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: V
^ -V
American iWatfjemattcal Series;
E. J. TOWNSEND
GENERAL EDITOR
COLLEGE ALGEBRA
sS\ BY
H^L^EIETZ, Ph.D.
A. E. CEATHORNE, Ph.D.
Assistant Professor of Mathematics
UNIVERSITY OF Il.l.IMUS
REVISED EDITION
NEW YORK
HENRY HOLT AND COMPANY
H-^L >?*/i
Ooptbioht, 1909, 1919,
BT
HENRY HOLT AND COMPANY.
PREFACE
This book is designed primarily for use as a text-book in the
freshman year of colleges and technical schools. Special atten-
tion is directed to the following features :
(1) The method of reviewing the algebra of the secondary
schools.
(2) The selection and omission of material.
(3) The explicit statement of assumptions upon which the
proofs are based.
(4) The application of algebraic methods to physical problems.
For the majority of college freshmen, a considerable period of
time elapses between the completion of the high school algebra
and the beginning of college mathematics. The review of the
secondary school algebra is written for these students. This part
of the book is, however, more than a hasty review. While the
student is reviewing a first course, he is at the same time making
a distinct advance by seeing the subject-matter from new view
points, which his added maturity enables him to appreciate. For
example, the functional notation, graphs, and determinants are
introduced and used to advantage in the review. The extensions
of the number concept receive fuller treatment than is usual in
a college algebra. The various classes of numbers from positive
integers to complex numbers are treated in the order in which
they are demanded by the equation.
The application of algebra in the more advanced courses in
mathematics has been an important factor in determining the
subject-matter. Not only are some of the topics usually treated
in the traditional course in algebra entirely omitted, but in each
chapter the material is restricted to the development of those
central points which experience has shown to be essential. While
a complete discussion of limits and infinite series does not prop-
erly belong in a course in algebra, it has been thought best to
111
421357
iv PREFACE
■
include an introduction to these subjects which covers in con
siderable detail only the theory necessary to a discussion of the
comparison and ratio tests. From the experience of the authors,
a great deal is gained by thus taking a very elementary first
course in limits and series.
While it is out of place in a book of this character to attempt
a critical study of fundamentals, great care has been taken to
point out just what is proved and what is assumed in so far as
a first-year student can be expected to appreciate the necessity
of assumptions.
Without trying to teach physics or engineering, many problems
are introduced in which the principles of algebra are applied to
physical problems, but no technical knowledge is assumed on the
part of the student. Rules for the mechanical guidance of stu-
dents in solving problems have been used sparingly.
The authors take great pleasure in acknowledging their in-
debtedness to their colleagues in the mathematical and engineer-
ing departments of the University of Illinois. We are indebted
to Professors Haskins and Young for suggestions during the
preparation of the manuscript as well as for a critical reading of
the manuscript; to Professors Townsend, Goodenough, Miller,
Wilczynski, Dr. Lytle, and to Professor Kuhn of Ohio State Uni-
versity for suggestions upon the manuscript ; to Professor Watson
for some of the practical problems ; and again to Professor Good-
enough for assistance in seeing the book through the press.
H. L. RIETZ.
A. R. CRATHORNE.
PREFACE TO REVISED EDITION
Suggested by experience in the class-room, a number of topics
have been simplified in treatment. In response to requests of
numerous teachers, several hundred new exercises and problems
have been introduced. The book has thus been freshened with
regard to both text and problems without however changing its
general character.
EL ±j. R.
January, 1919. A. R. C.
CONTENTS
CHAPTER I
INTRODUCTION
ARTICLE
1. Numbers .
2. Graphical Representation of Real Numbers
3. Greater and Less
4. Definitions and Assumptions
6. Derived Properties of the Numbers of Algebra
6. Positive Integral Exponents
l
7. Meaning of a«
8. Meaning of o*
v
9. Limitation on the Value of o«
10. Meaning of o°
11. Meaning of o m when m is Negative
12. Radicals
PAGE
1
1
3
3
7
10
12
12
12
13
13
14
CHAPTER II
ALGEBRAIC REDUCTIONS
13. Algebraic Expressions 18
14. Removal of Parentheses 19
15. Complex Fractions . . . . . . .19
16. Factoring 21
17. Radicals and Irrational Numbers 22
18. Reduction of Expressions containing Radicals to the Simplest Form 25
19. Addition and Subtraction of Radicals 26
20. Multiplication and Division of Radicals 26
21. Evaluation of Formulas 27
22. Imaginary Numbers 30
CHAPTER III
VARIABLES AND FUNCTIONS
23. Constants and Variables
24. Definition of a Function
31
31
VI
CONTENTS
ARTICLE PAGE
25. Functional Notation 32
26. System of Coordinates 33
27. Graph of a Function 35
28. Function defined at Isolated Points 36
29. Zeros of a Function 39
CHAPTER IV
THE EQUATION
30. Equalities
31. Definitions
32. Solution of an Equation ....
33. Equivalent Equations ....
34. Operations that lead to Redundant Equations
35. An Operation that leads to Defective Equations
36. Clearing an Equation of Fractions
41
42
43
44
45
46
47
CHAPTER V
LINEAR EQUATIONS
37. Type Form
38. Simultaneous Linear Equations ....
39. Graphical Solution of a System of Linear Equations
40. Determinants of the Third Order
41. Solution of Three Equations with Three Unknowns
49
50
52
53
54
CHAPTER VI
QUADRATIC EQUATIONS
42. Type Form
43. Solution of the Quadratic Equation ....
44. Solution by Factoring
45. Equations in the Quadratic Form ....
46. Theorems concerning the Roots of Quadratic Equations
47. Number of Roots
48. Special or Incomplete Quadratics ....
49. Nature of the Roots
50. Sum and Product of the Roots
51. Graph of the Quadratic Function ....
59
60
62
63
65
66
66
68
69
70
CONTENTS
Vll
CHAPTER VII
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS
ABTICLE
52. Type Form
63. Solution of Systems of Equations Involving Quadratics .
PAGE
75
76
CHAPTER VIII
INEQUALITIES
64. Definition . . .86
65. Absolute and Conditional Inequalities 86
56. Elementary Principles 86
57. Conditional Inequalities 88
CHAPTER IX
MATHEMATICAL INDUCTION
68. General Statement 90
69. Meaning of r' 92
60. Binomial Theorem ; Positive Integral Exponents ... 93
CHAPTER X
VARIATION
61. Direct Variation 97
62. Inverse Variation 97
63. Joint Variation 98
64. Combined Variation 98
CHAPTER XI
PROGRESSIONS
65. Arithmetical Progressions .
66. Elements of an Arithmetical Progression
67. Relations among the Elements .
68. Arithmetical Means
69. Geometrical Progressions
70. Elements of a Geometrical Progression
71. Relations among the Elements .
72. Geometrical Means
73. Number of Terms Infinite .
101
101
101
102
103
103
103
104
105
Vlll
CONTENTS
ARTICLE PAOH
74. Series 106
75. Repeating Decimals 106
76. Harmonical Progressions 107
77. Harmonical Means 107
CHAPTER XII
COMPLEX NUMBERS
78. Number Systems
79. Graphical Representation of Complex Numbers
80. Equal Complex Numbers ....
81. Addition and Subtraction of Complex Numbers
82. Multiplication of Complex Numbers
83. Conjugate Complex Numbers ....
84. De Moivre's Theorem
86. Roots of Complex Numbers ....
86. Division of Complex Numbers
110
111
112
113
114
115
116
117
119
CHAPTER XIII
THEORY OF EQUATION
87. The Polynomial of the nth Degree .
. 121
88. Remainder Theorem
. 121
89. Synthetic Division ....
. 122
90. Rule for Synthetic Division
. 124
91. Graphs of Polynomials
125
92. General Equation of Degree n .
. 126
93. Factor Theorem ....
126
94. Number of Roots of an Equation
126
96. Graphs of oo(x — n) (x — r 2 )- • • (x — r n ) .
128
96. Complex Roots
129
97. Graphs of f(x) when Some Linear Factors
are Imaginary .
130
98. Transformations of Equations .
131
99. Descartes's Rule of Signs .
135
100. Location of Roots by Graph
137
101. Equation in p-Form
139
102. Rational Roots
139
103. Irrational Roots. Horner's Method
141
104. Negative Roots
•
145
106. Summary
145
106. Algebraic Solution of Equations
148
107. The Cubic Equation .
149
CONTENTS
IX
ABTICLB PAGE
108. The Biquadratic Equation 151
109. Coefficients in Terms of Roots 153
110. Variable Coefficients and Roots 153
CHAPTER XIV
LOGARITHMS
111. Generalization of Exponents
112. Definition of a Logarithm
113. Derived Properties of Logarithms
114. Common Logarithms
116. Characteristic ....
116. Use of Tables ....
117. To find the Logarithm of a Given Number
118. To find the Number which Corresponds to a Given Logarithm
119. Computation by Means of Logarithms
120. Change of Base ....
121. Graph of logo £ ....
122. Exponential and Logarithmic Equations
128. Calculation of Logarithms
156
156
157
159
160
161
164
164
165
169
171
172
175
CHAPTER XV
PARTIAL FRACTIONS
124. Introduction
125. Case I
126. Case II
127. Caselll .
128. Case IV .
177
178
179
180
180
CHAPTER XVI
PERMUTATIONS AND COMBINATIONS
129. Introduction
130. Meaning of a Permutation
131. Permutations of Things All Different
132. Permutations of n Things not All Different
133. Combinations .
134. Combinations of Things All Different
135. Binomial Coefficients . . .
136. Total Number of Combinations
183
183
184
184
186
186
187
187
X CONTENTS
CHAPTER XVII
PROBABILITY
ARTICLE PA01
137. Meaning of Probability 190
188. Probability derived from Observation 191
189. Expectation of Money 191
140. Independent, Dependent, and Mutually Exclusive Events . 192
141. Repeated Trials 194
CHAPTER XVIII
DETERMINANTS
142. Extension of the Determinant Notation 196
(48. Properties of Determinants 198
144. Development by Minors 200
146. Theorem 203
146. Systems of Linear Equations containing the Same Number of
Equations as Unknowns 203
147. Systems of Equations containing More Unknowns than Equa-
tions 207
148. Systems of Equations containing Fewer Unknowns than Equa-
tions . 208
149. Common Roots of Quadratic and Higher Degree Equations in
One Unknown 209
CHAPTER XIX
LIMITS
160. Definition
161. Infinitesimals
162. Theorems concerning Limits ....
163. Both Numerator and Denominator with Limit
164. Infinity
166. Limiting Value of a Function ....
166. Indeterminate Forms
CHAPTER XX
INFINITE SERIES
213
214
215
216
217
218
218
167. Definition 221
168. Convergence and Divergence 221
CONTENTS
XL
Series with Positive Terms
ABTICLB
159. Fundamental Assumption
160. Comparison Test for Convergence
161. Comparison Test for Divergence
162. Summary of Standard Test Series .
163. Ratio Test for Convergence and Divergence
PAGE
223
224
227
228
229
Series with Both Negative and Positive Terms
164. Theorem ....
165. Ratio Test Extended
166. Alternating Series
167. Approximate Value of a Series
168. Power Series
169. Binomial Series
170. Exponential Series
171. Logarithmic Series
Answers .....
Index .....
232
232
234
234
236
238
240
241
243
265
A LIST OF SIGNS AND SYMBOLS
+, read plus. — , read minus.
X , or •, read times.
-*-, read divided by.
=, read is equal to.
=, read is identical with.
^=, read is not equal to.
->, read approaches.
<, read is less than.
>, read is greater than.
<, read is less than or equal to.
^, read is greater than or equal to.
a ! or [a, read factorial a.
( ) Parentheses.
t ] Brackets. Signs of aggregation. These signs are used
{ I Braces. >to collect together symbols which are to be
Vinculum. treated in operations as one symbol.
| Bar. J
a r , read a subscript r, or a sub r.
x 1 , x ,f •••, read x prime, x second ••• respectively,
lim x, read limit ofx.
<c->.oo, read x becomes infinite, or x increases beyond bound.
log a w, read logarithm ofn to the base a.
| a |, read absolute value of a.
a n , read a to the nth power, or a exponent n.
Va, read square root of a.
■\/a, read nth root of a.
f(x), <f>(x), etc., read "f" function ofx, " <f>" function ofx, etc.
n P r , read number of permutations of n things taken r at a time.
n C r , read number of combinations ofn things taken r at a time,
(x, y), read point whose coordinates are x and y.
zii
i
*
GREEK ALPHABET
Letters
Names
Aa
Alpha
B0
Beta
Ty
Gamma
A3
Delta
Ee
Epsilon
zr
Zeta
H,
Eta
90
Theta
Letters
Names
I t
Iota
Kjc
Kappa
AX
Lambda
M/x
Mu
Nr
Nu
E*
Xi
Oo
Omicron
Ilr
Pi
Letters
P/>
Z<r s
Tr
T v
Xx
Names
Rho
Sigma
Tau
Upsilon
Phi
Chi
Psi.
Omega
V
*s
COLLEGE ALGEBRA
CHAPTER I
INTRODUCTION
1. Numbers. In counting the objects of a group the child
makes his first acquaintance with numbers. These are the num-
bers called positive integers.
He next employs the number we call a rational fraction,
probably thinking of it first as an aliquot part, and later as
the quotient of two integers. Perhaps in the fall of a ther-
mometer below zero, the student had his first experience in the
use of negative numbers, even if he was not taught to use the
word negative. He may also come early to the convenient use of
the negative number to represent debit when the corresponding
positive number means credit.
To express the length of the diagonal of a square of side one
unit, or to find a number which multiplied by itself gives some
integer, not a perfect square, say 2, he uses a number which
is neither an integer nor a rational fraction, and employs a radi-
cal sign to represent it by writing V2 where V2 x V2 = 2.
Such numbers belong to a class of numbers known as irrational
numbers. (See p. 23.)
2. Graphical representation of real numbers. The four classes
of numbers mentioned in Art. 1 belong to the so-called "real
numbers " used in arithmetic and algebra. They may be repre-
sented by the points of a straight line as follows : Let X'X be
this line. (Fig. 1.) Choose a point on this line and call it the
zero point or origin. Adopt some unit of measurement OA:
Beginning at and proceeding in both directions, apply the
unit of measure to divide OX and OX' at equal intervals, thus
1
2 INTRODUCTION [Chap. I.
forming a scale of indefinite length. The positive and negative
integers may then be conveniently represented by the points
marking the intervals. Similarly, corresponding to any fraction
- (a and b integers), there can be constructed a point on X'X
b
such that the fraction denotes the distance and direction of the
point from 0. In fact, we assume that by means of this scale
we are able to represent conveniently all real numbers, and we
A
-• -• -7 -6 -5 -4 -3 -I -1 18 8 4 5 6 7 8 9
J" ' ' ' , ,'.' ',''''' U
P t 6 A P
Fiq. 1.
can say, to any point P on the line, there corresponds a number *
which indicates the distances and direction of P from 0, and con-
versely, we assume that to every real number there corresponds
a point of this line.
In addition to the real numbers, we shall find it desirable to
deal with so-called " imaginary " and complex numbers. A graphi-
cal representation is given for these numbers in Art. 80.
EXERCISES
1. What numbers are represented by the following points ?
(a) The point midway between 4 and 5.
(6) Points of trisection of the interval — 3 to — 4.
(c) Points of quadrisection of the interval 2 to 3.
2. State in words the position of points which represent J, $, — }, — 2, *■.
3. Suppose the scale of Fig. 1 represents the scale of a Fahrenheit ther-
mometer ; estimate the readings when the end of the mercury column stands
at Pi.t At P. At point midway between O and P. At point midway
between P and Pi.
4. A square piece of paper of side 4 is laid on Fig. 1 so that one corner
of the square is at O and the diagonal lies on the line OX. What number is
represented by the point at the other corner of the square which lies on OX ?
* For a more complete discussion, see Fine's Number System of Algebra,
Second edition, p. 41.
t Pi is read " P sub one. " See list of signs and symbols at the end of the
table of contents.
Arts. 2-4] DEFINITION AND ASSUMPTIONS 3
5. A circle of radius 1 rolls to the right along the line in Fig. 1, beginning
at O. What number is represented by the point at which the circle touches
the line after one complete turn ?
3. Greater and less. The terms greater than and less than which
are common to everyday life, when used in the technical sense
of algebra, are easily misunderstood. For this reason we point
out their geometrical significance. The real number A is said
to be greater than the real number B (written A > B) if the point
representing A falls to the right of the point representing B.
The number A is said to be less than the number B (written
A < B) if the point representing A falls to the left of that
representing B.
Exercise. Arrange the following numbers in descending order of
magnitude :
2, V2, -4, -VE, 3V5, -10, -20, J.
4. Definition and assumptions. Operations with numbers of
arithmetic suggest certain definitions and rules for algebra. The
student has probably performed algebraic operations according
to rules thus suggested by arithmetic without being conscious of
the assumptions which underlie these processes. We may now
proceed to a formal statement of assumptions made at the outset
in this algebra.
Letters are used to represent numbers; a number, which is
represented by a certain letter, is called the value of the letter.
In the following, let a, b, c represent any numbers.
The fundamental * operations of addition and multiplication of
numbers are subject to the following laws I-IX :
I. The sum t of two numbers is a uniquely determined number.
That is, given a and 6, there is one and only one number x
such that a + b =x.
II« Addition is commutative.
That is, a + b = b +a.
* The operations are fundamental in that no attempt is made to define
them. The "laws" are in the nature of assumptions since no attempt is
made to prove them.
t The sum is the result of adding ; the product is the result of multiplying.
4 INTRODUCTION [Chap. I.
III. Addition is associative.
That is, (a + b)+c = a+(b + c).
IV. If equal numbers be added to equal numbers, the sums are
equal numbers.
That is, if a = b 9
and c = d y
then a + c = b + d.
V. The prodtict of any two numbers is a uniquely determined
number.
That is, given a and b, there is one and only one number y
such that ab = y. In this case, a and b are said to be factors of y.
VI. Multiplication is commutative.
That is, ab = 6a.
VII. Multiplication is associative.
That is, (a6)c = a(bc).
VHI. Multiplication is distributive with respect to addition.
That is, a(b -f c)= ab -f ac.
IX. if egwaZ nwm&ers 6e multiplied by equal numbers, the prod-
ucts are equal numbers.
That is, if a = b,
and c = d,
then ac = bd.
The following laws X and XI lead us to definitions of subtrac-
tion and division, and enable us to give meanings to the symbols
0, -a, ?, 1, and \.
b b
X. Given a and b, there is one and only one number x, such
that x + b = a.
Subtraction is the process of finding the number x in x -f b = o.
In other words, to subtract b from a is to find a number x, called
the remainder, such that the sum of x and b is a.
Art. 4] DEFINITIONS AND ASSUMPTIONS 5
By X, the number x in x -f b = a exists when a = b. In this
case, the number x is called zero, and is written 0.
In symbols,
+ b = b. (1)
From X and the definition of 0, there exists a number x, such
that
x + b = 0.
In this case, x and b are said to be negatives of each other, and
x may be replaced by (— b).
If & is a positive number, a; is a negative number.
In symbols,
(-&) + & = 0, (2)
gives a definition of (— b).
XL GHven a arid b (6 =£ 0*), there is one and only one number
x, which satisfies bx = a.
Division is the process of finding the number x in bx = a. In
other words, to divide a by b is to find a number x, called the
quotient of a by 6, such that b multiplied by x gives a.
This quotient is often written -, and, when thus written, is
called a fraction.
In dividing a by b, the number a is called the dividend and b
the divisor, just as in arithmetic. Likewise, in the fraction -, a
is called the numerator and b the denominator.
By XI, the number x in
bx = a (bj=0)
exists when b = a, so that ax = a. In this case, the number a; is
called tm% and is written 1, that is,
- = !• (3)
a w
Further, by XI, and the definition of 1, there exists a number
x y which satisfies bx = 1 (6 ^= 0). This value of <b is called the
reciprocal of 6 and is written - •
b
* The sign =£ stands for " is no£ eguaZ to."
6 INTRODUCTION [Chap. I.
It should be noted that, by means of XI, a meaning is given
to unity and to the reciprocal of any number other than zero in
a manner analogous to that by which a meaning is given to zero
and to the negative by means of X.
The above propositions I-XI are stated for two and three num-
bers for the sake of simplicity. It can be proved from the given
assumptions that these propositions hold when three or more num-
bers are concerned in the process of addition or multiplication.
It is not to be inferred that these propositions I-XI are en-
tirely independent of one another, but rather that they constitute
convenient assumptions for the purposes of this course in algebra.
Although further assumptions are made later in the course, the
principles I-XI enable us to prove many important propositions
of algebra, and the wide application of algebra depends upon the
fact that many changes in the physical world take place in ac-
cordance with these laws.
As shown in the following exercises, the operations of algebra
are generalizations of the operations of elementary arithmetic.
EXERCISES
1. Show that 4x2, 9x7, 1 x - 1, - 4 x - 6, Ox- 2, J x - j, are
special cases of (x -f l)(x — 1) and of x(x — 2).
2. Show that 2x8x8, 5x6x11, J x J x V, -3x-2x3, — lx
O'xfi are special cases of x(x + f)(x +6) and of (x — l)x(x + 5).
3. Show that ^ = 0, 5| = 2, JL=-V5!=3, |=12, ^ = -8
x* — 4
are special cases of - % = x + 2.
x — 2
In the course of operations with the numbers of algebra,
the important question arises : Can any two given numbers be
added, subtracted, multiplied, or divided ? Our assumptions state
that the number HVHtom of algebra is such that this question can
be answered in the affirmative except in the case of division by zero.
Division by um> h (wvtmhd J)vm algebraic operations.
EXEROISES
1. Can any two jfiven numbers bo added, subtracted, multiplied, or di-
vided if the number §y*tt»m eonaku of positive integers only ? Illustrate
your answer.
Arts. 4, 5] DERIVED PROPERTIES OF NUMBERS 7
2. Can any two given numbers be added, subtracted, multiplied, or di-
vided if the number system consists of positive integers and quotients of
positive integers only ? Illustrate your answer.
3. Where is the flaw in the following ?
Let x = a (x:£0). (1)
Multiply both sides by x, x 2 = ax. . (2)
Subtract a 2 from both sides, x 2 — a 2 = ax — a 2 . (3)
Factor, (x — a) (x + a) = a(x — a). (4)
Divide both sides by x — a, x + a = a. (5)
But, by (1), x = a. (6)
By (5) and (6), 2 a = a. (7)
Hence, 2=1.
5. Derived properties of the numbers of algebra. From the fore-
going definitions and assumptions, the following propositions can
be proved. We shall present in detail the proofs of only a few
of them.
I. Adding a negative number (—a) is equivalent to subtracting
a positive number a.
That is, &+(— a)=6 — a.
To prove this, let b +(— a)= x. (I, Art. 4.) (1)
&+(-a)+a = s + a. (IV, Art. 4.) (2)
But ( - a) + a = 0. (Eq. (2), page 5.) (3)
From (2) and (3), b + = x + a. (4)
But 6 + = 6. (Eq. (1), page 5.) (5)
From (4) and (5), b = x + a. (6)
6 — a = a. (Def. of subtraction.) (7)
That is, b +(- a)=b-a.
n. Subtracting a negative number (—a) is equivalent to adding
the positive number a.
That is, b — (— a)= 6 + a.
in, 2%€ product of two numbers is when and only when at
least one of the numbers is 0.
Corollary. The quotient - is equal to when a is any num-
a
ber different from 0.
8 INTRODUCTION [Chap. L
IV. The product of a number a by a number (— b) is — a&.
To prove this, let a (— &)= x. (« exists by V, Art. 4.) (1)
Then a(- 6)4- «*> = x + ab. (IV, Art. 4.) (2)
a [(_ ft) + ft] = x + a6. (Vni, Art 4.) (3)
a-Q = x+ab* (Eq. (2),page5.) (4)
= x + a5. (in, Art5.) (5)
— ab = x. (Definition of negative.) (6)
From (1) and (G), a(— 6) = — ab.
V. jTO* product of ( — cO &V (— 6^ i* ab.
To prove this, lot
(-«)(-*) = *. (1)
(_ a )(_ fc) + «(_ 6)= a- - d6. (IV, Art. 4; IV, Art 5.) (2)
( _ {,)[( - <,) + a ] = x - «6. (VIII, Art. 4.) (3)
-ft-Osi- «k (Definition of zero.) (4)
« x - «6. (Why ?) (5)
af> - x. (Why ?) (6)
Krom (1) and (<>), (— <ri(— *)= rt &-
Tim Mtatoumnt that in multiplication We siTpu ^rive plus and
unlike signs givs mi nun includes IV and V.
VI. The quotient of two numbers is positive if the signs of the
dividend and divisor are alike; negative if they are unlike.
VI L A single parenthesis may be removed when it is preceded by
a positive sign without changing the signs of the terms within iL
VIII. -I single parenthesis may be removed* when preceded by a
negative sign if the signs of terms within it are changed.
That is, — {a + b — c + d - <>)= — a — b + c — d 4- e.
IX. The indue of a fraction is not changed by multiplying or
dividing both the numerator and denominator by the same number.
m , , . a ax
That is, T = r~
6 bx
* The • is a sign of multiplication, thus, 5 • 6 = 80.
-t Art. 5] DERIVED PROPERTIES OF NUMBERS 9
X. Changing the sign of either the numerator or the denominator
f of a fraction is equivalent to changing the sign of the fraction.
mi . . — a a a
That is, — — =
b b -b
XI. Adding two fractions having a common denominator gives
* a fraction whose numerator is the sum of the numerators and whose
I denominator is the common denominator.
That is, £ + ^ = ^±i.
c c c
Likewise, «_* = ^.
c c c
! XII. The sum and the difference of two fractions are expressed by
' a . c ad + bc - a x ad — be , . 7
r + -; = — 73 — , and — — - = — — — , respectively.
s b d bd ' 6 d 6d ' ^ *
m
a c
We can reduce — and - to a common denominator, since, by IX,
b d
a ad , c be
b bd d bd
By XI, we can complete the process.
XIII* The product of two fractions is a fraction whose numera-
tor is the product of the numerators and whose denominator is the
product of the denominators.
That is, - • i = ^.
' b d bd
XIV. To divide one fraction by another, invert the latter and then
multiply one by the oilier.
a
t^o* ;<, a t c b ad
Inat is, --s-- r- = — •
b d c be
d
Propositions III, XI, XIII, stated for two numbers in each
case, are readily extended to three or more numbers.
10 INTRODUCTION [Chap. I.
**. Po aiti f e integral exponents. Definition. — The expression
a* is read " a exponent x n or the " octh power of a." When x is
a positive integer, a* is a short way of writing a* a* a ••• to x
factors.
For, if in and n are positive integers, by the associative law of
multiplication,
a m a* = (a • a - a —• m times)(a -a* a ••• n times)
= a • a • a — m + n times
Illnstrations : 5 , -5 4 = 5 r , 3.3».3*.3*= 3 U .
Illustration : (3«)* = 3»
ID. (a&c — )*■ = a m b m € m «•
Illustration : (3 • 5 • 7 • 4)* = 3* • 5 s - 7* 4*.
IT. ("V^"-
Illustration :
W ?'
V.
Illustration :
a*
►-•»
* (*•>»)-
Illustration j 5. = I .
5* 5>
Art. 6] INDEX LAWS 11
EXERCISES
What numbers should be written in the blank parentheses in the fol-
lowing?
1. 8 • 8> • 8* = 8< >. 2. (2»)3 = 2< >.
3. (*T = 4<>. 4. 5 4 -a 3 -2 5 = 5(>.9(>.2(>.
\4V 5 • 9 . 22
5. /?\ 6 .!??)!. 2* = 2< >-3< >. 6. (3.5.9)2.(6.9)3=S ( >-5 ( >.
- cfl • a 8 A x 3 • x 7 ^
7. - — =- 8. - — — • 9.
a* (x&)2
Perform indicated operations and simplify when possible.
W)
10. mn^Y. 11. (U*»)\ 12. (2 a + 62),.
V mn / V(4ax) 3 / v '
13. a2».a*. 14. a"" 1 • a n +! • a n . 15. — •
a n
16. a2n. a 2. 17. *1 18 /«&_ 2 \\
19. /=i*S\-. 2() /ZL^) 2 \
21 14 m7n5 + 7 mW 22 w+tff +W) + W)-
7 m 6 n«
23. (a» + 6»)(a n - &»). 24. (x"" 1 + y* 1 " 1 )^"" 1 — y*"" 1 )-
25. (p»-2+p»-i)(p»-p»-3). 26. K a2a? V) n \ 3 .
27 /x^r+i\n # 28 /^2-iy.
V x*py J \ x2» )
»a /a n+1 \ n **a / 7 a* a n+1 \ x*
29. (- — ) -^-a». 30. f * 1 -^- Ji T-)-- ± -r
31. ( <**** \\ 32. (**)»- (*»)« .
V a» / (x n ) n — (y n ) n
33 8a»»-63» > 34 ^" + 1 ,
2 a n - 6 n x n + 1
35# [t^Y ' yZ^i' * 36# ( 3 l >n - 2 n )( 9 l> 2n + 6 P ,, tf*+ 4 2n )-
37. Establish index laws II, m, IV, V, VI, stating at each step the prin-
ciple used.
38. Translate the six index laws from algebraic symbols into English.
12 INTRODUCTION [Chap. L
i
7. Meaning of a*. The proofs of the above six laws of indices
assume that the exponents are positive integers. According to
the definition of a" (Art. 6), such an expression as a J has no mean-
ing whatever. If we use such expressions, we must first give
them a meaning. It is convenient to define them in such a way
that the laws for positive integral exponents hold also for frac-
tional exponents.
Assuming Law I, Art. 6, to hold, we shall have
at >ai* *a$ = a^ + * + J = a.
Assuming that some number exists whose third power is a, we
will denote it by a*. Another way of writing a* is y/a, which
is read " the cube root of a." In general, if q is a positive integer,
- - - _ + _.|._-|-... tog terms
a 9 • a 9 • a 9 ••• to q factors = a 9 9 q = a,
i
and a 9 means a number whose q\h power is a. Another way of
i _
writing a* is -v^a, which is read " the <?th root of a."
Thus, the fractional exponent - serves the same purpose as the
radical sign V .
8. Meaning of a*. According to Law I, Art. 6, if p and q are
positive integers,
- 1 - - + ! + -+... to p terms £
a 9 • a 9 ' a 9 ••• top factors = a 9 9 9 = a«,
p
and a 9 means the pth power of the qth root of a.
That is, a 7 =(Vay.
9 # It will be seen later (Art. 86) that any number a has q
distinct qth. roots. Thus, the number 4 has the two square roots
± 2. We shall, however, for the present, consider only the arith-
i
metical or positive value oi a 9 when a is a positive number.
- Arts. 7-11] MEANING OF a m WHEN m IS NEGATIVE 13
1
* With this limitation, it turns out that a 9 (a > 0) has one and only-
one value. For example, 9^=3, and not ±3. Likewise, \/81=3,
t and not ± 3. If both the positive and negative roots are meant,
r we shall write both signs before the radical.
!i Without this limitation, it will be seen from the following
£ illustration that the pth. power of a gth root of a number is not
necessarily equal to a given qth root of the pth power :
(4^) 4 = 16, while the square root of 4 4 may be either 4-16* or —16.
10. Meaning of a . In order that the first index law may hold
for an exponent zero, it is necessary that
~ a°*a m = a *"* = a m , (1)
or, a = 1, if a =£ 0.
That is, any number a with the exponent is equal to 1, provided
a=£ 0.
11. Meaning of a m when m is negative. Let m= —n, where n
is a positive number. By Law I, Art. 6, and Art. 10,
a n • ar n = a n ~ n = a° = 1. (1)
Hence, ar n = — , if a ^= 0.
7 a n
er = A- (2)
a -m v /
That is, in a fraction, any factor of the numerator may be trans-
ferred to the denominator, or vice versa, if the sign of the exponent
of the factor is changed.
22.3-4 7-1.52 2 2
Example: y-^ = ^^ = ^^^ = 2* . 7"» • 3 -*. 5*.
We have now found meanings for fractional, zero, and nega-
tive exponents consistent with the first law of indices. To give
logical completeness, it is necessary to show that the meanings
are consistent * with all the laws of indices, but we shall assume
* See Chrystal's Algebra, Fifth edition, Part I, p. 182.
14 INTRODUCTION [Chap. L
this with the exception of one case which as an illustration we
prove as follows :
To prove a 9 • b 9 = (ab) 9 f where p and q are positive integers, let
P 9
x=*a 9 'b 9 .
Then, x 9 = (a* • b' q )
= Uv (&V (Law III, Art. 6.)
= a p b p (Definition of fractional exponent.)
= {aby. (Law III, Art. 6.)
That is, x is the qth root of (aby, or
£
x = (db) 9 '
p p p
Hence, a 9 • b 9 =(aby.
We have now attached a meaning to the expression a* pro-
vided x is any integer, or rational fraction, positive or negative.
Moreover, we assume if a is positive and as is a rational number
that the number described by a" exists. It is possible also to
give a meaning to a* when x is not a rational number. However,
that is beyond our present purpose. (See Art. 111.)
12. Radicals. As stated in Art. 7, the nth root of a, written
_ i
Va, has the same meaning as a*. The number under the radical
sign is often called the radicand and the number which indicates
the root is called the index. Thus, in Va, a is the radicand and
n is the index.
The rules of operations with radicals may be obtained at once
from the laws of indices. Thus,
ii i
I. Va.>y6 = a n ^ = (a6)» = Va6.
Va
II.
Vb
a nja
Arts. 11, 12]
RADICALS
15
EXERCISES
Obtain identical expressions free from negative exponents and simplify.
1. Screes.
4. (9)-*.
7. air*.
10. (0.49)*.
aPx- 1
13.
OXr*
2.
5x3
5. (9-i) "1.
8. (I*)- 1 -
11. (0.49)"*.
14.
2 ar*b*<r*
5-ia*6-6c-i
3. (9-1)*.
6. (**)*.
9 (A)"*-
12.
2~* - 3 2 - 5-»
2-3 .3-2.5'
Perform the indicated operations and simplify.
16. (aHzcfy.
15. (a*)*.
18. (aoari • y*)-«. 1* ^J •
21. (- xfy3z-»)*.
23. (x* + y*) (x* - y*).
25. (a* + 6*) (a* -6*).
27. (x~* + y~*)(x~*-iT*).
29. (a*-a~*)(a* + a~* + l).
31. (a* + 6*)3.
33. (p- 2 -2p-i + l)(p-i-l).
35. (6x2y-i-15yx-i)-s-8x""*y~*.
17. (16xV")*
- 20. («»"* •
22.
(?)"*•
24. (ao + 6*) (a» -6~*).
26. x2y/x*y* + -rrY
\ x*y*/
28. (xO + x*)(x°-x* + x).
30. (a*+6*)(a* + 6*).
32. (x-2+x-i+xO)(x-2— x-i-fxO).
34. (x~* + x~*)(x* + x*).
36. (x-2 + x-i-56)^(x-i-7).
38. (x-*-16ir 4 )^(x-i + 2ir 1 )-
37. (m-3 + l)-*-(m-i + l).
39. (crio - a-5 + 1) -*- (a-2 - a-i + 1).
y 40. (X* + X 1 + X* + X + X* + 30) -*- (X* + ** + *"*).
/ 41. (a*6-i + 4a6'*-2o*-12aHl + 96)^(0*6"* + 2a*- 36*).
42. (a"*x~* + 2 + 2 o*x* + o*x*) + (a"* + x*).
16 INTRODUCTION [Chap.
Obtain identical expressions involving radicals instead of fractional €
ponents and simplify.
43. x*y* 44. «"*&*. *5. (a +6)*.
46. (a -&)"&. 47. (z».y- m )". 48. a-3&(a&-*)~"*.
49. 7aV*. ^ 50. [ae»(y««)*]*. 51. [(a"* • x~*z-i)*;
Obtain identical expressions involving fractional exponents instead
radicals and simplify.
x*y°
52. \/x*y*z*. 53. \/(ar*xr*)*. «*54. Va*Va«x».
55. V5»i5 -f \^x*2« - \/«62io. 56. -J*
'a
57. J^*\/— • 58 W 59 VS ^ ir -
60. -fcLje.
Simplify the following.
61. L *- 1 • a^J"- ^62. L(a*+«)»>-«. (a 2 *)*^-
63. VxVWfc. 64. (a«*.a».6*») .
• [W(£n
■
66. If x" = a*, show that x • f - ] = x a .
67. The number of revolutions per minute of a water-wheel is given
the formula .
Ft
n = 68.26 ~t»
where F is the fall of the water in feet, H the horse power. Calculate
when F = 16, H = 100.
68. We may write 0.000,010 in the form 16 • 10-«. Write the follow!
numbers in briefer form by the use of negative exponents: 0.000,00
0.000,000,003, 0.000,001,6.
69. Calculate the t< ances.
[(i)- 2 -(2)- 2 ]; [• »]5 [(*)"' -2«]; [(i)-2- ( _ 2 )-*].
Art. 12] RADICALS 17
70. If 3* and 2* are substituted for x in the expression
x 6 -x* — 5x* + 5x* + 6x,
show that the results reduce to the same number.
71. Two spherical particles each one gram in mass whose centers are one
centimeter apart attract each other with a force of 0.000,000,066,6 dyne.
Express this number as an integer multiplied by 10 with an exponent.
72. By the use of negative exponents express a micron as a part of a
meter. (A micron is one millionth of a meter. )
73. The numbers 6867, 5896, 4861, 3934 each multiplied by 10~ 10 give the
wave lengths in meters of deep red, yellow, blue, and ultra violet light
respectively. Express each wave length in microns.
74. Some authorities say that the mass of a hydrogen atom is 10~ 2 * gram.
How would this number be written in ordinary decimal notation ?
CHAPTER II
ALGEBRAIC REDUCTIONS
13. Algebraic expressions. In algebra, an expression is a
symbol or combination of symbols that represents a number.
Thus, x 2 -f y 2 — 25 and £ gt 2 + vt are expressions, if x, y, g> t y v,
represent numbers. If x = 4 and y = 6, the first takes the value
27. If g = 32.2, t = 10, and v = 5, the second has the value 1660.
For different values of the letters, an expression represents in
general different numbers. On the other hand, the same numbers
may be represented by many different expressions. For example,
x 2 — 4 and (a?— 2) (as -f 2) represent the same number. Expressions
which are equal for all values of the symbols for which the ex-
pressions are defined * are called identical expressions. The state-
ment that two identical expressions are equal is called an identity.
Thus, x 2 — 4 and (x — 2){x -f 2) represent the same number no
matter what value is assigned to x, and the statement
aj2_4=(aj-2)(a?4-2)
is an identity.
Two expressions may be equal without being identical. Thus,
for x = 2 or — 5, the expressions x 2 + 2 x — 1 and 9 — x are equal,
but they are equal for no other values of x, and hence are not
identical. Frequently, in problems with which we shall deal, the
work is made easier by replacing expressions by identical but
simpler expressions.
Exercise. Which of the following are identical expressions ?
22 x 2 3 xv
x 2 -y 2 , (x-y)(x + y), 4x-y, — — -, 4x(x + 2/), *x 2 + 4xy.
* This statement implies that we may not assign values to the letters which
1 x
make the members meaningless. Thus, = 1 4- is excluded when
18
Arts. 13-15] COMPLEX FRACTIONS 19
14. Removal of parentheses. A single pair of parentheses can
always be removed from an algebraic expression. However, if
the sign before it is -f, no change is necessary, but if — , the signs
of the terms within it must be changed (Art. 5, VII, VIII).
Expressions often occur with more than one pair of parentheses.
When one pair occurs within another pair, other symbols besides
( ) are used as follows : [ ] called brackets, \ j called braces,
and called the vinculum. All parentheses may be removed
by first removing the innermost pair according to the rule for a
single pair ; next, the innermost pair of all that remain, and so on.
EXERCISES
Remove the parentheses from the expressions of the first four exercises.
1. 8a+(56-a)-(2a-7&).
2. a — (6 — c + d + e).
3. 2x-[3x-{x-(2x-3x + 4)}- (5x-2)].
4. 3x-{y-|>-(x + y) -{-y-ty-x-y)}]}.
5. Find the numerical value of the expression in Exercise 4 when
x — 2, y = — 1.
6. Without changing the value of the expression
4c + 36 + 4a — 5c 2 + 4d 3 ,
write it with the last three terms in a parenthesis preceded by a minus sign.
7. Fill out the last pair of parentheses in a + 6— (c + d) = a — d
-( )•
8. Find the numerical value of
3 x - (6x + [- 4 x - y - x]) - ( -x - 3 y),
when x = 2, y = 4.
9. Annex a pair of parentheses preceded by a minus sign to 6 a + 6 b —
(a + 6) so that the resulting expression equals a + b.
10. Simplify {x(x + y) - y(x - 2^)}{x(x - y) + y(x + y)}.
15. Complex fractions. A complex fraction is one which has a
fraction in the numerator or denominator or in both numerator
and denominator. The rules for the simplification of arithmetical
fractions apply to algebraic fractions no matter how complicated
the numerator or denominator may be.
20 ALGEBRAIC REDUCTIONS [Chap.
The main principle is that (Art. 5, IX) the value of a fracti
is not changed by multiplying numerator and denominator by t
same number.
As illustrated by the following exercises on complex fractioi
a simplification is often brought about if we select for the numt
by which to multiply, the lowest common denominator of t
fractions which are in the numerator and denominator of the co
plex fraction.
Simplify:
1.
*-»
a a
_ a— 6 a + b
3. —^ —
b a
a — 6 a + b
!-«
5. £ •
1 + a
7.
X*
1-i
9.
m n
n in
(m+_n}2_ 4
97171
S
11.
EXERCISES
2.
1-1
X*
i-i
X
4
2ro
m 1
171
4m
6.
„ 1
x —
X
x + i+2
X
8.
x_y
y z
y z
10.
X
1
X* 1
y 2 g 12. — -
*_! + ! i--
y 2 ' y x 1 + ±
x
13. -V_5- . 14. *
y x y 1 X s
Arts. 15, 16] FACTORING 21
x-2+ — —
15. i+ 2 - 16. \ + *
X + 2+—L- 1 L_ l- *
x-2 1+x 1-x
1-1
1-
x a + x a — z
17. ,JL. 18. a ~ X a + X
_1_ a + x . a — a?
x 2 g — x g + x
x 2 -l —5-+ n
19.
X
x + i+1
X
21.
1 1
X X + ft
ft
1 1
23.
X* (x + ft)*
ft
E
on
t
20. "- 1 W+1
n n
n— 1 n + 1
x -f ft x
22. s+ft-1 «-l
24.
A-
1
r
8
r 2_
-s 2
1 +
7iJ
3p
Express the following as complex fractions and then simplify.
«... gx~ 2 «*% or" l 6*
27. -=— • 28. — — •
g*x-> g-*6 2
29. a " arl » 30. q ~ 2 - b ~ 2 .
car 1 — xr* a-i _ ftri •
31 g-i62x- 2 -2g-^V , 32 g&~ 2 + bar*
ar*xr* — 2 g-fy-» " g -2 — g- 1 ^- 2 + ft -2 "
*
33 x + ^-x 2 )* w 34< (g 2 - *')* a
x + x(a 2 - x 2 )"* x(g 2 — x 2 )""*
V^ 16. Factoring. Certain useful methods of resolving algebraic
expressions into their factors are illustrated in the problems
which are worked out in the following list.
22 ALGEBRAIC REDUCTIONS [Chap. II
EXERCISES
Separate each of the following into two factors.
1. 6 a 2 6x3 + 2 ab*x A + 4 abx*. 2. 10 m 4 n 2 — 16 m*n*.
Solution : 2 a6x 8 (3 a+ 6x+2 x 2 ). 3. 4 a 2 bc + 8 a 2 6 2 c 2 - 16 a'ft^.
4. 14 xyz — 7 xtyz + 28 xy 2 z»
^5. a 2 + a6 + ac + 6c.
Solution : a 2 + ab + ac + be = a(a + 6) + c(a + 6).= (a + c) (a + 6).
6. x 2 -f 6 x + xy + 6 y. 7. ax — 3 by + 6x — 3 ay.
8. xy + x — 2 y — 2. 9. m 2 — n 2 — (wi — n) 2 .
10. x 2 -2x — 8.
Solution : The result will be of the form (x + a) (x + 6). This expande<
is x 2 -f (a + 6)x + ab. Comparing with x 2 — 2 x — 8, we see that we can fine
factors of the given expression if we can find two numbers a and b such tha
a+ b = — 2, ab = — 8, or a = — 4, 6 = 2. Hence, x — 4 and x + 2 are factor
of x 2 — 2 x — 8.
11. x 2 — 3 x + 2. 12. 2 x 2 - 6 x + 4.
13. 2/ 2 -7s/ + 12. 14. a 2 + 7 a- 30.
15. s 2 -4s-60. 16. a 2 6 2 -7a6 + 10.
17, 8x 2 -18x + 9.
18. a3-&3.
Solution : a3 — &3 = (a _ 6) (a 2 + ab + 6 2 ).
„*» 19. a* - 27. 20. m* - 1.
21. 16 a 4 -6 4 . ^22. a 4 -81.
23. r« - ss. 24. 1 - 66.
25. 81s 4 -49* 4 . 26. a3-(a-&)3.
27. xty 2 — 9 a 2 6 2 . 28. x 2 + (6 - a)x — ab.
29. (3x-4) 2 -(2x-2) 2 . 30. a 2 + 2 a6 + 6 2 - c*.
/ 31. x 2 + y 2 -9x 2 y 2 + 2xy. 32. 64a3-27y3.
33. 1-343x3. 34. 27a3-8x*.
35. a3 + 63. 36. 27 + 8 x3.
^-37. 64 x6 + 126 y3. 38. x*y* + 216.
39. a* -6*. 40. (a + 6 + c) 2 -(a- & - c)».
41. x 4 + xV + y 4 . 42. x 9 - 8 x3 — xe + 8.
43. 1 + a 2 + a 4 . ^ 44. 16 x 4 + 4 a 2 6 4 x 2 + a 4 6*.
45. a 4 + 4. 46. (3 m + 2)3 + (2m + 1)».
17. Radicals. A rational number is one which can be repre-
sented as the quotient of two integers. In operations with radi-
Arts. 16, 17] RADICALS 23
cals we deal with irrational numbers ; that is, with numbers which
cannot be expressed as the quotient of two integers.* Any irra-
tional number can always be inclosed between two rational num-
bers which differ from one another by as small a number as
we please.
Thus, we may write,
1 <V2<2,
1.4 < V2 < 1.6,
1.41 < V2 < 1.42,
1.414 < V2 < 1.415,
Either of the sequences of numbers in the two outer columns
determines the square root of 2 in the same way that the sequence
.3, .33, .333, .3333, ••• determines the fraction £. Any irrational
real number may be determined in this way by a sequence of
rational numbers. Besides those numbers in which a radical sign
is used to express their exact values, there are many others which
belong to the class of irrational numbers. For example, the num-
ber it is an irrational real number determined by the sequence 3,
3.1, 3.14, 3.141, 3.1415, ....
Rules of operation with expressions involving radicals are given
in connection with the treatment of fractional exponents (Art. 12).
♦To show that V2 cannot be expressed as the quotient of two integers,
suppose it possible that V2 = — ,
n
7ft *
where — is a rational fraction in its lowest terms. At least one of the
n
numbers m or n is odd. Clearing of fractions and squaring both sides, we
get 2 n 2 = m 2 .
From this equation m 2 is an even number, hence m is an even number. If
m is even, m 2 contains the factor 4. Hence n 2 is an even number, and n is
7ft
itself even. This is contrary to the hypothesis that — is a fraction in its
lowest terms.
This proof is found in Euclid, and is supposed to be due to a much earliei
mathematician than Euclid.
24 ALGEBRAIC REDUCTIONS [Chap. II.
EXERCISES
Introduce the coefficients * of the following seven radicals under the radi-
cal sign.
1. 2 V8. Solution : 2 #3 = VI . V3 = #12.
2. SVll. 3. 2 #6. 4. 2 #7.
5. 2#13. 6. 7 #6. «-> 7. 4 #3.
8. Perform operations in Exercises 1-7, using fractional exponents in-
stead of radicals.
Change the following seven expressions into equivalent expressions having
no fractions under the radical sign.
9. J*. Solution: J?^ = #2 . #1 = 1 V6.
X 3 \3 V3 V3 V3 3
10. #J. 11. #|. 12. #f
13. 3#f m 14. 6#|. 15. 7vJ.
16. Change the radicals in Exercises 9-15 to fractional exponents and
reduce each to a fraction whose denominator is free from fractional expo-
nents.
Change the following eight radicals into equal expressions with as small
a positive integer as possible under the radical sign.
17. V8. Solution : #8 = VTT2 = #4 • #2 = 2 V2.
18. #82. ^19. Vl8. 20. V126.
21. #48. 22. #81. *• 23. #800.
24. #2187.
25. Perform operations in Exercises 17-24, using fractional exponents.
Change the following into equivalent expressions whose indices are the
smallest possible positive integers.
26. #4.
Solution : #4 = V VI = V2.
27. #8000. 28. #8l. ^ 29. #169. 30. #848.
31. #64.
32. Perform operations in Exercises 26-31, using fractional exponents.
i
* By the coefficient of a radical, we mean the number by which the radi-
cal is multiplied. In general, the coefficient of any symbol is the expression
by which the symbol is multiplied.
Arts. 17, 18] RADICALS 25
Change the following fractions into equivalent fractions, having no radicals
in the denominators.
33.
1+V2
Solution: 1 _ = 1 _ • * ~ V ^ = y/2 - 1.
1 + V2 1 + V2 1 - V2
34. — -• 35. — :' 36. ■ -• ^37.
^38. — r^ — = . 39. , 5 _ . 40. 4
36.
1
-V3
5
V6 V2 1-V3 ^ V8 + 3
V6 + V3 v/6-Vj V2-V3
41. Given y/2 = 1.4142, V3 = 1.7321, V5 = 2.2361, evaluate the expres-
sions in Exercises 33, 34, 36, 38 (to four significant* figures).
fl8. Reduction of expressions containing radicals to the simplest
form. An expression containing radicals is said to be in its
simplest form with regard to the radicals when, (a) there are no
fractions under the radical signs; (6) the radicand contains no
factor raised to a power whose exponent equals the index of the
root ; (c) the indices of the radicals are positive integers and as
small as possible ; (d) there are no radicals in the denominator.
EXERCISES
Reduce to the simplest form.
1. ^W* .
Solut,on: j^±Vj = ^V4 + 2V| = ^2 + j i V3
V3 V3 V3
= \/2 + $VS . V3^ ^108 4-2
V3 VS 8 '
2. v^. 3. V|. 4. — • 5. V1216. 6. y/32.
y/VL
* In giving a result such as 2.2361 to four significant figures, we give
2.236. In giving the same result to three significant figures, we give 2.24
rather than 2.23, for 2.24 differs less from 2.236 than 2.23 differs from 2.236.
In fact, it is usually desirable in giving any number of figures of an approx-
imate result, to find whether the next figure beyond those to be retained in
the result is less or greater than 5 ; for, if we should obtain a result 2.23 and
know that the next figure > 6, the result should be given to three significant
figures as 2.24
26 ALGEBRAIC REDUCTIONS [Chap. II.
7. In^f. 8. 3V * . 9. ^+*5-V8. la vi±VJ.
7 + V3 S 2 + 6VJ V2-V6+V3 Vf-VJ
Express in terms of radicals in simplest form.
11. 3* + 2* u- 2(6)* + (8)* b 1X g<-g-». ^ x-l__x*-l.
3* -2* / 3(6)* -2(3)* a*-a"i x*-l x*+l
19. Addition and subtraction of radicals. Radicals having the
same index and same number under the radical sign are called
similar radicals. Similar radicals can be combined in addition
and subtraction.
EXERCISES
Perform the operations indicated.
1. 2V6 + 8V6 + V24-2V64.
Solution : 2 V5 + 3V6 + V24 - 2 V64 = 2 VS + 3 V6 + 2>/6-6\/6= V6.
2. V3 + 4V3-2V3. 3. V24 + V64 - V96.
4. V48 xy* + yy/Wx + V8 x(x — • 9 y)\
5. \/3 + V27 + 3V / 27-v / 76 + v / 9. 6. VI28 - 2V60+V72 - Vl8.
7. feVa + xVa— y\/a3. 8. Vj + >/f - Vf .
t~^$. 2 Vf + J V60 + Vl6 + Vf ? 10. 6y/Ii* + 2 v / 2x'+ ^8x».
11. Sv/a + SVa + Va + SVa — 2>/a — 7^0.
12. 20(3)* - 4(3)*. 13. (243)* +(27)* +(48)*-8*.
14. (406)* -(180)*. 15. (136)* -(40)*.
20. Multiplication and division of radicals. In Art. 12 we have
seen that \/a • y/b = -\Vab and ^~ = yl- • In multiplying or
dividing radicals with different indices, the radicals should first
be reduced to the same index.
EXERCISES
Perform the indicated operations.
1. </S • V6.
Solution : ^8 = \/\7§ = \/9.
\/b = VyfiM = #126.
Henoe, \VS . V6 = #9 • vT26 = \/fl25.
Arts. 18-21] EVALUATION OP FORMULAS
27
2. V3.\/6. 3. V2a.v / 3o2. 4. 6V3-
6. W- #11. 7. 3v^a.4v^6.
9. (2V3 + 3V2)(3v / 3~V2). .10.
11. (Vm-fVn) 3 . 22.
13. (3VH-a-4Va)(vT+a+2\/a). 14<
15. y/X' y/y> y/z. 2.6.
70 #9
18.
17. _
19. a/2 . VI • #3.
21. <^ + *E
20.
22.
6Vf.\/2. x 5. 4vH.3^4.
8. (V7-l)(Vf+l).
(HV2-4Vl6)(V6 + \/6).
(V^12-vl9)("V^12 + Vl9).
(x 2 -x>/2 + l)(x2+»v / 2 + l).
V2*
4^9
■ •
2V3
-*- + #
. « + J|.
23
25. iia*.
3*
7*
27. — •
7*
24.
29. ^12 a* h- V8a»;
V28x
■ •
\/42x
7*
26. i--
7*
28. Wab -f- vl26 b*a.
30. V6ax 2 -*- #9 a^x 4 .
21. Evaluation of formulas.
1. The speed u in feet per second of a projectile of weight w pounds, and
diameter d inches is given by
1 = 1 ft* 1
v v 7000lo ,
where »o* is muzzle speed with which the shell is projected, and t is the
number of seconds after leaving the muzzle. Find v when Vo = 2800, t = 4,
d = 14, w = 1400.
