is x(n + l)2/n2, that is, (1 + 1/nyx.
Now if X be positive and less than 1, and any fixed quantity
k be chosen between x and 1, the ratio of convergence will be less
than k for all terms after the first one that renders
{l + l/nyx<k,
that is, for all terms after the n}^, where
n> y/x/{^k- y/x),
the square roots being taken positively. Hence the series is con-
vergent if ab V X < 1 .
If X = 1 the series is I2 + 22 + 32 + ..-, which is obviously diver-
gent, and it is therefore also divergent if x > 1.
When an infinite series is such that, after a finite
number of terms, its ratio of convergence u^^i/u^ is
always less than unity, but approaches unity as a limit,
the test by ratio of convergence fails. Further special
investigation is then necessary.
Ex. 4. The ratios of convergence of the series
i + i + i + i + ... + L+ 1 ■
2 3 4 n n+1
1 + 1 + 1+1+.. . +1 + }: + ...
22 3-' 42 n2 (w + 1)2 ^
are n/(n + 1) and ri- / {11 + 1)2 respectively, and each of them is
less than unity for all finite values of w, but approaches unity as a
limit when n is increased without limit. But, as will be immedi-
ately proved in the next article, the first series is divergent, the
second convergent.
CONYERGENCY AND DIVERGENCY. 455
318. The convergency or divergency of many series
may be determined by the methods of comparison,
explained in Arts. 315, 316, with the aid of the series
1/1* + 1/2* + 1/3* H- .-. as a standard. For this purpose
it is important to know for what values of k this standard
series is convergent.
VI. The series 1/1* + 1/2* + 1/3* + ••• is convergent
when k is greater than unity, and is divergent when k is
equal to or less than urdty.
(i.) If k>l, since each term is then less than the
preceding,
i + i<?
2* 3* 2*'
4* 5* 6* 7* 4*'
1. 1 . i . 1 . !_ , 1 . _! . A^-.:?
8* 9* 10* 11* 12* 13* "^ 14* 15* 8*"
and in general
i+_i_ + 1 <?:.
2nA^(2'* + l)* (2"+^-l)* 2«*
Hence, the entire series is less than
12 4 8 2**
— + — 4- — -I- — -4- ••• 4-—-+ •••
1* ^ 2* 4* 8* 2"* '
that is, it is less than
1+^+^-+-^ + -+^-+...
which is a geometrical progression whose common ratio,
l/2*~\ is less than unity, since A; > 1. The given series
is therefore converijent.
456 CONVERGENCY AND DIVERGENCY.
(ii.) If A; = 1, the series may be written in the form
l + i + (i + i) + (i + i + l + i)+-
V2«-i + 1 2"-^ + 2 ^ 2^y
with its terms so distributed into groups that each group
(within brackets) is greater than i Hence 2" terms of
the given series is greater than n-\-l terms of the series
^ + 2"+i + i+"*j *^^^^ ^s> greater than 1 + i n, which
may be made larger than any finite quantity by suflB.-
ciently increasing n. The series
is therefore divergent.
(iii.) If A;< 1, then each term of l/P+l/2*+l/3*+---
is greater than the corresponding term of the divergent
series 1 + i + ^ + •••. The original series is therefore
also divergent in this case. This completes the proof of
our theorem.
Ex. 1. The series
is divergent.
For since 2w/ (w^ + 1) > 1 /w for all values of n greater than 1,
.-. S2n/Cn2 + l)>Sl/w,
and S 1/n is divergent.
Ex. 2. The series 2 (w + 2)/(n^ + 1) is convergent.
For, (w + 2)/(w3 + i)< (n + 2)/n^<3n/n^ = Z/n\
.-. S (w + 2)/(n3 + 1)<3 S l/w2,
and S l/n^ has been shown to be convergent.
CONVERGENCY AND DIVERGENCY. 457
319. Def. If a series having both positive and nega-
tive terms remain convergent when all its negative terms
are made positive, it is said to be absolutely convergent.
The criteria of Arts. 315-318 are effective only when
applied to absolutely convergent series. And there is a
class of series, designated as semi-convergent, whose con-
vergency is due to the presence of negative terms, and
which become divergent when all the terms are taken
positively. Other tests, not yet considered, are then
necessary.
For example, as will be shown in the next article, the series
1 - ^ + i - i + ••• is convergent, although 1 +^ +^ +^ + ... is
divergent.
320. Test for semi-convergency.
VII. A series whose terms are alternately positive and
negative is convergent if each term is less than the preceding
and the n^^ term has zero as a limit when n increases ad
infinitum.
The general series conforming to these conditions is
U=Ui — 1*2 + ^3—^4 H ±^n T ^Wl ± •••>
with Wi > Wg > "^ > "^4 > ••• > ^n > Wn+1 > ••• > 0,
J limit r.
and „ = ^«n = 0;
and this we may write in either of the forms
U^— {lh— Us) — (^4 — y's) )
from which it is at once evident that the limit of the
series is greater than w, — Wj and less than u^, and
is therefore finite. It is also similarly evident that
abv (U — Un) is intermediate to abv (w„+i — Wn+2) a-iid
458 CONVERGENCY AND DIVERGENCY.
SihYUn+i (abv = absolute value of), and tliat therefore,
since limit u^ = 0, . ^
limit .:rx Txs ^ V-aT/V^
The series must consequently be convergent. /^
„r;(^-^")=o-
Ex. 1. The series 1 - 1/2 + 1/3 - 1/4 +.•• satisfies the condi-
tions of this test, and is therefore convergent.
Ex. 2. The series 2/1 - 3/2 + 4/3 - 5/4 + ... cannot be less
than 1/2 nor greater than 2, but it is not convergent, for the n^^
term, which is (?i + l)/w, has not 0, but 1, as its limit for w = oo.
321. Power Series. The infinite series whose general
term is a^x"" {a^ independent of x) is called a power series
in X. By Criterion IV. [Art. 317] this series is absolutely
convergent if abv xa^+i/a^ remain less than 1 in the
limit when n = cc, that is, if x satisfy the condition
abv cc < abv limit (a^/a^+i) when n = oo. Hence :
YIII. The power series
ao + aiX + aacc^ H f- a^af H
IS absolyitely convergent for all values of x that satisfy the
condition
abva!<aby^™^_^(a„/a„+0-
322. If a power series be only semi-convergent, since
the successive terms of every convergent series must ulti-
mately decrease in absolute value [Art. 313], the ratio
abv(iKa,i+i/a„) must ultimately still remain less than 1 for
finite values of n. But if abv(£ca„+i/a„) remain less than
1 in the limit when n = ao, the series is absolutely con-
vergent [Criterion VIII.]. Hence, semi-convergence can
CONVERGENCY AND DIVERGENCY. 459
only arise by the ratio of convergence being, in absolute
value, ultimately less than 1 for finite values of n, but
having the limit 1 when n = go.* Therefore :
IX. If the power series
tto + «ii» + ^2^ + • • • + o^n^;** 4- • • •
be convergent for a given value of x, it is absolutely conver-
gent for every numerically smaller value of x; and if it be
only semi-convergent for the given value, it is divergent for
every numerically larger value.
Upon the following corollary rests the proof of the
principle of indeterminate coefficients which will be dis-
cussed in the next chapter.
Cor. If the series CTq + %^ + ot2^+ ••• be convergent
for any finite value ofx,
Proof: From the identity
tto-f ttiJcH- a^x^ -\- ••• = ao-\-x(ai -\-a^-^a^-\- •••)
it is evident that for finite values of x that render
aQ + aiX-\- '" convergent, the series a^ -\- a.jX -\- • -- is also
convergent ; hence, by Criterion IX., aj -h ag^ + ••• is con-
vergent, that is, finite when a; = 0. Therefore, when
a;=0,
aiX -f ajX^ 4- ... = 0 X {finite quantity) = 0.
323. The Binomial, Exponential, and Logarithmic Series are
the three infinite series of greatest importance in mathe-
matical analysis. We may now test their convergency.
* Harnack, Differential and Integral Calculus, Cathcart's trans-
lation, § 44, page 73.
460 CONVERGENCY AND DIVERGENCY.
Ex. 1. In the binomial series,
mCm — l) m(m — l)---('m — n + 1)
l + mx+ 2! ^ + ••• + ^ ^^T! ^^" + ••*'
the number of terms is finite when m is a positive integer ; but
when m is not a positive integer, no one of the factors m, m — 1,
?/i — 2, ••• can be zero, and the series will not terminate. It is
with this second alternative that we are at present concerned.
The ratio of convergence is
Un+i m — n + 1
Un ~ n
and its absolute value, as n is increased, becomes more and more
nearly equal to x. If, therefore, abvcc<l, then, either from the
beginning or after a finite number of terms, abv(Wn+]/Wn) will be
and remain less than 1, whether the sign of Un-\-\/Un be positive or
negative, and, by Criterion IV. [Art. 317], the series formed by
adding together the absolute values of the terms will be conver-
gent. Hence, for all values of x numerically less than unity, the
binomial series is absolutely convergent.
If abv X be greater than unity, abv (Un+i/un) will ultimately
become and remain greater than 1 and, by Criterion V. , the series
will be divergent.
[It may be shown that this series is convergent when x = 1,
provided w < — 1, and when x = —l, provided m > 0. See
Treatise on Algebra, Art. 338.]
Ex. 2. The exponential series is
Its ratio of convergence is x/n, and this is numerically less than
unity for all terms after the first one that renders x numerically
less than n. This series is therefore absolutely convergent for all
finite values of x.
Ex. 3. The logarithmic series is
3 4 n
CONVERGENCY AND DIVERGENCY. 461
and its ratio of convergence is
\ n+l)
Un+i _ xn
tin n+1
This is numerically less than unity for all values of x that are
numerically less than unity, and for such values the series is there-
fore convergent.
If oj = 1, the series becomes 1 — ^4-^ — ;^+---, which is con-
vergent by Criterion VII.
If X = — 1, the series becomes -(l + | + i + i+ —)»^liich was
shown to be divergent in Art. 318.
Thus, the logarithmic series is absolutely convergent when x has
any value between — 1 and + 1 and semi-convergent when x = 1,
but divergent in all other cases.
324. Product of Series. If finite portions of the two
series
V=Vi-\-V2 + Vs+ ••• +v„-f ...
be multiplied together by the algebraic rule for the mul-
tiplication of two multinomials, a result will be obtained
which may be arranged in the form
P = UiVi + (^1^2 + U2V1) + (^^i-^s + WjjVa -f U3V1) -f- •••
4- {uiV^ + U2V^_-i -^ -f u„_iV2 + u^Vi) + ....
Whenever these three series are simultaneously conver-
gent, the third is defined as the product of the other two.
For our present purpose we assume that the first and
second series are absolutely convergent, denote their
limits by U and V respectively, and attach the usual
interpretations to U„, F„, etc. Let the third series be
denoted by P. It is required to prove the following very
important theorem :
462 C0NVERGEI5CY AND DIVERGENCY.
If U and V be absolutely convergent, then P will be a
convergent series whose limit is equal to the limit of the
product Ux V.
(i.) If all tlie terms in CT'and Fare positive, then the
product C/^n X V2n will contain all the terms of p2n pl^s
others, such as U2V2n, ^3'y2n-l7 ^3^2n} '^2n'^2nJ ^^C., aud P2n
will contain all the terms of the product Un X F„ plus
others, such as UiV2n, ^2^2n-i) ^3'^2n-2) etc."* Hence
U2nX V2n>P2n> ^4 X F„ ,
and this inequality remains true, however great a finite
value n may have. But in the limit, when n = oo,
JJ^n = Un= U, and F2„ = F„ = F Hence, when all the
terms are positive,
Ux V=P.
(ii.) If U and F contain both positive and negative
terms, change the signs of all negative terms and denote
the resulting series by C7' and F; and let P' be the
series formed from U and F' in the same way as P is
formed from C/'and F
Then U2n X F2n — An cannot be, in absolute value,
greater than U\n X V\n — P^'iny for the series which the
two expressions stand for differ only in the respect that
the former contains some negative terms that are posi'
tive in the latter. But, by the proof in (i.),
* The student can make this quite evident by writing out all the
terms of these products for small values of n.
CONVERGENCY AND DIVERGENCY. 463
hence it follows again that
Ux V=P*
This completes the proof of the theorem.
EXAMPLES liXXXIV.
Apply the proper test of convergency to each of the following
series :
1. 22. 33. 4 w(n+l)
1,1.1.. 1
X ^
x-\-\ x-f3 aj + 5 2c + 2n
• 1* -3* 6* (2n-l)*
6 1-2 2.3 3-4 4-5
*3.4 4. 56. 6 6.7
« 1 • 2 , 2.3 34 ^ 4.5 ^ '
32.42 42.52 52.52 62.72
7 a;-l y. — 2 x — 3 ,^ — ^ ■
x+1 x+2 x+3 x+4
The following expressions denote, in each case, the n*^ term of
an infinite series. By applying the proper tests, determine whether
the series thus indicated are convergent or divergent.
8. I 9. ?*(!L±il.
10.
1
/o. . u. — » ^« xu. •
2?i(2^i-l) 2" l + n2
* This result is also true when only one of the factor series
Z7, V, is absolutely convergent, the other being semi-convergent.
If neither U nor V be absolutely convergent, nothing is certain
about the convergency of P. [See Chrystal, Algebra, Vol. II.,
Chapter XXVI., § 14.]
464 CONVERGENCY AND DIVERGENCY.
l + n2
an2 + 6
, 1 + n
16.
;. ^ + 1.
17.
X"
n2
l + n2
L 1
18.
xn-l
an + b
n2
19. ^"
20.
l + x« '
21 (^ + 1)^^
22. -^:^.
INDETEKMINATE COEFFICIENTS. 465
CHAPTER XXXII.
Indeterminate Coefficients.
325. In Art. 148 it was proved that if a rational inte-
gral function of x vanish when x= a, it is divisible by
X — a without remainder. Let the function be
/ {x) = ax" + baf"-^ + caf'^ -\ [-1,
and let the division by a; — a be performed. It is clear
that the first terra of the quotient, namely, the term of
highest degree in x, will be ax""'^ ; and since dividend =
divisor x quotient,
.-. /(a;) = (a;-«)(aa;«-i-f-..).
Suppose that/(«) also vanishes whena;=/?, [/8 not =a],
then the product (a; — a ) x (aaj**"^ -f • • • ) , and therefore
either (x —a), or (ax""'^ + •••), must vanish when x = p.
But )8 — a is not zero ; hence {ax'*''^ + •••) vanishes when
Xz= f3, and is therefore divisible by x — ft-, and if the
division be performed, the first term of the quotient will
be ax'''^
.-. f{x) = (x-a){x- 13) {ax'*-' +...)•
In general, if there be r values a, /?, y, 8, ••• of a; for
which / (a;) vanishes, r repetitions of this process will
obviously produce
f{x) = {x-a)(x-(3){x-y)(x-8)"'{ax'*-^-\--.).
Finally, if ax"" -f- bx^~^ + ••• vanish for n different values
2a
466 INDETERMINATE COEFFICIENTS.
of X, it has n factors x — a, x — p, etc., and a final factor
ax^-'', or a. Therefore, under this hypothesis,
f{x) = a{x- a) {x _ ^) (x - y) ...,
in which there are n binomial factors.
Two or more of these binomial factors may of course
be identical. Should there be a factor of the r*^ degree,
say (x — ay, it is evident that the number of remaining
factors will he n —r.
326. A ratioyial integral function of the n*^ degree in x
cannot vanish for more than n values of x, unless the co-
efficients of all the powers of x are zero.
For, if/ (a;), being of the form
ax"" + hx""-^ + cx''-'^ H ,
vanish for the n values a, /5, y ..., it must be equivalent to
a{x — a){x — li){x — y)'-'.
If now we substitute for x any other value, k suppose,
different from each of the n values a, /3, y, etc. ; then
since no one of the factors k — a,7c — /S, etc., is zero, their
continued product cannot be zero, and therefore f{x)
cannot vanish for the value x = k, except a itself is zero.
But if a be zero, f{x) reduces to bx^~^ + ccc""^ + •••? and
is of the (n — 1)*^ degree ; and hence it can only vanish
for n — 1 values of x, except b is zero. And so on.
When all tlie coefficients are zero, the function will
clearly vanish for any value whatever of x.
327. The values of x for which the expression
is equal to zero, are the roots of the equation
ax"" + hx""-^ + ca;"-^ -\ = 0. [Art. 91.]
rNDETERMINATE COEFFICIENTS. 467
Hence, by Art. 326, an equation of the n'* degree cannot
have more than n roots, except the coefficients of all the
different powers of the unknown quantity are zero, in
which case any value of x satisfies the equation.
328. If the two expressions of the ?i*^ degree
aaf + baf"-^ + cx""'^ -\ ,
and psf + qx""-^ + rx"*-^ H ,
be equal to one another for more than n values of x, it
follows that the equation
ax"* + 6ic"-^ + c.'c"-^ -f = px"" + gic"-^ + rx""'- -\ ;
that is, the equation
(a—p)x*'-\- {b — q)x''-'-\- (c - r)x"-2 + ... =0
has more than n roots.
Hence, by Art. 827, the coefficients of all the different
powers of x must be equal to zero.
Thus « =i>j ^ = ?) c = r, etc.
Hence, if tivo rational integral functions of the n'* degree
in X be equal to one another for more than n values of x,
the coefficient of any poiver of x in one function is equal to
the coefficient of the same x>ower of x in the other.
This is the principle of indeterminate coefficients as
applied to rational integral functions.
329. When any two rational integral expressions,
which have a limited number of terms, are equal to one
another for all values of the letters involved, the condi-
tion of the last article is clearly satisfied, for the num-
ber of values must be greater than the index of the
highest power of any contained letter.
468 INDETERMINATE COEFFICIENTS.
Hence when any two rational integral expressions, which
have a limited number of terms, are equal to one another
for all values of the letters involved in them, we may equate
the coefficients of the different powers of any letter.
330. If the two infinite series
ao -{- a^x + a^ -\ \- a^x"" -\ ,
&o + b,x + b,x' +^.. + M" + •••,
be equal to one another for all finite values of x for which
they are convergent, then
tto = b^, Oi = bi, ", a„ = 6„, etc.
If there be a finite value of x for which the two series
are convergent, they are also convergent for a; = 0 [Cri-
terion IX., Art. 322], and when a; = 0, by the corollary
of Art. 322,
aiX + acpi? H = 0,
and
biX + 62^^ + • • • = 0 ;
.-. aQ — bQ.
We now have, for the finite values of x that make the
original series convergent,
a^x -j- a^x? H = b^x-{- b^y? -\ ,
that is,
«! + acpc + a^a^ 4- • . . == 6^ -|- ^2^^ + b<p(? -\ .
But these two series are convergent for all the values
of X that make a^ ■\- a^x -\- -- - and bx-\-b^x-\- -" conver-
gent [Art. 322], and hence, by again putting a; = 0, we
obtain
ai = 6i.
INDETERMINATE COEFFICIENTS. 469
By Recessive repetitions of this process we find ag = 62?
ttg = 63, and in general, «„ = 6„.
TMs is the principle of indeterminate coefficients as ap-
plied to infinite series.
APPLICATION TO INTEGRAL FUNCTIONS.
331. The principle of indeterminate coefficients was
applied, without formal notice of the fact, to Examples 3
and 4 of Art. 153. The following are other examples of
this class :
Ex. 1. Determine the values of c and d that will make
x^ + hx'^ + ex + d
a perfect cube for all values of x and b.
Assume x^ + bx^ + ex + d=(x + ky.
Then, since
(x + ^•)3 = x8 + 3 A:x2 + 3 k'^x + k^,
we must have
b = 3k, c = 3k^ d = k^, [Art. 328.]
whence
k = i,b, c = lb\ d=^b^
Ex. 2. Determine under what conditions ax' + 6x* + ex 4- d is
divisible by px^ + qx -{■ r without remainder.
If the division be performed, the quotient will consist of two
terms; namely, (a/p)x, and a second term independent of x.
Hence, the quotient is of the form (a/p)x + A;, where A; is a quan-
tity to be determined by the condition that the division shall be
exact. We have, therefore,
0x8 + bx^ + ex + d = (-x + k\ 0x2 ^qx + r)
+ (- + «A;^« + r*,
470 INDETERMINATE COEFFICIENTS.
and since this is to be true for all values of a;, the coefficients of
like powers of cc in the two members of this equation must be equal
[Art. 328] ; whence
aq ar
b=—-\-pk, c = ---\-qk, d = rk.
Replacing k in the first two of these equations by its value d/r
derived from the third, we have
bp = aq + dj)'^ /r, cp = ar + dpq /r.
These are the conditions necessary and sufficient in order that
the division may be exact.
Ex. 3. Transform x^ + pxy + qy^ into the "sum of two squares.
Assume u =(x + ■\/qy)h, v =(x — ■yjqy)k, and determine h and k
by the condition that
t|2 _j_ 1,2 = 3.2 _|_ pxy -I- qy^.
For this purpose we make the coefficients of xy and y'^ in
that is, in
(x2 + gy2) {h^ + A:2) + 2 y/qxyi^^ - k^) ,^^
equal to p and q respectively. We thus obtain
;i2 + A:2 = 1^ 2 V^C/i^ - A:2) = p,
or
yJ'lW 2y/ql' V2\V 2y/q)
Hence, h and k having these values,
x2 + pxy + qy'^ = {{x + ^qy)h}^ + {(x - y/qy)kf.
Ex. 4. Find the factors of
(62c2 _ a*) (6 - c) + (c2a2 - M)(c - a) + (a262 _ c*) («-&).
This expression vanishes if c = 5, or if a = c, or if 6 = a, and
therefore h — c, c — a^ and a — h are factors. [Art. 148. J
Also, since the expression is symmetrical and is of five dimen-
INDETERMINATE COEFFICIENTS. 471
sions, there must be a fourth symmetrical factor of two dimensions,
which must therefore have the form
Z(a2+ 62 + c2)+ M{bc + ca + ah),
The given expression is therefore identically equal to
(6 -c)ic- a) (a - h){L{a^ + 6^ + c^) + itf (6c + ca + ah)},
in which L and M are to be determined by equating coefficients in
accordance with the principle of indeterminate coefficients.
The coefficients of a* in the two expressions are respectively
- (6 - c) and - L(b -c), and the coefficients of # are b^ - c^
and &2 _ c2 _ jf (ft-2 _ c2) . hence L = l and M=0, and the re-
quired factors are (6 — c), {c — a), (a — b), (a^ + b'^ + c^).
EXAMPLES LXXXV.
1. Determine the value of k that will make a^ — «* + 2x3 + A;
divisible by ic* + x + 1 without remainder.
2. Determine the values of p, q, and r that will make sfi —
bpy? + bqx^ — r divisible by (a; — c)^ without remainder.
3. If ax^ + 6x + c and a'x'^ -\- 6'x + c' have a common factor of
the first degree, this factor, and also the remaining (second) factor,
in each case, will be rational in all the letters. Prove this by de-
termining the three distinct factors. [Compare Ex. 3, Art. 173.]
4. Prove that if
ax2 + 2 hxy +by'^-\-'2, gx + 2/y + c
be expressible as the square of a rational integral function of x and
y of the first degree, then
af = gh, bg = hf, and ch — fg.
6. Determine the relation between a and b that will make
(x + ay)" and x^ + bxy -\-y^ have a common factor of the first
degree in x and y.
6. Determine k such that x^ — y^ — x — Sy + k may be the
product of two rational factors of the first degree in x and y.
7. What value of k will make the three equations 2x — y +
3=0, x + y— 1 = 0, and x + ky + I = 0 simultaneous in x and y ?
472 INDETERMINATE COEFFICIENTS.
8. Supposing a and h to be given numbers, determine a bino-
mial expression (containing an arbitrary factor) which, when sub-
stituted for aj, will make ax + 6^ a perfect square (for all values of
the arbitrary factor).
9. Prove that if ax^ + ho(fi + ex + d is divisible by x^ — A;^, then
ad = he.
10. Find the factors of
4 a6c (a + 6 + c) 4- he (h^ + c^) + ca(c2 + a2) + ^^((^2 _}. 52).
11. Find the factors of
&2c2(& - c) + c'^a^ic - a) + a262 (a - 6).
12. Find the factors of
a^(h -c)+ 6*(c-a)+c*(a-6).
13. Find the cube root of
1 + 3x + 6x2 + 7 a;3 + 6 X* + 3x5 + x6.
APPLICATION TO PARTIAL FRACTIONS.
332. In Art. 165, the process of obtaining a single
fraction as the result of adding together any number of
given fractions was explained. It is sometimes important
to be able to perform the converse operation of decom-
posing a complex fraction into the sum of a series of
simpler partial fractions, and for this purpose the principle
of indeterminate coefficients is employed.
Ex. 1. Resolve ^^~ ^ into
(x-l)(x-2)
partial fractions.
The sum of the two fractions A/(x — 1), B/{x — 2) will pro-
duce a fraction whose denominator is (x — 1) (x — 2). Let it there-
fore be proposed to determine A and B such that
2x-5 ^ A B ^ ^(x-2) + g(x-l)
(X- l)(x-2) ~x-l x-2~ (x-l)(x-2)
INDETERMINATE COEFFICIENTS. 473
For this purpose it is obviously sufficient that
2x-5=(A + B)x-i2A + B).
Since the left member of this identity contains no power of z
higher than the first, both A and B are assumed to be constants,
and therefore, by the principle of indeterminate coefficients,
A+ B = 2, and 2 ^ + J? = 5,
that is, ^ = 3, and ^ = - 1.
. 2x-5 ^ 3 _ ^
(x-l)(x-2) x-1 x-2
Ex. 2. Resolve ^-^t into partial fractions.
x(x-l){x-2)
Assume ^l+i = d+_^ + _^ and reduce the
x(x-l)(x-2) X x-1 x-2
partial fractions to a common denominator. Then, the denomi-
nators being omitted, we have
x^ + l=A(x- l)(x - 2) + Bx(x - 2) -f Cx{x - 1).
We might now expand and equate coefficients, but it is simpler
to proceed as follows :
Since the identity is true for all values of x, put x equal to 0,
1, and 2 in succession. From these substitutions the following
results are at once obtained :
a; = 0.
1 = 2^,
A = h;
x = l,
2 =-5,
B = -2;
x = 2.
5 = 2(7,
C=|.
X2+1
1
2 . 5
x(x-l)(x-2) 2x x-1 2(x-2)
X^ 4- X^ -\- 1
Ex. 3. Resolve — — — — into partial fractions.
x3-3x + 2 .
The process of decomposition cannot be applied directly to this
fraction, because the numerator is not of lower degree than the
denominator. But, by division, any such fraction can be replaced
474 INDETERMINATE COEFFICIENTS.
by an integral function plus another fraction whose numerator is
of lower degree than the denominator. In the present case we find
a;3 + x^ + 1 _ J x'^ + 3 a; - 1
x3_3x + 2 a;3-3x + 2'
and we apply the process of decomposition to
(x^ + 3x - 1) /(x3 - 3x + 2).
Since x^-Sx + 2=(x- iy(x + 2),
this latter fraction, for auo;ht we as yet know, might be the result
of adding together three fractions of the respective forms
A/Cx-iy, B/(x-l), and C/(x + 2),
for (x— l)2(x + 2) is the least common multiple of (x — 1)^,
(x — 1), and (x + 2). Hence we assume
ic2 + 3x-l_ A B C
x3-3x + 2 (x-l)2 x-1 x + 2
and determine J, J9, and Cby the condition
x2.4 3 X - 1 = ^(x + 2) + ^(x - 1) (X + 2) 4- C{x - 1)2.
Then, putting x equal to 1, — 2, and 0 in succession, we obtain
the following sets of values :
x=l, 3 = 3^, ^ = 1;
a; = -2, -3 = 90, C = -\\
x = 0, -1 = 2^-2J5+C, S = f.
. x3 + x2 + 1 _ -^ ^ 1 ^ 4 1
x3_3x + 2 (x-l)2 3(x-l) 3(x + 2)
Should a factor of the form (x + A;)^ occur in the denominator
of the origijial fraction, three partial fractions ^/(x+yfc)^,
B/{x + A;) 2, G/{x + k) must be provided for, and so on.
2^.2 7 /v 2
Ex. 4. Resolve — into partial fractions.
x^ — 8
The denominator of this fraction may be resolved into the two
factors X — 2, x^ + 2 x + 4, or into the three factors x — 2,
X + 1 + iy/'^t X + 1 — iV^? and observing always the rule that the
INDETERMINATE COEFFICIENTS. 475
assumed numerator of any partial fraction must be lower in degree,
by one, than the denominator, we may assume either
a;2_7x-2_ A J D J D'
or
a;8_8 x-2 x+l-\- V'^ x + l- iV3'
x2-7x-2_ A , 5a; +C
a;3_8 x-2x2 + 2x + 4
Choosing the latter identity, we have
x2 - 7 X - 2 = A(x^ -{- 2x + 4)-\- Bx(x - 2)+ C(x-2),
and the following particular equations for the determination of
A, B, and C :
x = 2, -12 = 12^, ^=-1;
x = 0, - 2 = 4^-2 0, 0 = -l;
a;=l, - S = 7A-B-C, B = 2.
a;2-7x-2 1^ 2x-l
a^_8 a:-2 x^ -\-2x + 4:
The calculations for 2) and D' in the first identity will show
that
The student can easily verify these results.
EXAMPLES LXXXVI.
Resolve into partial fractions :
1 5x + i e g^-f-l
* x2 - 1 ' ' x(x- 1)2
2 x2 + 3 X + 3 , Y 2x + 1
' x2 + 3x4-2* ■ x(x;2-f 1)*
3 X + 7 g qx -f ft
* 3(a^ + 3)(xH-5)* ' x(x2 + 6)*
^ x2^15x + 8 g x2 - 4 X + 5
(X + 1)-^(X - 5)* ' (X - 1)2(X2 + 1)'
6 -4x + "B 10 7x2-llx-f 7
(X - 1) (X - 2) (X - 3)* ' (x - l)3(x + 2)*
476 INDETERMINATE COEFFICIENTS.
11.
5x^1
x^-1
16.
^3-1
X^+1
12.
4
x*-l
17.
x^ + ax+b
x%x'^ + b)
13.
2
X(X2 + X
+ 2)
18.
a;2 + l
x^ + 1
14.
X^-X
(aj2 + c) (0x2 + 1)
19.
a:2-l
X* + X2 + 1
15.
X2+1
_.
20.
X2+1
_.
x3(a^ +1) (« - 1)K«* + 1)
APPLICATION TO EXPANSION OP FUNCTIONS.
333. In the expansion of functions into infinite series
in ascending powers of the variable, only such values of
the variable are permissible as will make the series con-
vergent. For any other values, the limit of the series
cannot be placed equal to the given function, and if there
be no values of the variable for which the series is con-
vergent, the expansion is impossible.
For example, the process of division applied to
1 -7- (1 — £c) produces the series
l+cc-foj^-f-ic^-j ad inf.,
but the limit of this series is not equal to 1/(1 — x)
unless abv x<l.
The method of indeterminate coefficients, when applied
to the expansion of functions, does not prevent the
appearance of divergent series, and hence the results of
this method are conditioned upon the convergency of the
series it produces.
INDETERMINATE COEFFICIENTS. 477
Ex. 1. Expand (1 + x)/{\ 4- x"^) into an infinite series.
Assume J-+A - a + hx -\- cx^ ■\- dy? -\- '-.
1 + x^
Then, multiplying by 1 + aj^, we have
1 + x = a + &aJ + cx2 + (Zx3 + ex* + •••
+ ax2+ &x3 + cx* + —
= a + 6x + (a + c)x2 + (6 + d)x3 + -.
for values of x for which the series is convergent. Hence, by the
principle of indeterminate coefficients [Art. 330],
a = l, .6 = 1, a + c = 0, h-\-d = 0,
c = — 1, d = — 1, e = l, /= 1, etc.
.-. J-t4=H-x-x2-x8 + x4 4-a:^-a:« .
Ex. 2. Expand into an infinite series.
Decomposing this function into partial fractions, we find
2 _ 1 _^ 1 + x
-x8 + a2-x+l 1-x l+x2,
But, for values of x for which the series are convergent,
1
l + x + a;'' + a^ + x4 + x5 + ...
and }'^\ = 1 + X - x2 - x8 + X* + a^
1 -X
IjfX
1+X2
X3 + X2 - X 4- 1
2(1 + X + X* + a;6 + x8 + x9 + — )•
Ex. 3. Expand "^ into an infinite series.
(x - 2)2(2 x-1)
This fraction is equal to the difference
2 1__
(x-2)a l-2x*
478 INDETERMINATE COEFFICIENTS.
2
Assume = ao + uix + a2X^ + •-.
(x — 2)2
Then multiplying by 4 — 4 a; + a^'^, we have
2 = 4 ao + 4 aix + 4 a2X^ + 4 aga^s + ...
— 4 aox — 4 aix2 — 4 ^2x3 — •••
+ aox^ -\- aix^ H
for values of x that make the series convergent. Hence ^
4 Gfo = 2, 4 «i — 4 ao = 0, 4 02 — 4 ai + ao = 0,
and in general 4 a„ — 4 an-i + an-2 = 0.
From these equations the values of ao, ai, «2» ••• «n can he cal-
culated in succession. Thus
1 2 3 4
and in general, an = — J^-
2 1, 2^^3 4^3 + 4^4 + ...;
(x-2)2 2 22 23 2* 25
and then, since
— ^ = 1 + 2 X + 22^2 + 23a;3 + 2^x'^ + — .
1 — 2x
(x - 2)2(2 x-1) \2 I \2^ ) \2^ I
the general term of the series being ( — — 2"-^ J xP--'^.
Observe that for positive values of x every term of this series is
negative, but that the fraction is not negative unless x<.\.
334. Keversion of Series. Given
y = aQ-^a^x + a^'^-\-a^o?-\ , (i.)
we may propose to obtain a value of x in terms of y.
For this purpose we assume
a; = 60 + &i2/ + M' + %' + -, (ii.)
INDETERMINATE COEFFICIENTS. 479
and determine the coefficients b^, bi, bz,-" by the condi-
tion that the given value of y in (i.), when substituted
in (ii.), produces an identity in x.
The process is likely to be laborious, but its applica-
tion is sometimes successful and convenient, as, in Art.
359. The process is called the reversion of series.
Ex. Find a value of x that will satisfy the equation 2 x — ic* = y
for a given value of y.
Even powers of y will not appear in the result, and we may
assume
x = h{y + b0^ + b^y^ -\
= &i(2a;-ic3)4-&3(2ic-x3)8+ ....
Expanding (2x — x^y, (2x — x^)^ etc., and comparing coeffi-
cients, we obtain
^ 4 32 64 2048
This solution is effective for any value of y for which the series
is convergent.
EXAMPLES LXXXVII.
Expand the following functions into infinite series in ascending
powers of x and, when possible, obtain a formula for the general
term:
, 5x+ 1 g l + a;g
• -• • (l-x)»*
5x+l
1-X2
2
1-4x4-3x2
4-3x2-
-2x3
1-x-
2x3
1
(1 + xy
1
2. = 7.