826 V y
Find H when T = 400, w = 0.7, fir = 32.2, and v = 80.
3. r =
(p+5)c-»>
JB
Find T when a = 1222, b = 0.01869, JB = 86.24, v = 1.660, P= 11000.
* A letter with a subscript, say a*, is read, " a sub r."
28 ALGEBRAIC REDUCTIONS [Chap. IL
4. w = w' — s($o — «i).
Find w when
10' = 10.700,
« = 0.0060,
So = 8.5,
«i = 10.9.
5. The specific gravity S of a floating body is given by the expression
8 =
w x
where W\ is the weight of the body in air, w 2 is the weight of a sinker in
water, and wz is the weight in water of the body with sinker attached.
Determine the specific gravity of a body when by physical measurements
it is found that
wx = 16.60,
i0 2 = 182.3,
w 3 = 176 6.
*~ 6. One cubic centimeter of mercury at x° centigrade increases in volume
when heated to y° by an amount given by the following formula :
My - 3)
100
1 + 4*'
100
where A = 0.018. Find the increase in volume when the temperature is
raised from 12° to 126°.
7. The sag of an overhead trolley wire in an electric tramway is given
by the formula
where d is the number of feet in the sag, I is the number of feet of wire
between poles, L is the number of feet from pole to pole. Find the sag
when poles are 100 feet apart and length of wire is 100. 1 feet.
8. The time in seconds required for the discharge of water from one
vessel to another through an orifice in the side is
t _ 0.116. A-B. (J^-y4)
(A + B) . a '
where F and / are the differences in the heights of water in the two vessels
at the beginning and end respectively of the discharge, a is the area of the
orifice, A is the area of a horizontal section of the discharging vessel, and B
is that of the receiving vessel (measurements in inches) .
Find t when F= 128, /= 92, A = 96, B = 60, and the orifice is a circle
one inch in diameter.
Art. 21] EVALUATION OF FORMULAS 29
9. The area in square feet of the top of a well-designed chimney is given
by the formula .4 = 03 -2-
where Q is the quantity of coal in pounds used per hour and h is the height
of the chimney. What should be the area of the top of a chimney 160 feet
high which is connected with a furnace using 11,000 pounds of coal per hour?
• 10. To correct a barometer reading for temperature the following amount
is subtracted from the reading :
B m(t- 32)- s(t-G2)
1 + m (t - 32) '
where B is the barometer reading in inches, t the temperature in degrees
Fahrenheit, m = 0.00010, s = 0.00001. What is the corrected reading of the
barometer when the temperature is 76° and the barometer reads 29.96 ?
11. The quantity of water in cubic feet per second flowing through a
rectangular weir is given by the formula
Q = 3.33 • [i - 2 hlh%,
where h is the depth of water over the sill of the weir in feet, and L the
length of the sill.
Find Q, where L = 26, h = 1.6.
r 12. Let P be the day of the month, q the number of the month in the
year, counting January and February as the 13th and 14th months of the
preceding year, N the year, and n = I — — | — 2.
If P + 2g+ p2i!l] + * + [!]_„
be divided by 7, the remainder will be the day of the week of a given date
where Sunday counts as the first day. The expressions in brackets mean
the largest integer contained in the inclosed number. Verify this formula
for the present date.
13. Three equal uniform rods of weight w and the length I are jointed to-
gether to form a triangle ABC ; this triangle is hung up by the joint A,
and a weight W is attached to B and G by two strings of length — . The
compression in BC is given by *2
2V3 V3
Obtain x correct to nearest unit when TT= 1000, w = 600.
y 14. The area of a triangle whose sides are a, 6, c is given by the formula
V«(« — a) (« — b)(s — c)
where = a "*""*" c . Calculate from this formula the area of a triangle
2
whose sides are 6, 12, and 13 inches.
30 ALGEBRAIC REDUCTIONS [Chap.
15. The formula for the horse power H.P. of an automobile engin
by H.P.= Kanc ,
J (24) (83000)'
where P is the pressure in pounds per square inch, I is the length of str
of the piston in inches, a is the area of the end of the piston in square incl
n is the number of revolutions of the fly-wheel per minute, c is the num
of cylinders in the engine. How many H.P. are developed by a six-cylin
engine if P is 72, I is 6, a is 15.56, and n is 1200 ?
22. Imaginary numbers. The square root of a negative nu
ber is an example of an imaginary number mentioned in Art
To perform operations with square roots of negative numbe
replace any such number, say V— a, by i'Va, and operate witl
as with any other letter ; but replace i 2 in any expression by —
This method of operation is justified in Chapter XII.
EXERCISES
Perform indicated operations and simplify when possible by maki
t2=-l.
1. (l + i)(2-i).
Solution : (1+ 0(2 — i)= 2 + i— #,
= 2 + i + 1, since i* = — 1,
= 3 + 1
2. (.-Od + O. 3. (-I + if)(-|-^).
4. (-l + V-S)*. 5. (x + ai)(x-ai).
1+V3I
8. (24 + 7i)(24-7f). /*' lm _^Zi'
10. (l-i2)(l + i2). 11. (4 + 3i)2+(4-3f) > .
12. (a+6i)3 + (a-&i) 3 . 13. (3 + 4i)< +(3 - 4i) 4 .
- »• (-HfH-l+'fH-
15. (x — a — bi) (x — a + 6i).
CHAPTER III
VARIABLES AND FUNCTIONS
23. Constants and variables. A constant is a symbol which
represents the same number throughout a discussion. A variable
is a symbol which may represent different numbers in the discus-
sion or problem into which it enters. Many mathematical expres-
sions contain both variables and constants. Except in certain
geometrical and physical formulas it is customary to use the
letters a, b, c, ••• from the beginning of the alphabet for con-
stants and the letters ••• x, y, z, from the end of the alphabet
for variables.
Exercise. If A and B are two points in a plane and a point P moves in a
circle about A as a center, which of the distances PA, PB is constant?
which variable ?
24. Definition of a function. Many problems in mathematics,
physics, engineering, and chemistry involve two variables which
are so related that, a value of one being given, the other can be
found. The relation between the variables may be exhibited in
various ways. Sometimes the values of the variables are arranged
in the form of a table. For example, a life insurance agent refers
to a table to find the premium corresponding to a given age.
Here the two variables are " premium " and " age."
In algebra one variable may be connected with another in an
equation or one variable may be an algebraic expression contain-
ing the other. In the equation 3 a? — 5 y — 4, if a value be given
to x the corresponding value of y can be found. Thus if x = 0,
y = — £ ; if a? = 1, y = — £, and so on.
In evaluating the expression x 2 + x + 1, we find that x 2 + x + 1
= 1 when a; = 0, x 2 + x + 1 = 3 when x = 1 and so on. Fixing
the value of a? in the first illustration fixes the value of y ; in the
second illustration fixing the value of x fixes the value of x 2 + x + 1.
31
32 VARIABLES AND FUNCTIONS [Chap.
Definition. If two variables are so related that when a ?
of one is given, a corresponding value of the other is determi
the second variable is called a function of the first.
Thus in the equation Sx — 5 y = 4, y is a function of x.
expression x 2 + x + 1, and in general any expression contai
x, is a function of x. We may therefore and shall speak
" function of x" instead of "an expression involving the vari
x."
25. Functional notation. The symbols f(x) f F(x) f £(a?), •••
used to represent functions of the variable x. They are read,
function of x, " F" function of x, etc. If, in a discussion, f(x)
function x 2 + 3x — 1, then f(a) is a 2 + 3a — 1 and f(x + /
(x + h) 2 + 3(a> + h) - 1. Similarly, if
<f>(x) = x 2 + 5 x y
then <K2) = 22 + 5 . 2,
and $(1 + y) - (1 + y) 2 + 5(1 + y).
These illustrations bring out an important point in the functi<
notation, viz. : If the same functional symbol, say £( ), be i
more than once in any discussion, it stands in each case for
same operation or series of operations on the number or expres
contained in the parentheses of the functional symbol. 1
notation is very convenient because it enables us to indicate
value of the function for any values of the variable on whic
depends.
EXERCISES
1. The fact that the area of a circle may be calculated from a given ra
may be expressed in the functional notation by A = /(r) . Give the partic
form of /(r) in this illustration.
2. The volume, F, of a cube is a function of the edge x. Express
fact in functional notation.
3. The hypotenuse of a right angled triangle is of length 10. Express
length of one leg y of the triangle as a function of the length of the o
legx.
4. Boyle's law says that PV= C, where P is the pressure of a ga
its volume and C a constant for a given temperature. Express P as a f i
tion of V. Write in symbolic language the fact that O depends on
temperature T.
Arts. 24r-26] SYSTEM OP COORDINATES
33
5. If/(x) lBa5» + aj + l,flnd/(a),/(o),/(l),/(-l) f /(-10),/(a+l).
Find 0(2), 0(0), 0(-l), 0(1- V2),0(a+6).
6. Given 0(x) = x -±-^
x — 1
i 7. Given F(y) = ^-±1. Find F(o), F^Y F(Vx).
05 8. If /(x) =2x*-6x3-5x 2 + 5x + 3, show that /(l) =/(- 1) =
* /(3)=/(-i).
m 9. Given 0(x) = x 2 + x + 1. Find 0(x 2 ), 0(x + 1), 0(0(a)).
x -f- 1
10. Given y =/(x) = - . Show that/(y) reduces to x.
i 11. Given s = 0(«) =3j^. Find J-Y
*-f 2
12. Given s = 0(i) =
3* + 4*
Find 0(s) in terms of t
WT
26. System of coordinates. Let X'Xand F'Fbe two straight
lines meeting at right angles. Let them be considered as two
number scales with the point of intersection as the zero point of
each. Let P be any point in
the plane. From it drop per-
pendiculars to the two lines.
Let x represent the perpendicu-
lar to Y'Y, and y the perpen-
dicular to X'X. If P lies to
the left of F' F, x is to be con-
sidered negative. If P lies
above X'X, then y is positive.
It is clear that no matter where
P is in the plane, there corre-
sponds to it one and only one
pair of perpendiculars, x and y.
The lines of reference X'X and Y'Y are called the coordinate
axes, and their intersection is called the origin. The first line is
called the X-axis, and the second the F-axis. The perpendicular
to the X-axis from a given point in the plane is called the ordi-
nate or y value of the point. The perpendicular to the F-axis
is called the abscissa or x value of the point.
If we have two numbers given we can find one and only one
point P which has the first number for its abscissa and the second
34
VARIABLES AND FUNCTIONS
[Ch
. III.
for ita ordinate. If, for example, the numbers are 2 and — 5, m
measure from the origin, in the positive direction, a distance 2 on
the X-axis and at this point we erect a perpendicular and meas-
ure downwards a distance 5. We have then located a point
whose x is 2 and whose y is —5. This point may be represented
by the symbol (2, — 5). The symbol (a, b) denotes a point whose
abscissa is a and whose ordinate is b. The symbol J\a, b) is
sometimes used and is read, " the point P whose coordinates are i
and ft."
When a point is located in the manner described above, it is
naiil to be plotted. In plotting points and obtaining the geomet-
rical pictures we are about to make, it will be convenient to mc
coordinate paper, which is made by ruling off the plane into equal
squares with the sides parallel to the axes (Fig. 3). Then the
side of a square may be taken as the unit of length to represent a
number. To plot a point, count off from the origin along the
X-axis the number of divisions required to represent the abscissa
and from the point thus determined count off the number of divi-
sions parallel to the Faxis required to represent the ordinate.
n Arts. 26, 27]
GRAPH OP A FUNCTION
35
3
±
I
sett
Ml
f
It
EXERCISES AND PROBLEMS
1. Plot the points (3, 4), (- 3, 4), (-3, - 4), (3, -4), (6, 0), (- 5, 0).
2. Draw the triangle whose vertices are (3, — 1), (0, 5), (—4, — 2).
3. Draw the quadrilateral whose vertices are (2, —2), (—3, 4), (—6, —3),
(3, 4).
4. If a point moves parallel to the X-axis, which of its coordinates re-
mains constant ?
5. A line joining two points is bisected at the origin. If the coordinates
of one end are (12, 3) , what are the coordinates of the other end ?
6. A square of side 3 has one corner at the point (2, 1). If the sides of
the square are parallel to the coordinate axes, what are the coordinates of the
points that may be at the other corners of the square ?
7. Given a N. and S. line and an E. and W. line for reference lines
(Xand Y axes respectively), the following coordinates of points of a river
indicate its general course.
(0, - 1), (|, -2), (1, -2J), (3, - li), (3, 1), (4, 5), (6, 10), (- 1, 0),
(-2, 1), (-3, 2), (-SJ, 1), (-4, -1), (-5, -3).
Map the river from x= — 5 to x = + 5.
8. A square of side 4 has its center at the point (1,3)- What are the
coordinates of the corners (a) when the sides of the square are parallel to
the coordinate axes ; (6) when the diagonals are parallel to the axes ? (Give
answers correct to two decimal places.)
27. Graph of a function. By a method analogous to that em-
ployed in Prob. 7, Art. 26, a function may be represented with
reference to coordinate axes. This representation of a function
- is called the graph of the function. The graph of /(#) contains
all points whose coordinates are (x, /(#)) and no other points.
Example : Obtain the graph of f x + 4 for values of x between — 5 and+ 5.
Let/(x) = $ x -f 4. The object is to present a picture which will exhibit
the values of /(x) which correspond to assigned values of x Any assigned
value of x with the corresponding value of / (x) determines a point whose
abscissa is x and whose ordinate is/(x).
Assuming values for x and computing the corresponding values for/(x),
we obtain the following table.
X
4
1
2
7
2i
3
4
10
6
-l
5
2
+ j
-2
1
-3
-i
-4
-2
-6|-i
-11 ¥
+ i
2i
These corresponding values are plotted as coordinates of points in Fig. 4.
36 VARIABLES AND FUNCTIONS [Chap. IIL
It should be noted that there is no limit to the number of
corresponding values which we may compute and imagine plotted
in a given interval along the X-axis, and further that to small
changes in valuea of x there cor-
respond small changes in the values
oif(x). These facts suggest the
idea of a continuous curve to repre-
sent f(x) much as a continuous
curve is used in mapping a river.
(Prob. 7, Art. 26.)
It must not, however, be assumed
that all functions give continuous
-a-.Y graphs ; Article 28, below, considers
a graph made up of isolated points.
The important fact for this course
in algebra is that we may assume
a continuous curve for all functions
which are polynomials in x* and
for most other functions which occur in this course, although
the proof of continuity is beyond the scope of this book. That ia
to say, it is proved in higher analysis that a function of the type
otft" + rt^" -1 -(- — + a„ (n a positive integer)
has a continuous graph.
Hence, in finding the graph of a polynomial, when a sufficient
number of points are located to suggest the general shape of a
curve through them, draw a smooth curve through the points. In
particular, it is proved in analytic geometry that when n =1, the
graph of a function of this type is a straight line. Ill the problem
in hand, the graph is the straight line shown in Fig. 4.
28. Function defined at isolated points. Much use is made of
Bystems of coordinates in presenting statistical results when one
set is a function of the other.
The following infant mortality table is made up from the
• By a polynomial in x, we mean an expression of type
atfE" + o,a»-i + — + a,,
where n is a positive integer, and ft), a t , ■•• a. do not contain x.
Arts. 27, 28] CONTINUOUS CURVES 37
United States Life Tables of 1910. Out of 100,000 living new-
born babies in each class, it shows the number of deaths during
each month of the first year.
Month or
ES
f™
N*cmo
Males
Nboho
Fkhalis
City
Bowl
Males
1
4844
3787
7370
6380
4060
4570
9
1242
991
1977
1746
1370
997
S
1012
850
1831
1556
1091
822
4
863
740
1695
1394
941
699
S
750
648
1661
1252
835
505
673
578
1425
1134
755
515
7
010
526
1290
1036
604
450
B
553
486
1153
048
640
408
9
603
460
1037
874
586
363
10
457
421
937
800
637
325
11
12
420
399
300
35(1
857
802
725
663
496
406
206
277
Using the numbers of the months
responding numbers in the column headed
locate the upper
set of points in
Fig. 5. The ver-
tical unit is 500.
The lower set of
points is given
by the column
headed " Rural
Males." In Fig.
5 we thus present
to the eye the
relative infant
mortality of city
and country chil-
The graph in
this case is made
and the cor-
City Males" as
1
|
\
1
^
^
as
Jt*
.....
Bu
ii .1
it.:
-----
38 VARIABLES AND FUNCTIONS (Chap. I
up of 12 points. If we look upon the number of deaths as
function of the number of the month, this function is defined
only 12 points. The lines connecting the points in the figure e
not necessary, but aid the eye to take in the whole situatit
Where two sets of data are exhibited in the same diagram as
Fig. 5 the connecting lines prevent confusion of the two sets
points.
EXERCISES
Plot the graplis of the following (unctions.
3. +V2b-x>.
Solution : We flnil the following table.
4.0 4.8 4 ] 8 J 2.18 |
| Greater than 6
Imaginary
— 5 I Less than — 5
-
*
1
; 2.18 I I Imaginary
Plotting these points a
drawing a smooth cur
through them, we have Fig.
9. xl-s-lS.
10. it' -4.
11. 12+1- A
12. 2z»-IlX + 5.
13. x*-2JC + L
14. From the talife mi iwpi 37, Show graphically on the s
the infant mortality of " white innUs " mul of » iii>jrr\» males. "
16. Exhibit
"negro males'
.< muiio di:u,TiiHi the infant mortality ■
wwno diagram the infant mortality ■
■ii Arts. 28, 29]
ZEROS OF A FUNCTION
39
i
17. The breaking strength of ordinary manila rope is given by the
formula B = 7100 D 2 where B is the breaking weight in pounds and I) is
m the diameter of the rope. Exhibit this formula graphically for the diameters
$ h h b h h h *> H> !1> If, 1J, Ifc It, If 2 inches.
Is
Ji
18. The following table taken from a jewelry catalogue gives the price
of diamonds of the same quality for various weights. From this table give
* a graphical representation of the price of diamonds.
Weight in carats
Price in dollars
Weight in carats
Price in dollars
0.15
0.20
0.25
0.30
65
0.80
0.35
80
0.90
0.40
0.45
0.60
0.55
30
0.65
40
0.70
60
0.75
95
1.00
110
1.25
130
1.50
150
1.75
190
210
230
260
285
325
400
600
600
0.60
170
2.00
700
19. The morning and evening temperatures of a pneumonia patient were
as follows : 99°, 103.2°, 105°, 103.6°, 104.2°, 105°, 104°, 106°, 103°, 104.2°,
102.3°, 97.6°, 97.4°, 98.2°, 99°, 98.2°, 98.7°, 98.4°. Give a graphical repre-
sentation.
Hint : To save work in handling large numbers, 90 may be subtracted
-: from each of the above numbers and the differences plotted. Or " degrees
- of fever " may be plotted — that is, degrees above 98.6°.
20. A soldier under the U. S. War Risk Insurance plan contributes towards
8 his life insurance on a yearly term plan. This means that if he enters at
* age 21 he will pay in successive years while in the service the following
i amounts in premiums for $ 1000 of life insurance.
Year
Yearly premium in dollars
Year
1
2
3
4
5
6
7
7.80
7.80
7.80
7.92
7.92
8.04
8.04
8
9
8.28
10
11
12
13
14
8.16
8.28
8.40
8.62
8.64
8.76
Yearly premium in dollars
Represent graphically.
21. The postage on first-class mail matter is two cents per ounce or
fraction thereof. With weights for abscissas and number of cents for ordi-
nates exhibit this postage rate graphically.
29. Zeros of a function. By a " zero of f(x) " is meant a value
of x such that the corresponding value of f(x) is zero. Thus 3
and — 1 are zeros of the function x 2 — 2 x — 3, and ± 5 are zeros
of V25 — x 2 . Stated graphically, the " real zeros of f(x) " are the
40
VARIABLES AND FUNCTIONS [Chap.
abscissas of the points where the graph crosses the X-axis.
Figs. 6 and 7 the graphs and the zeros of V26 — 3* and jb* — '2x
are shown. One of the main problems of algebra is the devei
ment of methods for finding the zeros of functions. The graph:
solution of this problem, so far as real zeros are concerned, <
sists in finding where the graph crosses the X-axis. One of
advantages of the graphical method of dealing with function
that it presents to the eye the zeros of a function.
EXERCISES
Pint and find the real zeros of the following functions.
1, j-i_2x-S.
Solution : Compute the table.
x= I— sl-al— llo-rlals
-2j-3= 12
Plotting these points mid drawing :i smooth curve through them, we h
?'k. 7. The graph crosses the X-aiis at — 1, ami 3, which are there!
7 the zeros of x* - 2* - 3.
-! — - 2. 3x-5.
3. 4ie + 0.
4. jr*-«* + 5.
5. r= + ix.
6. Si*- 11* -4.
7. tir-S-a*.
8. x> + 3i=-x-3.
9. x»-26*.
10. Between what integers d<
■a.-li of the real zeros of x*+$ 3
11. Show that x' + x + l haa
„ CHAPTER IV
m
THE EQUATION
30. Equalities. A statement that one expression is equal to
another expression is called an equality. The two expressions
are called the members of the equality. There are two classes of
equalities, — identical equalities or identities, and conditional equali-
ties or equations. An identity is denned in Art. 13. It is there
^ stated that the two members of an identity are equal for all
3 values of the symbols for which the expressions are defined.
. Thus,
a 2 — a 2 =(a> — a)(x + a), 5a = 10a — 5 a
are identities. But in the equality x — 5 = 4, the two expressions
x— 5 and 4 are equal only when x has the value 9. An equality
of this kind, in which the members can be equal only for particu-
lar values of the letters involved, that is, are not equal for all
values, is sometimes called an equality of condition. In this
book we shall use the term equation to mean conditional equality.
When it seems necessary to indicate that an equality is an iden-
tity and not a conditional equality, we shall use the sign = in-
stead of the sign = between the members. But the sign = will
be used for both identities and equations when this usage can lead
ie to no confusion.
lb.
Which of the following equalities are identities ?
(a) x-10 = 0. (d) x2 + 4=(x + 2)2-4x.
(6) *l=J* = x _ a . W 5*2 + 2*-6 = 0.
v ' x + a
(c) x* + 2x=(x + 1)2-1. (/) r3^ =1 + x + x2+x3 + ilb-
In equalities (6) and (J) may x take all real values?
41
42 THE EQUATION [Chap. IV.
31. Definitions. In an equation there are some symbols whose
values are assumed known and others whose values are unknown.
These are spoken of as the knowns and unknowns. Following
the conventions of elementary algebra, the first letters of the
alphabet are used to' represent kn6wns, while the last letters
represent unknowns.
Any expression in the form
ao« n + a x x n ~ l + 02<B n ~ 2 + — + a n _i<& 4- a n >
where n is a positive integer, and a , a x , a 2 , •••, a n _i, a n represent
any given numbers, is called a rational integral expression in * x,
or a polynomial in x (cf. Art. 27). In other words, a rational in-
tegral expression in x is the algebraic sum of terms of the type
lex*, where a is restricted to take positive integral values. For
example,
2x* — 5 a; and f x 2 + 7 x — -J
are rational integral expressions in x.
As an extension of this definition, we define a rational integral
expression in x, y, z, •••as the algebraic sum of terms of the type
kx*yHi •••,
where a, ft, y, ••• are positive integers and k (called a coefficient)
does not involve x y y, z y •••. For example,
5 xhj + 3 xz + 3 x 2 — 1
is a rational integral expression in x> y> z.
By the degree of a term kx a y^zy in any letters x, y, z f ••• is
meant the sum a + fi -f- y + ••• of the exponents of the letters in
question. The degree of a rational integral expression is defined as
that of a term whose degree is equal to or greater than that of
any other term in the expression. Thus,
5afy + 3aa: + 3a 2 — i
is of degree two in x, one in y, one in z, one in y and z, two in x
and z, three in x and y, three in x, y, and z.
* By substituting the words ** function of " for " expression in " through-
out this article, we obtain the definition of an important class of f unctions.
Es Arts. 31, 32] SOLUTION OF AN EQUATION 43
if The statement that two rational integral expressions are equal
m is called a rational integral equation. By transposing terms, such
mi: an equation can manifestly be written in the form
l- f( x > y> z > •••) = °>
where f(x y y, z, •••) is a rational integral expression. The degree
of /(#, y, z, •••) in any letters is said to be the degree of the equa-
tion in those letters. In this course, the term degree is applied
to equations only when they are in the rational integral form.
? We sometimes speak of the degree of an equation without men-
ii tioning to what letters we refer. In this case, it is to be under-
3 stood that we mean the degree in all the unknowns.
x Equations of the first, second, third, fourth, and fifth degrees
6. are called linear, quadratic, cubic, quartic, and quintic equations
respectively.
EXERCISES
Give the degree of each of the following equations.
1. xfy 3 + 5 x = 0. 2. ax 2 + bx + c = 0.
3. - — — =1. 4. i/ B = a«.
a 2 b 2
5. Give the degree of the expression ax 4 — 4 mxPy 2 — 3 nxy 4- y 2 in x. In
y. In x and y.
6. Give the degree of the equation 1 x 4 — 4 ax?yz — 3 xyz + by 2 = 5 x 4 — 2 zty 2
in x. In y. In z. In y and z. In x and z. In x and y. In x, y, and z.
32. Solution of an equation. To solve an equation in one un-
known is to find values of the unknown that make the two mem-
bers equal. Any such value is said to satisfy the equation and
is called a solution or root of the equation. A solution of an
equation in more than one unknown is a set of values of the un-
knowns which satisfy the equation. Thus, x = 1, y = 2, is a
solution of y = x + 1.
To solve a system of equations in any number of unknowns is
to find sets of values of the unknowns which will satisfy the
equations. Any such set of values is said to be a solution of
the system of equations.
t
44 THE EQUATION [Chap. IV.
EXERCISES
1. Is 1 a root of x* — 5x + 4 = ?
2. Is 6 a root of x 2 — 7 x + 5 = ?
3. Is 4 a root of x 2 — 2x + 4 = 0?
4. Is 1 + K& = — 1) a solution of x 2 — 2x + 2 = ?
5. Is x = 6, y = 3, a solution of2x + 4xy = 28?
6. Is x = 1, y = 4, a solution of x 2 + 3 xy — y 2 = — 3 ?
7. Is x = 1, y = 2, z = 1, a solution of3x — 2y + 5z — 2 = 0?
33. Equivalent equations. Two equations or two systems of
equations are said to be equivalent when they have the same
solutions ; that is, when each equation or each system is satisfied
by the solutions of the other. Thus, the equations x — 2 = and
3 x — 6 = are equivalent, the second being derived from the
first by multiplying both members by 3. Again, the equations
x 2 — 5 x + 6 = 0, and — 10 x 2 = — 50 x + 60, are equivalent. The
second can be obtained from the first by performing the follow-
ing operations on both numbers.
(1) Multiply both members by — 10.
(2) Add — 60 x + 60 to both members.
It must not, however, be inferred when the same operation is
performed on the two sides of an equation, that there necessarily
results an equivalent equation. The following equations will
show that this is an unwarranted inference.
1. Consider the equation 3 x = x + 4. (1)
Square both members, 9 x 2 = x 2 + 8 x + 16. (2)
The equation (2) is satisfied by 2 and — 1, while (1) is satisfied by 2 and
not by — 1. Hence, (1) and (2) are not equivalent.
2. Consider the equation 3x + 2 = 5x — 8. (8)
Multiply both members by (x — 1) ,
(x - l)(3x + 2) = (x - l)(5x - 8). (4)
Equation (4) is satisfied by 1 and 5, while (3) is satisfied only by 6.
Hence, (3) and (4) are not equivalent.
3. Consider the equation Vl — x — x = — 1. (6)
First, add x to each member, then square both members. There
results 1 - x = 1 - 2 x + x 2 . (6)
Equation (6) is satisfied by x = and x — 1 ; but x = does not satisfy
(6). Hence, (5) and (6) are not equivalent.
Arts. 32-34] EQUIVALENT EQUATIONS 45
4. Consider the system of equations
IK ^
(7)
x + y = 15, 1
x — y = 5. J
Multiply the members of the first equation of (7) by x, the second by y.
There results
x(x + V)= 15 x, 1 , g)
y(x-y) = 5y. J
This system (8) is satisfied by the four pairs of numbers (10, 6),* (0, 0),
(0, — 5), (16, 0), but (10, 6) is the only one of these pairs which will satisfy
CO-
These simple examples show that the same operation performed
on the two members of an equation does not necessarily lead to
an equation equivalent to the original one.
It is manifestly important to know whether an equation is
equivalent to that from which it is derived ; and if non-equivalent,
whether it contains at least all the solutions of the original equa-
tion. If a derived equation contains all the roots of the original
equation and some others, we shall call it redundant. If the
derived equation lacks some roots of the original equations, we
shall call it defective. The student should always be on his guard
against treating two equations as equivalent simply because the
one has been derived from the other.
The following operations which the student has often per-
formed in elementary algebra lead to equivalent equations :
(a) Adding the same number to or subtracting the same num-
ber from both members.
(b) Multiplying or dividing both members by the same known
number provided this number is not equal to zero.
(c) Changing the signs of all the terms.
34. Operations that lead to redundant equations. The following
operations on the two members of an equation lead, in general,
to redundant equations :
(a) Multiplying both members by the same integral function of
the unknown.
* The notation (10, 6) means x = 10, y = 5. (See Art. 26.)
46 THE EQUATION [Chap. IV.
Example 1. Consider the equation 3 x + 2 = 5 x — 8. (1)
The solution is x = 5.
Multiplying each member hy x — 1, we have
(*-l)(3a; + 2) = (a>--l)(5a>--8). (2)
Equation (2) has roots 5 and 1, but 1 is not a root of equa-
tion (1).
Example 2. Consider the equation x — 1 = 0. (1)
The solution is x = 1.
Multiplying each member by x, we have
x 2 - x = 0. (2)
Equation (2) has roots and 1, but is not a root of (1).
(b) Raising both members to the same integral power.
Example 1. Take the equation 3 x = x + 4. (1)
Squaring each member, we have
9 a; 2 = a; 2 + 8 x + 16. (2)
Equation (2) has roots 2 and — 1, but — 1 is not a root of
equation (1).
Example 2. Take the equation — V# = 1. (1)
There is no value of x that satisfies (1).
Squaring both members, we have
x = 1. (2)
While 1 thus satisfies equation (2), it does not satisfy (1).
35. An operation that leads to defective equations. The follow-
ing operation leads, in general, to defective equations.
Dividing both members of an equation by the same rational
integral function of the unknown, when such function is a factor
of each member.
Example 1. Take the equation
(a? 2 - 6)(x - 2)= Sx - 6. (1)
The roots are 2, 3, and — 3.
Divide both members by x — 2 and we have
a; 2 -6 = 3, (2)
the roots of which are 3, and — 3.
That is, the root 2 is lost in dividing the members by x — 2.
A.rts. 34-36] CLEARING AN EQUATION 47
Example 2. Take the equation a? — 5 x 2 + 6 x = 0.
The roots are 0, 2, and 3.
Dividing members by x, we have
x 2 -5x + 6 = 0,
the roots of which are 2 and 3.
The root is lost in dividing by x.
36. Clearing an equation of fractions. We shall call an equa-
tion fractional only in case some of its terms are fractions with
unknowns in the denominators.
When a fractional equation is cleared of fractions, the resulting
equation is generally equivalent to that from which it is derived, but
it may be redundant.
Consider the equation
s-^- 4 * 2 -*- 5 *- 2 ^-^. (1)
aj2_3. T + 2 w
Clearing of fractions by multiplying both members by x 2 — 3 x + 2,
we obtain a* -3» + 2 ={x -2)(x 2 - 3s + 2),
which is satisfied by x = 1, x = 2, x = 3, while (1) is not satisfied
by x = 1, or x = 2, as can be shown by substitution.
In concluding this review of equivalent equations, it need
hardly be said that we have by no means exhausted all the types
of operations which it may be necessary to perform on the mem-
bers of an equation, but enough has been said about a few simple
operations to warn the student against proceeding blindly in de-
riving equations from a given equation. Unless the operations
on the members of an equation are known to lead to equivalent
equations, the student should never regard the solution as com-
plete until the test of substitution has been applied.
EXERCISES
1. Form equations by multiplying the members of 2 x = 6 by each of the
following expressions,
(a) x.
(6) x-3.
(c) ««-4.
What roots have the derived equations that are not roots of 2 x = 5 ?
48 THE EQUATION [Chap. IV.
2. Form an equation by multiplying the members of x 2 + 2 = 8 x by
x-8.
In respect to what root is the resulting equation redundant ?
3. What root has the equation x 2 — x = which is not a root of the equa-
tion derived by dividing the members of the given equation by x ?
4. Form an equation by dividing the members of
(x 2 -6)(x-2) = 3x-6,
by x — 2. What root has the given equation which is not a root of the de-
rived equation ?
5. Form an equation by squaring the members of x = 8 x — 4.
Show that 1 is a root of the derived equation but not of the given
equation.
6. Given an equation whose members are rational integral functions of x.
If you multiply the members by x — a where a is not a root of the given
equation, what root is introduced into the derived equation? Illustrate
with the given equation x = b.
7. If x — a is a factor of each member of a given equation, what root of
the given equation is, in general, lacking in the equation obtained by divid-
ing the members of the given equation by x — a ? Illustrate with the given
equation x(x — a) = b(x — a).
Reduce the following equations to rational integral equations and discuss
the question of equivalence.
8 5-3x _ 7-9x 9 3x~5 _ 3x-7
x+1 l + 3x* ' 4x— 7 4x — 6*
10. — — + — 1— = — 2 — 11. Vx^T _ i = VxZTi.
x 2 — 25 x + 5 5-x
12. Vx-1=-Vx^9". 13. Vx - 16 = 1 + Vx.
14. i-*±A.= x ~ 6 . 15. x = 15 - 7a .
2x+l x-2 x 2 -l 8(1 -x)
j i» x 2 — 5x4-4 «
CHAPTER V
LINEAR EQUATIONS
37. Type form. Any linear equation (Art. 31) in one
unknown
ax + b = 0, a =£ 0, (1)
can be put into the form x =
a
The point which represents on the line of Fig. 1 may be
a
conveniently regarded as the locus of equation (1) in one-dimen-
sional space.
A linear equation in two unknowns
ax + by + c = 0, b =£ 0, (2)
can be put into the form y = . . (3)
b b
Since in (3) we may assign to x any value and compute a cor-
responding value for y, the equation defines y as a function of x
Ln accordance with our definition of a mathematical function
^Art. 24).
The graph of a linear function has been discussed in Art. 27,
fcnd this graph may be conveniently regarded as the locus of
equation (2) in two-dimensional space.
The locus in two-dimensional space of an equation in two variables
consists of all points whose coordinates satisfy the equation and of
*uch points only.
EXERCISES
Plot the loci of the following equations.
1. x — y = 1.
Solution : Putting this in form (3), we obtain
y = x — 1.
49
50 LINEAR EQUATIONS [Chap. V.
The graph of the function x — 1 is shown in Fig. 8, and by definition this
la the locus of the equation. Since the graph of any rational integral func-
tion of degree one (Art. 27} ia a
straight line, the locus of any linear
equation ia a straight line.
^.2. 2x-y = l.
.-3. x~Sy = l.
4. -i x + 4 y = 4. ,
5. Hx+Zy-Q.
6. 7z-6;/ = 0.
1"
38. Simultaneous linear equations. ^ 1
Let a,x + bjy = c l>
■a i x + biy=c i ,
two linear equations in two unknowns. Multiply the mem:
bers of the first by b, and those of the second by — 6,. Adding
the members of the two resulting equations, we obtain
(afi, — a&jx =(6 a Ci — bfa),
or x= *&■-*>& provided a x b t - aA * 0.
In a similar manner, by multiplying the first and second equa-
tions by — dj and a! respectively, we obtain
v ^ a&-a&i provided a^-0^,^0.
afo — a^>i
We note that the denominators of the above fractions are alike.
This denominator may be denoted by the symbol
!«■ 6*1
which is called a determinant. Since it has two rows and two
columns, it is said to be of the second order. The letters o^, b\,
at, b t , are called the elements of the determinant, and a lt b±, oon-
stitute the principal diagonal. A determinant of the second order
then represents the number which is obtained by subtracting
Arts. 37, 38] SIMULTANEOUS LINEAR EQUATIONS
51
from the product of the terms in the principal diagonal, the prod-
uct of the other two terms. Thus, * *
1 2 <^J**i
x y
z w
= x w
y*>
3 4
= 4-6 = -2.
V
\J>
x =
c x b x
«1 Ci
c 2 b 2
y = -
<h c 2
«i&i
<x x b x
<h ^2
(h &2
Using the determinant notation, we may now write the solu-
tions of our equations in the form n
i
We note that the numerator of the solution for x is obtained from
the denominator by substituting in place of a u a 2 , which are the
coefficients of a; in the equations to be solved, the known terms c u
<%' In a similar manner, in the numerator of the solution for y
we replace b l9 b 2 by c u c 2 respectively.
EXERCISES
1. Solve : x + y = 3,
2x + 3y = l.
3
1
1
3
9-1 Q
x =
. — — 8,
1
1
3-2
'
2
3
*
1
3
2
1
!-6 r
y =
1
1
- = — — 5.
3-2
2
3
Solve the following pairs of €
jqi
lat
ions, using determinants.
■^ 2. 3x + 2y = 23,
>3. £ + K=ll,
&x-2y = 29.
2 5 '
• 4. 6x + 5y = 16,
5. 8x-6y = 9,
5x-12y=-19.
2x + 7|/ = 28.
c llx — 5y_ 3x + 2/
x+y y -x =9
22 32 •
3 ' 2
8x-3y = 23.
* + * + * = *.
2 9
52
LINEAR EQUATIONS
[Chap. V.
39. Graphical solution of a system of linear equations. As
stated in Art 37, the Ioc'ub of any linear equation in x and y is a
straight line. Any such equation is satisfied by an indefinitely
large number of pairs of values of x and y. That is, by the co-
ordinate? of all points on its locus. In the graphical solution of
the Bystem of two equations, we seek the coordinates of points
cojnmon to the loci of the two equations.
' As the loci are two straight lines, three cases arise:
• • (1) In general, two lines intersect in one and only one point
(2) Two lines may be parallel, and thus have no point in
(3) Two lines may be coincident, and thus have an indefinitely
large number of points in common.
Corresponding to these three cases, a system of two linear equa-
tions has, in general, one and only one solution, but it may have
no solution or an indefinitely large number of solutions. When
the loci are two parallel lines, there is no pair of numbers which
satisfies both equations, and the equations are said to be incom-
paiibte or inconsistent.
Referring to the expressions for x and y (Art. 38) in determinant
form, we see that there is one common solution of the equations
unless the determinant in the denominator is zero. When the
loci are two coincident lines, the two equations of the system are
equivalent.
7 o
V-
Find the solutions of the following eq vis-
ions by determinants and by plotting the
f Arts. 39, 40] DETERMINANTS OF THIRD ORDER
53
sri
4.
x-f 33/ = 10,
5.
2x-9j/ = ll,
3x-f 2y = 9.
3x+12y=15.
6.
2x — y= 1,
7.
3x-12j/=-12,
x + Sy = 11.
x + 8y = 20.
8.
6x + 5y = 16,
9.
* + *=8,
5x- 12i/=-19..
3 5
& V _!
9 10
10.
2x + 2y = 4,
11.
x-f y = 2,
3x + 3y = 6.
5x + 6 y = 20.
12.
9x-4y = 0,
13.
6x4- 7y = 3,
3x + 8y = 7.
3x+ 14y = 6.
14.
7x-j/ = 33,
12y-x = 19.
15. In solving a system of equations
<*>ix + b x y = Ci 1
a2X -f 62?/ = C2 J
by determinants, show that if the determinant in the denominator is zero,
and the determinant in one numerator is likewise zero, then the two equa-
tions are equivalent.
40. Determinants of the third order. The square array of nine
numbers with bars on the sides
*t h, ■•-.
i & -
is a convenient abbreviation for the expression
a)
and is called a determinant of the third order. As in the case of
the determinant of the second order, the letters a 1} b b etc., are called
the elements, and the letters a l9 6 2 , c$ form the principal diagonal.
The expression (1) is called the expansion or development of the
determinant. It is seen that each term of the expansion consists
of the product of three elements, no two of which lie in the same
row or in the same column. Any determinant of the third order
54
LINEAR EQUATIONS
[Chap. V.
may be easily expanded as follows. Rewrite the first and second
columns to the right of the determinant. The diagonals
a x b x c t
a x b x
<h b$ <H
«8 b 2
^3 &3 <*
<h \
running down from left to right give the positive terms. The
diagonals running down from right to left give the negative terms.
EXERCISES
Obtain the expansions of the following determinants.
13 4
2 7 3 =17. 5+3- 3-1 +4- 3- 2-4- 7- 1-8- 8- 1-6- 2. 8 = 1.
13 5
2.
5.
10 2 8
5 4
3 1 7
12 4
-2 -1 -4
3-3 5
1 -
1
1
2 3 5
3.
3 2
8
-4
5
•
4.
7 1 4
6 2 3
•
a b
x 1 3
x 2 as
•
6.
x y
•
7.
2x4
8.
1 1 1
u
v
6 8 1
x 4 x
41. Solution of three equations with three unknowns. Let the
three equations be
a x x + b$ + c x z = d u (1)
<W> + £#+<** = <*2> (2)
aa« + hy + c z z — d,. (3)
Multiplying (1) and (2) by b 2 and — b x respectively and adding,
we get
(a x b 2 — a 2 b x )x + (b 2 c x — b x c$z = d x b % — dj> x . (4)
Eliminating y in a similar manner from (1) and (3), we find
(a z b x — a x b z )x +(c z b l — b z c x )z = dfi x — d x bz. (6)
We now have two equations in two unknowns, x and z. Elimi-
nating z from these two, we find
[(a x b 2 — <hb\){^\ — b*Ci) — (a z b x — aA)^^ — &!<*)]«
=(dx& 2 - d 2 &i)(c3&i - & 3 ci)-(ds&i - d M(p2Ci - 61C-),
Arts. 40, 41] EQUATIONS IN THREE UNKNOWNS
55
which after soirfe simplification gives us,
^i^ + d^bzCx -f dzb x c 2 — d x b z C2 — d z b 2 c x — d^biCz
x =
«1&2 C 3 + 02&3 C 1 ~t~ ^l * ~~ a l&3C2 — «3&2 C 1 ~ ^2^1^
The denominator is the development of the determinant in
Art. 40, while the numerator is the same as the denominator with
a replaced by d. Hence we can write the solution for x in the
form
x =
di
bi
Cl
C&2
b 2
C2
<*3
bz
Cz
<h
*i
<h
«2
b 2
<h
«3
bz
cz
provided the determinant in the denominator is not zero.
In a similar way, we can find the values of y and z.
y =
a i di Ci
a\ bi di
02 C?2 C2
a 2 b 2 d>2
«3 <*3 C3
, z =
<h b z d z
C&J bi Ci
dl b x Cj
0*2 b 2 <%
«2 b 2 C2
<h b 3 c 3
(h b z C3
The denominators in the expressions for x, y, and z are the
same, while the numerators are obtained from the denominators
by replacing the coefficients of the unknown in question by the
known terms. For example, in the numerator of y } the knowns
d\, c?2, c? 3 replace b u b 2 , b z respectively.
Solve :
1.
EXERCISES
Solution :
« =
x — y — 2
: = -6,
2x + 2/ + z=0,
3x-5y + 8z = 13.
-6 -1 -1
Oil
13 -6 8
_-78
1 -1 -1
39
2 11
3-6 8
= -2,
56
LINEAR EQUATIONS
[Chap. V.
V =
1
-6
-1
2
1
3
13
8
_S9
1
-1
-1
39
2
1
1
3
-5
8
2. x + y-fz = 6,
3x — y + 2z = 7,
4x + 3y — z = 7.
4. 6x — 4y + 2z = 48,
8x + 3j/-4z = 24,
2x-6y + 8z= 19.
6.
8.
1
+1=
1,
X
y
1
+1-
1,
X
z
1
3/
z
1.
x + 2y — z = 1,
3x + 2i/-4z = 7,
x + 2y-z = 3.
z =
1
-1
-6
2
1
3
-5
18
1
-1
-1
2
1
1
3
-5
8
117 -8
3. 3x + 4y — 5z = 82,
4x-5j/ + 3z = 18,
5x-8j/-4z = 2.
5. 2x + 3i/ + z = 4,
x + 2y + 2z = 6,
6x + y + 4z = 21.
7. 2x + 5i/-3z= 17,
6x — 2y-5z= — 8,
3x + 7j/ + 4z=-18.
«4-i?/ + J2 = 10,
i(x + z) + j/ = 9,
J(x-z)-22/ + 7 = 0.
MISCELLANEOUS EXERCISES AND PROBLEMS
1. Show that
2. Show that
3. Develop
5. Solve for x
d\ b\ C\
G&2 &2 ^2
a>i 63 C3
#1 + 0*2 C\
61 + 6 2 c 2
ai
61
Cl
«i
&i
Ci
a2
&2
c 2
4. Develop
x 8
3 4
= 0.
6. Solve for x
7. Solve for x and y the system of equations
x
1
-1
1 y
-1 1
= 0,
2
X 1 1
y -1
-8 2 1
= 0.
= dl
62 C2
63 C3
+ 6l
c 2
c 3
Q>2
at
+ Ci
02 &2
(X3 63
'1
?2
=
«1 Ci
61 C 2
+
b 2
Cl
C2
9
X
2
V
1
1
1
X
4
2/
= 0.
Art. 41] EXERCISES AND PROBLEMS 57
8. A certain kind of wine contains 26 per cent of alcohol and another
kind contains 30 per cent. How many gallons of each must be mixed to
make 50 gallons of the mixture 27 per cent alcohol ?
9. What amounts of silver 72 per cent pure and 84.8 per cent pure
must be mixed to get 8 ounces of silver 80 per cent pure ?
10. The sum of the three angles of a triangle is 180° ; find the two acute
angles of a right-angled triangle if one of them is four times the other.
11. If the sides of a rectangular field were each increased 10 feet, the
area would be increased 14,500 square feet. If the length were increased
10 feet and the width decreased 10 feet, the area would be diminished
1700 square feet. Find the area of the field.
12. Two numbers are written with the same two digits ; the difference of
the two numbers is 46, and the sum of the digits is 9. What are the numbers ?
13. A six-figure number has 1 for the last figure. If this last figure is
removed and placed before the others, a new six-figure number is made whose
value is one third the original number. What is the original number ?
14. The planet Mercury makes a circuit about the sun in 3 months.
Venus makes the circuit in 7£ months. Find the number of months be-
tween two successive times when Mercury is between Venus and the sun.
15. In Wilson and Gray's determination of the temperature of the sun
the Fahrenheit reading of the temperature is 5652 more than the centigrade
reading. What is the centigrade reading ?
16. If h represent the height in meters above sea level, and b represent
the reading of a barometer in millimeters, it is known that b = k + hm,
where k and m are constants. At a height 120 meters above sea level the
barometer reads 761, at height 769 meters it reads 695. What is the formula
showing the relation between 6 and h ?
17. Two runners are practicing on a circular track 126 yards in circum-
ference. When running in opposite directions they meet every 13 seconds.
Running in the same direction, the faster passes the slower every 126 sec-
onds. How many minutes does it take each to run a mile ?
18. The relation between the boiling point w of water in degrees
Fahrenheit and h the height in feet above sea level is known to be of the
form x — wy = h, where x and y are numbers to be determined by experi-
ment. It is observed at the height 2200 feet that the boiling point is 208° F.
At sea level the boiling point is 212° F. What is the formula showing the
relation between w and h ?
19. It is required to find the amount of expansion of a brass rod for a rise
in temperature of one degree centigrade, also the length of the rod at a tem-
perature 0°. If c represent the expansion, and 6 the length required, it is
known that b = ct -f 6<h where b is the length of the bar at the temperature
t. When t = 20°, the length of the rod is 1000.22 j when t = 60°, the length
is 1001.65.
58 LINEAR EQUATIONS [Ch
20. A man has 985,000 at interest. For one put he receives S\
the other 4$>. His income from this money is $1900 per year.
the money divided ?
21. To find the average grade of a freshman in mathematics, his g
analytic geometry is multiplied by 5, his grade in algebra by 3, and hi
in trigonometry by 2, and the sum of the three products is divided
This gives 89 for the average grade. If the grades in analytic geomei
algebra had been interchanged, his average grade would have been 91.
three studies had all counted the same number of credits, his grade
have been 90. What are the grades in each of the three studies ?
22. A cistern is filled with three pipes. The first and second will 1
72 minutes, the second and third in 120 minutes, and the first and thir
minutes. How long will it take each of the pipes to fill it ?
23. Four numbers have the property, that when successively the a
of three is added to the fourth, the numbers 29, 23, 21, and 17
What are the numbers ?
24. Five numbers are arranged in order of magnitude. The difl
between any two consecutive numbers is the same number. The sum
numbers is 60. What positive integers satisfy these conditions ?
25. Some books are divided among 3 boys, so that the first had
than half of all the books, the second had one less than half the remi
and the third had 17. Find the number each received.
26. How many 5 per cent bonds of $ 100 each at 90 must I sell in
that by investing the proceeds in 6 per cent stock at 102 my income i
increased $ 800 ?
27. Between two towns the road is level for half the distance.
speeds on a bicycle are 3, 6, and 9 miles an hour uphill, on the levc
downhill, respectively. It takes 6} hours to go and 4f hours to r
What are the lengths of the level and inclined parts of the road ?
&
e
<^ CHAPTER VI
QUADRATIC EQUATIONS
42. Type form. Any equation of the second degree (Art. 31)
in one unknown x can, by transforming and collecting terms, be
written in the typical form
ax 2 + bx + c = 0,
where a, b 9 c do not involve x } and have any values with the one
exception that a is not zero. Since the result of multiplying the
members of an equation in this typical form by any given number
is an equation in typical form, the a, 6, c can be selected in an
indefinitely large number of ways. In particular, since we can
change the signs of all the terms in an equation, we may assume
that a is positive when it is a real number.
The function ax 2 + bx + c (a ^= 0) is called the typical quad-
ratic function.
EXERCISES
Arrange the following equations in the typical form and select a, b, and c
from the resulting equations.
1. 4 **-5+^ = 2*L 2 + m.
2 3
By transposing and collecting terms,
10 3
so that ° = ^r> & =-» c= — (5 + m).
2. x 2 +(2s + 6) 2 = 8x._
3 g + (»-l)' ali
9 16
4. a;2-2(&;--x + d2+--i = 0.
2 2 2
5. 4m*3» + 3Jfc 2 x 2 — 8mx + 3x — m + Jfc = 0.
6. (y + 3)»+(y-2)« = 0. 7. r2+(2r-7)=10.
8. (z + 2)»-(z-3)*-l=0. 9. u2+("iu + w)2 = ]fc.
59
HO QUADRATIC EQUATIONS [Chap.
43. Solution of the quadratic equation. A quadratic equal
may be solved by the process of " completing the square."
For oxumple, to solve 3ot?+5x — 2 = 0,
write tho equation in the form x* + \x = ■$.
Add (} • J) 1 bb |J- to both members, and the left-hand men
in a per loci squaw. That is,
<" ' (*+!)'' = *£• .
Kxtraot tho square root, a; + £ = ± £,
x = — 2, or £.
Moth of those values of x satisfy the original equation, as
Hon on Hulmti tilting them for x. Thus
•i(.. 2)« + 5(-2)-2 = 3. 4-10-2 = 0.
•Hi) 2 + r>a)-2 = 3.iH-|-2 = 0.
Apply thin method to the general quadratic equation
ax* + bx + o = 0.
TranHpoNn <* and divide by a,
a a
Add ( / ) to both members to make the left-hand membei
porfnot Nquaro,
ii VW a (,2aJ 4a* '
V 2a,/ 4 a*
Extract tho oquaru root,
,. , t) _ ± Vfi* - 4 oc
+ 2u- 2a~" '
or
or ff --6±V6»-4qc
2a
h Art. 43] SOLUTION OF THE QUADRATIC EQUATION 61
& The roots of the general quadratic equation
„ - 6+V6 2 -4ac ^ -6-V6 2 -4ctc
are * 1 = 2^ '* 2 = 2^
1 1 as may be verified by substitution. The expression
- b ± V&2 _ 4
ac
2a
may therefore be used as a formula for the solution of any quad-
ratic equation. Thus, to solve the equation
3 a .2 + 5 a j_2 = 0,
d: we substitute in the formula, a = 3, 6 = 5, c = — 2 and find
_ 5 + V 25 -4-3-(-2) = -5 + V 49 1
^6 6 3
Similarly
-5-V49 9
Sfy = = — Z.
EXERCISES
Solve the following equations by use of the formula, and verify by
substitution.
1. 2x 2 + x-l = 0. 2. 2x 2 -3x-2 = 0.
3. 4x 2 -9x + 2 = 0. 4. 8x 2 + x-2 = 0.
5. 2x 2 + 3x-9 = 0. 6. x 2 -6x-7 = 0.
7. 3x 2 + 8x-8 = 0. 8. x 2 + 3x+5 = 0.
9. 2x 2 -5x + 2 = 0. 10. 2x 2 + 7 = 4x.
11. 7x 2 + 7x + J=0. 12. « 2 + 12 = 8«.
_^.13. 3« 2 + 2 = 6*. 14. 3P + 2t = 4.
15. 7 j/ 2 + 9y - 10 = 0. ' ^ 16. 8 r 2 + r = 200.
17. 6x 2 + 5x = -l. 18. 9x 2 + 3x = 2.
.JL9. 7x 2 + 2x = 82. 20. x + 8 = 2g ~ 1 .
"* 2x-7 x-3
^ 21. 8«+ll+U«5. -22. *+* = £.
62
QUADRATIC EQUATIONS
[Chap. VL
33. x* + xV6-V6 = 0.
35.
1 x 2
1 1 4
x 1 -1
x 1 8
2 x S
10 3
= 0.
= 1.
37. x 2 — 2ux + 3x — 6a = 0.
39. x 2 + 2 <ix + to + 2 a& = 0.
90. Show by substitution that
are roots of ox 2 + bx -f c = 0.
3€. x* + ox — 2<* = 0.
3& 2x2 + 3»x-2m* = 0.
2a ' 2a
44. Solution by factoring. When the left-hand member of a
quadratic equation can be factored readily, this is the easiest
method of solution. Take, for example, the equation
The factors of the left-hand member are easily found. They an
(x + 3) and (x — 7), and we may write our equation in the form
(x + 3)(x-7) = 0.
Any value of x which makes either factor zero will satisfy tbi
equation. If x = — 3, we have
(_3 + 3)(_3_7)=0(-10)=a
Again if x = 7, we have
(7 + 3X7 -7) = 10- = 0.
Hen<\\ - 3 and 7 are solutions of the given quadratic
EXERCISES
Solvo tlio following by factoring.
1. x' Ux + U-0.
3. x* x 20 - 0.
3. 20x a + Ux 3-^0.
7. 2* 8 a 30 -0.
9. 7P-r lot- H -0.
Solve by any method.
11. 3X 2 - I6x:-4tf.
3. x2 + 8x + 7:=0.
-. 4. 4x2-f-4x + l = 0L
6. 3x a -13x = 10.
8. 2a 2 + a-3 = a.
10. tfy 2 + 35y-6=a
13. x* + 4>+¥=&
^rts. 44-45] EQUATIONS IN THE QUADRATIC FORM 63
13. x 2 -6x+10 = 0. 00* 14. 4x 2 -28x + 49 = 0.
15. 9x 2 -27x-70 = 0. 16. x 2 + x- a 2 - a = 0.
17. x 2 -|-(5-x) 2 =(5-2x) 2 . 18. x 2 -6x + 4 = 0.
19. 3x 2 + 2x + l = 0. 20. (l-e 2 )x 2 — 2mx + m* = 0.
21. Vx + 16 = x - 4.* 22. x + Vx + 6 = 14
^23. VlOx - 34 + 2 V^+l = V2(3x -f 36).
^ 24. V27 - x = Vx + 2 +.V3x + 8.
25. V4x-5+V2x-9 = 4.
26. -y* + 11 +^(*V+ 6x + 50 = 9.
. J^=^L = Vx^lQ.
\flB- 1ft
27
x-19
.28. Vl8x + 5~2V3x = V2.
^ 29. V2 Vx^l + V34-x = 9.
30. Vs-V2z + 1 = 1.
Solve the following equations to two significant figures.
31. x 2 - 1.83 x + 0.81 = 0. 32. x 2 - 0.91 x - 6.66 = 0.
33. 0.001 x 2 - 0.01 x- 0.1 = 0. 34. 2.1 x 2 + 10.3 x- 5.8 = 0.
^45. Equations in the quadratic form. If in an equation we
nay replace an expression containing the unknown by a new
etter and have a quadratic equation in that letter, then the origi-
lal equation is said to be in the quadratic form. Thus, in the
equation
a ._3-VaT =: 3-2 = 0,
f we let z=Va? — 3, we have z % — z — 2 = 0.
In 2ar* + aT* + l = 0,
J we let z = x*, we have 2 z 2 + z + 1 = 0.
* Hint : Square both sides. In solving such problems the results should
t>e tested in every case, remembering that the radical stands for the positive
iquare root of the number under it. Why should the results be tested ?
64 QUADRATIC EQUATIONS [Chap. VI.
EXERCISES
Solve the following equations and check results.
1. Vx + 10 + y/x + 10 = 2.
Solution : Let z = y/x + 10.
The equation then becomes
z 2 + z — 2 = 0,
or z = 1, or — 2.
Replacing z by its value in terms of x, we have
v^ + 10 = l,
or \te + 10 = - 2,
a; + 10 = 1, x = - 9.
s + 10= 16, s-6. .
Check : V— 9 + 10 + \^- 9 + 10 = 2J
1 + 1=2.
V6 + 10 + \^6 + 10 = 4 + 2,
Hence, the result x = — 9 satisfies the equation to be solved, but the result
x = 6 does not satisfy it.
2. x<- 18x2 + 36 = 0. 3# (4x 2 -3)2+(8x*-6)*=80.
-6. (x+IV+4x + -=12. — 7. Vx + 16 = 4.
\ 25/ X
8. x + Vx+ 6 = 14. 9. V2x+ 1 — x = J.
15
V+2x ^7-2* 2
10. Vx = 8-
Vx
12. 2x 2 -4x+3Vx2-2x+6 = 15. 13. x« + 7 a*x» - 8a« = 0.
^ 14. x 4 + 2x3-x2-2x-8=0.* 15. x 4 -8x3+28x*-28a& — 8 =
16. x3 + 7xi-8 = 0. -— 17. x-3-9x"* + 8=0.
18. x-6 + 31x-5-32 = 0. 19. 3v^2 + 6v^x-4 = 0.
-r20. x3-8 = 0. 21. ox^ + fcx^ + csO.
22. 2-6x-2-x-i = 0. .-23. x* - 4x- 2lVI = 0.
* Hint : Write the equation in the form
(x< + 2x3 + x 2 )- 2(x 2 + x)- 8 = 0.
Arts. 45, 46] ROOTS OF A QUADRATIC 65
24. _*_ + *l+i = l
a-2 + 1 x 2
25. (x+l)(x + 2)(x + 3)(a + 4) =24.*
46. Theorems concerning the roots of quadratic equations.
Theorem I. If r is a root of the equation
ax 2 + bx+c = 9 (1)
then x — r is a factor of the left-hand member. Conversely, ifx — r
is a factor of the left-hand member, then r is a root of the equation.
If r is a root of the equation,
ar 2 + br + c = 0, (2)
then ax* + bx + c = ax 2 + bx + c— (ar 2 + 6r-f c) (3)
= a(x 2 — r 2 ) + 6(a; — r) (4)
= (x — r) (ax + ar + 6). (5)
Hence, a? — r is a factor of ax 2 + bx + c.
If a; — r is a factor of ckb 2 -f for + c, then the substitution of r
for a? makes the factor x — r vanish, and r is a root of
aa* + bx 4- c = 0.
(Odd*
EXERCISES
Form quadratic equations of which the following are roots.
1. 3, 1.
Solution : When the right-hand member of the equation to be formed
is 0, the left-hand member has factors x — 3 and x — 1. Hence,
(x - 3)(* - 1) = a* - 4z + 3 =
is a quadratic equation with roots 1 and 3.
2. 3, - 2. 3. 7, 0.
4. V6, -6. 5. V5, -V6.
6. V6-1, VS+1. 7. t,t -i.
* Hint : Multiply first and fourth, and second and third factors together
and write in the form
[(as* + 6*) + 4] [(aJ» + 5*)+ 6]= 24.
f In these exercises, i* =— 1. (See Art. 22.)
66 QUADRATIC EQUATIONS [Chap. VI.
9. 2 + 6V3, 2-6V3.
11. 2, J.
13. 2, i.
n m
14. Verify by performing the indicated operations that
8. 4 + 3i,4-8i
io. l + '^.-l-
2 2 2
iV3
2
12. -, a.
a
V 2a /V 2a /
47. Number of roots. In order to avoid certain exceptions, an
equation f(x) = is said to have as many roots as /(») has factors
of the type x — r x where r x is any number. A factor x — r x
may be repeated. For example, if (x — r x ) 2 is a factor of f(x) f
we say that f(x) = has two roots equal to r v
We have shown that a quadratic equation has two roots. The
question arises : has it only two or may it have more ? This
question is answered by the following
Theorem II. A quadratic equation has only two roots.
Proof. Suppose there is, in addition to
— b + V& 2 - 4ac — b — V& 2 — 4ac
r = r 2 =
2a 2 a
a third root r 3 , distinct from r x and r 2 , of the equation
ax 2 + &b + c = 0.
By Ex. 14, Art. 46, aa? 2 + 6o? + c = a(<c — ^(a; — r 2 ).
Hence if r 3 is a root,
a(r z -r 1 )(r z -r 2 ) = 0.
But this is impossible since no one of these factors is zero.
(Ill, Art. 5.)
48. Special or incomplete quadratics. If b or c is zero in the
quadratic equation a# 2 4- foe + c = 0, the equation is said to be
incomplete.
Abts.4^48] SPECIAL OR INCOMPLETE QUADRATICS 67
I. When c = 0, ax 2 -f bx = is the typical form of the equation.
We can always write this equation in the form
x(ax + b) = 0.
Hence, the roots are and — . Conversely, if is a root of
a
a quadratic equation ax 2 + bx + c = 0,
then a-0 + &.0 + c = 0.
That is, c = 0.
Therefore, a gwadrafa'c equation has a root equal to zero when and
only when the equation has no knoivn term.
II. When 6 = 0, ax 2 + c = is the typical form.
In this case, x = ± \ — - •
* a
Conversely, if the roots of a quadratic equation are arithmeti-
cally equal, but opposite in sign, there is no term containing x in
the first degree ; for, if -f r and — r are both roots of
we have
ax 2
+ bx + c
= 0,
ar 2
+ br + c
= 0,
ar 2
— br + c
= 0,
2br
= 0.
r
+ 0,
b
= 0.
Since;
it follows th$tu
/Hence, a quadratic equation has two roots arithmetically equal
but 'Opposite in sign when and only when the term in x vanishes.
\
■ III. When 6 = 0, c =x 0, the typical form is a» 2 = 0. Both
\ roots of a quadratic equation are equal to zero when and only when
the tywm term and the term in x vanish,
\ .
68 QUADRATIC EQUATIONS [Chap. VI
EXERCISES
Determine k so that each of the following equations shall have one root
equal to zero.
1. 6x2_i6a; + 6-fc2 = o.
2. lOxz+Ha. + 2 A; -16 = 0.
3. 10y2_i6y + fc2_4fc + 3 = 0.
Determine k and m so that each of the following equations shall have two
roots equal to zero.
4. 5x 2 _i6ma; + fcc-4m + fc + 6 = 0.
5. 3z 2 — 8mz + 4kz + 6z + 4m + 2fc+ 1 = 0.
6. x2+(2m + l)x + 4fc2_|_2& = 0.
Determine k so that the roots may be arithmetically equal, but opposite in
sign.
7. x 2 + 3 kx + x + 7 = 0.
8. 2x 2 + fc2a.__4a._i_ 3_o.
Determine fc so that the sum of the roots may be 1.
9. kx 2 — 2 kx + x — 2 = 0.
10. 9x 2 — 4kx + 8x- 4 = 0.
Determine & so that the product of the roots may be 1.
11. 3&x2-25fc+fc + 8 = 0.
12. 2x*-6x + k 2 -2k-l =0.
49. Nature of the roots. In Art. 43, we found the two roots
of the quadratic equation
ax 2 + bx 4- c =
to be » 1= = — , 02 = —
to, _- a
In case a, 6, c are real numbers, the numerical character of these
roots depends upon the number b 2 — 4 ac under the radical sign.
An examination of x x and a^ leads at once to the following con-
clusions :
(1) If b 2 — 4 ac > 0, the roots are real and unequal.
(2) If b 2 — 4 ac < 0, the roots are imaginary and unequal.
(3) If b 2 — 4 ac = 0, the roots are real and equal.
It should be observed that if the coefficients are real and one
root is imaginary, then both roots are imaginary.
The quantity b 2 — 4 ac is called the discriminant of the equation
ax* + bx + c = 0.
Awrs. 48-60] COEFFICIENTS IN TERMS OF ROOTS 69
50. Sam and product of the roots. If we add together the two
roots of aaP + bx + c = 0, we have
— 6 + V& 2 — 4oc . — b — V& 2 -4oc b
x l + x l =i - 1 =
2a 2a a
If we multiply the two roots together, we have
a- a^ = / r -6 4-V6 2 -4ac^ f -6-V& 2 -4ac ^ = c
/-6 + Vfr 2 - 4 ac\ / -6-V6 2 -4ac \ = c
\ 2 a A 2a / a
Hence:
I. Tfce *t*m of the roots of a quadratic equation in x is equal
to the coefficient ofx with its sign changed, divided by the coefficient
of&.
II. The product of the roots of a quadratic equation in x is
equal to the known term divided by the coefficient ofx 2 .
EXERCISES
Determine the nature of the roots of the following equations.
I. x»+11x + 80 = 0. 2. x 2 - 8s + 25 = 0.
3. x»-16x + 64 = 0. 4. 2x 2 -33-2 = 0.
5. 4«*-28x + 49 = 0.
Determine the real values of & so that the roots of the following equations
may be equal.
6. s* + 8ite + A; + 7 = 0.
Solution : In order that the roots of this equation may be equal, we
must have 6* — 4 ac = 9 Jfc 2 — 4(fc + 7) = 0. Hence, k must be a solution of
the equation 9fc 2 — 4& — 28 = 0, or fc = 2, or — ^. Substituting these
values m the above equation, we get
x 2 + 6 x + 9 = (x + 3) 2 = 0.
. x* - V * - ¥ + 7 = I ( 9 x * - *2 * + 49) = 4 (3 x - 7) 2 = 0.
7. o&*+ite + 4 = 0. 8. fcc 2 + 6x+l = 0.
9. 4a£ + 12» + & = 0. 10. A: 2 x 2 + lOx + 1 =0.
II. (k+l)a£ + Jte + ifc+ 1 = 0. 12. x 2 + 12x + 8fc = 0.
13. (4x+ &) 2 =16x.
14. {^(l + m 2 ) + 2 fcmx + jfc 2 - r 2 = 0.
15. a?(mx + ifc) 2 + fc^x 2 = cW.
*r
QUADRATIC EQUATIONS
i and product of the roots of the following
17. ll-27*-I8z* = 0.
19. 6;B> + 7!E-62 = 0.
70
Determine by It
equations.
16. 7z> + 4z~3 = 0.
18. maH + 2a-6 = 0.
20. (1 -e*)&-2mx + m? = 0.
Determine the value of k in the following equations.
21. z 1 + kx - 5 = 0, where one root is — 5.
Solution : Let z x be the second root. The product of the roots of thi»
equation (II, Art. 60) is — 5.
Hence, —6*1 = — 6, otXi = l,
and, — 6 + ^i = — t, or & = 4.
22. i 1 + x — i t — 0, where one root is — 4.
23. x' - - x — k = 0, where the difference between the roots Is 9.
24. 15a:' + kx — 4 = 0, where one root is j.
25. kx' ! + i'j x + 6 = 0, where one root is 5 times the other.
26. 3x* — kx + 14 = 0, where the quotient of the two roots is ].
51. Graph of the quadratic function. In Chapter III we have
plotted certain quadratic functions. It can be shown, if a is
positive and different from zero, that
the graph of the function 03? + bx + c
has the same general characteristics
as the curve in Fig. 10. This curve
is called a parabola. The real roots of
the equation cu?+bx+c=0 are given
by the abscis-
f the
where
the curve
Arts. 50, 51] QUADRATIC FUNCTIONS 71
For we have shown that every quadratic equation has two roots,
real or imaginary. If the curve has no point in common with
the Xaxis, there is no real root of the equation. Hence both
roots are imaginary. If the curve touches the X-axis, both roots
of the equation are real and equal. These three cases are shown
in Fig. 11, where the graphs of x 2 — 2 x — 3, x 2 — 2 x 4- 1, and
x* — 2x + 5 are given.
EXERCISES
Construct the graphs of the functions in the following equations, and, by
measurement, determine the roots if they are real. Calculate the value of
the function for at least ten values of x between the limits given. Choose
the vertical unit so that the graph will be of convenient proportions for the
coordinate paper.
1. x 2 — 5x + 4 = 0, from x = to x = 5.
2. x 2 + x — 6 = 0, from x=— 4tox = 3.
3. 4x* + 12x + 5 = 0, fromx = — 4 to x = 1.
4. x 2 — 3x = 0, from x = — 1 to x = 4.
5. x 2 + 2x + 2 = 0, from x = — 3 to a; = 2.
6. x 2 — 6x + 11 = 0, from x = to x = 5.
7. 6 - Sx — x 2 = 0, from x = — 4 to x = 2.
8. — 3x2 + 2x — 4 = 0, fromx = -2 tox = + 2.
9. 4— 5x — x* = 0, from x = — 6 to x = 1.
10. x 2 - 1 = 0, from x =— 3 to x = + 3.
11. What are the general characteristics of the graph of the function
ax^+fcx + cifais negative ?
PROBLEMS
1. In the course of Steinmetz's solution of the problem of finding the
current strength in a divided electric circuit, it is necessary to solve the
equation
a2x 2 — as 2 + r 2 =
for a. His solution is
s 2 ± q 2
2x 2 '
where q 1 =V« 4 — 4r 2 x 2 . Verify the result.
2. If f 10,000 amounts to $ 11,130.25 when placed at compound interest
for two years, interest being compounded annually, what is the rate of
interest?
3. A rectangular court is 25 feet longer than it is wide and it contains
8760 square feet. What are its dimensions ?
72 QUADRATIC EQUATIONS [Chap. VI.
4. The altitude of a triangle exceeds its base by 48 feet. The area of
the triangle is 1387.5 square feet. Find the base and altitude.
5. If a ball is thrown upwards with a velocity ty, the distance d from
the earth to the ball after a given time t is given by the formula
d = vot - \ g&, (1)
where g = 32.2. The speed at the time t is given by
v t = v — gt. (2)
If the ball is thrown downwards with a speed t>o» the above formulas
become d = vd+lg&, (8)
*t = «o + 9t. (4)
If a ball is thrown upwards with a velocity of 50 feet per second, in what
time will it be just 30 feet from the ground ? Explain the two answers.
6. How long will it take the ball described in Problem 6 to reach the
ground?
[Hint : Put d = in formula (1).]
7. At what time is the velocity of the ball zero ?
8. How high will the ball rise ?
9. How does the time taken in rising to the highest point compare with
the whole time that the ball is in the air ?
10. How far does the ball rise in the second second ?
11. How long will it take a body to fall 400 feet, if it is thrown down-
ward with an initial speed of 50 feet per second ?
12. What is the velocity of a falling body at the end of 2} seconds if the
initial velocity be ?
13. If a body falls from rest, how far will it fall in the fifth second ?
14. The edges of a cube are each increased in length one inch. It is
found that the volume of the cube is thereby increased 19 cubic inches.
What is the length of the edges of the cube?
15. What is the area of a square whose diagonal is one foot longer than
a side ?
16. What is the area of an equilateral triangle whose side is. one foot
longer than the altitude ?
17. Show that the equation
x 2 + bx + c =
has one positive and one negative solution if c is negative.
18. In joining together two steel boiler plates with a single row of rivets,
the distance p between the centers of the rivets is given by the formula
p = 0.56 - + d,
Art. 51] PROBLEMS INVOLVING QUADRATICS 73
where t is the thickness of the plate and d the diameter of the rivet holes.
In a boiler the rivets are to be placed 1 J inches apart. If the thickness of
the plate is J inch, what is the diameter of the rivet holes ?
19. Graph on the same sheet the function 2x 2 -x + c, where c takes the
values 1, —3, 0, 10. What effect does changing the constant term in a
quadratic function have on the graph ?
20. Graph on the same sheet the function ax 2 — x + 3, where a takes the
values 5, 1, £, <fo. Decreasing the coefficient of x 2 has what effect on the
graph ? What is the effect on the roots of the quadratic equation ax 2 — x +
8 = 0, if a is made to approach ?
21. If 8 is the area in square inches of the flat end of a boiler, and t the
thickness of the boiler plate in sixteenths of an inch, then the pressure p
per square inch which the flat end plate can safely sustain is given by the
formula _ 200(t + 1)«
8-6
What should be the thickness to the nearest sixteenth of an inch of the
boiler plate for the end of a boiler 20 inches in diameter to sustain a pressure
of 100 pounds per square inch ?
22. In a group of points every point is connected with every other point
by a straight line. There are 300 straight lines. How many points are
there?.
23. A rectangular piece of tin is twice as long as it is wide. From each
corner a 2-inch square is cut out, and the ends are turned up so as to make
a box whose contents are 60 cubic inches. What are the dimensions of the
piece of tin?
24. Let h be the height and t the thickness (in feet) of a rectangular
masonry retaining wall. For very sandy soil with a grade angle of 20°, h
and t are connected by the equation
t 2 + 0.19 t-h- 0.18 £2 = 0.
What should be the thickness (to the nearest inch) of a retaining wall
four feet high ?
25. For loam, the equation in Problem 24 would be
& + 0.14 1- h-0.13h 2 = 0.
What should be the thickness of a retaining wall four feet high ?
26. A long horizontal pipe is connected with the bottom of a reservoir.
If fl" be the depth of the water in the reservoir in feet, d the diameter of the
pipe in inches, L the length of the pipe in feet, and v the velocity of the
water in the pipe in feet per second, then according to Cox's formula
Hd_ 4v 2 + 5v-2
L 1200
Find the Telocity of water in a 5-inch pipe, 1000 feet long, connected with
a r ee oTTO ir containing 49 feet of water.
74 QUADRATIC EQUATIONS [Chap. VI.
27. The so-called effective area of a chimney is given by
E = A-0.GV2, .
where A is the measured area. Find A when E is 30 square feet.
28. A stone is dropped into a well, and 4 seconds afterward the report
of its striking the water is heard. If the velocity of sound is taken at 1190
feet per second, what is the depth of the well ? (Use g = 32.2. See Prob-
lem 6.)
29. The electrical resistance of a wire depends upon the temperature of
the wire according to the formula
B t = B (l + at+b&),
where a and b are constants depending on the material, Bq is the resistance at
0°, and B t the resistance at t°. For copper wire a = 0.00387, 6 = O.OOO00597,
and Bo = 0.02067. At what temperature is the resistance double that at 0° ?
30. The radius of a cylinder is 10 and its height 4. What value can be
added to either the radius or to the height, and yet give the same increase in
volume ?
The following equations occur in some electrical problems.
31. g = — 5_. Solve for B.
32. t = a ( n ~~ n ') . Solve for (n-n').
l + &(n_n') 2
33 p = AW(r> + x>) m Solveforx .
r(Br — Xx)
34. In making war bread a mixture of rye and corn meal was used.
From a hundred pounds of rye flour a certain amount was taken and re-
placed by corn meal. Later, from the mixture the same amount was re-
moved and again replaced by corn meal. The resulting mixture was 16
parts rye to 9 parts corn. What were the proportions in the first mixture ?
CHAPTER VII
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS
52. Type form. The type form of a quadratic equation in
two unknowns is
Ax* + Bxy + Qy % + Dx + Ey + F= 0,
where at least one of the coefficients A, B, or C is not zero.
Such an equation is satisfied by an indefinite number of pairs of
values of x and y. If the pairs of real values of x and y which
satisfy the equation be considered as the coordinates of points,
and these points plotted on coordinate paper, they will lie on a
curve called the locus of the equation.
EXERCISES
Arrange the following equations in the typical form, find values of A, B,
C, D, IS, and F. Find at least 4 pairs of values of x and y which satisfy
each equation.
1. (x + y) 2 = 3x 2 -y + 2.
Solution : Written in the typical form this becomes
-2x 2 + 2xy + y 2 + 2/ — 2 = 0,
where A = — 2, B = 2, C = 1, D = 0, E= 1, F = -2.
Substituting x = 1 in the equation, there results,
2/2+3y-4 = 0,
or y = 1, or — 4.
Hence, x = 1, y = 1 ; x = 1, y = — 4 are two pairs of values satisfying the
equation.
Putting x = 2 in the equation, we find in the same way two other pairs of
valuer g== 2, y= ~ 5 + V ^ ;x = 2,y = ^^=^J, etc.
2. z* — zy + tf = Z1. 3. 3 x 2 - 2 y2 = 4.
4. x* + y 2 = 25. 5. x 2 + y 2 = (x + y + l) 2 .
6. xy = 2x-y + 9. 7. (x + y)(3 -x) = (x -y)(2 + y).
75
76 SIMULTANEOUS QUADRATICS [Chap. VII.
53* Solution of systems of equations involving quadratics. Al-
though there are an indefinite number of pairs of values of x and
y which satisfy one quadratic equation, there are not more than
four pairs which satisfy two non-equivalent quadratic equations.
If these values are real numbers, they are the coordinates of
points lying on the locus of each of the given equations. Hence,
the real solutions of a pair of simultaneous quadratic equations
can be represented graphically by the points of intersection of
the loci of the two equations. The general problem of similiter
neous quadratics is that of finding pairs of values of x and y
which satisfy two equations of the form,
A& + B x vy + C&* + D 1 x + E$ + F l = 0]
Apt + B&y + C# 2 -f D& + E# + F t = J '
where the coefficients may have any values. As illustrated in
the following example, this general problem involves the solu-
tion of an equation of the fourth degree.
Solve z' + 2/' + a-9 = 0,l
x 2 + 22/ 2 -3y-8 = 0. j
Subtracting the second from the first, we have
-2/ 2 + 32/ + x-l=0,
or x=l— Sy + y*.
Substituting in the second equation, we have
(1 - Sy + y2)2 + 2 y2-3y-8 = 0,
or y 4 -6y3+ 13y2-9y-7=0.
At this stage of our progress in algebra, we cannot solve a
general equation of the fourth degree, hence we cannot proceed
with the solution of this problem. Although we cannot solve
the general problem of simultaneous quadratics, yet there are
some types of these equations which can easily be solved. We
take up a few of the most important.
Case I. When each equation is of the form
Ax 2 +Cy 2 + F=0.
If, instead of x and y, we consider x 2 and i/ 1 as the unknowns,
the method of solution is that for linear equations.
Asr. 63] SOLUTION OF SIMULTANEOUS QUADRATICS 77
EXERCISES
Solve
1. 16as* + 27y* = 576,
Solution : Solving for as* and y* we have
1576 27!
8 = ±S.
i
Hence, we find the following four solutions,
(3, 4), (-3,4), (8, -4), (-3, -4).
To show theee solutions graphically, we plot the loci of the t>
Solving each fory, we have y = ± \ ~ ,
y = ± V25 - as*.
The first equation ban for its locus
the oval-ehaped figure A, B, C, D,
called an ellipse, the second, the
circle (Fig. 12). The points of in-
tersection represent graphically the
four pedis of solutions. If the loci
of two equations do not intersect,
the solutions of the equations will
be found to be imaginary.
Plot and solve :
2. 9** + 26^ =
as* + v* = 18.
3. 9j* + 35y* =
x» + y* = 9.
5. 8as* + 25y* =
x' + tf' = 25.
Obtain to
8. 4.8x* + 9.1 p*
x» + y*=9.8.
iiiiii
mm
Fio. ]2,
4. 9a* + 25y* = 226,
as* + »* = 4.
6. 9 a;' + 25?/* = 225, 7. 4 s* - 9
26 a; 1 + 9 y* = 226. as* + y* -
figures the solutions of the following :
9. l.fl a» + 0.21 »« = 8.6,
x*-y* = -4.1.
* The graph of 9 as* + 25 p* = 226 is an ellipse of breadth 10 and height 6.
Its position with respect to the axes is similar to that of the ellipse in Fig.
12. The graph of x 1 + y* ~ 18 is a circle of radius 4, and center at the in-
n of the axes.
78 SIMULTANEOUS QUADRATICS
Cask II. When one of the equations is linear.
EXERCISES
Plot and solve :
1. x' + y* - 8 x — i y — 6 = 0,
8 x + 4 y = 5.
Solution : From the linear equation,
5-4y
3
Substituting In tie quadratic equation and reducing, we h
6y* + iy -28 = 0.
Solving, we obtain y = 2, or — if.
Substituting these values, in the linear equation, we find
x = -l, orV-
The solutions are then (- 1, 2), (¥i - V>-
4 5
(x + yy = 200-z.
0. (8a ! + !/)(i/ + 2) = (* + 5)(3 i i
12. 0.1y + 0.125z = y-z,
y-0.6x = <i.lbxy-Sx.
The loci of these two equations
are shown in Pig. 13. The circle
with its center .at the point (4, 2)
and radius 6 is the locus of the firat
equation, while the straight line
AB is the locus of the second
equation.*
2. xi + yi — bx-ty — 6 = 0,
Sa; + 4y+6 = 0.
3. x i + y i — ix-iy~S = 0,
33 + 4^ + 16 = 0.
4. # + ?/> = 68, 5. xy = 30,
x + y = 10. * + y = 18.
6. xp = 40, 7. ay — 6* = 1,
&-y = 8. 7*-y = l.
9. y>-4z = 0,
EC - y + 1 = 0.
■">' 11. B | l- »g + » |
2(e + l0 =*(4z-y).
13. 0.3a + 0.126y = 8a-y,
3x — 0.5y = 2.25ay + Sy.
• The remaining two lines of the figure belong to the next two exercises.
In Ex. 2, where the line touches (is tangent to) the circle, only one pair at
numbers satisfies the system, and the solutions, arc said to be equal,
Art. 53] SOLUTION OF SIMULTANEOUS QUADRATICS 79
Obtain to two significant figures the solutions of the following :
14. ay =-0.24, 15. y 2 - 4.1 x = 0.38,
3.1 « — 0.63 y = 4.3. x - 0.37 y = 0.29.
find the values of a, o, c, or r in the following exercises so that the
straight line which is the locus of the first degree equation
(1) cuts the other locus in two distinct points,
(2) is tangent to the curve,
(3) fails to meet the curve.
16. x 2 + y* = r 2 ,
3 x + 4 y = 6.
The locus of the first equation is a circle with center at the origin and a
radius equal to r.
Solution : From the second equation, we have
3
Substituting in the first, we find
25 y 2 - 40y + 25 — 9 r 2 = 0.
Solving for y, we obtain, y = *0±30vV 2 -l m
If y is real, r 2 — 1 must be equal to or greater than zero.
Furthermore, if r is any number greater than 1, the two looi intersect in
real and distinct points.
U r as 1, there is only one value for y, and the line is tangent to the circle.
If r < 1, the line does not intersect the circle.
17. a^ + y^r 2 , 18. x 2 + y 2 = 25,
x + y = 10. 3 x + 4 y = c.
19. «* + j/ 2 =26, 20. x 2 + y 2 = 25,
ax + 6y = l. 7x — by = 3-
21. For what values of o in terms of r and m does the system of equations
y = mx + 6,
X* + y 2 = r 2
have equal solutions ?
22. Determine the relation between a, 6, and k such that the system
y = mx + &,
has equal solutions.
a* 6*
SIMULTANEOUS QUADRATICS [Ch.
. VII.
Case III. When all the terms which contain (he unknowns are
of the second degree.
EXERCISES
Solve the following pairs of equations.
1. z» + 8 xy = 28,
Solution ; Let y = m
From the first equation,
x and substitute in both equation*
st' + Bin*'
= 88,
whence
From the second equatic
X*
x*+ mW
1 + 3 m"
= 20,
whence x*
Equating these values of z 1 , we obtain
~1 + bi»'
28
l+3m
20
l + m»'
Clearing of fractions and reducing, we obtain
or
7 m> - 15 m + 2
= 0,
= 2, or*.
Substituting these values of m in x*
we find, for m = 2,
a?
= 4,* = ±S,
:±]V10 = ±0.68+.
The solutions are therefore
(2, 4), (-2, -4), ({VlO,
jVlO), (-Jv'lo, -I-n/10).
The loci of the two equations
of this exercise are shown in
>-jf Fig. 11. The geometrical Inter-
pretation of the substitution
y ■= mx is also shown ill the
■ & + zv + V = %
& + y* = is.
Abt.63] SOLUTION OF SIMULTANEOUS QUADRATICS 81
4. 2y* — 4ajy + 3a> 2 = 17, 5. <c 2 + y 2 = 65,
y« — x? = 16. sy = 28.
6. x J + xy + 2y 2 = 74, 7. <c 2 — 4y 2 = 9,
2 aj2 + 2 ay + y 2 = 78. xy + 2 y 2 = 3.
8. *±I + «Z* = 1° 9 . x 2 -x y + y 2 = 21.
y 2 - 2 xy + 15 = 0.
«— y x + y 3 '
« 2 4- y* = 46.
10. 4a*-2a& = & 2 -16, 11. x 2 + 5ey = 4,
6 a* = 7 aft — 36. y 2 + xy = 1.
Find to two significant figures the solutions of the following :
12. a> 2 + 1.6 y 2 = 8.8, 13. x 2 + 0.12 xy= 104,
ay = 2.3. y 2 + 1.4zy = 21.
Case IV. When the equations are symmetrical.*
The typical form of a symmetrical quadratic equation in two
unknowns is
A(x* +y*) + Bxy + D(x + y) +F=0.
EXERCISES
Solve the following systems of equations :
1. a 2 + y 2 + x + y = 8, (1)
xy + a + y = 5. (2)
Solution : Let x = u + u, y = u — r. Substituting in the two equations
we obtain after reductions
M 2 + r 2 _f_ u _ 4 ? (3)
u 2 -d 2 + 2u = 5. (4)
Eliminating t? 2 by adding we obtain an equation in u,
2m 2 +3u = 9,
from which u = { or — 3.
The four solutions of (3 ) and (4) are then
.(i tt, (I, - i). (-3, iV2), (-3, -tV2).
From the first pair
a> = u + fl=f + l = 2,
y = u — u = J — J = l.
In the same way we find from the other pairs,
= _3_fV2l
x = 11 x=-3 + iV2| x=-3-iV21
y = 2j' y=-8-ivSj y=-3 + iV2J
* An equation is said to be symmetrical with respect to x and y whenever
interchanging x and y leaves the equation unchanged.
82 SIMULTANEOUS QUADRATICS [Chap. VII.
2. x + y — 2xy = 6, 4. x 2 + y* + a5y + x + y=17,
x 2 + 6 xy + y 2 — 4 x - 4 y = 6. x 2 + y* — 8xy + 2x + 2y = 9.
3. 3x 2 + 3y 2 = 8(x + y)-l, 5. 3x(l + x) + Sy(l + y)= 64,
xj/ = x + 2/ + l. 2x + 2y + xy = 16.
6. x 2 + y 2 = 13,
xy = 6.
Many systems of equations of degree higher than two, and
systems containing three or more unknowns may be solved by
combinations and variations of the above methods, but these four
cases do not by any means include all the simultaneous equation?
whose solution can be reduced to the solution of the quadratic
Usually the solution of such a system is in the nature of a puzzle
for which no special rules can be given. Whatever method be used
it must be kept in mind that the ultimate test of a solution is
substitution in the given equations. Many equations coming
under the cases mentioned here may be solved more easily by
other special methods. For example, Exercise 6 under Case IV
may be done as follows :
Given x 2 + y 2 = 13,
xy = 6.
Multiply the second equation by 2, add this result to and subtract it from
the first equation, and thus obtain the system
x 2 + 2 xy + y 2 = 26,
x 2 - 2 xy + y 2 = 1.
Extracting the square root of each, we have
x + y = ± 6,
x-y = ±l.
From these equations we find the same four results which were obtained by
the general method for solving symmetrical equations.
EXERCISES AND PROBLEMS
1. X + y - = ^, 2. 3x-i=4,
y x 3 y
y
2
3 1_1 = 5 *• x + y + 2y 2 = ll,
' x y % 3x-2y-2y» + 9 = 0.
i + -i=233.
x 2 y 2
Art. 53]
EXERCISES AND PROBLEMS
83
5. xy — x = 12,
xy4-3y = 36.
7. (x4-l)(y + 2) = 28,
9. x(y - 4) = 14,
y\x + 1) = 33.
11. (x-l)(y4-6)=80,
(x4-4)(y-2)=39.
13. x 2 - 4 xy + 3 y 2 = 0,
x 2 4- y 2 = 5(y4-6).
15. (x-4y-3)(x4-y) = 0,
x 2 4-y 2 = 22 + 4xy.
17. (x-y) 2 4-(x-y) = 6,
x* = y(2x4- l)4-2x- 18.
Hint : Solve the first equation for
{x - y).
19. x» - y» = 19,
as- y = 1.
21. x*4-2xy4-y 2 =(x4-y) (3x4-1),
X 2 — y 2 + 2xy4-l = 0.
23. x 2 — yy/xy= 14,
y 2 — x Vxy = — 7.
25. x 2 — xy + 1/ 2 = 3 a 2 ,
x4-y = 3(x — y).
27. x 2 + y 2 = 13,
y 2 + 2 2 = 25,
s 2 + x 2 = 20.
29. x(y -f z) = 8,
y(s + x) = 18,
z(x + y) = 20.
31. xy + zu = 17,
xz + yu = 13,
xtt + yz = 11,
x + y+3 + t« = ll.
6. 2y 2 --4xy4-3x 2 = 17,
y 2 — x 2 = 16.
8. x 2 + y 2 = 8 y + 1 - 2 xy,
2(x + y) 2 = 10x + 9y + 3.
10. 3(x 2 -y 2 ) = 2x4- 17,
x 2 - y 2 + x 4- y = 18.
12. (x + y) 2 + (x + y)=12,
3 x 2 4- y 2 = x 4- y + 4.
14. 2(x+l)(2/-4) = y + l,
(3x-l)(2y-9) = 6.
16. x 2 — xy — 2y 2 = 0,
x 2 + 2 y 2 - 3 x = 12.
18. x 3 4- y 3 = 28,
x + y = 4.
Hint : Divide the first equation by
the second.
20. x 3 - y 3 = 63,
x 2 + xy + y 2 = 21.
22. v^-Vi/rr^Cx-y),
xy = 36.
24. x 4 — y +Vx 2 — y = 20,
x 2 — y 2 = 544.
26. x 2 + y 2 = a 2 ,
Vx 4- Vy = \/a.
28. x 2 + 2y 2 -z 2 = 5,
2x + y + z = 6,
x + 4y — z = 5.
30. x + y = 11,
z + w = 10,
xy = zu,
x 2 + y 2 + z 2 + u 2 = 125.
32. x 2 + y 2 4- z 2 4- u 2 + « 2 = 19,
x + y + z = w + t> — 1,
X + V = Z + M,
x + y 4- v = 5,
y 4- z = t>.
84 SIMULTANEOUS QUADRATICS [Chap. VII.
Eliminate x and y from the following equations.
33. x 2 + y 2 = a 2 , 34. x 2 + y 2 = a 2 ,
x + y=6, x 2 -y 2 =6 2 ,
x — y = c. xy = c 2 .
35. x 2 + xy = a 2 , 36. -- + |- = 2,
2/2 + xy = 6 2 ,
x 2 -f 2/ 2 = c 2 .
a 2 ft 2
xy = a&,
a«x 2 + ft 2 ?/ 2 = c*.
37. A circular track is built so that the width of the track is ^ of the
inside diameter. After construction it was found that the area of the track
was 2564 square yards. What are the inside and outside lengths of the
track ?
38. The fence around a rectangular field is 1400 feet long. The diagonal
of the field is 600 feet long. What are the dimensions of the field ?
39. The diagonal of a rectangular parallelopiped is 14 inches long. The
sum of the three dimensions is 22. The reciprocal of one dimension is one
half the sum of the reciprocals of the other two. What are the dimensions
of the solid ?
40. A father divided % 1000 between his two sons and kept it for them at
simple interest until called for. At the end of 3 years, one son called for all
the money due him and received $665.50. At the end of 4 years the other
son received % 576 as his share. How was the money originally divided and
what rate of interest did the father pay ?
41. A few days after the outbreak of the war a 25-pound bag of sugar
cost the retailer 77J cents more than it did just before the outbreak. For
$ 165 a grocer received 1550 pounds less sugar after the outbreak than he
would have received before for the same amount. What was the price before
and after the outbreak of the war ?
42. A silver wire 1000 millimeters long is to be covered with a layer of
gold until the diameter of the wire is increased one tenth. It is found that
4 cubic millimeters of gold is needed. What is the diameter of the wire ?
43. A road between two towns is 33 miles long. At eight o'clock, from
each of the two towns, a traveler starts toward the other and walks at a uni-
form speed. At eleven o'clock they meet, but one traveler arrives at his
destination 1 hour and 6 minutes earlier than the other. How many minutes
are needed for each to walk 1 mile ?
44. A man divides a tract of land into city lots. He sells the lots all at
the same price and realizes $4800. If the number of lots had been one
greater, and the price $8 per lot cheaper, he would have received the same
amount of money. How many lots were there and what was the price per
lot?
Art. 53] EXERCISES AND PROBLEMS 85
45. A circular cylinder is inscribed in a sphere of radius 10. The total
surface of the cylinder is half the surface of the sphere. What is the radius
and what is the altitude of the cylinder ?
46. The diagonals of the three faces of a rectangular parallelepiped which
meet in a vertex of the solid are 10, 12, 14, respectively. What is the
volume of the solid ?
47. The floor of a rectangular room contains 273 square feet, one wall
189 square feet, and the adjacent wall 117 square feet. What are the dimen-
sions of the room ?
48. Psychologists assert that the rectangle most pleasing to the human
eye is that in which the sum of the two dimensions is to the longer as the
longer is to the shorter. If the area of a page of this algebra re mains
unchanged, what should its dimensions be ?
49. An aeroplane, flying 75 miles per hour and following a long straight
road, passed an automobile going in the opposite direction. One hour later
it overtook a second automobile. The automobiles passed ouch other when
the aeroplane was 100 miles away. If both automobiles travel with the same
speed, how far apart were they when the aeroplane passed the second one.
and what was their speed ?
50. Two students attempt to solve a problem that reduces to a quadratic
equation. One in reducing has made a mistake only in the constant term of
the equation, and finds 8 and 2 for the roots. Tin; other makes a mistake
only in the coefficient of the first degree term, and finds - - i) and 1 for
roots. What was the quadratic equation ?
51. The radius of the front wheel of a carriage is 6* inches less than that
of the rear wheel. If the front wheel makes 80 more revolutions than the
rear wheel in going a mile, what is the circumference of each wheel t/> Mm
nearest inch?
52. Two polygons have together 12 sides and V,) diagonals. How many
aides has each ?
53. A military ambulance traveling 12 miles per hour tx-ut on ahead u
motor cyclist messenger who could travel with twice the *|*<d of the amhu
lance. A half-hour later it was found necessary to revise the hh'hkuw, nod
a second motorcyclist was sent to overtake the first and to give him tin; n<-w
message. The second messenger returned to the amhuJa/i'n; In foity live
minute*. What was his average h\**A d taring that time V
CHAPTER VIII
INEQUALITIES
54. Definition. The expressions " a is greater than b " (a > b)
and " c is less than d " (c < d), when a, 6, c, d, are real numbers,
mean that a — b is a positive number and c — d is a negative num-
ber. Such expressions are called inequalities. Two inequalities
a > b, c > d, which have the signs pointing in the same direction,
are said to be alike in sense. If the signs point in opposite direc-
tions, as a ><*b, c<d, they are said to be different in sense. The
expression a ^ b is read " a is less than or equal to 6," and
a ^ b is read " a is greater than or equal to b."
55. Absolute and conditional inequalities. We have seen that
there are two kinds of equalities, identical and conditional equali-
ties. Corresponding to these there are two kinds of inequalities.
An inequality such as a 2 4- b 2 > — 1, which is valid for all real
values of a and b, is called an absolute inequality ; while an in-
equality such as x — 4 > 0, which holds only when x is greater
than 4, is called a conditional inequality. In a. conditional in-
equality the letters cannot take all real values.
56. Elementary principles. The following elementary princi-
ples, which follow at once from the definition of an inequality,
aiust be observed in dealing with inequalities.
I. TJie seiise of an inequality is not changed if both aides are in-
creased or decreased by the same number. In particular, the sense
is not changed if we transpose a term, changing its sign.
Let a > b.
Then a — b = n 9 where n is positive,
and a + fc-6-fc = «,
or (a + k) - (b 4- k) = n.
Hence, a 4- A; > b 4- k.
86
Arts. 64-56] ELEMENTARY PRINCIPLES 87
II. The sense of an inequality is not changed if both sides are
multiplied or divided by the same positive number.
JUL The sense of an inequality is reversed if both sides are mul-
tiplied or divided by the same negative number.
The proofs of II and III are very similar to the proof of I.
EXERCISES
1. If a and b are not equal, show that a 2 + b 2 > 2 ab.
Solution : (a — ft) 2 > 0, since the square of any real number is positive.
That is, a 2 — 2 ab + b 2 > 0.
By Principle I, a 2 — 2 ab + b' 2 + 2 ab > + 2 ab,
or a 2 + b 2 > 2 ab.
2. Show that ° "*" > Va6, if a and b are positive and unequal.
Show that the following inequalities subsist, the letters representing dis-
tinct positive numbers.
3. £±i > 1°*. 4. a 2 +b 2 + c 2 >ab + ac + be.
2 a + b
5. —Jr- > "~T — 6 - <* + ->2, ifa=£l.
7. a* + 3&»>2&(a+fc). 8. ? + ->2.
6 a
9. If 6, a\ and / are positive, and - < - < - , show that a + c + e lies
. . a^c b d f & + <*+/
oetween - ana -•
6 /
10. Given a* + b 2 = 1, c 2 + d 2 = 1, show that a& + ccf < 1.
11. Show that if we denote by | a |,* | b |, etc. the numerical values of a,
6, etc., then
|a+6|^|o| + |6|,
|a-6|g|o| + |6|,
|a-6|>|a|-|6|.
12. Given a < a\ < A, b\ > 0,
a < 02 < -4? &2 > 0,
a < a n < 4, 6„ > 0,
show that
a(pi + 6| + — + &n) < <*1&1 + «2&2 + — + «A < -4(&i + &2 + — + &»).
• The expression | a| is often read "absolute value of a," meaning the
numerical value without regard to sign.
INEQUALITIES
{Chap. VIIL
57. Conditional inequalities. By transposing terms every in-
equality may be reduced to an inequality of the form P>0, or
P < 0. If one or both sides involves a
variable, say x, it can be put in one of
the two forms f(x) > 0, or /(*)<0. In
this connection the most important prob-
lem is to find the range of values of the
variable for which the inequality holds.
In the case of linear inequalities the solu-
tion is easy. Thus, to find the values of
x for which the inequality
3x + 19>12-ie
holds, all the terms can be transposed to
the left-band side, and there results
Hence the
inequality
in question
holds only
Fio. 15. for
■ >-*
Graphically,
8 x + 19 > 12 - x
for those values of x for which the
graph of the function
3a + 19-12 4-a;==4a; + 7
lies above the X-axis (Fig. 15).
The graph is of great service in
determining the values of x for
which one function of x is greater
or less than another function. Thus, to find the range of values
of x for which
2x*-3as+8>3 a + 2a; + 4,
we transpose all terms to one side and have
x 3 — 5jc + 4>0.
Y—
- t
- t-
. jz
- u .
_ j
. r
_ 7
\
- t -
. j .
-J- *
_ L
1
\
t
Abt. 57]
CONDITIONAL INEQUALITIES
89
The graph of this function is shown in Fig. 16. It crosses the
X-axis at 1 and 4, and for x > 4, or x < 1, the function a 2 — 5 <c-f-4
is positive; while for 4 > x > 1, it is negative; hence,
2 a 2 - 3 a; -f 8 > a? 2 4- 2 x + 4
for a? > 4, and a? < 1, while
2aj 2 -3a? + 8< a 2 + 2 a + 4
for 4 > a? > 1.
EXERCISES
For what values of x do the following inequalities hold ?
x 3
1. 7x-8>82.
2.
1 2
>0.
3.
x 3
1 2
4 3
7 x
5. 8x*-llx>4.
7. (x-6)(x-2)>(x-l)(x-3).
9. 2 -*=l<0.
x + 8
U. ax + &>0.
4. x 2 <4.
6. x + ->2.
x
8. 2x 2 -3x- 10 >x 2 .
,~ a — x ^a 2 — x 2
a + x a 2 + x 2
f>0,
12. x 2 -4x + 3{ =0,
<0.
14. ^±_f<0; >0.
x — 3
x 1 4
3 x -5
4 2 3
>
-1 x 1
1 3 2
x —3 2
<0,
13. aafl + bx + c { =0,
<0.
x 3 -1
13-2 1 -2 >0. 16.
2 3 x
17. Given ai > a 2 > 03 > ••• >a n , and a r > x > a r+ i ; show that
(x — ai)(x — a 2 )(x — a 3 ) ... (x — a n )
represents a positive number when r is even, and a negative number when
r is an odd number.
18. Show that (x — a{)(x — a 2 ) 2 is positive if x>ai, and negative if
■y ■
CHAPTER IX
MATHEMATICAL INDUCTION
58. General statement. Many important theorems in algebra
can be proved by a method called mathematical induction. The
method may be best explained after applying it to a simple
example.
Let it be required to show by mathematical induction that the
sum of the first n odd numbers is the square of n. It should be
noted that the nth odd number is 2 n — 1. We have then to
prove
1.+ 3 + 5 + - 4- (2 n - 1) = n 2
for all positive integral values of n.
We note that the theorem is true for n = 1 and n = 2. Assume
it to be true for n = r. We have then
1 + 3 + 5 + - 4- (2r - 1) = r 2 . (1)
Adding the next odd number, that is, 2 r 4- 1 to both sides of (1),
we obtain
1+3+5+ «. + (2r-l)+(2r+l)=r 2 +2r+l=(r+l)», (2)
which states that the sum of the first r + 1 odd numbers is
(r + 1) 2 . Hence the theorem is true for n = r + 1, if it is true for
n — r. We know it is true for n = 2, hence it is true for
n = 2 + 1 = 3. Since it is true for n = 3, it is true for n = 4
and so on. Therefore, the sum of the first n odd numbers is the
square of n.
In general, an argument by mathematical induction consists
of two necessary parts.
90
Abt.58] METHOD OP MATHEMATICAL INDUCTION 91
First. To show by mere verification that the principle in
question is true for some particular case, preferably for n = 1,
or n = 2.
Second. To show that if it is true for the case n = r, it is true
for the case n = r + 1.
It is no proof simply to show that a theorem is true in a num-
ber of cases. For example the above theorem is not proved by
showing that it is true, for n = l,n = 2,n = 3,na4 and so on
for a definite number of cases. The second part of the proof is
necessary.
A celebrated example illustrating this point is the expression
n*—n + 41. From the table we see that n 2 — n 4- 41 is a prime
n =
1
41
2
43
3
4
53
5
61
6
71
7 '
83
8
97
9
113
10
131
11
151
12
173
n> — n + 41
number for all integral values of n up to 12. The table could
be continued up to n = 40 and the lower row would still contain
nothing but prime numbers. However we have no proof that
n* — n + 41 is prime for all integral values of n. To prove this
it is necessary to take the second step in the proof by mathe-
matical induction, i.e. to prove that if r 2 — r + 41 is prime, then
(r + 1) 2 — (r -f 1) -f 41 is prime. But this is impossible. In fact,
when n = 41, we have
n * _ n + 41 = 1681 = 41 2 ,
a number which is not prime.
Again, it is no proof simply to show that if a statement is true
for n = r it is true for n ==r + 1. „ For example, assuming that
the sum of the first r even numbers is an odd number it follows
that the sum of the first r -f 1 even numbers is odd. Though the
second part of the proof of this statement by mathematical in-
duction can be correctly presented, we know the statement to be
false. The first part of the proof is lacking.
In a proof by mathematical induction both parts are necessary
and are of equal importance.
92 MATHEMATICAL INDUCTION [Chap. IX
EXERCISES
Prove by mathematical induction.
1. 2 + 4 + 6+... +2n = n(n + 1).
Solution : For n = 1, we have
2 = 1-2.
Hence, the statement, 2 + 4+6 + — +2n = n(n + 1), is true for n = 1,
and the first part of the proof is satisfied. Suppose it is true for r numbers.
That is, assume that
2 + 4 + 6+ ... +2r = r(r + l). (1)
Adding the (r + l)th term to both sides,
2 + 4 + 6 + ... + 2 r + (2 r + 2)= r(r + 1}+ 2 r + 2.= (r + l)(r + 2). (2)
By comparing (1) and (2), it is seen that (2) can be obtained from (1) by
replacing r by r + 1. That is, if the formula holds for r integers, it holds for
r + 1 integers. But we know the formula is true for n = 1, hence, it is true
for n = 2, and so on. Hence, for any integer n,
2 + 4*+ 6 + ... + 2 n = n(n + 1).
2. 1 +2 + 3+ .- +n = -(n + 1).
#3. 3 + 6 + 9 + 12+ ... +3n = 8n ( n+1 ) .
^4. 12 + 22 + 32+ ... +n2 = |n(n + l)(2n + l).
5. 22 + 42 + 62+ ... +(2n)2= 2n ( n + 1 )( 2n + 1 ).
o
6. -JL 4. _L_ + _L_ 4. ... to n terms = n
1-2 2-3 3-4 n+ 1
7. x — y is a factor of x n — y n if n is any positive integer.
Hint : By verification we see that
ajr+l — yr+l = X (X r — 2T) + W(X — y).
If x — y is a factor of x r — y, then it is a factor of xr+ l — y+i.
8. x 2n — y 2n is divisible by x + y if n is any positive integer.
9. 13 + 23 + 33+ ... + n3 = n2 ( n + 1 ) 2 = (1 + 2 + 8 + •« +n)».
10. 2 + 22 + 23 + ... + 2» = 2(2» - 1).
11. gn "~ fln = x n "i + ax n ~ 2 + -. + a n ~*x + a n ~i.
x — a
59. Meaning of r!. The symbol r!, read "factorial r,"* is
used to indicate the product 1 • 2 • 3 ••• r. Thus,3 ! = 1 • 2 • 3 = 6;
7! = 1.2.3.4.5.6.7 = 5040.
* The symbol [r is often used to represent factorial r.
Arts. 58-60] BINOMIAL THEOREM 93
EXERCISES
Evaluate the following expressions :
,9! « 81 Q 3151 A 31 + 5 1
61 71 41 41
5. -£~ 6. Prove — ^ — = r. 7. — ^ — =?
3161 (r - 1) 1 (r - 2) 1
60. Binomial theorem; positive integral exponents. By multi-
plication, we find (a + xy = a* + 2ax + x*-
(o + x)* = a 8 4- 3a?x 4- 3 aa? 2 + a- 3 = a 3 4- 3 a 2 a 4- ^-^ a» 2 + a 8 .
(a -f- a?) 4 = a 4 4- 4 a?x + 6 a 2 « 2 4- 4 aa 3 4- «*
= a 4 4- 4 a 3 « 4- ^ a 2 ^ 2 4- 4 '^' 2 aa- 3 4- x 4 .
If w represents the exponent of the binomial in any one of the
above three cases, we notice :
(1) The first term is a n .
(2) The second term is na n ~ l x.
(3) The exponents of a decrease by unity from term to term
while the exponents of x increase by unity.
(4) If in any term the coefficient be multiplied by the expo-
nent of a and divided by the exponent of x increased by unity,
the result is the coefficient of the next term.
For n < 5, we may then write
(a + x) n = a* 4- na*-^ 4- n \ ~ ' a n ~ 2 x* 4- «.
. w(n— 1) ••• (n— r4-2) n _ r+1 ,_. , , n
4- -A l k^ — x_ J. a n r +i x r-i+ ... + x * m
(r - 1) !
Here the question naturally occurs : Does the expansion hold
for n > 5 ? It can be shown by mathematical induction that it
holds for any positive integral value of n.