1 - 4 X 4- 3 x2
3 4-3x2-2x« g
1 4-a;
x2 + 15x + 8
(X + l)2(x - 6)
6. — -• 10.
1
11. Prove that, if n, r = positive integers, the coefficient of
xn+r-i in the expansion of (1 + x") /(I - x)2 is (n + 2 r).
480 INDETERMINATE COEFFICIENTS.
APPLICATION TO SUMMATION OF SERIES.
33^. If the general term of a series be given as a
rational integral function of the number of terms, the
sum of n terms may be found by the method of indeter-
minate coefficients.
Denoting the nth term of the series by u^, let
Sn = % + «^2 + W3 H h W„,
Sn-\ = Wi + ^2 + ^3 H f- ^n-lj
then, for all values of n,
If, now, u^ be given in the form
w„ = tto + otin 4- a2n^ -\ ,
and we assume that
^S'n = &o + bin + 62^^ + M^ H
for all values of n, where bo, &i, 62? ••• ^-re as yet undeter-
mined coefficients, then
Sn-j = &0 + bi{n - 1) + b,{n - 1)^ + -,
and
S^-S^.i = h{n-(n-l)l + b,\n'-(n-iyi + ...
^bi-h b2(2n - 1) 4- bs(3n^ - Sn + 1)
4- 64(47^3 - 67i2 + 4n - 1) -V ...,
and we have the identity
tto + ttin + «2^^ 4- •••
\^ ^bi-h-^-bs 4-(2&2-3&3+-)^
+ (363-664+ •••)^'+-.
INDETERMINATE COEFFICIENTS. 481
from which we obtain the values of bi, 62? ©tc, by equat-
ing the coefficients of like powers of n.
Suppose, for example, that w„ is of a degree in n not
higher than 2. Then
&4=&5 = &6= ••• = 0,
and &i — &2 + ^3 = <^o > 2 62 — 3 63 = dj , 3 63 = a2,
and from these equations are derived in succession
Sn = h + («o + !«! + ia2)n + ^(tti 4- a2)w^ + Jagn^
The value of bo may then be found by putting w = 1 in
this formula.
Observe that if n^ be the highest power of n contained
in S„, the difference S^ — *S'„_i will contain 71^"^ but not
n^. Hence, the degree {in n) of /S^ is higher by 1 than the
degree ofu^.
Ex. 1. Find the sum of the first n terms of the series whose n^
term is 1 — w + ri^.
Here ao = 1, ai = — 1, 02 = 1 ;
= &o + fn + |n8;
and since ^1 = 1 = 6© + 1, that is, 60 = 0,
... >S„ = }n(n2 + 2).
Ex. 2. Find the sum of the first n terms of the series 3 4- 4 +
6 + 9 + 13 + 18+ ....
The differences of successive pairs of terms of this series form
the arithmetical progression 1, 2, 3, 4, ... ; that is,
4-3=1, 6-4=2, 9-6 = 3, etc.
2h
482 INDETERMINATE COEFFICIENTS.
Hence, denoting the successive terms of the given series by wi,
M2» Us, W4, ••• Un, we have
Ml = 3, W2 = Ml + 1,
us = ui + l + 2,
M4 = wi + 1 + 2 + 3,
«n = Wi + l + 2 + ...4- (n-l)
= Wi + Kw-l)(2 + w-l-l) [Art. 259]
= Mi+ ^W (W — 1).
This, the n^^ term of the given series, is therefore 3 — ^w+^w^.
Hence, in the formula for Sn, we have
ao = 3, ai = — I, a2 = ^,
5n = 60 + (3 - i + tV) »* + K- i + D w2 + iw«
= 60 + ¥ »i + i w^ ;
and since /S'l = 3 = 60 + 3, bo = 0.
Let the student verify the results of these two examples by use-
ing the method here explained, but without substituting in the
formula.
336. The foregoing method may be generalized and
made applicable to power series in which the coefficient
of a;** is a rational integral function of n.
For the summation of such a series it will be found
sufficient, when the (n + 1)*^ term is
to assume
S,^, = (60 + &in + ••• + hn')x-+' + k
and determine 60? ^\y"'^h ^^^ ^ from the identity
An example will best show how the method is applied.
INDETERMIKATE COEFFICIENTS. 483
Ex. 1. Find the sum of the first w + 1 terms of the series
1 + 2a; + 3x2 + 4x3 + ....
The (w + l)th terra of the series is (n + 1) x", and if aS^„+i denote
the sum of the first w + 1 terms, then
(n + l)x^ = S„+i - Sn.
Assume Sn+i = (a + 6w)x"+i + k,
where k is independent of n, and a, b are as yet undetermined
coefficients. Then
S„ = {a-\- bin - l)}x« + k,
and Sn+i- Sn = {a(x-l)-\-b + 6(x - l)n}x»,
whence w + 1 = a(x — 1)+ 6 + &(x — l)n.
Equating coefficients in this equation, we now have
a(x- 1)+ 6 = 1, 6(x-l)=l,
K l^l-6x-2
or 0 = , a =
x-1 x-1 (x-l)2
Hence 5„,, = { ^ + -^ }..+• + *,
for all values of n. We now determine k by putting n = 0 in this
expression for Sn+i, and by this substitution we obtain
that is. k =
(X - 1)2
Hence finally, Sn+i = { -^^ + -^ }a;«+i + — ^,
I (X — 1)-^ X— IJ (X — 1)2
which is the required result.
EXAMPLES LXXXVIII.
Each of the following expressions is the nth term of a series
whose first term is got by putting n = 1. Find the sum of the
first n terms of each of the series thus indicated.
1. 2n2-l. 3. w(n + l)(7i + 2). 6. p + g(n - 1).
3. n(n+l). 4. n^-n + 1. 6. —
4n2 — 1
484 INDETERMINATE COEFFICIENTS.
Each of the following expressions is the (n + 1)''^ term of a
series whose first term is got by putting n = 0. Find the sum of
the first n + I terms of each of the series thus indicated.
7. (p + qn^x"^. , 10. (p + qn + rn^)x'^.
8. (w2 - w + l)x^. 11. (2'» - 2 w + l)ic«.
9. {2{n + ly - l}x^. J2 _wMi2jL+2_2n
' (n + l)(w + 2)
Find the sum of the first n terms of each of the following series :
13. 1 +2x4-5x2+ 10x3 + 17x4 + 26a:5+....
14. 1 . 2 . X + 2 . 3 . x^ + 3 . 4 . x3 + 4 . 6 . X* + ••..
15. 1 + 2x + 4x2 + 7x3 + llx* + 16x5 + ....
16. 1 + 2x + 6x2 + 13x3 + 23x* + 36x5 + ....
PERMUTATIONS AND COMBINATIONS. 485
CHAPTER XXXIII.
Permutations and Combinalions.
337. Definition. The different ways in which r things
can be taken from n things, regard b^ing had to the order
of selection or arrangement, are called the permutations of
the n things r at a time.
Thus two permutations will be different unless they
contain the same objects arranged in the same order.
For example, suppose there are four objects, represented by the
four letters a, 6, c, d.
If we take the four objects one at a time, we have the four per-
mutations a, &, c, d.
If we take the objects two at a time, we have twelve permu-
tations
a?), ac, ad, ba, be, bd, ca, cb, cd, da, db, dc.
Now to find the whole number of the permutations three at a
time we can proceed thus. Take any one of the permutations
two at a time and place after it either of the two letters which it
does not contain ; then all the permutations so obtained will be
different, for either we shall have used a different permutation of
two letters or else the final letters will be different ; moreover,
since all the possible permutations two together have been used,
no permutation of three letters can have been omitted. Hence
the number of permutations of 4 things 3 together = number of
permutations of 4 things 2 together x 2 = 12 x 2 = 24.
We can by similar reasoning find the permutations of n things
r together.
486 PERMUTATIONS AND COMBINATIONS.
The number of permutations of n different things
taken r at a time is denoted by the symbol ^P^.
338. To find the number of permutations of n different
things taken r at a time, where r is. any integer not greater
than n.
Let the different things be represented by the letters
a, b, c, '-'.
It is obvious that there are n permutations of the n
things when taken one at a time, so that „Pi = n.
Now, if beside any one of the different permutations
(r — 1) at a time, we place any one of the n — (r — 1)
letters which it does not contain, we shall obtain a per-
mutation of the n things r at a time. We thus obtain
n — {r — l) = n — r-\- 1
different permutations r at a time from every one of the
different permutations (r — 1) at a time.
Hence „P, = „P,li x (n - r + 1).
Since the above relation is true for all values of r, we
have in succession
nPr-l=nPr-2Xin-r + 2),
nPr-2=nPr-sX{n-r-h3),
,P3 = nAx(il-2),
Also „Pi = n.
Multiply all the corresponding members of the above
equalities, and cancel all the common factors ; then
„P, = 7i(n-l)(n-2)...(7i-r + l),
r being the number of factors on the right.
PERMUTATIONS AND COMBINATIONS. 487
If all the n things are to be taken, r is equal to n, and
we have
„P„ = n(7i-l)(7i-2)...3.2.1.
Def. The continued product 71(71 —1 ) (n — 2) ... 3 • 2 • 1
is denoted by the symbol \n, or by n ! The symbols \n
and n ! are read ' factorial n.' Thus [4 = 4 ! =4 . 3 . 2 • 1.
Ex. 1. In how many different ways can 6 boys stand in a row ?
The number required = ePe = 6 • 5 • 4 • 3 • 2 • 1 = 720.
Ex. 2. How many different numbers can be formed by using
three of the figures 1, 2, 3, 4, 5 ?
The number required = 6P3 = 5 • 4 • 3 = 60.
Ex. 3. Show that 10P4 = 7P7,
10P4 = 10 -9 . 8 . 7 and 7P7= 7 ..6 . 5 • 4 . 3 .2 . 1.
339. To find the number of permutations of n things
taken all together, when the things are not all different.
Let there be n letters ; and suppose p of them to be
a's, q of them to be 5's, r of them to be c's, and so on.
Let P be the required number of permutations.
If in any one of the actual permutations, we suppose
that the a's are all changed into p letters different from
each other and from all the rest; then, by changing only
the arrangement of these jp new letters, we should, instead
of a single permutation, have p ! different permutations.
Hence, if the a's were all changed into p^ letters differ-
ent from each other and from all the rest, the 6's, c's, etc.,
being unaltered, there would be P xpl permutations.
Similarly, if in any one of these new permutations we
suppose the 6's are all changed into q letters different from
each other and from all the rest, we should obtain q ! per-
mutations by changing the arrangement of these q new
488 PERMUTATIONS AND COMBINATIONS.
letters. Hence the whole number of permutations would
now be
Pxpl xql
By proceeding in this way we see that if all the letters
were changed so that no two were alike, the total number
of permutations would be
Pxpl xql xr\ ....
But the number of permutations of n different things is n !.
Hence P xpl X ql X rl x -•- = nl;
nl
.-. P:
plqlrl
Ex. 1. Find the number of permutations of the six letters
aaabbc taken all together.
The number = — — — = 60.
3!2! 1!
Ex. 2. How many different numbers can be formed by the
figures 2, 3, 3, 4, 4, 4, 4 ?
7 '
The number required = '— = 106.
^ 4!2!
\
340. Def. The different ways in which a selection
of r things can be made from n things, without regard
to the order of selection or arrangement, are called the
Oombinations of the n things r at a time.
Thus two combinations will be different unless they
both contain precisely the same objects.
For example, suppose there are four objects, represented by the
four letters a, 6, c, d. The different combinations one at a time
are a, b, c, d ; the combinations two at a time are ab, ac, aj?, 6c,
bdy cd ; the combinations three at a time are abc, ubd, acd, bed ;
PERMUTATIONS AND COMBINATIONS. 489
and there is only one combination, namely ahcd^ when aU the
letters are taken.
The number of combinations of n things r together is
denoted by the symbol „0^.
341. To find the number of combinations of n different
things taken r at a time.
Since every combination of r different things would
give rise to r\ different -permutations, if the order of the
letters were altered in every possible way, it follows that
„axr! = „P, = n(ii-l)(n-2)...(n-r + l).
Hence ^o^^^(^- 1)(^- 2)-(^ -^+ 1).
1.2.3...r
By the following method the number of combinations
of n different things r at a time can be found indepen-
dently of the number of permutations.
Let the different things be represented by the n letters
a, &, c, d, .••. 4 ^v
In the combinations of the n letters r together the
number in which a particular letter occurs must be equal
to the number of ways in which r — 1 of the remaining
n — 1 letters may bif5(pelected. Hence in the whole
number of combinations r together every letter occurs
n-\Cr-i times, and therefore the total number of letters
employed is n-\Gr-i X n; but, since there are r letters in
each of the „0^ combinations, the total number of letters
employed is „C^ x r.
Hence r x ,,(7^ = ?i X „_iC^_i.
Since the above relation is true for all values of n and
r, we have in succession
490 PERMUTATIONS AND COMBINATIONS.
(r - 1) X , ia_i = {n - 1) X ._2a-2
(r - 2) X n-20r-2 = {u - 2) X n-A-^
2 X „_,+2C2 = (n - r + 2) X n-.+iCi.
Also n-r+iCi = n — r 4- 1.
Multiplying corresponding members of the above
equalities and cancelling the common factors, we have
r ! X „a = 71 (71 - 1) (n - 2) . .. (n - r + 1),
that is .0. = ri(n-l)(n-2)...(n-r4-l),
r !
By multiplying the numerator and denominator of the
fraction on the right by {n — r) ! , we have
r _^(y^-l)(^-2)-»-(n-r + l)(n — r)!_ * n
r\{n — r)\ r\{n—r)\
Ex. 1. How many groups of 3 boys are there in a class of 15 ?
The number = 15C3 = ^^'^^'^^ = 455.
1 • 2 • 3 ^^
Ex. 2. There are 6 candidates for 4 vacancies, and every
elector can vote for any number of candidates not greater than
the number of vacancies. In how many ways is it possible
to vote ?
An elector can vote for 4 candidates in ^d ways, for 3 candi-
dates in eCs ways, for 2 candidates in 6 02 ways, and for 1 candidate
in eCi ways. Thus the whole number of ways in which an elector
can vote is
6C4 + 6C3 + 6^2 + eCi = 16 + 20 + 15 + 6 = 66.
( '
PERMUTATIONS AND CpifBiNATIONS. 491
342. From the formula obtained in the last article^
we have
r - ^-
{n — r)\r\
I SO that the number of combinations of n things r together
is equal to the number of combinations of n things n —r
together.
The above proposition follows, however, at once from
the fact that whenever we take r out of n things we
must leave n — r, and if every set of r things differs in
some particular from every other, the corresponding set
of n — r things will also differ in some particular from
every other. Hence the number of different sets of r
things must be equal to the number of different sets of
n — r things.
343. To prove that JO, + „C,_i = nH-i<^r-
The total number of combinations of (71 + 1) things r
together can be divided into two groups according as they
do or do not contain a certain particular thing. The
number which do not contain that particular thing is
the number of ways in which r of the remaining n things
can be taken, which is „(7^; and the number which do
contain the particular thing is the number of ways in
which (r — 1) of the remaining n things can be taken,
which is J3r-\' Hence ,,+iC^ = J^, + rfir-v
Thus, if n = 7 and r = 4, we have
7-6.5-4-3-2.1^6-5.4-3.2>l 6..5.4.3-2-1
4. 3. 2. 1-3. 2-1. 4-3.2. 1-2.1 3.2.1.3.2-^'
492 PERMUTATIONS AND COMBINATIONS.
Note. — The above important proposition may also be proved
from the general formula. Thus
vf^r + n^r— 1
_ n{n -V){n- 2)...(n - r + 1> n{n - l)(n - 2) — (w - r + 2)
1.2.3...r 1.2.3...(r-l)
^ n{n - l)(n - 2). -(71 -> + 2) .. .
1.2.3...r ^^ -r y-r 5
_ (w + l)w(n - l)(n - 2)---(n -7-4-2)
1.2.3...»-
344. Greatest Value of „C7^
T'rom Art. 341 we have
y^n — In — 2 ^yi-r + 1
. n - n y n-r + 1
r
Hence rflr^nCr-i according as '^^— r. 1 ; that is,
according as r ^ i (n + 1) .
Thus the number of combinations of n things increases
as r increases, st) long as r is less than -^-(w + l). If
r = ^{n 4- 1), then ^Cr = nC^-u but r cannot be equal to
1(91 + 1) unless n is odd. If r > i(w + 1), the number
of combinations diminishes as r increases.
Thus, if n is even, „C^ is greatest when r — ^w, and if
n is odd, „(7^ =nC'r-i when r = |^(n + l),^and these values
are the greatest.
Ex. 1. Find the greatest value of gCV-
^ The greatest value is when r = 4, and the value is
8 . 7 » 6 • 5 ^ ^Q
^ 1.2.3.4
PERMUTATIONS AND COMBINATIONS. 493
Ex. 2. Find the greatest value of ^iCr.
The greatest values are when r = 6 and r = 5, and the values are
ii.io.9.8.7^^ea
1.2.3.4.5
EXAMPLES LXXXIX.
1. Find^sPg. 2. FindioP^. 3. Find gP^.
4. How many different numbers can be formed by using the
five figures 1, 2, 3, 4, 5?
6. How many different numbers can be formed by using one or
more of the figures 1, 2, 3, 4, 5, 6 ?
6. In how many ways can 10 boys be put in a row so that two
particular boys should not be together ?
1/ 7. Of how many different things are there 720 permutations
when taken all together ?
8. Find the number of permutations of all the letters of each
of the words success, Mississippi, and algebraic.
1 9. In how many ways can the nine letters aaaabbbcc be ar-
ranged in a row ?
\ 10. How many different numbers can be formed by using the
seven figures 1, 1, 1, 2, ^, 3, 3 ?
^ 11. How many permutations can be made with the letters of the
word essences, and how many begin with e and end with s ?
I 12. Find ^Pn, having given that nPg = nP^ x 2.
13. Find n, having given that „Pg = nPj x 12.
14. Find ^^C^, 15O12, and ^^0^^.
15. JKnCs 3
16. If ^,o_
17. How many different numbers of three figures could be made
liy taking three of the digits 0, 1, 2, 3, 4 ?
494 PERMUTATIONS AND COMBINATIONS.
18. If 2nC'3 = nCa X 12, find n.
19. If 2nC, = nCi X 24, find n.
j 20. Numbers are formed by writing the five figures 1, 2, 3, 4, 5
in every possible order. How many of these numbers are greater
than 23,000 ?
J
21. In a certain town there are five letter boxes. In how many
ays can a person post three letters ?
( 22. How many different sums could be made with 2 pennies,
3 shillings, and 4 sovereigns ?
I 23. How many different sums could be made with 4 pennies,
6 shillings, and 5 sovereigns ?
24. Show that n+iCr+l = n^+l + 2 nCr + nCr-l-
25. Show that in gC^ the number of combinations in which a
particular thing occurs is equal to the number in which it does not
occur.
26. Show that in ^^C^ the number of combinations in which a
particular thing occurs is one-third of the whole number of the
combinations.
27. Show that in 3„Cn the number of the combinations in which
a particular thing occurs is one-third of the whole number of the
combinations.
28. There are 10 candidates for 6 vacancies in a committee ;
in how many ways can a person vote for 6 of the candidates ?
29. At an election there are 5 candidates and 3 members to
be elected, and an elector may give 1 vote to each of not more
than 3 candidates ; in how many ways can an elector vote ?
30. In how many ways can a picket of 3 men and an officer be
chosen out of a company of 80 men and 3 officers ?
31. In how many w-ays can a cricket eleven be chosen out of
30 players ; and in how many different ways could two elevens be^
chosen to play a match with one another ?
PERMUTATIONS AND COMBINATIONS. 495
. -; 32. Out of 9 children, of whom 5 are boys and 4 girls, in how
X many ways can 4 be chosen, 2 being boys and 2 girls ?
33. In how many ways can 2 ladies and 2 gentlemen be chosen
to make a set at lawn tennis from a party of 4 ladies and 6 gentle-
men?
34. In how many ways can 8 children form a ring ?
y 35. There are n points in a plane, no three of which are in the
'"same straight line ; and the points are joined in pairs by straight
lines which are produced indefinitely. How many straight lines
are there, and how many triangles are formed by them ?
/
496 THE BINOMIAL THEOREM.
CHAPTER XXXIV.
The Binomial Theorem.
345. The binomial theorem for a positive integral
index was proved, by induction in Art. 205. The follow-
ing proof makes use of the principles of the combinatorial
analysis, explained in Chapter XXXIII.
Suppose we have n factors, each of which is a + 6. If
we take a letter from each of the factors of
(a + 6)(a + 6)(a + 6)-"
and multiply them all together, we shall obtain a term of
the continued product ; and if we do this in every possible
way we shall obtain all the terms of the continued product.
Now we can take the letter a every time, and this can
be done in only one way, hence a** is a term of the
product
The letter b can be taken once, and a the remaining
(n — 1) times, and the number of ways in which one h
can be taken is the number of ways of taking 1 out of n
things, so that the numbet is „(7i; hence we have a
second term
Again, the letter h can be taken twice, and a the
remaining {n — 2) times, and the number of ways in
which two 6's can be taken is the number of ways of
THE BINOMIAL THEOREM. 497
taking 2 out of n things, so that the number is ^Cg; hence
we have a third term
And, in general, b can be taken r times (where r is any
positive integer not greater than n) and a the remaining
(n — r) times, and the number of ways in which r 6's
can be taken is the number of ways of taking r out of n
things, so that the number is ^C^ ; hence we have, as the
(r+l)*^term, ^^^-^
The letter h can be taken every time in only one way ;
hence we have the term &", which agrees with the result
obtained by putting r = ?i in „C^ • a" '"6'*, since „(7„ = 1.
Thus {a-\-h){a -\-h) {a -\-h) "• to n factors
= a" + nCi • a"-^6 + nC2a" ^6=^ + ••• + nC^a^'-'-ft" + ... + 6«.
Hence, when n is any positive integer, we have
{a+by = a« + nOi • a'^-'b + A • «""'^' -}-•••
This proves the binomial theorem for a positive
integral index.
The series on the right is called the expansion of
{a + by.
If in this we Substitute the known, values [Art. 341]
of nGi, „C2, etc., we obtain the form in which the theorem
was written in Art. 205, namely :
(a + 6)" = a^ + na''-'b + '^i'^-^) a^-^b- + ...
2i
498 THE BINOMIAL THEOREM.
346. General Term. Any term of the expansion oi
(a + 6)'* will be found by giving a suitable value to r in
^(^ _ 1) (n - 2) . .. (n - r + 1) ^n-rj^r^
rl
This is called the general term of the expansion. It
should be noticed that it is the {r-\-iy^ from the beginning.
347. Some applications of the binomial theorem :
If we put w = 2 in the binomial formula, we have
(a + 6)2 ^ «2 + 2 ab + b^.
If we put n = 3, we have
(a + by = a^ + ^a'^b + — «62 + js
1 1 • iS
= a^-{-Sa^b + Sab^-\-bK
If we put w = 4, we have
(a + by = a* + ^a^ft + f^a^S^ + f44«&' + ^*
1 x • 2 1 • 2 • o
= a* + 4 a36 + 6 a^-b^ + 4 aft^ + b\
If we put 2 X for a, and —3y for b, and « = 5, we have
(2x-Syy=(2xy + 6(2xy(-Sy) + ^(2xy(i-Syy
+f^(2a^)^(-3y)« + ^;^;^;^(2x)(-y)H(-3y)5
= 32x5 - 240 a;*?/ + 720 ic^ _ io80 x'V + SlOiicy* _ 243^5.
348. In the expansion of (a + 6)" by the binomial
theorem, the (r + 1)*^ term from the beginning and the
(r-\-iy^ term from the end are respectively
^CrCi'^'-'b' and JJ^_,a'}f-\
But „C, = „0„_, , for all values of r. [Art. 342.]
THE BINOMIAL THEOREM. 499
Hence in tlie expansion of (a + 6)'', the coefficients of
ayiy tivo terms equidista^it from the beginning and the end
are the same, so that the coefficients in order are the same
when read backwards as when read forwards.
349. If in the formula of Art. 345 we put a = 1 and
b = x, we have
1-2 rl{n—r)\
This is the most convenient form of the binomial theo-
rem, and the one which is generally employed.
It should be noticed that the above form of the bino-
mial theorem includes all cases; if, for example, we
want to find (a 4- 6)", we have
(a + 6)"= { a(l + 1) }"= a"(l + ^)"= a'(l + «^ + etc.)
= a" + wa"-^6 -f etc.
350. Greatest Term. In the expansion of (l + x)" by
the binomial theorem, the (r-f-1)"^ term is formed from
the r**^ by multiplying hj x(n — r + 1)/^.
Now x(n — r-\- l)/r = x\(n-\- l)/r — Ij, and (n-\-l)/r
clearly diminishes as r increases; hence x{n — r-\-l)/r
diminishes as r is increased. If a;(n — r + l)/r be less
than 1 for any value of r, the (r + 1)!** term will be less
than the r^\ In order, therefore, that the r*^ term of
the expansion may be the greatest, we must have
n — r + 1 ^ , n — r — l-\-l
X < 1, and X ^ > 1.
r ' r — 1
(n-[-l)x , ' (n-\-l)x ^
Hence r>-^-^-p^, and r<\_^( +1.
500 THE BINOMIAL THEOREM.
If r = (n 4- l)x/{n + 1), then x(n — r -{-l)/r= 1, and
there is no one term which is the greatest, but the r'^
and (r 4- 1)*^ terms are equal and are greater than any
other terms.
The absolute values of the terms, in the expansion of
(1 + a;)* will not be altered by changing the sign of x;
and hence the r*^ term of (1 — x)" will be greatest when
the r^^ term of (1 -|- a;)" is greatest.
Ex. 1. .Find the greatest term in the expansion of, (1 + x)!''
when x = ^.
The r^^ term is the greatest if r>J^ and r<i^^ + l. Hence
the third term is the greatest.
Ex. 2. Find the greatest term in the expansion of (2 + 3xy^
when x = ^.
(2 + Sxy^= {2(1 + |a:)P= 2i2(l + ?^xy^.
Hence the f^ term is the greatest if r > -3/- and r < ^^~ + 1. Hence
the sixth term is the greatest.
The greatest coefficient of a binomial expansion, i. e. the
greatest value of ^C^, was found in Art. 344.
EXAMPLES XO.
1. Write out the expansions of :
M'^e+ij- f(«oe^H-i)'-(..-i)'.
/ 2. Find the 11th term in the expansion of (4 a; ] •
, / 1\20
I 3. Find the 17th term in the expansion of / a;2 \ .
(/73\ 2n
CC2 \ .
THE BINOMIAL THEOREM. 501
5. Write down the twq middle terms in tlie expansion of
1 >^ 2n+l
^+-,
6. Find the term in the expansion of (a + xy which has the
greatest coefficient. _^
7. Find tlie terms in the expansion of ^^^aJ)^ which have the
greatest coefficients. ^B
8. Find the greatest term in the expansion of (1 + 2 x)^ when
x = i.
9. Find the greatest term in the expansion of (1 + 3x)i8 when
x = l
10. Find the greatest term in the expansion of (3 + 4 xy^ when
11. There are two consecutive terms in the expansion of
(5 X + 7)2* which have the same coefficient : which are they ?
12. Write down the coefficient of iC" in the expansion of
(ax - 6)".
13. Sliow that the coefficient of x** in the expansion of (1 + a;)^'*
is double the coefficient of x" in the expansion of (1 — x)2«-i.
14. Show that the middle term of (1 + x)2» is
1.3.6...(2n-l)o.^
nl
I 16. Employ the binomial theorem to find 99* and 999^
351. Properties of the Coefficients. We now proceed to
consider some properties of- the coefficients of a binomial
expansion. For this purpose we write it in the form
(1 + xy = Co + c^x + c^+ '" + cX + ••• + c^a;**,
where
nl
Co = c„ = 1, Ci = c«_i = n, and c^ =c„_^ = — - — - — -.
ri{n — r)!
I Put ic= 1 in the above formula; and we have
2" = CoH-Ci + C2+ — +c^+---c».
602 THE BINOMIAL THEOREM.
Thus the sum of the coefficients in the expansion of (1 + x)"
is 2".
II. Again, put a; = — 1 ; and we have
( 1 - 1 ) '^ = Co - Ci H- C2 - Cg + • • • ;
.-. 0 = (Co + C2 4- C4 + ••• ) - (ci + C3 + cs 4- •••)•
Thus the sum of the coefficients of the odd terms of a bino-
mial expansion is equal to the sum of the coefficients of the
even terms.
III. Since c^ = c„_^ [Art. 342], we may write the bino-
mial theorem in either of the following ways :
(l + xy = Co + CiX-^c^-\ h cX H c^x%
or (l^-a;)"=c„+c„_la;+c„_2a;24-...c„_X^-•••^-Cla;"~^+Coa;^
The coefficient of mf" in the product of the two series
on the right is equal to
Co' + Ci2 4-C22 + -4-c„2.
Hence* Cq^ + Ci^ H + c^^ -] cj is equal to the coeffi-
cient of x"" in (1 + xy X (1 + xy, that is in (1 + xy^", and
this coefficient is
(2n)\
nln\
Hence the sum of the squares of the coefficients in the
expansion of (1 -\-xy is equal to ^ ^—- <
n\n\
Ex. 1. Show that
ci + 2 C2 + 3 cs + ••• + rcr + '" + wc„ = n2»-\
* See Art. 328.
THE BINOMIAL THEOREM. 503
Cl + 2 C2 + 3 C3 4- ••• + rcr + ••• + ncn
1.2 1.2.3 ^'^r!(w-r)!
\ ^^ ^^ 1.2 (r-l)!(n-r)! /
= n(l + l)«-i
= n2«-i.
Ex. 2. Show that
Co-^Ci +ic2 - ... + (- l)n-^!!_-^_
2 3 n+ln+1
2 ^3 1.2 4 1.2.3 ^ ^ n+1
- ^ ^ (1-1)"+1
n+ 1 n+l
1
n + 1*
352. To /lid ^Ae continued product of n binomial factors
of the form x-\- a, x -\-h, x + c, etc.
It will be convenient to use the following notation :
Si is written for a + 6 + c + .••, the sum of all the let-
ters taken one at a time. /S'gis written for ab-{-ac-\-bc-\ — ,
the sum of all the products which can be obtained by
taking the letters two' at a time. And, in general, S^ is
written for the sum of all the products which can be
obtained by taking the letters r at a time.
504 THE BINOMIAL THEOHEM.
Now if we take a letter from each of the binomial
factors of
{x -\-a)(x-\- b) (x -\- c) (x -\- d) " -,
and multiply them all together, we shall obtain a term of
the continued product ; and, if we do this in every pos-
sible way, we shall obtain all the terms of the continued
product.
We can take x every time, and this can be done in only
one way ; hence x^ is a term of the continued product.
"We can take any one of the letters a, b, c, • • •, and x
from all the remaining n — 1 binomial factors ; we thus
have the terms ax''-\ bx''~\ ca;'*"^ and on the whole
^1 . x^-\
Again, we can take any two of the letters a, b, c, •••,
and X from all the remaining n—2 binomial factors ; we
thus have the terms abx^'"^, acx""'^, etc., and on the whole
8, . x^-\
And, in general, we can take any r of the letters a, 6,
c, •••, and X from all the remaining n — r binomial fac-
tors ; and we thus have S^ • a?""''.
Hence {x-\-a){x-[-b){x-{-c){x-\-d) "'
= a;" + ^1 • x^-^ + Si ' aj"-2 H \-S^' aj"-*" + -..,
the last term being abed---, the product of all the letters
a, b, c, d, etc.
By changing the signs of a, b, c, etc., the signs of S^^ S^,
/S's, etc. will be changed, but the signs of S2, S^, Sq, etc,
will be unaltered. Hence we have
{x — a) {x — b) {x — c) (x — d) '"
= a;" - aSi . a^"-^ -f /Sa • aj«-2 - /Sg . a;«-3 + ...
+ (- l)*-^, .»"-'• H h (- l)"a6cd ....
THE BINOMIAL THEOREM. 505
353. The Multinomial Theorem. The expansion of
can be found by the method of Art. 345.
We know that the continued product
(a + 6 + C + •••)(« + ^ + c4----)(ot + & + c +•••)•••
is the sum of all the partial products which can be
obtained by multiplying any term from the first multi-
nomial factor, any term from the second, any term from
the third, etc. If there are n of the multinomial factors,
every term of the required expansion must be of the n^^
degree, and, therefore, all the terms must be of the form
a'^b'd '", where each of r, s, t, etc. is zero, or a positive
integer, and r + s+^-f".=n.
Now the term O'd'cf ••• will be obtained by taking
a from any r of the n multinomial factors, which can be
done in „CV different ways ; then taking b from any s of
the remaining ?i — r factors, which can be done in „_^(7<,
different ways ; then taking c from any t of the remain-
ing n — r — s factors, which can be done in „_^_gO, differ-
ent ways ; and so on. Hence the total number of ways
in which the term (X''6V • • • will be obtained, that is, the
coefficient of the term in the required expansion, must be
that is,
n\ ^^ (n-r)! ^^ {n-r-s)l ^^ ^ n\
rl{n—r)l s\{n—r—s)l tl{n—r—s—t)l rlsltl---'
Hence the general term in the expansion of
{a + 6 + c-f '"Y is
— — -— — a'^b'^,
rlsltl'"
606 THE BINOMIAL THEOREM.
where each of r, s, t, etc. is zero or a positive integer, and
r + s-\-t-\ = n.
Ex. 1. Find the coefficient of abc in (a + 6 + c)^.
Here n = 3, r = s=^t = l. Hence the required coefficient is
Ex. 2. Find the coefficients of a*6, a^b^, and a^b'^c in (a+fe+c)^.
We have the terms
^a% -^aW, and -^l—a%H.
4!1! 3!2! 2!2!1!
Hence the required coefficients are respectively 5, 10, and 30.
EXAMPLES XCI.
In the following examples co, Ci, C2, are the coefficients of jc^,
a;!, x^ ••• in the expansion of (1 + x)^.
1. Prove that Ci — 2 C2 + 3 cs 4- (— l)"-inc„ = 0.
2. Prove that Co - 2ci + 3c2 + ( - l)''(n + 2)c„ = 0.
3. Prove that co + -ci + -C2 + — -f- -^—Cn = ^"^^ ~ ^'
2 3 71 + 1 n + l
4. Prove that ^ + 2^+ 3^+ .•• + n-^ = -n(n + 1).
Co Ci C2 C„_i 2
6. Provethat^« + |+|+"-« + ...=-^. "
13 6 7 n + l
6. Prove that
Co + cix + 2 Cix"^ + ••• + rCrX*" ••• + wc„x" = 1 + naj(l + a;)"-^.
7. Prove that CoCi+CiC2 + C2C3+ h Cn_iCn =, -.'T -x.'
(n + l)!(/i — 1)!