Assume
(a + x) m = a" 4- mar^x 4- m ( m "" ^ a m -*x* 4- -
. m(m — 1) ••• (m — r 4- 2) „+! __i , , ^
4 — * i — ^m ■ — L a wrrr + 1 7r* + ...4- x m .
(r-1)!
94 MATHEMATICAL INDUCTION [Chap. IX.
Multiply both members of this assumed equality by a -f x, and
we obtain
(a + «) m+1 =
a m+i + ma m x + ... + m(m - l)-(m - r + 2) ^-h-,^ + ... + ^,
(r-1)!
+ a?X + - + ffl ^ m " 1 >'< m ' r + 3 -). qm-H-agr-l + ... + maa?" + *"«
(r-2)!
= a«+i+(m + l)gm x + ... + (m + l)m ... (m - r + 3) g^,^ + ...
v y (r-1)!
+ (m + l)aa; m + * m+1
This expansion is the same that -would be obtained by substi-
tuting m-+l for m in the expansion of (a 4- x) m . Hence, if the
expansion is true for n = ra, it is true for n = m -f 1. Since we
know it is true for n = 2, it is true for n = 3, and so on. Hence,
when n is any positive integer,
(a + a?) w = a n + na"" 1 * + n ^7 ^ a*-*^ + -
+ n(n- !)•••(*- r + 2) ^.^^..i _ ^
(r - 1) !
This expansion of a binomial is called the binomial theorem.
In the expansion of (a + x) n , the rth term is
n(n - l)(n - 2) - (n - r + 2) an -r+ w-i
(r-1)!
which may also be written
n\
ar^+hr*.
(r - 1) ! (n - r + 1) !
The term involving of is
n(n-l)(n-2)...(M-r + l) ^.^ = n] tf ^ >
r! rl(n — r)l
Each of these terms is sometimes called the general term of the
binomial expansion.
In this chapter the exponent n in the binomial expansion is '
limited to positive integral values, but no assumption has been
made with regard to a or x, so we are at liberty to use the expan-
sion no matter what sort of numbers a and x may be. Thus, in
Abt. 60] BINOMIAL THEOREM 95
(26 — 4r 2 ), a = 2b and x = — 4^. In a later chapter it will be
shown that the expansion may be interpreted so as to hold when
n is a negative or fractional number, but in that case the number
x must lie between — a and 4- a.
EXERCISES
Expand
1. (2x-Sy*y.
Solution :
(2» - 3y*)* = (2x)* + 4 (2x)3 (- 3 y*) + 6(2x) 2 (- 3y*f + 4(2x)(- 3y3)3
+ (_ 3y3)4 = 10^4 _ 96x3^3 + 216x2|/« - 216a^ + 81 y* 2 .
2. (a + x)«.
4. (a + 6)7.
6. (J + 2 a)*.
8. (x-3y 2 ) 5 .-
K>-
3.
(a — x) 6 .
5.
(2 - X)5.
7.
(1 + ay.
9.
(a + Vo>.
11.
KH-
;)'•
13.
( Vx — Vy) 8 .
15.
(o^ + a*) 6 .
.
17.
(3 - l)i.
19.
(a + 6 + o)3.
Hint : Consider (a
-h 6) as repre-
senting one number.
10.
12 l a H 4
12 b\
16. (x-i-y*) 4 .
VV6 Vx/
-.(vs+.+g 1 . «.(f-i +1 )'
22. Find the seventh term in the expansion of (x* — 2y 2 ) n - i / A
Solution : The rth term is given by the expression ^ ' *- CA,XJ
n(n-l)-..(n--r + 2) .^ ,
(r-1)!
Here n = 11, r = 7, a = x*, x = - 2 y 2 , n - r + 2 = 6.
Substituting these in the expression for the rth term, we have
11.10.0.8.7.6 (a ,fy ( _ 2y2 y = 29668 xV-
6 !
s**3. Find the fourth term of (a -4 &)". ~) (^X — /J #
> 24. Find the eleventh term of (2 x - j/)". / ^ -^ Z 7 .
^ 25. Find the middle term of (x 2 + 3 y 2 ) 8 . | 7££ ^^
^ 26. Find the fourteenth term of (a + &)**.
96 MATHEMATICAL INDUCTION [Chap. IX.
27. Find the eighth term of
x a
Va y
is
28. Find the sixth term of (xVy + yVz) 9 .
29. Find the middle terms of (1 — a*) 7 .
30. Number the terms of the expansion of (a + 6) 6 , and with these num-
bers as abscissas and the coefficients of the corresponding terms as ordinates,
plot points.
31. On the same coordinate paper on which points described in Problem
30 are plotted, graph points for the expansion of (a + 6) 10 .
32. Use the binomial theorem to find (102) *.
Hint: 102 =(100 + 2).
Use the binomial theorem to find \f
33. (99)«. . • 34. (61)5. 35. (.08)«.
36. (1.1) 10 , correct to four significant figures. »
37. ( 1 . 1 ) 15 , correct to four significant figures.
CHAPTER X
VARIATION
61. Direct variation. In Chapter III we have seen that if y is
a function of x, then in general y changes when x changes. We
might say that y varies when x varies, but the word " varies " has
come to have a more restricted meaning when used in this con-
nection. Each of the statements " y varies as x" " y varies di-
rectly as x" " y is proportional to a?," " y is directly proportional to
x" means that the ratio of y to x is constant. That is,
£ = Jc or y = kx.
x
The constant k is called the constant of variation. The expres-
sion " y varies as x " is often written
y oca?.
The area of a circle varies as the square of its radius. That is,
A oc f 4 , or A = kr 2 , if A represents the area and r the radius. Here,
k, the constant of variation, is equal to ir. If a train moves with
a uniform speed, the distance s traversed varies as the time t
That is, 8 oc t , or s = kt.
62. Inverse variation. Each of the statements "y varies in-
versely as x" " y is inversely proportional to x " means that y is
equal to the product of the reciprocal of x and a constant. That
is,
k
x
The volume of a gas varies inversely as the pressure. That is,
k
v = - t
if v represents volume and p pressure.
97
98 VARIATION [Chap. X.
63. Joint variation. The statement "z varies jointly as x
and y" means that z equals the product of xy and a constant
That is, z = Jcxy.
The distance which a train moving with a uniform speed trav-
erses varies jointly as the speed and the time, or d = kvt, where
d is the distance covered, v the speed, and t the time. In this
case k = 1, if v and d are measured with the same unit of length.
64. Combined variation. The statement "z varies directly as x
and inversely as y " means that z varies jointly as x and the re-
ciprocal of y. That is,
2 = — •
y
The attraction F of any two masses mx and 7% for each other
varies as the product of the masses and inversely as the square
of the distance r between the two bodies. That is,
j P __ km 1 m 2 i
9*2
EXERCISES AND PROBLEMS
*
Write the following statements, 1 to 8, in the form of equations.
1. y varies as x, and y = 64, when x = 2.
Solution : y = kx.
Substituting 64 for y and 2 for a, gives
64 = 2 k, or k = 32.
Hence, y = 32 x.
2. y is directly proportional to x, and y = 18 when jc = 6.
3. s varies as t\ and « = 64 when t = 2.
4. p varies inversely as u, and v = 128 whenp = 16.
5. z varies jointly as x and ?/. When x = 2, and y = 3, it is found that
z = 120.
6. z varies directly as x and inversely as y. When x = 2 and y = 8, it is
found that z = 120.
«
7. The volume V of a sphere varies directly as the cube of its radius r.
Arts. 63-64] EXERCISES AND PROBLEMS 99
8. The volume V of a circular cylinder varies jointly as its altitude h and
the square of its radius r.
9. The number of feet a body falls varies directly as the square of the
number of seconds occupied in falling. If the body falls 16.1 feet the first
second, how many feet will it fall in ,5 seconds ? In 10 seconds ?
^f^ 10. The safe load of a horizontal beam supported at both ends varies
jointly as the breadth and square of the depth, and inversely as the length
^between supports. If a 2 x 6 inches white jtfne joist, 10 feet long between
supports, safely holds up 800 pounds, what is the safe load of a 4 x 8 beam
« of the same material 15 feet long ?
11. The pressure of wind on a sail varies jointly as the area of the sail
and the square of the wind's velocity. When the wind is 15 miles per hour,
the pressure on a square foot is one pound. What is the velocity of the
. wind when the pressure on a square yard is 25 pounds?
12. The pressure of a gas in a tank varies jointly as its density and its
absolute temperature. When the density is 1 and the temperature 300°, if
the pressure is 15 pounds per square inch, what is the pressure when the
density is 8 and the temperature 320° ?
13. Write in the form of an equation the following physical law : the
bend 6 of a rod supported at both ends varies directly as the weight m hung
at its middle point, directly as the cube of the length I of the rod between
supports, inversely as the width w of the rod, and inversely as the cube of
the depth d.
' 14. A beam 15 feet long, 3 inches wide, and 6 inches deep when sup-
ported at each end can bear safely a maximum load of 1800 pounds. What
is the greatest weight that can safely be placed on a beam of the same mate-
rial 18 feet long, 4 inches wide, and 4 inches deep ? (See Prob. 10.)
15. A plank 10 feet long, 10 inches wide, and 2 inches thick is supported
at both ends. A weight of 180 pounds hung at the middle makes it bend 3
inches. How much will the plank bend if placed on edge ? (See Prob. 13.)
16. The area of the top of a well-designed chimney varies as the quantity
of coal used per hour and inversely as the square root of the height of the
chimney. The top of a 150-foot chimney connected with a furnace using
11,000 pounds of coal per hour is 27 square feet. What should be the area
of the top of a 100-foot chimney connected with a furnace using 2500 pounds
of coal per hour ?
17. If a heavier weight draws up a lighter one by means of a cord passed
over a pulley, the number of feet passed over by each weight in a given
time varies directly as the difference of the weights and inversely as their
sum. If 5 pounds draw up 3 pounds 16.1 feet in 2 seconds, how high will 10
pounds draw 9 pounds in 2 seconds ?
100 VARIATION [Chap. X.
18. The time of a railway journey varies directly as the distance and
inversely as the speed. The speed varies directly as the square root of the
quantity of coal used per mile and inversely as the number of cars in the
train. In a journey of 82 miles in } hour with 12 cars, J ton of coal is
used. How much coal will be consumed in a journey of 64 miles in 2 hours
with 10 cars ?
19. A paper disk is placed midway between two sources of light which
are 12 feet apart. If the amount of light falling on the disk varies inversely
as the square of the distance from the source of light, show that if the disk
is moved parallel to itself a distance 2V3 feet, the whole amount of light
falling on the disk is trebled.
20. How far must the disk in Problem 19 be moved from a point midway
between the two lights so that the total amount of light on the disk is
doubled ?
21. A solid spherical mass of glass 2 inches in diameter is blown into a
hollow spherical shell whose outer diameter is 4 inches. If the volume of a
sphere varies as the cube of the diameter, what is the thickness of the shell ?
22. Kepler's third law states that the square of the number of years it
takes a planet to revolve about the sun varies directly as the cube of the dis-
tance of the planet from the sun. Let the distance from the earth to the
sun be 1. How long would it take a planet whose distance from the sun is
100 to complete one revolution ?
23. The number of inches a body falls in one second varies inversely as
the square of the distance from the earth's center. At the surface of the
earth a body falls 103 inches in one second. How far would it fall in one
second if it were as far away as the moon ? (The distance of the moon from
the earth's center may be taken as sixty times the radius of the earth.)
24. The crushing load of a solid square oak pillar varies directly as the
fourth power of its thickness and inversely as the square of its length. If a
four-inch pillar 8 feet high is crushed by a weight of 196,820 pounds, what
weight will crush a pillar of the same wood 6 inches thick and 12 feet high ?
CHAPTER XI
PROGRESSIONS
65* Arithmetical progressions. An arithmetical progression is
a sequence of numbers each of which differs from the next
preceding one by a fixed number called the common difference.
Thus,
2, 4, 6, 8, ...
is an arithmetical progression with the common difference 2. In
the arithmetical progression
10, 8, 6, 4, 2, ...
the common difference is — 2.
The numbers of the sequence are called the terms of the pro-
gression.
66. Elements of an arithmetical progression. Let a represent
the first term, d the common difference, n the number of terms
considered, I the nth, or last term, and s the sum of the sequence.
The five numbers a, d, n, Z, and s are called the elements of the
arithmetical progression.
67. Relations among the elements. Since a is the first term, we
have, by definition of an arithmetical progression,
a + d = second term,
a + 2 d = third term,
a + 3 d = fourth term,
• • • •
o+(n- l)d = nth term.
That is, I = a + (n - l)ef. (1)
101
102 PROGRESSIONS [Chap. XI
The sum of an arithmetical progression may be written in each
of the following forms :
s = a+(a + cT) + (a + 2d) + ... +(Z- 2d) + (l-d) +1,
8=l+(l — d) + (l-2d)-\ h(a + 2d) + (a+d)+a.
By addition
2s=(a+l)+(a + r) + (a + t) + --- + (a+l) + (a + r)+(a + l)
= n(a + 1).
Therefore, s=3(a+J). (2)
Whenever any three of the five elements are given, equations
(1) and (2) make it possible to find the remaining two elements.
Exercise. Establish formulas (1) and (2) by mathematical induction.
68. Arithmetical means. The first and last terms of an arith-
metical progression are called the extremes, while the remaining
terms are called the arithmetical means. To insert a given number
of arithmetical means between two numbers it is only necessary
to determine d by the use of equation (1) and to write down the
terms by the repeated addition of d.
EXERCISES
Find I and s for the following seven sequences :
I. 2, 11, 20, ... to 10 terms.
Solution: l = a + (n— l)d.
Here, a = 2, d = 9, n = 10.
2 = 2 + 9- 9 = 88.
2 V J
= 6(2 + 83) = 425.
2. — 2, —6, —8, •«• to 20 terms. 3. 3, 7, 11, ••• to 16 terms.
4. 6, 1, — 3, to 20 terms. 5. $, ^ i, to 8 terms.
6. 7, *£, V, ... to 16 terms. 7. - J, - J{, - fj, ••• to 17 terms.
8. Given a = 19, d = — 2, s = 91 ; find n and L
9. Given d = 4, n = 15, I = 69 ; find a and s.
10. Given I = — J, n = 14, d = \ ; find a and 8.
II. Given a = 12, I = - 76, a =— 620 ; find w and d.
12. Given a" = 7, 2 = ^, d = J ; find n and *.
Arts. 67-71] RELATIONS AMONG THE ELEMENTS 103
13. Given a = — 4, 1 = — 64, n = 21 ; find d and s.
14. Given d = l,l = fy 8 = 1 V" a J ^d a ^ w -
15. Given i = 50, n = 50, a = 1275 ; find a and d.
16. Given d = 10, n = 10, s = 10 ; find a and J.
17. Given a = 7, n = 7,s = 7; find d and 2.
18. Insert 6 arithmetical means between 3 and 8.
Solution : We have to find d, when a = 3, I = 8, and n = 6 + 2 = 8.
Since l = a+(n — l)d,
we have 8 = 3 + 7d, or d = $.
Hence, the 6 arithmetical means between 3 and 8 are
V, V* Vi Vi ¥i ¥•
-~»19. Insert 8 arithmetical means between 2 and 14.
_ 20. Find the arithmetical mean between 10 J and 16.
-w 21. Insert 9 arithmetical means between 1 and — 1.
68. Geometrical progressions. A geometrical progression is a
sequence of numbers in which the same quotient is obtained by
dividing any term by the preceding term. This quotient is called
the ratio. Thus,
3, 6, 12, 24, ...
is a geometrical progression with a ratio 2.
70. Elements of a geometrical progression. The elements are
the same as those for an arithmetical progression with one
exception. Instead of the common difference of an arithmetical
progression, we have here a ratio represented by r.
71. Relations among the elements. If a represents the first
term, then
ar = second term,
ar 2 = third term,
ar* = fourth term,
• • • •
ar n-l = n ^j 1 t erm#
That is, I = ar»-i. (1)
By definition,
st = a + ar + ar 2 + ar z + ... + ar n '\ (2)
Then, ar = ar + ar 2 + ar* + ••• + ar n ~ l + ar n . (3)
104 PROGRESSIONS [Chap. XL
Subtracting members of (2) from members of (3), we have
sr — 8 = ar n — a.
Hence, , = «£!=_« = «0=J3 . (4)
V — 1 1 — V
Since I = ar n_1 , (4) may be written in the form
• = r - (5)
r — 1
Here, as in an arithmetical progression, whenever any three of
the five elements are given, relations (1) and (5) make it possible
to find the other two.
Exercise. Establish formulas (1) and (4) by mathematical induction.
72. Geometrical means. The first and last terms of a geo-
metrical progression are called the extremes, while the remaining
terms are called the geometrical means. To insert n geometrical
means between two given numbers is to find a geometrical pro-
gression of n+2 terms having the two given numbers for extremes.
EXERCISES
1. Given a = 1, r = 3, n = 9 ; find I and 8.
2. Given a = — 2, r = 3, n = 8 ; find I and a. ',/-.* \
3. Given a = J, r = ^, n = 10 ; find 2 and 8.
^ 4. Given a = 0, r = — 2, n= 8 ; find 2 and 8.
5. Given .s = 242, a = 2, n = 5 ; find r and I.
6. The 3d term of a geometrical progression is 3, and the 6th term is
81. What is the 10th term ?
7. What is the 7th term of a geometrical progression whose first term is
2 and third term 3 ?
8. What is the sum of the first 5 terms of a geometrical progression
whose first term is 2 and third term 8 ?
9. The first term of a geometrical progression is 3, and the last term 81.
If there are four terms in the geometrical progression, find the common ratio
and the sum of the series.
"10. Insert one geometrical mean between 7 and 343. *
11. Insert two geometrical means between 2 and 1024.
12. Insert four geometrical means between 12 and J.
13. What is the eighth term of the progression a 2 , ab, &*, ••• ?
14. If each term of a geometrical progression is multiplied by the
number, show that the products form a geometrical progression.
(
Arts. 71-73] NUMBER OF TERMS INFINITE 105
73. Number of terms infinite. Consider the geometrical pro-
gression
h h h &> '•••
It may at first thought appear that the sum of the first n terms
of this progression could be made to exceed any finite number
previously assigned by making n large enough. That this is not
the case and that the sum can never exceed unity, will be seen
from the following illustration. Conceive a particle moving in a
straight line towards a point one unit distant in such a way as to
describe -J- the distance in the 1st second, % the remaining distance
in the 2d second, \ the remaining distance in the 3d second, and
so on indefinitely. This is represented in Fig. 17.
The distance AB represents one unit of distance. In the first
second the particle moves from A to P v In the 2d second it
moves from P x to P 2 , and so on. The total distance traversed by
the particle in n seconds is given by the sum
■J- + i + i + ••• to n terms,
which sum cannot exceed nor equal 1, no matter how many terms
we take, but can be made to differ from 1 by as small a positive
number as we please by making the number of terms large enough.
A
Fia. 17.
Thus, when n = 10, the sum is |£ff (Prob. 3, Art. 72). In
this illustration, 1 is said to be the limiting * value of the sum
of the first n terms of the series. If s n represents the sum of the
first n terms of the series, we write
lim -i
8 — n-^oo 8 n — *■>
which reads, " the limit of s n as n increases beyond bound is 1." t
* For definition of limit, see Art. 150.
t The symbol "iv->co " stands for **n increases beyond bound," or
"+* equivalent " n becomes infinite. "
106 PROGRESSIONS [Chap. XL
The limit s is called the sum of the geometrical progression
with infinitely many terms.
For any geometrical progression in which" the ratio is less than
1, the above argument can be repeated, and it can be shown that
there is a limiting value to the sum of the first n terms of such a
series. In Art. 71 , we have shown that the sum of the geometrical
progression
a -+- ar + ar 2 -+- ••• + ar" -1
, a(l — r n ) a ar*
is given by 8 n =-^ / = - _
1 — T 1 — T 1 — T
.We may then write
ni m *. = nlToo T^ «!T« t^ • (See Art. 152.)
It will be proved in the chapter on Limits (Art. 154) that
lim _^L = o when|r|<l.
Hence, 8= Hm *n=— — •
n->oo 1 — r
74. Series. A series is an expression which consists of the sum
of a sequence of terms. Thus, the indicated sum of the terms of
a progression is often called a series.
A finite series is one which has a limited number of terms.
An infinite series is one in which the number of terms is infinite ;
that is, the number of terms has no bound.
75. Repeating decimals. Repeating decimals furnish good
illustrations of infinite series which are at the same time the sum
of the terms of a geometrical progression with infinitely many
terms. For example, .33«S33 ••• may be written as the series
.3 + .03 + .003 + .0003 + ».,
where a = .3 and r = .1. The limit of the sum of n terms of
this series as the number n increases indefinitely is ■$-. Again,
.9828282 ... may be written
.9 + .082 + .00082 + ...
Arts. 73-77] HARMONICAL MEANS 107
where the terms after the first form a geometrical progression
in which a = .082 and r = .01.
The expression "limit of the sum of n terms of the infinite
series as n increases beyond bound " is often abbreviated by say-
ing merely " sum of the infinite series."
EXERCISES
Find the sum of the following infinite series :
*• | + | + i+-. -2. 1 - J + 4-^+ ».. 3. f-J + J .
4. £4.-12 + 6-3+ .... 5. 4-3 + f . 6. - 4 - \ - ^ .
Find the limiting value of the following repeating decimals :
7. .636363.... 8. .44444.... 9. .83333-...
10. .363636-... ^11. 40.909090-... 12. .54128128....
76. Harmonical progressions. Three or more numbers are said'
to form a harmonical progression if their reciprocals form an arith-
metical progression. The term " harmonical " as here used comes
from a property of musical sounds. If a set of strings of uniform
tension whose lengths are proportional to 1, £, \, \, £, i be
sounded together, the effect is harmonious to the 6ar. The sequence
i l i i i ...
is a harmonical progression since the reciprocals form the arith-
metical progression
1,2,3,4,5,....
77. Harmonical means. To find n harmonical means between
two numbers, find n arithmetical means between the reciprocals
of these numbers. The reciprocals of the arithmetical means
are the harmonical means.
EXERCISES
'— 1. Insert three harmonical means between 3 and 12.
""* 2. Insert two harmonical means between 2 and J.
^ 3. Insert four harmonical means between ^ and £.
4. What is the harmonical mean between a and b ?
„ 5. Show that 4, 6, 12 are in harmonical progression, and continue the
series for two terms in each direction.
L
108 PROGRESSIONS [Chap. XI.
6. If -4, G, and H stand respectively for the arithmetical, geometrical,
and harmonica! means between two numbers a and 6, show that G 2 = AH.
7. The numbers a, 6, c are in arithmetical progression, and 6, c, d are in
harmonical progression ; prove that ad = be.
PROBLEMS
1. What distance will an elastic ball traverse before coming to rest if it
be dropped from a height of 20 feet and if after each fall it rebounds one
third of the height from which it falls ?
2. If a falling body descends 16^ feet the first second, 3 times this dis-
tance the next, 5 times the next, and so on, how far will it fall the 30th sec-
ond, and how far altogether in 30 seconds ?
3. Assume that a baseball will fall 16 feet the first second, 48' the next,
80 the next, and so on. A baseball was dropped from the top of Washington
Monument, 550 feet high, and caught by an American League catcher.
About how fast was the ball falling when caught ?
4. A swinging pendulum is brought gradually to rest by friction of the
air. If the length of the first swing of the pendulum bob is 80 centimeters,
and the length of each succeeding swing is ^ less than the preceding one,
what is the distance passed over in the fifth swing ?
5. What is the total distance passed over by the pendulum bob de-
scribed in Problem 4 in 6 swings ?
6. A ball rolls down an incline of 20 degrees, 6.47 feet the first second,
and in each succeeding second, 10.94 feet more than in the preceding second.
How far will it roll in 10 seconds ?
7. Two straight lines h and l 2 meet at the point 0. From a point in h
drop a perpendicular to l 2 . From the foot of this perpendicular drop a per-
pendicular to Zi and so on indefinitely. If the lengths of the firat and second
perpendiculars are 6 and 5 respectively, what is the sum of the lengths of all
the perpendiculars ?
8. In a raffle, tickets marked 1, 2, 3, 4, etc., are shaken up in a hat and
drawn by the purchasers one at a time. The price of a ticket is the num-
ber of cents corresponding to the number on the ticket. If the raffled article
is worth $ 10, what is the least number of tickets which will insure no loss to
the vendors of the tickets ?
9. A person contributes one cent and sends letters to three friends ask-
ing each to contribute one cent to a certain charity and to write a similar
letter to each of three friends, each of whom is to write three lettera, — and
so on until ten sets of letters have been written. If all respond, how much
money will the charity receive ?
Art. 77] PROBLEMS 109
^. 10. Twenty-five stones are placed in a straight line on the ground at in-
tervals of 4 feet. 10 feet from the end of the row is a basket. A runner
starts from the basket and picks up the stones and carries them, one at a
time, to the basket. How far does he run altogether ?
11. An employer hires a clerk for five years at a beginning salary of
$ 600 per year with either a rise of $ 100 each year after the first, or a rise of
$ 26 every six months after the first half year. Which is the better propo-
sition for the clerk ?
12. The population of a town is 10,000. It loses annually 2 per cent of
its population by deaths and gains 3 per cent by births, and at the end of each
year, it has gained 100 people as a result of movings into and away from the
town. What will the population be in 10 years ?
13. Find the limit of the sum of the series
H 1- = h ... where n>0.
n + 1 (n+1)* (n+ 1)3
14. What is the sum of the first n odd numbers ?
15. What is the equation whose roots are the arithmetical and the har-
monical means between the roots of x 2 — 16 x + 48 = ?
16. if — i — _, form an arithmetical progression, show that »,
y — x 2y y — z
y, and z form a geometrical progression.
17. What is the number which added to each of the numbers 1, 3, 2,
produces a geometrical progression ?
18. The fourth term of a geometrical progression is 108, the seventh
term is 864. What is the 10th term ?
19. There are four numbers in arithmetical progression. When these
numbers are increased by 2, 4, 8, 15 respectively, the sums are in geometri-
cal progression ; find the f our numbers in arithmetical progression.
20. Three numbers whose sum is 18 are in arithmetical progression. If
you multiply the first by 2, the second by 3, and the third by 6, the resulting
products form a geometrical progression. Find the three numbers.
21. Find the present value of ten annual payments of $ 1000 each made
at the ends of the next ten years, if money is worth 6 per cent compounded
annually.
I - .
CHAPTER XII
COMPLEX NUMBERS
78. Number systems. If our number system consisted of zero
and positive integers only, the solution of an equation such as
3.r ■ - - = would be impossible; for no number in the system
considered satisfies this equation. We can extend the number
system so as to include the class of numbers to which the solution
belongs. These new numbers are the rational fractions.
While the solution of 3x — 2 = is possible in a number system
composed of zero, positive integers, and rational fractions, the
solution of an equation such as x -+- 4 = is impossible. To meet
the demands of such equations, we find it expedient again to ex-
tend the number system so as to include the negative numbers.
In a number system thus extended an equation ax + b = 0, where
a and /> are any integers or fractions, has a solution.
The solution of an equation such as x 2 = 2 demands a further
extension of our number svstem. It must be made to include
irrational numbers, that is, numbers which cannot be represented
by the- quotient of two integers (see p. 23). But the number
system thus extended is not sufficient to meet all the demands of
the equations met in algebra. In this number system it is im-
possible to solve certain quadratic equations, for example, the
equations a? 2 -f 1 = and x 2 — 6 x + 13 = 0. It is necessary again
to extend the system so as to include numbers of the- form aF\-bi 3
where a and b are real numbers, discussed in Art. 1 ; and where t
is a symbol whose square is — 1, that is, i = V— l'(see Art. 22).
These numbers are usually willed complex numbers and sometimes
imaginary numbers. When a = they are called pure imaginary
numbers.
The term " imaginary number " is here used in a technical
sense. The numbers are imaginary in the same sense that a
110
lllj/
Arts. 78, 791 GRAPHICAL REPRESENTATION
fraction, a negative number, or an irrational number is imaginary
for a number system consisting of positive integers.
At this point the question may be asked, — In working with
this new system which includes complex numbers, may we not
find it necessary to add new numbers, at present unknown, just
as we found it necessary to add fractions, negative numbers, and
irrational numbers to our system of positive integers ? The
answer to this question is that the system of complex numbers
is sufficient to meet the demands of the equation.
While we have seen that the solution of equations with integral
coefficients demands fractious, negative numbers, irrational num-
bers, and complex numbers, it is not to be inferred that all
numbers are roots of equations with integral or rational coeffi-
cients. For example, the irrational number ir cannot be the root
of an equation with rational coefficients. The proof is beyond
the scope of this book.*
Jf9. Graphical representation of complex numbers. We have
Been that all real numbers may be represented by points on a
straight line. The complex
number x + ^depends on two !
real nnmben Qt and y,)and may
be represented graph ieal ly by
a point in a plane. Two lines,
X'X, Y' Y, are drawn perpen-
dicular to each other and in-
tersecting at 0, Fig. 18. To
represent the number 2 + 3 i,
measure off on X'X to the
right the distance 2, and up
the distance 3. In general, the
graph of the number x + »/ is
the point whose coordinates
are (x, y). The line X'X is
often called the axis of
"reals" and the line Y'Y the axis of imaginaries.
• See Klein, Famous Problems
Beniati and Smith, p. 68.
i Elementary Geometry, translation by
112
COMPLEX NUMBERS
[Chap. XII.
x+iy
>X
Fio. 19
*It is often convenient to represent complex numbers by
another method. Connect the point
which represents x + iy with the origin
as in Fig. 19. Let the length of this
line be r. The point can then be repre-
sented by giving the length r and the
angle 0. From the figure
x s= r cos $ ,
• y = r sin $ 9
x* + yp = ri
Hence, the number x + iy may be written in the form
■ x + iy = r(cos fl4-is injflty
This form is called the polar form of a complex number. The
angle is called the argument or amplitude, the length r the
modulus or absolute value of the complex number. It should be
noted that the complex numbers include all real numbers. In
Fig. 18, the real numbers are represented by points on the line X'X.
The pure imaginary numbers are represented by points on the line
TY.
80. Equal complex numbers. If two complex numbers a + bi
and c + di are equal, then a = c and b = d. For, if
a + bi = c 4- di, (1)
by transposing, a— c = (d — b)i. (2)
Unless a — c = d — b = 0, we should have a — c, a real num-
ber, equal to (d — b)i, an imaginary number.
Conversely, if a = c, and b = eZ,
a + bi =■ c + di.
Hence, when any two expressions containing imaginary and reed
terms are equal to each other, we may equate the real parts and the
imaginary parts separately.
In particular, ifa + bi = 0,a = and b = 0.
* The remainder of this article and the articles marked * may be omitted
by those who have not studied trigonometry.
Abts. 7&-81] ADDITION AND SUBTRACTION 113
EXERCISES
Represent graphically the following numbers and in each case find the
argument and the modulus.
1. 2-3i
Solution : The number is represented in Fig. 18.
The modulus r is given by
r = y/x\+ y* = V4 + 9 = Vl3.
To find the argument 6 we have
tan0=£ = -?, = arctan-?: sin0= ?-•
* % 2' V13
2. -2-Si 3. -2 + 3£ 4. 1 + i.
5. 1-i. 6. 6-4i. 7. - t.
8. 4i. 9. -6*. 10. f i.
11. 4 + Oi. 12. -5. 13. jj-i.
14. 0.7 + 1.1 i.
3 2
Write the following complex numbers in the form x + iy.
15. 3(cos 80° + t sin 30°). 16. 2(cos60° + t sin 60°).
17. 2(cosl50° + isinl50 o ). 18. 4(cos90° + i sin 90°).
19. (cos 225° + * sin 226°). 20. 6(cos0° + tsinO°).
21. f (cos 270° + i sin 270°). 22. If (x + 1) + %{y - 1) = 0, what
are the values of x and y ?
What must be the value of x and y in order that the following equations
may be true ?
23. « + y + i(x-y)=2 + 4i. 24. 2x+7y+ i(3x-2i/) = -3-t.
25. 2** + fl* + {(x-y)=l + t. 26. 3x + xyi + 2y — ix — 3y-6=0.
27. ** + «B + ty + V 2 = 130 + 8 {(as - y).
81. Addition and subtraction of complex numbers. We assume
that the number i like other numbers obeys all the laws of algebra.
Given two complex numbers a+bi, c+di, we may write the sum
and difference.
Thus, (a + &0-f( c + di) = (a + c) + (6 + d>',
(a + bi)-(c + di) = (a - c) + (6 - d)i.
Hence, to add (or subtract) complex numbers, add (or subtract)
the real and imaginary parts separately. The result is a complex
number.
COMPLEX NUMBERS
[Chap. XII.
Fig. 20.
To add two complex numbers, a + bi and c + di, graphically,
we represent the numbers as points A and B in Fig. 20. Connect
each point with the origin 0.
Complete the parallelogram,
having OA and OB for adja-
cent sides. The vertex P
represents the sum of the
two given complex numbers.
For, from the figure the
coordinates of the fourth
vertex are OQ and QP. But
OQ=OD + DQ = a + c,
QP= QR + BP=b + d.
Hence, P represents the
point (a + c) + (6 + d)i which is the sum of a -f- bi and c -f di.
To subtract one complex number from another graphically, say
c + di from a + bi, we graph the points which represent — c — di
and a -f bi, and proceed afiLJpr a dditi on
EXERCISES
Perform the following operations algebraically and graphically.
-1. (l + t) + (2 + 3t). 2. (2 + 2i) + (l-3i).
3. (2 + t)-(l + 4Q. 4. (3-6i) + (3 + 6i).
5. (3-6t)-(3 + 6i). 6. (- 3- 4f) + (6 - t).
7. (0 + 3t) + (l-4i). 8. (6 + 0t) + (-8+7t)-(4+2t).
9. (4 + i)'-(3 + 3t) + (l-t). 10. (4 + 3i)-(4 + 3t).
* 82. Multiplication of complex numbers. Let a + ib and c — id
lWany t"70 complex numbers. Since i obeys all the laws of alge-
bra, we have
(a -f ib)(c -f id)= ac -f- ibc + tad + Pbd =(ac - - 6d)+ t'(6c •+- ad).
The result is a complex number. To multiply two complex num-
bers graphically, let the two numbers a + ib, c -f- id be represented
by the points P x and P 2 (Fig. 21). Eeducing to the polar form,
we have
a + bi = ?*!(cos Oi + i sin ${),
c + di = r 2 (cos 2 + i sin 2 ).
Arts. 82-83] CONJUGATE COMPLEX NUMBERS
115
By actual multiplication,
(a 4- bi)(c + di)
= fy^cos 0i c 08 02 4 *( sm ^l cos 02 + cos 0i sin 02) — si 11 0i sin 02]
= ^,[008(0! + 2 )+ < sin(^ + 2 )].
Hence, the modulus of the product
of two complex numbers is the product
of their moduli and the argument is
the sum of their arguments.
The point P, which represents
(a 4- bi)(c + di) may then be con-
structed by drawing through O a line
making an angle = $ t + 2 with the
line OX, and constructing on this
length a^egment OP, whose length
is rtf*
Fig. 21.
83. Conjugate complex numbers.
Numbers which differ only in the
sign of the imaginary parts are called
conjugate numbers. Thus, 3 + 2 i and 3 — 2 i are conjugate. Since
(a 4- bi) + (a — bi)= 2 a,
{a + bi)(a-bi)=a* + b\
and (a + bi) — (a — bi)=. 2 bi,
we see that the sum and the product of two conjugate complex
numbers are real numbers, but the difference of two conjugate
complex numbers is an imaginary number.
EXERCISES
Multiply both analytically and graphically, finding the arguments and
moduli of the products.
1. (3 + V3i)(2 + 2i).
Solution.
(3+V3t)(2 + 2f)=6+6i + 2V3i + 2\/3i2 = 6-2\/3 + «(6 + 2\/3).
Putting the numbers in the polar form, we have,
3 + V3 i = 2 V3(cos 30° + t sin 30^},
2 + 2 i = 2 V2 (cos 45° + i sin 46°).
COMPLEX NUMBERS
[Chap. XII
Fio. 22.
6. (l + i)2(2-2\/8i).
8. (-2-2i)(2 + 2i).
10. (0 + 3i)(2 + i).
Hence,
n = 2 V3, r 2 = 2\/2, 0j = 80°, fc = 46°.
The modulus of the product is, then,
nrz = 4V6,
and the argument is 75°.
Let Pi [and P 2 in Fig. 22 represent the two
given numbers. Through draw a line mak-
ing an angle of 75° with the line OX. On this
line measure off the distance
OP = 4V8.
The point P then represents the product of the
two numbers.
- 2. (3+V3i)(2 + jV3i).
• 3. (l+V3t)( 4 + *V'8i).
- 4. (v^2+V^2)2.
5. (1 + i) 4 .
7. (-2 + 2i)(2 + 2i).
9. (l + i)(l+2i)(l+3i).
11. (0 + 2i)(0-2i).
J* 84. De Moivre's theorem. If two complex numbers are equal,
then as a special case of Art. 82, we have
r(cos -f i sin 0)r(cos + ism 0) = r 2 (cos 4- 1 sin0) 2
==r 2 (cos204-tsin2tf).
Multiplying both sides of this identity by r (cos 0+i sin 0), we have
r 3 (cos -f < sin Of = r*(cos 3 + 1 sin 3 0),
and it can easily be proved by mathematical induction that
[r (cos -f i sin 0)] n = r n (cos n$ + i sin n 0\ y>
where w is any positive integer. . '.' >^ l '
This relation is known as De Moivre's theorem, and holds also
for fractional values of the exponent.
To prove the theorem when the exponent is the reciprocal of a
i
positive integer, consider the expression (cos + i sin 0)* in which
n is a positive integer.
Aura. 83-85] ROOTS OF COMPLEX NUMBERS 117
L
Ijet = w<£,
i i
then (cos + i sin 0) n = (cos n^ + i sin w <£)*
= [(cos <f> + t sin <£)"]* = cos <£ + 1 sin <£
8
= cos- + tsin -•
w »
De Moivre's theorem thus gives an easy method of finding any
power or any root of a complex number.
The proof can be readily extended to the case of an exponent
which is any rational fraction.
*A 85. Roots of complex numbers. From Art. 84, the nth root
of x + iy is
I }. 1/ q 0\
(x + iy) n = [r (cos + i sin 0)] w = r n l cos - + i sin - ) •
\ n nj
If m be any integer, cos (0 + m 360°) = cos 0,
sin(0 + m36O°)=sin0.
We may then write
i i
(*+^=[^(costf-fisin^)] n = [r|cos(0 + m36O o )
i
+ isin(0 + m36O°)}]»
Jr^c 9 + m 360V . . $ + m 360°"]
= r" cos — ■ f-isin — - •
L n n J
If now we let m take the values 0, 1, 2, 3, •••, n — 1, we find w
results, all different numbers whose nth powers are x 4- iy. We
may then state the following
Theorem. Any number has n distinct nth roots.
EXERCISES
Using De Moivre's theorem, find the indicated powers and roots.
1. (8 + V5i)*.
Solutio* : Writing 3 + VS i in the polar form,
3 + V3i = 2V3(cos30° + isin30°).
By De Moivre's theorem, <?} - A
(3 + <\/3f)« =[2\/3(co83a° + isin30°)]«
= 144(cos 120° + i sin 120°)
= 144(-H-iy5f) ><>3
= -72 + 72V5i ; n ^
118
COMPLEX NUMBERS
[Chap. XII.
* 3. (3 + V3t)».
J 5. (i + iVSi) 6 -
41. (2 + 2t)«.
</s. (l + o 10 -
/2. (3+y3i)».
y 4 . (-i-iV3i)«.
/ 6. [2 (cos 10° + i sin 10°)]».
9. #- 2 + 2 i.
Solution : Writing — 2 + 2ji in the polar form, we have
- 2 + 2 i = 2 V% (cos 135° + i sin 135°).
Hy I)e Moivre's theorem,
^-2+2i==(-2+2i)*=[2V2{cos(136 +m360°)+t8in(136 +m360 )}]*
= V2 [cos (45° + m 120°) + { sin (45° + m 120°)].
- ^ For m = 0, 1, and 2, this expression
reduces to
1 + i, V2(cos 166° + i sin 185°),
811(1 V2 (cos 286° + i sin 286°)
respectively. Any one of these three
>X numbers is a cube root of — 2 + 2 i.
The points Pi, P& P 8 representing
these three numbers lie at equal
intervals on a circle of radius V2
(Fig. 23).
10. v^2 + 2i.
Fig. 23.
12. Vs + VS i.
J 14. \/G4(eosG0 + tsin60°).
J 16, VcbsaOO^H- i sin 300°.
18. y/l
Hint : Write in the form \/0 + i.
19. vm
11. V-8 + 8V31
J 13. \/64(cos 60° + i sin 00°).
/l5. ^64(cos60 + <sin6a o ).
17. ^-4-4i
20. Vi. 21. #1. 22. v^S. 23. yffil
Find all the roots of the following equations and represent them graphi-
cally.
24. x s - 27 = 0.
Hint : x* = 27. The roots of the equation are then the three cube root 4
of27+0i.
25. x* - 1 = 0. 26. x* - 32 = 0. 27. x* - 1 = 0.
28. x*-16 = 0. 29. x«-l=0. 30. x*-l=0.
Arts. 85, 86]
DIVISION
V419
4 86. Division of complex numbers. The quotient of two com-
plex numbers may be obtained as follows.
a + ib a + ib c — id
~~* — ^^^-^^^ • — — ^— ^
c + id c + icZ c — id
__ ac -f- bd — i(acZ
~~ c 2 + d 2
-be)
c 2 + d 2 c 2 + d 2
This is a qomplex number. Writing the two given complex
numbers in the polar form, we have
r^coB^ + t sinflj)
r f (cos 2 + i sin $ 2 )
— Tir 2 (co8 $ x + i sin flQCcos fl 2 — i sin #2)
r 2 2 (cos 6 2 + i sin 2 )(cos #2 — * si* 1 #2)
= tyjeos jB x — fl 2 ) + i sin (flx — fl 2 )]
™ r 2 2 (cos 2 2 + sin 2 2 )
- = - l [cos (fix - 2 ) + 1 sin (0,
r 2
Hence, the modulus of the quotient of two
complex numbers is the quotient of their
moduli, and the argument is the difference of
their arguments.
If, in Fig. 24, P x and P 2 represent the
points a + ib and c + id respectively, the
point P which represents the quotient
"*" may be constructed by drawing
c-\- id
through a line making an angle 0=0 1 —0 2
with the line OX, and constructing on this
line a segment OP, whose length is ^ •
*2
- 40].
Fig. 24,
120
COMPLEX NUMBERS
[Chap. XII.
Solution :
EXERCISES
Find the quotients of the following pairs of numbers, analytically and
graphically.
1. (4 + 4i) + (2 + jV3i).
4+4t _ 4 + 4i 2-#>/8t _ 3+V5 . 3-VS 1
2 + jvlH 2+}V3i 2-}VSt 2 2
Writing both numbers in the polar form, we obtain
4 + 4 f = 4\/2(cos 46° + i sin 46°),
2 + f a/3 t = }V3(cos 30° + i sinW).
n = 4V2, r 2 = 1 V3, $i = 45°, ft = 80°.
The modulus of the quotient is then — = V6,
r*
and the argument is $ = $\ — $% = 15°. Let
Pi and P a in Fig. 25 represent the two given
numbers. Through draw a line, making
an angle of 15° with the line OX. Measure
off on this line the distance OP = \/6. The
point P then represents the quotient of the
two numbers.
Fra - ^ 2. (-2 + 2\/3i) + (l+>/8i).
3. (_j«jV3 0-(-l + lV3i). 4. (4 + 40 + (l-i).
5. (l + i)2+(2-2\/3i).
7. 2-r(l + t).
9. (l + i) + (l-i).
6. -4i+(-2 + 2i).
8. (-2 + 2i) + (-40-
10. (S + Vso + CVS+so.
Graph the following complex numbers and their reciprocals.
11. 1 + t. 12. 8 + V3i. 13. 2-2i. 14. 8 + 7i. 15. -3 + 6i.
CHAPTER XIII
THEORY OF EQUATIONS
87. The polynomial of the nth degree. The general form
(Art. 27) of a polynomial of the nth degree is
atfv n + a x x n ~ l -f a^c n ~ 2 -f ••• + a
nJ
where n is a positive integer, a , a lf ••-,«„ are independent of a>,
and Oq =£ 0. The polynomial of the second degree has been dis-
cussed in connection with the quadratic equation.
When f(x) is used in this chapter, it is to be understood to
mean a polynomial in x.
EXERCISE
By comparing the following polynomials with the general form, deter-
mine n, Oq, cii, ••♦, a n .
(1) /(*) = *«* + *«* +8.
(2) /(«)= ♦ a 8 + £* 4 + 10s 2 + is*.
(8) f(x) = (i + V3)x5 + 5 x 2 + 10.
88. Remainder theorem. Iff(x) is divided byx—r,the remain-
der %sf{r).
Given f(x)= ajxr + OjX 71 ' 1 + ... + a n _ x x + a n , (1)
Then, /(r) = a«r* + a^" 1 + •» + a^r + a B , (2)
and f(x)-f(r) = o^s* - r-) + a 1 (aj"- 1 -r"- 1 )+ ••• +a„_ 1 (s-r). (3)
But since oj — r is a factor of each of the expressions x n — r n ,
05""" 1 — r"" 1 , •••, x — r (Ex. 11, Art. 58, p. 92), it is a factor of the
right-hand member of (3).
If we should inclose the whole right-hand member of (3) in a
parenthesis, we could remove the factor x — r, and call what re-
mains inside the parenthesis Q(x).
Then we have from (3),
f(x)-f(r)=(x-r)Q(x). (4)
121
122 THEORY OF EQUATIONS [Chap. XIII.
Transposing /(?•) and dividing by x — r, we have
x— r x—r
which is what was to be proved.
Corollary. If f(x) vanishes when x=s r, then f(x) is exactly
divisible by x — r.
EXERCISES
1. Show, by the remainder theorem, that x n — a* is exactly divisible by
x + a if n is even.
2. Show that x n + a n is divisible by x + a if n is odd.
3. Without performing the division, find the remainder when x* — 3 x 2 +
2 x - 1 is divided by x — 2.
4. By means of the remainder theorem, find a value for k such that
x* + 3 kx 2 + 6 x + 2 is divisible by x - 1.
89. Synthetic division. The operation of dividing a polyno-
mial by a binomial x—r can be performed rapidly by means of
a process called synthetic division. This rapid division, com-
bined with the remainder theorem, gives a convenient method of
evaluating/ (x) for different values of x.
For example, divide
5 a* - 6 a? + 8 a; 2 - 24 x-6 by ae- 2.
By the ordinary method
5x*- 6^ + 8 a? 2 - 24 a-6 la -2
5a^-10af> 5a*+4a* + 16a5 + 8
4 x* + 8 x 2
4 a? - 8 <c 2
16a; 2 - 24a;
16 x 2 - 32 a:
8a;-
8a;-
6
16
+ 10
Arts. 88, 89] SYNTHETIC DIVISION 123
Manifestly, the work can be abridged by writing only the coef-
ficients, thus,
5_ 64-8- 24 — 611 — 2
5
-10
4-44-8
4-4-8
4-16-
4-16
-24
-32
4-8-
4-8-
6
16
5 + 4 + 16 4- 8
4-10
Since the coefficient of x in x — r is unity, the coefficient of the
first term of each remainder is the coefficient of the next term to
be obtained in the quotient. Further, it is not necessary to write
the terms of the dividend as part of the remainder, nor the first
term of the partial products.
The work thus becomes :
5_ 64-8-24-611-2
-10
+ 4
-8
+ 16
—
32
+
8
-16
+ 10
We may omit the first term of the divisor and write the work
in the following more compact form :
5. 64-8-24- 6 [-2
_ IQ _ 8 - 32 - 16
If we replace — 2 by + 2, we may add the partial products to
the numbers in the dividend. Then, we have :
5- 6+ 8-24- 6[2_
+ 10+ 8 4-3 2 4-16
5 + 4 4- 16 4- 8 4- 10
The quotient is 5 a? 4- 4 x 1 4- 16 x 4- 8, and the remainder is 10.
124 THEORY OF EQUATIONS [Chap. XIII.
90. Rule for synthetic division. To divide f(x) by x — r,
arrange f(x) in descending powers of x, supplying aU missing
powers by putting in zeros as coefficients.
Detach the coefficients, write them in a horizontal line and in the
order a^ a ly a 2 , •••, a n .
Bring doicn the first coefficient a ; multiply a, by r, and add the
product to Oi ; multiply this sum by r, and add the product to a*
Continue this ftrocess ; the last sum is the remainder, and the pre-
rnliiig sums are the coefficients of the powers of x in the quotient,
arranged in descending order.
PuooF of Rtle. This rule may be established by mathematical in-
duction.
liy long division,
ao/ n + aiX" -1 + a+x n --+ »- + a < a»-*+a >+ ig»-«- 1 + ■«■ + a,|x— r
|ao* n_1 +Oi+aor)z"-* + — + (<*.-i + m*_i + — + r*-*at)z*-*
dpi* — (iprx"- 1
(di + a r)x n - 1 + aaX" -2
(tti +_apr)^ , '" 1 _-_(air + a o. r2 ) xn ~ 2
We note that the coefficient of z*~* in the quotient is formed according
to the rule. Assume that the coefficients in the quotient down to that of
x"-« are formed according to the rule. On this hypothesis, proceed by long
division to find the coefficient of x* - * -1 in the quotient. This may be ex-
hibited as a continuation of the division above as followB :
(a*-i + ra^z H — + r»- 1 a )x n -* +1 + a g x n ~' + a^+iX*-*- 1 + — +a*
(a,-i + ra,_ 2 + -•• + fig^x*-** 1 - (ra 9 . Y + r^a^i + — + fOtjg*^
(a, + ra^.i + r*a^. 2 + — + fao)x"-« + a^iX—*-* + ...+«;
This shows that if the coefficients in the quotient down to that of x*-* are
formed according to the rule, the coefficient of the next lower power is
formed according to the rule. Hence, the rule is established.
EXERCISES
1. Divide x* + 3 x 3 — 5 x + 3 by x — 4 by synthetic division.
Solution : 1 + 3+0— 5+ 3|_4
+ 4 + 28 + 112 + 428
1 + 7 + 28 + 107 + 431
The quotient is x 3 + 7 x 2 + 28 x + 107 and the remainder is 4S1.
2. Divide 3x 3 + 5x 2 + 2x+ 1 by x + 5. (In this case r = — 6.)
3. Divide 5x 3 — «6x + 3 by x + 2.
4. Divide7x<-3x 2 -2byx + J. 5. Divide x* - 125 hf « — 6.
Ait-re. 90. 91] GRAPHS OF POLYNOMIALS
125
01. Graphs of polynomials. When the coefficients of f(x) are
real numbers, the march of the function for different values
of x can be clearly presented by the use of the graphic methods
explained in Arts. 27-28. To any value assigned to x, there cor-
responds one and only one value of the polynomial f{x). This is
sometimes expressed by saying that /(x) is single valued. The
fact that the graph of f(x) is a continuous curve (see Art. 27)
makes it of much service in the theory of equations.
EXERCISES
Construct the graphs of the following
functions and locate their real zeros approxi-
mately (to within 0.6).
1. /(g)s3>-6s) + llB — «.
As pointed out in Art. 89, synthetic divi-
sion furnishes a convenient method of
evaluating /(s) for different values of z.
Thua/(0.5) is obtained as follows :
1_8 + H_ 8 1 0.5
0.5-2.75 + 4 125
1 - 6.5 + 8.26 - 1.875"
Hence,/(0.5)=- 1.875.
In this way the following values are
obtained:
/(-
/(1.6)= 0.375.
/(2)=0.
/(2.5) = - 0.375.
/(3)=0.
/(4)=0.
/(5) = 24.
u Fig. 2<! ; it presents
/(-l) = -24.
/(-0.6) = -13.125.
/(0)=-e.
/(0.5) = - 1.875.
/(1) = 0.
The graph Is shown ii
to the eye the following facts :
(1) /(z) has zeros at 1, 2, and 8.
(2) f(x) is positive when 2 > x > 1, and
when x > S.
(8) /(*) is negative when i <1 and when
8>ie>2.
2. as»-flit*+8a8.
3. X*-1H-
H4z
5. x*-2l , -7ir ! + 8* + 12.
T. 8*»-4*>-12i« + 8.
126 THEORY OF EQUATIONS [Chap. XIII.
02. General equation of degree n. By equating to zero the gen-
eral polynomial of the nth degree, we obtain what is known as the
general equation of the nth degree in one unknown. That is to say,
a^fc n 4- a x x n ~ l + a& n ~ 2 -f- ••• •+• a n =
is the general equation of the nth degree.
The principal object of this chapter is to present methods
which aid in determining exactly or approximately the real roots
of special numerical * equations included under this type. It
is largely for this purpose that we discuss the graphs of poly-
nomials. The zeros of the polynomial are the roots of the equa-
tion formed by equating the polynomial to zero. The real roots
of the equation may then be looked upon geometrically as the
abscissas of the points of the X-axis where the graph of the poly-
nomial cuts this axis.
93. Factor theorem. Ifr is a root of the equation f(x) = 0, then
x — r is a factor off(x).
Since a zero of f(x) is a root of the equation f(x) = 0, this
theorem follows directly from the corollary to the remainder
theorem (Art. 88).
Exercise. State and prove the converse of the factor theorem.
94. Number of roots of an equation. Every equation, /(*)= 0,
of the nth degree has n roots and no more.
To prove this proposition we assume the fundamental theorem
that every equation, f(x) = 0, has at least one root. More ex-
plicitly, we assume that
There always exists at least one number, real or complex, which
will satisfy an algebraic equation of the nth degree, whose coefficients
are any real or complex numbers, t
* The term ** numerical equations " is here used to indicate that the coeffi-
cients are not literal.
t This fundamental theorem was first proved by Gauss in 1797. For proof,
see Fine's College Algebra^ p. 688.
Arts. 92-94] FUNDAMENTAL THEOREM 127
Let ri be a root, then (Art. 93), x — r A is a factor of /($) and
/(«)-(*- r,)/i(<»), (1)
where /i(«) is a polynomial of degree n — 1, beginning with the
term Oo^*" 1 . By the theorem assumed, f(x) = has at least one
root. Let r 2 be a root ; then
fi(x) = (x-r 2 )f 2 (x),
and f(x) = (x - r x ) (x - r 2 )f 2 (x), (2)
in which f 2 (x) is a polynomial of degree n — 2, beginning with the
term OoOJ*" 2 . Continuing this process, we separate out n linear
factors with a quotient Oq, so that
f(x) = a (x - r x ) (x - r 2 ) ... (a - r n ) (3)
where r x , r 2 , • • •, r w are n roots of /(a:) = 0.
If f(x) = has another root different from any of these, let r
denote such a root. Then, from (3),
<h(r - n) (r — r 2 ) — (r — r n )= 0.
But here we should have the product of factors equal to zero when
iko one of the factors is zero. As this is impossible (III, Art. 5),
there are not more than n roots of f(x) = 0. Fence, every equation
of the nth degree has n roots and no more Furthermore, every
polynomial of the nth degree is the product of n linear factors.
It is not, however, possible, in general, to determine these factors
if n exceeds 4 (see Art. 106). Two or more of the n roots of
f(x) = may be equal to each other, in which case the equation
is said to have multiple roots. If (x — r) m is a factor of f(x),
then f(x) = is said to have m roots equal to r. Thus, (x — 2) 8 =
has five equal roots, and (x — l) 2 (a: — 3) (x — 4) 3 = has two roots
equal to 1, one equal to 3, and three equal to 4.
Cobollabt I. If two polynomials of degrees not greater than n
are equal to each other for more than n distinct values of the variable,
the coefficients of like powers of the variable are equal.
Let a<fc n + aiX"' 1 + ••• + a n = bfP n + b x x n - 1 -\ + b n (4)
for more than n values of x.
Prom (4),
K - & )af + ((*! - bjx"' 1 + . . . + (a H - b n ) = 0. (5)
128 THEORY OF EQUATIONS [Chap. XIII.
Then a^ - b Q = 0,
<*!-&! = 0,
• • • •
a n -6 n = 0.
For, if any coefficient in (5) were not equal to zero, we should
have an equation of degree equal to or less than n with more than
n roots, which is contrary to the theorem just proved.
Hence, a = 6 , a x = b lf —, a„ = b n .
Corollary II. If two polynomials of degrees not greater than
n are equal for more than n distinct values of the variable, they are
equal for all values, and the equality is an identity.
95. Graphs of a,QX n + axx™- 1 + ... + a n
= a (i» - r\)(jv — ra>-(s» — r n ),
We assume that a , a x , •••, a n are real numbers, and further for
convenience of expression that a is positive, although this is
not a necessary limitation. In Art. 91 the graphs of a few poly-
nomials are plotted. Some important properties of these graphs
appear when the polynomial is separated into linear factors. We
cannot at this point make the treatment so complete as later, but
we can well consider two important cases :
1. When the factors x — r ly x — r 2 , •••, x — r n are aU real and
distinct
Arrange the factors so that r x > r 2 > • • • > r n _! > r w . When x>r x
all the factors are positive and the graph is above the X-axis.
Fig. 27.
When r x >x> r 2 , one factor is negative and the graph is below the
X-axis. When r 2 >x> r 3j two factors are negative, and the graph
is again above the X-axis. Continuing this process, we see that
the graph crosses the X-axis at the n points, x = r u 2=1*3, •••,
Aetb.W-96] COMPLEX ROOTS 129
x = r n9 and we obtain a general notion of the nature of the curve.
See Fig. 27.
2. When the factors are real but some of them repeated.
To discuss the graph in this case, take for example,
f{x) = ao(x - r x )\x - r 2 ) (x - r 3 ) 5 ,
and let n > r 2 > r 3 .
Since the factors x — r 2 and x — r 3 occur to powers with odd ex-
ponents, it follows as above that the curve crosses the X-axis at
x = r 2 and x = r z . But it does not cross at x = r u since the sign
of f(x) is the same when x > r x as when ^ > a? > r 2 , and the curve
touches the axis at a? = r v In general, if a factor x — r occurs to
a power with an odd exponent, the curve crosses the X-axis at
a? = r, while if it oc-
curs to a power with
an even exponent, it
merely touches the
X-axis at x=r. See
Kg. 28. FlG - 28 -
Another case is discussed in Art. 97, where imaginary factors
occur.
96. Complex roots. If a complex number a -f bi is a root of an
equation f(x) = with real coefficients, the conjugate complex number
a — bits also a root.
By the hypothesis, a + bi satisfies the equation
a<p n -|- a x x n ~ l + ... + a n = 0. (1)
Put a + bi for x in this equation and expand the powers of a + bi
by the binomial theorem. Represent the real part of this ex-
pansion- by P and the imaginary part by iQ. Then
P + tQ = 0. (2)
Whence, P = and Q = (Art. 80). (3)
In the binomial expansion of a + bi with any exponent, imagi-
nary terms occur when and only when a term involves an odd
power of bi as a factor. The result of substituting a — bi instead
of a 4- W is clearly obtained by replacing i by — i in (2). We
obtain thus p _ iQ.
THEORY OF EQUATIONS
, by (3),
P = and Q -
ue,
P-iQ = 0,
a - bi is
a root of (1).
97. Graphs of /(x) when so me linear factors are Imaginary. Since
imaginary factors of f(x) occur in conjugate pairs when the
coefficients in f(x) are real, it follows that in this case / '(as) may
be regarded as the product of a$, of real linear factors of the type
x — )■, and quadratic factors of the type
(«-«)•+ P=(.-a-M)(»-a+M),
where a, 6, and rare real numbers. Whenall the roots off(x)=0
are real, the polynomial f(x) is the product of real linear factors,
but if/(a;)=0 has imaginary or complex roots, /(as) contains
quadratic factors of the type (x — a)* + 6* which cannot be sepa-
rated into real linear factors.
In Art. 95 the graph of f(x) is discussed when the polynomial
is the product of real linear factors, and it is shown that, corre-
sponding to each linear factor x — r, the graph meets the X-axis
at x = r. It should now
be noted that
(x-a)* + b*>0,
for all real values of x,
and there ia, therefore,
corresponding to quad-
ratic factors of fix), no
- intersection of the graph
with the X-axis.
Exam pl« : Graph
/ (z) = X* - 7 X* - 4 z» + 78 *
= x<a: + 3)(z>-10z+26)
= *(z + 8)[(*-6)Hl].
Corresponding to the linear
factors as and x + 3, the graph
* la **■ intersects the X-axis at x =
and X = — 3 respectively (Fig. 29). Corresponding to the quadratic factor
x 1 — 10a: + 20 there ia no intersection with the X-axis. (In Fig. 89 one
horizontal space represents one unit, while one vertical space represents:
twenty u
a.)
ABTa.9fr-9g] TRANSFORMATIONS OF EQUATIONS 131
EXERCISES
1. If ri, f2, •••* r n are roots of an equation, show that
( x _ri)(x-r 2 )(x-ra)... ( x - r n )=0
is the equation or an equivalent equation.
2. Form equations which have the following roots and no others,
(a) 2, 3, 5.
(6) l+2i, l-2i, wheni*=-l.
(c) 1+V2, 1- v/2,3.
(d) v% -V2, V3, -V3.
(c) 1, - 2, 3, 0.
</) 2+ >/3, 2-V3, -2+V3, -2-V3.
3. By means of the theorem concerning the number of roots otf(x) = 0,
show : (1) if /(x) = be multiplied by a polynomial in x, the resulting equa-
tion has more roots than / (x) = ; (2) if / (x) = be divided by a polyno-
mial in x, which is a factor of /(x), the resulting equation will have fewer
roots than/(x) = 0.
4. Plot the graphs of the following :
(a) /(x) = (x- l)2(x-3)2.
(6)/(x) = x(x-l)(x-4).
(c)/(x) = (x-l)(x + 2)(x + 7).
(d)/(x) = (x + 5)(x-6)2.
(e) /(x) = (x-2)2(x + 2).
5. Show that 3 and J are double roots of
9x5 _ si X 4 + 58x3 + 58x 2 - 51 x + 9 = 0,
and find the other root.
Separate the following polynomials into real linear and quadratic factors
and plot the graphs.
6. x*-l. 7. x3+ 1. 8. x4-l.
9. x»-l. 10. x 4 + 4x 8 + 3x 2 -4x-4.
11. Show that an equation /(x) = of odd degree and with real coeffi-
cients has an odd number of real roots.
m
98. Transformations of equations. Frequently an equation can
be solved more readily, or its properties can be discussed better,
by transforming it into an equation whose roots are related to
those of the given equation in some specified manner. For ex-
ample, in solving numerical equations for their real roots (Arts,
132 THEORY OF EQUATIONS [Chap. XIH
102-105), we shall have use for the following transformations of
/(x)=0:
x = — , x = — a/, and x =a x 9 + h.
m
x*
If we make x = —(or x 9 = mx) in /(»)= 0, we obtain an equa-
m
tion in x* whose roots are m times those off(x)r= 0. If we make
x = — x* in/(a?)= 0, we obtain an equation in a/ whose roots are
equal in absolute value but opposite in sign to those of /(«)= 0.
If we make x = x' -f h in f(x) = 0, we obtain an equation in a/ each
of whose roots is less by h than the corresponding root of /(»)= 0.
Those transformations can be performed rapidly by means of
the following rules :
1. To obtain an equation each of whose roots is m times a cor-
responding root of f(x)=0: multiply the successive coefficients be-
(jinning tcith that of a?"" 1 by m, m 2 , m 8 , •••, respectively. *
For example, to find the equation each of whose roots is double
the roots of the equation a 4 — £ x* + 3 x* + 1 =0, we make m = 2,
and obtain
x* - 2(4 a*)+ 22(3 a*) + 2 8 (0 . a>)+ 2* = 0,
X i _ 8 a? + 12 a* + 16 = 0.
a/ .
To establish this rule, substitute x = — in
m
/(«)= Oofl5 n + a^*- 1 + a& n -* H h a n _!a; + a, = 0. (l)
The result of this substitution is
, aV , /V\ n ~ l . /V\ n ~ 2
or a$ n 4- maxaj' 11 " 1 4- m 2 a 2 a?' n ~ 2 + ••• + w"'^ + m"c^ = 0, (3)
after multiplying by the constant m n . The rule is thus estab-
lished.
2. To obtain an equation each of whose roots is equal in absolute
value to a root off(x) = 0, but opposite in sign : change the signs of
the odd degree terms inf(x)= 0.
* In carrying out this rule any missing power of x should he supplied with
zero as a coefficient.
Abt.98] TRANSFORMATIONS OF EQUATIONS 133
For example, the roots of the equation
x 4 - 2 x* - 13 a 2 + 14 x + 24 =
are 2, 4, — 1, — 3, and the equation with roots — 2, — 4, 1, 3 is
x 4 + 2 x* - 13 jb 2 - 14 x + 24 = 0.
The rule follows at once from rule 1, by making m = — 1.
3. To obtain an equation each of whose roots is less by h than a
corresponding root of a given equation f(x)=0: divide f(x) by
x — h and indicate the remainder by R n . Divide the quotient by
x— h, and indicate the remainder by R n -\. Continue this process
to n divisions. The last quotient, c^, and the remainders, R l} R 2 , • • •>
S n are the coefficients of the transformed equation. Tlie new equa-
tion is then,
orf* 4- Rix'*- 1 4- R*x' n -* 4- • • • R n -&' 4- R n = 0.
The division should be performed by the method of synthetic
division.
For example, find the equation each of whose roots is less by 2
than the roots of the equation
The work is as follows :
1-4-3+ 2[2
+ 2 - 4 - 14
1-2
4-2
-7
-12
R$ = — 12,
1 +
1
-7
R* = — 7,
1 + 1
i*i= 1,
a = 1.
The required equation is
rf + a£ — 7a>-12 = 0.
■
To establish the rule, substitute x = x' + h in
a&r + ap"- 1 H h a n _ Y x + a n = 0. (1)
This gives the equation in x'
a/af + hy + a x (x' + h)»- 1 + . . . + a^s' + A) + a. = (2)
134 THEORY OF EQUATIONS [Chap. XIII.
whose roots are less by h than those of (1). Expanding the bi-
nomial powers and arranging in powers of x\ we may present the
result in the form
a<p' n + Axri*- 1 + A^'"- 2 H h A n _ x x' + ^ = 0. (3)
If in (3), we make x t = x — h, we obtain
a (x - h) n + A x (x - ft)"- 1 + A 2 (x - &)»-* + • • • + A n _ x (x - h)
+ A n = (4)
which is the same as equation (1) arranged in powers of x — h.
From the form of equation (4), it follows that A^ is the re-
mainder when f(x) is divided by x — h\ A n _ x is the remainder
when the quotient of the last-named division is divided by x — h ;
continuing this process to n divisions, A x is the last remainder,
and Oq is the last quotient. That is,
A x = R u
which establishes the rule.
EXERCISES
Obtain equations whose roots are equal to the roots of the following equa-
tions multiplied by the number opposite.
1. x3 + 2»2_7 x _i=o. (6) 2. &3----5-=o. (5)
3. x4_10x2_3x-2 = 0. (-1) 4. x*-3x 2 + 10 = 0. (-2)
Obtain equations whose roots are equal to the roots of the following equa-
tions multiplied by the smallest number which will make all the coefficients
integers and that of the highest power unity.
5. *3-2*2 + is-10 = 0. 6 2s< + 3a*-6**+6*-l=0.
8. x3 + x* + ?-§? = 0.
7. 3x3 + 4 = o. 6 86
9. Obtain equations whose roots are equal in absolute value but opposite
in sign to the roots of equations given in Exs. 1-4.
Obtain equations whose roots are equal to the roots of the following equa-
tions diminished by the number opposite.
Abtb.98,99] DESCARTES'S RULE OF SIGNS 135
10. 2x*-8x 2 + 4x-5 = 0. (2)
Solution : We apply transformation 3, Art. 98. By synthetic division,
this gives 2+ 0- 3+ 4- 6|2
+ 4+ 8 + 10 + 28
2+ 4+ 6 + 14 + 23 i* 4 = 23,
+ 4 + 16 + 42
2+8 + 21 + 56 R z = 56,
+ 4 + 24
2 + 12 + 45 R 2 = 45,
+ 4
2 + 16 R 1 = 16,
ao = 2.
Hence, 2x 4 +16x 8 + 45x 2 + 56x + 23 = 0is the required equation.
11. x 2 -7x + 6 = 0. (3)
12. x3-7x + 7 = 0. (1)
13. x3-27x-86 = 0. (3)
14. x* - 6 x< + 7.4 x* + 7.92 x 2 - 17.872 x - 0.79232 = 0. (1.2)
15. 2x 4 + 16x 3 + 45x 2 + 56x + 23 = 0. (-2)
99. Descartes's rule of signs. In a polynomial arranged in
descending powers of x, if two successive terms differ in sign
there is said to be a variation in sign. For example,
x* — 4 x* +- 3 x 2 + 4 x — 5
has 3 variations of sign as is shown more clearly by writing down
the signs H h H • Multiply this polynomial by a? — 1.
There results x i — 5x 4 -{-7x z -\-x 2 —9x-\-5
with 4 variations of sign. This last polynomial has one more
positive zero (see Art. 95) than the first. If increasing the num-
ber of positive zeros of a polynomial always increases the number
of variations in sign by at least one, then the number of positive
roots is never greater than the number of variations of sign.
Theorem. The number of positive roots of an equation f(x) =
does not exceed the number of variations of sign off(x), nor does the
number of negative roots exceed the number of variations of sign of
This is Descartes's rule of signs.
136 THEORY OF EQUATIONS [Chap. XIII.
The part which relates to positive roots will be established by
showing that whenever a positive root r is introduced into an
equation, there is added at least one variation of sign. Let
f(x) = be an equation of degree m. It is only necessary to
show that (x — r)j{x) has at least one more variation of sign than
f(x). Group the terms of f(x) between consecutive variations of
sign in brackets. In general, for a function of degree m, we have
f(x) = [&o& w + bix m ~ l + ••• + bjS"-*]
± IK-f? + K-+0- 1 + - + » J,
where b , b lf b 2 , ••• b m are positive.
Multiplying this function first by x, then by — r, and adding,
we obtain
(x - r)f{x) = [&oa w+1 + (&i - V)«" ± -]
- C(Vi + rbj**-* ± -]
+ W g +i + rb g )x»-*± •••]
± [(&— + *-+d*+ l ± -]
Trb m .
It will be noticed that the coefficients of the first terms in the
several brackets, that is, b 0> (p^x+rbj, (b^x +!•&,), —, are all posi-
tive. Hence, the signs between the brackets remain unchanged.
The signs within the brackets are uncertain, but however they
may occur there is at least one variation between the first term
of one bracket and the first term of the next bracket. Hence, as
far as the terms in the brackets are concerned the number of
variations remains the same or is increased. But there is added
the variation caused by the term T rb m whose sign differs from
the sign of the last bracket. Therefore, there is at least one
more variation in (x — r)f(x) than in f(x).
The part of Descartes's rule which relates to negative roots
follows from the fact that the roots of /(— x) = are equal in
absolute value but opposite in sign to the roots of f(x) = 0.
Arts. 09, 100] LOCATION OF ROOTS BY GRAPH 137
EXERCISES
Find the maximum number of positive and of negative roots, and any
other information about the nature of the roots of
1. x» + 6x — 7 =0.
Solution : There is one variation in sign, hence, there is not more than
one positive root. /(— x) = — x 3 — 5 x — 7 with no variation in sign, hence,
there are no negative roots. Since there are three roots of the equation, two
are imaginary and one positive.
2. 3 x* + x 2 + 2 = 0.
3. x 6 + 1 = 0.
4. x* + 3x 2 +l =0.
5. x» + 4x 2 + x = 0.
6. x» — 1 = (n odd).
7. x" — 1 = (n even).
8. Given that the roots of 6 x 8 — 3 x 2 — 4 x + 11 = are all real, deter-
mine the signs of the roots.
9. Show that the equation 7 x 6 — 2 x 2 — 2 x + 4 = has at least two
imaginary roots.
10. Show that the equation x 8 + 6x 3 + 4x — 10 = has six and only six
imaginary roots.
100. Location of roots by graph. If the real roots of an equa-
tion /(a:) = are greater than a and less than b, these roots are
said to be contained in the interval a to b along the X-axis. The
number a is a lower limit and the number 6 is an upper limit of
the interval.
We are concerned with the graph of f(x) chiefly throughout an
interval along the X-axis which contains the roots of f(x) = 0.
To avoid the labor of plotting the graph outside of this interval,
it is desirable, in evaluating f(x) for x = b by synthetic division,
to know whether b is greater than any root. The following cri-
terion will be found helpful.
If all the sums are positive in the synthetic division by x — b
(b positive), then b is greater than any root. For, a greater number
than b would make the sums still greater. For example, to show
that 6 is greater than any real root of
138
THEORY OF EQUATIONS
[Chap. XIII.
we divide by x — 6 by synthetic division,
1-5 + 3-42-50 [6
6 + 6 + 54 + 72
1 + 1 + 9 + 12 + 22
and we observe that a number greater than 6 would increase each
sum.
To find a lower limit of the negative roots of f(x)=0, it is only
necessary to find as above, by synthetic division, an upper limit
of the positive roots of /(— x) = 0.
After the graph of f(x) is plotted throughout an interval which
contains the roots, at least the approximate values of the real
roots of f(x)=0 are presented geometrically. The following
principle aids in locating roots during
the process of plotting the graph.
If f(a) and f(b) have contrary signs,
the equation f(x) = has at least one real
root between a and b.
For the points P x and P 2 (Fig. 30)
which correspond to x = a and x = b are
on opposite sides of the .X-axis, and any
continuous curve connecting P x and P 2
crosses the X-axis at least once between
a and &. Since, to every intersection of the graph with the X&xis
there corresponds a real root of the equation (Art. 95), we assume
this principle.
Fig. 30.
EXERCISES
1. Show that x 3 + 8x-7=0 has a real root between and 1, and that
the other roots are imaginary.
2. Show that — 1 is a lower limit of the roots of
x 4 - 5 x« + 3 x 2 — 42 x — 50 = 0.
3. Show that 2 is an upper limit of the roots of a; 4 — 2 x 8 +8 x*— 5 x -f-1 =0.
Find the integral part of each real root of
4. x 3 + x 2 - 2 x - 1 = 0. 5. x 8 + 2 x + 6 = 0.
6. x»-2x-5 = 0. 7. x3 + 2x-6 = 0.
8. x 4 - 12x« + I2x-S = 0. 9. 8x3-36x* + 46z-15 = 0.
10. x3 - 3 x 2 - 4 x + 11 = 0. 11. x» - 2 x - 6 = 0.
Abts. 100-102] RATIONAL ROOTS 139
101. Equation in the p-iorm. An equation
x n + Pix n ~ l + P& n ~ 2 + - +p n = 0, (1)
where p u p 2) •••,!>« are an y numbers, may be called the j>form of
the equation of the nth degree.
The general equation,
<*(F n + (hP n ~ l + atfc*- 2 H f- a n = 0, (2)
can clearly be reduced to the p-form by dividing its members
by clq. The j>form . is more convenient for the statement of
certain theorems (Art. 102) than the form (2).
Exercise. Reduce 6a; 3 — 3 sc 2 + 2 x — 5 = to the p-f orm.
102. Rational roots. Any rational root of an equation f(x) =
in the p-form with integral coefficients is an integer and an exact
divisor ofp n .
To prove this theorem, suppose, if possible, that - is a root of
b
f(x)= 0, where - is a fraction in its lowest terms. Then,
Multiplying (1) by b n ~\ we obtain
% + Picr"i+p*ar*b + - +2Via& n-2 +2> n & n - l = 0,
o
or T = - C^"" 1 +iV* n "" 2 & + - +p n -ia&*- 2 +p n b n ~ l ). (2)
All the terms of the right-hand member of (2) are integers,
while the left-hand member is a fraction in its lowest terms.
Hence, the hypothesis that -isa root leads to an absurdity.
Next, suppose that c is a root, where c is an integer. Then,
c» +P1C- 1 +p&-* + - +2> n _ 1 c +p n = 0. (3)
Transposing p n and dividing through by c, we obtain
c n ' 1 +2>ic n ~ 2 +P2C"" 3 + .- +P.-1 = -&. (4)
c
140 THEORY OF EQUATIONS [Chap. XIII.
*
Each term of the left-hand member of (4) is an integer.
Hence, & is an integer; that is, c is an exact divisor of p n .
c
To obtain the rational roots of an equation in p-form with integral
coefficiently it is only necessary to test whether the integers which are
the exact divisors ofp n satisfy the equation.*
EXERCISES
Find the rational roots of the following equations.
1. x 8 - 9x* + 23x- 15 = 0.
Solution : By Descartes's rule of signs, this equation has no negative
roots. Hence, we need try only 1, 3, 5, and 15. By synthetic division,
1 - 9 + 23 - 16[i
+ 1- 8 + 16
1 _ 8 + 15 +
The depressed equation is x 2 — 8 x + 15 = (x — 5) (x — 8) = 0. Hence, 1, 3,
and 5 are the roots.
2. 108x3-54x2 + 45x-13 = 0.
Solution : In the p-form this equation is
Transform (1) into an equation whose roots are 6 times those of (1). This
gives
X 3 _ 3 X 2 + i5 X _ 26 = 0. (2)
The rational roots of (2) divided by 6 give the rational roots of (1). By
Descartes's rule, (2) has no negative roots. Hence, we need try only 1, 2,
13, 26. Depressing the equation,
1 _ 3 + is _ 26[J_
+ 1 - 2 + 13
Hence, 1 is not a root.
1-2 + 13-13
l-3+15-26[2
+ 2 — 2 + 26
1 _ l + 13 + o
The depressed equation x 2 — x + 13 = has no rational roots. Hence, 2 is
the only rational root of (2) and } is the only rational root of (1).
y3. x<-15x2 + 10x + 24=0. 4. x»- 4x« +2x- J = 0.
%. 5. x* + 3x 2 -4x- 12 = 0. 6. 2x» + x* + 2x + 1 = 0.
* If p n is a large number, this method is too long to be practical.
Abts. 102, 103]
IRRATIONAL ROOTS
7. 8z" + 8;e> + a:-2 = 0.
\9. x<-**-8x«-14s + 80 =
Vll. i&-tt& + 2,7x-li> = <
13- I* -46** + 403 + 84=^0.
8. 4a*-8s" + 6x-l = 0.
Ao. 10s*+17i>-16;b' + 2:c=0.
12. x»-4s*-17!C + e0 = 0.
yi*. 8a:'-2a!>-4a + l = 0.
IS. 24x*-4ie>-2ic>
13 x
16. I08a"-64a!» + 46a:-13=0.
17. a»-8i< + 15z' + 20zi-7ez + 4i
103. Irrational roots. Horner's method. The irrational roots
of a numerical equation can be obtained to any desired number
of decimal places by a method of approximation called Horner's
method. The method can be best explained by first applying it
to an example. In case an equation has some rational roots, it
should always be depressed by removing such roots before con-
aidering irrational roots.
Example : Find the real roots of
x* - 2 3* + 4 x* - 15 x + 14 = 0.
1. Test for rational roots as in Art. 102.
It results that 2 is the only rational root
1-2+4- 15 + 14(2
+ 2 + 0+ 8-14
1+0+4-
The depressed equation is
x* + ix-
= 0.
(1)
(2)
1
f
1 I I
2. Test for the interval which contains the real
roots. From Descartes' b rule, equation (2) has not
more than one positive root, and it has no negative
root. Furthermore, 2 is greater than any root
(Art. 100>
3. Plot a? + 4 x - 7 from x = to x =2. V "° " 3L
The graph (Fig. 31) shows that 1 is the first figure of the root.
4. Transform to diminish roots by 1 ; or graphically, change the
origin to the point marked 1. The numerical work is as follows:
142 THEORY OF EQUATIONS DChaf. XIII.
1+0+4-7L1
+1+1+5
1-h 1 -ho' —2
+ 1+2 '
1 + 2+7
+ 1
1 + 3
The first trans formed equation is then
x* + 3z!* + 7*! - 2 =0. (3)
This equation has a root between and l r since (2) has a root
between 1 and 2. By evaluating f( x r ) = x t a -+- 3 jtj* + 7 Xi — 2 for
successive tenths (0.0. 0.1. 0.2. — . 0.9), we find that this function
is negative when x x = 0.2 and positive when x x = 0.3l Hence, (3)
has a root between 02 and 0.3. An approximation to this root is
given by neglecting the second and third degree terms in (3) and
solving
7^-2=0.
The root of this equation between and 1 is x 1 = 0^-»-. It
is important to observe from the graph of f(x) that the sign of
the known term in each transformed equation is to be the same
as that of the original equation after the rational roots have been
removed.
Transforming (3» into an equation whose roots are less by 0.2,
we have
1
i
- 2 0-2
0.2
0.64
1.528
1
3.2
:.64
- 0.472
0.2
0.68
1
3.4
0.2 :
8.32
1 36
or xf + 3.6 xj - $.32 r* - 0.472 = (4)
as the second transformed equation. The root of equation (4)
which we seek lies between and 0. 1. Nectar ting powers of Xf
higher than the first, it appears from ttu* equation
8.32 jew- 0.472 =
Art. 103] IRRATIONAL ROOTS 143
that X2 lies between 0.05 and 0.06. That the root is in this in-
terval may be tested by evaluating x£ + 3.6 x£ + 8.32 x^ — 0.472
for X2 = 0.05 and 0.06.
Transforming (4) by synthetic division into an equation whose
roots are less by 0.05, we obtain
a? s s + 3.75 a** + 8.6875 x^ - 0.046875 = 0. (5)
Neglecting powers of a? 8 higher than the first, it appears from
the equation
8.6875 x z - 0.046875 =
that x^ lies between 0.005 and 0.006.
Transforming (5) by synthetic division into an equation whose
roots are less by 0.005, we have
a> 4 8 + 3.765 x? + 8.725075 x 4 - 0.003343625 = 0. (6)
The root of this equation between and 0.001 can be obtained
at least as far as the first figure by neglecting powers of x A above
the first. This gives
a? 4 = 0.0003+
Transforming (6) into an equation whose roots are less by
0.0003, we obtain
xf + 3.7659 x b 2 + 8.72733427 x b - 0.000725763623 = 0.
The root of this equation between and 0.0001 can be obtained
at least so far as the first significant figure by neglecting powers
of asg above the first. This gives
a? 6 = 0.00008+.
Taking the sum of successive diminutions of the roots of (2),
we obtain as the approximate value of the root sought
x = 1.25538 + .
The preceding work of transformation may be compactly ar-
ranged as follows :
144
THEORY OF EQUATIONS
[Chap. XIII.
1
1
1
2
1
8.6
0.06
3.66
0.05
3.70
0.05
4
1
-7 11.
6
5
2
-2
1
8
0.2
7
0.64
-2
1.528
1
3.2
0.2
7.64
0.68
- 0.472
1
3.4
0.2
8.32
It!
8.82
0.1825
- 0.472
0.425125
10.05
8.5025
0.1850
- 0.046876
8.6876
1
8.75
0.005
8.6875
0.018775
- 0.046875
0.043631376
1
3.755
0.005
8.706275
0.01880
-0.003343625
1
3.760
0.005
8.726076
10.
1 8.765
0.0003
8.725075
0.00112959
-0.008848625
.002617861377
1 3.7653 8.72620459
0.0003 0.00112968
- 0.000725763628
1 3.7656
0.0003
8.72733427
I0J
1 3.7659 8.72733427 -0.000725768628
0.00008 0.0003012784 0.000698210843872
1 3.76598 8.7276356484-0.000027662779128
0.00008
The heavy type indicates the successive transformed equations.
The process can evidently be continued to find the root to any
required number of decimal places.
If a root of an equation is known to be small, one important
point to note is that such a root can, in general, be well estimated
by dividing the known term, with its sign changed, by the coeffi-
cient of the first degree term. The coefficient of the first degree
term is, for this reason, sometimes called the trial divisor in ob-
taining approximate roots. A still better estimate of a root can,
Arts. 103-105] SUMMARY 145
in general, be obtained by dropping terms of degree higher than
the second, and solving the quadratic.
When an equation has more than one irrational root, each is
treated separately as we have treated the single irrational root
in this example.
If two roots of an equation f(x) = are nearly equal, their
separation may become laborious, but the separation may be
accomplished by assigning values to x sufficiently near each other
in plotting the graph of f(x). For example,
4 a? - 24 a? 2 + 44 x - 23 = 0,
has two roots between 2 and 3. By assigning successively the
values x = 2, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3, in plotting
the graph we find that one of these roots is between 2.2 and 2.3,
while the other is between 2.S and 2.9.
104. Negative roots. The negative roots of f(x)=0 are
obtained by finding the positive roots of /(— a?)=0, and changing
their signs. It is therefore sufficient to discuss the method of
obtaining positive roots.
105. Summary. In solving a numerical equation /(«)= for
all its real roots, the following rules may be found helpful in
systematizing the work :
1. Test for rationed roots; and if any exist, depress the equation
by removing the corresponding factors.
2. Determine an interval which contains the irrational roots.
3. Plot the depressed polynomial throughout this interval to locate
the roots.
4. Apply Horner's method to find the irrational roots. To do
this, fix the attention upon some positive root whose location is
known to be between two consecutive integers. Obtain by synthpHc.
division (Art. 89) an equation whose roots are less than those of the
given equation by the smaller of these two integers. The new equa-
tion has a root between and 1. Locate this root between two suc-
cessive tenths; and decrease the roots by the smaller of these tenths.
The equation thus obtained has a root between and 0.1. Locate
this root between two successive hundredths, and again decrease tJie
146 THEORY OF EQUATIONS [Chap. XIII.
roots by the smaller of these hundredths. Continue this process to
any required number of decimal places.
Add together all the diminutions of the roots to obtain the re-
quired root
If more than one root is contained between two consecutive integers,
separate them by means oftlue location principle.
5. Treat negative roots in the same manner as positive roots after
changing f(x) = intof(— a?) = 0.
EXERCISES AND PROBLEMS
Find a positive root of each of the following equations to two decimal
places.
1. x 3 -3x 2 -2x + 5 = 0. 2. x 4 - 8** + 14** + 4s— 8 = 0.
3. x& + 12 x 4 + 69x3 + 150 X 2 + 201 x - 207 = 0.
4. x3- 100 = 0. 5. x*- 1000 = 0.
Find the rational roots, and the value of irrational real roots to two
decimal places, of the following equations.
6. x3-3x 2 -2x + = 0. 7. x3 + 30x-420 = 0.
8. x3-8x 2 -4x + 13 =0. 9. x 3 + 3x 2 + 4x + 6 = 0.
10. x 8 + 4x 2 -5x-20 = 0. 11. 3x*-.2x»-21** — 4x + ll =0.
12. 2x 4 - 12x3 + 12x- 3 = 0. 13. x*-.3x-l=0.
14. x 4 -3x3 + 3 = 0. 15. x« + 4a5* + 4a& + a = 0.
16. An open box is made of a rectangular piece of tin 10 inches by 20
inches by cutting equal squares from the corners and turning up the sides.
Find (to two decimal places) the side of a square cut out if the volume of
the box is 187 cubic inches.
17. A piece of property can be bought for $7550 cash or $8000, payable
in four equal annual instalments of $2000 each, the first instalment being
paid at once, and the remaining instalments at the ends of 1, 2, and 8 years.
What, yearly rate of interest compounded annually gives the two offers equal
present values ?
18. A sphere of yellow pine 1 foot in diameter floating in water sinks to a
depth x given by 2 x 8 - 3 x 2 + 0.9657 = 0.
Find the depth to 3 significant figures.
19. A sphere of ice 1 foot in diameter floating in water sinks to a depth x
given by the equation 2 x 8 — 3 x 2 + 0.693 = 0.
Find the depth to 3 significant figures.
Art. 105] EXERCISES AND PROBLEMS 147
20. A cork sphere 1 foot in diameter floating on water sinks to a depth x
given by the equation
2 a* _ 3 X 2 + o.24 = 0.
If the sphere is 2 feet in diameter, the immersed depth is given by
2 x*~ 6x 2 +1.92 = 0.
Find the depths to 2 significant figures.
21. The width of the strongest beam which can be cut from a log 12 inches
in diameter is given by the positive irrational root of the equation
X* — 144 x + 665 = 0.
Find the width to 3 significant figures.
22. The speed in feet per second of a 1-inch manila rope transmitting
4 horse power, under a tension of 300 pounds on the tight side, is given by the
equation
«3 - 19200 v + 211200 = 0.
Find the velocity to 3 significant figures.
23. The diameter of a water pipe whose length is 200 feet, and which is to
discharge 100 cubic feet per second under a head of 10 feet, is given by the
real root of the equation
x 6 -38x- 101 = 0.
Find the diameter to 3 significant figures.
(Merriman and Woodward, Higher Mathematics, p. 13.)
24. The algebraic treatment of the trisection of an angle whose sine is a
involves the solution of the cubic equation
4x 3 = 3x — a.
The unknown, x, is the sine of one third the given angle. When a= JV2,
, find x to 3 significant figures.
25. A vat in the form of a rectangular parallelopiped is 8 x 10 x 12 feet. If
the volume is increased 600 cubic feet by equal elongations of the dimensions,
find elongations in feet to two decimal places.
26. In Problem 25, if the volume is increased by elongations proportional
to the dimensions, find each elongation.
27. From the American Report on Wholesale Prices, Wages, and Trans-
portation, for 1891, the median wage is given in dollars by \ of a value of x in
the equation
2561} = Oq + a\X + a& 2 + a 3 x 3 4- a 4 x 4 ,
where ao = 6972^, a x = - 657^, a* = - 33 J{, a* = -»#, a 4 = - rf T . Find
the median wage correct to mills.
148 THEORY OF EQUATIONS [Chap. XIII.
106. Algebraic solution of equations. In Arts. 102-105, methods
are discussed by which we obtain approximately the real roots of
numerical equations. We turn now to a brief consideration of
equations with literal coefficients.
Solving such an equation consists in obtaining an expression in
terms of the coefficients which satisfies the equation. In other
words, it consists in finding a formula which gives the roots in
terms of the coefficients. For example, the roots of the typical
quadratic
0^+^+0 = 0,
— b ± V6 2 — 4 ac
are — •
2a
The roots of an equation are functions of the coefficients, and it
is important to inquire into the character of these functions.
The solution is said to be an algebraic solution, if these functions
of the coefficients involve no operations except a finite number of
additions, subtractions, multiplications, divisions, and extractions
of roots.
The algebraic solution of an equation is often called the solution
by radicals.
In Arts. 107, 108, the general cubic
OqSc 3 H- a x x 2 + a^x + og = 0,
and the general biquadratic
a^x 4 4- tti^ 3 4- a&? + a z x + a 4 = 0,
are solved by radicals.
The algebraic solution of the general fifth degree equation
a x* 4- aix* 4- a 2 « 3 4- <h% 2 4- a 4 je -f a 6 =
engaged the attention of mathematicians during the eighteenth and
the first quarter of the nineteenth century. In 1826 Abel proved
that the typical fifth degree equation has no algebraic solution.
Since that time a branch of mathematics, known as the theory of
substitution groups, has been much developed. While a treat-
ment of substitution groups is beyond the scope of this book, it
Awns. 106, 107] THE CUBTC EQUATION 149
may be stated that, by means of this theory, it is shown that
no typical equation
ewe* + a^- 1 H h a n ^x + a n ±s
has an algebraic solution if n exceeds 4 ; and necessary and suffi-
cient conditions that an equation has an algebraic solution are
established.
107. The cubic equation. The general cubic equation is
afl? + a x x 2 + a 2 x -f ag = 0. (1)
By making x = y --£*-, (2)
oa Q
equation (1) is transformed into
* , + !s af ae ' + i$-s3+;-* (3)
which has no term of the second degree.
Let 3g= 3<¥8-ai 8 (4)
3a 2 v '
Ik Oq 6 OOq £ CLq
Then (3) takes the form,
y* + 3Hy+G = 0. (6
Now assume y = u* + v*, (7)
tod -H* = uv. (8)
From (6), (7), and (8), - # = u + v. (9)
•RKirnimtiing « from (8) and (9), we have
U 2 + q u -^3 = 0, (10)
and solving this quadratic in u, we find for a solution,
.--g+yg+iip. (ii)
From (8) and (11) we have
^-g^^-g -y^+ ig- 8 . (i2)
i* 2
150
THEORY OF EQUATIONS
[Chap. XIII.
The double sign before the radical in the solution of the quad-
ratic in u is omitted because taking the negative sign before the
radical would simply interchange the values of u and v. Since
y = U* + v i f
the three values of y are :
yi = ut -,
1 H
y 2 = wu » —
wu
V
2 1 -H"
(13)
where u* is any one of the three cube roots of u, and to is a com-
plex cube root of unity (Art. 85).
Exercise. Test the solution by substitution of these values of y in (6).
By means of (2) and (13), the roots of equation (1) are *
Xi = U
u i 3a
= wu* — l
X2 = WU 1
WU
i 3a '
Xs=vflu% - — ^-«
W 2 u i 6 Oq
(14)
s
When the coefficients of the equation are real numbers, the
numerical character of the roots depends upon the number under
the radical sign in (11) and (12).
When G 2 -f 4 Il z is negative, u is a complex number. In this
case, to obtain y from (7) would involve the extraction of the
cube root of complex numbers. As we have no general algebraic
rule for extracting such a cube root, the case in which G* + 4 J5P
is negative is called the irreducible case. These roots may, how-
ever, be obtained by a method involving trigonometry (see Art. 85).
Even when G 2 -f 4 H 3 is positive, the solution presented above is
not, in general, so well adapted to obtaining real roots of numeri-
cal equations as the methods of Arts. 102-105.
* This solution of the cubic is due to Tartaglia, but was first published by
Cardan (1645).
Arts. 107, 108] THE BIQUADRATIC EQUATION 151
108. The biquadratic equation. The general biquadratic
Ooaj* + a x x^ + age 2 + a z x + a 4 =
may ,be written in the p-f orm (Art. 101) as
x 4 + PiX? +p 2 x 2 + p z x + p 4 == 0. (1)
Adding (mx + b) 2 to both members of (1), we have
x 4 + p x x* + (i> 2 + »i 2 )« 2 4- (i> 3 + 2 m&)x- + p 4 -f 6 2 = (ma; + 6)2. (2)
Assume the identity,
« 4 +i>i« 8 +(i>2+w 2 )« 2 +(i> 3 +2 m6)a;+i) 4 +62=(a J 2 + |L a . + g )2 # (3)
Equating coefficients of like powers of x, we have
»i 2
!>2 + m2 = ^- + 2< ? , (4)
p 8 + 2 mb = p x g, (5)
jp 4 + & 2 = ? 2 . (6)
Eliminating m and 6 from (4), (5), and (6), we obtain
Q>i 2 + Sq - ±p 2 )(q* -p A ) = ( Pl q -p z y, (7)
or 8g«-4j) 2 9 2 4-(2i> 1 i>3-8i9 4 )g + 4p 2j p 4 - i ) 1 2p 4 _ i ,32 = 0. (8)
This is a cubic in q. Since the general cubic is solved by radi-
cals in Art. 107, we may assume a value of q known. When q
is known, the values of m and b are obtained from (4) and (6).
From (2) and (3), we have
(x* + ^x + q)* = (mx + 6)2, (9)
which is equivalent to the two quadratic equations
x 2 4- Qx + q - mx — b = 0,
and x 2 4- 4^# -f q -f m# + 6 = 0.
The solutions of these two quadratics give the four roots of (1).*
♦This solution is due to Ferrari and was first published by Cardan (1545)
152 THEORY OF EQUATIONS [Chap. XIIL
EXERCISES
1. By Tartaglia's method, solve x 3 — 4x 2 + 6x — 4 = and verify the
results by substitution.
Solution : Here Oo = 1, a\ = — 4, a% = 6, a% = — 4.
From (4) and (5), Art. 107,
From (11), u = 10 + 6 ^ t ■
v ' 27
3
From (14) the roots of the given equation are
2, 1 + i, 1 - i
Substitution for x shows that each of these numbers satisfies the equation to
be solved.
2. Solve x*-6x 3 + 12x 2 -20a; -12 = 0. (1)
Solution : Adding (mx + ft) 2 to both members of this equation gives
X 4 _ 6x8 +(12 + m 2 )x 2 +(2 mb - 20)x + 6* - 12 =(»w + by. (2)
Assume the identity
X 4 _ o j.3 + (12 + m 2)a;2 + (2 m6 - 20)x + & 2 - 12 =(** - 3x + g) 1 . (3)
Equating coefficients, we obtain 12 + m 2 = 9 + 2 g, (4)
2m&-20=— 6g, (5)
6 2 - 12 = g 2 . (6)
Eliminating m and b from these three relations, we have the cubic
g 3 -6g 2 + 42g-68 = 0. (7)
This cubic has a root q = 2. From (4), (5), and (0), the corresponding
values of m 2 , ft 2 , and mb are
7/j 2 = 1, 6 2 = 16, mb= 4. (8)
From (2), (3), and (8), (x 2 - 3 x + 2) 2 = (x + 4) 2 . (9)
This equation is equivalent to the two quadratic equations
x 2 -3x + 2-(x + 4) = 0, (10)
and x 2 — 3x + 2 + x + 4 = 0. ■ (ix)
The roots of (10) are 2 ± V0, and those of (11) are 1 ± iy/&. These four
values satisfy the given biquadratic.
Solve the following equations, and verify the results by substitution.
3. x 3 + 4 x 2 + 4 x + 3 = 0, 4. 2 x 3 - 3 x 2 + 2 X - 8 = 0.
5. 3x 3 -2x 2 -6x + 4 =0. 6. 8x»- 28x 2 + SOx — 9 = 0.
7. x 4 + x 3 - x 2 - 7 x - 6 = 0. 8. x 3 — 2 x 2 + 8 = 0.
9. x»+ 12x 2 + 57x + 74 = 0. 10. 4x 4 -36x»+45xH-64x-81=0.
Arts. 108-110] VARIABLE COEFFICIENTS AND ROOTS 153
109. Coefficients in terms of roots. Let r i9 r 2 , • • •, r n be the roots
of x n + P&*- 1 + p&r-* + • • • + p n = 0, (1)
Then, from Art. 94,
*• +P1X 11 - 1 +i> 2 « n " 2 H h p n = (» - r i)( x - r 2 )--(x- r n ),
= x n — (r x + r 2 H f- r n )» n - 1 + (ly, + ^r, H h r^irja"-*
- (Wi + • ■ • + r n _ 2 r n _ x r n )x n - z + ...+(- l) n r x r 2 r z • • . r n , (2)
by actual multiplication of the binomial factors of the second
member.
Equating coefficients in (2) (Art. 94, Cor. I), we have
—Pi = r l + r 2 + - +r n}
p 2 = r x r 2 + r x r z -\ \- r n _ x r n ,
—Pi = Ws + - + r«-2n»-ir„>
(■
That is, — p x =
-i>8 =
*V
(^)
sum of the roots,
sum of products of roots taken two at a time,
sum of products of roots taken three at a time,
(— l)"p n = product of the roots.
If certain relations among the roots are given, the expressions
(A) of the coefficients in terms of the roots may aid in solving
the equation.
110. Variable coefficients and roots. Given that r x , r 2 , — , r n
are the roots of
atfc n H- aiX n ~ l + a 2 x n ~ 2 -f — h a n = 0,
(i)
relations (A), Art. 109, may be written in the form
a
n)
— = n^2 + nr, + ••• + r n _ x r n ,
a.
(-l)^ = ri¥ 3...r s .
W
154 THEORY OF EQUATIONS [Chap. XIII.
If clq remains fixed, it follows from (B) that
a n — ►* 0, if one root approaches zero ;
a n — > and a n _x — > 0, if two roots approach zero ;
a H — >» 0, a n _! — >» 0, — (&„_,+! — > 0, if r roots approach zero.
In certain problems of analytic geometry, it is desirable to
know the character of the coefficients of (1), if some of the roots
become indefinitely large.
In (1), put x = — . This gives
i?° +-^ + _Sl.+ ... +**=! + a^-a,
x x n x x n ~ l Xf" 2 x 1
or, a -f «i»i + «2»i 2 H h a n -i x i n ~ l + aW = 0. (2)
For our purposes, we may define a> = oo (read, x becomes infi-
nite) as — when x x — > 0. The conditions which make x = oo are
x x
then precisely the conditions which make Xi-+> 0. Hence, from
(2), if a n remains fixed, it follows that
if a — >• 0, one root of equation (1) becomes infinite ;
if a© -> 0, and a x — > 0, two roots of equation (1) become infinite ;
••...«••••
if a — >0, cii— >0, — a r _i— >Q, r roots of equation (1) become in-
finite.
EXERCISES
1. Solve x 3 — 2 x 2 — 4 x + 8 = 0, the sum of two of the roots being 2.
2. Solve2x 3 — 3x 2 + 2x — 3 = 0, the sum of two of the roots being zero.
3. The roots of x 3 -6x 2 + 3x + 10 are in arithmetical progression.
Find them.
4. Solve x 3 — 8 x 2 + 6 x + 50 = 0, two of the roots being equal,
5. Obtain the roots of ax 2 — 13 x + 1 = 0, to three decimal places, when
a = 10, 1, 0.1, 0.01, and 0.001.
6. Obtain the roots of x 2 — 13 x + a = 0, to three decimal places, when
a = 10, 1, 0.1, 0.01, 0.001.
* The symbol — > is read " approaches."
Art. 110] EXERCISES 155
7. Determine m and b so that the quadratic
8x 2 — (mx+ ft) 2 — 4x + 2 =
shall have two infinite roots.
8. Determine m and b so that the cubic
x* + x(mx + 6) 2 — 3x 2 + 5 =
shall have two infinite roots.
9. Determine m and b so that the quadratic
9x 2 -16(mx+&) 2 = 25
shall have two infinite roots.
10. Determine m and b so that
8x* + 34x(mx + 6)+ll(mx + 6) 2 — » + 21(mx + 6) =
shall have two infinite roots.
■J*
"OHAPTKR XIV
LOGARITHMS
111* Generalization of exponents. In Art. 6, a* is defined when
a; is a positive integer ; and a meaning is obtained (Arts. 7-9) from
the laws of exponents for a* when x is any rational number.
Thus, 4 6 = 4 • 4 • 4 • 4 • 4, and 8 J is the square of the cube root of 8.
But no meaning has been obtained thus far for a? when x is ir-
rational ; for example, 4^ is thus far undefined. We have,
however, denned V2 as the limit of a sequence of rational num-
bers
1, 1.4, 1.41, 1.414, 1.4142, - . ..
When a variable, z, taking this sequence of rational values, ap-
proaches V2 as a limit, it can be proved that a'(a < 0) has a
limit, and we define d^* as this limit. In accordance with this
illustration, a x (a > 0), if x is irrational, is defined as the limit of
a* when z approaches x.*
112, Definition of a logarithm. Ifa* = y(a > 0, a gfc 1), then x is
said to be the logarithm of y to the base a, and this is written
» = l0( Ja V-
The two equations a* = y (1)
and x=logjf (2)
thus mean exactly the same thing ; and the terms logarithm and
exponent are equivalent.
We shall assume in what follows that :
1. Corresponding to any two positive numbers y and a (o^fel)
there exists one and only one real number x such that a 9 = y.
This assumption is sometimes expressed by saying that any
* a* can also be defined, consistently with the laws of exponents, as the
limiting value of an infinite series of positive integral powers of X (Art. 170).
_ 156
Arts. 111-113] PROPERTIES OF LOGARITHMS
157
positive number has one and only one logarithm, whatever positive
number is the base (unity excepted).
2. The laws of exponents (Art. 6) which apply to rational expo-
nents are also valid when irrational exponents are involved.
EXERCISES
1. log 2 8 = ? log c a = ? log 4 2 = ? log 4 256 = ? log 6 4 = f
2. Fill out the following table.
Babe
Number
Logarithm
81
4
10
5
23
i
2
A
3
*V
-
32
-5
113* Derived properties of logarithms.
1. . The logarithm of a product equals the sum of the logarithms of
Us factors.
Let log a u = x and log a v = y, (1)
then, a* = u, a v = v } (Definition of logarithm.)
and uv = a x+u . (Art. 6 and Art. 112, Assumption 2.)
Hence, log a uv = x -f- y>
that is, \og a uv = log a u + log rt v.
Similarly, log a (uvw) = log a u + log rt v + log a w,
and so on for any number of factors.
logio 256 = logio 3 + log 10 5 + log 10 17.
a* =
■u, a v =
v,
u _
■a* - *.
lo&
V
:a?-y,
log 2 =
log a w-
log a v.
158 LOGARITHMS [Chap. XIV.
2. The logarithm of a quotient is equal to the logarithm of the
dividend minus the logarithm of the divisor.
As above, let log a u = x and log a v = y,
then,
and
Hence,
that is,
Example : logio |f f = log 10 625 — logw 188.
3. The logarithm of u v is equal to v times the logarithm of u.
To prove this, let x = log a u or a* = u. (1)
Then, from (1), u v = a vs .
(Law of Indices, and Art. 112, 2.)
Hence, log CT u v = vx = v log a u. (2)
Example : log 10 (257) * = J logio 257.
Making v = n and u = - respectively, we have
n
(a) The logarithm of the nth power of a number is n times the
logarithm of the number.
(b) The logarithm of the real positive nth root of a number is the
logarithm of the number divided by n.
EXERCISES
Express the logarithms of the following expressions in terms of the loga-
rithms of integers.
i. *io g ^i.
9*6*
Solution : log — = log v^8 - log 9* - log 6* (1 and 3, Art 118.)
9*6*
= Jlog8-ilog9-ilog6. (8, Art 118.)
2. log?- 2 . 3. log ^ . 4. kg-$L
33 * VT5V& 11*28*
* When in a problem the same base is used throughout, it is customary sot
to write the base.
Arts. 113, 114] COMMON LOGARITHMS 159
Express the logarithms of the following in terms of the logarithms of
prime numbers.
5. log («»)* 6. log-OKI-.
(36)2(72)* (75)2(12)2
7. {pg ( 10 °) 2 . 8. log(V2^r2^6).
(20)*(75)*
9. log (v^2 VOQ ^25). 10. Prove that log a 1=0.
Given logio 2 = 0.3010, log 10 3 = 0.4771, logio 7 = 0.8451 , find the logarithms
of the following numbers to the base 10.
11. 12. 12. 30. 13. 42. 14. 420.
15. 180. 16. 900. 17. 343. 18. W-
19. }|." 20. ai,. 21. t^ t . 22. V504.
23. \/29±. 24. VJ. 25. vT7l5. 26. #43218.
114. Common logarithms. While any positive number can be
used as the base of some system of logarithms, there are two
systems in general use. These are the common or Briggs's system
and the natural or Naperian system. In the common system the
base is 10, while in the natural system the base is a certain irra-
tional number e = 2.71828 •••. It may be stated that the com-
mon system is adapted to numerical computation, while the
natural system is adapted to analytical work.*
In the following discussion of common logarithms, log x is
written as an abbreviation of log 10 x.
Since, 10° = 1 10-* = 0.1
10 1 = 10 10- 2 = 0.01
102 = 100 10- 3 = 0.001
103 = 1000 10- 4 = 0.0001
• • • • • • •
it follows that
logl =0 logO.l =-1
log 10 =1 log 0.01 =-2
log 100 =2 log 0.001 =-3
log 1000 = 3 log 0.0001 = - 4
* The notation In x for log e x and log x for logio x is frequently used when
both kinds of logarithms appear in the same problem.
160 LOGARITHMS [Chap. XIV.
So far as these powers of 10 are concerned, it may be observed
that the logarithm of the number becomes greater as the number
increases. In accordance with this observation, we may assume,
if a < x < b, that
log a < log x < log b. (1)
For example, log 100 < log 765 < log 1000,
or 2 < log 765 < 3.
When the logarithm of a number is not an integer, it may be
represented at least approximately by means of decimal fractions.
Thus, log 765 = 2.8837 correct to four decimal places.
The integral part of a logarithm is called the characteristic and
the decimal part is called the mantissa. In log 765, the charac-
teristic is 2 and the mantissa is 0.8837. For convenience in
constructing tables, it is desirable to select the mantissa as posi-
tive even if the logarithm is a negative number. For example,
log | = - 0.3010 ; but since - 0.3010 = 9.6990 - 10, this may be
written log -J- = 9.6990 — 10 with a positive mantissa. The fol-
lowing illustration shows the method of writing the characteristic
mantissa :
log 7185 = 3.8564
log 718.5 = 2.8564
log 71.85 = 1.8564
log 7.185 = 0.8564
log 0.7185 =9.8564-10
log 0.07185 = 8.8564 - 10
115. Characteristic. With our decimal system of notation,
the characteristic in the case of the base 10 is very easy to deter-
mine by a simple rule. Herein lies the advantage of this base.