8. Prove that C0C2+C1C3+C2C4 + ••• + c„_2Cn =
(n+2)!(n-2)!
THE BINOMIAL THEOREM. 507
9. Prove that C(? - cx^ -^ c^^ h (- V)^Cr? is equal to 0 if
n '
n be odd, and to — — ^^ — if n be even.
ln\\n\
10. Prove that
(ji — \)\n\
11. Prove that
(2 71-1)!
(w-l)!(w-l)!
12. Prove that
13. Prove that
1 Cl J C2 , /- l>vn C** _ Wj
X x+l x-{-2 ^ x+n x(x+l)(x-\-2) -.{x+n)
14. Show that, if there be a middle term in the expansion of
(1 + a:)", its coefficient will be even.
THE BINOMIAL THEOREM: ANY INDEX.
/ 354. In Arts. 345, 349, we derived the series
1-2 r !
as the expansion of (1 + a;)**, but on the hypothesis that
n was a positive integer ; and under this limitation the
series necessarily terminates at the (n-\-iy^ term, for
the (n + 2)"'^ term and all subsequent terms contain
w — n as a factor.
When n is not a positive integer, the series is endless,
since no one of the factors n, n — 1, n — 2, etc., can in
that case be zero. Hence, to prove the binomial theorem
for a fractional or negative index is a proposition in
508 THE BINOMIAL THEOREM.
infinite series, and any demonstration that may be pro-
posed can only be valid under conditions that render the
infinite series convergent. This limitation must he pre-
supposed in all the demonstrations that follow.
f 355. Before proceeding to give a proof of the bino-
mial theorem when the index is fractional or negative,
we give some examples of its application. In these
examples x must be understood t6 have only such values
as make the series convergent.
/ Ex. 1. Expand (l+cc)-i by the binomial theorem. Putn= — 1
in the formula, and we have
(1 + ,)-! = 1 +(- 1). + (- r>(- 2),. ^ (- IX- 2X-3)^3 + ...
1 ' Z L • Z ' o
^(-l)(-2)-(-r)^^...
rl
= 1 - a; + a:'-^ - a:3 + ...+(- lya?- -f ....
I Ex. 2. Expand (1 + x)-^.
We have
(1 + ,)-. = 1 +(_ 2). + tl21(^,. + t.2X-3X^^3 ^ ...
i. • Z 1 • J • o
+ (-2)C-3)(-4).-C-r-l)^. _^ _
rl
= l-2a;-f 3a;2-4a:3 + ... 4-(_ l)r(,. + i)a:'- + ....
/ Ex. 3. Expand (1 + x)-K
(1 + ,)-3 = 1 + (_ 3). + £^|K^ x» + £^8K^^iK^.3 + .,
r\
= J_(l . 2 - 2 . 3ic + 3 . 4ic2 - 4 . 6a:3 + ...
• +(-iy(r + l)(r + 2)a?- +...}.
THE BINOMIAL THEOREM. 509
Ex. 4. Expand (1 - x) K
(1 - x)-^ = 1 +(_ i)( _ aj)+ (-^X-l) (_ xy
1-2
+ (-i)(-f)(-f)(-:,)3+... + (-^)(-|)-(-i-^+l)(_a;W
-1 I ^x I ^'^x'^ I ^•^•^x'^ 1 I l-3-5-(2r-l)
-^^2''^2.4'' +2.4.6'^ + + 2.4.6...2r "^^ *
356. Euler's Proof of the Binomial Theorem. For our
present purpose we shall employ the functional symbol
f{n) to denote the series
1 + "^+ lT2 ^+ 172^ ^^+-'
that is, the entire series if n be positive and integral, or
its limit [Art. 310] if n be fractional or negative. In
this general form, with n unrestricted in value, the series
is called the binomial series.
Let there be three binomial series formed as follows :
/(n)=l+n«+?^i^^ x' + ... (i.),
/(m)= l + mx + V^i^ZzH-^ + ... (ii.),
. \ ^ . / . \ I (7n-{-n)(m-{-n—l) o , .... ^
and let x be assumed to have only such values as will
make them absolutely convergent. We may then multiply
(i.) and (ii.) together, and their product will be a converg-
ent series [Art. 324].
Now, if the series on the right of (i.) and (ii.) be mul-
tiplied together, and the product be arranged according
510 THE BINOMIAL THEOREM.
to ascending powers of x, the result must involve m and
n in the same manner whatever their values may be.
But, when m and n are positive integers, we know that
/(m) is (1 -f x)'^ and that f{n) is(l + x) "", and the prod-
uct of /(m) and f{n) is therefore in this case (1 + a;)"'+%
which again, as m + n is a positive integer, is f{m -f n).
Hence when m and n are positive integers, the product
/(m) X f{n) is /(m + n) ; and as the form of the product
is the same for all values of m and 7i, it follows that
/(m) X f{n) = /(m + n) (a),
for all values of m and 7i, for which (i.) and (ii.) are abso-
lutely convergent.
For the purposes of this proof we need not enquire
what values of x, m and n will render the series under
discussion absolutely convergent. But see Art. 323,
Ex.1.
By continued application of (a), we have
J(rn) X f{n) xf{p) X -" =f{m -{-n) xf(p) X —
=/(m +^+P + ---)-
Now let m = n = p = • • • = r/s, where r and s are posi-
tive integers ; then taking s factors, we have
/(- I x/|-)X •••to s factors =/( -\ — | — tosterms);
\sj \sj \s s p
•■{
n\\\ =/w-
But since r is a positive integer, /(r) = (1 -f- a?)'.
.-. (1 +(.)'= j/*"
••• (i+'»)^=/(;
THE BINOMIAL THEOREM. 511
This proves our theorem for any positive fractional
exponent.
Next, assuming that the binomial theorem is true for
any positive index, it can be proved to be true also for
any negative index.
For, from (a),
/(-n)x/(n)=/(-n + 7i)=/(0).
Hence, as /(O) is clearly 1, we have
•^^"""^ ""7^ ^ al^r ''''''^ "" '' positive,
= (l + a;)-.
Thus (1 4- «)"'*=/(— n), which proves the theorem for
any negative index.
Ex. 1. Show that the coeflScients of 2c«-i in the expansions of
(1 — a:)-" and (1 + x)2«-2 are equal, n being any positive integer.
The coefficients are
(;-nX-n-l)...(-n-n+2), .,., ^^^ (2n-2)! ^,,^
(w-1)! ^ ^ (n-l)!(n-l)!
The former = n(n + l)...(2n - 2) ^„., ^ ^2n-2)l
(n-1)! (n-l)!(n-l)I
/ Ex. 2. Find -^101 by the binomial theorem.
V101= V{100(1 + jh)} = 10(1 + xk)*
= 10(1 + .005 - .0000125 + .0000000625} = 10.04987562....
Ex. 3. Find the coefficient of x^ in the expansion of
(1 - 2 x + 4 x2)-2
according to ascending powers of x.
512 THE BINOMIAL THEOREM.
(1 - 2 X + 4 x2) -2= {1 - 2 a;(l - 2 x)]-^
= 1 + 4 x(l - 2 a:) + 12 x\l - 2 x)2 + 32 x^(l -2xy+"'
= 1 + 4x + 4x2 - 16x3 + ....
Ex. 4. Show that, if (1 + x + x2 + x^)-^ be expanded in a series
of ascending powers of x, the coefficients of x^ and x^ will be zero.
(1 +X + x2 + x3)-2=/lj:i^y^=(l-X)2(l -x4)-2
= (l-2x + x2)(l+ 2x^ + 3x8 + 4x12 4-...).
EXAMPLES XCII.
1 1. Expand (1 — x)~* to 5 terms.
; 2. Expand (1 + 2x)~* to 5 terms.
' 3. Expand (2 — x)-^ to 6 terms.
4. Expand (1 — 3x)~^ to 5 terms.
3
5. Expand (1 — 5x)^ to 5 terms.
Find the general term in the expansion of each of the following :
- 6. (l-x)-5. 7. (1-x)-". 8. (1-x)'^ 9. (l + x)2.
10. (l+x)"3. IL (l-2x)"l 12. (l + 3x)"3.
13. Find the coefficient of x^ in the expansion of
according to the ascending powers of x. "^ '^ ''
14. Expand (1 — 3x)^ to 5 terms, and write down the general
term in its simplest form.
16. Find the coefficients of x^ and x* in the expansion of
(1 -2x + 3x2)-i.
THE BINOMIAL THEOREM. 613
16. Find the coefl&cients of x^ and x* in the expansion of
17. Find the first negative term of (1 + ^xy.
18. Find the first negative term of (1 + Sx)'^.
I 19. Find each of the following to four places of decimals by
means of the binomial theorem :
(i.) ^110, (ii.) ^130, (iii.) ^630.
20. Find the coefficients of ofi and x^ in the expansion of
(1 + X + X' + X8)».
21. Show that in the expansion of (1 + x + x^)-! in a series of
ascending powers of x, the coefficients of x^ and x* are zero.
22. Show that in the expansion of (1 + x +x2 + x'- + x^ + x*)"'*
in a series of ascending powers of x, the coefficients of x* and x^
are zero.
2
614 EXPONENTIAL AND LOGARITHMIC SERIES.
CHAPTER XXXV.
Exponential and Logarithmic Series.
357. The Exponential Series is an infinite series whose
first term is 1 and whose {n + 1)*^ term is x^'/n !. It is
absolutely convergent for all finite values of x [Art. 3:
Ex. 2]. Let its limit be denoted hj f{x) ; then the tw
exponential series
/(x)=i+x+i;+fi+- +!;+•.. (i.)
may be multiplied together and their product will be a
convergent series [Art. 324].
Compare the terms of this product with the terms of
the series
/(a^+2/)=l+(aj+2/) + ^^±p^+ •- + ^^±^
The coefficient of a;'"^/" in f(x) x f(y) is clearly
l/{mln\) and in f(x + y) the term (x + ?/)'"+"/ (m + n) !
is the only one in which x^y"" can occur, and by the bino-
mial theorem for a positive integral index, the coefficient
of x'^y'' in {x + yy+''/(m + n)! is
1^ . 1"^, that is, 4^.
(m + n) ! mini mini
^^
p
EXPONENTIAL AND LOGARITHMIC SERIES. 515
Hence, whatever x and y may be, the coefficients of
y,myn ^j.^ y^^^ ^ J^y^^ ^^^ ^j^ y^^ _j_ ^^ g^j.g ^j^g same, however
large m and ?i may be, that is, whatever (integral) values,
between 1 and + oo, m and n may have. It follows that
the series (iii.)? with its terms expanded, is identical
with the series formed as the product of the series (i.)
and (ii.) ; hence
/Wx/(2/)=/(a; + 2/), (iv.)
for all values of x and y.
By repeated application of (iv.) we obtain
/W xf(y)xm X ... =f{x + y) x/(0 X ...
=/(a; + 2/ + ^ + -).
Therefore, if z be any positive integer,
/(I) x/(l) x/(l) X to z factors
=/(l + 1 + 1 + ... to z terms),
that is \f(l)l'=f(z).
But /(l) = l + l+l + l+. ..+ !+...
— ^ zl 61 nl
= e. [Art. 305.]
... e*=/(0)=l + z + — + ...+^ + ...,
J\ ) ^ 2! n\
if z be a positive integer.
If z he a positive fraction, say z =p/q, where p and q
are positive integers, then
f(t\ X f(^\ X f(^ xtoq factors
\qj \qj \qj
=f(^^+\^\ + '''^oqi..u..y
516 EXPONENTIAL AND LOGARITHMIC SERIES.
• = e^, \; p is Si positive integer ;
Hence e'=f{z), for all positive values of z tjotl
integral and fractional.
If z be negative, say z = — z', then by (iv.),
/(-^')X/(2')=/(0),
and y(0) is obviously equal to 1.
.*. /( — z') = — — = — , '.' y is positive.
f{z') e^' ^ ^
Hence, for all rational values of z, be they integral or
fractional, positive or negative,
«' = /(.) = !+. + |^+-+J + -.
Should z be irrational, we replace it by a rational
fraction, which may be so chosen as to differ from the
given irrational value by an arbitrarily small quantity.
[See Art. 249.]
358. We may now derive an infinite series, in ascend-
ing powers of x, for the exponential f auction a^.
Let A = In a ; then a = e\ and a^ = e'^* ;
where A depends only upon a. This theorem, by means
of which a"" can be expanded in a series of ascending
powers of x, is sometimes called the exponential theorem.
EXPONENTIAL AND LOGARITHMIC SERIES. 517
359. The Logarithmic Series. Let a = e^, so that
X = In a.
Then a== = e^^ = e^^°«.
Hence, from Art. 357, we have
^x ^ ^\na =l-fajlna + i (a;lna)'^ + —^ (xlnay + ....
Now put a = l-\-yj then we have
(l+2/)'==l + ajln(l4-2/) + |yla;ln(l+.v)r + -.
Now, provided y he numerically less than unity, (1 + y)*
can be expanded by the binomial theorem; we then have
^ , , x(x — l) » , x(x — l)(x — 2) , ,
l-hxy-\- ^^^ ^y^-\- ^ -^,2.3 y^^'"
= l4-a;ln(l + 2/)+||{a;ln(H-2/)j=^+-.
Equate the coefficients of x on the two sides of the last
equation. [Art. 330.] We thus obtain
\n{l + y) = y-yl + t-t^....
This is called the logarithmic senes.
360. In order to diminish the labour of finding the
approximate value of the logarithm of any number, more
rapidly converging series are obtained from the funda-
mental logarithmic series.
Changing the sign of y in the logarithmic series
lna + y) = y-l + t-y^ + ..., (i.)
we have ln(l — 2/) = — 2/ — ^ — ^ — ^ ... (ii.)
2 3 4
618 EXPONENTIAL AND LOGARITHMIC SERIES.
Hence lni±^ = ln(l + y)- ln(l - y) [Art. 297, lY.]
Put - for i±l, and therefore *^?i^:^ for y: then we
n 1 — 2/ m + n
have from (iii.)
71 ( m + n 3 \m + ny 5\m-\-nJ )
361. We are now able to calculate logarithms to base
e without much labour.
Put m = 2, n = 1, in formula (iv.); then
ln2 = 2U + l.l+l.l + .4,
13 3 33^5 35 /*
from which it is easy to obtain the value
In 2 = .693147....
Having found In 2, we have from (iv.)
ln§ =
2
-{hi
.1 + 1.
53 6
h-
•}'
.-. In3-
-ln2 =
.405465...
Hence
ln3 =
In 2 + .405465...
=
.693147...
+ .405465...=
1.098612....
Again,
putting m =
= 5, w = 3,
we have
H-
-IM'
■h-l
h-
.
=
= H^-^8
"^1280
f 1 +•■
28672
••}
=
= .510826..
. ;
EXPONENTIAL AND LOGARITHMIC SERIES. 519
.-. In 5 = 1.098612 ... + .510826 ...
= 1.609438 •.. ;
.-. In 10 = ln5 + ln2
= 1.609438... + .693147.. .
= 2.302585 ....
By such processes as these the natural logarithm of any number
can be found to any requisite degree of approximation.
362. In Art. 302 it was shown that In 10 is the recip-
rocal of the modulus of the system of logarithms whose
base is 10. Hence the modulus of common logarithms is
M= 1/lnlO = 1/2.302585...
= .434294....
EXAMPLES XOIII.
31^5! 71 91^
'• ' = ^+2T + 3l + 4l+*-
2 ^31 5! 7!
5. (i + .).^ = i+|^ + 3^2^4^^
6 1 = -L_lJ_ + J_+J__L
■ 2 1.3 3.5 6.7 7.9 *
7. ln2 = -^-f--^+ J-+ J-+..
1.2 3.4 5.6 7.8
520 EXPONENTIAL AND LOGARITHMIC SERIES.
8. In2=- + — J^ — + ^
2 1.2.3 3.4.5
X l2a;+l 3
1
6.6.7
1
+
3(2x+l)3 5(2x+l)5
10. ln-^ = 2|— ^— + i ^ + 1 ^
11. \nx =
1_^1 a;2
(2:
1 ^1 a:3
1)3 5(2aj2_i)6
■■•}•
1
+
12. in«+^^ 2ax
a - X a2 4. a;2 3
X + 1 2 (X + 1)2 3 (x + 1)»
1/ 2 ax y I 1/ 2 ax Y
13. -logx = -^{^ + l(^V+...}.
lnalx+1 3Vx + l/ J
14. K
where
prove that
« = -i + ^\^r3,
M + r 4- w = c*,
w + 0)1? + w^w? = e««>==,
w + w^v + wto = 6"'=*,
and that w^ + ^3 + t(;3 — 3 m2;w7 = 1.
15. Eind the expansions in ascending powers of x for m, v,
and w of example 14.
LOGARITHMIC COMPUTATION. 521
CHAPTER XXXVI.
Logarithmic Computation.
363. The logarithms used in all theoretical investi-
gations are natural logarithms; but for the purpose of
making approximate numerical calculations, for reasons
that will shortly appear, logarithms to base 10 are
always employed. On this account logarithms to base
10 are called common logarithms.*
We have shown, in the preceding chapter, how natural
logarithms, or logarithms to base e, can be found ; and
having constructed a table of natural logarithms, the
logarithms to base 10 are obtained by multiplying the
former by the modulus of the latter, that is, by the con-
stant factor ^''log e [Art. 291, VI.], whose numerical value
has been found to be
''loge = -^=AS^29-". [Art. 362.]
In 10
common logarithms.
364. In what follows the logarithms must always be
supposed to be common logarithms, and the base 10 need
not be written.
* Also Briggian logarithms in honour of Henry Briggs, of Oxford,
who first constructed tables of such logarithms.
522 LOGARITHMIC COMPUTATION.
If two numbers have the same figures, and therefore
differ only in the position of the decimal point, the one
must be the product of the other *and some integral
power of 10, and hence from Art. 297, III., the logarithms
of the numbers will differ by an integer.
Thus log 421.5 = log 4.215 + log 100 = 2 + log 4.215.
Again, knowing that log 3 = .30103, we have
log. 03 = log(3 ^ 100) = .30103 - 2.
On account of the above property, common logarithms
are always written with the decimal part positive. Thus
log .03 is not written in the form -1.69897, but 2.30103,
the minus sign referring only to the integral portion of
the logarithm, and being written above the figure to
which it refers.
Definition. When a logarithm is so written that its
decimal part is positive, the decimal part of the logarithm
is called the mantissa, and the integral part the character-
istic.
365. The characteristic of the logarithm of any number
can be written down by inspection.
First let the number be greater than 1, and let n be the
number of figures in its integral part ; then the number
is clearly less than 10" but not less than 10**-^ The
logarithm of the number is therefore between n and
71 — 1 ; thus the logarithm is equal to n — 1 + a decimal.
Thus the characteristic of the logarithm of any number
greater than unity is one less than the number of figures in
its integral part.
For example, 235 is greater than 10^ but less than 10^. Hence
log 235 = 2 + a decimal, so that the characteristic is 2.
LOGARITHMIC COMPUTATION. 523
Next, let the number be less than 1. Express the
number as a decimal, and let n be the number of ciphers
before its first significant figure. Then the number is
less than 10"** and not less than 10~"~\
Hence, as the decimal part of the logarithm ^ must he
positive^ the logarithm of the number will be
— (n + 1) + a decimal fraction,
the characteristic being — (n +1).
Thus, if a number less than unity he expressed as a
decimal, the characteristic of its logarithm is negative and
greater hy one than the numher of ciphers hefore the first
significant figure.
For example, .02 is greater than 10-2 i,ut less than lO-i ; hence
log .02 is — 2 + a decimal, the characteristic being — 2. Also
.00042 is greater than 10-* but less than 10-3 ; hence log .00042 is
— 4 + a decimal, the characteristic being — 4.
366. Conversely, if we know the characteristic of the
logarithm of any number whose digits form a certain
sequence of figures, we know where to place the decimal
point.
For example, knowing that log 1.1467 = .0594408, we know that
the number whose logarithm is 3.0594498 is 1140.7, and that the
number whose logarithm is 4.0594498 is .00011467.
367. Tables are published which give the logarithms
of all numbers from 1 to 99999 calculated to seven places
of decimals : these are called " seven-figure " logarithms.
For many purposes it is, however, sufficient to use five-
figure, or even four-figure, logarithms.
In all Tables of logarithms the mantissae only are given,
for, as we have seen, the characteristics can always be
written down by inspection.
524 LOGARITHMIC COMPUTATION.
In making use of tables of logarithms we have, I. to
find the logarithm of a given number, and II. to find the
number which has a given logarithm.
I. To find the logarithm of a given number.
If the number have no more than five significant fig-
ures, its logarithm will be given in the tables. But if
the number have more significant figures than are given
in the tables, use must be made of the principle that
when the difference of two numbers is small compared
with either of them, the difference of the numbers is
approximately proportional to the difference of their
logarithms.
An example will show how the above principle, called
the Principal of Proportional Differences, is utilized.
Ex. 1. To find the logarithm of 1.14673.
From the tables we find that the log 1.1467 = .0594498, and that
log 1.1468 = .0594877, and the difference of these logarithms is
.0000379. Now the difference between 1.14673 and 1.1467 is
f'oths of the difference between 1.1468 and 1.1467. Hence to find
log 1.14673 we must add to log 1.1467 three-tenths the difference
between log 1.1467 and 1.1468, that is we must add
.0000379 X -^^ = .0000113.
Hence log 1.14673 = .0594498 -h .0000113 = .0594611.
II. To find the number which has a given logarithm.
Ex. 2. Find the number whose logarithm is 2.0594611.
We find from the tables that log 1. 1467 = .0594498 and that
log 1.1468 = .0594877, the mantissa of the given logarithm falling
between these two.
Now the difference between .0594498 and the given logarithm
is .0000113, and the difference between the logarithms of 1.1467
LOGAEITHMIC COMPUTATION. 525
and 1.1468 is .0000379. Hence, by the principle of proportional
differences, the number whose logarithm is .0594611 is
• 1.1467 + l^f X .0001 = 1.1467 + .00003 = 1.14673.
[N.B. The approximation can only be relied upon for one
figure.]
Thus .0594611 is log 1.14673, and therefore 2.0594611 is
log 114.673.
Ex. 3. Find ^100, having given log 4.6415 = .6666584
and log 4.6416 = .6666677.
log ^100 = J log 100 = § = .6666666.
Now ' log 4.6416 = .6666677,
and log 4.6415 = .6606584.
Hence log ^100 - log 4.6415 = .0000082,
and log 4.6416 - log 4.6415 = .0000093.
Hence ^100 = 4.6415 + ff of .0001 = 4.64159.
COMPOUND INTEREST AND ANNUITIES.
368. The approximate calculation of compound inter-
est for a long period, and also of the value of an annuity,
can be readily found by means of logarithms.
All problems of this kind depend upon the three fol-
lowing. [The student is supposed to be acquainted with
the arithmetical treatment of interest and present worth.]
I. To find the amount of a given sum at compound
interest, in a given number of years and at a given rate
per cent.
Let P denote the principal, n the number of years, ^ r
the interest of ^ 1 for 1 year, and A the required amount.
Then the interest of P for 1 year will be Pr, and there-
fore the amount of principal and interest at the end of
526 LOGARITHMIC COMPUTATION.
the first year will be P(l + r). This last sum is the
capital on which interest is to be paid for the second
year; and therefore the amount at the end of the second
year will be [P(l + r) ] (1 + r) = P(l + ry. Similarly,
the amount at the end of 3 years will be P(l + i-y,
and at the end of n years will be P(l + r)".
Thus Az=P(l^ry.
Oor. If the interest be paid and capitalized half-yearly,
it can be easily seen that the amount at the end of n
years will be P[lH--] •
Ex. Find the amount of $100 in 25 years at 5 per cent per
annum.
Since the interest on .$ 100 is $ 5, the interest on $ 1 is $ ^j^.
Hence the amount of $1 at the end of the first year will be
f (1 4- ^V)' ^"d at the end of 25 years will be $(1 + ^)^.
Hence ^A, the amount required, will be f$ 100(1 + ^)^.
Hence ' log ^ = log 100 + 25 log |^.
From the tables, we find that
log21 = 1.3222193, and log20= 1.3010300.
Hence log^ = 2 + 25(1.3222193 - 1.3010300)
= 2.5297325.
Now, from the tables, we find log 338.63 = 2.5297254
and log 338. 64 = 2.5297383.
Hence log A - log 338. 63 = . 000007 1 ,
and log 338.64 - log 338.63 = .0000129.
Hence A = 338.63 -f tV^ of .01
= 338.63 + .005 = 338.635.
Hence the amount required is $338,635.
LOGAEITHMIC COMPUTATION. 527
II. To find the present value of a sum of money which
is to he paid at the end^ of a given time.
Let A be the sum payable at the end of n years, and
let P be its present worth, the interest on $ 1 being $ r
per annum. Then the amount of P in w years "must be
just equal to A.
Hence, from I., P= A{l-\- r)~".
Ex. What is the present worth of $ 1000 payable at the end of
100 years, interest being at the rate of 5 per cent per annum ?
P= 1000(1 + ^^)-^^ = 1000(f^)-i«>.
Hence log P = log 1000 - 100 (log 21 - log 20)
= 3 - 100(1.3222193 - 1.3010300)
= 3 -2.11893 = .88107.
Now, from the tables, log 7.6045 = .8810707.
Hence the required present worth is $ 7.6045.
III. To find the present value of an annuity of^A pay-
able at the end of each of n successive years.
If the interest on $ 1 be $ r ; then from II.
the present value of the first payment is ^(1 + r)"^,
second . . . A(l-\-r)-^,
. n*^ . . . . A{l + r)'\
Hence the present value of the whole is
^5(l + r)-^ + (l + r)-2+...+(l-fr)-«|
= ^(l + r)-^;-[^ + y: = jjl-(l + r)-j.
Ex. Find the present value of an annuity of $ 100 to be paid
for 30 years, reckoning interest at 4 per cent.
528 LOGARITHMIC COMPUTATION.
The interest on $ 1 is $ jf q, therefore 1 + r = 1.04.
Hence P = ^{1 - (1.04)-30}
= 2500(1 -(1.04) -30}.
Now log(1.04)-so = - 30 log(1.04) = - 30 x .0170333
= -.510999 = 1.489001.
From the tables, log 3.0832 = .4890017 ; hence 1.489001 is
log .30832.
Hence P = 2500 (1 - .30832) = 2500 x .69168
= $ 1729.20.
EXAMPLES XCIV.
1. Write down the numbers whose logarithms are 5.3010300
and 5.3010300 respectively, knowing that log 2 = .3010300.
2. Write down the numbers whose logarithms are 3.2990931
and 4.2990931 respectively, knowing that log 1.9911 = .2990931.
3. Having given log 46854=4.6707467 and log 46855 =4. 6707559,
find log. 0468546.
4. Having given log 58961 =4.7705648 and log 58.962 = 1.7705722,
find log. 005896 14.
6. Find -^29, having given log 29 = 1.4623980,
log 19611 = 4.2924776, and log 19612 = 4.2924997.
6. Find ^100, having given log 19307 = .2857148, and
log 19308 = .2857373.
7. Having given log 2 = .3010300, log 3 = .4771213,
log 17187 = 4.2352001 and log 17188 = 4.2352253,
find ^15 to 5 places of decimals.
8. Having given log 3.4277 = .5350028, log 32483 = 4.5116561,
and log 32484 = 4.5116695, find v^034277 to 6 places of decimals.
LOGARITHMIC COMPUTATION. o29
In solving the following problems use the Table of Logarithms,
p. 670, and carry out the results to four places only.
9. Find the amount of $ 1 in 100 years at 5 per cent compound
interest.
10. Show that a sum of money will be more than doubled in
18 years at 4 per cent compound interest.
11. Find the amount of i^SOO in 10 years at 4 per cent com-
pound interest, the interest being payable half-yearly.
12. What is the present value of ^1000 which is to be paid
at the end of 15 years, reckoning compound interest at 3 per
cent ?
13. What is the present value of an annuity of $ 500 that ceases
at the end of 25 years, interest being reckoned at 6 per cent ?
14. If the average death-rate per annum in a city be 1^^ per
cent, and the average birth-rate be 2f per cent, and if there be no
increase or decrease in the population by migration, in how many
years will the population be doubled ?
15. The sum of ^ 100 was deposited in bank at compound inter-
est on the second of January every year during ten years, the last
deposit being made in January, 1880. Then on each succeeding
second of January (beginning in January, 1881) during ten years,
$ 100 was withdrawn. Interest being reckoned at 6 per cent, what
amount remained on deposit January 1st, 1891 ?
16. Twenty annual payments of $500 each are deposited with
an assurance company for the benefit of a person to whom, be-
ginning with the twentieth year, the entire amount paid in, to-
gether with accruing interest, is to be returned in 40 equal annual
payments. Reckoning interest at 5 per cent, what should be the
amount of each annual payment ?
Compute the following :
17. ^32.4.
19. ^99.83.
21. </91/^0.
18. ^8.934.
2l
20. ^92x^77..
22. </(67/9).
530 CONTINUED FRACTIONS.
CHAPTER XXXVII.
Continued Fkactions.
369. An expression of the form a-\
d
e + etc.
is called a continued fraction. This is usually written in
the more compact form
In the general case the numerators b, d, f, •" and the
denominators c, e, g, -" may be severally either positive
or negative ; but the continued fractions of greatest
importance, and those to which the present discussion
will be confined, have the form
^1111
in which b, c, d, e, -" are positive integers and a is either
an integer (positive or negative), or zero.
The numbers a, b, c, d, ••• are called the first, second,
third, etc., partial quotients respectively.
A continued fraction is said to be terminating or non-
terminating according as the number of partial quotients
is finite or infinite.
CONTINUED FRACTIONS. 531
The following examples will show how continued
fractions may be produced.
Ex. 1. Convert 74/26 into a continued fraction.
Reduced to its lowest terms this fraction is 37/13. The process
is then as follows :
E = 2 + -^ = 2+ ' =2-,^ '
13 ^13/11 i,_J_ 1 ,_1_
11/2 6 + i
Ex. 2. Convert (—x^-\-x^-\-l)/(x'^-\-l) into a continued fraction.
By the ordinary process of division we have
- CC8 + X2 + 1 , , , X , , 1
X
Ex. 8. Convert y/7 into a continued fraction.
Since 2 is the largest integer less than y/7, we write
V7 = 2+(V7-2) = 2 + ,
^^ ^ l/(V7-2)*
and we repeat this process upon the successively occurring surd
forms as many times as may be necessary. Thus, segregating in
each case the largest integer less than the fractional surd form, we
obtain in succession
1
:ZL±l = i^2ZL^ = i^ L
V7-2 3 3 3/(v7-l)
3 ^V7 + l_.^ V7-l_^ 1
V7-1 2 2 2/(v/7-l)
^ =:V7 + 1^1 V7-2^^ 1
V7-1 3 3 3/(V7-2)
^ =^7-l_2 = 4+V7-2 = 4 4- ^
V7-2 ^ "^ l/(V7-2)
The fractional surds now recur in the order in which they have
appeared in this cycle, and the continued fraction is therefore non-
terminating. Writing in the successive partial quotients we have
-/7=2+ — — — — ^^ ad inf.
532 CONTINUED FRACTIONS.
By reason of the recurrence of partial quotients in
continued fractions of this class, they are called recurring
or periodic continued fractions.
370. Corresponding to every real number, commensu-
rable or incommensurable, there exists a continued fraction,
either terminating or non-terminating, to which the number
is equal.
For, let X be the given number. Then, if a be the
greatest integer that does not exceed x, we may write
X = «i + ^,* (i.)
where Xi is a positive number not less than 1, and where
tti is a positive or negative integer, or possibly zero, this
last case occurring whenever X<1.
A repetition of this process with Xi gives
X, = a,-\-^, (ii.)
where a^ is the greatest integer that does not exceed X2
and X2 > 1 ; and n — 1 such repetitions result in the gen-
eral form
-X'„_i = a„ + — , (iii.)
* For example
where J / < 1 + V^ < 4,
1
and l~V5=-2+(3-V5) = -2-H^3^^^^^^,
where ~2<1-V^<-1-
CONTINUED FRACTIONS. 533
with the conditions : a„ = a positive integer not >X„ and
X„ ^ 1, for all integral values of n.
Substituting from (ii.) in (i.) we now obtain
and replacing Xg in this result by the expression for X^
derived from (iii.) by making 71 = 3, we have
«2+ %-f Xs
and in general, by further repetitions of this process,
X = a +J_ JL _1_...J:..
^ ^ ^2 + ^3 + «4 + X„
This is the required continued fraction.
In this result X„ may be an integer for a finite value
of n and the continued fraction therefore terminating, or
X„ may never be an integer, however large n may be, in
which case the continued fraction will be non-termi-
nating.
371. This process of finding partial quotients will be
recognized as identical with that of finding the greatest
common divisor in arithmetic. Now when this process
is applied to two integers that have no common divisor
greater than 1, the last divisor is 1, and the last remain-
der is 0. Hence :
Every commensurable number may be converted into a
terminating continued fraction.
On the other hand, since every terminating continued
fraction may obviously be converted into an ordinary
534 CONTINUED FRACTIONS.
fraction with commensurable numerator and denomi-
nator, and no such fraction can be equal to a surd,
it follows that :
A surd cannot be equal to a terminating continued
fraction.
372. The series of terminating continued fractions
«ij 0.1 + — , «! H ; , etc.,
formed, as here indicated, out of one, two, three, etc., of
the partial quotients of
,111
^2 + 0^3 + ^4 +
are called convergents ; ai is the first convergent, ai + l/a^
is the second convergent, and so on. Written as ordinary
fractions, the first three convergents are easily found to be
a^a^ 4- 1 aia^z + ^3 + <^i
The first convergent, but none of the others, may be zero.
For the determination of convergents of higher order
special methods, to be explained presently, are employed.
373. The convergents of a continued fraction of the
form tti H ••• (tti, a^, a^, -" all positive) are
^2 + % +
alternately less and greater than the fraction itself.
For, the first convergent is too small because the frac-
tional part • • • is omitted : the second convergent
a, H — is too large because in the fractional part a por-
CONTINUED FRACTIONS. 635
tionof the denominator, namely, , is omitted; and
tbe third convergent is too small because the part omitted
belongs again to the numerator. And in general, in con-
structing a convergent, the part of the original continued
fraction omitted belongs to the numerator or denominator
according as the convergent is of odd or even order.
Hence the theorem.
374. To prove the law of formation of the successive
convergents.
Let the continued fraction be
.111
^2 + % + ^4 +
The first three convergents are
«! aiCtg + l (aia2 + l)a3 + ai
1 ttg ajj^a + l
the third of which may be formed from the first and
second by the following rule: To obtain the numerator
(or denominator), add the numerator (or denominator)
of the first convergent to the product of the numerator (or
denominator) of the second by the third partial quotient.
Denoting the successive convergents by Pi/q^, Pi/qz,
Pz/^z-) "• Pn/Qni assume that this law applies to the forma-
tion of the n^^ convergent.