If y is a number which has n digits in the integral part) then
10- 1 = y < 10", (1)
and by Art. 114, (1), n — 1 < log y < n.
Hence, log y = n - 1 +/,
where /is positive and less than 1.
Hence, to find the characteristic of the common logarithm of a
number which has an integral party subtract 1 from the number of
digits in the integral part.
Arts. 114-116] USE OF TABLES 161
If y represents a decimal fraction, we may move the decimal
point ten places to the right, and apply the rule just stated to the
integral part of the number so formed, provided we subtract 10
from the resulting logarithm. That is,
lQio y= yi
y 10 10 10 10 '
where log y = log y x — log 10 10 ,
= log y x - 10.
The result so obtained could manifestly also be obtained by the
following rule :
To find the characteristic of the common logarithm of a decimal
fraction, subtract from 9 the number of ciphers between the decimal
point and the first significant figure. From the number so obtained
subtract 10.
If two numbers contain the same sequence of figures, and
therefore differ only in the position of the decimal point, the
one number is the product of the other and an integral power
of 10, and hence, by Art. 113, the logarithms of the numbers
differ only by an integer. Thus,
log 3722 = log 37.22 + log 100
= log 37.22 4- 2.
Hence, the mantissa of the common logarithm of a member is in-
dependent of the position of the decimal point In other words, the
common logarithms of two numbers which contain the same
sequence of figures differ only in their characteristics. Hence,
tables of logarithms contain only the mantissas, and the computer
most find the characteristics by the foregoing rules.
116. Use of tables. On pp. 162, 163, a " four-place " table of
logarithms is given. In this table, the mantissas of the loga-
rithms of all integers from 1 to 999 are recorded correct to four
decimal places. "Five-place," "six-place," and "seven-place"
tables are in common use, but this four-place table will serve for
our present purposes.
Methods by which such a table can be made will be discussed
after applying the logarithms found in the table to purposes of
162
LOGARITHMS
[Chap. XIV.
N
10
0000
1
2
3
4
5
6
7
8
9
0043
0086
0128
0170
0212
0253
0294
0334
0374
11
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
12
0792
0828
0861
0899
0931
0969
1004
1038
1072
1106
13
14
1139
1461
1173
14<)2
1206
1523
1239
1553
1271
1584
1303
1614
1335
1644
1367
1673
1399
TW5~
WQ
"1732
15
1761
1790
1818
1847
1875
1903
1931
1950
1967
2014
16
2041
2068
20i«
2122
2148
2175
2201
2227
2253
22tf)
17
2304
2330
2355
2380
240?
2430
2455
2480
2504
2529
18
2553
2577
2601
2625
2648
2672
2695
2718
2742
2765
19
2788
2810
2833
2856
2878
2900
2923
2945.
2967
2989
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
8201
21
3222
3243
3263
3284
3304
3324
3340
3365
3385
3404
22
3424
3144
3464
3483
3502
3522
3541
3560
3579
3596
23
3617
3636
3655
3674
3692
3711
3729
3747
3766
3784
24
3802
3820
3838
3856
3874
3892
3909
3927
3945
3962
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
26
4150
4166
4183
4200
4216
4232
4249
4265
4281
4298
27
4314
4330
4316
4362
4378
4393
4409
4425
4440
4456
28
4472
4487
4502
4518
4533
4548
4564
4579
4594
4609
29
4624
4639
4654
4669
4683
4698
4713
4728
4742
4757
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
31
4914
4928
4942
4955
4969
4983
4997
5011
5024
5038
32
5051
5065
5079
50<)2
5105
5119
5132
5145
5159
5172
33
5185
5198
5211
5224
5237
5250
5263
5276
5289
5302
34
5315
5328
5340
5353
5366
5378
5391
5403
5416
5428
35
5441
5453
5465
5478
5490
5502
5514
5527
5039
5551
36
5563
5575
5587
5599
5611
5623
5635
5647
0658
5070
37
5682
56<H
5705
5717
5729
5740
5752
57<i3
5775
5786
38
57<«
5809
5821
5832
5843
5855
5866
5877
5888
5899
39
5911
5922
5933
5944
5955
5966
5977
5988
5999 -
6010
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
41
6128
6138
6149
6160
6170
6180
6191
6201
6212
fflflfk
42
6232
6243
6253
62(53
6274
6284
6294
6304
6314
6325
43
6335
6345
(5355
6365
6375
6385
6390
6405
6415
6425
44
6435
6444
6454
6464
6474
6484
6493
6503
6513
6022
45
6532
6542
6551
6561
6571
6580
6590
6599
6809
6618
46
6628
6(537
6646
6656
6665
6675
(5684
6693
6702
6712
47
6721
67.50
6739
6749
6758
6767
6776
6785
6794
6803
48
6812
6821
68:50
6839
6848
6857
6866
6875
6884
6893
49
6902
6911
6920
6928
6937
(5946
6955
6964
6972
0981
50
6990
6998
7007
7016
"7024
7033
7042
7050
7009
7067
51
7076
7084
7093
7101
7110
7118
7126
7135
7148
7102
52
7160
7168
7177
7185
7193
7202
7210
7218
7228
7280
53
7243
7251
7259
7267
7275
7284
7292
7300
7306
7816
54
7324
7332
7340
7348
7356
73(54
7372
7380
7888
7896
Art. 116]
LOGARITHMS
163
N-
1
2
3
4
5
6
7
8
9
55
7401
7412
7419
7427
7435
7443
7451
7459
7466
7474
56
7482
7490
7497
7505
7513
7520
7528
7536
7543
7551
57
7559
7566
7574
7582
7589
7597
7604
7612
7619
7627
58
7634
7642
7649
7657
7664
7672
7679
7686
7694
7701
59
7709
7716
•
7723
7731
7738
7745
7752
7760
7767
7774
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
61
7853
7860
7868
7875
7882
7889
7896
7903
7910
7917
62
7924
7931
7938
7945
7952
7959
7966
7973
7980
7987
68
7993
8000
8007
8014
8021
8028
8035
8041
8048
8055
64
8062
8069
8075
8082
8089
8096
8102
8109
8116
8122
65
8129
8136
8142
8149
8156
8162
8169
8176
8182
8189
66
8195
8202
8209
8215
8222
8228
8235
8241
8248
8254
67
8261
8267
8274
8280
8287
8293
8299
8306
8312
8319
68
8325
8331
8338
8344
8351
8357
8363
8370
8376
8382
69
8388
8395
8401
8407
8414
8420
8426
8432
8439
8445
70
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
71
8513
8519
8525
8531
8537
8543
8549
8555
8561
8567
72,
8573
8579
8585
8591
8597
8603
8609
8615
8621
"8627
78
8633
8639
8645
8651
8657
8663
8669
8675
8681
8686
74
8692
8698
8704
8710
8716
8722
8727
8733
8739
8745
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
76
8808
8814
8820
8825
8831
8837
8842
8848
8854
8859
77
8865
8871
8876
8882
8887
8893
8899
8904
8910
8915
78
8921
8927
8932
8938
8943
8949
8954
8960
8965
8971
79
8976
8982
8987
8993
8998
9004
9009-
- 9015
9020
9025
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
81
9085
9090
9096
9101
9106
9112
9117
9122
9128
9133
82
9138
9143
9149
9154
9159
91(55
9170
9175
9180
9186
88
9191
9196
9201
9206
9212
9217
9222
9227
9232
9238
84
9243
9248
9253
9258
9263
9269
9274
9279
9284
9289
85
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
86
9345
9350
9355
9360
9365
9370
9375
9380
9385
9390
87
9395
9400
9405
9410
9415
9420
9425
9430
9435
9440
88
9445
9450
9455.
9460
9465
9469
<H74
9479
9484
9489
89
9494
9499
9504
9509
9513
9518
9523
9528
9533
9538
90
9542
9547
9552
9557
9562
9506
9571
9576
9581
9586
91
9590
9595
9600
9605
9609
9614
9619
9624
9628
9633
92
9638
9643
9647
9652
9057
9661
9666
i)671
9675
9680
98
9685
9689
9694
9699
9703
9708
9713
9717
9722
9727
94
9731
9736
9741
9745
9750
9754
9759
9763
9768
9773
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
96
9823
9827
9832
9836
9841
9845
9850
9854
9859
9863
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
98,
9912
9917
9921
9926
9930
9934
9939
9943
9948
9952
99
9956
9961
9965
9969
9974
9978
9983
9987
9991
9996
164 • LOGARITHMS [Chap. XIV.
arithmetical calculation. In order to use the tables we must
know how to take from the tables the logarithm of a given num-
ber, and how to take from the tables the number which has a
given logarithm.
117. To find from the table the logarithm of a given number.
EXAMPLES
1. Find the logarithm of 821.
Glance down the column headed N for the first two significant figures,
then at the top of the table for the third figure. In the row with 82 and the
column with 1 is found 9143.
Hence, log 821 = 2.9143.
2. Find the logarithm of 68.42.
This number has more than three significant figures, so that its logarithm
is not recorded in the table. It may, however, be obtained approximately
from logarithms recorded in the table by a process of interpolation. In this
process, it is assumed that to a small change in the number, there corre-
sponds a change in the logarithm which is proportional to the change in the
number. This assumption is called the principle of proportional parts. As
in Ex. 1, we find that the mantissas of 6840 and 6860 are 8861 and 8867,
respectively. The difference between these two mantissas is 6. Since 6842
is two tenths of the interval from 6840 to 6850, by the principle of proportional
parts, we add to 8351 , „ 6 — 1+
Hence, log 68.42 = 1 .8352.
118. To find from the table the number which corresponds to a
given logarithm.
EXAMPLES
1. Find the number whose logarithm is 2.4675. The mantissa 4675 is not
recorded in the table, but it lies between the two adjacent mantissas 4609 and
4083 of the table. The mantissa 4669 corresponds to the number 298 and
4083 corresponds to 204. The number 4675 is ft of the interval from 4669 to
4083. By the principle of proportional parts, the number ^vhose mantissa
is 4075 is 2930 + ft X 10 = 2034+.
Hence, log 293.4 = 2.4075.
2. Find the number whose logarithm is 9.3025 — 10.
From the table, log 0.2000 = 9.3010 - 10
log 0.2010 = 9.3032 - 10 c ' '
Difference = 0.0022
(9.3025 - 101 - ^9.3010 - 101 = 0.0015.
By the principle of proportional parts, the number is
0.2000 + J J \ 0.0010 = 0.2007. . ■* "
Abts. 116-119] COMPUTATION WITH LOGARITHMS 165
EXERCISES
Obtain, from the table, the common logarithms of the following.
1. 163. 2. 80. 3. 999.
4. 1.41. 5. 0.00785. 6. 6563.
7. 7.864. 8. 3.142. 9. 0.5236.
10. 1.732. 11. 0.8665. 12. 0.0298.
Obtain, by means of the table, the numbers whose common logarithms are
the following.
13. 2.7182. 14. 9.8532-10. 15. 3.1416.
16. 0.5236. 17. 7.8321-10. 18. 4.2631.
19. 8.5432-10. 20. 1.4142. 21. 0.4343.
119. Computation by means of logarithms. The application of
logarithms to shorten calculations depends upon the properties
of logarithms given in Art. 113. By means of logarithms labo-
rious multiplications and divisions may be replaced by additions
and subtractions ; and involution and evolution may be replaced
by multiplication and division.
EXAMPLES
1. Find the value of N = 6,82 ° x 8,e74 to four significant figures.
2.851 s °
log 6.320 = 0.8007
log 8.674 = 0.9382
log (6.320) (8.674) = 1.7389
log 2.851 = 9.4550
log .AT =1.2839
iST= 19.23.
In using logarithms, much time is saved and the liability of error is decreased
by making a so-called form for all the work before using the table at all.
Thus, in Example 1, the u form " is
log 6.320 =
log 8.674 =
log (6.820) (8.674) =
log 2.851 =
\ogN =
N =
166 LOGARITHMS (Chap. XIV.
2. Make a form for evaluating JTs foSS ) ^ j» jg*.
log 6.85 =
log (6.85)* =
log 8.542 =
log #8.542 =
log (6.85)* #£642 =
log 65.27 =
logV66\27 =
logN =
N =
3. Evaluate N= #- 68.61.*
log 58.61 = 1.7680 n
log #58.61 = 0.5893 n
^ = -3.885.
EXERCISES AND PROBLEMS
Evaluate to four significant figures by logarithms.
1. (0.2386)3. 2. 0.0631 x 7.208 x 0.5127.
3 0.3384 4 2563 x(- 3.442)
' 8.659 ' 714.8 x 0.6110 "
5. V64K3. 6. (0.9828)*.
? 763.2 x 2.163, a ^ SSBm
986.7
9. # - 0.62305. 10. Y ' 0001289 -
, #0.0008276
13. #0.7684. 14. (0.008648)*.
15. (0.4754)*(0.6782)*. 16. V48 °0 .
(1.06)*
17. J(3.142)(0.5236), ^ / -Q.03206
* (85*) 18 - A/ 7.962 '
19. #3.1416 x (16)3. 20. #185« - 112*.
Hint : 1852 - 1122 = (135 + H2) (185 - 112).
* When a number is negative, find its logarithm without regard to sign,
writing n after a logarithm that corresponds to a negative number to as to
keep the negative sign in mind.
Art. 119] PROBLEMS 167
21. ^2I0« - 1672. 22. v^lOO.
23. (-0.03674)*. 24. (J)*.
2- 12(0.5236* 26. ( V 5K^).
(-52.36)*' ( ^)(^>
27. The time £ of oscillation of a simple pendulum of length I feet is given
in seconds by the formula , — j —
'32.16
Find the time of oscillation of a pendulum 3.326 feet long. (Take
ir = 3.142.)
28. What is the weight in tons of a solid cast-iron sphere whose radius is
5.343 feet, if the weight of a cubic foot of water is 62.355 pounds and the
specific gravity of cast-iron is 7.154 ?
29. Find the volume and surface of a sphere of radius 14.71.
30. The stretch of a brass wire when a weight is hung at its free end is
given by the relation m gi
~^r*k J
where m is the weight applied, g = 980, I is the length of the wire, r is its
radius, and & is a constant. Find k for the following values : m = 944.2
grams, I = 219.2 centimeters, r = 0.32 centimeter, and S = 0.060 centimeter.
31. Find the length I of a wire which stretches 5.9 centimeters for a
weight of 1826.5 grams hanging at its free end, the diameter of the wire being
0.064 centimeter, and k = 1.1 x 10 12 .
32. The weight P in pounds which will crush a solid cylindrical cast-iron
column is given by the formula
P= 98,920^-,
where d is the diameter in inches and I the length in feet. What weight will
crush a cast-iron column 6 feet long and 4.3 inches in diameter ?
33. For wrought-iron columns the crushing weight is given by
(KM
P = 299,600 V-.
What weight will crush a wrought-iron column of the same dimensions as
that in Problem 32 ?
34. The weight Wot one cubic foot of saturated steam depends upon the
pressure in the boiler according to the formula
330.36'
where P is the pressure in pounds per square inch. What is W if the pres-
sure is 280 pounds per square inch ?
168 LOGARITHMS [Chap. XIV.
35. By using the formula given in Problem 84, find the pressure in a
boiler when a cubic foot of steam weighs just one pound.
36. The diameter in inches of a connecting rod depends upon the diam-
eter D of the engine cylinder, I the length of the connecting rod, and P
the maximum steam pressure in pounds per square inch, aooosding to Mark's
formula /— ■=.
d = 0.02758V D.f- VP.
What is d when D = 30, I = 75, and P= 150?
37. For D = 10, I = 60. a table in Kent's Pocket Book, based upon the
formula in Problem 36, gives d = 2.14 inches. What was the m«rimmn
steam pressure used ?
?=^ffW2y,
38. The discharge of water from a triangular weir is given by
15
where c is a constant 0.502. g is the acceleration due to gravity 82.2 feet per
second, and H is the water head. Find q when H = 0.3 foot.
39. Given pv lM = 400 as the relation between pressure and volume of air
expanding under certain conditions. Findp when v = 24.
40. The number, n, of vibrations per second made by a stretched string
is given by the relation .. r^r-
where I is the length of the string, M the weight used to stretch the string,
m the weight of one centimeter of the string, and g = 960. Find n, when
M = 6213.0 grams * = *^-9 centimeters, and m = 0.00670 gram.
41. What must be the weight per centimeter length of a wire which is
70.05 centimeters long and is stretched by a weight of 4406.6 grams, in order
that it may vibrate 178 times per second ?
42. The work in foot pounds done during the adiabatio expansion of a
gas from pressure p\ to pressure p» is
A-l
"*= "*£*-£)']
where r t is the original volume of the gas and k is a constant. Find W for
k = 1.41, pi = 00, p* = 15, ri = 3.5.
43. If 8 1500 is placed at 3 per cent interest, converted annually, what
will it amount to in 10 years ?
Hint : In n years S I will amount to "? (1.03)".
44. What will s 10.000 amount to if left at interest for 10 yean at 4 per
cent: (irt converted annually? (b) converted semiannually? (c) con-
verted quarterly ?
Arts. 119, 120] PROBLEMS 169
45. What sum of money left for 20 years at 5 per cent, converted annu-
ally, will amount to $ 10,000 at the end of that time ?
46. If $1 had been kept at interest 5 per cent, compounded annually,
from the beginning of the Christian era to the present time, how many digits
would occur in the integral part of the accumulated amount when expressed
in dollars ?
47. The formula y = 0.0263 x 11 gives the relation between y and x when
x stands for the stress in kilograms per square centimeter of cross section of
a hollow cast-iron tube subject to tensile stress, and y for the elongation of
the tube in terms of ^ cm. as a unit. Compute y when x = 101.8.
48. The formula y—ks*g<*, where log fc=5.03370116, log «= —0.003296862,
log g = — 0.00013206, logc = 0.04579609, gives the number living at age x in
Hunter's Makehamized American Experience Table of Mortality. Find, to
such a degree of accuracy as you can secure with a four-place table of loga-
rithms, the number living (1) at age 10, (2) at age 30.
49. Given that one kilometer is equal to 0.6214 mile. Find the number
of miles in 2489 kilometers.
50. Given that one kilometer equals 0.6214 mile, and that the area of
Illinois is 56,625 square miles. Express the area of Illinois in square kilo-
meters (to four significant figures).
120. Change of base. The logarithm of a number y to the base
b is equal to the product of its logarithm to the base a and the loga-
rithm of a to the base b.
That is, log 6 y = log c y • log 6 a. (1)
Let u = log a y and v = log 6 y. (2)
Then, a« = y, b* = y, (3)
and a u = b v . (4)
a = 6 s , (5)
- = log 6 a,
u
v=sti log 6 a. (6)
From (2) and (6), log 6 y = log„ y log 6 a. (7)
Example : log 10 128 = log 2 128 log 10 2.
By making y = b in (7), we obtain
1 = log a b log 6 a.
That is, 10*^ = ^. (8)
170 LOGARITHMS [Chap. XIV.
The number log 6 a is often called the modulus of the system of
base b with respect to the system of base a.
In Art. 115, attention is called to the advantages of 10 for the
base of a system of logarithms to be used in numerical calculations.
For analytical purposes, as will appear in the calculus, it is con-
venient to use natural logarithms. This system has for its base
an irrational number e = 2.71828 • •. In the chapter on Infinite
Series, there will be given a series from which this approximation
to e is obtained, and another series from which the logarithm of
a number to the base e can be obtained to any number of decimal
places. It turns out that
log. 10 = 2.3026,
and log 10 e = , * = 0.4343.
log. 10
By (1), log 10 y = log. y log l0 e,
= 0.4343 log. y.
The number log 10 e = 0.4343 is the modulus (to four significant
figures) of common logarithms with respect to natural logarithms.
EXERCISES
1. Given log. 2 = 0.6931, find log 10 2 and compare the result with the
value in table, p. 162.
2. Given logio 2 = 0.3010, find log, 4.
3. Given logio 3, find log5 3.
Hint : By Art. 120, log 5 3 = logio 3 • log 6 10,
= logio 3
logio 5
4. Given logio 3, find logs 81.
Find the logarithms of the following numbers.
5. 10 to the base 2. 6. 10 to the base 4.
7. 10 to the base 3. 8. 10 to the base 5.
9. 75 to the base 3. 10. 13 to the base 20.
11. 130 to the base 20. 12. 1300 to the base 20.
Arts. 120, 121] EXPONENTIAL EQUATIONS
171
121. Graph of y = log a 05(a>l). A general notion of the
value of the logarithm of any number can be easily fixed by
Tef erence to the graph of y = log c x. This graph is also the graph
of x = a v . In the graph (Fig. 32) we take a = e = 2.718 • • •, but
T
__ .^-^_^^^«— . ^^_«m^hhw | L I i ■ ■ ■ — — ■ ■ -■ ■ i.- ■ -^ ■ " v ~ a ^ B ^^^^^^H ^^^^^^^^BV ^^^^^^^^^^« h i ■
I—- .^K^-MH^^^^^H— ■^_ H ^ B __^^^^_M» — 1 . l-i^^^^^^^— — ■■ ■ ■ !■ — ^^^^^^^— — . — *— »^^^-> -^^^^^^^^^^^^^— ■B^^M^^B^^^^^^— ^^^^^^KMMB^HM IH^M.
Fig. 32.
the general form of the curve is not changed if a be given any
other positive value greater than 1. If the student retains this
picture, he should find it easy to keep in mind the following facts
when the base is greater than unity.
1. A negative number does not have a real number for its loga-
rithm.
2. The logarithm of a positive number is positive or negative
according as the number is greater than or less than 1.
3. If x approaches zero, log x decreases without limit.
4. If x increases indefinitely, log x increases without limit.
1. Plot the graph of y
2. Plot the graph of y
Hint :
3. Plot the graph of x
4. Plot the graph of x
EXERCISES
logio x by using tables to find logio x.
;log 6 x.
10g5*=l<>g!Of.
logio 5
■ k>g 6 V-
:10g 2 y.
172 LOGARITHMS [Chap. XIV.
122. . Exponential and logarithmic equations. An equation
' which involves the unknown or unknowns in the exponents is
often called an exponential equation. Thus, 2* = 16 is an expo-
nential equation in 2. In this simple example, the value of x can
be obtained by inspection ; but a table of logarithms is, in general,
of value in solving exponential equations.
Such equations arise in a variety of problems. For instance,
at compound interest, the amount of one dollar at a nominal rate
of 0.05 per annum is (see Problem 43, p. 168)
Wl + ^dollars,
in which t is the time in years and n is the number of times per
year that interest is converted. We may also write
H( 1+ TrT-
When n is increased beyond bound, the interest is said to be con-
verted continuously. It turns out that, in this case (see Art. 170),
where e is the base of natural logarithms.
Example : What will $ 1000 amount to in one year at 6$ interest con-
verted continuously ?
Solution : Let S be the amount of $ 1000 at the end of a year, then
S = 1000 e<> ■«
log S = log 1000 + 0.05 log e = 3.0217,
S = $ 1051.
An equation which involves the logarithm of an expression
that contains an unknown is sometimes called a logarithmic equa-
tion. Thus,
\og l0 2x = 3
is a logarithmic equation. To solve this equation, we may write,
from the definition of a logarithm,
2 x = 10 s = 1000.
Hence, x = 500.
Art. 122] PROBLEMS 173
EXERCISES
AND PROBLEMS
Sob
e the
following equations for
X.
1.
6» =
10.
Solution
•
•
Since 6*
= 10,
logio 6*
= logio 10 = 1.
x logio 5
X
= 1.
1
logio 5
= 1 =1.481.
.6990
2. 2«*6»-i = 4**3*+ 1 .
Solution : logio 2 s * S 2 *-* = logio 4 te 3*+*,
3x logio 2 + (2x- 1) logio 5 = 5x logio 4 +(*+ l)logi 3
= 10 x logio 2 + (x + 1) logio 3.
Transposing and collecting terms, we have
x(2 logio 5-7 logio 2 - log 10 3) = logio 3 + logio 6.
logio 3 + logio 6
~ 2 logio 6-7 log 10 2 -
logio 3
0.4771 + 0.6990
~ 1.3980 - 2.1070 - 0.4771
= -0.9916.
3. 16 = log w x*.
Solution :
16 = logio a 2 ,
(1)
From (1),
X* = 1016,
(2)
•
X = ± 108.
(3)
4. 11* = 7.
5. (0.3)* = 0.8.
6. 3** = 632.
7. 5<**-*> = ^.
8. 21**- & = 9261.
9. In a geometrical progression, I = ar"- 1 , solve for n in terms of a, Z,
and r.
10. In a geometrical progression, a = , solve for n in terms of a,
r, and «. ~~
Solve for x and y the following systems of equations.
11. 6»+v = 82, (1)
3*-v = 4. (2)
Solution : From (1) and (2),
(x + y) log 6 = log 82, (3)
(x-y)log3 = log4. (4)
174 LOGARITHMS [Chap. XIV.
Solving the linear equations (3) and (4) for x and y, we get
21og6 21og8' KJ
y 2 log5 21og6 k }
Complete by computing the value of (5) and (6) to three decimal places
by the use of tables.
12. 2'+" = 18, 13. 4*+» = 6*»,
3* = 2v. log (x + 1)= log (y - 3).
14. In how many years will $ 1000 amount to $ 2000 at a nominal rate of
0.00 per annum, (1) when interest is converted annually, (2) when it is con-
verted quarterly, (3) when it is converted continuously ?
15. Solve for x the equation e*+ er* = y ; (a) when y = 2, (6) when y = 4.
16. If fluid friction be used to retard the motion of a flywheel making
Vo revolutions per minute, the formula V= Vo er* gives the number of revo-
lutions per minute, after the friction has been applied t seconds. If the con-
stant k = 0.36, how long must the friction be applied to reduce the number of
revolutions from 200 to 50 per minute ?
17. The pressure, P, of the atmosphere in pounds per square inch, at a
height of z feet, is given approximately by the relation
P = P <r*»,
where Po is the pressure at sea level and A; is a constant. Observations at
sea level give P = 14.72, and at a height of 1122 feet, P = 14.11. What is
the value of k ?
18. Assuming the law in Problem 17 to hold, at what- height will the
pressure be half as great as at sea level ?
19. If a body of temperature Ty° be surrounded by cooler air of tempera-
ture T °, the body will gradually become cooler and its temperature, T°, after
a certain time, say t minutes, is given by Newton's law of cooling, that is,
T= ro+CT!- r )<r*,
where k is a constant. In an experiment a body of temperature 66° C. was
left to itself in air whose temperature was 15° C. After 11 minutes the tem-
perature was found to be 25°. What is the value of k ?
20. Assuming the value of k found in Problem 19, what time will elapse
before the temperature of the body drops from 26° to 20° ?
21. Solve the equation log„ (3 x + 1) = 2 for x.
22. Solve the equation logio (x 2 — 21 x) = 2 for x.
Arts. 122, 123] CALCULATION OF LOGARITHMS 175
23. In solving an important problem in the elements of mechanics, it
turns out that
t = l! ta + v^L+S (1)
k t?o
where s is the distance traversed by a moving point in time t. It is, in
general, more useful to have s in terms of t than t in terms of 8. Hence,
express s in terms of t in equation (1) .
123. Calculation of logarithms. At this point the inquiring
student will naturally bring up the question as to how the loga-
rithms of numbers are computed so as to make a table of logarithms.
Logarithms were invented by Napier about the year 1600 and
common logarithms by Briggs a little later. The invention grew
out of the comparison of two series of numbers — the one in
arithmetical progression and the other in geometrical progression.
The following theorem lies at the foundation of the early methods
of computing logarithms :
If a series of numbers are in geometrical progression, their corre-
sponding logarithms are in arithmetical progression.
Let the numbers in geometrical progression be
a, ar, ar 2 , ar 3 , —, ar n ~\ (1)
Then, log a, log ar, log ar 2 , log ar 3 , •••, log ar n ~ l (2)
are in arithmetical progression.
In this arithmetical progression, the first term is log a, and the
common difference is log r. The following example illustrates the
use of this principle in calculating logarithms. Given log 1 = 0,
log 10 1000 = 3, the geometrical mean between 1 and 1000 is
V1000 = 31.62. Then 1, 31.62, 1000 is a geometrical progression,
and 0, 1.5, 3 is the corresponding arithmetical progression, so
that 1.5 = log 10 31.62. Next, insert a geometric mean between
1 and 31.62, also between 31.62 and 1000. This gives
1, 5.624, 31.62, 177.8, 1000 as the geometrical series,
and 0, 0.75, 1.5, 2.25, 3 as the corresponding logarithms.
We could next insert between any two of these numbers a geo-
metrical mean, and find its logarithm. By continuing this process,
we could insert means until the numbers would differ by as little
176 LOGARITHMS [Chap. XIV.
as we please. This method of calculating logarithms has the dis- '•
advantage of giving the logarithms of numbers spaced unequally,
since the numbers are in geometrical progression.
Another method of obtaining logarithms, which has many ad- j
vantages over the one just given, is discussed briefly in Art. 171 j
of the chapter on Infinite Series ; but its more complete treatment
belongs to the calculus.
CHAPTER XV
PARTIAL FRACTIONS
124. Introduction. Early in the study of algebra we added
together algebraic fractions and found the sum to be a single
fraction whose denominator is the lowest common multiple of the
denominators. Thus,
6 3 = 9a? + 15
a> + l » + 2 aj2 + 3aj + 2*
It is often necessary to perform the inverse operation, that is,
to decompose a given fraction into a sum of other fractions (called
" partial " fractions) having denominators of lower degree. Thus
9 T» 11
it is easily shown that can be decomposed into -f
a 2 — 1 8+1 x—1
An algebraic fraction is said to be proper when its numerator
is of lower degree than its denominator. In this chapter it is
necessary to consider only proper fractions ; for if the degree of
the numerator is not lower than that of the denominator, the
fraction may be reduced by division to the sum of an integral
part and a proper fraction. Thus,
3if 4 -3aj 2 + 2x o o , 2
— : = 6 X 1 -f
X
X 2 —l X 2 — 1
We shall assume the possibility of decomposing any proper
fraction whose denominator contains factors prime to each other
into the ^partial fractions of the types
A B Cx + D Ex + F
x — a (x— a) p) x' 1 -f mx + n (x 2 + mx -f- n) qf
where A, B, C, D, E, F are constants, p, q positive integers, and
z*-\~mx + n an expression without real linear factors.* With
this assumption we shall show how to decompose certain classes
of fractions.
• See Chrystal's Algebra, Fifth edition, Part I, Chapter VIII.
177
178 PARTIAL FRACTIONS [Chap. XV.
125. Case I. When the denominator can be resolved into factors
of the first degree, all of which are real and different.
Example: Resolve "~ x "*" — into its simplest partial
fractions. x ""
The sum of three fractions
x 1 — x 1 -fa;
will give a fraction whose denominator is x — x*. We, therefore,
try to determine A, B, and C so that
x — a 3 x 1 — x 1 + a?
__ A(l - x) (1 + s)+ Bx(l + s)+ Ox(l — x)
~~ x(l + x)(l-z)
Then, 1 -x + 6x*=A(l--x)(l + x) + Bx(l + x)+Ox(l-x). (1)
The two members of (1) are equal for all values of x except pos-
sibly for x = 0, x = 1, x = — 1. Hence, by Art. 94, Corollary II,
they are equal for these values. In (1), making
x = 0, we obtain -4 = 1;
making x = 1, we obtain B = 3 ;
making x = — 1, we obtain = — 4.
Therefore, ±zJH±™ = 1 + _§ i_.
a; — a 3 a; 1 — a? 1-1-05
The values of A, B, and C could also have been obtained by
arranging the right-hand member of (1) in powers of x and equat-
ing coefficients of like powers (Art. 94, Corollary I) ; thus,
l-a; + 6a; 2 = ^l + (J3 + G)x +(- A + B- C)&.
A = l,
2?+C = -l,
-A + B-C=6.
These equations when solved yield A = l, B = 3, «= — 4.
In resolving a fraction into partial fractions, for every factoi
(x — a) occurring in the denominator there is a single partial f rao-
tion of the form where A is a constant.
x — a
K* I 1
Exercise. Resolve — ^ — into partial fractions.
x2-2x-35
Arts. 125, 126] INTRODUCTION 179
126. Case II. When the denominator can be resolved into reed
linear factors, some of which are repeated.
Example : Resolve ^— — ~~ ^ into its simplest partial
fractions. \ "~ /
The sum of four fractions
A.B^ C , D
x x 2 x—i (x—iy
will give a fraction whose denominator is x\x — l) 2 ; we therefore
try to determine A, B, C, D so that
6aj3-8s 2 ^4a; + l_^ , B , C , D
x 2 (x — l) 2 x x 2 x — 1 (x — lf
Then,
6^-8^- 4a? +1 = Ax(x - 1) 2 + B(x - l) 2 + Cx*(x- 1) + Dx 2
= (A + C)x*+(-2A + B-C+D)x 2
+(A-2B)x + B.
Equating coefficients of like powers (Art. 94, Corollary I) we
have
A+C=6,
-2A + B-C+D = -8,
A-2B = -4,
B=l.
Solving these equations for A, B, C, D y we find
A = -2, B = l, (7 = 8, D = -5.
Hence, 6^8^-4,-fl^ 1 8 5
2
X*(X — l) 2 X X 2 X — 1 (X — 1)
In this case, for every factor (x — a) which occurs r times there
are r partial fractions of the form
A\ A 2 ^ A r
x-a' (x-a) 2 ' "' (x-a) r>
where A^ A2, ••• , Ay are constants.
Exercise. Separate 8g8 + 4g2 + 8g + 2 into partial fractions.
z(x + 3)3
180 PARTIAL FRACTIONS [Chap. XV.
127* Case III. When the denominator contains quadratic
factors which are not repeated and which cannot be separated into
real linear factors.
Example: Resolve — — — - into a sum of partial
fractions. (* + * + 1)(* + 1) V ^
Let 3*-2 _ Ax + B , C .
(x 2 + x + l)(x + 1) 3* + a: + l s+1
Then, 34> — 2 =(-4oj+ #)(« + 1)+ G(p + x + 1),
= (^t + C)<e* +(-4 + 5+ C> + B + C!
Equating coefficients of like powers, we have
A+C=3,
A + B+C=Q,
B+C = -2,
whence, A = 2, B = - 3, (7=1 ;
and «* = » ^ 2— » ' 1
(a^ + aj + lXai+l) a^ + a+l a? + l
In this case, for every factor x 2 -f- mx -f n occurring once, there
is a single partial fraction of the form "*" - — , where A and
£ are constants. * + m* + n
Exercise. Resolve 8a?3 ~ 4a;2 + 6g ~ 1 — into partial fractions.
(x 2 + x + l)(x* — x + 1)
128. Case IV. W7ien £fte denominator contains quadratic
factors which are repeated.
Example: Resolve X \ + ^~?f~ bx 7* into partial fractions.
Let a? 4 + a? 3 -2a? 2 -5a;-4 _ A Bx+C f Jag+Jg
(a>-l)(a* + a* + l)* «-l rf + *+l («f+» + l)»'
then,
^+a5'-2a^-6»-4 = ^(a?+aj+l)H(i^ + 0)(»-l)(rf+» + l)
+ (Zte + .B)(a>-l)
=(A + B)x*+(2A+Cy+(3A + D)at
+ (2A-B-D+E)x + (A-C-E).
Arts. 127-128] INTRODUCTION 181
Equating coefficients, we have :
A + B=l,
2A4-C=1,
3^1 + Z) = -2,
2A- B-D + E = -5,
A-C-E = -4.
Solving these equations for A, B, C, D, E 9 we find A = — 1,
£ = 2, = 3, D = 1, E = 0. Hence,
a^-ho 8 — 2a* — 5a — 4_ 1 2s + 3 a;
(a;-l)(a; 2 + a; + l) 2 a;-l a^ + a; + 1 (x 2 + x + l)*
In this case, for every factor (x 2 + ma; + n) occurring r times,
there are r partial fractions of the form,
A x x + B x A& + B 2 A r x + B r
a* 2 -f mx + n (a; 2 + mx + n) 2 (a; 2 + mx -f n) r
where -4 X , A 2 , •••, -4 r , JBj, -B 2 > •••, B r are constants.
Exercise. Resolve 2g4 ~ 6g3 + 15g2 + 21g + 16 into partial fractions.
x(x2 - 2x + 4)2 -
EXERCISES
Resolve into the simplest partial fractions :
2. ' * + 4 3.
2x-x 2 -
X*
2x —
5
(X - 1) (X
-2)
X2 + 1
X 2 — x— 6 2x-x 2 — x 3 x 2 — 2x — 3
5x l — 6x — 5 2x— 5 - x 2 + l
3* *: zrr - : ~* ©•
(x - l)3(x + 2) (x - 1) (x - 2) x(x - 1) (x - 2)
? 5x + l 8 x2 + l x2-4x + 5
x*-l # ' x(x-l)2' * (aj_i)2(x2 + l)
10 x* + x 8 + 2x 2 -x-l X1 x + 4
X 8 — X
12 5x2 -f 8x + H
(x2 + l)(x + l)(x-3)'
14. PC 2 *" 1 ) - 15.
x(x + 2)(x - 3) (x2 _ 4x + 3)(x - 2)
16. — *— . 17. -i-. 18. -*±I~
x» + 4* a 3 — 1 x'(x — 1)
19. 1 + 7 — * 20. 2x + 5
(X-
-1)(X2-
2x2 — 5;
5x + 6)
c + 7
(X-
-3)(x2-
X 2 -f x-
■ 3x + 2)
-3
(1 + 8 x) 2 (l - 10 x) (x - l)3(x - 3)
182
PARTIAL FRACTIONS
[Chap. XV.
21.
23.
25.
27.
29.
31.
33.
35.
37.
5-9*
(l-3x)»(l+x)
x* — 2x» + 3s 2 — x
(x-l)(x»-x+l)» #
2x« — 8x l — 7x + 1
x 4 + a? — x — 1
3x-l
x»(x + l)*'
x3 -2x»-6x-21
x*-4x-6
(5 j3-8a 2 — 4a; + 1
x*(x - 1)* ■ "
45 + 86 as -a;*
x*-6x*-27*
x*-2
(x* + x + l)(x* + x + 2)*
17 - 11 x + 7 a*
(l+x + x*)(l-x)»'
22.
24.
26.
28
32.
34.
36.
38
9 — 2x
(x + 2)(x* — 2x+6)
6x» + 2s» + 2x — 2
X«-l
4x»-18s»
— ^^— ■■ ■»■ ■•
(2x-3)*
20x»-2s»
x*-16
30 5x» — 4x+16
' (x-8)(x*-x + l)*\
3x» + 19x» + 35x
(* + 2)»
x» + s + 2
(x-l)*(x*-x + l)*
x» + 2as» + 2
(x*+l) 2
4x* + 8x» + 6as» + 6as + 5
(8x + 2)(x* + l)*
CHAPTER XVI
PERMUTATIONS AND COMBINATIONS
129. Introduction. Two positions are to be filled in an office
— one that of stenographer and the other that of messenger.
There are 12 applicants for the position of stenographer, and 3
for that of messenger. In how many ways can the two positions
together be rilled ?
The position of stenographer can be filled in 12 ways, and with
each of these there is a choice of 3 messengers. Hence, the two
positions can be filled in 12 x 3 = 36 ways.
This example illustrates the following
Fundamental Principle. If one thing can be done in m dif -
ferent ways; and if, after this is done in one of these ways, a second
thing can be done in n ways, thenHie two together can be done in
the order stated in mn ways.
For, corresponding to each of m ways of doing the first thing,
there are n ways of doing the second thing. In other words,
there are n ways of doing the two together for each way of doing
the first thing. Hence, there are in all mn ways of doing the
two things together.
A convenient and evident extension of the fundamental princi-
ple may be stated in the following form :
-JfcMlk<LtMjHl-Wn be done in m x waySj aj&econd in miwaySjja third
in m a w ay s, an d so on } the number of different ways in which, they
can be done when taken all togdhexJ^tha order stMed-is>.m t mtf*f*'».
130. Meaning of a permutation. Each different arrangement
which can be made of all or part of a number of things is called a
permutation.
By the expression " number of permutations of n things taken
r at a time " is meant the number of permutations consisting of r
183
1 y "2—
184 PERMUTATIONS AND COMBINATIONS [Chap. XVL
things each which can be formed from n different things. Thus,
the permutations of the letters abc taken all at a time axe
abc, a cb, b a c, b c a, c a b 9 cb a
The permutations of the four letters abed taken three at a
time are ;
abc
b a c
cab
dab
a cb
b c a
c b a
db a
a c d
bed
cb d
db c
ad c
b d c
cdb
deb
ab d
bad
cad
da c
ad b
b d a
cda
d c a
131. Permutations of things all different. The special cases
just considered lead us to the problem of deriving a formula for
the number of permutations of n things taken r at a time. The
symbol n P r is used to represent this number.
The number of permutations of n different things taken r at a
time is
n P r = 7i (n — 1) ••• (n — r + 1).
The number n P r required is the same as the number of ways
of filling r different positions with n different things. We may
represent the n things by a 1? a 2 , •••, a n and ask how many permu-
tations of r letters can be formed from them. For the first place
there is a choice of n letters, for the second a choice of n — 1, for
the third of n — 2, and so on. For the rth place there is then a
choice of n — r + 1 letters. It follows (Art.. 129) that
w P r = n(n — 1) ••• (n — r + 1). (1)
When r = n, (1) becomes
n P n = n(n — 1) •••2' 1 = nU (2)
Tliat is, the number of permutations of n things taken n at a time
is n!
132. Permutations of n things not all different. Consider the
number of permutations of the letters in the word book. It
gives no new permutation to interchange the o's. Let P be the
Arts. 131, 132] EXERCISES AND PROBLEMS 185
number of permutations. If we should replace oo by dissimilar
characters o x o 2 , there would be 2 ! permutations of o x o 2 corre-
sponding to each of the P permutations. But if the letters were
all different the number of permutations would be 4 !. Hence,
4! = 2! P, P = |^=12.
' 2!
This example illustrates the
Theorem. If P is the number of permutations of n things taken
all at a time, of which n x are alike, n 2 others alike, n z others alike,
and so on, then ?
p — lii
w 1 !n 2 !w 3 ! •••
To establish the theorem, suppose we should replace rij like
things by n x unlike things, there would be P • n x ! permutations
obtained from the original P permutations. In each of these
permutations there would be n? things alike, and n z others alike.
Similarly, replacing the n 2 like things by n 2 dissimilar things, we
get P • «x ! • 7*2 ! permutations in each of which there would be n 8
alike. Continuing this argument, we find that the number of
permutations of n things taken all at a time, when n x are alike,
n 2 others alike, n z others alike, and so on, is given by
n!
P =
n x I n 2 1 n 3 !
EXERCISES AND PROBLEMS
1. How often can 6 ball players take seats together on a bench without
sitting twice in the same order ?
2. In how many different orders can the colors violet, indigo, blue, green,
yellow, orange, and red be arranged when taken all together?
3. How many different permutations can be made of the letters of the
word "stone " when taken 3 at a time ?
4. Five different positions are to be filled, and there are 20 applicants
each applying for any one of the positions. In how many ways can the
positions be filled ?
5. In how many ways can ten books be arranged on a shelf if the places
of two of them are fixed ?
6. Given n Pt = 6»p8, find n. ^
186 PERMUTATIONS AND COMBINATIONS [Chap. XVI.
7. How many permutations can be made of the letters of the word
44 Illinois " ? Of the word " Mississippi " ?
8. In how many ways may a party of 8 people take their places at a
round table ?
9. How many different combinations may be struck from 8 bells if
only 3 are struck at one time ?
10. Four persons enter a railway carriage in which there are 6 seats. In
how many ways can they take their places ?
11. Find the number of permutations of letters in the word "level."
12. How many different numbers of six figures each can be formed by
permuting the figures 233455 ?
13. Write all the permutations of the letters abed, when taken (1) two at
a time, (2) four at a time.
133* Combinations. A set of things or elements without refer-
ence to the order of individuals within the set is' coiled a com-
bination.
Thus, dbc y acb, bac, bca, cab, cba are the same combination.
By the " number of combinations of n things taken r at a time "
is meant the number of combinations of r individuals which can
be formed from n things.
Thus, the combinations of a b c taken two at a time are ab,
ac. be. ? v '-^
134. Combinations of things all different. Let m C r denote the
number of combinations of n things taken r at a time Then a
formula can be derived for n C r by establishing the relation be-
tween n G r and n P r .
Take one combination of r things ; with this r ! permutations
can be made. Take a second combination ; with this r ! permu-
tations can be made. There are thus r ! permutations for each
combination. Hence, there are in all n C r rl permutations of n
things taken r at a time. That is,
„C r .r! = „P r ,
p
whence n O r = !L — r •
m r\
.> v
Arts. 132-136] COMBINATIONS 187
Since . n P r = n (n - 1) • • • (n - r + 1), (Art. 131)
we have wCr = n(n-l)...(n-r + l) ,
Multiplying numerator and denominator by (n — r) !, we get
r - n '
r ! (n — r) !
n:
f
Since „C„_ r = > i- »- 1 )-( r + 1 ) = ,
(n — r) ! (ra — r) ! r !
it follows that the number of combinations of n things taken r
at a time is the same as the number taken n — r at a time.
135. Binomial coefficients. It may be noted that the formula
for n C r is the coefficient of the (r + l)st term of the binomial
expansion (a + x) n . The binomial theorem for positive integral
exponents may therefore be written in the form
(o+*)"=«" + »Cia»- 1 a?+ ll 2 a"-W+ - + n C n ^ax n ~ l + n C n x\
136. Total number of combinations. The total number of com-
binations of n things taken 1, 2, 3, •••, w at a time is 2 n — 1. If we
write the binomial theorem as in the last section, we obtain
(1 + *)• = 1 + n O x x + n C&> + .- + n CUs»-i + n C n x».
Putting x = 1, we get
2 n — 1 = n Ci -h W C 2 H + n^n-l + nPn*
EXERCISES AND PROBLEMS
1. A woman with 10 friends to invite can have how many dinner parties
with 6 guests without having the same company of 6 twice ?
2. A man and his wife wish to invite 4 men and 6 women to dinner,
but find they can entertain only 6 guests at one dinner. In how many
ways can they invite 3 men and 3 women out of this group ?
3. A man has 6 friends. In how many ways can he invite one or more
of them to dinner ?
4. How many different assemblages of 1000 persons can be selected
from an assemblage of 1002 persons ?
5. In a certain town, there are 4 aldermen to be elected, and there are
S candidates. How many different tickets can be made up ?
188 PERMUTATIONS AND COMBINATIONS [Chap. XVI.
6. Find 20^17 J 12 CV •
7. Given n C 4 = 210, find n.
8. Given n P r = 272, and n C r = 136, find n and r.
9. How many different sums of money can be formed with a penny,
a nickel, a dime, a quarter, and a half dollar ?
10. How many different sums may be formed with a penny, a nickel, a
dime, a quarter, a half dollar, and a dollar ?
*
11. There are five letter boxes in a town. In how many ways can a
person post two letters ?
12. In how many ways can 5 books be selected from a set of 11 ?
' 13. A committee of 6 is to be chosen from 7 Englishmen and 4 Americans.
If the committee is to contain at least 2 Americans, in how many ways
may the committee be chosen ?
14. Trove that n C r = n C n _,..
15. Make use of the theorem «£>=*»£*-«■ to evaluate 100 Cm-
16. Out of 15 consonants and 4 vowels how many words can be formed
each containing 3 consonants and 2 vowels ?
17. How many straight lines can be drawn through pairs of points selected
from 10 points no three of which are in the same straight line ?
18. In how many ways can a pack of 52 playing cards be divided into 4
hands, the order of the hands, but not the cards in the hands, to be regarded?
19. There are 5 trails to the top of a mountain. In how many ways may
a person go up and return by a different trail ?
20. In how many ways can 8 books be arranged on a shelf so that two
particular books will not be together ?
21. How many baseball teams of 9 men each can be chosen from 15
players of whom 8 are qualified to play in the infield only, 5 In the outfield
only, and 2 in any position (battery included in infield) ?
22. How many different combinations can be formed with the following
weights ?
1 decigram, 1 gram, 1 10-gram,
1 2-decigram, 1 2-gram, 1 20-gram,
1 3-decigram, 1 3-gram, 1 80-gram,
1 5-decigram, 1 5-gram, 1 50-gram.
23. On how many nights may a different guard of 4 men be posted out
of 16 soldiers ? On how many of these nights will any particular man be
on guard?
Abt. 136] EXERCISES AND PROBLEMS 189
24. With 4 white balls, 6 black balls, and 9 red balls, how many different
combinations can be formed each consisting of 1 white ball, 3 black balls,
and 5 red balls ?
25. A town which has 11 physicians, 13 teachers, and 8 lawyers can form
how many committees each consisting of 3 physicians, 4 teachers, and 2
lawyers?
26. A company consists of 100 soldiers. In how many ways is it possible
to leave 60 men to garrison a fort and to divide the remainder into two
scouting parties of 20 men each ?
CHAPTER XVII
PROBABILITY
137. Meaning of probability. If a bag contains three white
and live black balls and one ball is drawn out at random, what is
the probability that this ball is white ?
The event in question is said to happen if a white ball is drawn,
and to fail if a black ball is drawn. The number of ways in
which the event may happen is 3, and the total number of pos-
sible ways in which it may happen and fail is 8. For this reason,
$• is said to be the probability of drawing a white ball. This
illustrates the following definition of probability :
If all the happenings and failings of an event can be analyzed into
r + s possible ivays each of which is equally likely; and if in r of
these ways the event will happen, and in s of them fatly the proba-
IT
bility that the event will happen is and the probability that it
s r + s
will fail is •
r + s
Corollary. The sum of the probability that an event will hap-
pen and the probability that it will fail is 1, which is the symbol for
certainty.
In applying the definition of probability, the fact should not be
overlooked that the ways are assumed to be "equally likely."
To illustrate the need of precaution in this matter, consider the
following
Example : What is the probability that a man, A, in good
health will die within the next 24 hours ?
We might argue that the event can happen in only one way
and fail in only one way, and that the probability that A will die
in the next 24 hours is therefore |. What is the flaw in this
argument ?
190
Arts. 137-139] EXPECTATION OF MONEY 191
The expression "equally likely" indicates that we have no
more reason to expect the event to take place in one way than in
any other.
138. Probability derived from observation. If it be observed that
an event has happened m times in n possible cases (n a large num-
ber) ; then, in the absence of further knowledge, it is assumed that
the best estimate of the probability that the event will happen on a
m
given occasion in question is —, and that confidence in this estimate
increases as n increases.
Such estimates of probability are of much practical value in
insurance and statistics. For example, according to the American
Experience Table of Mortality, of 85,441 men living at the age of
30, the number living ten years later is 78,106. The probability
that a man aged 30 will live ten years is taken to be %g|j ^.
139. Expectation of money. If p is the probability that a person
wiU win a sum of money m, we may define his expectation aspm.
PROBLEMS
1. A bag contains ten times as many white balls as black balls, and one
ball is to be drawn out at random. What is the probability that the ball
drawn is white ?
y 2. Five coins are tossed. What is the probability that exactly two of
them are heads ?
Solution : Since each coin can fall in two ways, the five can fall in
2 5 = 32 ways. The two coins can be selected from the five in 5C2 = 10
ways. Hence, the probability is Jj.
3. From a bag containing 6 black and 4 white balls, 3 are drawn at
random. Find the probability that 2 are black and 1 is white.
4. If from a suit of 13 cards, 2 cards are drawn, what is the probability
that an ace and a king are drawn ?
5. A gambler is to win $ 30 if an ace is thrown with a single die ; what is
the value of his expectation ?
6. According to a mortality table, it appears that of 100,000 persons at
the age of ten years, only 69,804 reach the age of fifty. Find the proba-
bility that a child aged ten will reach the age of fifty years.
192 PROBABILITY [Chap. XVII.
7. From a committee of 3 sophomores, 4 juniors, and 6 seniors, a sub-
committee of 4 is selected by lot. Find the probability that it will consist :
(1) of 2 juniors and 2 seniors ; (2) of 1 sophomore, 1 junior, and 2 seniors ;
(3) of 4 seniors.
8. From a bag containing 10 five-dollar bills and 20 two-dollar bills, I
have the privilege of drawing a bill at random. What is the value of my
expectation ?
140. Events of a set are said to be independent or dependent
according as the occurrence of any one of them does not or does
affect the occurrence of others in the set. They are said to be
mutually exclusive when the occurrence of any one of them on a
particular occasion excludes the occurrence of any other on that
occasion.
Independent events. The probability that all of a set of independ-
ent events will happen on a given occasion when all of them are in
question is the product of their separate probabilities.
Let P be the probability that all events of the set will happen
and 2h, lh,"*Pr be their separate probabilities. It is to be
proved that P = p x p 2 ••• p r - Suppose the event corresponding to
p x can happen in a x ways and fail in 6j ways : the event corre-
sponding to p 2 can happen in a 2 ways and fail in &j ways ; and
so on.
Then > *-sh^« ft -;*+S' ~» A -S+r
By the fundamental principle (Art. 129) all the separate events
can happen together in a x a 2 ••• a r ways out of
(«i + &i) (a 2 + b 2 ) ». (a r + 6 r )
possible ways of happening and failing.
Hence P = W«*r
nence, r (a> + ^)(a 2 + & 2 ) ... (a r + & r )
= PlP2-Pr-
Dependent events. If the probability of a first event i*P\> and if
after this has happened, the probability of a second event is ft; then
the probability that both events will happen in the order specified is
Pi P* The extension to any number of events is obvioua
Arts. 139, 140] EVENTS 193
Exclusive events. If the separate probabilities of r rnuttiaUy ex-
clusive events be Pi,P2> " m ,p r 9 the probability that one of these events
will happen on a particular occasion when all of them are in ques-
tion tepi+p* H hlV
This proposition may be regarded as an immediate consequence
of the definition of probability for mutually exclusive events.
To illustrate, the probability of throwing an ace or a deuce in
single throw is clearly £ + £ = £.
PROBLEMS
1. If the probability is \ that the age of a man selected at random from a
group of men is between 20 and 26 years, and \ that it is between 25 and 35,
what is the probability that his age is between 20 and 35 years ?
2. What is the probability of throwing an ace with a single die in two
trials?
3. The probability that A will live ten years is \ and the probability that
B will live ten years is J. What is the probability that they will both live
ten years ?
4. Find the probability of drawing 2 white balls in succession from a bag
containing 5 white and 6 black balls if the first ball drawn is not replaced
before the second drawing is made.