Then the hypothesis is
Now we may pass from the n^^ to the {n-\-iy^ con-
vergent by changing a„ into a„ + l/a„+i ; and hence, if
the n*^ convergent be
«nPn-l+Pn-2
536 CONTINUED FRACTIONS.
the (n + 1)*^ convergent will be
(g, + l/n+l)Pn-l -\-Pn-2 ^ qn+l(^nPn-l +^-2) + Pn-l
(a„ + l/a,+i)5n-l + qn-2 «n+l(««gn-l + g„-2) + ^n-l'
and by the hypothesis this is equal to
f^n+APn + Pn-l
Hence, if the law be valid for the formation of the
71*^ convergent, it is also valid for the formation of
the (n + 1)*^ convergent. But the third convergent is
formed in accordance with this law ; hence the law . is
valid for the formation of every convergent. [Art. 145.]
375. Properties of Oonvergents. In proving the proper-
ties of convergents we consider the continued fraction to
be
,111
and denote the n^^ convergent by p^/Qn-
I. If Pn/<Jn ^6 the n'* convergent to a continued fraction,
then
PrSln-l- Pn-iqn = {-^Y-
For, by the law of formation of convergents just
proved,
PrSln-X -Pn-l(ln = (^nPn-l + Pn-2)qn-l- Pn-l{<^nqn-l+ Qn-2)
= (- 1) (Pn-lQn-2 -Pn-2gn-l)
= (- l)'(i>n-2g«-3 -Pn-S^n-a)
= i-ir-\P2q,-Piq2)-
CONTINTJED FRACTIONS. 537
But p^i - Piq2 = (aitta + 1) - a^a^ = 1 = ( - l)^;
.*. Pn^n-l -Pn-iqn = ( " 1)"
II. Each convergent is in Us lowest terms.
For, any common divisor of p„ and q^ will divide
PnQn-i — Pn-iQni ^liat is, unltj ; and hence 1 is the only
such divisor.
III. The difference between two successive convergents is
a fraction whose numerator is unity.
For Pn ^ j>n-l ^ PnQn-l ^ Pn-lQn ^ ^
TV. Any convergent is nearer in value to the entire con-
tinued fraction than the immediately preceding convergent,
and therefore nearer than any preceding convergent.
For, the continued fraction itself is obtained from the
{n -\-iy^ convergent by putting a„+i H ••• in
the place of a„^.i ; and hence, if F denote the continued
fraction
Un+X-\- -'APn^-Pn-X , ,
Qn+l + ^qn
Qn + Qn-l
where X stands for , a positive quantity
less than unity.
Hence
p Pn+l ^Pnfl + \Pn Pn+1
Qn+l Qn+1 + Ag„ qn+1
^ HPnQn+l -Pn+lQn) ^ ( - 1)"X ,
538 CONTINUED FRACTIONS.
and
Qn Qn+l H- ^9 Qn
^ (Pn+ign - Pngn+l) ^ ( - 1)"^^
9n{qn+l + AgJ qn(qn+l + ^Qn) '
But X is less than 1 and is positive, and q^+i is greater
than g„ ; hence
TJ7 i^n+1 ^ TjT Pn
V. The convergents of odd order progressively increase,
but are always less than the continued fraction; and the
convergents of even order progressively decrease, but are
always greater than the continued fraction.
This proposition is a direct consequence of IV. and of
Art. 373.
Note. — In enumerating the orders of the convergents of the
continued fraction ai H , it is customary to regard ai as the
az +
convergent of first order, even when it is zero.
VI. The difference between a continued fraction F and
its n*^ convergent pj ^n ^^ ^^ss than 1/q^n ^^^^^ greater than
1/2 9 V
For, by IV.,
Ji fsj — ~
qn qniqn+i + MnY
where A. is positive and less than unity. Hence
9ng«+i qn qniqn+i + qnY
and since g„+i > g„,
1 >. XT P.
'•^^^^^1^2?
n+l
CONTINUED FRACTIONS. 539
Another form of this test of error is sometimes useful.
Since
5'n+l = ttn+l^n + ^n-l
that is, q^qn+i < a^+iQ^,
VII. Thus, any convergent which immediately precedes
a large partial quotient is a near approximation to the
value of the continued fraction.
Ex. 1. Assign a limit of the error involved in assuming the
fourth convergent as an approximation to the value of
J J i__i i_ JL J_ J 1
2+1 + 3+15+2+1 + 3+15+"*'
The first four convergents are
1 3 P3^3 + 1^4
I* 2' Qs 2 + 1 3'
16
11'
and by VI. F-—< — - — = -^'
3 £2-
2' gs
£4_3
.3 + 1
2 + 1
.4 + 3
94 3
.3 + 2
1
15 . 112
[Art. 374.]
Thus, the error in taking \^ as the value of the continued
fraction is less than 1/1815, or .00055 +.
Ex. 2. Find a series of fractions which converge towards the
true value of y/2.
Converting y/2 into a continued fraction, we have
V2 = l + V2-l = l+ J, -,==1+ ^
V2 + 1 2 +
V2 + 1
1 + ^J-JL....
2+2+2+
540 CONTINUED FRACTIONS.
and the first six convergents of this are ^
1 3 2-3 + 1^7 2-7 + 3^17
l' 2' 2 • 2 + 1 5' 2 . 5 + 2 12' "
2-17 + 7^41 2-41 + 17 ^99
2.12 + 5 29' 2.29+12 70'
the last of which differs from y/2 by a quantity that is less than
1/(2 X 702), or 1/9800.
376. A non-terminating continued fraction of the form
ai H has a true limit, namely, _ {Pn/%) •
In V. of the preceding article it was shown that
P2n^ ■pi^P2n+l
Q2n Q2n+1
for all values of n (P2n/Q2n and i?2n+i/^2n+i representing
the convergents of even and odd orders respectively),
and in VI. it was also shown that
and F-?^<^.
Q2 n+1 Q 2 n+1
Adding the corresponding members of the last two
inequalities together, we have
P2n ^2«+l .J_ , _J_
— ~~ ^^ 2 ' 2 '
92 n Q2n+1 Q. 2n 9 2 n+1
Now, q2n and q^n+i are the sums of products of integers
whose number increases without limit with the increase
of w, and we may therefore make q^n and q^n+i as large
as we please; that is, we may make the difference
i?2n/^2n " i>2 n+i/g2 n+1 as Small as wc please, by sufficiently
increasing n.
CONTINUED FRACTIONS. 541
Hence, P2n/%n ^^^ P2n+i/Q2n+i coiiverge to a common
value when n = gc, the former through increasing, the
latter through decreasing, values [Art. 375] ; and since
the continued fraction is always intermediate in value
to any two consecutive convergents [Art. 373], '
T-j limit / , X limit / , ^
or, what is the same thing.
EXAMPLES XCV.
Convert the following numbers into continued fractions :
1. |. 3. Uh 8. -I 7. V17. 9. Va2 + 1.
2. |. 4. .31. 6. -n. 8. 2V3. 10. ^^ - n^ + n + 1
Calculate the successive convergents of :
2+ 3+ 2+ 2
12. l+^J_J_-JLl.
2+ 3+ 4+ 5+6
13. i + A_JL^J_JL
3+6+7+9+11
14. 1+-L 1111
3+1+15+1+3
111
15. If Pr/Qr denote the r^ convergent of — ,
a+ 6+ a+ 5 +
show ih2Ltp2n+2 = P2n + ?><Z2n and q^+o = ap2n + (ab + l)q2n-
16. If Pr/Qr denote the r*^ convergent to any continued fraction,
prove that
Pn+l - Pn-l _Pn,
Qn+l - Qn-l Qn
542 CONTINUED FRACTIONS.
377. Kecurring Oontinned Fractions. In Art. 369 a con-
tinued fraction whose partial quotients recur in a definite
order presented itself as the equivalent of a quadratic
surd. The following examples show how we may pass
from the recurring continued fraction to the quadratic
surd.
Ex. 1. Convert into a quadratic surd.
Let the fraction be denoted by x. Then
a + X
whence cc^ _|_ qj^ _ 1 _ 0.
The roots of this quadratic equation are
2 2
If a be positive, the continued fraction is positive and the posi-
tive value of the radical must be chosen.
Ex.2. Convert into a quadratic
surd. 2+1 + 3 + 2+1 + 3 +
Let X denote the value of the continued fraction. Then
^ = _1_J ^
2+1+ 3+ic'
and the three convergents of this are
3 + x + l 4 + a;
h h and
9 + 3X + 2 ]l + 3ic
TT * 4 + a;
Hence
ll + 3x
or 3 ^2 + 10 X - 4 = 0,
and the required surd is the positive root of this quadratic
equation ; that is,
CONTINUED FRACTIONS. 543
Ex. 3. Convert 2+ — -^ -^-^-i--^ -^-^..- into a
1+5+2+1+3+2+1+3+
quadratic surd.
Let y denote the entire continued fraction, and x the part which
is recurring ; then
= J_J_J_
^ 2+1+3 + **'
and 2/ = 2 +
5 + x
The value of x was found to be — ^ + ^^^ in Ex. 2, and the
o
value of y, obtained from the second equation, is
^ 6 + a;
or, when x is replaced by its numerical value and the denominator
is rationalized,
119 + V37.
^ 44
378. Every recurring continued fraction is equal to one
of the roots of a quadratic equation whose coefficients are
rational.
Let y denote the entire continued fraction, and x the
recurring part ; then y and x have the respective forms
,11 111 ...
,11 1 11 r'\
x = ai-\ > (ii.)
in which all the letters are integers, and all, except
possibly a, are positive.
lip/q and p^ /q' be the convergents to (i.) correspond-
ing to the partial quotients h and Tc respectively, then
y=—, —y (ill.)
^ q'x-\-q ^ ^
(iy.)
644 CONTINUED FRACTIONS.
whence, by solving for x,
p-qy .
X ,
q'y -p'
and if r/s and r'/s' be the convergents to (ii.) corre-
sponding to the partial quotients «„„i and a^ respectively,
then
X = , ^ ■• (v.)
s'x+s ^ ^
Eeplacing x in (v.) by its value in terms of y in (iv.),
we obtain
P — qy r'(p — qy)+r(q'y-p')
q'y -p' s'{p - qy) + s{q^y -p')
a quadratic equation in y.
Cleared of fractions, equation (v.) is
s'y? -\-{s — r')x — r = (}'j
and, since {s — r'Y — 4(— rs') is positive and — r is neg-
ative, its roots are real, one being positive, the other
negative. Hence the values of y, as shown by equation
(iii.), are both real.
Ex. 1. Convert 2-ir— ^ — —"-
^1+2+1+2 +
and _i+A_J_JLiJ....
^3+ 1+ 24- 1+ 2+
into quadratic surds, j
Let , 0^ = 2+^ A...,
and y = -^-^^T^~-""
3+ 1+ 2 +
Then x = 2 + -l-l, ^^ ==^ i + i J_ 1;
1+ X 3+ 1+ a;
CONTINUED FRACTIONS,
from which, by writing down the successive convergents, we derive
*=-
whence, by solving for x,
a;2 _ 2 ic - 2 = 0,
and, by solving for y,
yi -2y-2 = 0.
Thus X and ?/ are the roots of the quadratic equation
«2_2«-2 = 0,
and since x must be positive and y negative,
. •. X = 1 + VS, y = l-y/3.
Ex. 2. Express the positive root of x^ — 2 a; — a^ = 0 as a con-
tinued fraction, a being a positive integer.
If a, — /3 denote the roots of this equation, a its positive, — /3 its
negative root, then
x^-2x- a2=(x - o)(a; 4- /3),
and this identity makes it at once clear that x^ — 2x — a^ will be
negative for all positive values of x less than a. Hence the largest
integer less than a is the largest integral value of x that will make
a;2 — 2 X — a^ negative. This value is easily seen to be 1 + a, and
we therefore write
a = l + a + a — 1— a.
But a2 - 2 a - a2 = 0, and therefore
a2 _ 2 a + 1 - a2 = (a - 1 - a) (a - 1 + a) = 1,
whence o = 1 + a + - — ~
a ~1 + a
Then
a-l + a = 2a + } ,
a - 1 + a
and therefore
2a + ^- 2a+2a +
a — l + a
2m
546 CONTINUED FRACTIONS.
In the larger text-books of algebra it is shown that any-
quadratic surd can be developed into a recurring con-
tinued fraction. [See Treatise 07i Algebra, Art. 367.]
EXAMPLES XCVI.
Convert the following surds into continued fractions :
1. 3 + V5. 4. 1 - v/7. 7. ^V'^'^^.
V2 - 1
2. 3-V5. 5. i(v3-l). 8. 2 + \/l + a2.
3. 1+V7. 6. 2+-L. 9. 2-VrT^2.
Convert the following continued fractions into surds :
10. *_L J_ J_ J_ ....
2+ 3+ 2+ 3 +
jj * 1 1 1 1 1 1
2+2+1+2+2+1 +
12. 3 + *-l- J- JL....
5+ 5+ 5 +
13. l + J_*A.^^....
2+ 1+ 1+ 1 +
14. l+A_*_^A_^JL..
2+ 2+4+2+4 +
15. %_!+ 1 1
a+l+ a-l+ a + l+ a— 1 +
Express the positive root of each of the following equations as a
continued fraction :
16. 0^2 -2a;- 1 =0. 18. a;2-2a;-9 = 0.
17. a;2 _ 2 ic - 2 = 0. \Q, x^ - 4:X + ^ - a^ = 0.
* The asterisk indicates the beginning of the recurring part.
CONTINUED FRACTIONS. 547
20. Prove that
\a+b+c-\-d+ l\ b+ c+ d-i- J
21. Show that a is a root of the quadratic
a — a — a —
equation x^ — ax+ 1 =0.
22. Show that the negative root of the equation x^—2x — a^=0j
expressed as a continued fraction, (a = a positive integer) is
1 1 1
a +
-2a+ -2a+ — 2a +
23. Express both roots of the equation x^ — ax — a%^ = 0 in the
form of continued fractions, a and h being positive integers.
548 SCALES OF NOTATION,
CHAPTER XXXVIII.
Scales of Notation.
379. In arithmetic, any number whatever is repre-
sented by one or more of the symbols 0, 1,2, 3, 4, 5, 6, 7,
8, 9, called figures or digits, by means of the convention
that every figure placed to the left of another represents
ten times as much as if it were in the place of that other.
The cipher 0, which stands for nothing, is necessary
because one or more of the denominations units, tens,
hundreds, etc., may be wanting.
The above mode of representing numbers is called the
common scale of notation, and 10 is called the radix or
base.
380. Instead of 10, any other number might be used
as the base of a system of numeration, that is, of a sys-
tem by which numbers are named according to some
definite plan, and of the corresponding scale of notation,
that is, of a system by which numbers are represented
by a few signs according to some definite plan.
Thus 4235 is said to be written in the scale of seven, if
3 stands for 3x7, 2 for 2x7x7, and 4 for 4x7x7x7,
so that every figure placed to the left of another repre-
sents seven times as much as if it were in the place of
that other.
SCALES OF NOTATION. 549
In general, any number N is expressed in the scale of
r when it is written in the form '"d^d^d^dQ, where each of
the digits df^did^d^'" is zero, or a positive integer less
than r, and where d^ stands for c^o units, d^ stands for
di X r, dz stands for c^a X r x r, and so on. Thus
]Sr=do-^dir-\-d2r^-\-'-'.
381. Any positive integer can be expressed in any scale
of notation.
Let N be the number, and let r be the radix of the
required scale.
Divide N by r, and let Qi be the quotient and d^ the
remainder.
Then N= d,-\-rx Q,.
Now divide Qi by r, and let Qa be the quotient and di
the remainder.
Then Qi = d^-\-r x Q^] .'. N= d^ + rd^ + r'Q^.
By proceeding in this way we must sooner or later
come to a quotient which is less than r ; let this be after
n divisions by r. The process is now complete, and we
have
N=^ do + d,r + ^27-2 + ... + d,r-,
so that the number would in the scale of r be written
d„'"dsd^ido.
Each of the digits do, dy, c?2, • • • is a positive integer less
than r, and any one or more of them, except the last, d„,
may be zero.
550
SCALES OF NOTATION.
Ex. 1. Express 1062 in the scale of 7.
The quotients and remainders of the successive divisions by 7
are as under :
1062
151 remainder 5 = dQ
. . . 4 = c?i
. . , 0 = do
21
Thus, 1062 when expressed in the scale of 7 is 3045.
Ex. 2. Change 2645 from the scale of 8 to the scale of 10.
Since 2645 = 2x83 + 6x82 + 4x8 + 5
= {(2x8 + 6)8 + 4)8 + 5,
the required result may be obtained as follows : Multiply 2 by 8,
and add 6 ; multiply this result by 8, and add 4 ; then multiply
again by 8, and add 5. This process is clearly applicable in all
cases.
Ex. 3. Express 2156 in the scale of 5. Ans. 32,111.
Ex. 4. Express 34239 in the scale of 11.
Ans. 237^7, where t is put for 10.
Ex. 5. Express \^ as a fraction in the scale of 4. Ans. j^.
Ex. 6. Change 31426 from the scale of 8 to the scale of 4.
Ans. 3030112.
We may first express 31426 in the scale of 10, as in Ex. 1, and
reduce the result to the scale of 4, as in Ex. 2. The result may,
however, be obtained by one process, as under :
31426
4
6305 remainder 2
4
4
1461 1
314 1
4
63 0
4
14 3
3 0
Thus, the number required is 3030112.
SCALES OF NOTATION. 551
Explanation. We first divide 3 eights plus 1 by four, giving
quotient 6, and remainder 1 ; we then divide 1 eight plus 4 by
four, giving quotient 3, and remainder 0 ; and so on.
382. It would be a good exercise to perform all the
ordinary rules of arithmetic with numbers expressed in
various scales.
Ex. 1. Add 2345, 6127, and 1503. [Scale 8.]
Ex. 2. Subtract 3154 from 4021. [Scale 6.]
Ex. 3. Multiply 234 by 466. [Scale 7.]
Ex.4. Divide 22326 by 315. [Scale 8.]
Answers. 12177, 423, 150663, 66.
383. Eadix Tractions. ' Eadix fractions in any scale
correspond to decimal fractions in the ordinary scale ;
thus .abc •••, in the scale of r, stands for
To show that any given fraction may he expressed by a
series of radix fractions in any proposed scale.
Let F be the given fraction; and suppose that when
expressed by radix fractions in the scale of r, we have
i^=.a6c--., that is i^=- + -, + -,+ •••,
r r^ r^
where each of a, b, c, -" is zero or a positive integer less
than r.
Multiply by r ; then
r r-
552 SCALES OF NOTATION.
Hence a must be equal to the integral part, and
b c
- + -- + • • • must be eq ual to the fractional part of F xr.
r r-
Let jp\ be the fractional part of Fr ; then
r IT
Multiply by r again, then as before h must be equal to
the integral part of F^ x r.
Thus a, 6, c, ••• can be found in succession.
Ex. 1. Express y^^^ by a series of radix fractions in the scale
''^^' ^\x6 = 3 + tV; TVx6=:0 + i; 1x6 = 2.
Hence -302 is the required result.
Ex. 2. Change 431.45 from the scale of 10 to the scale of 4.
The integral and fractional parts must he done by separate
processes.
4
4
• 4
4
431
.45
107 .... 3
26 .... 3
4
1.80
6 .... 2
1 .... 2
4
3.2
red result is 12233.130.
4
0.8
384. Theorem. Any number expressed in the scale of r
is divisible by r — 1, if the sum of its digits is divisible
by r — 1.
Let J^ be the number, S the sum of the digits, and let
the digits be do, di, d^, etc.
Then N= do + d^r + d^r^ -] h dj-^
and S =dQ-\-di + cZg + • • • + (^n-
Hence JSf-S=d,{r-l)+d^Cr''-l) + "'-^d„(r''-l),
SCALES OF NOTATION. 553
Now each of the terms on the right is divisible by
r - 1. [Art. 147.]
Hence iV— ^ is divisible hj r — 1, and therefore when
S is divisible by r — 1, so also is JH.
As a particular case of the above, any number expressed
in the ordinary scale is divisible by 9 when the sum of
its digits is divisible by 9.
385. Theorem. Any number expressed in the scale of r
is divisible by r -{- 1 when the difference between the sum of
the odd digits and the sum of the even digits is divisible by
r + l.
Let N= do + d^r + d^r^ + d^r^ -\ ,
and D = dQ — di-\-do — ds-\ .
Then N-D=di(r + 1) + d^ii^ - 1) + ^3(7^ -j- 1) + •••.
Now each of the terms on the right is divisible by
r + 1. [Art. 147.]
Hence, JV— D is always divisible by r + 1, and, there-
fore, when D is divisible by r -f 1, so also is N.
As a particular case, any number expressed in the
ordinary scale is divisible by 11 when the difference
between the sum of the odd and the sum of the even
digits is divisible by 11.
EXAMPLES XCVII.
1. Express 2156, 7213, and 192457 in the scale of 6.
2. Change the following numbers from the scale of 7 to the
ordinary scale : 2135, 4210, 30012, 123456.
3. Change 8152 and 23678 from the scale of 9 to the scale of 12.
4. Express 23.42 and 123.45 in the scale of 5.
20
554 MISCELLANEOUS EXAMPLES VII.
5. Multiply 2.31 by 1.25, the numbers being expressed in the
scale of 6.
6. Show that the numbers represented in any scale by 121,
12321, and 1234321 are perfect squares.
7. Find the values of a and 6 in order that 215al463 may be
divisible by 9 and by 11.
8. Find the values of a and h in order that 516a72456 may be
divisible by 99.
9. Find the scale in which 314 is represented by 626.
10. Find a number of two digits in the scale of 5 which is
doubled by reversing its digits.
11. Find a number of two digits in the scale of 8 which is
doubled by reversing its digits.
12. Find a number of two digits in the scale of 7 which is
trebled by reversing the digits.
MISCELLANEOUS EXAMPLES VII.
1. Solve the equation x-^^a = 6 + ^/(ax^ 4- c).
2. Show that (x -h y + zy^+^ - x'^"+i - y^»+^ - z^^+^ is divisi-
ble by (y -^ z) (z + x) (x -\- y) , and find the quotient when n = 1,
and when n = 2.
3. Show that (x + yy -x^ -y^ = S xy{x + y),
and that (x -\- yY — x^ — y^ = b xy{x + y) (oj^ ■\- xy -\- y^).
4. Simplify «^ + ^- + ^'
{a-b){a-c) (6-c)(6 — a) {c—a)(c-b)
6. Simplify ^ + ^« + «^
(a— 6)(a — c) (b — c){b-a) {c—a)(c—b)
6. Simplify ^ + i- -f ^
aCa-&)(a— c) b{b—c)Q) — a) c(c— a)(c-6)
MISCELLANEOUS EXAMPLES VII. 556
7. Simplify
1
a2(a-6)(a-c) b\b-c)(ib-a) c\c-a){c-h)
8. Simplify ^±^ + £±^ + 9i+l
(a-&)(a-c) (6-c)(6-a) {c-d){c-h)
9. Simplify ^±^ + ^±^ + « + ^
a(a-6)(a-c) 6(6-c)(6-a) c{c-a){c-h)
10. Show that "^ ■ ^
+
(a-6)(a-c)(a; + a) (6_c)(6-a)(x+6)
1 1
(c-a)(c-6)(a; + c) (x + a)(x + 6)(x + c)
11. Show that ^ + ^
+
(a-6)(a-c)(x + a) (6 - c)(6 - a)(x + 6)
c —X
12. Show that
(c-a)(c-6)(x + c) (x+a)(x+ 6)(x + c)
a2 , 62
(a-6)(a-c)(x + a) (6 - c)(6 - a)(x + 6)
{c-a){c-h)(x-\-c) (x+ a)(x + 6)(x-t-c)
13. Simplify (6-^-c^)«+(c^-a^)« + K-6-^)«.
(6-c)8+(c-a)8 + (a-6)»
14. Prove the following :
(i.)a + 6- «' ■ *'
a — 6 6 — a
(ii.)a+6+c=- «^ + ^' • ^'
aii.) a + 6 + c+c? =
{a-h){a-c) (b-c)(b-a) {c-a){c-b)
{a-b){a-c)(a-d) (b-c){b-d)ib-a)
+ . ,..^\. ,.+ ^^
(c-d)(c-a)(c-6) (d-a)(d-6)(d-c)
556 MISCELLANEOUS EXAMPLES VH.
a3 , 68
15. Show that
(a-6)(a-c)(a-d) (b - c)(b - d)Cb - a)
c3 , # ^^
(c-£Z)(c-o)(c-6) (c?-a)((?-&)(<Z-c)
16. Show that ^^^ ■ ^^«
(a-6)(«-c)(a-d) (&- c)(6-d)(6-a)
(^«6 _l abc _ J
(c-d)(c-a)(c-6) ((?-a)(d-6)(d-c)
17. Show that ^ (x.- ay + ^ (x-b)^
+ 7 ^7 -(X-<))2 = X2.
(c — a) (c — 6)
18. Prove that (b + c)2 + (c + ay + (a + 6)^ - (c + a) (a + 6)
- (a + 6)(& + c) - (6 + c)(c -f a) = a2 + 62 + c2 - 6c - ca - a5.
19. Show that (6 + c)3+(c+a)3+(«+6)^-3(6 + c)(c + a)(a + 6)
= 2(a3 + 63 + c3-3a6c).
20. Show that (b + c - ay + (c + a - by +(a + b - cy
-3(6 + c -a)(c + a-6)(a+ 6-c) = 4(a3 + 63 + c3-3a6c).
21. Show that (ic2 + 2?/0)3 + (y2 + 2 2x)3 + (^2 + 2xi/)3
- 3(x2 + 2 yz) (y^ + 2 zx) (z^ + 2xy) = (x^ -\- y^ + z^ - S xyzy.
22. Show that, if cc2 _^ ^2 _|_ ^2 _ (^y^ -{- zx + xy), then x — y — z.
23. Prove the following :
(i.) if 2(a2 + 62) = (a + 6)2, then a = 6 ;
(u.) if 3(a2 4. 62 + c2) = (a + 6 + c)2, then a = 6 = c ;
(iii.) if 4(a2 + 62 + c2 + cZ^) = (a + 6 + c + (Z)2,
then a = h = c = d\
and (iv.) if n{a'^ + 62 + c2 + ...) = (« + 6 + c + -O^,
then a = 6 = c = •••,
w being the number of the letters.
24. Prove that, if the sum of two positive quantities be given,
their product is greatest when they are equal to one another.
MISCELLANEOUS EXAMPLES Vn. 557
25. Prove that, if the product of two positive quantities be given,
their sum is least when they are equal to one another.
26. Show that the least value of x + - is 2, the least value of
X
4 9
X + - is 4, and the least value of x + - is 6, x being real and
X X *
positive.
27. Show that ^^—^ — ^ "*" cannot be greater than 7 nor less
than I, for real values of x.
28. If the sum of a given number of positive quantities is fixed,
their continued product will be greatest when they are all equal.
29. If the continued product of a given number of positive quan-
tities is fixed, their sum will be least when they are all equal.
30. Prove that, if ^^ ~ ^ = ^^ ~ ^ , then will each fraction be
y + z 0 + X
equal to ^V ~ ^ ^ and also equal to x-\- y -\- z.
x + y
31. Prove that, if a, 6, c, d be all real and if
(a + by + (6 + c)2 + (c + dy = 4(a& -hbc + cd),
then a = b = c = d.
32. If «^-^^ = ^^ - ^^ , then will a = 6, or c = tf,
a-b - c + d a — b — d-{-c
or a + b = c + d.
33. Show that, if x, y, z be determined by the equations :
(a - a)2x + (a - pyy + (a - yyz = (a - 5)2,
(6 - a)2x + (6 - /3)2?/ + (6 - 7)2^ = (6 - S)^,
(c _ a)2x +(c - /3)2?/ +(c - 7)^0 =(c - 5)2;
then will (d - a)2x +(d- /3)2y + (c? - 7)2^ =(d- 5)2,
where cZ has any value whatever.
34. If the equation — ^— + — ^ = -^— + — ^ have a pair of
x+a x+6 x+c x+d
equal roots, then either one of the quantities a or 6 is equal to one
558 MISCELLANEOUS EXAMPLES VII.
of the quantities c or d^ or else - + i = - + -. Prove also that the
a h c d
roots are then -«,-«, 0 ; - 6, - 6, 0 ; or 0, 0, - ^^ •
« + 6
35. If 2(ic2 + r,^i2 _ a:a;') (?/2 + yn _ ^y') = ^^2^2 + ^/2y/2^ then will
jc = oj' and y — y' .
36. Show that
{y -\- z)(z + X) (x + y) -\-xyz = xyz(x -\- y ■\- z)l- + -^ - )-
37. If a, &, c, cc are all real quantities, and
{cfl + 62)x2 - 2 6(a + c)a; + 62 _^ c^ = 0,
then a, 6, c are in geometrical progression and x is their common
ratio.
1 - a;2*
38. Show that (1 + x) (1 + x^) (1 + x^)— to n factors
1-x
39. If s = a 4- & + c, prove that
(as + he) {hs + ca) (cs + ah) = (6 + c)2(c + ay {a + 5)2.
40. If a + 6 + c + d = 0 ; then will
(a3 + 53 _|. c3 + #)2 = 9(6cd + c(?a + dah 4- a&c)2
= 9(&c — ad){ca — hd){ah — cd)
= 9(6 + cy\c + a)2(a + 6)2.
41. Show that a(a + c?)(a + 2(Z)(a + 3d) + d^ is a perfect
square.
42. Prove (by the method of indeterminate coefficients) that
the necessary and sufficient condition in order that
ax'^ + 2hxy + by^ + 2gx + 2fy-\-c
may be resolved into two factors, rational and of the first degree
in X and y, is
abc -\-2fgh- ap - hg^ - ch'^ = 0.
,43. Find (by the method of indeterminate coefficients) the cube
root of each of the following expressions :
(i.) x^ - 24 x^y + 192 xy^ - 512 yK
(ii.) x6 - 9 x5 + 33 X* - 63 x^ + 66 x2 - 36x + 8.
(iii.) 8x6 - 36 x& + 102x4 - 171 x^ + 204x2 _ 144x + 64.
MISCELLANEOUS EXAMPLES VII. 559
44. Find the limit of (« — a;)^ - 1 ^^^^ x = 0.
X
45. Find the limit ot { x V~^ when x = 1.
i-D-
46. Prove that ^^^^^^ (1 + x + 2 ! x2 + 3 ! a^ + - + n !a?»)= oo
for all finite values of x.
47. Expand ^-^ + ^f-\-^ + ^ into an infinite series.
(1 + x)%l - x)8
48. Expand (1 + ac)'"^ and (1 + ic)-'»/2 into infinite series.
49. Expand ^e'' + e-^) and ^(e* — e-') into infinite series.
50. Resolve
1 4- ai + (1 + ai)«2 + (1 + ai)(l + a2)a3 + -
+(1 + ai)(l + as) - (1 + an-i)an
into n binomial factors.
51. Find the sum of the first n 4- 1 terms of the series
1 I ^ I (gi + ^)^ I (ai + g) (a2 + x)x
. (ai + a;)(a2 + a;)(a8 + x)x ^
aia2(Xsci4
52. Show that the sum of the first n + 1 terms of
J ■ m . m(jm + 1) . m{m + l)(m + 2)
"^ 1 ■*■ 2 1 3 !
m(m + l)(m + 2)(m + 3) ^
4!
•g (m + l)(m + 2)(m + 3) ■.■ {m^-n- 1).
n!
63. Show that the sum of the first n + 1 terms of
1 w ■ m(m - 1) m{m - l)(m — 2)
12! 3!
m(m - l)(m - 2)Cm - 3) _
4 !
(1 - w)(2 - m)(3 - m) ••• (n - m)
nl
560 MISCELLANEOUS EXAMPLES VII.
54. Find the n*^ term and the sum of the first n terms of the
series i + 3 + 7 4. 13 + 21 + 31 + 43 + -..
55. Find the sum of the first n terms of the series
3 + 4 a: + 6 a;2 + 10 x3 + 18 x* + ....
56. Convert a +V2 + a^ into a continued fraction.
57. Convert the positive root of ax^ — ahx — 6 = 0 into a con-
tinued fraction.
Denoting by Pi/Qi, P2/Q2, •••, Pn/Qn, the successive con-
vergents of ai + hi
a2 + b2
«3 + •••
(Pi = «!, ^1 = 1), prove the following relations :
58. P„ = GnPn-l + bn-lPn-2,
Qn = ClnQn-l + hn-lQn-2'
59. Pn+lQn - PnQn+l=(- 1)*'&1&2&3 - K.
60. -Pw+l ^»—( 2\n^l^2"'ftn
Qn+l Qn QnQn+1
61. f^ = ai + ^ - -^^ + -" +(- 1)- ^'^^ "' ^"'
Pn
DETERMINANTS.
661
CHAPTER XXXIX.
Determinants.
h
c
i
c
a
1
a
h
b'
c'
c'
a'
a'
b'
386. If the matrices
[Art 173.]
be used to denote the cross-product differences
6c' — 6'c, ca' — c'a, ab' — a'b,
the values of x and y derived from the simultaneous
equations
ax -^ by + c = 0,
a'x + b'y + c' = 0,
may be presented in the form
[Art. 172.]
These cross-product differences whose matrices have
two rows and two columns, are called determinants of the
second order.
Ex. 1. The condition that the equations
ax + by = 0, a'x + b'y = 0,
may be simultaneous in x and y is
b c
c a
b' c'
, y =
c' a'
a b
a b
a' V
a' b'
a b
a' b'
= 0.
562
DETERMINANTS.
For, if the equations be simultaneous, then
b'
a'
whence (a6' — a'b)y = 0,
and unless ?/ = 0, we must have ab' — a'b = 0.
Ex. 2. Write down the solution by determinants of the simul-
taneous equations
(C+Vy^ = 0, 3a; -y + 2 = 0.
Solution : '^:.
y =
2
-1
1
2
-1
2 " 3
-.1
-1
1
1
2
2
3
•
3
-1
4-1
3-2^5
1-6 7*
Ex. 3. Let the three equations
aix + biy + ciz = 0,
a2X + b2y + C2Z — 0,
azx + bzy + czz = 0,
be regarded as simultaneous in x, y. z; multiply them by
biCz — &3C2, &3C1 — &1C3, 61C2 — 62C1,
respectively ; add the resulting equations together ; show that the
new coeflBcients of y and z vanish ; and thus derive, as the condi-
tion of simultaneity,
ai
62 C2
- a2
bi ci
+ 053
6i ci
bz cz
63 Cz
62 C2
= 0.
387. When its three matrices are replaced by their
equivalent cross-product differences,
62C3 — 63C2, 61C3 — 63C1, 61C2 — 52^1,
DETERMINANTS.
563
expression
tti
62 c,
— 0.2
h c.
+ ^3
61 Ci
h Cg
h C3
^':i C2
becomes
ttj^aCs — Cti&gCg — a2^iC3 + ^26301 4- tta^iCa — a362Ci-
It is called a determinant of the third order, and its
matrix form is
ttj 61 Cj
0,2 62 ^2
ttg 63 Cg
We shall frequently denote this determinant by the
abbreviated notation [^aib.jC^'].
Let it now be required to solve the simultaneous equar
tions
a^x -f biy + CiZ + di = 0,
a.^-\-b^ + c,js + d2 = 0,
«3^ + ^32/ + C32; + f^3 = 0.
When the left members of these equations are multi-
plied by
62C3 — 63C2, — 61C3 4- &3C1, and biC^ — b^i
respectively, and the resultant equations are added
together, as in Ex. 3, Art. 386, the new coefficients of
y and z become
^1(^2^3 — te) — h{biCs — fcgCi) + ^3(^1^2 — &2C1) = 0
and
^1(6263 — ^3^2) — C2(&iC3 — 63C1) + c^(biC2 — 62C1) = 0
respectively, and there remains
0,
«1
&i
c
+
a2
h
C2
tts
h
C3
d,
h
Cl
d.
b,_
C2
d.
b>
Cs
664
DETERMINANTS.
whence, in the abbreviated notation,
Eepeating this process for the purpose of eliminating
in succession z, x and x, y, we obtain
[ai&sCg]'
In these fractional values of the unknown quantities
we observe that the common denominator is the deter-
minant [ai^a^s]? and that the numerators are three deter-
minants which may be derived from [^162^3] by writing
(in the matrix form) d^did^ in the place of aia2a^, &i&2^3?
and C1C2C3 respectively.
388. A determinant is fully described as a quantity
when its terms with their proper signs are written down;
but, for the purpose of discovering its properties, it is
regarded as evolved from a principal term of the form
dib^Cgd^ • • • Z„ by keeping the letters in fixed order and per-
muting the suffixes 1, 2, 3, • • • n in all possible ways, or
by keeping the suffixes in fixed order and permuting the
letters in all possible ways, and then giving to the terms
thus obtained the positive or negative sign according as
the arrangement of the suffixes (or letters) in such term
is derived from that of the principal term by an even or
an odd number of inversions of order ; whereby we define
that an inversion of order occurs as often as a suffix (or
letter) precedes another suffix (or letter) which in the
principal term it follows.
DETERMINANTS.
565
For the determinant of the third order this method of evolving
the several terms is indicated in the following scheme. The
letters are supposed to stand in the order a be, and only the suffixes
are written down.
oral order.
1 inversion.
2 inversions.
3 inversions.
Sign
123
__^
+
123
132
—
123
213
.
_
123
213
231
+
123
132
312
+
123
132
312
321
_
For example, there are three inversions in 3 2 1 because 3 pre-
cedes both 2 and 1, and 2 precedes 1.
389. Def. A determinant of the n^ order is the alge-
braic sum of the n I products that may be evolved from
a principal terra aib^ajd^^'-lr^ by keeping the letters in
fixed order and permuting the suffixes 1, 2, 3, •••n in all
possible ways ; and any term is preceded by the positive
or negative sign, according as it is derived from the
principal term by an even or an odd number of inver-
sions of the natural order of the suffixes 1, 2, 3, ••• w.
The matrix of this determinant of the n^ order is
«1
h
cr-h
ttg
b.
c,...k
as
63
Cs-h
•
:
; i
«„
K
C.-ln
This determinant is frequently denoted by the abbrevi-
ation [_aib2Cs " ' l^'] ', sometimes also by 2 ±aib^s"'L'
The n^ quantities aj, bi, Cj, aj? ^2i"'K 2,re called the
elements of the determinant, the n\ products aA^^s • • • ?„ ,
bm
DETElRMINAN^S.
cb^iC^"'ln, etc., are its terms, the diagonal line aih^f^-'-l^,
extending downwards from the upper left-hand corner,
is the principal diagonal, and the product ai&^Cs---^? whose
factors are the elements of the principal diagonal, is the
principal term.
In the present discussion we shall be chiefly concerned
with determinants of the second and third orders.
390. It follows from our definition that every term
of a determinant of the n^^ order contains all of the n
letters a, 6, c, ••/, and all of the n suffixes 1, 2, 3, •••n,
but none of the letters or suffixes twice repeated ; hence,
every term contains one and only one element from each
row, and one and only one element from each column of
the matrix.
391. If in a matrix of a determinant the letters a, b,
c,"'l and the numbers 1, 2, 3, ---n be interchanged,
statements concerning rows become statements concern-
ing columns, and vice versa; hence, a determinant in the
matrix form is not altered by changing its rows into col-
umns and its columns into rows.
Thus
and
Unless the contrary is explicitly stated, determinants
will henceforth be thought of as given in the matrix
form.
ai &i
=
tti a2
0^2 &2
bi 62
J
Oi 6i Ci
=
tti tta as
Ctg ^2 ^2
bi 62 ^3
as
h c.
Ci C2
C3
'I
DETERMINANTS.
567
392. When given in their matrix forms, determinants
of the second and third orders are easily expanded into
the algebraic sums of their terms by the rule which is
expressed in the following formulae :
a
I &i
=
= aifta ■
- «2&1,
»
a^ 62
tti 61 Ci
= «!
62 C2
-a2
&1 c,
+ a3
61 Ci
«£ ^2 C2
h Cs
&3 C3
62 C2
^3 &3 C3
Ex. 1. Expand
2 3 1
6 2 3
3 12
By the rule of expansion this determinant is equal to
2 3
1 2
3 1
1 2
+ 3
3 1
2 3
= 2(4 -3)- 5(6-1)+ 3(9
= 2 - 26 + 21 = - 2.
2)
Ex. 2. Expand
a
h
9
h
b
f
9
f
c
= A.
By the rule of expansion we have
A = aibc -P)-h{hc - fg) + g{hf^ bg)
= abc + 2fgh - aP - bg^ - ch\
EXAMPLES XCVIII.
1. Count the number of inversions in 2143, 3421, 4321, and
53412.
2. If abode be regarded as the standard order, count the number
of inversions in acebd, cdeba, and dbace.
568
DETERMINANTS.
3. What are the signs of the terms hdk^ ahf, and Mc in
a
b
c
d
e
f
9
h
k
4. Write, with their appropriate signs, all the terms of the deter-
minant [aih2Czd{\.
5. In the determinant of the fifth order [^aih^Czd^e^l determine
the signs of the terms asbiCzdse^, aihsCidiCs, a^h^C\d^e2^ and
056403(^261 •
Evaluate the following determinants :
6.
1 2
4 5
3
6
7 8
9
•
7.
1
-2
3
-2
3
-4
3
-4
6
8.
1 3
3 5
5 7
5
7
9
9.
1 0
0 5
3
0
7 0
9
.
J
10.
—
1
0
0
0
-1
0
0
0
-1
11.
0
a
b
,
h
0
a
a
b
0
•
12.
a
b
c
b
c
a
c
a
b
.
13.
a
b
c
b
b
b
c
b
a
.
14. Show that
as bs cs = —
ai bi Ci
=
&i ai ci
a2 ?>2 C2
a2 &2 C2
62 a2 C2
ai bi ci
as bs cs
bs as Cs
16. Show that
1 -i
i 1
= 2, where i = y/— 1.
DETERMINANTS. 569
393. Properties of Determinants. An interchange of any
two adjacent rows (or columns) cf a determinant changes
its sign, but affects its value in no other way.
For, in the determinant [aA^s •••?„], the interchange
of any two adjacent rows (or columns) merely inter-
changes two adjacent suffixes (or letters) in every term,
that is, produces one inversion of suffixes (or letters) in
every term without disturbing the order of the letters
(or suffixes), that is, changes the sign of every term, and
affects the terms in no other way.
394. An interchange of any tivo rows (or columns) of a
determinant changes its sign, but affects its value in no
other way.
For, suppose k rows to lie between the two rows that
are to be interchanged. These k rows, together with the
two rows to be interchanged, form a series of A: + 2 rows.
Then we may make the interchange by passing the upper-
most row of the series downwards over k rows, and then
passing the lowest row upwards over A; + 1 rows. The
total number of interchanges of adjacent rows is there-
fore 2k + 1, an odd number, and each such interchange
changes the sign of the determinant. Hence the total
effect is a change of the sign of the determinant.
The proof for the effect of the interchange of any two
columns is identical with the foregoing.
395. A determinant, in which any two rows (or columns)
are identical, is equal to zero.
For, if two rows (or columns) be identical, the deter-
minant is unaltered, either in sign or magnitude, by the
570
DETERMINANTS.
interchange of these two rows (or columns). But, by-
Art. 394, the interchange of any two rows (or columns)
of a determinant changes its sign. Hence the determi-
nant in question is not altered in value by changing its
sign, and its value must therefore be zero.
Ex. 1. Determine the value of
1
a aP-
1
b &2
1
c c2
Two rows of this determinant become identical and its value
zero when a = b. Hence it must have a — b as a factor [Art. 148].
For a like reason it contains b — c and c — a as factors. But the
determinant is of the third degree in a, 6, c. Hence, denoting its
value by A,
A = Z(6-c)(c-a)(a-&)
where Z is a number.
But the principal term of the determinant is bc^, and this is the
only term in which bc^ occurs, and the coefficient of bc^ in
L(b — c) (c — a) (a — b) is L ; .•. L = 1, -,
A=(&-c)(c-a)(a-6).
Prove that
i37(l-/3)(^-7)(7-l)i
ai37(tt-i3)(/3-7)(7-a).
and
Ex. 2.
Ex. 3. Prove that
1
/3
7
1
/32
y2
1
^3
7»
/3 7
./3«
396. If all the elements of one row, or of one column,
of a determinant be multiplied by the same quantity, the
determinant itself will be multiplied by that quantity.
DETERMINANTS.
571
For, every term of the determinant contains one
element, and only one, from each row and from each
column [Art. 390] ; and hence, if all the elements of
one row, or of one column, be multiplied by the same
quantity, every term and, therefore, the sum of all the
terms, will be multiplied by that quantity.
397. A determinant in which any tivo rows, or any two
columns, differ only by a constant factor, is equal to zero.
This theorem is a direct consequence of Arts. 396, 397.
For example :
0.
ma na d
= mn
a a d
nib nb e
b b e
mc nc f
c c f
398. Def. When any number of columns and the same
number of rows of a determinant are suppressed, the
determinant formed by the remaining elements is called
a minor determinailt.
The minor determinant obtained by suppressing one
column and one row is said to be of the first order, or to
be a first minor; similarly the result of suppressing two
columns and two rows is called a second minor ; and so on.
The result of suppressing the column and row through
any element a; of a determinant A is called the minor
corresponding to that element and is denoted by A,.
Thus the first minors of A = ai b\ c\
Gi 62 C2
as ^3 • cs
corresponding to the elements ai, 61, ci,
respectively.
&2 C2
ai C2
aa 62
Aa,=
, Aft =
, Ac =
68 Cs
\
as Cs
1
as 6s
572
DETERMINANTS.
It is evident that when, in the first row or first column,
all the elements except the first vanish the determinant
reduces to the product of the non- vanishing element and
its corresponding minor.
ai &i Ci
= «!
62 C2
0 &2 C2
63 C3
0 63 Cs
Thus
399. To expand a determinant of the n^^ order in terms
of n of its first minors.
Since every term of the determinant [afiz'^s •*• ^n] con-
tains one and only one of the elements a^, ag? ctg, ••• a,^
[Art. 390], we may write
= a^Ai 4- ^2^2 + 0^3-^3 -\ 1- a„ A? («)
a,
61
c,..-h
012
h
Ci — k
a.
h
Cs-ls
a„
bn
Cn-L
ai
bi ci
az
62 C2
as
63 C3
in which none of the coefficients Ai, ^2^ ••• ^„ contain
any of the elements ai, a^^ ••• a„.
For example,
ai(&2C3 — 63C2)+«2(&3Cl — 61C3) + «3(&lC2 — &2C1).
The quantities A^, Az, -" A^ are called the co-factors of
ai, ttg, ••• a„ respectively. We require to determine these
co-factors in terms of the minors of a^, (Xg, ••• a„.
Por this purpose let the severalrows ag \"'h^ % ^3 ••* ^sj
• • • a„ 6„ • • • ?„, in the matrix form of the determinant
[ai&2C3-"^„]? be alternately moved into the place of tho
DETERMINANTS. 573
first row. Then the determinant, without being changed
in magnitude, assumes successively the forms
...(_l)«-l[aAC2f^3-^n-l];
for, when any row, say the fc'**, is moved into the place of
the first row, it passes over k — 1 rows above it, and by
such a transference the determinant does or does not
change sign according as k is even or odd [Art. 393].
Hence (a) may have the following several forms :
[aibiCsd^ • • • Z„] = a^Ai + «2-42 H h a„ A, (a)
- [0261(53(^4 • • • ^n] = dlA + ^2^2 + — + a^A, 08)
[036102^4 •••?„]= ai^i + 02^2 H h a« A, (y)
— la^jbic^ds •••?«] = tti^i + «2^2 H h «« A> (S)
(_ 1 ) -1 [a„6iC2 • • • Z„-i] = cti^i + ^2^2 H h «« A- (^)
Now, in (a) put 02= a3= ••• = a„ = 0 ; then [ai^a^s • • •/„]
becomes cii[62C3 •••?„], that is, OiA^ [Art. 398], and there-
fore
ai^i=aiA^^.
In (/3) put ai= a3= 04= ••• = a„= 0 ; then [a26iC3 •••?„]
becomes a2[6iC3 •••/„], that is, a2^a,> ^°^ therefore
«2^2 = — <*2^a,-
From (y), (8), and (X), we obtain in like manner,
03^3 = as^^,
674
DETERMINANTS.
Hence, if A stand for the original determinant
If columns and rows be interchanged, we obtain
A = aAa, - b,\ + cAc, ( - 1)"-'^A^,
or, if the first and second columns be interchanged,
A = - bA,^ + ^2^5^ - b,\ +...+(- ^rK\ ;
and it is now evident that the expansion may be made in
terms of the minors of any row or any column.
400. If the elements of one column (or row) of a deter-
minant be multiplied in order by the co-factors of the corre-
sponding elements of any other column {or row), the sum
of these products will be zero.
In the determinant
ai
6i-..A;i...
ag
bz-'h"'
«3
bs"'ks"'
let the elements ki, kz, k^,'-' be multiplied in order by
the co-factors Ai, A^, A^, •••of the elements ai, ag, 0^3, •••,
and the products added together. The result is
kiAi -\- k2A2 + k^A^ + • • •;
DETERMINANTS,
whose matrix form is
k.
6, •••*;, •••
h
hi — k^--
h
h-h-
575
[Art. 399.]
But this is zero, since two of its columns are identical.
[Art. 395].
The result is the same when these combinations are
made with any two columns or rows ; for any column or
row may be moved into the place of the first column
without changing the magnitude of the determinant.
401. If each element of any column (or row) be a bino-
mial the determinant can be expi^essed as the sum of two
determinants of the same order.
For, if Ai, A2, As,'" be the co-factors of the elements
of the first column of
tti -h «! bi Ci .
0.2 + "2 ^2 ^2 •
ag + Og 63 Cg .
= A,
then, by Art. 399
A = (aj + ai)^i + (a2 + «2)A + (% + 03)^3 + —
= tti^i -f a2^2 + «3^3 H h «l^l + «2^2 + (h^z +
tti 61 Ci
0^2 ^2 ^2
% ^3 ^3
+
«1 61 Ci . .
"2 &2 Cg . .
«3 &3 Cg . .
576
DETERMINANTS.
402. A determinant is not altered in value by adding to
the respective elements of one column {or row) the same
multiples of the corresponding elemeyits of any other column
{or roiv).
For, if in the expanded form
aiAi + a2^2 + a^As + ... = A,
of tlie determinant
a, bi-
..fc,...
^2 h
..fc^.-.
a, 63.
..^3-
0 .
•
. .
.
we replace ai, as, a^, •••by ai + mki, ag + m7c2, a^ + mk^,"-
m being any number, this expanded form becomes
(% + mki)Ai 4- («2 4- wfcg) A + («3 + 'mks)As H
= A + m {kjAi + A;2^2 + hA^ -\ ) .
But by Art. 400
kiAi + A;2^2 + Ms + ••• = 0 ;
ai-\-mki bi"'ki"
a2 + mk.2 b2"'k2"
a^ + mk^ bs*"ks-
Ex. 1. Show that
1 1 3
12 2
1 5 -1
= 0,
DETERMINANTS.
677
By adding the second column to the third the determinant
becomes
= 0.
1 1 4
= 4
111
1 2 4
1 2 1
1 5 4
1 5 1
Ex. 2. Show that a 6 1
h a 1
a + h 0 1
By adding the second column to the first the determinant
becomes
= 0.
a + h b 1
= (a + 6)
1 b 1
b -\- a a 1
1 a 1
a + b 0 1
1 0 1
Ex. 3. Show that b + c c a — c = ab(b — c).
2b b a-b
b-^ a a a
Subtract the second column from the first and add it to the
third ; the determinant becomes
b c a
= ab
1 c 1
b b a
1 b 1
b a 2a
la 2
Subtract the second row from the first ; then
ab
1 c 1
1 b 1
1 a 2
=:ab
0 c-6 0
1 b 1
1 a 2
= ab{b — c).
403. To express the product of two determinants {of
the second order) as a third determinant of the same order.
As a preliminary step in the demonstration, we tirst
establish the following lemma :
578
DETERMINANTS.
If ai bi m n
a2 h p q
0 0 ai A
0 0 a2 ft
then, whatever he the values of m, n, p, q,
A = Ai • Aa.
= A,
a, 6i
a2 h.
ai ft
a2 ft
= A^
= A,
For, 'A = aj
h V q
0 ai ft
0 a2 ft
a2
6i m n
0 ai ft
0 a2 ft
= a-J)2^2 — (^^A2 = ^1 • ^2-
Q.E.D.
Now in A put m = q= —1 and n=p = 0, so that
ttj 6i — 1
a2 62 0
0 0 ai
0 0 a2
0
-1
ft
ft
Then multiply the first two rows by ai, ft respectively,
and add the products to the third row; and again,
multiply the first two rows by a2, ft respectively, and
add the products to the fourth row. The result of these
operations is
A =
a2
«!«! + a2ft
a^a-i + a2ft
l>2
&!«! + &2ft
&1«2 + &2ft
-1 0
0 -1
0 0
J 0 0
=
by the lemm
-1 0
0 -1
a. But
aiCJa + a2ft
-1 0
0 -1
&1«1 + ^sft
&1«2 + &2^2
= 1;
DETERMINANTS.
579
cfri 6,
«i ft
=
aa h
«2 ft
ax h\ C\
and A2 =
ai /3i 71
Qi 62 C2
a2 ^2 72
as &3 C3
as /33 73
aicti H- cisft &i<^i + ^2/?!
aia2 + ^2/^2 ^1*^2 4- ^2ft
Note. — This method of proof is general, and may be applied to
determinants of any order. For determinants of the third order
the result is as follows :
If Al:
then
A1A2 = aitti + a2/3i + a37i ?>iai + b^^i + b^yi ciai + C2/3i + C371
ai02 + a2i32 + 0372 &ia2 + &2i32 + &372 Cia2 + C2i32 + C372
aias + a2^3 + 0373 bias + b^^s + &373 Cias + c^^a + C373
For the proof in detail of this more general case the student is
referred to Smith's Treatise on Algebra, Art. 430.
By virtue of the fact that a determinant is not altered in magni-
tude by changing the order of its rows or columns, or by inter-
changing columns and rows, the product of two determinants may
be exhibited in various forms.
404. To solve the simultaneous equations,
aiX-\-biy-\-CiZ-\-dit = kif
ttgX + b2y + cji -\- d^ = kzf
a-sX + 63?/ + c^z + d.t = ^3,
a^x + 642/ + c^z + d^jt = ^4.
Multiply these equations in order by A^, A^, A^, A^, the
co-factors of cii, a^, a^, a^ respectively in the determinant
laib./^sd^'], and add the resulting equations together. Then,
by Art. 400, the new coefficients of y, z, and t vanish, and
there remains
^a^Ai -f ttaA + «3^3 + (iiAi)tc = kiAi -h A;2^2 + hA^-^k^A^,
that is, laib2Cs,d4]x = Ikib^c^d^].
580
DETERMINANTS.
By using as multipliers the co-factors of b^, 62? ^3? ^4,
we obtain in a similar manner
and the symmetry of these relations shows that we shall
have for the values of z and t
laib2Csdi\z = [aib^hd^'],
[0162^3(^4]^ = \^aib2C.Jc^'].
This method is obviously general, and readily appli-
cable to simultaneous equations with any number of
unknown quantities.
405. Eesultants. A resultant, or an eliminant as it is
also called, is a determinant the vanishing of which is
the necessary and sufficient condition in order that two
or more equations may be satisfied by the same value or
values of the unknown quantities involved. Thus the
resultant of the equations
ax-\-by = 0,
a'ic + 6'?/ = 0
is aV — a'& [Ex. 1, Art. 386], and the resultant of the
equations
a^x + b^y -f c^z = 0,
ttgic + boy + c^z = 0,
o^^ + hy + C32; = 0,
is tti bi Ci
C(/2 O2 C2
^3 Oq C3
[Ex. 3, Art. 386.]
DETERMINANTS.
581
Ex. 1. Find the resultant of the equations
aic2 + 6x + c = 0,
a'«2 + b'x + c' = 0.
Multiplying each equation by x we obtain the set of four equa-
tions.
ax^ + 6x2 + ex + 0 = 0,
0 x8 + ax2 + 6x + c = 0,
a'x^ + 6'x2 + c'x + 0 = 0,
0x» + a'x2 4- b'x + c' = 0,
which may be treated as simultaneous equations in the four quan-
tities x**, x2, X, 1.
Multiply these equations in turn by the co-factors of the fourth
column of the determinant formed by the coefficients a, b, c, 0, etc.,
and add the resulting equations together. Tlien, by Art. 400, the
new coefficients of x*, x^, and x all vanish and there remains
a 6 c 0 =0.
0 a 6 c
a' h' c' 0
0 a' V d
This determinant is the resultant required and its vanishing is
the condition necessary (it may also be proved sufficient) in order
that the equations
ax2 -f 6x + c = 0, a'x2 + 6'x -f C = 0
may have a common root. [Compare Ex. 3, Art. 173.]
Ex. 2. Find the condition necessary and sufficient in order that
the equations
ax8 + 6a;2 + ex + (i = 0,
px2 + gx + r = 0
may have a common root.
682
DETERMINANTS.
Multiplying the first equation by x and the second by cc^ and x
we obtain the set of five equations
ax^ + hx^ + cx^ -{- dx =0,
ax^ + 6x2 + cic + (? = 0,
px^ -\- qx^ + rx^ = 0,
px^ + qx'^ -\- rx =0,
• px^ + qx-\-r = 0.
If these are to be satisfied by the same value of x we must have
a h c d (i =0,
Q a h c d
p q r 0 0
0 p q r 0
0 0 p q r
which is the required condition.
The method of elimination employed in these examples is
called Sylvester's dialytic method.
EXAMPLES XCIX.
1. Transform a h c
d e f
g h k
into an equivalent determinant having : (1) de/as its first column,
(2) cfh as its first row, (3) adg as its third row.
2. Solve by determinants
3x + 2?/ + 2 = 0,
5x+12y-l=0.
3. Solve by determinants
5x + y - 40-3=0,
2x4-3^ + 70 + 4 = 0,
3x -22/+ 50- 7=0.
DETERMINANTS.
583
4. Find a value of k such that the equations
2a;-3y + l = 0,
Zx - by -\-k = (i,
x + y -7 =0,
may be simultaneous in x and y.
5. Solve by determinants
x-2y-3z -\-2t = 2,
5x-{-y-\4tz + 6t = 3,
6x-Sy -h z+ 10« = 0,
7x-2y-\-2z + 2t=-2.
Evaluate the follov\ring determinants
6. 8 2 8 10.
11
8
2
8
7
5
6
9
4
1
a — h
X
X
X
b-c
X
X
X
c — a
0
h
9
h
0
f
9
/
0
11.
1 4-a
1
1
1
1 + 6
1
1
1
1 + c
9.
14.
4 3 7
2
18.
1 a a2 a8
6 0 2 1
1 6 62 6«
6-210
1 c c2 c8
3 6 4 3
'
1 d (22 (^
•
X a — b
13.
b — c c b
— ax c
- c c — a a
b — c X
•
- b —a a — b
le f oUov^ing ide
ntities :
b-\- c
a - c
a-b
= 8 abc.
b-c
c-i- a
b-a
c -
-b
c - a
a -\-b
584
DETERMINANTS.
15.
16.
17.
18.
&+c+2a h c
a c + a + 26 c
a b a + b + 2c
= 2(a + 6 + c)8
aixi + biX2 aivx + bxy^
a^xi + b^Xi a^yi + &22/2
1 X x^.
•
1 2/ 2/2
1 5? 2!2
a\ a^ as
6i 62 &3
Ci C2 C3
and therefore,
ai bi ci 2
^2 ^2 C2
as bs cs
Ai Bi Ci
=
A2 Bi Ci
As Bz Cs
aixi + biyi aiX2 + &12/2 I
a2a:i + &22/I «2a;2 + ^22/2 I .
(w + x)2 (v + xy (w + a-) '2
(W + 0)2 (V + 0)2 (to + 0)2
«! ^2 «3
61 62 &3
Ci C2 C3
^1 ^1 Oi
A2 B2 C2
^3 -B3 O3
where ^1, J5i, ^2? etc., are the co-factors of ai, 5i, a2, etc., in the
determinant [ai&2C3].
19. If X + y + 0 = aic + 6y + C0 = 0, prove tl^^t
X y z =0.
cab
b c a
20. Find the condition necessary and sufficient in order that
ax2 + 6x + c = 0 and x^ = l
may have a common root.
21. Determine X and Fin terras of «, 6, x, and y, such that
a 6
•
a; 2/
=
X Y
b a
y X
Y X
RATIONAL FACTORS AND HIGHER EQUATIONS. 585
CHAPTER XL.
Rational Factors and Higher Equations.
406. To Factor 7? -\-f + :^ -Zxyz. In Art. 126 it was
seen that this ternary cubic is identically equal to
(« + ?/ + z) {x^ + if + z^-yz-zx- xy\
so that one of its linear factors is x -\-y -\-z. Arrange
the terms of the second factor as a quadratic in x, and
add and subtract {y -\- z)y4, thus :
a;2_ (,jj^z)x + ^1±^ - {y-±ll + f+z'-yz. [Art. 122]
In more condensed form this is
or, as the difference of two squares,
(-^)'-e^-T)"
Its two factors, rational and linear, in x, y, z, are now
easily written down [Art. 115]. When simplified, by
collecting coefficients, they are
x-\- (x)y -\- <D% x -\- (Jy -\- (oZf
586 EATIONAL FACTORS AND HIGHER EQUATIONS.
where, for brevity, w and a>^ are written in place of
-i + iV^=^ and -^-^V^
which the student will recognize as the two complex
cube roots of unity. [Art. 188.]
Hence the three rational factors of the first degree of
Q^-^-f-^-z'^—Zxyz are x + y-{-z, x + o}y-\-(o% x+oy^y+wz,
where to and o>^ are the complex cube roots of unity.
407. To Pactor the Cubic in x: To factor the special
cubic in a;, of the form a^+qx-\-r, add and subtract
— y^ — z^ — SxyZf producing the identity
a^ -{- qx -\- r = a^ — y^ — :^ — S xyz + y^+ :^ -\-xCSyz-hq) + r
and determine y and z by the conditions
. f + ^ = — r,Syz = — q.—'
Then at once [Art. 406]
a^ + qx + r = a:^ — y^ — ^ — S xyz
= (x — y — z)(x — wy — mh) (x — <i?y — ws;),
where w, <o^ are the complex cube roots of unity.
From the two equations in y and z is obtained, by
eliminating z^ the quadratic equation
or 2/6^^^_|_^0^
whence 2^ = _r + ^(J + g,
RATIONAL FACTORS AND HIGHER EQUATIONS. 587
408. The general cubic expression x^ -\- p7? -\- qx-\-r
may be reduced to the special form y^ -\- Qy -\- B. By
easy reductions we obtain the identity
(y + hy ^p(y-^hy-^q(y + h)-\-r = f-{-(p -[-Sh)y'
^(q-\-2ph-^Sh^y^r-\-qh-^ph^-\- h\
The coefficient of y^ will vanish if h = -^^p'^ hence
the substitution x = y — \p will effect the required trans-
formation. The new coefficients are
R = r-\pq-\-^p\
and the identity sought for is
y?^px'J^qx + r={x + \pf+Q{x-\-\p) + R.
The factors of this general cubic may now be written
down in accordance with the formula of the last article.
Thus
a?-\-p7?+qx-\-r
= (x-\-\p-T-Z)(x-\-\p-iaY-u?Z)(x-\-\p-iJ^T-ioZ),
where r« = -:? + A/^' + ^
2"^VT'^27'
Z3 = -^-J^V^.
2 A' 4 ^27
409. The three factors of the last article yield at once
the roots of the cubic equation
Q? + p7? -f ^-a; + r = 0.
588 RATIONAL FACTORS AND HIGHER EQUATIONS.
These roots are
x = -lp-\-<oY-j-(x)^Z,
When p — 0, the first of these expressions for the
value of X becomes
which is known as Oardan's Pormula, for the solution of
the equation
410. To Factor the Biquadratic in x : Let it be assumed
that the general biquadratic
s^-^pof + qx^ -^rx±s
can be expressed as the difference of two squares of the
form
(x^ + ax-j-ipy-(bxA-oy,
and let the values of a, 6, c, p be sought which will make
such a reduction possible. Expanded and arranged ac-
cording to the powers of x, this difference becomes
a;* + 2aaj3 + (a^ -f p - 62)ic2 4. (^^ _ 2 bc)x-\- ip' - c^.
This will be identical with the given biquadratic expres-
sion if
2a=p, a^ -\- p — b"^ = q,
ap — 2bc = ry ^p^ — (^=:s,
BATIONAL FACTORS AND HIGHER EQUATIONS. 589
and these four equations are sufficient to determine a, h,
c, and p. Eliminating a, we have
2bc = ^pp — r, 4:b^=p^ — 4:q + 4:p, c^ = \p^ — s]
whence {p^ — 4:q -\- 4: p) (\ p^ — s) — (^pp — rf = 0,
or p^ — qp^ -{- (pr — 4:S)p—p^s-\-4:qs — r^ = 0.
Thus the solution of a cubic equation (the reducing cubic
it is called) is a necessary part of this transformation.
When one root of this cubic equation is found, the values
of a, b, and c are at once known, and the biquadratic
form is then expressible as the product of the two quad-
ratic factors
lay^^(a-\-b)x + ip + cl'\x'-\-(a-b)x-{-ip-c\, .
each of which may be resolved into the product of two
factors of the first degree by the method of Art. 122.
411. The roots of the biquadratic equation
iC* -i-pa^ + qa^ + ^^ + ^ = 0
are now obtained by solving the two equations
a^ 4- (a 4-^)2; + ^p + c = 0,
a^-{-(a-b)x + ^p-c = 0.
These roots then are
x = -^(a + b±V(a^y-2p-4:c)y
oc = -^(a-b± V(a-by-2p-\-4:c),
a, 6, c, and p being obtained by means of the equations
P^ — QP^ + (P^ — 4: s)p—ph-\- 4. qs — 7^ = 0,
a = ^p, 6 = ^Vi>^ — 4g + 4p, c = iV/3^ — 4s.
690 EATIONAL FACTORS AND HIGHER EQUATIONS.
The general algebraic solution, in terms of radicals, of
equations of higher degree than the fourth has been
shown to be impossible.
.EXAMPLES C.
Solve the following cubic equations :
,^ 1. 2 x3 + X - 3 = 0. ^^ 4. a;3 + 12 5c - 12 = 0.
2. x^-60x- 189 = 0. 6. x^-6 a^x - 9 a^ = o.
^3. x^-Qx-Q = 0. 6. x3-6ax2-6a%-2a3 = 0.
Transform the following expressions into the products of quad-
ratic factors : >
7. 9x4-27x3 + 18x-4.
8. X* + 4 x3 + 4 x2 - 5 X - 12.
9. x* - 3 x3y + 2 xV - »^?/^ - 2/*»
10. x^ + 4 x^ - 2 x2 - 12 X - 6.
11. 4 x* + 8 x^i/ - 4 xr/3 - yK
12. Show that
ax3 + 3 hx^y + 3 cxy'^ + dy^ =p(x + XyY + g(x + fiyy,
.. . 1 F + ^A 1 -F-yA
and p, g = K« + ^«^-4J2'VA)» i(a-Va2_4^V^);
where i^, H=ad — be, ac — &2,
and A = (ad - 5c)2 - 4 (ac - 62) (bd -c^).
412. To Factor the General Quadratic inx, y:
aa:^-\-2hxy + %' + 2^a;+ 2/^ + c.
This expression, arranged according to descending powers
of X and multiplied by a, has the form
(axy-^2{hy-\-g)ax + a(by' + 2fy-{'C).
BATIONAL FACTORS AND HIGHER EQUATIONS. 591
It may here be regarded as a quadratic in arc, and may be
factored by the method of Art. 122. Thus, adding and
subtracting (Tiy -\- gfy ^^ ^^^^
{axf ■\-2Qiy^ g) ax + Qiy + gf- (hy+gy+a(by'-h2fy-\-c)
= (ax + hy + gy - IQi' - ah) f -2 (af- hg)y + ^^ _ ^^
= {ax ■i-hy-\-g + ^D) (ax-{-hy + g — y/D),
where D stands for (h^ — ab) y^ — 2 (af— hg) y -\-g^ — ac.
If these factors are to be rational integral functions of
X and y of the first degree, D must be a perfect square,
the condition for which is
(h^ - ab) if - ac) - (af -hgy = 0, [Art. 179]
or a^hc + 2 afgh — a^f^ — abg^ — ach^ = 0 ;
or, since a is not zero,
abc + 2fgh-af - bg'-ch^ = 0.
It is easy to verify that
abc + 2fgh - af - bg^-ch^ =
[See Art. 385, Ex. 2.]
This expression (function of the coefficients) is called
the discriminant of ax^ -\- 2 hxy + • • • + c, and is usually
denoted by A. The first part of our conclusion now is :
If aa^ 4- 2 hxy + • • • + c is resolvable into linear factors,
then necessarily A = 0.
But conversely, if A = 0, then, taking positive values
only of V/i^ — ab and ^/g^ — ac,
a
h
9
k
b
f
9
f
c
af—hg = V/i^ — ab • -Vg^ — ac, if af— hg > 0,
(af— hg) = -y/h^ — ab • VgT—aCf if af—hg<, 0.
592 EATIONAL FACTORS AND HIGHER EQUATIONS.
In the former case
in the latter
D = ( V^' - ah . y +^g^-aG)\
and a (ax^ + 2 hxy -\- by^ -\- 2 gx + 2fy + c)
is (ax + % 4- gy — (Vh^ — ab • y — -Vg^ — ac)^,
if af— hg > 0,
but is (ax + hy -\- gY — ( V/^^ — ab • y + V^^ — ac)^,
if af—hg<0.