5. One purse contains 9 coins consisting of 2 dimes, 3 quarters, and 4
half dollars. If one coin is drawn at random from the purse, what is the
probability of its being either a quarter or a half dollar ?
6. In a bag are 4 white and 6 black balls ; find the chance that out of 5
drawn, 2 and only 2 are white.
7. Find the probability of throwing at least 8 in a single throw with two
dice.
8. A traveler has three railroad connections to make. If the probability
is } that he would make any particular connection taken alone, what is the
probability of his making all three connections ?
9. The probability that a man of a certain age will die within 20 years is
0.2, and that his wife will die within that time is 0.15. What is the proba-
bility that at the end of 20 years (1) both will be dead ? (2) both will be
living ? (3) the man will be living and his wife dead ? (4) the man will
be dead and his wife living ?
t
>4
194 PROBABILITY [Chap. XVII.
141. Repeated trials. If p is the probability that an event tvtit
happen in any single trial, then n C r p r q n ~ r is the probability that this
event will happen exactly r times in n trials, where q = l—p is the
probability that the event will fail in any single trial.
For, the probability that it will happen in r specified trials and
tail in the remaining n — r is p r q n ' r (Art. 140), and r trials can be
selected from n trials in n C r ways. These ways being mutually
exclusive, we have, by Art. 140, that the probability in question
is nC r p r <r~ r -
It will be observed that n C r p r q n ' T is the (n — r+ l)th term of
the binomial expansion of (p + q) n .
We next inquire into the probability that an event such as is
described above happens at least r times in n trials. The event
happens at least r times if it happens exactly n, n — 1, n — - 2, •••,
or r times in n trials.
Hence, we have the following
Thkokkm : The probability that an event wiU happen at least r
times in n trials is p H + n C n ^ip n ~ x q + n C , I ,_2.P*~V + *" +«C»-rl' r 9"" r '
This expression is the first n — r -f- 1 terms of the binomial ex-
pansion of (p -f- q)\
PROBLEMS
1. In tossing a coin, what is the probability that in six tosses (1) exactly
three result in heads ; (2) at least three result in heads ?
2. According to the American Experience Table of Mortality, out of
100, 000 persons living at the age of 10 years, 01,914 are living at the age of
21 years. Each of live boys is now 10 years old; what is the probability
that exactly four of them will live to be 21 ? That at least four of them will
live to be 21 ?
3. A's chance (probability) of winning any single game against B is f.
Find the chance of his winning at least three games out of seven.
4. In tossing ten coins, what is the probability that at least four of them
will be heads ?
5. Find the expectation of a man who buys a lottery ticket in a lottery of
100 tickets where there are four prizes of $ 100, ten of $60, and twenty of 16.
6. If, in the long run, one vessel out of every 60 is wrecked, find the
probability that of 6 vessels expected (1) exactly 6 will arrive safely, (3) at
least 6 will arrive safely.
Art. 141] PROBLEMS 195
7. Which is the greater, the probability of throwing at least one ace in
six trials of throwing a die, or the probability of throwing at least one head
of a coin in two trials ?
8. According to the American Experience Table of Mortality, out of
89,032 persons living at the age of 25 years, 26,237 will be living at the age of
76. A husband and wife are 25 each at the date of marriage ; what is the
probability that they will live to celebrate their golden wedding ? What is
the probability that at least one of them will be living 50 years after the
marriage?
9. A machinist works 300 days in a year. If the probability of his
meeting with an accident on any particular work day is T u^, show that the
probability of his entirely escaping injury for a year is approximately }.
(Use logarithms.)
10. A card is to be drawn from a whist deck and replaced by a joker,
and then a second card is to be drawn. What is the probability that both
cards drawn will be aces ?
11. A bag contains 6 balls. A part or all of the balls are drawn. Under
the condition that the ways in which r balls can be drawn from 6 balls are
equally likely for r = 1, 2, 3, •••, 6, what is the probability of drawing an
even number of balls ?
12. An Italian nobleman, interested in gambling, had, by continued ob-
servation of a game with three dice, noticed that the sum 10 appeared more
often than the sum 9. He expressed his surprise at this to Galileo and asked
for an explanation. Find the probability of (1) the sum 10, (2) the sum 9,
and explain the difficulty of the nobleman.
13. A and B take turns in throwing with a single die, A throwing first.
The one who throws an ace first is to receive a prize of $ 66. What are the
values of their expectations ?
14. In the population of continental United States as given in the census
of 1900 there were 75,994,575 persons of whom 64,763 were blind and
89,287 were deaf. What is the best estimate from these figures of the proba-
bility that a person chosen at random from such a population would be (1)
blind ? (2) deaf ? (3) both blind and deaf if the two were independent ?
Estimate to the nearest integer the number in the total population that
would be both blind and deaf if it were correct to assume the two defects
independent.
CHAPTER XVIII
DETERMDfAlfTS
142. Extension of the determinant notation. Determinants of
the second and third orders were used in Chapter V in the solution
of systems of linear equations in two nTiif-ttrfffl ifiiknimrtitt ^ ^nd
a determinant of the second order was so. dsfiwdw.tibAt.th9 pair
of values
lei V
X —
Cl
V
«1
M
flj
*>,
y=
satisfies the system of equations,
provided
*'0
(1)
00
Analogously, a determinant of the third order was so defined
that the set of values
35= -
dx
h
Cl \
dj
b t
Cj
d s
b,
%•
<h
bi
?l
a,
b t
c s
a»
h
Cs
y —
a x di Ci
a 2 d* <%
03 d s C3
<h
&i
<a
a 2
b,
Ci
03
&s
Cs
z =
<h
h
d,
a»
b,
*
<h
h
*,
<h
h
<i
Of
bt
<*
<h
6.
«t
(3)
196
s
Art. 142]
EXTENSION
(4)
satisfies the system of equations Q '
ap + ba + cg^dt,
provided I c*i l>, c : i
[«» i>a Cal
The determinant notation is extended in the present chapter to
the solution o£ systems of linear equations containing more than
three unknowns, and to certain problems of elimination.
It will be observed that each term (e.g. a,£» t and a s b&) in the
expansions, > . .
1 =oA-oA» (5)
]<* *\ -■"
Ui b* ?« Pf= «Ac» + o^gc, + a&c, - a^A - aACs - «iV» (6)
of determinants of orders 2 and 3 respectively, consists (except
for sign) of the product formed by taking one and only one ele-
ment from each row and column. This fact suggests the exten-
sion of determinants to represent certain expressions in n*
elements by m eans of an array,
6 a C, d A •■■ ?a
&* <* <*»•■■ k
o 4 c ( dt •■- 1,
a. K
<L - L
CO
where the expansion is to consist of terms which are products
formed by taking one and only one element from each row and
column, and where the signs of terms are to be consistent with
the special cases of n = 2 and n — 3.
A determinant such as (7) which has n rows and m columns is
called a determinant of the nth order. The diagonal afi^ ■■■ Z„ is
called the principal diagonal.
DETERMINANTS [Chap. XVIII.
To fix the signs of terms in the expansion of a determinant
of any order, the notion of an inversion is introduced. If, in an
arrangement of positive integers, a greater precedes a less, there
is said to be an inversion. Thus, in the order 12543, there are
three inversions :. 5 before 4, $ before 3, 4 fjefore 3". In 23M576,
there are four inversions. When applied to any term in the
expansion of a determinant such as (7), we say there is an inver-
sion if the order of the subscripts presents an inversion when the
letters (apart from subscripts) have the order abed ••• I of the
principal diagonal. With respect to determinants of orders 2 and
3, it may be observed that the number of inversions is even when
the term is positive, and that the number of inversions is odd
when th e term is neg ative.
Consistently with these conditions, we lay down the following
Definition. A square array of n 2 elements, such as has been
considered in the cases n = 2 and n = 3, is called a determinant of
the nth order. It is an abbreviation for the algebraic sum of all the
different products that can be formed by talcing as factors one and
only one element from each column and each row of the array, and
giving to each term a positive or a negative sign according as the
number of inversions of the subscripts of tlie term is even or odd,
ivhen the letters have the same order as in the principal diagonal.
It may be added that if in any case the number of inversions
in the principal diagonal is different from zero, the sign of a term
is + or — according as the number of inversions in its subscripts
differs from the number in the principal diagonal by an even or
odd number. Since the subscripts fix the signs of terms, it may
appear necessary to carry subscripts along in any numerical case,
but we shall derive other modes of expansion (Art. 144) which
make this unnecessary. We shall, in general, use the Greek let-
ter A to represent a determinant.
143. Properties of determinants. The following theorems em-
body the most important properties of determinants.
I. TJie expansion of a determinant A of wder n contains n ! terms.
Since the number of terms is the same as the number of
permutations of the subscripts 1, 2, 3, •••, n, the number is n\
(Art. 131).
Abt. 1431
DEFINITION
199
II. If in a determinant A corresponding rows and columns are
interchanged, the expansion is unchanged.
Thus,
HI. If two rows (or columns) of a determinant A are interchanged,
the sign of 'the determinant is changed.
Let us take for simplicity a determinant of the third order, but
the argument used will clearly apply to any determinant. Thus,
<h
*>l
Cl
«1
a 2
<h
Og
&2
c 2
=
&1
h
h
«8
h
%
<a
c 2
Ps
a,
&i
Cl
a 8
\
c 8
<h
b 2
c 2
= —
a 2
h
c«
Og
h
%
Oi
6.
<h
In the first place, interchanging two adjacent rows will simply
interchange two adjacent subscripts in each term of the expan-
sion. This will change the sign of every term of the expansion.
Consider next the effect of interchanging any two rows (or
columns) separated by m intermediate rows. The lower row can
be brought just below the upper one by m interchanges of adja-
cent rows. To bring likewise the upper row into the original
position of the lower row, m + 1 further interchanges are neces-
sary. Hence, interchanging the two rows in question is equiva-
lent to 2m + 1 interchanges of adjacent rows. Since 2m + 1 is
an odd number, this process changes the sign of the determinant.
IT. If a determinant A has two rows (or columns) identical, its
value is zero.
If we interchange two rows, we obtain, by III, — A. But since
the interchange of two identical rows does not alter the deter-
minant we have A = — A
that is, 2 A = 0,
or A = 0.
Y. If all the elements of a row (or column) of A are multiplied by
the same number m the determinant is multiplied by m.
For, one element from the column multiplied by m must enter
into each term of the expansion of A.
200
DETERMINANTS
[Chap. XVIII.
Yl. If one row (or column) of A has as elements the sum of two
or more numbers, A can be written as the sum of two or more deter-
minants. That is,
A =
«i4- a i + a i" &i Ci
a 2 + a 2 +a 2 " b 2 c 2
Os + as' + tts" & s <*
a x bi Cj
a/ bi Cx
—
a 2 o 2 c 2
a s 6 3 Cg
+
aj b 2 c?
a* 63 Cz
+
a," 6 2 Cs
a," 63 Cj
This theorem is evident for this special case, since, each term
in the expansion of A is evidently equal to the sum of the corre-
sponding "terms of the three determinants. Similarly, we can
prove the general case.
VII. The value of any determinant A is not changed if each
element of any row (or column), or each element multiplied by any
given number, m, be added to the cowesponding element of any other
row (or column).
By V and VI,
Oj + ma s 02 a 3
b x + mb s b 2 b 8 = 6 2 b 2 b z + m'
Ci + mc 3 c% c y
+ 0,byIV.
a x
a* 2
<h
&i
b*
h
Ci
c,
• ^3
<h
a s
<h
&i
h
h
<h
Of
«3
h
a,
Ot
<*
c%
«•
'3
Likewise, the theorem can be proved for a determinant of any
order.
144. Development by minors. If we suppress both the row
and column to which any element, say c k , of the determinant
belongs, the unsuppressed elements form a determinant called
the first minor of c k , and which we shall denote by the capital
letter O k . Thus, in 7 ,
a 2 o 2 Cj
#3 ^3 ^3
the minor of b 9 is
a 3 C3
This notation means that the determinant is multiplied by
Art. 144]
DEVELOPMENT BY MINORS
201
A determinant A may be expressed in terms of the elements
c& <fc •••, c n of a column (or row) and their first minors as follows :
Form the product of each element such as c k in the column by the
corresponding minor C k . Give each of the products thus formed a
positive or a negative sign according as the sum of the number of the
row and the number of the column containing c k is even or odd, and
take the algebraic sum of these results. This sum is equal to A.
Thus,
Oi bi Ci
Oj b 2 c 2
a a b» Ct
l 3
'3 ^3
&*
c 2
»i
<h
h
Ci
= a x
m
— a 2
+ «3
&8
<*
h
<h
l>2
c*
If we can establish this theorem, we have a systematic method
for expanding any determinant, since the first minors of A are
again determinants which can be expressed in terms of their own
minors. This process can be continued until we have the expan-
sion of A.
The proof of the theorem involves two steps :
(1) The coefficient of the leading element a x in the expansion
of A is the minor, A l9 of a v For, A x is a determinant of order
n — 1 in elements b 2 , b 3 , ••• b n , ••• and its expansion therefore con-
tains a term for each permutation of 2 3 4 ••• n. As to the
signs of terms, the number of inversions is not changed by pre-
fixing a x .
(2) The coefficient of any element c k in the expansion of A is
its minor C k with a + or a — sign, according as the sum of the
number of the row and the number of the column containing c k
is even or odd. If c k is in the Ath column and &th row, we can
bring it to the leading position (column 1, row 1) without disturb-
ing the relative positions of elements not found in column h or
row k. This is done by interchanging the column in which c k
stands with each preceding column in turn until c k is in column 1,
and the row in which c k stands with each preceding row in turn
until c k is in row 1. In making these changes, the sign of the
determinant is, changed, h — l-f& — 1 = &+&— 2 times (Art. 143,
III). Hence, if A r denotes this determinant with c k as the leading
letter, a' = <- !)*+*-* a = (- 1)*+*A.
202
DETERMINANTS
[Chap. XVIII.
Let C k be the minor of c k in A r . By (1), the sum of the terms
in the expansion of A' which contain c k is CjjCf^ Since the
minor of c k in A' is the same as in A, the coefficient of c k in the
expansion of A is ( — l)***^ This establishes the second step.
A =
1. Develop
Solution :
112
2 3
3 2 1
1-11
2. Develop
A =
1
4
1
= 1
A =
EXERCISES
112 1
2 3-4
3 2 10
1-11 1
2 3-4
2 10
-111
-0
-1
12 1
2 10
-111
12 1
2 3-4
2 1
2 12 1
3 3 14
4 4 3 2
6 6 2 1
+ 8
1 2
2 8
-1 1
1
-4
1
= 48.
Hint : Subtracting column 2 from column 1, we have by VII, Art 148,
A =
112 1
3 14
4 3 2
6 2 1
3 14
=
4 3 2
6 2 1
3. How many inversions are there in the arrangement 14528687?
4. Develop
1 2
3 4
1 1
2 1
1
-1 2
2 1
3 1
5. Develop
2 7 6 6
117 8
16 3 4
4 7 6 6
6. Show that A =
1
a
a*
1
b
6 2
1
c
= (a-6)(6-c)(o-a>
Hint : When a = 6, two columns are identical so that A vanishes, and
by the factor theorem, Art. 93, a — b is a factor of A.
-Vara. 146, 146] SOLUTION OP EQUATIONS
203
1
a a 2
a 8
7. Factor A =
1
1
b b 2 W
c c 2 c 8
into linear factors.
•
1
d d 2 d 8
1
a b
8. Show that A =
1
a 2 b 2
= a6(a~6)(a-l)(l-6).
1
a 8 6 8
a
6 c
a a a
9. Factor A =
a 2
6 2 c 2
10. Factor A =
abb
a 8
6 8 c 8
a b c
145. We shall now establish a theorem of determinants which
is useful in performing the eliminations required in the solution
of equations in two or more unknowns.
Theorem. In developing a determinant by minors with respect
to a certain column (or row), if the elements of this column (or row)
are replaced by the corresponding elements of some other column (or
row), ike resulting expression vanishes.
For example, we have, by Art. 144,
A =
«1
hi
Ci di
02
b 2
C 2 U>2
<h
h
Cj a 3
a 4
h
c 4 d 4
= a l A 1 — a 2 A 2 + a 3 A 3 — a A A^.
We are to prove that
b x A x — b^A 2 4- b 3 A 3 — b A A 4 = 0. (1)
The left member of (1) is equal to the expression of the de-
terminant derived from A by replacing the column of a's by the
ft's with corresponding subscripts. But this gives a determinant
with two columns identical, which therefore vanishes (Art. 143,
IV). The same method of proof can manifestly be applied to a
determinant of any order.
146* Systems of linear equations containing the same number of
equations as unknowns. In Chapter V, we used determinants to
express the solution of simultaneous equations containing two and
three unknowns. We are now in a position to make use of deter-
minants to solve a system of n linear equations in n unknowns.
204
DETERMINANTS
[Chap. XVIII.
For simplicity of notation, take n = 4, and consider the system
of equations
age -f- &# + c^ + dju) = ft*
a& 4- b$ + P^ + d 3 w = k^
a A x + b$ + c& + d+w = A;*,
a)
(2)
(3)
(4)
to be solved for x, y, z, and w if a solution exists. It is conven-
ient to call the determinant of the coefficients of the unknowns,
h Ci <*i
A =
d*
d s
Q>2 ®2 ^2
<h h C 8
a 4 6 4 c 4 d A
the determinant of the system of equations.
Case I. When A =£ 0,
As above, let -4,, .4 2 , •••, B 1? B 2 , ••• be the minors of a b a„ •••,
&i> &2> ••" respectively. Multiplying both members of (1), (2), (3),
and (4) by A lf — A% A& and — A^ respectively, we obtain
A x OiX- + A 1 b 1 y + A &Z -f- A^w = AJkto (5)
— ^a^ — AJbgj — A&& — A^dyo = —AJc* (6)
^a» + <AJ>& + Agp& + A^ 3 w = Age* (7)
— -4 4 a 4 # — AJj^y — .4 4 c 4 z — A A d A w = — -4^ (8)
Adding (5), (6), (7), (8), we obtain A for the coefficient of
x (Art. 144), and zero for coefficients of the other unknowns
(Art. 145). That is,
Similarly, A - y = - 2*^ + ^A - % 4- BJc* (10)
A-z = C& - CJc 2 + Cfo - Cfa (11)
and A-t0 = -A^ + A*»--<BA + Afa- (12)
If, in A, we replace the a's by fc's and expand, we have the
right-hand member of (9). Similarly, replacing the Vb 9 tfs, and
d's respectively by fc's, we have the right-hand members of (10),
(11), and (12). It follows that
Art. 146]
SOLUTION OF EQUATIONS
205
x=
z =
h h ci
<?i
«1
«/j Cj Uj
1C2 0% c%
d 2
a 2
A>2 ^2 ^*2
#8 O3 c$
C?8
<h
A?3 C3 U3
k 4 b 4 c 4
d 4
■> y =
a 4
& 4 cfc d 4
A
A
ty 61 ki
<h
«i
&i (a A?j
0% 0% 1C2
d t
«2
#2 Cg A?2
0$ 8 AJ3
d$
«8
^3 C 3 ^3
a 4 b 4 k 4
d 4
• tn —
a 4
b 4 c 4 Jc 4
is a solution, and the only solution, of (1), (2), (3), (4).
The following rule may then be applied to obtain the solution
of any system of n linear equations containing n unknowns when
A, the determinant of the system, is not zero :
Any unknown is equal to a fraction whose denominator is the
determinant of the system, and whose numerator is the determinant
formed from the determinant of the system by substituting for the
coefficients of the unknown sought the corresponding known terms
with that sign attached to each known term which it has when on the
side of the equation opposite the unknowns.
Case II. When A = 0.
If a solution exists when A = 0, it cannot take the preceding
form, since division by zero is excluded from algebraic operations.
While the theory becomes too complicated in this case to be
presented in full here, certain particular cases may well be
considered.
As a rule (subject to certain exceptions), a system of equations
has no solution when A = 0. For example, the system
3a?+4#=5,
6x+8y=9
has no solution. Likewise the system
x + y — z = 5,
4cX + y — 2z = 9,
5x + 2y-*3z = l
has no solution.
206
DETERMINANTS
[Chap. XVIII.
A system may, however, have an infinite number of solutions
when A = 0. For instance, the equations
«+y-«=0, (13)
±x + y-2z = 0, (14)
5x + 2y-3z=0 (15)
constitute such a system. These equations are manifestly satis-
fied by x = y = z = 0. To obtain other solutions solve (13) and
(14) for x and y in terms of z. This gives
» = i«, * = **• (16)
These values of x and y satisfy (15) as well as (13) and (14).
Hence any value assigned to z with the corresponding values
x and y obtained from a; = £z, y=*\z satisfies (13), (14), and
(15). Since z may have any value, there is an infinite number of
solutions of the system in question.
Systems with an infinite number of solutions may be more
generally illustrated by the homogeneous * equations
a x x + b$ + c x z = 0, (17)
«2# 4- b$ + c& = 0, (18)
a& + b<& + c& = 0, (19)
a a 6, Cx
when A= a 2 b 2 c 2 =0, (20)
G&3 O3 C3
but some minor of A is not zero, say
«2
61
.*0.
(21)
To prove that (17), (18), (19) have an infinite number of solu-
tions, substitute in (19) the values
x =
— Cfi
bi
<h
-c,*
— c&
b 2
-> y =
a 2
— CgS
«1
&i
<*i
&i
<h
b 2
ttg
&2
* A homogeneous equation is one in which all the terms are of the same
degree in the unknowns.
Arts. 146, 147] SOLUTION OF EQUATIONS
207
which satisfy (17) and (18) when condition (21) is fulfilled.
This substitution gives, after clearing of fractions,
^ 5j Cj
— «a 8 — zb s + zc 3 = z a 2 o 2 c 2
C 2 2 fl 2 C 2 fl 2 v 2
#3 ^8 ^3
which, by (20), vanishes whatever value be assigned to z. Hence,
z can take an infinite number of values, each of which with the
corresponding x and y satisfies (17), (18), and (19).
147. Systems of equations containing more unknowns than equa-
tions. Consider first the single equation
3aj + 5#-6 = (1)
with two unknowns. It is manifest from our work on loci of
equations (Art. 38) that there are an infinite number of pairs of
values of x and y which satisfy this equation.
Consider next the two equations,
3a;-4y-2s + 1 = 0, (2)
4a> + 3y-3-6 = (3)
with three unknowns.
We may solve (2) and (3) for x and y in terms of z. This
gives
x =
2
z-1 -
-4
z + 6 3
3 -4
4 3
lQs + 21
25
«
y =
3 2z-l
4 z + 6
3 -4
4 3
^ -5s + 22
25
(6)
Any value assigned to z and the corresponding x and y obtained
from (4) and (5) satisfy (2) and (3). Hence, the system has an
infinite number of solutions.
The main point to be brought out by these illustrations is that,
in general, from n equations containing more than n unknowns,
we may solve (Art. 146) for some selected n of the unknowns in
208
DETERMINANTS
(Chap. XVIII.
terms of the remaining unknowns. We are then at liberty to
assign any values to these remaining unknowns, and thus obtain
an infinite number of solutions. The problem in the exceptional
cases in which it is impossible to solve for a selected set of n un-
knowns is too complicated to be treated here.
148* Systems of equations containing fewer unknowns than
equations. Consider the equations
a 1 aj + % + c 1 = 0, (1)
a& + b& + C2 = 0, (2)
a& + 1> s y + Cz = 0. (3)
In oider that these three equations may be consistent,* it is neces-
sary that
a?= —
Ci
&i
c 2
b 2
«1
h
a 2
b*
y=-
a 2 C2
9
<h b*
<ii b x
a 2 b 2
*o,
which satisfy (1) and (2) shall also satisfy (3). This requires
that
— a
3
Ci &i
a x c
C2 b 2
— 60
C&2 C2
«i &i
u 3
a x b x
C&2 6 2
a 2 b 2
+(?s==a
Clearing of fractions, and interchanging columns in
obtain
- & 3 + c a
«3
&1 «1
6 8 Ca
c^ c 2
Oj 6 2
=0,
we
or,
a x bi C]
c^ 6 2 C 2
<h &3 C S
= 0, (Art 144)
(*)
* Two or more equations are consistent (Art. 89) when they have
mon solution.
Arts. 148, 149] SOLUTION OF EQUATIONS 209
as a condition to be satisfied in order that equations (1), (2), and
(3) be consistent. Stated in words, in order that three linear
equations in two unknowns have a common solution, it is neces-
sary that the determinant formed of the coefficients of the un-
knowns and of the known terms vanish.
The method used for three linear equations in two unknowns
can clearly be extended to any number n of linear equations in
n — 1 unknowns. It results that the determinant formed of the
coefficients of the unknowns and of the known terms must vanish in
order that the n equations in n — 1 unknowns have a common
solution.
While the vanishing of this determinant is a necessary condi-
tion for the existence of a common root, it is not a sufficient con-
dition as is shown by the following example.
Take the system of equations
a + y-4 = 0, (5)
2x + 2y + 5 = 0, (6)
x + y-6 = 0. (7)
Here,
11-4
2 2 5
11-6
= 0,
but any two of the equations are inconsistent.
In establishing the above necessary condition, we assumed
that two of the equations have a solution. This condition is
satisfied by no two of equations (5), (6), (7).
149* Common roots of quadratic and higher degree equations in
one unknown. Consider first the system,
aa; + & = 0, a^=0, (1)
a'x* + b'x + c' = 0, a'=£0, (2)
consisting of one linear and one quadratic equation. In order
that (1) and (2) have a common root, it is necessary and suffi-
cient that the solution x = — which satisfies (1) shall also sat-
es
isfy (2). This requires that
210
DETERMINANTS
[Chap. XVTII,
^-^ + c' = 0,
a'
a
or
a'b 2 -dbb' + a 2 c' = 0.
(3)
The relation (4) among the coefficients is the condition that (1)
and (2) have a common root.
The left-hand member of (4) may be put into determinant form
as follows: Multiply (1) by x, and the resulting equation in
combination with (1) and (2) gives the system
ax + b = 0,
ax 2 4- bx = 0,
a'x 2 + b'x -f c' = 0,
(1)
(6)
(2)
which should be thought of as linear equations in two unknowns,
x and x\ From Art. 148, it is necessary that
a
b
a
b
a'
b'
c'
=o,
(6)
in order that (1), (2), and (5) have a common root. But (6) is
merely (4) written in determinant form, as can be easily verified.
Consider next the equations
aar> + &a; + c = 0, a^=0, (7)
a'xs + b'x* + c'x + d' = 0, a'*=0. (8)
If we multiply (7) by x and by x 2 , and (8) by a?, to form addition^
equations, we obtain
ax 2 -f bx + c = 0,
ax 3 -f bx 2 + ex == 0,
ax 4 -f bx* -f c# 8 = 0,
a'a 3 + 6 V 4- c ( x + d' = 0,
a'x 4 4- 6'z 8 4- c'x 2 4- d'z = 0,
which can be treated as linear equations in four unknowns, x, a? f
a 8 , and x 4 . From Art. 148> it is necessary that
Art. 149] COMMON ROOTS OF EQUATIONS
211
a b c
a b c
a b c
a' V c' d'
a' V c' d'
= 0,
CO
in order that (7) and (8) have a common root. Moreover, when
condition (9) is satisfied, (7) and (8) have a common root, but the
proof* is beyond the scope of this book.
A relation such as (6) or (9) which results from eliminating
the unknowns from systems of equations is often called the
eliminant of the equations.
The method of elimination employed in the last two examples
is called Sylvester's method of elimination. It consists in form-
ing from the given equations additional equations by multiplying
the given equations by successive powers of x until we have one
more equation than powers of x. These powers of x are then
treated as distinct unknowns and the eliminant is obtained as in
Art. 148.
EXERCISES
Solve by using determinants.
1. 4 x — 3 y — 6,
8x + y = 17.
3. I-I-Ul,
x y z
M-Ui,
x y z
-L+-L + 1-
4 x 4y 2z
= 1,
2. 3z + 4y-22 = 5,
4x — 3y-f 82= — 4,
2x + 8y-Sz — 5 = 0.
4. 3x + 2y + 4z — w— 13 = 0,
5x + y — z + 2xo — 9 = 0,
2x + 3y-7 z + Sw— 14 =0,
4x-4y + 3s — 6w-4 = 0.
5. Find a value of k such that
kx — 3 y — 5 = 0,
$x + y - 17 =0,
kx+ 2y— 10 =
are consistent equations. Can k take more than one value ?
* For proof, see Bocher's Introduction to Higher Algebra, pp. 200-202.
212 DETERMINANTS [Chap. XVIIL
6. Discuss the number of values of x, y, z which satisfy
x + 3 y — « = 0,
-2y + « = 0,
&x + y + 2z = 0,
and find the ratios x : y : z of corresponding values (apart from as = y =2=0).
7. Eliminate it from the equations
Wz - 2 fce a + 1 = 0,
, , x 2 + k-Skx =
by Sylvester's method.
8. Find the eliminant of ax 2 + bx + c = and as 8 = 1.
9. Are the equations a; 2 + 33 + 2 = and x* + &z* + 9z + 2 = con-
sistent ?
10. Determine b so that
3se 2 -8x — 3 = 0,
x*-bx*-x-6 =
have a common root.
11. By Art 148, it is shown that
that the two equations
02 &2
= is a necessary condition
aix + b x = 0, (d! ^= 0)
<*2X + &2 = (d2 # 0)
be consistent. Show that this condition is also sufficient for this special
case.
CHAPTER XIX
LIMITS
150. Definition. If a is a constant and x is a variable ivhich
assumes in order a given sequence of values in such a manner that
| a — x | * becomes and remains less than any assigned number
d (d > 0), then x is said to approach a as a limit
We have had many illustrations of limits in elementary mathe-
matics. Thus, in geometry the area of a circle is considered as
the limiting value of the area of the inscribed regular polygon as
the number of sides is indefinitely increased. Here the terms of
the "given sequence of values" are the areas of the inscribed
polygons .as the number of sides is increased. Again, as we
annex 3's to the decimal .3333 •—, its value runs through the
sequence of numbers .3, .33, .333, etc., which can be made to
approach as near to £ as we please. In the geometrical pro-
gression l + | + £ + £+...,
S n} the sum of the first n terms, runs through the sequence
i, h h ¥. ->
and approaches the limiting value 2.
The essence of the definition of a limit lies in the words " be-
comes and remains less." For example, if x runs through the
sequence of values h - £, f , - f , f , - f , ...,
the difference 1 1 — x | takes on the values
h h h h h 4 -
and becomes less than any assigned number but it does not re-
main so. In this case we cannot say that x approaches 1 as a
limit.
To indicate that x approaches a as a limit, we use the notation
x—>a or lim x = a.
j
ei notation \a — x\ means the absolute value of a — z. (Cf . footnote,
213
214 LIMITS [Chap. XIX.
151. Infinitesimals. A very important class of variables
which are assumed to run through a sequence of values consists
of those which have the limit zero. They are called infinitesimals.
The area between a circle and the inscribed regular polygon as
the number of sides increases, the weight of the air in the re-
ceiver of a perfectly working air pump, and the difference 2 — S n ,
where S n is the sum of the first n terms of the series 1 + £ +
\ 4- •••, are examples of infinitesimals.
Theorem. If u— >0 and v->>0, and X and Yare always nu-
merically less titan some positive constant k 9 then Xu + Yv — >Q.
In other words, if u and v are infinitesimals, then Xu + Yv is an
infinitesimal.
Let d be any positive number however small. Since lim u = 0,
and lim v = 0, \u\ and | v \ will ultimately become and remain less
than — - • For these values of u and v, we have
2k '
|X«|<^,
Yv \ <L 2T>
whence, \Xu\ + \ Yv\< ^ X \+^ d -
By hypothesis, \X\ + \Y\<2k 9
hence, |Xw| + | Yv\<d.
But the absolute value of a sum is never greater than the stun of
the absolute values of the numbers. (See Art. (56) Exercise 11.)
Hence, | Xu + Yv\ \ <Xu\ + \Yv\
and \Xu+ Yv\<d.
Since d may be chosen as small as we please,
Xu+ Fv->0.
This theorem may be extended to any number of variables.
Corollary. Ifu-^0 and v -> 0, and G is a constant, then
Xu + Yv + (7-> O.
Arts. 151, 152] THEOREMS 215
Examples: If u— >0 and v— >0
then (1) 7w + 3t?->0,
(2) a + 6 — u — v — > a + 6,
(8) a& — (a — u)t> — &u — > ab.
15&. Theorems concerning limits. The following theorems fol-
low directly from the theorem of Art. 151.
Theorem I. The limit of the sum of two variables is the sum of
their limits.
Let the variables be x and y, and let
lim a? = a, lim y = b.
Then, x = a — u, y=sb — v,
where w->0, v->0.
Adding, we have x + y = a + b — (u + v).
From the corollary of Art. 151, a + 5-(w + v)->0,
or lim (x + y) = a -f- b = lim x -f- lim y.
Corollary I. The limit of the sum of any finite number of
variables is the sum of their limits.
Corollary II. The limit of the difference of two variables is
the difference of their limits.
Theorem II. The limit of the product of two variables is the
product of their limits.
Using the notation of Theorem I,
xy =(a — u)(b — v) = ab — [(a — u)v + bu].
From the theorem of Art. 151,
(a— u)v -f bu— >0.
Hence, lim xy = ab = lim x lim y.
Corollary I. The limit of the product of any finite number of
variables is the product of their limits.
Corollary II. If n is a positive integer,
lim x n = a n = (lim x) n .
Corollary III. If c is any constant,
lim ex = c lim x.
216 LIMITS [Chap. XIX.
153. Both numerator and denominator with limit zero. If both
the numerator and the denominator of a fraction - approach the
y
limit zero, we have a rather curious result, as is shown by the
cases which occur in the following example.
In the fraction - let y approach through the sequence of
values ^
111 1
— , — « — « •••• — ••••
2» 2*' 2» 2»
Let x approach through one of the four sequences :
111 1
(a)
(*>)
V £' 4»' '"' 4~»' """
1 1
V2' VI' V8' ' V2='
• ••-
( c ) 2' 2*' 2*' "*' P ""' ( * = any constailt )
m\ 1111 . 1
W 2'~*' 2*'~~W'' ± 2^' ,%
Case (a). We have here lim - = lim ^ ^ I^q^
y n->oo 1 n-^oo2»
(See Art 164.)
/>»
Case (6). Here - passes through the sequences of values
y
V2, V2*, V23,-, V2v-
which increases without limit.
Case (c). In this case lim - = Jc.
y
or
Case (d). Here - takes alternately the values + 1 or — 1 and
approaches no limit.
We see then that if x and y both approach as a limit, their
ratio may approach any number whatever including 0, may in-
crease without limit, or may oscillate between two fixed numbers.
Arts. 153, 164] INFINITY 217
154* Infinity. If the numerator of the fraction is constant, or
has the limit a(a =£ 0), while the denominator has the limit 0,
then - increases without limit and is said to become infinite.
y
This is usually expressed by writing - =oo .
if
It is not, however, to be inferred that infinity is a limit. The
as
variable - in the case just given approaches no limit. If z is a
if
variable which increases without limit, the various expressions
" lim z = oo," " z— >oo," " z = oo," should not be read " z approaches
infinity " or " z equals infinity," but " z becomes infinite," " z in-
creases without limit." Infinity is not a number in the sense in
which we are using the term.
Theorem I. lim 1 = 0.
Let d be any assigned small positive number. Let n > - where
— d
1 d 1
x is any number greater than 1. Then - < - < d. That is, -
* n — x n
becomes and remains less than any assigned number.
Theorem II. If I r I < 1, ^ r" = 0.
Since | r | < 1, it can be written in the form | r | = — - — where h
is positive. Hence, "■"
1 1
r n \=-
(1 + h) n 1 + nh + positive terms
(By Binomial Theorem.)
11
Therefore, \r n \< —^ — < — .
1 -{-nh nh
By Theorem I of the present article and Corollary III, Art. 152,
lim J_ A
n->-oo nh
Hence, ^ |r*|=0.
Since r* = ± | r* |, we have
r n = 0.
218 LIMITS [Chap. XIX.
Corollary. If\r\<l, ^ -^- = 0. I
11 n->»ao 1 — r
Exercise. Let y approach through the sequence
0.1, 0.01, 0.001, — .
Show that the fraction - may be made to approach any number as a
V
limit, may increase without limit, or may oscillate between two numbers.
155. Limiting value of a function. Let f(x) represent any
function of x. If x runs through any sequence of values and
approaches a limit a and at the same time f(x) takes on a
sequence of corresponding values such that
lim/(a;)=^,
we may abbreviate and write
which reads, " As x approaches a through any sequence of values,
f{x) approaches the limit A " ; or, more briefly, " The limit of
f(x), when x approaches a, is A."
K /(«)= I 1 i a /W»
the function is said to be continuous for x = cl
156. Indeterminate forms. To find the value of the fraction
when x = 2, we substitute and find the value to be 4.
x 2 + x - 2
x-1
But when a; = 1, by substitution we find -, a meaningless symbol.
x z 4- x — 2
We may write — = x + 2,
but since division by zero is excluded from our operations, this
simplification does not hold for x = 1. But for every other value
of x y however near to 1, the division is possible. Hence, letting
x approach 1, we have
Urn a*+*-2 _ lim
a-»l 1 =x-»l(* + ^) = <*.
Arts. 154-166] INDETERMINATE FORMS 219
Although substitution of x = 1 in — — — — - gives us a meaning-
x — 2
less symbol, it is convenient to assign a value to the fraction.
Whena ? = l > we^n e g2 + g 7 2 tobe^ 1 a?2 + a? 7 2 = 3. Giv-
ing this value to the fraction makes — — — - — = x -f 2 true for
x — 1
all values of x. In general, if f(x) is a fraction which for x = a
r
takes the form -, we define/(a) to be x \^ a f( x )-
The student should note that this is not a necessary definition
of /(a), but merely a convenient one. The convenience arises
from the fact that with such a definition of /(a), the function
becomes continuous at x = a.
We wish sometimes to find the limit of the value of a function
as the variable increases without limit. The following example
illustrates the method.
Find the limit of ^-i for a?->oo .
3a? + 2s-l
By the theorems on limits this will take the meaningless form
sg-, but dividing numerator and denominator by x 2 , we can write
the fraction as
1+-
« 2
x x 2
2 2 1
and since — , -, — are infinitesimals by Theorem I, Art. 154, we
a? 2 x a? 2
have - for the limit of the fraction.
oo
J
other such forms which may arise are • oo and oo — oo , but the
expressions which give rise to these forms may be reduced to
the form -, as shown in the following examples.
The symbols -, ^ are called indeterminate forms. Among
220 LIMITS [Chap. XIX.
Example 1 : (x 2 -f x — 2) • takes the form • oo when x = 1. For
x — 1
any other value of x we may write
( X 2 + x-2). _L_ = x2 + g - 2 .
a — I a> — 1
TT««™ nm /~5 i «. ON ! "HI /X* + X- 2\
Hence, ^ (s' + s- 2) ■ ^ = ^ ^ g _ i ) = 8.
Example 2 : — = + takes the form oo — go when x = 8.
x 2 - 9 x(x 2 - 9)
For any other value of x, we may write
x-1 x 2 + x - 6 = - 2(x - 3) = - 2
x 2 -9 x(x2-9) x(x 2 -9) x(x + S)'
Hence, *» ( *^±-* 2 +*-G\ = Mm -2 = _1.
s->3 ^ X 2 _ 9 x (x 2 - 9) / «->3aj(x +8) 9
EXERCISES
What values should be given to the following expressions in order to make
them continuous for values of x indicated ?
1.
X3-1
X- 1
when x = 1.
2.
x«-4
when x = 2.
3.
x - 3_
Vx~V3
when x = 3.
4.
1-x*
1-x*
when x = 1.
5.
x 3 -f a 8
x 2 — a 2
when x = — a.
6.
<* *>■.!.
whenx = 2.
7.
1 2
— when x = 0.
8.
x 2 2x —
1
■ when as = 1.
x x(x + 2) (x-1) 2 0&-1) 1
9. x °~ + x + 1 whenx=-l.
x+ 1
As x increases without limit find the limits of the following fractions.
10. *±i. 11. -2— 13. «*- 8g .
x x* + l 2x» + 6x»
CHAPTER XX
INFINITE SERIES
157* Definition. Let u u u 2y •••, w n , ••• be any unending sequence
of real numbers positive or negative. The expression
«i + w 2 -+- ^3 + •• + u n + u n+1 + .»,
when the terms are formed according to some law of succession, is
called an infinite series.
In the discussion of geometrical and harmonical progressions
we have met such series, for example,
and l +£ + £ + £+....
158. Convergence and divergence. In the series
Ui + U 2 + V 3 + - + M» + —,
let S n represent the sum of the first n terms, that is,
Si = u u
S 2 =Ul-\- U 2 y
$3 = ^1 + U 2 + U Z ,
S n =
: U x + W 2 + —
+ tt B ,
For
example, in
the series
given
in Art. 157,
s 3 =
i+i+i+i+
we have
'! + * = *,
; l + i + i = i.
s n =
■}+Wi+
221
= 2- 1
2 n_ i
222 INFINITE SERIES [Chap. XX.
In the series 1+2+3+4H — , we have
$, = 1 + 2 + 3 = 6,
£ n = l + 2+3+...+n = !(l+n).
In the series 1 — 1 + 1 — 1 + 1— •••, we have
£i = l,
#3 = 1,
S n = 1 or 0.
These three examples illustrate three cases which may occur.
I. S n approaches a limit as n increases without limit. In the
first example above, S n is never greater than 2, no matter how
large a number n represents and approaches 2 as a limit, when n
increases without limit.
II. S n is numerically larger than any assigned number for a
sufficiently large value of n.
This case is illustrated in the second problem.
III. S n remains finite but does not approach a limit as n in-
creases without limit.
This case is illustrated in the third series, where S n may have
either of the values or 1, according as n is even or odd.
Series which come under Case I are called convergent series
and are by far the most important. Series which are included in
Cases II and III are called divergent series. We have then the
Definition. When in an infinite series the sum of the first n
terms approaches a limit as n increases without limit, the series is
said to be convergent. Otherwise it is divergent
The limit of the sum of n terms of a convergent series, written
lim S n , is often called the sum* of the series. In connection
* The word * ' sum " is here used in a purely conventional sense. It 1b not
to be understood as the sum of an infinite number of terms, but as the limit
of the sum of n terms as n increases without limit.
r >
% Am*. 168, 159] FUNDAMENTAL ASSUMPTION 223
with convergent series we shall also use the expression " limiting
value of the series " to mean lim S n .
Many important mathematical investigations and the solution
of many practical problems require the use of infinite series and
depend upon the question of convergence. The central problem
of this chapter will be the problem of deciding whether a given
series is convergent or divergent. It will be found convenient to
take up first those series in which the terms are positive numbers.
SERIES WITH POSITIVE TERMS
159* Fundamental assumption. An infinite series of positive
terms is convergent if S n is always less than some definite number,
however great n may be.*
Let IT be a number such that S n <K for all values of n. Since
the series contains positive terms only, S n is a variable which in-
creases as n increases. Since it can never attain so great a value
as Kj we assume that there is some number less than K which
S u approaches as a limit.
To illustrate this assumption, consider the series
o. ! ,-j, 1
O S x S 2 S 4 2 X
Fiq. 33.
and take points on the line OX (Fig. 33), to represent S iy S 2 > £ 3 , •••
so that the measure of OS x , is S ly of OS 2 , is S 2f etc.
^=1 + 1 = 1.2500,
*s = l+^ + |,= 1.2870,
S,= l+± + | + ±=1.2909,
• For proof see Pierpont's Theory of Functions of a Real Variable, Art.
100. Fine's College Algebra, p. 59, Art. 192.
224 INFINITE SERIES [Chap. XIX.
We can show that the sum of n terms of this series is less than 2.
(See Art. 160, Example 1.) Hence, according to the assumption of
this section, there is some point not farther to the right than 2
which S H approaches as a limit when n->oo.
An analogous assumption exists for a series all of whose terms
are negative. An infinite series of negative terms is convergent,
if S n is always algebraically greater than some definite number,
however great n may be.
EXERCISES
1. It can be shown that the sum of the series,
1 + _L + JL + J_ + .I. + ... + 1 + ... •
T l! 2! 8! 4! T ^(n-l)! T '
is always less than 3. Illustrate the assumption of this section graphically
by means of this series.
2. Illustrate graphically the assumption for a series of negative terms by
means of the series,
2« 3» 4*
160. Comparison test for convergence. Consider the series of
positive terms
t*i + *2 + w» + ••• + n m + ••••
If from some term on, the terms of this series are equal to or less
than the corresponding terms of a known convergent series,
i'i + ^2 + v* H h v H + —
of positive terms, then the u series is convergent.
Let S n = it! + w 2 + - + w„
and S n '= v i + v 2 + — + v»*
Suppose that after the Arth term
then, if S k = u x + u 2 + — + u ta
and S k = v x + v 2 + — + v*,
we have S m -S k < SJ - S k ',
or S n < SJ - 8 k ' + S k < lim SJ - S h ' + 8*
* For meaning of 1 !, 2 !, 3 !, etc., see Art. 69.
Arts. 159, 160] COMPARISON TEST 225
Since by hypothesis lim S n ' exists, it follows that S n is always
less than a definite number, and by the assumption of Art. 159
the series u x + u^ + u z + ••• -f u n -f ••• is convergent.
Example 1 4. Prove the series
2 + 1+— + A + A + ... + 1 + ...
to be convergent.
Solution : For purposes of comparison take as the v series the geometri-
cal progression 1+1 + I + ... +_!_ + ...
2 2 2 2*- 1
which we have shown to be convergent (Art. 167). Write the given series
under the comparison series :
l+i+- + — +••• +-i- H ,
2 2 2 2 3 2 n "i
2+1 +i+— + — + + ••-.
T 2» 3» (n-l)»-i
After the third term, each term in the second series is less than the corre-
sponding term just above it. That this is true for every term after the third
is shown by examining the two nth terms. If n > 3, then
l <-L.
(n- l)n-i^2*-i
Beginning with the fourth term, the sum of n terms of the first series is
always less than £. Hence, the sum of the second series can never exceed
2 + 1 H f- - = 3|. In comparing two series it is not sufficient to compare
a few terms at the beginning of the series. The nth terms should be com-
pared.
Example 2 : Test for convergence the series
<L» + *i* 8 x 1 + JL + _J_
1 1 1 3 2-32 3.33
If we neglect the first three terms of this series and prove the remainder
to be convergent, the given series must converge. For, if the series beginning
with the fourth term has a sum, the sum of the entire series will be the sum
of terms after the third plus 3 . 3 3 -f 2 • 3 2 + 3 = 102. Beginning then with
the fourth term and comparing with the series,
1 + 1 + 1 + -
which is known to converge to } (Art. 73) , we have
1 +- + — + — + — +-^— H — ,
8 3* 8» 3»-i '
and 1 + - + — — + — — H — H 1- —.
32. 32 8- 33 ^( n -l).3»-i
226 INFINITE SERIES [Chap. XX.
Each term in the second row is equal to or less than the corresponding
term in the first. Hence, the second series converges to some number not
greater than } and the sum of the entire series in question is not greater than
103.5.
Iii testing for convergence it is often convenient to omit a finite
numlwr of terms as in the above example. That this is always
justified is shown by the following
Thro rem. The convergence of a series is not affected by neglect-
ing a finite number of terms.
For the sum of the terms neglected is a definite number which
added to the sum of the new series gives a definite number for
the sum of the entire series.
EXERCISES
Tost the following series for convergence :
1. l+-L. + _i- + _^+ ....
2-2 32» 4-2»
2. 1+1+1 + ! + ....
2 3* 4»
3. - 1 — + — +-L+-L+ ....
1.2 2-8 8-4 4-6
Solution : Write S* in the form
M"K-l) + (r3 + " + (K-±i)"-:h-
Hence, n-*ao & » = 1 '
4. i + L + _L + Jl+ ....
2! 3! 4!
5. l+- + i + -+— -
2 2 8* 4*
6. 1+ I+i+!+ ....
2» 3» 4»
7. 1 + — + -f + — , wherep^2.
2* 8"
8. _*- + _*!_+_?!_+... , where x<l.
12 2-3 34
9. 1 +1 + 1+ ..., wherep>l.
Abts. 160, 161] COMPARISON TEST 227
Solution : Write down the inequalities,
1+1 2 _ 1
2* 3* 2* 2*- 1 '
4f 5p 6p 7p 4J" 4*- 1 '
1,1, , 1 . 8_ 1
8p 9* 16* 8* 8* -1
Add the members of the inequalities, thus
1 4- 1 + 1+ ... <_i_ + _L + _L+ ....
k 2" 3" 4? 2"" 1 4P -1 8P -1
The right-hand side of this inequality is a geometrical progression whose
£~ ratio is , which is<l whenp>l. Hence, the series is convergent.
2p~~ i
i This is a useful series for testing other series.
t 10. l+.£+*i + ^ + ..., where a^l.
I 2^ 3* 4*
161* Comparison test for divergence. Given the seines of posi-
tive terms.
U X + U2+ .- +U n + ...
If from some term on, the terms of the series are equal to or
greater than the corresponding terms of a known divergent series of
positive terms.
• • • .
Vj 4- v 2 H M„ +
then the given series is divergent.
Let S n = u x + u 2 + - + u ni
and /S , n f = v x + v 2 H h v n .
After the fcth term, suppose
w *+l > Vft+i, Mji+i > V k+2 , '~,U n ^V ni
then *. - S t 5 «.' - *'i
or #»>#»' — #*' + #*•
By hypothesis >S Jl f increases without limit as n increases. Hence,
if n is made large enough, S n will exceed any given number, and
the u series is divergent.
228 INFINITE SERIES [Chap. XX.
A useful oomparison series for divergence is the harmonica!
8eries l + i + i + i+-,
which can be shown to be divergent by means of the inequalities :
1 + ±>1,
Adding members of the inequalities, we have
l+£+i+i+" >l + i + i + i+-.
But the right-hand member of this inequality can be made as large
as we please. The series in question is therefore divergent.
Any geometrical progression a + ar + ar* + ••• in which the
ratio is greater than 1 can be shown to be divergent by compari-
son with the series a + a4-a+ — ; and such a geometrical pro-
gression is often useful as a comparison series in testing for
divergence.
EXERCISES
Prove the following series divergent.
1. 1+2 + 8+4+.... 2. 1+-L + -L + -L+....
V2 V5 VI
3. 1 H 1 \- — !-••• where p is positive and less than 1.
2p 3* 4p
4 - H+S + I + - 5 - 1 +i+5+f+-
162. Summary of standard test series. When any new series
has been shown to be convergent or divergent, we evidently in-
crease our supply of series for comparison purposes, but a few
standard series are so useful as to deserve special mention.
For convergence :
1. a + ar + ar 2 + ... + a?" 1-1 + ••• (r < 1).
1,1,1,, 1,
2. !-•••+■ — - 4- ••••
1.2^23 3-4^ w(w + l) r
Abts. 161-163] RATIO TEST 229
For divergence :
1. a + ar + ar 2 + — + ar n ~ l -f — (r > 1).
2 - 1+4 + 1 + 7+ •'■+-+"••
2 3 4 n
163. Ratio test for convergence and divergence. Another im-
portant test for the convergence or divergence of an infinite series
is the so-called ratio test.
Theorem. If, as n increases beyond bound, the ratio -^-tl ap-
u
proaches a limit X, the series of positive terms n
^i 4- ^2 + ••• + u n + •••
is convergent if X < 1 and divergent if\ > 1. If X=l, this test fails.
1. X < 1. Since lini -^ = X, we can make -^ differ from X
<*» K
o- , , , X
O 7i r i
Fig. 34.
by as small a number as we please. Hence, if we plot values of
!^±1 on the line OX, as n increases the points representing ^n±l
will concentrate about the point X. If n is taken large enough,
they will lie to the left of the point r, where X < r < 1. For
these values of n, we have
U n+1
— < r, u n+1 < ru n ,
^T 1 <r, u n+2 < ru n+1 < rhi n ,
,T±3 <r , u n+z <r*u n ,
^n+2
• •••••
Since r < 1, the series
rw n + r 2 u n -f r 3 ^ + ••• = u n (r -f r 2 + r 8 H — )
is convergent. But each term of the series
w B+1 + u n+2 + w n+s + .»
230 INFINITE SERIES [Chap. XX.
is less than the corresponding term of the ru series. Hence, by
Art. 160, the series u x + u^ + ••• u n + ••• is convergent.
2. X > 1. In this case the points representing -^ will ulti-
mately concentrate about the point X on the line OX 9 and if n is
n 1 ' I X
i r X
Fia. 36.
large enough, they will lie to the right of the point r, where
l<r<A. Then
or
and
u n
^
'9
w n+ l
>™ n ;
^„+2
>
r 2 w n ,
^n+S
>
*X>
Therefore, since the series
ru n + r 2 u n + r*u n + ...
is divergent for r > 1,
the series u x + u 2 + u z + ••• + u n + •••
is divergent (Art. 161).
3. X = 1. If lim H=±i = 1, this test fails. This is illustrated
in the two series,
1.2 2.3 T 34 T '
and 1 +o + 5 + 7+"'-
Z O 4:
The first has been shown to be convergent (Art. 160), the sec-
ond divergent (Art. 161), but for each lim-^ = 1.
Art. 163] RATIO TEST 231
Exampub 1 : Consider the series
2 + 22 + 21 + i**""'
Here, u^x = 2-±I, M n = ^,
tin+i _ n + 1 2 B _ n-f 1 _ n 1 1 1
t^ ~ 2 n +i " n 2n ~~2n + 2ri 2 2n*
lim^ = lim(I + J_A = I.
Hence, the series is convergent.
Example 2 : Consider the series
2 2 2 2 s 2 4
2»+i 2 n
Here, u^ =__, ^ =___,
wi = 2n+1 , (^ + i) 2 =2 /n + iy
w„ (n + 2) 2 ' 2» \n + 2J '
lim?^±l = 2.
Hence, the series is divergent.
EXERCISES
Apply the ratio test to the following series :
1. i+-L+i_ + _L+.... 2. 1+A+A + A+...
2! 3! 4! 2 2* 2» 2«
3. 1 + 21+32 + 4! H + AL + ^L +
2! 3! 4! 100 1002 100*
«. « 2 2 2 s 2 4
Op Op ^p
6. 1 H \- — A \- •• • » where p may have any value.
2! 3! 4!
7. i+Jl+jL + .... 8. 1+2+4 + 8+....
3 ! 6 !
..!+£+£+£+....
2! 3! 4!