These differences of squares are immediately resolva-
ble into the product of rational factors of the first degree
in X and y. Our result is :
A rational integral function of the second degree in x, y
may be resolved into the product of two rational integral
factors of the first degree if and only if the discriminant
be zero.
Carried to its conclusion, the factoring process gives
us the formula
ajx' + 2hxy + by'' + 2gx + 2fy + G
= - [ax + (^ + V/i^ — ab)y-\-gT '\/g^ — ac]
X [ax 4- (h—-\/h^ — ab)y + g ± V/ — ac],
in which the upper signs are to be taken if af—hg> 0,
the lower if af— hg < 0.
Ex. 1. Factor 2x^ - xij - y^ -{- Sy -2.
Here h^ — ab = f , g"^ — ac = 4, af — hg = 3. The upper signs
are to be taken, and the factors are 2 aj + y — 2, x — ?/ + 1.
RATIONAL FACTORS AND HIGHER EQUATIONS. 593
Ex. 2, Factor 2 x'-^ - 2 icy - y^ - 2 x - 2 y - 1.
Here Ti^ - ab = S, g"^ - ac = S, af— hg = — 3. The lower signs
are to be taken, and the factors are
(V3 + l)aj + 2/ + l, iy/S-l)x-y-l.
EXAMPLES CL
Find the factors of such of the following expressions as are
factorable ;
1. x2-2/2_aj + 3y-2.
2. 2x^-2y^-x-Sy-l.
8. 2 x2 - 3 xy - 2 1/2 4- 3 x + 4 y - 2.
4. x^ + xy-y^ + Sx-1.
6. x'^-4xy + Sy^ + 2x-2y.
6. 2x^-7xy + Sy^-3x~y-2.
7. 2x^ + bxy-}-2y^-x + y-l.
8. 2x^-Sxy-2y^ + x-\-lSy-16.
9. x2 - y2 + (1 _ c) x + (1 + c) y - c.
10. 2 x2 + 2 xy + y2 + 6 a; + 6 y + 9.
11. x2 - xy + 2/2 - X - y 4- 1.
12. 2x2-4xy + y2_2/y-/2.
Determine the value, or values, of c,f, a, and g which will make
the following expressions separable into factors of the first degree
in X and y :
13. x2-y2_7a; + 9y-c.
14. x2 - 5 y2 + 3 a; + 2/y + 1.
15. grx2-y2+25rx-3y-l.
16. 0x2 - 2 xy + 3 y2 _ X 4- 2 y + 2.
17. ax2 + 2 xy - y2 + 3 X - 4 y - a.
18. Find the relation between / and b which will make
x2-by^ + Sx-}-2fy + l
resolvable into factors of the first degree in x and y.
2q
594 RATIONAL FACTORS AND HIGHER EQUATIONS.
413. Simnltaneoiis Equations. If each of a set of simul-
taneous equations of higher degree can be factored, their
solution may be reduced to that of a set of simultaneous
equations of lower degree. Suppose that
are simultaneous equations in x and y, and that <^ and \f/
can be resolved into factors of the first degree, say
<l> = l'm'n, i{/ = p'q.
Since </> and j/r are simultaneously zero if, and only if, a
factor in each is zero [Art. 127], all the solutions of
cfi = 0, \l/ = 0 will be obtained by solving, in turn, the fol-
lowing pairs of equations of the first degree :
Z = 0, j9 = 0; m = Oj p = 0'j n = 0, p = 0;
1 = 0, q = 0'j m = 0, q = 0'j n = 0, q = 0;
and obviously there can be no others. The two groups
of equations, the <^-j/^ group and the l-m-n-p-q group, are
called equivalent systems.
Def. Two systems of equations are equivalent when all
the solutions of either are identically the same as those of
the other.
Ex. 1. Solve the simultaneous equations ic^ _ ^2 _ 2 a; -j- 1 = 0,
x2 - 4a;^ + 3?/2 + X - 31/ = 0.
The factors of the first are x + y — \ and x — y — 1] of the
second, x — ^y and cc — y + 1. Hence an equivalent system is
(1) a; + 1/ - 1 = 0, x-^y = 0
(2)x + y-l=0, x-y-\-\ = (i
(3) x-y -1=0, x-Sy = 0
(4) X - y - 1 = 0, x-y+l = 0
the solutions of which are respectively
a;, y = i i ; 0, 1 ; I, i ; and 00, co.
RATIONAL FACTORS AND HIGHEB EQUATIONS. 595
Ex. 2. Solve the simultaneous equations
z^ + y^ - 10 = 0, x^-2xy -{-y^-2x + 2y = 0.
Since x^ - 2 xy + y"^ - 2x + 2 y = (x - y - 2) (x - y), an equiva-
lent system here is
(1) x-y-2 = 0, x2 + 2/2_io = 0; *
(2) x-y = 0, x2 + y2_io = o.
The solutions are easily found to be :
for the first set, x, y = 3, 1, or — 1, 3 ;
for the second set, x, y = y/6, y/6, or — ^^5, — ^5.
414. The foregoing method of solving a pair of simul-
taneous equations, <^ = 0, «/' = 0, fails unless either <f> or
i}/, or both, can be resolved into integral factors of the
first degree, and it is not usually true that an integral
function of the second (or higher) degree in x and y has
such factors.
When <^ and ij/ are both of the second degree, however,
it is always possible to find a quantity k, such that the
quadratic function (fi-\-k\p shall be resolvable into two
factors of the first degree. For this purpose we have
only to put the discriminant of <^ + kx^/ equal to zero and
solve the resulting equation in k [Art. 412]. We then
substitute the system <^ = 0, <^ + fci/^ = 0, for the system
<^ = 0, i/^ = 0.
Suppose a value of k found for which the discrimi-
nant vanishes ; then
<t> -\- kip = I • m,
where /, m are of the first degree, and the system of two
sets,
(1) <^ = 0, Z = 0; (2) <^ = 0, m = 0;
is now equivalent to the one set, <^ = 0, ^ = 0.
596 RATIONAL FACTORS AND HIGHER EQUATIONS.
Let us apply the method to an example.
Ex. 1. Solve x"^ + 2xy - y"^ - 2 X + 2 y + 1 = 0,
a;2 + 2/2 _ 5 = 0.
The function <(> + k\}/ in this case is
x^ + 2xy - y^ ~ 2x + 2 y + 1 + k(x^ + y'i - 5),
or (i^k)x'^ + 2xy + (k-l)y'^-2x + 2y + (il-5k);
and its discriminant is
(A; + 1)(A; - 1)(1 - 5A:)- 2 -(1 + A;)-(A:- !)-(! - 5A:),
or -^k^^k^-^-Sk-i.
This is zero if A; = 1. For this value of A:, 0 + k<p becomes
2x'^ + 2xy -2x + 2y -4,
or in its factored form,
2(x + l)(x + y-2).
Our problem is now reduced to the solution of
x + 1 = 0, x2 + y2 _ 5 = 0,
from which we find
x,y=-l,2, or a;, y = - 1, - 2,
and of X + y - 2 = 0, a;2 + 2/2 _ 5 _ 0,
whose solutions are
x,y = l + ^V6, 1-^V6, or x,y = l-^y/6, l + jVC.
Ex. 2. Solve ic2 + 2/2 + 4 a; - 1 = 0,
a;2 _ ^2 _ 2 y _ 1 = 0.
The factored form of ic2 — 2/2 — 2 2/ — 1 is
(x-y-l)(x + y + l),
and the discriminant of
ic2 4- 2/2 + 4 X - 1 + A: (x2 - 2/2 - 22/ - 1)
is O.F + O.A;2 + 3A;-6.
RATIONAL FACTORS AND HIGHER EQUATIONS. 597
Hence (x — y — l)(x + y -\- 1) = 0,
r.2 ^ y2 ^ 4x - 1 + ^(x^ - y^ -2y - 1) = 0
is an equivalent system. The last equation reduced is
4 a;2 _ 2/2 + 6x - 5 2/ - 4 = 0,
or in its factored form,
(2x - y - l)(2x + y + i) = 0.
Hence the following is another equivalent system :
(l)ic-y-l = 0, 2a;-2/-l=0; x,y = 0,-l;
(2) x-y-l = 0, 2x-\-y-\-i = 0; x,y = -l,-2;
iS) x + y+\l = 0, 2x-y-l=0; x,y = 0, -1;
(4:)x + y + l=0, 2x-\-y + 4 = 0; x,y = -3,2.
The solutions are written at the right of each pair of equations.
EXAMPLES OIL
Solve the following systems of simultaneous equations :
1. jc2+2«2/ + y2-3x-3y+2=0, 2x:2-Sxy-\-y^-5x-\-3y-\-2=0.
2. Sx^-{-Sxy-Sy^=0, ix'^-ixy+y'^-4x+2y=0.
3. x'^-^2xy-Sy^-6x+6y=0, xy + y-2=0.
4. x^+xy-Sy^-Sx+4y-\-9=0, xy-2z=0.
6. 2x2-4a:y-j/2_8x+8y+33=0, a;2-4aj+i/2-21=0.
6. x'^-j-xy + y'^+x-\-y-S=0, xy-{-2x+2y-^i=0.
7. x^-\-xy-~y^+ax—ay+c=0, xy — ax+ay+c=0.
8. x^+Sxy+2y^+x+y=0, a;«+y«+(a;+y)(a;-2)(2/-l)=0.
9. 2x^+xy-4x-l=0, y'^-^2xy-^y+l=0.
10. x?-[-y^-\-l-Sxy=0, xy+6=0.
11. x^+y^+z^—Sxyz=0, ax-\-by-{-cz=l, lx+my+nz=0.
12. Solve x^ + px^ + gx2 4- rx 4- s = 0 by assuming x2 = y, and
assigning to X a value that will make
y2 + pxy -\. qy ^ rx + s + \{x^ - y)
break up into linear factors in x and y.
598 horner's method.
? + ^>0;
CHAPTER XLI.
horner's method.
415. The cube roots of 2/^ and z^ in Cardan's formula can-
not be found by the arithmetical cube-root process unless
4 ' 27
and even when this condition is satisfied the various re-
ductions are frequently quite laborious. In the solution
of numerical equations a process of direct numerical
approximation is found to be more expeditious. Horner's
Method, first published in 1819, based on Horner's process
of synthetic division [Art. 83], is applicable to finding
the real roots of equations of any degree.
As a preliminary step in establishing Horner's method
we require the proposition of the next article.
416. To transform a multinomial in x into a multi-
nomial in (x — a). J"
For our present purpose it will be sufficient to apply
this transformation to a multinomial of the fourth
degree.^*
When the binomials (a + ^)*, (a-\-hy, {a + lif, in the
expression
M=za{a + hy-\-p{a-\- lif + g(a -t- hf + r (a H- /i) + s,
* See Art. 446 for a discussion of the general case.
hoener's method. 599
are expanded, and the coefficients of the several powers
of h are collected, the result must be of the form
in which ijg, R^, i?i, Rq involve a but not h. li x — a
replace h in these two equivalents of Jf, the result Is the
identity
ilif = aic* 4- pic^ -\- qx^ -\- rx -{- s
^a{x-ay -\- Rz{x-ay + Ri{x-af -\- R^{x-a) -\- Rq.
Let M be divided by a; — a, giving as quotient Mi ; the
remainder is Rq and
Jfi = a (ic - af -\-R^{x- af + i?2 (a; - a) + R^j
Jf= Jfi(a;-a) + i?o.
Similarly, let Jfs, J^fg, Jf4 denote the successive quotients
of Ml hy x—a, M2 by ic— a, and M3 by a;— a, respectively ;
then
Mi = M2{x — a) -f i?i,
M2 = Ms(x — a) -\- R^f
Ms = J/4 (a; - a) + i?3,
3/4 = a.
Thus the coefficients Rq, Ri, R^, R3, of the transformed
multinomial, are the successive remainders in these
several divisions, and a is the last quotient, namely, the
quotient which does not contain x — a.
It is obvious that, although the multinomial is here
of the fourth degree, this process of transformation is
applicable to multinomials of any degree, and that the
rule may be stated in the following general terms :
Divide the given multinomial by x — a, and set aside the
remainder as the teryi not containing x — a. Divide the
600 HORNER'S METHOD.
quotient of this division by x — a, and set aside the remain-
der as the coefficient of x — a. Repeat this process until
there is a quotient which does not contain x — a. This last
quotient will be the coefficient of the highest power of x—a.
Ex. 1. Transform x^ — 11 x^ -\- 15 x -\- 27 into a multinomial in
(x - 9).
Divide the given multinomial by x — 9, using detached co-
efiBcients. It is convenient in this work to place the divisor at the
"^^*' 1 -11 15 27 [9= a
9 -18 -^
- 2 - 3 0 = Bo
_9 63
7 60 = i^i
_9
1 16 = i?2
Thus, the transformed multinomial is
(X - 9)3 + 16(a; - 9)2 + 60(a; - 9);
and it is now obvious that 9 is a root of the equation
a;3 _ 11 a;2 _|. 15 a; + 27 = 0.
Ex. 2. By trial it is found that the equation
4ic8-9x2-14a; + 4 = 0
has a root between .2 and .3 ; find this root.
Transform Ax^ — 9 x^ — 14t x + 4: into a multinomial in (x — .2).
4 ^9.0 -14.00 4.000 [^
^ - 1.64 -3.128
-8,2 -15.64 .872 = Bq
' \ ^ - 1.48
** *' *_7.4 -17.12 = i?i
.8
4 -6.6 = B2
The transformed multinomial is
4 y8 - 6.6 2/2 _ 17.12 y + .8J2^ r^ x
^.
Horner's method.
601
For a trial value of y, neglect the first two terms and write
- 17.12 y + .872 = 0 ; y = .05.
Divide 4 y' - 6.6 y'^ - 17.12 y + .872 hj y - .05.
-6.6
-17.12
.872
.2
-6.4
- .32
-17.44
-.872
0
.2
-6.2
- .31
-17.75 = i?i
.2
1:25
0 = i?o
4 -6 =i?2
The transformed multinomial is
4 ;s8 _ 6 ^2 _ 17.75 z; z = y-..05 = x- .25.
The equation in x can now be written in the form
4Cx - .25)8 _ 6(x - .25)2 _ i7.75(a; _ .25) = 0,
of which .26 is at once seen to be a root.
The two parts of this work can be placed together, and the
whole compactly arranged as follows :
|.26 = tt
^0
-9.0 -14.00
.8 - 1.64
4.000-
-3.128
-8.2
.8
-15.64
- 1.48
.872
- .872
-7.4
.8
-17.12
- .32
0
-6.6
.2
-6.4
.2
-6.2
.2
-17.44
.31
-17.75 =
I
h
-6
J?2
For the completed value of a to be an exact root, it is
necessary and sufficient that Rq = 0, and the smaller Hq
is, the nearer does a approximate to a root.
602
HORNER S METHOD.
Ex. 3. The equation
X* - 16 x3 + 64 x2 - 60 = 0
has a root between 1 and 2. Find an approximate value for this
root.
-16 64 0
1 -15 49
-60 11.127
49
-15 49
1 -14
49
35
-11.0000
8.6081
-14
1
35
-13
84.000
'2.081
- 2.39190000
1.76817136
-13
1
22.00
- 1.19
t
86.081
1.963
- .623728640000
.622246188641
1 -12.0
.1
20.81
- 1.18
88.044000
.364568
- .001482451359
■
-11.9
.1
19.63
- 1.17
88.408568
.359944
-11.8
.1
18.4600
- .2316
88.76851200
.12380066
0
3
-11.7
.1
18.2284
- .2312
88.892312663
.123236869
1-11.60
.02
17.9972
- .2308
89.01554953201
)0
-11.58
.02
17.766400
- .080591
-11.56
.02
17.685809
- .080542
-11.54
.02
17.605267
- .080493
1-11.520
.007
17.52477400
1
-11.513
.007
-11.506
.007
-11.499
.007
I -11.4920
The next figure in this root is easily found to be zero.
HORNEK S METHOD.
603
Ex. 4. Find the cube root of 5.
The equation to be solved is
efficients of x^ and x are zero.
x3 — 5 = 0. Note that the co-
The process of solution is as
follows :
5
11.71
4.000
3.913
2
1
3.00
2.59
- 0.087000
87211
3.0
7
5.59
3.08
+ .000211
3.7
7
4.4
7
8.6700
511
8.7211
512
5.10
1
8.7723
6.11
1
6.12
1
5.13
This result is too large by .000211 ; that is (1.71)8 = 5 + .000211.
Expressed in another way, .000211 is the result of substituting 1.71
for a: in xS- 5. [Art. 149.]
For methods of abbreviating Horner's method the student is
referred to the larger works on the theory of equations.
EXAMPLES cm.
Apply Homer's method to finding, either exactly or approxi-
mately, the real roots of the following equations :
1. a:2_4aj-}-2
2. a;8-2 = 0.
0.
3. x3-6a;-6 = 0.
4. 2 «8 _ 7 a;2 + 2 X + 6 = 0.
604 HORNER'S METHOD.
5. 4a;3- 18a;2 + 23x- 7.5 = 0.
6. 2x8-85x2-85x-87 = 0.
7. CK* - a;3 - 2 a:2 + 3 cc - 3 = 0.
8. o^-x^ + x^-Sx^ + Sx-S=0. (Resolve into factors.)
9. x8 - 3 a;3 - 2 a;5 + 6 = 0. (Resolve into factors.)
10. x3 + 12x-12 = 0.
BOOTS AND COEFFICIENTS. 605
CHAPTER XLII.
Roots and Coefficients. Symmetric
Functions.
417. The general rational integral function of a;, of
degree n, is
f{x) = ttoa;" + CLi^'^ H h a^-iX + «„•
It is sometimes desirable to discuss it under the form
P(X) = «" +i)ia;"-l ^ [- Pn-l^ -\- Pny
in which the coefficient of a;** is unity. The two forms
are directly connected with one another by writing
Then f(x) = aoP(x),
and the roots of the two equations,
f(x) = 0, P(x) = 0,
are the same.
In the present discussion it is assumed that there
exists at least one number, real, imaginary, or complex,
which, when it is substituted for x in P(a;), will make
P(x) = 0.* Such a number is called a root of the equa-
tion P(x) = 0 [Art. 327].
* See the larger works on the Theory of Equations, for a proof
of this theorem.
606 EOOTS AND COEFFICIENTS.
Under this assumption it follows [Art. 325] that any
multinomial P(x) of the 'n}^ degree has, as one of its
forms,
F(X) = (X- aXx - I3)(X -y)...(x- v),
where a, ft y, ••• v, are the n roots of P(x) = 0.
. If we wish to express f(x) in this form, we write
f(x) = af,{x - a)(x -/3)'..(x — v).
This product-form of a multinomial calls to our atten-
tion the important fact, that the problem of finding the
roots of an equation f(x) = 0 is essentially identical
with that of resolving the multinomial f(x) into its
linear * binomial factors. Since there are just n of these
factors [Art. 325], an equation of the n^^ degree has just
n roots.
418. By expanding the product-form of f(x) or P(x)j
we can obviously express the coefficients a^, aj, ••-, or
Pif P2> ••*? as functions of the roots a, ft y, ••• v. Thus
the expanded form of (x — a){x — /5)(a; — y) is
a^ - (a -f )8 + y) 0^ + (/8y + ya -f- a/3) a; - a^y.
Hence,^ if the cubic equation be
a^-\-px^ + qx + r=Of
and a, ft y, be its roots, the following relations exist
between the coefficients and the roots :
2) = - (a -f ^ 4- y), 5' = iSy + ya -f aft r = — aj8y.
* First degree factors. ^
ROOTS AND COEFFICIENTS. 607
Passing to the equation of the n*^ degree, P(x) = 0,
and denoting its roots by a, y8, y, ••• v, and writing
Sa = a + )8 + y + SH \-v
= sum of the roots,
^j8 = a/3 + ay + aS H h av + )8y + /SS + •••
= sum of products of pairs of roots,
Sa^y = a)8y + a^8 + .•• + a^v + ayS + —
= sum of products of roots in threes,
a^y ... V = product of all the roots,
the expanded form of (a; — a) (a? — P)(x — y) -" (x — v) is
05" - JC'-^Sa + iC"-22ay8 + a;«-32a^y + .- + a^y ... v;
and in order that this may be identical with
af* +i>ia5"-^ +p^-^ +P3«""^ H \-Pn,
there must exist between pi, p2f p^, ••- Pnt and a, ^, y, ... v,
the following relations [Art. 328] :
Pi = - 2a, Pa = Saft Pg = - ^^7, — 2>« = ± o.Py — v.
r/iese equations state the simple law that connects the
coefficients and the roots of the general equation of the
n^ degree.
Since there are n of these equations, and just n roots,
the enquiry naturally arises: why is it not possible to
find a, ft y, ••• V, by some process of elimination? The
answer is, that any such process necessarily leads back
to the equation of the n*^ degree with which our investi-
gations began. Nevertheless, the knowledge of these
608 BOOTS AND COEFFICIENTS. ^
relations between roots and coefficients is an important
aid in the solution of equations.
419. Symmetric Functions. The expressions %a, %a^,
^aySy, etc., are evidently such that the interchange of any
two of the letters leaves the expressions themselves un-
changed ; they are therefore symmetric functions of the
roots. [Art. 150.] Such functions are important in
the general theory of higher equations.
Any symmetric function of the roots can be expressed
in terms of the coefficients of the given equation. A
formal proof of this statement is not presented here;
a few examples, as verifications of particular cases, must
suffice for this elementary presentation.
Ex. 1. Sa2 = (Sa)2 - 2 SajS ^
Ex. 2. Sa8 = (Sa)8 - 3 Sa2j3 - 6 Sai57.
Also Sa8 = Sa2 . Sa - Sa2/3.
/. 2 Sa3 = 3 Sa2 . Sa - (Sa)3 + 6 Sa/Sy
= - 3i)i(i>i2 - 2i)2) + Pi^ - eps
= - 2pi3 + 6piP2 - Qps.
,\ Sa3 = _ pi8 + Spip2 - 3p$. '
Ex. 3. Sa2^ = Sa2 . Sa - Sa^
= - Pi(Pi^ - 2i)2) + i?!^ - 3i)ii?2 + 3i)8
= -PlP2+Sp3.
Ex. 4. To solve the cubic equation by the aid of symmetric
functions.
BOOTS AND COEFFICIENTS. 609
Denote by a, /3, 7, the roots of the general cubic equation
x^ + px"^ -i- qx + r = 0,
and take note of the relations
a + /3 + 7 = ~p,
/Sy + 7a + a/3 = g,
a/37=-r. [Art. 418.]
Then, denoting by 1, w, w2^ i^^q three cube roots of 1, and by
m and n, the trinomials a + w/3 + u^ and a + w^^ + ory respec-
tively, we have three linear equations
a + /3 + 7=-p,
a + «/3 + u^ = niy
a + w2/3 + ary = w,
from which, with the help of the relations 1 + w + w^ = q, w^ = 1,
we obtain, by easy eliminations,
a = i(-P + w + n),
/S = \(—p + w^m + wn),
7 = ^( — p + wm + w^n).
The problem of solving the cubic is now reduced to the finding
of m and n.
The product of the two expressions for m and n gives [Art. 406]
rnn = a2 + /32 -f 72 - (/37 + 7a + ap)
= (a 4- /3 + 7)2 _ 3(/37 + ya + ajS)
= p2_3g;
and the product of the three expressions for a, /3, 7 is [Art. 406]
apy = — r = ^^ ( — p^ + m^ + ?i* + Sjpmn) ;
and from the last equation we find
w8 + w8 = p3 _ 3p(p2 _ 3 ^) _ 27 r
= -2i?8 + 0pg-27r.
2s
610 ROOTS AND COEFFICIENTS.
For the sake of brevity let
/, gf = - 2p3 + ^pq -27r, p^-Sq;
then m^ -{■ n^ =f, mn — g^
f and g being known quantities, and the solution of these equa-
tions in m and n gives immediately
The real cube roots of these two quantities (numbers) can be
found as soon as the numerical values of / and g are given, pro-
vided
and the roots of the cubic are then obtained in the form
— p + m + n — p -\- ttfVt 4- (tin — p + (^m + to^w
3 ' 3 ' 3 *
But if /2 < 4 g^, real cube roots of
Kf+^f'-^9^) and i(f-Vrz-4g^) .
cannot be obtained by the usual arithmetical process. The follow-
ing trigonometrical method is then effective.
Write, for the moment,
h=y/ig^-f^ so that Vp - 4 g» = ih,
where i stands for V— 1 and h is real. Then
and since (^)% (-^V = ^: = 1,
we may write —^—= = cos 3 d, — — = = sin 3 d,
2V^ 2V^3
ROOTS AND COEFFICIENTS. 611
The substitution of cos 3 0 and sin 3 6 for their equivalents in the
expressions for m^ and # now gives
m3 = V^(cos 3 ^ + I sin 3 ^),
n3 = Vg^(cos Sd-isinSe),
and by the taking of cube roots
m = Vg(cos 6 + i sin ^),
n = Vg(co8 e — i sin 6).
EXAMPLES CIV.
Verify the following relations. Take note that they have refer-
ence to an equation of the w'*^ degree having n literal coefl&cients
Ph P2» •••» Pm and n roots o, /3, •••, p, and that
pi = - Sa, p2 = SajS, ps=- 2a/37, ^4 = Sa/375.
1. pi2 = Sa2 + 2 SajS.
2. P18 = - Sa* - 3 Sa2^ - 6 Sa/37.
3. pi;)2 = - Sa2/3 - 3 So/37.
4. pi* = Sa* + 4 Sa^iS + 6 Sa2/32 + 12 Sa2/37 + 24 Sa/S78.
5. PiPs = Sa2/37 + 4 Sa/375.
6. ^2^ = Sa2/32 + 2 Sa2/37 + 6 Sa/975.
7. I)i2p2 = 2a8/3 + 2 Sa2/32 + 5 Sa2/37 + 12 Sa/378.
8. Sa* = pi* — 4ipi^p2 -\-2p2^ + 4pip3 - ip^.
9. Sa3/3 = pi2p2 - 2 p2^ - pip3 + 4 Pi-
le. Sa2j32 = p2^ - 2piP3 + 2 Pi-
ll. Sa2/37 =piP8 — 4p4.
12. Sa8/375 =pi2p4-2p2P4.
/ 13. Sa2^2^2=^g2_2p2p4.
^12 KOOTS AND COEFFICIENTS.
14. Denote by a, ^, 7, 5 the four roots of
x^ -{- px^ + qx"^ -i- rx + s = 0,
let A = l(ap + yd), B = i(a7 + )85), C= l(ad + ^37),
and prove the following relations :
A-j-B + C=lq,
BC + CA + AB = l(pr - 4 s),
ABC=i(r^+p2s-iqs).
15. Verify that the values 2A,2B,2C, of Ex. 14, are the roots
of the reducing cubic of Art. 410.
GRAPHICAL METHODS.
613
CHAPTER XLIII.
Graphical Methods.
420. Graphs. If f(x) be any rational integral function
of Xj and it be assumed that
then to every value of x there corresponds one value of
y. Let the corresponding values of x atod y be repre-
sented by lengths laid off on two fixed straight lines
which intersect at right angles in a point 0, the values
of X on the horizontal line OX, the values of y on the
vertical line 0 Y. Lines extending from 0 F to the right
shall be regarded as positive, those extending to the left
negative ; and lines extending from OX upward shall be
regarded as positive, those extending downward negative ;
and these positive and negative
lengths shall correspond respec-
tively to positive and negative
values of x and y.
Suppose OM, ON (Fig. 1) to
represent a pair of values of x,
y ; through M, N draw straight
lines MP, NP, parallel respec-
tively to OY, OX, and intersecting in P. To this pair
of values of x, y there now corresponds the point P, and
Y
N
y
p
0
X ^
/I X
Fig. 1.
614 GRAPHICAL METHODS.
to every pair of values x, y there corresponds one, and
only one, such, point. It is customary to designate it as
the point (x, y).
If X, y be now supposed to undergo simultaneous and
continuous change, but always in such a way as to
satisfy the equation
the point P will move in its plane and trace out a line,
in general a curved line, or curve. This curve is called
the graph of the function f(x), and y =f(x) is said to be
its equation. Every rational integral function has such
a graph.
The nomenclature of analytic geometry, as applied
to this graphical representation, is convenient and is
as follows : x, y are the rectangular co-ordinates of
the point (x, y), x its abscissa, y its ordinate; OX,
OY are the co-ordinate axes, and 0 is the origin of
co-ordinates.
Other forms of graphical representation may be em-
ployed, but the foregoing is the most useful for our
present purposes.
421. Given any function f{x), an indefinite number of
isolated points on its graph may be determined, through
which the curve may be roughly drawn free-hand. First
assign a succession of small positive and negative values
to X (the exaniples here given will not require large
values), then compute the corresponding values of f{x).
A few examples will illustrate the process.
GRAPHICAL METHODS.
615
Ex. 1. Construct the graph of x"^.
The equation y = x^ yields the following series of corresponding
values of x and y, tabulated in vertical columns :
(-3,9)
y =
= aj2
X
y
-3
9
-2
4
-1
1
0
0
1
1
2
4
3
9
These seven points (0, 0), (1, 1), (-1, 1), (2, 4), (-2, 4), (3, 9),
( — 3, 9) serve to determine the general trend of the curve, which is
as represented in Fig. 2. Other points can be easily interpolated.
Ex. 2. Construct the graph of x"^ -2x- 1.
The equation y = x^ — 2x — I yields the following system of
corresponding values, and the graph is represented in Fig. 3.
y = x^-2x-l
e2,7)\
/a7)
(-1.2)\
/
K3.2)
- \
(0,-1)
0
/
X
-13
(1,-
-3)
X
y
-3
14
-2
7
-1
2
0
-1
1
-2
2
-1
3
2
4
7
5
14
Fig. 8.
616
GRAPHICAL METHODS.
It is obvious from an inspection of the figure that there is a
value of X between — 1 and 0, and another between 2 and 3,
which will make a;^ — 2 cc — 1 vanish. Hence there is one root of
the equation x^ — 2x — 1=0 between — 1 and 0 and another be-
tween 2 and 3.
The curves of Figs. 2 and 3 have axial symmetry with respect
to a straight line. The axis of symmetry in Fig. 2 is OF, in Fig. 3
it is a line parallel to OF at unit's distance to the right of the
origin. The graph of every function of the form ax^ + fta: + c has
such an axis of symmetry.
Ex. 3. Construct the graph of x^ — 9 a:.
In order that the values of y may be conveniently small we
make y = \{x^ — 9x).
y^\{x?-^x).
(-2,5y>
y
\
U
A
0
^(3,-5)
X
y
-4
-14
-3
0
-2
5
-1
4
0
0
1
- 4
2
- 5
3
0
4
14
Fig. 4.
This curve has symmetry With respect to the origin,
points are represented in the figure at 3, — 3 and 0.
The root
It will be observed that root-points, that is, points whose
abscissae are roots of the equation f(x) = 0, lie at the
Intersection of the graph and the a;-axis.
GRAPHICAL METHODS. 617
EXAMPLES CV.
Construct the graphs of the following functions :
1. 2x-l. 4. ic2-2jc-4. T. x^-7^-\-x-\.
2. (x-l)2. 5. x^-2x. 8. X*- 6x^ + 6.
3. ^2 - X + 1. 6. a;3 - 2a; + 1. 9. x^ - 3x2 _ 6a;.
Note. — The graph of example 9 consists of two distinct parts.
BINOMIAL EQUATIONS.
422. In the foregoing graphical scheme only real roots
are represented in the diagram. In general, for the
graphical construction of complex roots a plane is nec-
essary, and the analysis for the general case is very
complicated. But the root-points of so-called binomial
equations, whose form is
aaj'* -f 6 = 0,
may be constructed as points on the circumference of a
circle, whose radius is the positive numerical ?i*^ root
of ± b/a, of + b/a if b/a be negative, of — b/a if b/a be
positive.
Let a be this numerical n^^ root; the equation then
has the form
a;" ± a** = 0,
and there are two distinct cases, a;" = a", and x^ = — a**,
which present two corresponding types of solution :
(1) X = a Vl ; (2) X = a^-^.
Since an equation of the n^^ degree has n roots, there
are n values of Vl, as also of -\/—l, which are the n
roots of 2;" = 1, or of 2;" = — 1. These roots having been
618
GRAPHICAL METHODS.
Fig. 5.
found, the n roots of x'' = a"" or of x"" = — a" are the n
values of aVl, or of aV— 1. The problem of solving
ax'' + 6 = 0 is thus reduced to that of solving equations
of the form 2" = ± 1.
423. A complex quantity x-{-iy (i = V— 1) contains
two numbers, x, y, which may be regarded as the rec-
tangular co-ordinates of a point
. (Fig. 5). Thus interpreted, every
complex number has assigned to
it a point, its graph, in a plane,
and to every point there corre-
sponds a complex number, its
affix. The points we are about
to consider lie upon the circumference of a circle of unit
radius, a U7iit circle, whose centre is at the origin ; their
co-ordinates, therefore, satisfy the condition
'X^ + f = l.
\
If two complex numbers x + iy, x' -\- iy', whose parts
X, y, x', y\ satisfy this condition, be multiplied together,
the result will be of the form u + iv, that is,
{x-\- iy)(x' + iy') = u-{- iv,
where u = xx' — yy', v = xy' -f x'y,
and it can be verified at once that
u^^v^^ix'-h y") (x'^ + y"^ = 1.
Hence, if the graphs of a series of complex numbers lie
upon the unit circle, so does also the graph of their product.
GKAPHICAL METHODS.
619
If there be n equal factors, the product has the special
form {x + lyY and hence, in particular.
If {x-\- iyY = u -{-iv and oi? -\-]f — 1, then w^ + i;^ = 1
and the graph of u-\- iv is on the unit circle.
424. Given x-\-iy and a?' + iy\ it is required to con-
struct the product (x + iy){x' -\- iy').
Let OM, MP=x,y,
OR, BQ = x', y\
(Fig. 6) and let u, v, be the hor-
izontal and vertical co-ordinates
of Q, so that
ON, NQ = u, V.
Then, in the similar triangles
QSR, OMP, and SON, POM,
QS : y' =zl:x, SN: u = y : x,
1
Fig. 6.
and
v=:qS + SN=^l{y^ + uy).
X
.'. y' = vx — uy.
Also,
x'^=l-y'^ = {u^-\-v^{7?-\-f)-iyx-uyy
= (ux -f vyy.
.'. a;' = wa;-f vy.
In the figure here constructed it is obvious that the
plus sign for this value of a' must be chosen. It can be
shown that this choice is necessary in all cases.
By forming the expressions xx^ — yy', x'y -f- xy', we
now have
xx' — yy' = ux^ -\- uy^ = u,
xy' -f x'y = vy^ -(- va^ = v.
620 GRAPHICAL METHODS.
But {x + iy)(x' + iy^) = xx' — yy' -\- i (xy' + x^y).
.'. (a? + iy)(x' + iy') = w + iv,
which identifies Q as the required graph of the given
product.
Thus it has been shown that the graph of the product
(x + iy) (x' + iy') is constructed by first determining P
as the graph ofx-\- iy, in the usual way, and then con-
structing the point (x', y'), using OP, instead of OA, as
the axis of abscissae. By repeating this process the
product of any number of factors may be constructed.
It is now evident that the graph of the series of consecu-
tive powers,
(x + iyj, {x + iy)\ (x + iyf, ..•(«; + lyy,
is a series of points distributed at equal intervals over the
circumference of the unit circle.
For, the constructions for these powers present a
series of congruent right triangles OMP, ORQ, etc., and
the arcs AP, PQ, etc., are therefore equal.
425. Any root-point of the equation z"" = 1, or of the
equation z^ = — 1, is on the unit circle.
For, let to be the root in question, and let {x, y) be the
corresponding root-point. Then
o) = a? -f I?/
where x' = xj-yjiy? + 2/^),
and we may write
where u-\-iv = (x' + iy')\
GRAPHICAL METHODS. 621
But this demands that u^-\-v^=l [Art. 423] and that
v = 0, u = ±l [Art. 234].
426. If 0) be a root of z'^ = 1, any integral power of w
is also a root; for then, k being any integer,
(a>*)" = (a>»)* = 1* = 1.
Ifoibea root o/ 2** = — 1, any odd integral power of w
is also a root; for then, k being any odd integer,
(o>*)'* = (a)")* = (-l)* = -l.
Ifo/' = l, and ifti/" be the lowest power ofo) that has this
value, 0), 0)^, o>^, • • • (!)"■, are all distinct, and are the n roots
of z^ — 1. For, if any two are equal, as w^, o>*, where I
and k are less than n, then a>'~* = 1, which contradicts
the hypothesis.
7/" to** = — 1, and ifw'* be the lowest power of o) that has
this value, w, w^, o>*, •••w'^""^, ai-e the n roots 0/2** = — 1.
For, tu^" = + Ij and this is the lowest power of a> that
has the value 4- 1 ; thus o, <o^, w^, ••• o)^", are distinct, and
therefore also w, w^, <o^, ••• a>^**~\
427. The root-points of 2'*= ± 1 may now be definitely
placed on the circumference of the unit circle.
I. When 2" = + 1. For all values of n, 1 is a root.
Beginning with (1, 0) as a first root-point, the whole
series of n root-points are distributed at equal intervals
over the entire circumference.
II. When 2" = — 1. For no values of n is 1 a root.
If 2n points, of which (1, 0) is one, be distributed at
equal intervals over the circumference and be numbered
622
GRAPHICAL METHODS.
consecutively 1, 2, 3, 4, •••, 2w, and be so adjusted that
2 71 is assigned to the point (1, 0), the odd numbers will
mark the positions of the n root-points.
Ex. 1. Solve the equation z^ = —\.
Divide the circumference into six equal parts and number the
division-points 1, 2, 3, 4, 5, 6, assign-
ing 6 to the point (1, 0). The root-
points are at the positions 1, 3, 5. The
points numbered 1, 5 have a common
.abscissa which is obviously ^ ; and
their ordinates are
Hence the three roots are
Ex. 2. Solve the equation z^^ = 1.
If an inscribed regular decagon and an inscribed regular hexa-
gon have a common vertex at A (Fig. 8) and their adjacent ver-
tices be at P and Q respectively, PQ is
the side of a regular pentadecagon. Take
AB = PQ ; then A and B are two adja-
cent vertices of a regular pentadecagon.
Let the co-ordinates of P, §, and B be x,
y, u, V, and x', y'.
Then m, v = i, JV3»
x,y = i(l + V5), i V(10 - 2 V5),
and x', y'= ux -\- vy, vx — uy. [Art. 424.]
Therefore the co-ordinates of the root-point of i^^^ = 1 which
is adjacent to A are
Fig. 8.
iC'
= H1+V5+V(30-6V5)},
2/' = HV3+Vl5-V(10-2V5)},
and the corresponding root of z^^ = 1 is
(0 = x' + iy'.
The fifteen roots of z^^ = 1 are therefore w, w^, w*, ♦••, w^^
GRAPHICAL METHODS. 623
Binomial equations of the form 2" = ± 1 are a special
type of reciprocal equations. [See Art. 449.]
EXAMPLES CVI. V-
Solve the following equations ;
1. z^ = 27. 6. z^ = --l. 9. 2^8 =-16.
2. ^4 = 4. 6. ^ = 5. 10. 2:i'> = 1024.
3. z^ = -l. 7. 06 = -27. IL 2io = -16.
4. 05 = 32. 8. 08 = 625. 12. 0I6 = 1.
Verify the following statements :
13. 02n+i = 1 has the real root + 1, but no others.
14. 02n+i = _ 1 has the real root — 1, but no others.
16. 02n = 1 has the two real roots + 1 and - 1, but no others.
16. All the roots of z-** = -1 are complex.
17. If m be prime to n, the equations 0"* = 1 and 0" = 1 have
no common root except unity.
18. If n be a prime number and a any one of the complex roots
of 0" = 1, then a, a% o^, •••, a^^ are the n roots of 0** = 1.
624
DERIVATIVES.
CHAPTER XLIV.
DERIVATIVES.
428. The Difference Katio. Let f(x) denote the rational
integral function
The graph of this function is a curve, ^nd
is its equation; values of x are abscissae, values of y
ordinates. For a second pair of co-ordinates x^, y^ there
exists the similar equation
2/i=/(^i)-
Let a?!— ic and yi—yhe denoted
by Ax and Ay respectively.*
Pi
1
y
Ay
AX
M
Fig. 9.
Then
and
Ay=f(x + Ax)-f(x)
Ay f(x-\-A*x)~f(x)
Ax Ax
This fraction is called the difference-ratio of the function
f(x). In the figure (Fig. 9) with OX, 0 Y as axes and
* A and the letter following it are regarded as inseparable.
Thus Ax X Ax = (Ax) 2, and this cannot be written A^x^, Similarly
Ax X Ay cannot be written A'^xy.
DERIVATIVES. 625
PSPi a segment of the curve, the co-ordinates of P and
Pi are
OM, MP, 0M„ M,P, = X, y, x„ y^
and ^ = ^.
Ax PR
In the triangle PRPi, right angled at R, the ratio
RPy/PR, that is, the ratio of the opposite side to the
adjacent, is called the tangent of the angle RPPi, and is
designated by the abbreviation tan RPP^ ; hence
^ = tan i2PPi.
Ax
Thus, the difference-ratio, corresponding to any two
points on the curve, is equal to the tangent of the angle
which the secant line through the points makes with the
aj-axis.
429. The Derivative. Let the values Xi, y^ now approach
the values x, y ; then Ax and Ay both approach zero, and
the difference ratio tends to assume the form -, whose
limit, however, we may propose to determine. In the
figure. Pi approaches P, the secant line turns about P
and approaches coincidence with the tangent line at P.
Denoting by <^ the angle which this tangent line makes
with the aj-axis, we have, in the limit, when Aa* = 0 and
Ay = 0,
limit A^ ^ limit f(x-^Ax)-f(x) ^ ^^^
This limit is a new function of x. It is called, the
derivative of f{x) with respect to the variable x, and is
2s
626 DERIVATIVES.
denoted by f(x). It is equal to the tangent of the angle
which the tangent line at the point {x, y) makes with the
X-axis. To differentiate a function is to find its derivative.
Ex. 1. Find the derivative of oj^.
The difference-ratio is
fix + Ax) -/(x) ^ (X -^ Ax)2 -x^^.^^ ^. Ag,
Ax Ax ^Ax
Discarding the unity factor, and passing to the limit, we have
/'(x)=2x.
In the differential calculus the derivative is a function
of far-reaching importance, and in that subject it calls
for extended discussion and the setting up of a large
number of general and special formulae. Here we shall
only need to consider a few special cases. In particular,
we need to be able to write down the derivative of a
rational integral function.
430. Derivative of a Constant. Since a constant does not
undergo any change in consequence of the changes of
other quantities, its difference-ratio is always zero, and
hence its derivative is also zero.
431. Derivative of a Sum. Let /^(x) and f2{x) be two
different functions, and let
f(x)=Mx)+Mx).
The difference-ratio of this sum is
f(x -\- Aa?) -f{x) ^Mx + Ax) -f /2 (a; + Ao;) -/^ (x) -f^ (a?)
Ao; Aa;
_/i(a; -f Ax) -fi(x) f^jx + Aa;) -f^jx)
Ax Ax
DERIVATIVES. 627
Hence, passing to the limit for Ax = 0,
That is, the derivative of the sum of any two functions
is equal to the sum of their derivatives.
In this proof it has been taken for granted that the
limit of the sum of any two quantities is equal to the
sum of their limits.
432. Derivative of a Product. Let f (x) and /g (x) be any
two functions, and let
f(x)=Mx).Mx).
The difference-ratio of this product is
f(x + Ax) -fix) ^ /i (a; + Aa;) ./^(a; + Ax) -f^(x) -f.jx)
Ax Ax
^ [f(x + Ax) -f(x)^f2(x + Ax)
Ax
lA{x + Ax)-f,(x)^f{x)
"^ Ax
Hence, passing to the limit,
/' (x) =Aix)A'(x) +Mx)f,'ix).
That is, the derivative of the product of any two quan-
tities is equal to the second into the derivative of the first
plus the- first into -the derivative of the second.
Here again it has been taken for granted that the limit
of the sum of two quantities is equal to the sum of their
limits.
628 DERIVATIVES.
If a be a constant, we have, as a corollary of the last
formula, that if
f{x) = af^{x),
then f(x)=af^{x).
Hence, the derivative of the product of a constant and a
function is equal to the constant, multiplied by the deriva-
tive of the function.
433. Derivative of a Power. Let f(x) = x\ The differ-
ence-ratio is
f(x + Ao;) —f(x) _(x-\- Axy — x""
Ax ~ Ax
But (x + ^xy = a;" + nx^-^Ax + MAx^j
where JHf is a multinomial in x and Ax. Hence
f(x-\-Ax)—f(x) . ^ , . , -.^ . Aaj
^ ^ "^ — ^^-^^ = (nx^'-^+MAx) -— .
Ax ^ ^ Ax
This last expression differs from na;"~^ by an arbi-
trarily small quantity when Ax is made sufficiently small.
Therefore, in the limit, when Aic = 0,
f'(x) = nx'^~\
This proof presupposes that n is a positive integer,
though the formula is true for all values of n.
Thus, the derivative of the n^^ power of the independent
variable (n = a positive integer) is n times the (n — 1)*^
power of the variable.
This rule is also applicable to finding the derivative of
(x ± ay. [See Ex. 2 of Art. 435.]
DERIVATIVES. 629
434. Derivative of f(x-\-a). If a be a constant, the
difference between two values of x-\- a is equal to the
difference between the corresponding values of x, for a
does not change. Thus
A (a; -h a) = Ax,
and if /(a; + a) be any function of a; -f a, then
Af{x + a) =/[a; + a 4- A(a; + a)]-/(a; + a)
=zf(x-{-a + Ax) -fix + a) ;
that is, the difference-ratio, and therefore also the deriva-
tive, of f{x -h a) is the same whether a; + a, or a; alone,
be regarded as the independent variable. Hence the
following theorem :
The derivative of f{x + a) with respect to x = the deriva-
tive of fix -\- a) with respect to x+a=f\x + a). [See
Ex. 2, Art. 435.]
435. The foregoing formulae include all the cases of
differentiation that will present themselves in the follow-
ing pages, but a few examples are added in further illus-
tration of the subject.
Ex. 1. Find the derivative of the quotient of any two functions.
Then fix)'f2{x) = fiix),
and, by differentiation [Art. 432],
fix) -Mx) -^f^'ix) .fix) =/l'(«),
6S0 DERIVATIVES.
But /(x) =
/2(^)
Ex, 2. Find the derivative of {x — a)^.
The difference-ratio is
{x + Ax - ay- {x - ay _ n{x- ay-^^x + M^y?'
Ax ~ Ax '
where iW is a multinomial in (x — a) and Ax. In passing to the
limit, with Ax = 0 everything disappears except n{x — a)"-^, which
is the required derivative.
Ex. 3. Find the derivative of a".
The difference-ratio is
= a*
Ax Ax
In Art. 307, Ex. 1, it was shown that
limit «^i:zl = lna.
AX:^0 Ax
Hence, derivative of a== = a* In a.
Hence also, derivative of e' = e*.
EXAMPLES CVII.
Find the derivative of each of the following functions :
1. x2 + x3. 8. x"'(x'* + a). jg J^.
2. 3x2 + 2x3. 9. (x+l)3. ^^
3. (x+l)(x2-2). 10.
X + 1 14.
x'^
4. 5x2(3x+l). ^ 15. I
6. X2 + 2 &X + C. -1 17 b\Qg rf
12. i + a;. ^' ^
7. ax« 4-6. ic 18. ln(x + a).
DERIVATIVES. 631
436. If fix) have the factor (x + a)", the derivative of
fix) has the factor (x + a)"~^.
For, if /(a;) = (ic + a)"<^(a;),
then f'(x) = n(x-\-a)''-^<ti{x) + (x + a)''<f>'(x) '
=ttB + a)"-i \n<f, (x) + (a; + a) cf>\x)l,
437. Equal Boots. The following is an immediate cor-
ollary of the foregoing theorem :
If f(x) = 0 have k roots equal to a, /' {x) — 0 has k—1
roots equal to a.
It is evident, from the form of /' (a;), that x -\- a will
not appear as a factor in /' {x) unless it appears at least
twice as a factor in f{x). Hence the necessary and suffi-
cient condition that fix) = 0 shall have equal roots is
that fix) and /' (x) shall have a common factor involv-
ing x.
To find the equal roots, obtain the highest common
factor of fix) and /' (a;) and solve the equation obtained
by putting this factor equal to zero.
Ex. The H.C.F. of jc* - Qx^ + 4 a; -I- 12 and its derivative
4 X* — 18 X 4- 4 is found to be x — 2. Hence (x - 2)* is a factor
of X* — 9x2 -f- 4x -f- 12, the other two factors of which are then
obtained by division and inspection. They are x + 1 and x + 3.
438. Maxima and Minima. If fix) increase, attain a
value /(a) and then decrease, /(a) is said to be a maxi-
mum value of fix). If fix) decrease, attain a value
632
DERIVATIVES.
/(a?)
Fig. 10.
/(/8) and then increase, f(/3) is said to be a minimum
value of f(x).
Suppose f(x) to be a rational integral function of x,
and let its graph be constructed. It may have maximum
and minimum ordinates, as at A
and B (Fig. 10).
As the point (x, y) passes along
the curve from left to right, x
increases and Ax is positive.
During this movement f(x) in-
creases or decreases according
as A?/ is positive or negative,
that is, according as the difference-ratio Ay/Ax, and
therefore the derivative f'(x), is positive or negative.
In fact, briefly,
f(x) is an increasing or a decreasing function according
as f (x) is positive or negative.
If, then, f (x) pass from positive to negative values,
becoming zero at the instant of change, f{x) will first
increase and then decrease, will attain in fact a maximum,
at the instant when / {x) changes sign. Similarly, f{x)
will attain a minimum when /' (x) passes from negative
to positive values. Hence
If f (a) = 0, and if in passing through this value f (a)
pass from positive to negative values, then /(a) is a
maximum.
■ Jf f (o.) = ^? ^1^^ if ^^ passing through this value /'(a)
pass from negative to positive values, then /(a) is a
minimum.
DERIVATIVES.
633
The theory of maxima and minima is an important
aid in tracing graphs of functions.
Ex. 1. Construct the graph of aj^ — 3 x, determining the maxi-
mum and minimum ordinates.
The derivative of x^ - 3 a; is 3 ic2 - 3, and the roots of 3 x2_3 =: o
are 1 and — 1. The maximum and minimum ordinates, if there
be any, must therefore be
13_3.i=_2 and (_ l)3_ 3(_ i) = 2.
But it is easy to see that x^ — 1,
the variable part of the derivative,
changes from a negative to a posi-
tive quantity when x passes through
1 , and from a positive to a negative
quantity when x passes through —1 ;
hence —2 is a minimum ordinate and
2 is a maximum ordinate.
The graph of the function is rep-
resented in Fig. 11.
{-2,-2)
Ex. 2. Find the maximum and minimum values of x^ — 9 x.
The derivative of x'^ — 9 x is 3 (x^ — 3) and the values of x
which make it vanish are — V3 and + \/3. The corresponding
maximum and minimum ordinates are 6\/3 and —6 VS. The
graph is that of Fig. 4, Art. 421.
439. Successive Derivatives. The successive results
obtained by repeating the process of differentiation are
called successive derivatives ; if the process be once, twice,
• •• (n — 1) times repeated, the result is the second, third,
• • • ii**" derivative. These successive derivatives of an
J{x) are denoted by f'(x), f'(x), f'(x)y •••/'*X^)? respec-
tively. For example, if
634 . DERIVATIVES.
f{x) =x%
then f(x) = nx^-^j
f"(x) = n{n - l)x''-^,
f"'(x) = n(n - l)(n - 2)af-«,
f(-\x) = n{n - l)(n - 2) ... 3 • 2 • 1,
and all the higher derivatives are zero. f(x)j f'(x)j
••'P''\x)j are also called the first, second, -^ n^ derived
functions of f(x).
EXAMPLES CVIII.
Find the third derivative of each of the following functions :
1. _^!_. 2. -^L-. 3. ^'
x^ - a2 x^- a2 ic2 _ ^2
Find the n}^ derivative of each of the following functions :
4. -J_. 6. i+^. 6. (1+^.
1 —X I —X 1 —X
Solve the following equations, each of which has equal roots :
7. x3 + 15 x2 - 33 X + 17 = 0.
8. «* - 6 x3 + 11 ic2 - 4 a; - 4 = 0.
9. x5 - 3 x* - jc3 + 7 ic2 - 4 = 0.
10. x6_i2a:2-16 = 0.
11. 2 x* - 9 x3 + 6 x2 + 20 X - 24 = 0.
12. x5 - 3 x4 + 5 x3 - 7 x2 + 6 X - 2 = 0.
Note. — In examples 11 and 12 find the highest common factor
of the first and the second derivatives.
DERIVATIVES. 635
Construct the graphs of the following equations :
13. y = x^-3x^-9x-5. 16. y = x^- 2x^^-5.
14. xy -x"^- y-l=0. 17. y = 2 yfi - 6 xf^ -\- 5 x'^ -\- 1.
15. x2 + ?/2 - 2 cc = 0. IS. xy - y + 1 = 0.
19. It is required to construct, with the least possible amount
of sheet metal (thickness of metal given), a cylindrical vessel,
open at the top, which shall contain a given quantity of liquid.
Determine (in terms of its volume) the exact (interior) dimensions
of the vessel.
636 Taylor's theorem.
CHAPTER XLV.
Taylor's Theorem.
440. Let/(aj) denote the rational integral function
Then f(x -j- h) is a rational integral function of h and
can be arranged in ascending powers of h, thus :
f(x + h) = A + A^ + A2h^ H + AJi%
in which A, ^i, etc., involve x, but not h. Differentiate
each member of this equation with respect to h, regarding
X as constant, and repeat the process on each successive
equation until /^"^^ + ^0 ^^ obtained. The results are
f(x +h) = A + 2 A,h + 3 A,h' + -,
f'(x -^h) = 2 A + 2 . 3 ^/i 4- 3 • 4 AJv" + ••-,
f"(x + h) =2.3^ + 2.3.4^/1 + 3.4.5^/124-...,
/(")(a; + /i) = 1.2.3...nA.
In these equations put 7i = 0 ; then
/'(^) =2 A, f"(x)=l-2.3A„
/(-i)(a!) = (n-l)!A-i, /">(«) = »! A-
637
Thus the values of Aq, A^, A2, • • • A^ are known, and
f(x + h) =f{x) + hf(x)+^f'(x) + •.. +2/"X«^)-
This formula is known as Taylor's Theorem. It here
appears in a special form, applied to a rational iiitegral
function. The values of f'(x), f'(x), etc., are now found
by differentiating successively the original multinomial,
f(x) = aox"" + ch^f-^ H h a^-^x + a„,
and its derivatives. Thus,
f(x) = noox''-^ + (n — l)(Xia;"-2 H h a*-^
f\x) =n(ri-l)aoa;"-2+... -f2a''-2,
/(-)(a;) = 1.2.3...7iao.
The student may verify these results by writing x-\-h
for x in.aoo;** + ajOj"-^ + \- a^, developing the binomials
(x + hy, (x 4- hy~'^, etc., and collecting the coefficients
of the various powers of h.
Ex. 1. Substitute 1 + h for x in o^ — Sx^ + a; + 2 and develop
the result in terms of powers of h.
Let fix) = x8 - 3x2 + X + 2.
Then /'(x)= 3x2 - 6x + 1, /"(x)=6x-6, /'"(x) = 6,
and all the higher derivatives are zero.
The substitution of 1 in these functions gives
/(1)=1, /'(l) = -2, /"(!)= 0, /•"(!)= 6.
.'.f{l + h)=l-2h + h^
= (l-h){l-h-h^).
638 Taylor's theorem.
Ex. 2. Develop (1 + x)^ in ascending powers of x.
Let /(x) = x^.
Then /'(x) = 5x*, /"(x) = 20x3, /"'(x)=60x2,
/iv(x) = 120 X, and /^(x) = 120.
.•./(l + x) = /(l) + x/'(l)+ixV"(l)+-
= 1 + 5X + -2f X2 + -6^X3 + V^X* + X^
= 1 + 5 X + 10 x2 -f 10 x3 + 5 x4 + x5.
441. Oontinuity. It has been seen [Art. 284] that
variables and functions may change continuously, or
discontinuously. The continuity of a function /(a?), for
a given value of x, is tested by observing the behavior
of the difference
f{x^K)-f{x\
where fi is an arbitrarily small quantity and either pos-
itive or negative. If this difference remain permanently
smaller than an arbitrarily small quantity 8, while h
passes, through either positive or negative values, into
the value zero, the function fix) is continuous for the
value X. Thus y? is always a continuous function ; for
{x-\-}if-y?=^2xh^h?,
and this is zero when /i = 0 whether li pass into zero
through positive or negative values. But l/(a; — 1) is a
discontinuous function at a; = 1 ; for
1 1 ^ -Ti
x-\-}i — \ x — 1 {x — V){x^'h — V)
and if a; = 1 this is — /i/(0 • /i), and is — oo when h ap-
proaches zero through positive values, but becomes 4- oo
when /i passes into zero through negative values.
Taylor's theorem. 639
442. Oontinuity of /(«). Suppose the variable x to
undergo continuous change, and let a and a-\-h be any
two finite values of x. If f{x) be a rational integral
function of a;, then by Taylor's theorem,
/(a + h) -/(a) = A/'(a) +|^/"(a) + •.• +^/<"'(<')-
If h be arbitrarily small, the difference /(a + h) —f{a)
is evidently itself arbitrarily small. Hence, by the cri-
terion of Art. 441, the function f(x) changes continu-
ously when X undergoes continuous change.
In this proof a and a-\-h have any finite or zero values.
Hence the rational integral function f(x) is continuous for
all finite and zero values of the variable.
This theorem shows us that the graph of a rational
integral function is a continuous line.
Corollary. Tlie derivatives of a rational integral func-
tion (being themselves rational and integral) are continuous
for all finite and zero values of the variable.
443. Rolle's Theorem. Iff(x) assume the same value for
two different values, a and b, of x, the derivative f (x) will
vanish for some value of x between a and b.
As X passes continuously from a to 6, f(x) passes con-
tinuously from f{a) to f{b) [Art. 442]. But, by hypothe-
sis, /(a) =f{b), and since f(x) varies, it must attain, in
the interval a, b, at least once, either a maximum or a
minimum value [Art. 438]. Therefore /'(a;) must vanish,
at least once, for a value of x between a and b.
Hence also, Iff{x) — 0 have two distinct real roots, a and
b, f (x) = 0 will have at least one real root between a and b.
This is EoUe's Theorem.
640 Taylor's theorem.
EXAMPLES CIX.
Show that the following functions are discontinuous for the
value x = 1 :
1. -^ — 2. — -. 3. In (X - 1).
x^-1 (X - 1)2 ^ ^
Show that the following functions are continuous for all finite
values of x :
4. (x-a)". 5. —^ — 6. a^.
^ ^ x2 + 1
Develop the following functions into infinite series in ascending
powers of x, by Taylor's theorem, and assign the range of values
of X for which the series are convergent :
Y x + 1
■ X- 1
9. 1 .
V(^ + 1)
11 1
(X - 1)2
8. e-+\
10. a^-\
12. In (x.+ 1),
13. Show, by the aid of Rolle's theorem, that for the equation
/(x) = 0, of the third degree, to have all its roots real and unequal,
it is necessary and sufficient that the roots a, j3, of f'(x) = 0, shall
be real, and that
/('^)-/(^)<0.
14. Apply the theorem of example 13 to x^ + gx + »' = 0, and
show that the condition there stated is fulfilled if
4 ^3 _i_ 27 7-2 < 0.
TRANSFORMATION. 641
CHAPTER XLVI.
Transformation.
444. When a multinomial in x is so changed as to be-
come a multinomial in some function of x, or of another
variable, it is said to be transformed. The general type of
all the important transformations in the elementary theory
of equations is the so-called homographic transformation,
whereby x is replaced by a function of the form ^^ \^-
yZ -^6
This general type of transformation is a combination
of the three special types
z H- hj kz, and - ;
z
for, identically
yZ-^8 y y^ Z -\- S/y y Z-\-h'
where k = PI ~^ and h = S/y ; and the successive sub-
stitutions ^
X^ ^^^ Z ~\~ flm X2 ^— — J X^ — f€X2) X^ — fl/g ~|~ —
^ y
lead directly to
a , k
y Z + h
The effect of the three special types of transformation on
a function is to change it to a function of a; -f- h, or of kx,
2x
642 TRANSFOEMATION.
or of -; and on an equation the effect is : (1) to increase
its roots by h, or decrease them by —h\ih be negative ;
(2) to multiply its roots by k, or divide them by 1/k if
A; < 1 ; or, (3) to change the roots into their reciprocals.
The three types will be taken up separately.
445. To transform f(x) into a function ofx — a.
In Taylor's theorem applied to the function /(a + h)
[Art. 440], let ic — a take the place of h ; then
f{a + X- a) =f{x) =/(a) + (x - a)/(a)
+ ^>(a)+-+^^/»'(a),
which is f(x) expressed in terms of the powers of x — a.
It is the transformed function required; the new coeffi-
cients are functions of a.
In Art. 416 it was seen how these coefficients may be
computed by simple and successive division by x — a,
and the rule for the computation is there given.
The proof of the rule was given for a multinomial of
the fourth degree. The student will now find no difficulty
in rehabilitating the proof for the general case.
Ex. 1. Transform x* + 11 x^ + 41 a:2 + 61 x + 30 into a function
of X + 2.
The new coefificients are 1, 3, — 1, — 3, 0, and the transformed
function is
{x + 2)4 + 3 (x + 2)3 - (x + 2y-S(x + 2).
This transformation was constantly used in the applica-
tions of Horner's method, and the student is referred to
Art. 416 for other examples.
It was also used in Art. 408 for getting rid of the
TRANSFORMATION. 643
second term of the general cubic oc^ -\- px^ + qx -\- r, and
it may be used with similar effect on any multinomial.
Ex. 2. Transform ax* + bx^ + cx'^ + dx + e into a function of
X — a and assign a value of a that will make the coefficient of
(x — a)3 vanish.
Here /(a) = aa"^ + ba^ + ca^ + da + e,
/'(a) =4aa3 + 36a2 + 2ca + (i,
/"(a) =12aa2 + 6&a + 2c,
/'"(a) = 24aa + 6 6,
/iv(a) = 24a.
Hence the transformed function is
a (X - a)*+(4 aa + 6)(x - 0)84- (6 aa^ 4- 3 6a + c)(x - a)2
+ (4 aa8 + 3 6a2 + 2 ca + d) (X - a)
+ aa* + 6a3 -f ca"^ ^ da + e.
The coefficient of (x — o)^ will vanish if a =— b/(4a). The
third term will drop out if a satisfy the equation
6 aa2 + 3 6a + c = 0,
the fourth term if
4 aa8 + 3 6a2 + 2 ca + (^ = 0.
When the multinomial of the n^ degree,
f(x) = aox"" + ttio;"-^ H h a^_ix + ««,
is transformed in this fashion, the coefficient of (a;— a)"~^
in the transformed function is
T-^^TTT /"-'^ (a) = oi + na^
{n - 1) !
the coefficient of (x — a)"~^ is
1 f ("-2) (a\ = a„ 4- (n - V\ a,a 4- ^(^~^)
/("-2) (a) = a, + (n - 1) a^a + '1111^^ a^\
(71-2)!" '" ^ ^ '' ^ ' 2
644 TRANSFORMATION.
and so on. Any term in the transformed function may
therefore be made to disappear by solving an equation in
a, of the first degree if it be the second term that disap-
pears, of the second degree if it be the third term, and
so on.
446. To transform f(x) into a function of kx.
To achieve this transformation we have only to replace
ic by - • kx. If the given function be
a(fi(f + ajps''-'^ H h a„_ia; + a^
the transformed function is
^ (kxy + y^, (kxy-' -\-'"-\-^'Qcx) + a„.
By means of this transformation troublesome frac-
tional coefficients may be got rid of.
Ex. 1. Transform x^ -\- ^x"^ — ^x -{- ^^ = 0 into an equation
whose coefficients are all integral.
Replace x by kx and multiply by k^. The equation becomes
where y=:kx; and if A; = 6,
this is y3 _}. 10 2/2 _ 20 2/ + 9 = 0.
447. Note the special case k = — l. We here substi-
tute — (— 0?) for X and obtain
(- 1)X(- «')"+(- l)"-'«i(- ^y-' + «n-i(- ^) + a„,
or ± ao2/" =F ai^/""^ ± a„-i2/ + «„,
TRANSFORMATION. 645
wherein y — — x. Observe that the signs of the terms
become alternately plus and minus. Applied to an equa-
tion, this transformation produces a new equation whose
roots are those of the old with their signs changed.
It is useful in the computation of negative roots by
Horner's method; the transformation takes place, of
course, before the computation begins.
A simple rule for writing the transformed equation in
this case is obvious. It is :
Change the sign before every other terrriy beginning with
the seco7id.
448. Eeciprocal Equations. It is evident that a function
of X cannot, by any transformation, be written as a func-
tion of 1/x, and that therefore the third type of trans-
formation [Art. 444] necessarily introduces a new variable
Zy connected with the old by the relation z = l/x.
Applied to an equation, say the general equation
aoa?" -f aix""-^ + h a„_ix -f a„ = 0,
this transformation, x = 1/z, after multiplication by 2",
leads to a reciprocal equation
aX + ««-i2;""^ -\ 1- ^iZ + a© = 0,
and the two equations obviously stand to one another in
the following relation of reciprocity : The coeflBcient of
af in either is the same as the coefficient of a?""'" in the
other, and the roots of the one are the reciprocals of the
roots of the other.
449. An equation is its own reciprocal, that is, self-
reciprocal, if it remains unchanged when x is replaced
646 TRANSFORMATION.
by l/x, after its terms have been multiplied by the
proper power of x. Thus, a^ + 2a^ — 2a7 — 1 = 0 is a
self-reciprocal equation, or simply a reciprocal equation."*
If the general equation of the n^^ degree satisfy this
test, the two equations of the last article must be identi-
cal, the necessary and sufficient condition for which is
[Art. 328]
^ = _^ = _^ = ... = ^!^ = ^.
This condition may be expressed more briefly thus,
^ = ^'- = e, [r = 0,l,2,...7i]
where e is either +1 or — 1, but is the same for all the
ratios.
This formula furnishes the complete criterion for a recip-
rocal equation.
Ex. 1. 3x*-f6a;3 — 5x — 3 = 0isa reciprocal equation ; for
-3-5 6 3
Ex. 2. 3x*-|-5x3-}-2ic — Ssc- 3 = 0is not a reciprocal equa-
tion ; for
A^A = ^ = ^ = _i,but2 = + i.
-3-5 5 3 2
This equation becomes reciprocal when —6x and — 3 are
changed to + 5 a: and + 3.
* The term reciprocal equation, without the prefix self, usually
means, in mathematical literature, what I have here called a self-
reciprocal equation.
TRANSFORMATION. 647
Ex. 3. Solve the equation
This is a reciprocal equation. It may be written in the form
or, again, in the form
3(. + iy_5(x + l)+l=0,
1 3 1
a quadratic equation in x + -, whose roots are -, -. The problem
is now reduced to the solution of the two equations
^ + 1 = 3 Ul.
X 2' X 6
The four roots of these equations are
450. Reciprocal equations have the following proper-
ties, which are easily recognized as consequences of the
equations aja^_^ = a^^^ja^ = c. The formal proofs are
suggested as exercises for the student.
I. When c = + 1.
The coefficients of terms equidistant from the first and
the last are equal ; i.e. a^ = a„_^.
If the degree of the equation be odd, it has — 1 as a
root. Its left member is then divisible hj x -\-l, and the
quotient, equated to zero, is a reciprocal equation of even
degree.
648 TRANSFORMATION.
11. Whene = -1.
The coefficients of terms equidistant from the first and
the last are numerically equal with contrary signs; i.e.
a, = - a„_,.
If the degree of the equation be even, the coefficient of
the middle term is zero.
If the degree of the equation be odd, it has + 1 as a
root, and its left member is divisible by x — 1. The
quotient, equated to zero, is a reciprocal equation of even
degree.
If the degree of the equation be even, it has both + 1
and — 1 as roots, and its left member is divisible by
a^ — 1. The quotient, equated to zero, is a reciprocal
equation of even degree.
EXAMPLES ex.
Apply the transformations indicated, and solve completely the
following equations :
1. x^ -S ax^ + Sa^x + a% = 0.
Transformation : x = a(l + 0).
2. (1 + c^)x^ - 3 ax2 + 3 a2x - a3 = 0.
Transformation : x = — - —
1 + z
3. Solve completely the equation
(1 + c3)x8 - 3(a + &c3)x2 + 3(a2 + i^cS-^x - (a^ + b^c^) = 0.
4. Apply the transformation x = ° "^ ^^ to the equation
x^ -\-px^ + qx + r = 0,
and assign to a and /3 such values as will reduce this equation to
the form
Az^ + B = 0.
TEANSFOEMATION. 649
Discuss the following special cases :
(1) pq = 9r; (2)Spr = q^; (3)3g=p2;
(4) (pq -dry- 4(3 pr - q^) (Sq- p"^) = 0.
Transform the following equations into others, in each of which
there shall appear no term of the second degree :
5. a;3 + 3a:2-9x + 1 =0. 7. x4-8x8+24 x2-28x-85=0.
6. 3x3+15x2+25x+c=0. 8. x^-Sx^-{-lSx^-lbx-^6=0.
Transform into equations whose coefficients are all integral the
following :
9. x3 + 1 x2 + i X + f = 0.
10. X* + j\X^ + 7^X2 + 3j|^X - 5^2 = 0.
Solve the following equations :
11. 5x3 + 3x2- 3x-6 = 0.
12. x* - 2 x8 - 6 x2 - 2 X + 1 = 0.
13. X* - 2 ax8 + 2 ax - 1 = 0.
14. x« - 3 x5 + 2 x* - 2 x2 + 3 X - 1 = 0.
16. x4 - 2 x8 - 50 x2 + 2 X + 1 = 0.
16. x' - 4 x6 - X* + x8 + 4 x2 - 1 = 0.
17. Prove that every reciprocal equation through division (if
necessary) by x + 1, or x — 1, or x2 — 1, may be reduced to one
which is of even degree and whose coefficients satisfy the condition
Qr = an-r-
650
CRITERIA FOR REAL ROOTS.
CHAPTER XLVII.
Criteria for Real Roots.
. 451. If f(x) be a rational integral function, and if f(a)
and f(P) be of opposite sign, f(x) will vanish for so7ne
value of X between a and jB.
For, when x passes, by continuous change, from a to p,
fix) varies continuously [Art. 442], and therefore passes
through all intermediate values between /(a) and f{(S).
One of these intermediate values must be zero, since /(a)
and /(/8) are of opposite sign.
The graph of the function makes the truth of this
proposition evident to the eye.
If MA, NB (Fig. 12) be the ordi-
nates /(a), /(^), the one positive,
the other negative, the curve, in
passing from A to B, will cross
the a;-axis at some point whose
ordinate f{x) is zero. There may
be several such points within the
interval MN.
Fig. 12.
Ex. Find limits between which the roots oix^ — Zx'^ — x + \ = Q
lie.
If/(x) = xs - 3 x2 - ic + 1, then
/(-I) = -2, /(t))=l, /(I) = -2, /(3) = -2, /(4)=13.
Hence one root lies between — 1 and 0, a second between 0 and
1, a third between 3 and 4.
\
CRITERIA FOR REAL ROOTS. 651
452. If /(a) and /(/?) have opposite signs, f{x) = 0 has
an odd number of real roots between a and /3, but if f(a)
and f(l3) have like signs, f(x) — 0 has either an even num-
ber of real roots between a and y8, or none.
If A and B be the points whose ordinates are /(a)
and /(^) respectively, then the graph of f{x), in passing
from A to B, will cross the aj-axis an odd number of times
if A and B be on opposite sides of this axis, but an even
number of times, or not at all, if A and B be on the same
side of the a^axis.
453. Every equation of odd degree has at least one real
root, positive or negative, according as the constant term is
negative or positive.
For, if f(x) be of odd degree, and j>„ be its constant
term,
/(+Qo)=+QO, /(0)=p„, /(-Qo)=-oo;
and when p„ is positive, f(x) = 0 has a real root between
0 and — cc, but when p^ is negative, the root is between
0 and H- oo .
454. Every equation of even degree, whose constant term
is negative, has at least two real roots, one positive, the
other negative.
For, if f(x) be of even degree, and p^ be its constant
term,
/(+<^)=+^, /(0)=p„ /(-c») = + ^;
and since p^ is negative, f{x) = 0 has one real root be-
tween 0 and -|- oo, and a second between 0 and — oo.
652 CRITEKIA FOR REAL ROOTS.
455. If an equation with real coefficients have the com-
plex root, a + ih, it also has the conjugate complex root,
a - ip:
Assume /(a 4- ifi) = 0. By Taylor's theorem,
/(„ + h) =/(a) + hf(a) + |l/'(a) + ....
The value of /(a + ifi) is obtained from this formula
by writing ift for h] and since the even powers of
i = ± 1, and its odd powers = ± i, the result is of the
form
/(a + ip) = A-{- iB,
where A and B are real quantities. By hypothesis,
/(a4-W = 0;
.-. ^ = 0 and J5 = 0. [Art. 234]
But /(a — ip) can be obtained from /(a + i^) by chang-
ing i into — i ; therefore,
f{a -ib) = A-iB = 0',
that is, a — i/S is a root of f(x) = 0. Thus, complex roots
always occur in conjugate pairs.
456. If an equation with rational coefficients have the
irrational root, a + -^(3, it also has the conjugate irrational
root, a — -y/jS.
By writing ^/3 in place of i^, the proof of this propo-
sition becomes identical with that of the preceding
article.
CRITERIA FOR REAL ROOTS. 663
457. An equation whose coefficients are integers, and the
coefficient of whose highest term is unity, cannot have a
rational fraction as a root.
For, if possible, suppose the equation
a;" = — p^a;"-^ — pg^""^ — * • • — i>«
to have the root a/^, in which a and (3 are prime to one
another. Then, substituting a/p for x, and multiplying
by ^-\
^ = -p,a--'p -p^^-'p' p„l3--\
an impossible result, since an irreducible fraction cannot
be equal to an integer.
458. Descartes' Rule. When in a sequence of terms, or
expressions, their signs, H , are ranged in their proper
order, they are said to present a series of permanences and
variations; a permanence when two like signs are in
juxtaposition, as + +, or , a variation when two
unlike signs are in juxtaposition, as + — , or — -|-.
Thus the sequence + — — 4- -f — has two perma-
nences and three variations. The following theorem,
known as Descartes' Kule of Signs, counts the number of
variations presented by the terms of an equation in order
to give information concerning the number of its real
roots. It refers only to equations with real coefficients.
An equation f(x)=0 cannot have more positive real roots
than f(x) has variations of sign, nor more negative reaX
roots thanf{— x) has variations of sign.
654 CRITERIA FOR REAL ROOTS.
Consider any multinomial having both permanences
and variations, as + ^ — — — + + — -|- — _ -f.
We introduce a positive root by multiplying the multi-
nomial by a binomial whose signs are + — . The signs
involved in this multiplication range themselves thus :
+
+
—
—
—
+
+ -
- + -
—
4-
+
—
H-
+
—
—
—
+
-f -
- 4- -
—
+
—
—
+
+
+
— -
- 4- -
+
+ -
4-±-T=F4-±-H =FH
The double signs indicate ambiguities, uncertainties.
Here are the signs of the multiplicand and product
arranged in groups, separated by asterisks, to call atten-
tion to the variations in the multiplicand :
4-± -TT -h ± - + - T + -
Note the following particularities :
(1) In the product, every ambiguity, or group of
ambiguities, has adjacent to it, -|- on one side, — on the
other.
(2) Therefore, every sequence of signs enclosing such
an ambiguity (or group), as-f ± — , or— q= q= -|-,
presents at least one variation, in whatever way the
ambiguities are interpreted.
(3) To every variation in the multiplicand corre-
sponds, in the product, either a variation, or an
ambiguity.
(4) A new variation (not represented in the multi-
plicand) is introduced at the end of the product.
CRITERIA FOR REAL ROOTS. 655
It follows at once that the variations in the product
exceed those in the multiplicand by at least one.*
But we may suppose that all the positive real roots
of an equation are introduced in the manner above in-
dicated, namely, through multiplication by a binomial
whose signs are -\ — . Hence an equation cannot have
more positive real roots than its terms have variations
of sign.
Since the negative roots of f(x) = 0 are the positive
roots of f(—x) = 0, the second part of Descartes' rule is
merely the first part restated for the equation J{—x)=0.
Thus the above demonstration suffices for both parts of
the rule.
Ex. 1. The equation a^ + 5a;2-f3x — 7=0 has certainly one
positive and one negative real root, by Art. 454.- For positive
values of x its signs are + + H — , and for negative values + H .
Hence, by Descartes' rule, it has only one positive, and only one
negative real root.
Ex. 2. The equation x^ -^ qx + r = 0^ if q, r be both positive,
has only one real root. For if x be positive, the signs are + -f +>
and if x be negative, they are 1-. The real root is negative.
459. The following propositions are consequences of
Descartes' rule of signs. The proofs of them, here
omitted, are suggested as exercises for the student.
1. Whe7i all the roots of an equation f(x) = 0 are real,
the number of its positive roots, and of its negative roots,
is exactly equal to the number of variations presented by
f(x) and by f{— x) respectively.
* Possibly by more than one, always by an odd number.
656 CRITERIA FOR REAL ROOTS.
J
2. If the terms of f(x) are all positive, the equation
f(x) = 0 has no positive real root.
3. If all the coefficients of the even poiuers* of x in
f(x) are positive (or all negative) and all the coefficients of
the odd powers of x are negative (or all positive), the equa-
tion f(x) = 0 has no negative real root.
4. If f(x) contain no odd powers of x and the coeffir
cients are all positive^ (or all negative), the equation
f(x) = 0 has no real root.
5. If f(x) contain no even powers * of x and the coeffi-
cients are all positive (or all 7iegative), f(x) = 0 has the real
root x = 0, but no other.
EXAMPLES OXI.
One root of each of the following equations is indicated ; find
all the others :
1. 60a;8-55a;2 + 34a; + 6=:0; f(l-V^).
2. «* + 4 a;3 - 5 x2 + 6 X - 1 = 0 ; i (1 - V^=^).
3. 4x*-9x2+'3x + l=0; ^(l+\/2).
4. a;5-4ic4 + 9x3-x2 + 4x-9 = 0;2+ V^^.
Assign a superior limit to the number of real roots which each
of the following equations may have, under the conditions enu-
merated :
6. x^ + qx + r = 0;q>0.
6. x^ + qx + r = 0; g < 0.
7. £C* + 5x2 + rx + s = 0 ; Q' > 0.
8. x^ ± (ax^ + 6) = 0 ; a > 0, 6 > 0.
9. x^ ± (ax2 _ 6) = 0 ; « > 0, 6 > 0.
10. x"^ + ax* + 6x3 + c = 0 ; a, b, c unrestricted.
* It is understood that the constant term is the coefficient of the
even power x^.
CRITERIA FOR REAL ROOTS. 657
Determine, under the conditions indicated, the exact number
of the positive and negative real roots of each of the following
equations :
11. x/^ + (qx^ + rx)-s = 0; g > 0, r > 0, s > 0.
12. x* ± (qx^ -\-rx)-s = 0; g > 0, r > 0, s > 0.
13. x^ + (a - 2)x2 - 2 (a - 2) X + 2 a = 0 ; a>2.
14. x2«+i + ax2«-i + 6x2 + 1 = 0 ; a>0, b>0.
15. ic2« + ax* + 6x3 - 1 = 0 ; qj > o.
For each of the real roots of the following equations find con-
secutive integers between which the root lies :
16. 3 x3 - 8 x2 + X + 2 = 0. 18. x5 + 5 x2 - 3 = 0.
17. x4 + 3x2 -60 a: + 34 = 0. 19. x5 + 5x2 + 3 = 0.
20. x7 + 7 x4 + 3 x3 - 1 = 0.
2n
658 Sturm's theorem.
CHAPTER XLVIII.
Sturm's Theorem.
460. The process*- through which an accurate account
of the character and position of the roots of an equation
having real coefficients may h6 rendered was discovered
by the French mathematician Charles Sturm, who made
known his discovery to the Academy of Sciences at Paris
in 1829.* The results of this investigation are embodied
in what is known as Sturm^s Theorem.
461. Let the rational integral function of x, of the n^^
degree, with real coefficients, be denoted by X, its first
derivative by Xi. The equation X = 0 shall have no
multiple root, and consequently X and X^ shall have no
common factor [Art. 437]. Divide X by Xj, representing
the quotient by Qi, the remainder by — Xg; then divide
Xi by Xg, representing the quotient by Q2, the remainder
by — X3. Eepeat this process of dividing the last divisor
by the last remainder with its signs changed, until a numeri-
cal remainder — X^ (not containing x) is obtained. These
successive divisions produce j)—l quotients, Qi, Q2,'" Qp-i,
and j9 — 1 modified remainders, X2, Xg, ••• X^, viz. the true
* Published in Memoires presentes . . . par des Savants Etran^
gers, Paris, 1836.
stukm's theorem. 659
remainders with all their signs changed. The series of
expressions,
(i). X, Xi, X2, Xg, ••• Xp,
are called Sturm's Functions. Since in the division process
dividend = divisor x quotient + remainder,
these functions and the quotients Qi, • • • Q^_i satisfy the
identical relations :
X =XM-x^
Xi = X2V2 — x^j
(ii.) X, =X,Q,-X^
Xp_2 = Xp_iQp_i — Xp.
The essential properties of Sturm's functions which
render them valuable for our purpose are the following :
1. They are all (except the last) continuous functions
of a; [Art. 442].
2. The last, Xp, being independent of x, preserves its
algebraic sign unchanged whatever variations x may
undergo.
3. Two consecutive functions cannot become zero for
the same value of x ; for if two consecutive functions be
zero, all are zero, and X, Xi have a common factor.
4. If, for a given value of x, any one function, as X^,
vanish, the two adjacent functions, X^.i, X^+i (on either
side of X^), have opposite signs ; for if X^ = 0, then
5. Whenever x, while increasing, passes through a
root of X = 0, the ratio X/X^ changes from a negative
to a positive value.
660
A formal proof of the fifth proposition is required:
Since X and Xi cannot both vanish for the same value
of X, zero can be neither a maximum nor a minimum
value of X [Art. 438], and therefore, in passing through
zero, either X is first negative, then zero, then positive,
or it is first positive, then zero, then negative. In the
former case, X is an increasing function, and Xj is posi-
tive ; in the latter case, X is a decreasing function, and
Xi is negative [Art. 438]; and therefore in both cases
the ratio X/Xi passes from a negative value through
zero to a positive value.
462. Sturm's functions are applied to the determinar
tion of the number of the real roots of an equation,
X = 0, that lie between any two numbers, a and ^
LP > a], in the following manner :
In each of Sturm^s functionSy
X, Xi, X2, ••• Xp_i, Xp,
substitute a for x, producing p -f 1, positive and negative,
numbers. Count the variations in sign that occur in pass-
ing from the first to the second, to the third, . . . to the last,
of these numbers, and denote the number of variations by
v^. Substitute similarly p for x, count again the variations
in sign, and denote their number by v^. The difference
is the exact number of the real roots of X= 0, that lie
between a and ft. This is Sturm's Theorem.
To prove this theorem, let x be supposed to pass by
continuous increase from a to p. Since X, Xi, X2, •••,
Xp_i, are all continuous functions of x, none of them can
x„
^r+l
. In fact,
thef(
Before the change.
^.-1
Xr
Xr^i
(1)
—
+
+
(2)
—
—
+
(3)
+
+
—
(4)
+
—
—
Sturm's theorem. 661
change sign without passing through zero, and the sign
of Xp is invariable. But when any one except the first
becomes zero, say X^ = 0, the two adjacent functions,
X^_i, X^^i, have opposite signs, and a change of the sign
of X^ from -f to — , or from — to +, cannot affect the
number of variations presented by the three terms X^_^
possible alternatives are:
After the change.
Xr-i X^ X^+i
- - +
- -h +
+ - -
+ + -
In every instance, one permanence and one variation.
Hence no variations are gained or lost through any
changes of sign that can occur in the functions Xi, Xg,
-Xs) "'j Xp.
Let X now pass through a root of X=0. At this
instant X changes sign, and the ratio X/X^ passes from
a negative to a positive value. Hence only four different
groupings of the signs of X, X^ X2, before and after x
passes the root of X = 0, are possible, namely :
Variations
Before. After, lost.
(1) + - + - - + 1
(2) + - - _ _ _ 1
(3) - + 4- + + 4- 1
(4) - + - + + _ 1
In every case one, and only one, variation is lost ; and,
since the vanishing and consequent changing of sign of
662
STURM S THEOREM.
the ratio X/X^ presents the only readjustment of signs
that can cause the disappearance of variatigjis in sign,
no such disappearance will take place except when x
passes a root of X = 0. Hence the number of real roots
of X = 0 between a and y8 is exactly v^ — v^. q.e.d.
Hence, the total number of real roots of an equation^
X=0, is
463. If fractional coefficients make their appearance
in Sturm's process, they may be got rid of by introducing
or removing suitable positive factors, as is done in finding
highest common factor [Art. 135]. When the process is
thus adulterated, Sturm's functions are correspondingly
modified, but without losing any of their fundamental
characteristics described in Art. 461.
Ex. 1. Determine the number of the real roots of the equation
a;3 - 5 jc + 5 = 0.
We have X= x^ — 5 x + 5, Xi = 3x2 — 5. The division pro-
cess, for the determination of X2, X3, etc., may be exhibited as
follows :
X =
x^ - 5x+ 5
3
3x3-15x + 15
3x8- 5x
3x2- 5
2
lOx-15
X2= 2x- 3
6x2- 10
6 x2 - 9 X
9x - 10
9x - ^
Xi
3x+f = $2
'-■^+ 10 = -| = X8
Sturm's theorem.
Thus Sturm's functions (modified), four in number, are
x3_5x + 5, 3x2-5, 2x-3, - |.
The following table of variations shows that the equation X= 0
^ only one real root, a negative one, between — 2 and — 3 :
Value of %. Sequence of Signs. Variations.
-00 _ + _ _ 2 *
+ 00 + + + _ 1
-3 _ + _ _ 2
-2 + + _ _ 1
The following is a graph of the function X:
y = aj8 — 5x + 5
X
y
-4
-39
-3
- 7
-2
7
-1
9
0
5
1
1
2
3
3
17
Fig. 13.
Until some test is applied, it is doubtful whether the curve
crosses the horizontal axis between the abscissae 1 and 2. The
test by Sturm's functions shows that it does not. By Horner's
process, the real root is found to be — 2.627+.
By the criterion for maxima and minima, [Art. 438], the co-
ordinates of the elbow-points, J., i?, in the graph, are — f,
5+10v'i^5, and f, 5 — lO^j^, and this result also shows that
the graph does not cross the a;-axis a second time.
464. In order that all the roots of an equation of the
wth degree may be real and distinct, it is necessary and
664 Sturm's theorem.
sufficient that the series of Sturm's functions shall lose
exactly n variations when x passes from — oo to -f oo.
There must therefore be just n + 1 Sturm's functions,
whose degrees are respectively n, n — 1, "• 2, 1, 0, alter-
nately even and odd; and this series must present, for
ic = — 00 ^ variations, for ic = -h oo n permanences, since
we must have
"W-oo — y+00 = n.
But this requires that the coefficient of the first term
(term of highest degree) of every one of the functions
X, ••• X„ shall be positive. Hence:
In order that an equation of the n^^ degree may have
all its roots real and distinct, it is necessary and sufficient
that its Sturm^s functions shall he n-\-l in number, and
shall have positive numbers only as coefficients of their
terms of highest degree.
Ex. 1. The equation x^ — 6x + 3 = 0 has all its roots real;
between what integers do they lie ?
Sturm's functions (modified) are :
a;3_5a; + 3, 3a;2 + 6, lOx-9, \5^;
four in number, coefficients of terms of highest degree all positive.
The sequences of sign for aj = — 3, — 2, 0, 1, 2 are :
Value of X.
Sequence of Signs.
Variations.
-3
- + - +
3
-2
+ + - +
2
0
+ - - +
2
1
- - + +
1
2
+ +. + +
0
Thus one root
lies between - 2 and — 3,
another between 0
and 1, a third between 1 and 2.
Sturm's theorem. 665
465. If among Sturm's functions an intermediate one,
as X^, is found, which cannot change its sign when x
ranges through the values to be assigned to it, the func-
tions which follow it may be disregarded ; the computa-
tion may stop with X^. The functions X, Xi, Xg,."- X^
then determine the number and position of the real roots ;
for they have all the fundamental properties which suf-
fice to prove Sturm's theorem.
Ex. 1. Thus, for the equation x^ + Bx^-\-k = 0 the only func-
tions that need be considered are
X=(i(^ + 6ci^ + k, Xi = 6(x^+ 3 x2),
for X* 4- 3 x2 cannot change its sign for real values of x. The appli-
cation of the test gives
V-oo — V+oe =1 — 0 = 1.
The equation has only one real root.
EXAMPLES CXII.
Determine, by Sturm's theorem, the number and position of
the real roots of the following equations :
1. x8-3x2 + 3a:-ll = 0. 6. x'^ + ix^ + ix^ + 48 = 0.
2. x^-Zx^-3x + ll=0. 6. x5-5x-l = 0.
3. .x3 - 2 x2 -}- X - 1 = 0. 7. x* + 3 x* -f 3 x2 -I- a2 = 0.
4. 2x8-7ax2+10a2x-6a3=0. 8. x"^ - 7 x^ 4- 1 = 0.
9. Show that, if a and h be positive real numbers,
x^ -1- 5 ax^ + 5 6x -f c = 0
has one, and only one, real root.
10. Show that, if a, m, and n be positive real numbers,
X2m+1 _|_ ax2'»+i 4-5=0
has one, and only one, real root.
666
MISCELLANEOUS EXAMPLES VIII.
11. Show that two of the roots of x'^— 4iX + Sa = 0 are neces-
sarily imaginary, and that the other two are real if a<l, or
a = 1, imaginary if a > 1.
12. Show that
x4 _ 2 x3 - (2 (z - 3) a;2 + 2 (a - l)x + (a - 1) (a - 2) = 0
has all its roots real if a > 1, or a = 1, all its roots imaginary if
a<l.
13. Show that
x* - 2 (a + l)x3 + (a2 + 4 rt + 1) x2 - 2 (a3 + 1) a; + a2 = 0
always has two real and two imaginary roots except when a^ = I.
14. Obtain Sturm's functions for the cubic x^ + qx -\- r = 0, and
show that the condition, necessary and sufficient, that the three
roots of this equation shall be real is 4:q^ -{■ 27 r^<iO.
15. Making use of the results of Example 14 and of the trans-
formation of Art. 408, obtain Sturm's functions for the general
cubic x^ + px^^ -{- qx -}- r = 0.
16. Obtain Sturm's functions for the equation x"" — x-{- 1 =0
and show that it has only one real root if n is odd, none if n is even.
MISCELLANEOUS EXAMPLES VIII.
1. Verify that
1 a a2
=
1 a
be
-
1 6 62
1 b ca
1 c c2
1 c ab
.
2. Prove that
- 6c 6c + 62 6c + c2
,
ca + a2 — ca ca + c2
a6 + a2 a6 + 62 - aU
= 2
(b + C)2 C2 62
c2 (c + ay a2
= 2 (6c
62
(
a -
hby
3. Solve the equation
x — a x — b X — c
x — b X — c X — a
X — c x — a x—b
= 0.
MISCELLANEOUS EXAMPLES Vm. 667
4. Verify tliat
a-^ a^ a^
a\bi aihi a^hz
6i2 bi^ 632
= (a2&3-
- azbi) (asfti - ai&s) (ai&2
a2&i).
Find the rational linear factors of the following expressions :
5. x\y + 2!) + y\z + x) + z\x + ?/)-(«/ + 0) (2 + x) (x + t/).
6. X3 + y3 + ^3 _|_ 2 x?/0 - X2(j/ + 2;) - 2/2(5; + x) - z\x + y).
7. 2(1/202 + ;22aj2 ^. ^.2^2) _ a;4 _ y4 _ ;j4.
8. Show that each of the linear factors of
X2(X2 - 4 yt - 22) _ y2(y2 _ 4 X0 - «2)
+ 22(^2 _ 4 y^ _ a;2) _ ^2(^2 _ 4 2;x - y2)
is of the form % -^ y\ -\- z>? -\- 1\^, where X* - 1 = 0.
9. Show that if ax^ + 2hxy + by^ -^2 gx -^ 2fy + c he factor-
able and one factor is x + Xy + /*, the other is
ax-\-(2h — Xa)?/ + 2 g — fxa.
10. Prove that if x* + px^ + qx^ -\- rx -\- s = 0 have three equal
roots, q"^ - 9 pr -{- 12 s z= 0.
11. Given, that a, /3, 7 are the roots of x^ +px2 + gx + r =.' y,
write the equation whose roots are :
(1) a2, )82, 72; (2) l/a2, I//32, I/72.
12. Given, that a, /S, 7 are the roots of x^ 4-l>x2 + gx + r = 0,
find the roots in terms of o, /3, 7, of
r2x8 + (2 qr -p^)x^ + (q^ - 2pr)x + r = 0.
13. The roots of x^ +px^ + gx + r = 0 being a, j3, 7, express the
cyclo-symmetric function a2/3 + /32«y _|. ^2^ Jq terms of p, g, and r.
14. Solve the equations z^ = 1 and 2;2o z= — 1.
15. Show that ax" + & = 0 may be expressed in the form of a
reciprocal equation.
16. Given x2 — 2;x + 1 = 0 and f(z) = x> + 1, prove that all the
roots of f{z) = 0 are real and are situated between + 2 and — 2.
668 MISCELLANEOUS EXAMPLES YUL,
17. Construct the graph of (x^ — l)/x^.
18. Find the maximum and minimum values of
2 x3 + 3 x2 - 36 X - 10.
19. Determine the shape of the rectangle of maximum area that
caii be inscribed in a circle of giren radius.
20. Determine the shape of the rectangle of maximum area
that can be inscribed in a given square.
21. Determine the shape of a triangle of given area and mini-
mum perimeter. •
22. Solve completely the equation
(1 - c4)x4 -iax^ + Q d?-x^ - 4 a^x + a* = 0.
23. Solve completely the equation
(1 - c*)x4 - 4 (a - 5c*)x3 ^^{a^- hH'^)x^ + 4 (a^ - h^cS)x
4-(a*-&4c4)=o.
24. Find the relation between the coefficients of
X* + px8 + gx2 + rx + s = 0,
y/Ji^ich will make it possible to reduce this equation to the form
25. Solve the simultaneous equations
— a a — b a — c
y — a y — b y — c
26. Prove that all the roots of the equation
X — a X — /3 X — y X— V
are real.
MISCELLANEOUS EXAMPLES VIH. 669
27. Show that the three roots of
— a^(x — a) — h'^{x — jS) — c'^(x — 7) — 2 ahc = 0 are real.
28. Prove that
/W^^_. _1_ .^_ , . ^_ .
f(x) x-a'^x- ^^x-y'^ '^ x-v
where a, /3, 7, ... v, are the roots of f{x) = 0.
29. Given, thsU f(x) = ^^~^^, and that /(«) (x) = Xn = n*!^
derivative of /(x), verify that
(l-x2)/(x)+2wx/(x)=0,
(1 - x^)Xn" - 2 xX„' + n(n + 1)X„ = 0,
where X„", X„' = first and second derivatives, respectively, of X„.
30. Apply the transformation x = "^ "^ ^ to ax^ + 2bx + c
yz+ S
and, denoting the transformed function by a'z^ + 2b'z + c', prove
that
a'cf - 6'2 =(a5 - pyy(ac - 62).
31. Apply the transformation x =^£±E to ax^+Shx^+Zcx-td;
yz+8
and, denoting the transformed function by a'z^ + 3 b'z"^ + Sc'z -^ d',
prove that
a'2d'2 + 4 a'c'3 + 4 d'6'8 - 3 6'2c'2 - 6 a'b'c'd'
= (ad - pyy(a'^d^ + 4 ac3 + 4 d&3 - 3 62c2 _ 6 abed).
670
TABLE OF LOGARITHMS.
N.
0 12 3 4
6 6 7 8 9
d.
10
12
13
0000 0043 00S6 0128 0170
0212 0253 0294 0334 0374
41
0414 0453 0492 0531 0569
0792 0828 0864 0899 0934
1139 1 173 1206 1239 1271
0607 0645 0682 0719 0755
0969 1004 1038 1072 1 106
1303 1335 ^3^7 1399 1430
38
35
32
'14
1461 1492 1523 1553 1584
1761 1790 1818 1847 1875
2041 2068 2095 2122 2148
1614 1644 1673 1703 1732
1903 1931 1959 1987 2014
217;^ 2201 2227 2253 2279
^0
28
26
17
18
19
20
21
22
23
2304 2330 2355 2380 2405
2553 2577 2601 2625 2648
2788 2810 2833 2856 2878
2430 2455 2480 2504 2529
2672 2695 2718 2742 2765
2900 2923 2945 2967 2989
25
24
22
3010 3032 3054 3075 3096
3118 3139 3160 3181 3201
21
3222 3243 3263 3284 3304
3424 3444 3464 3483. 3502
3617 3636 3655 3674 3692
3324 3345 3365 3385 3404
3522 3541 3560 3579 3598
37" 3729 3747 3766 3784
20
19
18
24
3802 3820 3838 3856 3874
3979 3997 4014 4031 4048
4150 4166 4183 4200 4216
3892 3909 3927 3945 3962
4065 4082 4099 41 16 4133
4232 4249 4265 4281 4298
18
17
16
27
28
29
30
31
32
33
4314 4330 4346 4362 4378
4472 4487 4502 4518 4533
4624 4639 4654 4669 4683
4393 4409 4425 4440 4456
4548 4564 4579 4594 4609
4698 4713 4728 4742 4757
16
15
15
4771 4786 4800 4814 4829
4843 4857 4871 4886 4900
14
14
13
13
4914 4928 4942 4955 4969
5051 5065 5079 5092 5105
5185 5198 5211 5224 5237
4983 4997 50" 5024 5038
5119 5132 5145 5159 5172
5250 5263 5276 5289 5302
34
35
36
5313 5328 5340 5353 5366
5441 5453 5465 5478 5490
5563 5575 5587 5599 561 i
5378 5391 5403 5416 5428
5502 5514 5527 5S39 5551
5623 5631 5647 5658 5670
13
12
12
37
38
39
40
5682 5694 5705 5717 5729
5798 5809 5821 5832 5843
591 1 5922 5933 5944 5955
5740 5752 5763 5775 5786
5855 5866 5877 5888 5899
5966 5977 5988 5999 6010
12
II
II
6021 6031 6042 6053 6064
6075 ^85 6096 6107 61 1 7
"
N.
0 12 3 4
5 6 7 8 9
d.
TABLE OF LOGARITHMS.
671
N.
0 12 3 4
6 6 7 8 9
d.
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
11
59
60
61
62
63
64
^1
69
70
6021 6031 6042 6053 6064
6075 6085 6096 6107 61 1 7
10
10
10
10
10
9
9
9
9
6128 6138 6149 6160 6170
6232 6243 6253 6263 6274
6335 6345 6355 6365 6375
6435 6444 6454 6464 6474
6532 6542 6551 6561 6571
6628 6637 6646 6656 6665
6721 6730 6739 6749 6758
6812 6821 6830 6839 6848
6902 691 1 6920 6928 6937
6180 6191 6201 6212 6222
6284 6294 6304 6314^6325
6385 6395 6405 6415 6425
6484 6493 6503 6513 6522
6580 6590 6599 6609 6618
6675 6684 6693 6702 6712
6767 6776 6785 6794 6803
6857 6866 6875 6884 6893
6946 6955 6964 6972 6981
6990 6998 7007 7016 7024
7033 7042 7050 7059 7067
9
8
8
8
8
8
8
8
7076 7084^023.7101 7110
7160 7168 7177 7185 7193
7243 7251 7259 7267 7275
7324 7332 7340 7348 7356
7404 7412 7419 7427 7435
7482 7490 7497 7505 7513
7559 7566 7574 75^2 75^9
7634 7642 7649 7657 7664
7709 7716 7723 7731 7738
7118 7126 7135 7143 7152
7202 7210 7218 7226 7235
7284 7292 7300 7308 7316
7364 7372 7380 7388 7396
7443 7451 7459 7466 7474
7520 7528 7536 7543 7551
7597 7604 7612 7619 7627
7672 7679 7686 7694 7701
7745 7752 7760 7767 7774
7782 7789 7796 7803 7810
7818 7825 7832 7839 7846
7853 7860 7868 7875 7882
7924 7931 7938 7945 7952
7993 8000 8007 8014 8021
8062 8069 8075 8082 8089
8129 8136 8142 8149 8156
8195 8202 8209 8215 8222
8261 8267 8274 8280 8287
8325 8331 8338 8344 8351
8388 8395 8401 8407 8414
7889 7896 7903 7910 7917
7959 7966 7973 798o 7987
8028 8035 8041 8048 8055
8096 8102 8109 81 16 8122
8162 8169 8176 8182 8189
8228 8235 8241 8248 8254
8293 8299 8306 8312 8319
8357 8363 8370 8376 8382
8420 8426 8432 8439 8445
7
7
6
6
6
8451 8457 8463 8470 8476
8482 8488 8494 8500 8506
6
N.
0 12 3 4
5 6 7 8 9
d.
672
TABLE OF LOGARITHMS.
N.
0 12 3 4
5 6 7 8 9
d.
70
71
72
n
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
96
97
98
99
100
8451 8457 8463 8470 8476
8482 8488 8494 8500 8506
6
8513 8519 8525 8531 8537
8573 8579 8585 8591 8597
8633 8639 8645 8651 8657
8692 8698 8704 8710 8716
8751 8756 8762 8768 8774
8808 8814 8820 8825 8831
8865 8871 8876 8882 8887
8921 8927 8932 8938 8943
8976 8982 8987 8993 8998
8543 8549 8555 8561 8567
8603 8609 8615 8621 8627
8663 8669 8673 8681 8686
8722 8727 8733 8739 8745
8779 8785 8791 8797 8802
8837 8842 8848 8854 8859
8893 8^99 8904 8910 8915
8949 8954 8960 8965 8971
9004 9009 9015 9020 9025
6
6
6
6
6
6
6
6
5
9031 9036 9042 9047 9053
9058 9063 9069 9074 9079
5
5
5
5
5
5
5
5
5
5
9085 9090 9096 9101 9106
9138 9143 9149 9154 9159
9191 9196 9201 9206 9212
9243 9248 9253 9258 9263
9294 9299 9304 9309 9315
9345 9350 9355 936o 9365
9395 9400 9405 9410 9415
9443 9450 9455 9460 9465
9494 9499 9504 9509 9513
9112 9117 9122 9128 9133
9163 9170 9175 9180 9186
9217 9222 9227 9232 9238
9269 9274 9279 9284 9289
9320 9325 9330 9333 9340
9370 9375 9380 9385 9390
9420 9425 9430 9435 9440
9469 9474 9479 9484 9489
9518 9523 9528 9533 9538
9542 9547 9552 9557 95^2
9566 9571 9576 9581 9586
5
9590 9595 9600 9605 9609
9638 9643 9647 9652 9657
9683 9689 9694 9699 9703
9731 9736 9741 9745 9730
9777 9782 9786 9791 9795
9823 9827 9832 9836 9841
9868 9872 9877 9881 9886
9912 9917 9921 9926 9930
9956 9961 9965 9969 9974
9614 9619 9624 9628 9633
9661 9666 9671 9675 9680
9708 9713 9717 9722 9727
9754 9759 97^3 9768 9773
9800 9805 9809 9814 9818
9845 9850 9854 9859 9863
9890 9894 9899 9903 9908
9934 9939 9943 9948 9952
9978 9983 9987 9991 9996
5
5
5
5
5
4
4
4
4
0000 0004 (X)09 0013 0017
0022 cx)26 0030 0035 0039
4
N.
0 12 3 4
5 6 7 8 9
d.
\ ■:
.^S.
' J-
..4- I
.^10.
$\
m
.*.-,
l-S^'-
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