232 INFINITE SERIES [Chap. XX.
SERIES WITH BOTH POSITIVE AND NEGATIVE TERMS
164. Thus far we have considered only series whose terms are
all positive or all negative. The following theorem will throw
light on the convergence of series whose terms are not all of the
same sign.
Theorem. An infinite series of real terms which are not all of
the same sign is convergent if the series formed by making aU the
terms positive is convergent
After all the minus signs have been changed to pins signs,
let the series be Wl 4. ^ + ^ -j. ....
By hypothesis, this series is convergent and therefore has a
limiting value S. The sum of the first n terms of this series is
then less than S. Hence, the sum, S H , of the first n terms of the
original series is numerically less than S. Let these n terms con-
sist of p positive and q negative terms. If P p be the sum of the
positive terms and N q the sum of the negative terms, then
S n = P p -N 9 .
But P p and N q are always less than S. Hence, by Art 159,
P p and N q approach fixed numbers P and N respectively as n
increases without limit. Then
m S n = P— N, a definite number.
and the series is convergent.
165. Ratio test extended. The ratio test can readily be ex-
tended to series whose terms are not all of the same sign. Since
a series of positive terms is convergent if
lim w^+i 1
it follows, from the theorem just proved in Art. 164, that any
series is convergent if the numerical value of -2±! is less
than 1. That is, if n "*°° M -
lim
u
n+l
«*,
<1.
To extend the ratio test for divergence we need the following
important
Arts. 164, 165] RATIO TEST EXTENDED
233
Theorem. In any convergent series, the limit of the nth term as
increases without limit is zero. That is,
lim
n-
00 U n
0.
For, we may always write u n = S n — S n _ v
As n increases both S n and 8 n _i approach the same limit S ; hence
(Art. 152),
n
ir„ «»= n i m w (*. -*»-.)= « - * = o.
While it is necessary that lim u n = in order that a series be
convergent, it is not sufficient. That is, lim u n may equal zero
and the series be divergent. Exercises 2, 3, 4, 5, p. 228, furnish
iples of such series.
> 1, the nth term cannot approach zero as a
If
lim
n->oo
u
n+l
u.
limit, hence, the series is divergent. We may then write the
ratio test for any infinite series
u x + U2 + t*3 + —
as follows :
** n->co
v
if
lim
n->oo
lim
n->co
< 1, the series converges.
> 1, the series diverges.
= 1, the test fails.
Example : Test for convergence and divergence the series
1 - 2x + 3ff 2 - 4x 3 -h ....
Here,
and
m„+i
u
n
n+ l)x n |_|/ n+ 1 \
nx n ~ l \ n J
x
lim
w„+i
u.
= x
Hence, if x lies between + 1 and — 1, the series is convergent. For
| x | > 1, the series is divergent. The interval between + 1 and — 1 is called
the interval of convergence of the series, and is represented graphically by
the heavy part of the line in Fig. 36. For the points 1 and — 1 the test tells
xm nothing.
DIVERGENT
CONVERGENT
DIVERGENT
6.
Flo. 36.
234 INFINITE SERIES [Chap. XX.
166. Alternating series. A series whose terms are alternately
positive and negative is convergent if each term is less than the pre-
ceding term, and if the nth term approaches zero as a limit when n
increases without limit.
Let the series be
Ui — U2 + U z — M 4 + - +(— 1)"~X ± —,
where u u t^, t^, — are positive,
and U2 < u u w, < t^, —,
and where m *!™V u n = 0.
Let n be an even number. We may then write S n in the form
Since each parenthesis contains a positive number, S n is positive
and increases as n increases. We may also write S n in the form
Since the parentheses are again positive,
S n < tii.
By the assumption of Art. 161, as n increases beyond bound,
JS n approaches a limit S. But
hence, lim S n+1 = lim S n + lim te^.
By hypothesis, lim u n+i = 0.
Hence, lim S n+i = lim S n = S,
and the series is convergent.
167. Approximate value of a series. In the case of some series,
for example, a geometrical progression, we are able to find exactly
the limiting value of the sum of n terms as n increases without
limit, but with many series we must be content to find an approxi-
mation to the limit, say correct to a certain number of decimal
places.
Abts. 166, 167] APPROXIMATE VALUE OF A SERIES 235
Example. Calculate
1-1 + I-1 + 1+...
3! 5! 7! 9r
correct to four decimal places.
1 = 1.00000 _ 1 = _ 0.16667
i = 0.00833 3 !
6m - — = -0.00020
-=- = 0.00000 l !
9! 1
1.00833 11 !
_ -A, = - 0.00000
- 0.16687 - 0.16687
0.84146
To four decimal places then the sum is 0.8415. But the ques-
tion arises as to just where we must stop adding terms. Even if
-— has no significant figure in the fifth place, we are dropping
an infinite number of terms, a number of which when added to-
gether may affect the result materially. In the case of an alter-
nating series, this question is easily answered by the
Theorem. The sum of the first n terms of an alternating series
differs from the sum of the series by less than the (n + T)th term.
Let S represent the limit of the sum of the series, S n the sum
of the first n terms, and R n the remainder. For n, even or odd,
we have
^» ! = U n+1 — U n+2 + ttn+3 — -.
From Art. 166, the sum of this alternating series, whose first
term is w B+1 , is less than the first term. Hence,
\S-S n \ = \R n \<u n+1 .
EXERCISES
Test the following series for convergence and divergence.
1. 1 — i + i — i+ ••. 2. 1 — ^-+-±z ^=+-.
V2 V8 V4
3. l+l + ! + i+.... 4. -J— + 1 + — * — + ....
2 3* 4» 1+2 1+22* 1 + 3. 2»
236 INFINITE SERIES [Chap. XX
5. 1 L_ + — I L_+..., (s>o, a>0).
x x + a x + 2a x + Sa v .
6 . 1+ 2! + 3i + £+....
21314!
7 1 I X I X I X I •
1.1 1.02 1.003 1.0004
io. ; + *&±p + «(» + wy 2) + .... (8>0) .
Compute, correct to four decimal places.
12. i_I.± + I..L_l.J_+....
2 2« 3! 25 6! 27 7 1
13. i_I + JL_l + ....
14 1 1 1 1 1 1 1 1
13 2 3* 3 33 4 3*
Write down the first five terms of the series in which u» has the following
^,1 ues, and investigate the series for convergence or divergence.
15. u n = — • 16. 11* = — •
n 3» ^ n*
On 1
17. u n = • 18. u.=
1 + 2» +1 1 + nVH
For what values of x are the following series convergent t
19. _JL+_J_ + ^L_ + .... 20. ? + - + -+•••
1+x* 2 + 2x* 3+3x* 2 4 6
21. -J— + 1- 1 , + *- 1
2x + l 3(2x+l)3 6(2x+l)*
22. 2 * + 3-2a; 2 + 4.3x«+ ....
23. 1+1,+ 1 •
x + 2 (x + 2)2 (x + 2)3
168. Power series. The series
Oq -f a x x -f atf? + ••• + a M 05* H — ,
in which Oq, a u a,, — are independent of x y is called a power aeries.
Such a series may converge for all values of x, may diverge for
all values of x except for x = 0, or it may converge for some
values of x and diverge for others.
Abts. 167, 168]
POWER SERIES
237
t
i
Theorem. If in the series
Oo 4- a x x + ajjOJ 2 +
£ the ratio
<*»+!
a.
approaches a limit r, Men Me series converges /or
l-Kt
, and diverges for \x\>
1_
r
This result follows directly from Art. 165. Applying the ratio
te8t lim w»+i = lim a n+ i* n+1 _ ry
n->-co m n->oo <* a? n
n n
From Art. 166, the series converges if | rx I < 1, that is, for
1
x\<
~Fot \x\ > — the series diverges. If r = 0, the series converges
for all values of x. For x = - , the test fails ; the series may or
r
may not be convergent for this value of x.
Cobollaby. If a power series is convergent for x = b, it is con-
vergent fir every value ofx numerically less than b.
EXERCISES
m
Hud the interval of convergence of the following series. Exhibit the
results graphically.
JL. — r • + + ••••
2 2* 2*
n,,^^. n* „ (n+1) 2 q w +i (n + 1)2
Solution: 0, = -, o„ +1 = i__^, ^_ = L__A.
lim Ogii^ lim (n + 1) 2 -. lim
n->oo
a* n->-oo 2n*
n->oo
/ n* 2 n 1 \ = 1
\2» 2 2n2 2n2J 2
The series is therefore convergent for | x | <2 (Fig. 37).
DIVERGENT
CONVERGENT
DIVERGENT
-a
r
Fio.
37.
2
2. l + 2*+(2*)« + (2«)»
-j_ . ...
3.
1+? + ^
3 32
X^X
33
4. l + * + 2!a? + 8!x3+ ....(x=jfc0). 5. x+^+^x
« I o !
238 INFINITE SERIES [Chap. XX.
©. 1 + 5* + -t — 1-— — h ••••
2! 3!
•? i o*_i_ 2 ' 3 ** 2.8 4 ,., , 2 ■ 8 . 4 • 5^ ,
7. l-2* + —tf -_«,« + _ _ «•+....
By division, expand the following fractions into series and test for con-
vergence.
8. — ?—- 9. * ■ 10. — i— • 11. *
1-* 1-4* 2-3* 0+s) 8
The expansions of sin *, cos x, arc sin *, and arc tan * are given below.
Find the interval of convergence of each of the series.
12. 8 inx = x- — + — - — +....
3! 5! 71
13. cos*=l-^+?i-^ + ...
2! 4! 6!
1* . arc sin * = * A — « — — ._. — u _._._.__ 4. ...
2 3 2 4 5 2 4 6 7
15. arc tan* = * — — + — — — + ....
3 6 7
16. From the series in Exercise 12, find the value of sin 6° to four decimal
places.
Hint : Express 5° in radians. ,
169. Binomial series. The power series
., , , m (m — 1) „ , m (m — 1) (m — 2) . ,
1 + mx H ±— — t-x 2 H * -^ igf + ....
is called the binomial series. If m is a positive integer, the series
ends with the (m -f l)th term and has been shown to be the ex-
pansion of (1 -f- x) m . If m is not a positive integer, the series
is an infinite series, but it can be shown that it converges towards
(1 + x) m when x has any value which makes the series convergent.
In other words, it can be shown that for these values of x, the
binomial expansion holds for any exponent, integral or fractional,
positive or negative.*
*For a proof of the binomial expansion for any exponent, see Fine's
College Algebra, p. 663.
Arts. 168, 169] BINOMIAL SERIES 239
The binomial series is convergent for [ x | < 1, and divergent for
| x | > 1. For, we have
• __ m(m — l)(m — 2) — (m — n )
(n + 1) !
__ m(m — l)(m — 2) — (m — n + 1)
w!
a n n + 1 '
m _
1
lim
a»+i
lim
m
— n
lim
n
n-^oo
««
n->«oo
n
+i~
n^co 1 +
1
n
=-1.
Hence, from Art. 168, the series converges for — 1 < x < 1.
In expanding (b + $) m for fractional or negative values of m,
we may write it in the form b m l 1 4- - J . The expansion will
or
then proceed in powers of -, thus :
- < 1, that is, the interval of con-
b\
vergencefor the expansion of (6 -+ x) m is the interval between —b
and + b (Fig. 38).
DIVERGENT CONVERGENT DIVERGENT
I o
-b 5 b
Fia. 38.
EXERCISES
Expand the following binomials to five terms and indicate the interval
for which the expansion holds.
1. (1 + 3*)-*. 2. (l + 2x)i
3. vT+x". 4. (2-3x)~*.
5. (2 + *)*. 6. *
VI -8x
7. (6-4x2)"*. 8. (l-x)~*.
9. (l + 4x*)-i. 10. (3a + 2x)*.
240 INFINITE SERIES [Chap. XX.
Extract the following roots to four places of decimals by the binomial
expansion.
11. ^65.
Solution : \^66 = (4» + 1)*= 4^1 +—\
= 4 (l + l.I_i.l.l + ...)
\ 3 4» 8 3 4? J
= 4(1 + 0.00621 - 0.00008 + ...)
= 4.0207+
12. V6. 13. Vl6. 14. VD52.
15. #998. 16. #66. 17. #180.
18. #731. 19. #268. 20. #53l6.
170. Exponential series. The power series
is called the exponential series. It is convergent for all values of
x. For, we have
hence, ^ ^ = 0.
n->-Qo a
It can be proved that
lim
(*+ir-*+i+fi+ij+
where e = l + l+i + -i+...
= 2.71828,
correct to five places of decimals.
This number e is the base of the natural system of logarithms
Let a* = e\
then x log, a = h.
Hence,
Abfs. 169-171] LOGARITHMIC SERIES 241
171. Logarithmic series. The power series
Jo n
is called the logarithmic series.
fat
L
Since
a n+1 = ± ,
n -f- 1
<*« =
71
if
a-
i
we have
i
!
i
and the
series is convergent for x | < 1.
DIVERGENT
CONVERGENT
DIVERGENT
-i 6
1
Fig. 39.
It will be shown in the calculus * that this series converges to
log, (1 -f x) for any value of x for which the series is convergent.
The series can then be used to find logarithms of numbers to the
base e. Thus,
log.(f)=log,(l+i)
= i_(i)_ 2 ,(i)!_(i)_ 4 . ...
2 2 3 4
The logarithmic series can be used to calculate logarithms of
positive numbers less than 2. However, it converges so slowly
that it is not well adapted to numerical computation.
To derive a more convenient series for the calculation of natural
logarithms, we proceed as follows : f
log,(l+s)=s--| + |-^+....
yw9 /W^ /m4
Hence, log e (l -x) = -x---- — --+....
Ii o 4
By subtraction, log, (1 -f x) — log, (1 — a;) = log, ^
1 — SB
=2 (* + f + ! + -)
* Townsend and Goodenough's First Course in Calculus, p. 326.
t For a more detailed discussion, see Osgood's Introduction to Infinite
Series, pp. 23 and 44.
s
242 INFINITE SERIES [Chap. XX.
Let L±» = 5L±i,
1 —x m
whence x = — - — •
2m + 1
We have then
log M + l = g / 1 , 1 . 1 . \
m \2m + l 3(2m + l)» 5(2i»+l)« /
or logJm + 1) = log e ra -f 2( — 1 1- + ...V
5A. j be \^2m + 1^3(2m+l) 3 5(2m + l)5^ )
If m = 1, we have
log.2 = +2 Q + l + ^. 6+ ...) = 0.6931 + ....
Letting m = 2,
log.3 = log.2 + 2(I + -L+...)
= 0.6931 + 0.4055 + ... = 1.0986 + ....
In this way the logarithm of any number to the base e may be computed.
From Art. 120, if a is any positive number, we have
logio a = i°&£ = — L- . log. a.
10g e 10 l0g e 10
Hence, if we have computed the logarithm of a number to the base e, we
can find its logarithm to the base 10 by multiplying by . To five sig-
! log. 10
nificant figures, , ^ 1f) = 0.43429.
In computing a table of logarithms we need compute the logarithms of
prime numbers only. The logarithms of composite numbers may then be
found by means of the theorems on logarithms.
EXERCISES
1. By computing the logarithms of 2, 3, 5, 7, construct a table of logarithms
to the base 10 for the numbers 1 to 10.
2. Using the series for log e (1 + x), compute log e f to three places of
decimals.
3. Find log e 16 and log e 17 to four significant figures.
ANSWERS
[The answers to some exercises are intentionally omitted. ]
Article 6. Page 11
1. 5. 2. 9. 3. 6. 4. 3, 2, 3.
5. 5, 4. 6. 12, 5. 7. a. 8. 1.
9. -. 10. m%7. 11. 68
y* 16 a 2 * 6
12. 8 ofi + 12 a 2 6 2 + 6 ab* + 6*. 13. aP». 14. a*".
15. a n . 16. a 2n + 2 . 17. a 2 "- 2 .
»• ©■
4"y» 16 V n ro
22. x*y 2 +xy + 1. 23. a 2n — 6 2n . 24. x 2n ~ 2 — y 2n ~ 2 .
25. |? 2 »- 2 —p2n-5 _|.p2n-l_p2n-4 # 26. a 6n ~ 9 • £l2n-3 . y9n-3 #
27. a^ n • J« n . 2/» 2 . 28. x<\ 29. 1.
80. 7a4 "" t g ' g "" 3 - 31. a*». 3 2 ^ ,
15
88. 4 a 2n + 2 a n 6 n + 6 2n . 34. x 2n — x n + 1.
85. (a 2 + ab + 6 2 ) (x + y) 2 . 36. 27p3n _ 8 q zn t
Article 12. Pages 15-17
1.
5x 8
2.
b.ato*.
3.
f
4.
b
5.
3.
6.
f-
7.
i
8.
«•
9.
1-
10.
0.7.
11.
If
12.
3*
2-6 2
13.
17.
X
•— •
a
2xy 3 .
►
14.
18.
10 67
•
a 7 c 3
X 2
— •
V
15.
19.
J.
a 0.06
X 0.14
16.
20.
a*b*c.
x*y*
J
21.
— a;i V«" M .
22.
p -o.u
23.
x— y.
24.
1-6.
25.
«*-
6*.
26.
x*y J + x%* .
27.
X" 1 — y~K
28.
1 +x*.
243
244 ANSWERS
29. a -l. 80. a + 2aH* + 6. 81. a + Sahl + Sah$ + 6.
a
S3, x-« + ar» + 1. 88. jr» - 3jr* + 3i>-i — 1.
84. 2 + x* faf*. 8ft. 2xV"*- 5y*x~*.
86. ari + 8. 87. m-* - wri + 1.
88. ar 3 — 2 ar V 1 + 4 ar-ty-* — 8 ?-*.
89. O" 8 + cr 7 - cr 5 - cr 4 — or 3 + or 1 + 1.
40. x + A 41. aV* + 2 a^ - 3 b\. 48. x~* + a* -haici
48. VSty*. 44. -^- 48. #(a + b)K
Va
46. — l 47. ^L. 4g. — = .
49. 7. 60. yVx*s. §1.
52. x*y*f W. ar<z~l. M. oW.
55. xV 56. £-*_. §7. 2«*.
58. 2*. 59. a*zny. 60. y **** *
61. a***- 1 . 82. a * . 68. x*.
64. a»" • 6.
85. x^». • 67. 170.4.
68. 10-8, 8 • 10-», 16 . 10-7. 69. 3}, - \ % 0, 8|. 71. 666 . 10-w.
78. 10-«. 78. 0.6867, 0.5896, 0.4861, 0.3934 microns
74. 1 preceded by twenty-four zeros after the decimal point.
Article 14. Page 19
1. 5 a + 12 6. 8. a-b+c + d — e. 8. 6x + 2.
4. r + 2 y. 5. 0.
6. 4c + 36-(5c*-4a-4d s ). 7. a-d — (c-6).
8. 20. 9. 6a+56-(a+6)-(4a + 36>
10. (*« + *■)».
Article 15. Pages 20-21
1. *_. 2 . *±1. 8. 2a * *
z-y x a* + & s 2a% — 1
5 . k=JL e. *=I. 7. e±£±l. t. «-i*,
a x+1 x sx + V*
ANSWERS
245
9.
tn + ft
10.
m — n
18. x + y.
17. x 2 -x + l.
, 1
x+ 1
14. J3L.
2ox
18.
21.
84. -
x(x + ft)
1
a 2 + x 2
22.
11. « + y.
is. £rJ.
05 + 2
19. X-l.
1
r«(r -f «)
25.
(x-l)(x+ft-l)
E
27. -.
a
3r+ 2ft
28. a 2 6. 29. x.
28.
26.
12. « + 1.
16. 2.
20. n.
2x + ft
x 2 (x + ft) 2 '
6p+ 14 J?
a+ 6
80.
81. aWLziW.
Vy3-2X3 /
88. (a 2 -x 2 )*.
82. a -f- 6.
a 2 - x 2
84.
x
Article 17. Pages 24-25
2. V99. 3. \/ia
7. ¥m. 10. £^26.
14. SvOl. 15. V42.
21. 2v1J. 22. 3^3.
28. v^9. 29. \/l3.
85. 2v^4.
88. 3^5- 3 V3.
4. \/TT2.
11. i\/260.
18. 2^4.
23. 20\/2.
30. V7.
l +VF
5. vl04.
12. |V28.
19. 3V2.
24. 3v^3.
81. 1/8.
6. v^068.
18. v"81.
20. 5V5.
27. V20.
84. Z@.
36.
-2
89. $0/20+ V2).
= 0.4142.
37. 15-5V8.
40. -i(^6 + V2i),
41.
1+V2
1
V6
1
1 -V3
4
I V2 - V3
= 0.4472.
= -1.3660.
= 12.585
Article 18. Pages 25-26
2. ^^300. 8. *V35. 4. ^\/363. 5. 9>/l5. 6. V2.
7.
26-7V3
23
8. T >V(5-2V3). 9. ^ + Vl5 . 10. -5.
11. 5 + 2V3". 12. » -A" 18. Va+1+^. 14. x* + 2.
33 ^ a
as
246 ANSWERS
Article 19. Page 26
2. 3V3. 3. V6. 4. xy/Sz. 5. 9\/3. 6. V2. 7. (b+x-ay)y/a.
8. jVg-JvS". 9. yy/b-X 10. 9v^2"i. 11. 7Va-2v / a
12. 16(3)*. 13. 16(3)4. 14 . 3 (6) i 15 . 5 i
Article 20. Page 27
2. #675. 3. atyVla. 4. 100. 5. 24^6. 6. #77. 7. 12^0^.
8. 6. 9. 12 + 7V6. io. 2V3-Via 11. m + Sv^m^n -f 3Vmn*+7i.
22. $V5. 23. ^. 24. J^IM^x. 25. 6*. 26. 7*.
12. 5. 13. 3-5a + 2Va(l + a). 14. x*+l. 15. Vz*y«z\
16. W. 17. 5^4. 18. #3. 19. 2\/l08. 20. 2 #3. 21. a/ x .
_ ^
a
27. 7~* = ^76. 28. ^ = \Wa. 29. iv^a. 80. v^Oa^.
"5 6
Article 21. Pages 27-30
1. 2288 feet per sec. 2. 25.30. 3. 500.7. 4. 10.71. 5. 0.743.
6. 0.0205. 7. 1.9 feet. 8. 9.39. 9. 26.95. 10. 29.83. 11. 154.
13. 1077. L4. 30. 15. 50.9.
Article 22. Page 30
2. 5. 3. 1. 4. -2-2iV3. 5. x 2 + a 2 . 6. —i. 7. 7.
8. 625. 9. i 10. 0. 11. 14. 12. 2a(a* - 3 6*). 13. 1054.
14. 0. 15. (x - a)2 + 62.
Article 25. Pages 32-33
G
1. A = wr*. 2. V = f(x). 3. 2/ = VlOO - x*. 4. P = - , C=/(*).
5. « 2 + a + 1, 1,3, 1, 91, <*2+ 3 a + 3. 6. 3,- 1, 0, 1- V2, a + b + * .
a+ 6 — 1
7. - 1, *-±^-, ? + A. 9. x 4 + &2+ i, X 2 + 3x + 3,
1 — X 2 X — 1
x< + 2x 3 + 4x2 + 3x + 3. 11. 1+2 * . 12. 7 * + 10 .
3 + 4t 15J+22
Article 26. Page 35
5. (- 12, -3).
6. (-1, 4), (2, 4), (5, 4), (5, 1), (5, - 2), (2, - 2), (- 1, - 2), (- 1, 1).
8. (a) (-1,5), (3,5), (3,1), (-1,1).
(b) (1, 5.83), (3.83, 3), (1, 0.17\ (- 2.83, 3).
ANSWERS 247
Article 29. Page 40
2. f . 8. - {. 4. 1, 5. 5. 0, - 4.
6. 4,-}. 7. 1,5. • 8. 1,-1,-3. 9. 0,5,-5.
10. and 1 ; — 1 and ; — 6 and — 5.
Article 36. Pages 47-48
1. (a) 2 x 2 = 5 x with the additional root 0. (6) 2 x(x — 3) = 5(x — 3)
with the additional root 3. (c) 2 x(x 2 — 4) = 5(x 2 — 4) with the additional
roots 2 and — 2. 2. (x 2 + 2) (x - 3) = 3 x(x — 3). This equation is re-
dundant with respect to a root 3. 8.0. 4. (a) x 2 — 6 = 3 or x 2 = 9 ;
(6) 2. 5. 8x 2 - 24x + 16 = 0. 6. a. 7. a. 8. 7 x - 1 = 0.
9. 7x=12. 10.-3x + 8 = 0. 11. x - 5 = 0. 12. x - 25 = 0.
13. x-64 = 0. 14. x 2 -5x-14 = 0. 15. (x - l)(7x 2 - 16x - 15)=0.
16. 7x-75 = 0. 17. x = l.
Article 38. Page 51
2. x = ty, y = J. 3. x = 16, y = 15. 4. x = 1, y = 2. 5. x = 7,
y = 2. 6. x = 7, y = 11. 7. x = 6, y = 12.
Article 39. Pages 52-53
2. x=l, y = 1. 8. x = 3, y = 4. 4. x = 1, y = 3. 5. x = 1,
x = — 1. 6. x = 2, y = 3. 7. x = 4, y = 2. 8. x = 1, y = 2.
9. x = 18, y = 10. 10. An infinitely large number of solutions. 11. No
solution. 18. x = J, y = J. 13. x = 0, y = $. 14. x = 5, y = 2.
Article 40. Page 54
2. 154. 8. 41. 4. 39. 5. 15. 6. axv + byu. 7. x 2 -30x + 40.
8. 0.
Article 41. Page 56
2. x = 1, y = 2, z = 3. 8. x = 10, y = 8, z = 6. 4. x = 10, y = 2,
z = 3. 5. x = 2, y =— 1, z = 3. 6. x = 2, 2/ = 2, z = 2. 7. x= — 4,
y = 2, z = — 5. 8. No solution. 9. x = 7, y — 4, z = 3.
Miscellaneous Exercises and Problems. Pages 56-58
8. 0. 4. 2y — 2x. 5. 6. 6. x = J. 7. x = 2, y = 5.
8. 30 gallons of 25 per cent alcohol and 20 gallons of 30 per cent alcohol.
9. 3 oz. and 5 oz. 10. 18° and 72°. 11. 512,000 sq. ft. 12. 27 and 72.
13. 428,571. 14. 5 mo. 15. 6900. 17. 5 min. 29£ sec;
6 min. 45 sec. 19. c = 0.036, b = 999.5. 20. $20,000,
$15,000. 21. Analytic geometry 85, algebra 95, trigonometry 90.
248
ANSWERS
22. 2, 3, and 6 hr. 23. 12, 9, 3, 21. 24.
25. First 8, second 15. 26. 1020 bonds.
87. Going 9 miles uphill, 13J miles level, 4£ miles
downhill.
Article 44. Pages 62-63
f 2, 7, 12, 17, 22.
0, 6, 12, 18, 24.
4, 8, 12, 16, 20.
6, 9, 12, 15, 18.
8, 10, 12, 14, 16.
10, 11, 12, 13, 14.
1. 7,2.
8. -1,-7.
3. 5, - 4. 4.
-i-
5. -M-
6. 5, - f
7. i, - 4. 8.
1, -|.
9. -2, f
10. - 6, J.
11. 7, -2. 18.
-2ift.
13. 3 ± i.
14. h I
15. y, - J. % 16.
a, — a — 1.
17. 5, 0.
18. 8±V6.
19. _! ± iV2. 20.
m m
_ «
3 3
1 + e 1-e
21. 9.
22. 10.
83. 5. 24.
2.
25. 5.
26. 50.
87. 16,23. 28 .
&> f •
29. 9,33.
30. No solution.
81. 1.6- or 0.75. 32.
3.1 or -2.2
33. 16 or 6.2.
34. 5.4 or -0.51.
Article 45.
Pages 64-65
2. ± 2, ±
3. 8. ±iV7, ±L
4. ±2, ±2t.
5. 1, - 6,
- 5 ± WS
2
6. 1, 1, -8±2>/2.
7. 0.
8. 10.
9. f, - J. 10. 25, 9.
11. ±j.
12. - 1, 3.
18. -2a, a, £(- 1 ± iV3),
a(l ± iV3).
y A -1±VI3 -l±iV3
2 2
15. 2 ± 2 i, 2 ± V5.
16. 1.
17. 1,1.
fc iV3.
18. 1. 19. (_l±$v/21)«.
20. 2, - 1 ;
2! ^/-6iV62-4ac
2a
22. 2, -I
23. 0, 49.
24. 1.
25. 0, - 5,
_ 5 + V- 15
•
2
Article 46.
Pages 65-66
2. x 2 — x -
-6 = 0.
8. x*— 7x = 0.
4. x 2 + (5
-V5)x-5V5 = 0.
5. x 2 - 5 = 0.
•
6. x 2 - x •
2V5 + 4 = 0.
7. x 2 + 1 = 0.
8. X s — 8* + 25=5 0.
10. x 2 + x + 1 = 0.
12. *- (*+*)* +!
a
1.
k= ±VE.
8.
k = 3, or 1,
5.
fc = — 1, m
7.
A: = -J.
9.
fc = l.
11.
fc = 4.
ANSWERS
9.
X 2 -
- 4x
- 71 = 0.
11.
X 2 -
-$x
+ 1=0.
= 0.
13.
X 2 -
•
\ mn /
Article 48.
Page 68
2.
A: =
= 8.
249
+ 1=0.
= i-
4. m =— I, fc=— 8.
6. A; = 0, or — J, m =
8. fc=±2.
10. A; = 3.
12. k = 3, or - 1.
f
Article 50. Pages 69-70
1. Real and unequal. 2. Imaginary.
4. Real and unequal.
8. fc = 9.
11. Jfe= — 2, or — f.
14. k = ± rvT+m 2 .
w. - J, - H-
2 m m 2
5. Real and equal.
0. fc = y.
12. t = f
20.
1- e 2 ' 1-e 2
24. fc = -4.
15. k = ± y/a*m?+ b\
18. -*, -1.
m m
22. A: = 3.
25. fc = l.
8. Real and equal.
7. fc = ±4.
10. k = ± 5.
18. fc=l.
16. - *, - f
23. A; = 20.
26. A: = ±13.
2. 5£ per cent.
4. 33.89, 81.89.
6. 3.1+ sec.
8. 38.8+ ft.
Problems. Pages 71-74
3. 50 x 75.
5. 0.8+ sec. or 2.3~ sec.
7. 1.5+ sec.
9. One half the whole time.
10. 4.9 ft. Ball begins to fall before the end of the second second.
11. 3.7" sec. 12. 80.5 ft. per sec. 18. 144.9 ft.
15. 5.828+.
24. 1 ft. 4 in.
27. 33.47.
14. 2 in.
18. 0.625.
83. 14 x 7.
26. 8 ft. per sec.
16. 12 + 7V3 = 24.125.
22. 25.
25. 1 ft. 2 in.
28. 238 ft if time is measured to nearest tenth of a second.
250 ANSWERS
29. 198°-. 80. 5. 31. jg = 1* V l - 4 flgjg.
2g
32. »-tf= ± 2Tb » 33. x= ^^
34. 1 part of corn to 4 parts rye.
Article 53. Case I, Page 77
2. (W7, |), (|V7, -J), (- |V7, J), (-|V7, -J).
8. (0, 3), (0, - 3).
4. (^vi, fvii). ( 5 -V5, -I vh). (-^vs, f vn).
(-£v*-fv5i).
5. (5, 0) L (-6,0). __ __ __
6. (HV34, «VS4), (JW34, -«V34), (- tfV34, ^34)^
(-JJv/34, -J|VS4).
7- (*V85, ^ Vol), (ftVBs, - A Vol), (- AVBK, AV5T),
(-*V86,-aV5I).
8. (0.32, 3.1), (-0.32, -3.1), (0.32, -3.1), (-0.32, 3.1).
9. (1.1, 2.3), (- 1.1, 2.3), (1.1, -2.3), (-1.1, -2.3).
Case II, Pages 78-79
2. (1,-2).
8 / _i_8tV6 -18 + 6iV6 \ / JZ J_+8rv/6 -18-6iV6 \
4. (3, 7), (7, 3). 5. (4, 9), (9, 4).
6. (8, 5), (-5, -8). 7. (1, 6), (- *, -2).
8. (4, 10), (100, - 110). 9. (1, 2).
10. (1, 1), (-J, -J). 11. (2, 3), (-f|).
12. (0, 0), (4, 5). 18. (0, 0), (- 1, -2.4).
14. (1.4, -0.17), (0.034, -7.1). 15. (1.1, 2.2), (0.03, -0.71).
17. r>5V2, r = 5V2, r<5V2. 18. c<25, c = 25, c>25.
19. Two points for any real value 20. Two points for any real value
of a. of b.
21. b = ± rVl + ra 2 . 22. k=± y/afim* + b*.
Case III, Pages 80-81
2. (4, 1), (- 4, - 1), (14, - 4), (- 14, 4). 8. (- V5, V6), (V6, - V6).
4. (8,6), (-3, -6), ({,¥),(-*, -¥)•
5. (4, 7), (-4, -7), (7, 4), (-7, -4).
ANSWERS 251
6. {8, 5), (- SV- 5), (8, - 5), (-8, 5).
^V21, *V 21 )» (- $ V2I, - W 21 )-
^6, 3), (- 6, - 3), (6, - 3),_(- 6, 3).
;4, 6), (-4,-6), (3V3, V3), (-3V3, -V3).
< 2 ,4), (-2,-4), (ISivioi. ^-VfolV (zL^vior, ^VToiV
* ' " v ' " \101 ' 101 ) \ 101 101 /
;*V5, |V5), (-*VS, iV5).
;2.7, 0.85), (2.7, -0.85), (-2.7, 0.85), (-2.7, - 0.85).
10, 1.4), (- 10, - 1.4), (11, 17), (- 11, - 17).
Case IV. Page 82
5,0), (0,5), (1,-4), (-4,1).
3 2^ (2 S"i Z -l + ^23 -l-iV23 \ / -l-iV23 -1-HV23 x
' h ^ ' h \ 6 ' 6 J' V 6 ' 6 J'
S, „, (1, S), ( -» + V«B -lO-VSO^
(
-19-V309 - 19 + V3 09\
8 ' 8 j'
2, 8), (3, 2), (- 5 + iy/U, - 5 - iVIT), (- 5 - iVfi, - 5 + iVIl).
3,2), (2,3), (-3, -2), (-2, -3).
Exercises and Problems. Pages 82-85
8, 1). (- -fr, ~ «)• 2. (2, J), (- j, - J).
A. ». (- i, - A)- *• (1, 2), (- A, - V).
2, 7), (18, |).
3, 5), (-3, -5), (!,¥),(_!, _Y).
6. 2), (3, 5). 8. (2, 3), (- tf» - Jft).
2, 11), (J, 12). 10. (5, 4), (V-, - ¥)•
9, 5), (-^, -V).
1,2), (i, J), (-i + iV3, -3-tV3), (-l-iV3, -3+i-v/3).
« <n / 9 s \ /5+V285 5 + V266\ /6-v / 265 5-V265Y
°' 2) ' 1 _ 2' "I)' ( — ' r~ j' {— I" ' 4— /
2, 5), (J, V)- _
7, 1), (- 49, - 18), (ivll, - iVll), (- iVll, iVll).
4,2), (-2,-1), (1=^, =±±2%), (U^n, =I^H).
-4, -6), (5,3), (=i^M, =«±2^l),
- 9-Viil -3-\/J4i \
"~2 ' 2 )'■
('
252 ANSWERS
18. (3, 1), (1, 3). 19. (3, 2), (- 2, -3). 90. (4, 1), (- 1, -4).
21. (0, 1), (2, 5), (iV2, - iv/2),(- 1V2, *V§).
22. (9, 4), (4, 9), (6, 6), (- 6, - 6). 28. (4, 1). 24. (5, 0), (- 6, 9).
25. (2a, a), (- 2 a, - a). 26. (0, a), (a, 0).
27. (2, 8, 4), (2, - 3, 4), (2, -3,-4), (2, 3, - 4), (- 2, 3, 4),
(-2,-3, 4), (- 2, 3, - 4), (- 2, - 3, - 4).
28. (2, 1, 1), (V, - 10, - V). 29. (1, 8, 6), (- 1,-3,-6).
80. (8, 8, 6, 4),* (8, 3, 4, 6), (8, 8, 4, 6), (8, 8, 6, 4).
81. (1, 2, 3, 5), (2, 1, 5, 3), (J, f, |, |), (J, *, | f J), (3, 5, 1, 2), (5, 3, 2, 1),
(i, I. i. 1), «, I, Si 1).
82. (1, 1, 2, 2, 3),t (2, 1, 1, 3, 2). 88. 6* + c* + 2a 2 .
84. a4-6* = 4c 4 . 85. a* + & 4 = c«(a 2 + & 2 ). 86. a4 + &4 — c«.
37. 628.5, 653.3. 88. 300 and 400. 89. 12, 6, and 4.
40. $ 550, $450 ; 7 %. 41. 4.4 cents, 7.5 cents.
42. 0.1556. 48. 10 and 12.
44. 24 lots, $200 per lot. 45. Radius, 2VB, altitude, &V&.
46. 467.8. 47. 9 x 13 x 21 ft.
48. 5.4 x 8.15 inches. 49. 100 miles, 25 miles per hour.
50. x 2 - 10 a; + 9 = 0. 51. 12 ft. 11 in., 16 ft. 1 in.
52. 7 and 5. 53. 36 miles per hour.
Article 57. Page 89
1. x>5. 2. x>f. 8. x<9.
4. — 2 < x < 2. 5. x > 4 and * < — J. 6.x positive but j= 1,
7. x<|. 8. x>5andx<— 2. 9. — 8<a><|.
10. 0<x<a, ifa>0;a<x<0, ifa<0.
11. x > — , if a is positive; x < — , if a is negative.
a a
12. x 2 - 4x + 3 > 0, if x > 3 or < 1.
= 0, if x = 1 or 3.
< 0, if 1 < x < 3.
18. ax 2 -f bx + c > 0, if a is positive and x =£ — — ±vo — 4ac ^^ ^
tween these values.
ax* + to + c = 0,ifa ; = - 6 + V6 *- 4ac -
2a
ax 2 -f 6x+ c < 0, ifaisa positive and x lies between """ "^^ ~ •
2a
and^- V * 2 - 4ac
2a
* (8, 3, 6, 4) means 05 = 8, y = 8, «=6, t*=4.
t (8, 1, 2,»2, 3) means aj=l, j/=l, e=2, w=2, « = 3.
ANSWERS 253
14. ?-±_§>0,ifx>8or<-5; ^±^<0, if - 5 <»<3.
x— 3 x — 3
15. x < — 4, and x > 4. 16. For all real values of x.
Article 59. Page 93
1. 604. 2. 8. 3. 30. 4. V- 5- 84 - *• r * - r -
Article 60. Pages 95-96
6. fa + a + 6 a 2 + 16 a 3 + 16 a 4 .
8. a 5 — 15 xfy 2 + 90 xty 4 — 270 x*y* + 405 xy* — 243 y 10 .
a. a« + 6 a 5 V& + 15 a 4 6 + 20 a*b Vb + 15 a 2 6 2 + 6 ab*Vb + & 8 .
* A ~4 . j 9 . a ■ 4 . 1 11 12 . 40 . 12
10. x 4 + 4x 2 + 6 + — + — • 11. H-r + — •
x 2 x 4 e e 8 e 5
15. a 4 6 4 - 8 gW + 24 a 2 6 2 - 32 aft + 16.
13. x 4 — 8 xM + 28 x*y — 56 xM + 70 x 2 y 2 - 56 x$^ + 28 xy* - 8 x^ + y 4 .
14. 91J. 15. a9 + 6a^ + 16a^ + 20a^~ + 15a 1 ^~ + 6a^ a + a 4 .
16. x- 4 — 4ar 3 y* + ex-ty* — 4x-ty + 2/*.
17.82. W . *-«^+ »*-»***+ 16^- •«$* + »!.
19. a» + 8a J 6 + Sa& 2 + &» + 3a«c + 6a&c + 3 6 2 c + 3ac* + 3 6c' + c».
so. .i + e. + u.+ Mxl+^ + A + iJ + l + i..
». ^ + i^l,3 + 2 j4 + 4^8 x + 1 J l_^ 16 24
81 27 27 S3 3 x 3x 2 x 3 x 4 x»
32 16 32
3x7 x" x? '
28. - 14,080 cflb*. 24. 2489344x7^0. 25. 5670 xfy».
J*!-M*. 27. - J»L*W*. 28 ' 126 * ¥ ^
5 ! 13 ! 7 ! 6 ! e i
29. -35 a*, +35 a*.
32. 11,040,808,032. 88. 941,480,149,401. 34. 345,025,251.
85. 0.886842380864. 36. 2.594. 87. 4.177.
Article 64. Pages 98-100
2. y = 3x. 3. 8 = 16^. 4. p=?^?.
v
5. 2 = 20xy. 6. ^ = 180x^ 7 V=kr*
254 ANSWERS
8. V=khr* 9. 402.5 ft. 1610 ft. 10. 1896+ lb.
= vhr*
fan 73
11. 25 mi. per hour. 12. 48 lb. 13. b = ^--
14. 888$ lb. 15. 0.12 in. 16. 7.5 sq ft.
17. 3.4- 18. f^yT. 20. 3V5±Vl7.
21. 0.087- in. 22. 1000 yr. 23. 0.05+ in.
24. 441,720 1b.
Article 68. Pages 102-103
2. l- -59, s= -610. 3. 1 = 59, 8 = 465. 4. I = - 71, s =-660.
5. I = A, 8 = 3. 6. Z = y, s = 142. 7. I = J}* * = Vf-
8. n = 7 or 13, I = 7 or - 5. 9# a = 3? 8 = 455.
10. a = -7, 8=^J^5. 11. n = 20, d=-4.
1 2 ■
12. n= 16, 8 = 142. 13. d=-3, 8 =-714.
14. a = -$, n = 23. 15. a=l, d=l.
16. a = -44, 1=46. 17. J =-5, d=-2.
19. 5, 8, 11. 20. 13J. 21. j, I, f, i, 0, - J, - $, - f, -*.
Article 72. Page 104
1. I = 6561, 8 = 9841. 2. I = - 4374, 8 = - 6560.
3- *=:nk¥» *=t8H- 4 - * = -768, 8 =-510. 5. r = 3, 1 = 162.
8. .62.
11. 16, 128.
6.
6561.
7. y.
9.
r = 3, 8 = 120.
10. 49.
2.
f
18. ?.
a*
1.
f 2. f.
Article
3. f
7.
A- «• f
9. j.
•»•
4. 16. 5. V 1 - •• -¥•
10. a- ". 4°W- *»• tttt*
Article 77. Page 107
1. 4, 6. 2. i, f 3. ,V, J, h i- *. -*f&
Problems. Pages 108-109
1. 30 feet. 2. 948} J and 14475 feet.
3. Between 185 and 190 feet per second.^ 17 4"""
4. 19.683. 5. 122.853. 6. 647. 7. 36.
8. 45. 9. $885.73. 10. 2900 feet:
ANSWERS 255
11. The latter proposition is worth $25 per year more to the clerk.
12. 12,092. . 13. -• 14. n*.
n
15. x 2 -14x + 48=0. 17. -J. 18. 6912.
19. 6, 8, 10, 12. 20. 3, 6, 9. 21. $7721.73.
Article 80. Page 113
8. r = Vl3, = arc tan f . 3. r = Vl3, = arc tan — §.
4. r = y/2,6 = -- 5. r = V2,0 = --.
4 4
6. r = ViT, = arc tan — $. 7. r = 1, = — - •
8. r = 4, 0=-. 9. r = 6, = ---
'2 '2
10. r = 5,0 = -. 11. r = 4, = 0.
2 2
12. r = 5, = it. 18. r = JV13, = arc tan — $.
14. r = Vl.7, 0=arctanJ^. 15. £\/3 + fi.
16. l + tV3. 17. -V3 + t. 18. 4i.
19. -Jv^-ivli. 20. 6. 21. -li.
82. g=— l,y = l. 28. x = 3, y = — 1. 24. x =— J|, y = — |fe.
85. x = 0, y =- 1, or x = f y =-*J.
86. x = 0, y = — 5, or x = 2, y = 1.
87. x = 9, y = 7, or x = — 9, y = - 7.
Article 81. Page 114
1. 3 + 4t. 2. 3-i 3. l-3i 4. 6. 5. -lOi.
6. 2-5i." 7. \-i 8. - l + 5i 9. 2-3i. 10. 0.
Article 83. Page 116
8. 4 + 4iV8, = -, r = 8. 8. 1? V3i, 6 =1, r = i?V3.
3 3 2 3
4. 4i,0=£,r = 4. 5. -4, = *-, r = 4.
6. 4\/3 + 4i, = £, r = 8. 7. - 8, = it, r = 8.
6
8. -8i, = 5^, r = 8. 9. -10, = *-, r = 10.
10. 6i, = - , r = 6. 11. 4, = 0, r = 4.
2
256 ANSWERS
»
Article 85. Page 118
2. -432 + 144iV3. 3. 24tV8. 4. 1/ 5. 1
6. 512 1 7. -128-128i. . 8. 82 i.
10. - 1 + i, V2(co8 16° 4- i sin 16°), V^cos 256° + i sin 266°) .
11. 2 + 2tV3, -2-2iV3.
12. \/12(cos 15° + i sin 15°), \/l2(cos 196° + i sin 196°).
18. 4V3 + 4i, -4V3-4i.
14. 4(cos 20° 4- i sin 20°), 4(cos 140° + % sin 140°), 4(cos 200° + * sin 260°) .
15. 2(cos 10° + i sin 10°), 2(cos 70° + i sin 70°),
2(cos 180° + isin ISO 5 ), 2(cos 190° 4- isin 190°),
2(cos 250° + i sin 260°), 2(cos 810° + i sin 310°).
These results are represented graphically by the six points at equal inter-
vals on a circle of radius 2, beginning with $ = 10°.
16. \ V3 + £ i, cos m° + i sin 66°,
cos 102° + i sin 102°, cos 138° + i sin 138°,
cos 174° + isin 174°, - J V3 - J i,
cos 246° + i sin 246°, cos 282° + i sin 282°,
cos 318° + i sin 318°, cos 354° + i sin 364°.
These results are represented graphically by the ten points at equal inter
vals on a circle of radius 1, beginning with 6 = 30°.
17. 1 + i, V2(cos 117° + isin in°),
V2(cos 189° + i sin 189°), V2(cos 261° + i sin 261°),
\/2(cos 333° + i sin 333°).
18. cos 18° + i sin 18°, i,
cos 162° 4- i sin 162°, cos 234° 4- i sin 234°,
cos 306° + i sin 306°.
19. |VJ+|i, -fVS+fi, -3?. 20. ±2.
21. 1, -i + £V3i, _i-£V3i. 22. 2, -l + iVS, -l-iV'k
28. 3(cos 22J° 4- i sin 22J°),
3(cos 112J° + i sin 112i°), 3(cos 202|° + i sin 202*°),
3 (cos 292 J° + i sin 292J°) .
24. 3, -14- fiVS, - f -|iV3.
25. 1, cos 72° 4- i sin 72°,
cos 144° 4- * sin 144°, cos 216° 4- i sin 216°,
cos 288° 4- i sin 288°.
26. 2, 2(cos72°+ i sin 72°),
2(cos 144° 4- * sin 144°), 2(cos 216° + i sin 210°),
2 (cos 288° 4- i sin 288°).
27. See answer to 21. 28. ±2, ± 2i.
29. 1, i + *iV3, -i4-iiV3, -1, -i-JiVS, i-itVS".
80. ±1, ±h ±iV2, ±£V2t.
ANSWERS 257
Article 86. Page 120
8. l+i'VS. 3. -i + i*V3. 4. 4i. 5. -jV3 + Ji. 6. -l+£
7. 1-i. 8. -J-Ji. 9. t. 10. |v/3-ii.
Article 88. Page 122
8.-1. 4. fc=-8.
Article 91. Page 125
2. 0, 2, 4. 3. 1, 2, 4. 4. Zero at 1, between 2.5 and 3, and between
— 2 and — 1.5. 5. Zeros at — 2, — 1, 2, 3. 6. Zero between 1.5 and
2, and between — 2 and — 2.5. 7. Zero between — 1.5 and — 1, between
— 1 and — 0.5, between and 0.5, and between 2.5 and 3. 8. Zero at 2,
between 0.5 and 1, and between 2.5 and 3.
Article 97. Page 131
8. (a) x3 - 10x 2 + 31 x - 30 = 0. (6) x 2 - 2 x + 5 = 0.
(c) x 3 -5x 2 + 5x + 3 = 0. (d) x 4 -6* 2 + 6 = 0.
(e) x 4 - 2x 3 - 5x 2 + 6x = 0. (/) x 4 - I4x 2 + 1 = 0.
Article 98. Pages 134-135
1. x3 + 10x 2 -175x- 125 = 0. 2. x3-z 2 -40 = 0.
8. x 4 - 10x 2 + 3x-2 = 0. 4. «3 + 6x 2 -80 = 0.
5. x 3 -6x 2 + 3x-270 = 0. 6. x 4 + 3x 3 - 12x 2 + 20x - 8 = 0.
7. x« + 108 = 0. 8. x 3 + 6x 2 + 6x - 498 = 0.
11. x 2 — x-6 = 0. 12. x 3 + 3x 2 -4x+ 1 = 0.
13. x 3 - 9x 2 — 90 = 0. 14. x 5 - .7 X 3 + 2 x - 8 = 0.
15. 2x 4 -3x2 + 4x-5 = 0.
Article 99. Page 137
8. 4 imaginary roots. 3. One negative and 4 imaginary roots.
4. 1 negative and 2 imaginary roots. 5. 6 imaginary roots. 6. 1 posi-
tive and n — 1 imaginary roots. 7. 1 positive, 1 negative, and n — 2
imaginary roots. 8. 2 positive and 1 negative roots.
Article 100. Page 138
4. 1. •••, —0. •••) — 1. •••. 5. — 1. •••• 6. 2. •••. 7. 1. •••$
8* — 1. •••, 11. ••*, 0. •••, 0. •••• 9. 0. •••, 1. •••, 2. •••. 10. — 1. •••j
1. •••) o. ••• 11. 2.
258 ANSWERS
Article 102. Pages 140-141
3. — 1, 2, 3, - 4. 4. J. 5. 2, — 3, — 2.
6. -i- 7. -J. 8. I,},*.
9. 3. 10. 0. 11. 1.
12. - 4, 5, 3. 13. 2, 6, - 7, - 1. 14. J.
15. No rational roots. 16. J. 17. 1,2,-2,8,4.
Article 105. Pages 146-147
1. 3.13-. 2. 5.24-. 3. 0.64. 4. 4.64. 5. 3.98. 6. 3, 1.41,
-1.41. 7. 6.17. 8. 2.36-, 2.69, -2. 05". 9. -2.21. 10. 2.24-
11. 3.01,0.63,-2.02,-0.95. 12. 5.83,0.27,0.93,-1.03. 13. 1.88,
-0.35,-1.53. 14. 1.18,2.87. 15. -3. 16. 1.73. 17. 4.00 per cent.
18. 0.606. 19. 0.860. 20. 0.32, 0.64. _21. 8.86. 22. 11.07. 23 2.92.
24._a_259. 25. 1.46. 26. 2 v^J* - 8 = 1 .20, Jv^ - 10 = 1.50,
3v^|* - 12 = 1.80. 27. $ 1.536.
Article 108. Page 152
8. - 3, w, a* 4. f , i, - i. 5. f , + V2, - V2.
6. J, J, §. 7. 2, - 1, - 1 ± iV2. 8. - 1, 1 ± 1«V2.
9. -2,-5±2 iV3. 10. 4, 1, 3 ± 3 V2.
Article 110. Pages 154-155
1. 2, 2, - 2. 2. f , ± i. 8. - 1, 2, 5. 4. 6, 6, — 2.
5. For a = 10, { 1.218, 0.082 } ; a = 1, { 12.923, 0.077 }.
a = 0.1, / 129 * 923 ' 1 a = 0.01, { 1299.923, 0.077},
' \ 0.077, /' l s
a = 0.001, {12999.923, 0.077}.
6. For a = 10, { 12.179, 0.821 }, a = 1, { 12.923, 0.077 },
a = 0.1, {12.992, 0.0077}, a = 0.01, {12.999,0.0008},
a = 0.001, {13.000-, 0.000+}.
7. m=±\/3, b=T |V3. 8. m = ±t, b = T^- 9. me±f>d=0.
10. m = -3, 6=-2, or m = -^, 6 = ^ f .
Article 113. Pages 158-159
2. 21og2-3 1og3. 8. \ log 13 - J log 10 - 1 log 48.
4. \ log 25 - l log 11 - J log 23.
5. i log 5 + I log 7 - 4 log 2 - 4 log 3 - J log 2 - 1 log a
6. - y. log 2 - V log 3 - § log 6 - J log 11.
ANSWERS 259
7. 31og2 + ^log5-|log3. 8. ^log 2 + Jlog 3 + flog 7.
9. S log 2 + J log 11 + J log 3 + f log 5.
11. 1.0791. 12. 1.14771. 18. 1.6232. 14. 2.6232.
15. 2.2764. 16. 2.9542. 17. 2.5353. 18. 1.3313.
19. -0.1370. 20. -2.4013. 21. -3.0124. 22. 1.3512.
0.8228. 24. -0.1605. 26. 0.6469. 26. 0.7726.
Article 119. Pages 166-169
1. 0.01359. 2. 0.2332. 8. 0.03906. 4. 24.15.
5. 8.641. 6. 0.7038. 7. 1.673. 8. 0.4852.
9. -0.4704. 10. 0.2979. 11. 0.03229. 12. 0.7295.
18. 0.9630. 14. 0.3857. 15. 0.4420. 16. 64.94.
17. 0.2917. 18. -0.1606. 19. 6.636. 20. 27.89.
21. 25.31. 22. 3.162. 28. 0.0696. 24. 0.9047.
25. -0.8646. 26. 1.639. 27. 1.01 sec. 28. 142.5 tons.
29. Volume = 13,330, Surface = 2719. 80. 1061 . 10*.
81. 11,660. . 82. 834,200. 88. 1,476,000. 34. 0.608.
85. 476. 86. 4.578. 87. 100 pounds.
88. 0.125 cubic foot. 39. p = 4.629. 40. 177.6.
41. 0.0068. 42. 24,470. 43. $2014.
44. (a) 14,790. With a seven-place table the following
(6) 14,860. more accurate results are obtained :
(c) 14,860. (a) 14,802.4.
(6) 14,859.4.
45. $3767. (c) 14,888.7.
46. 41 digits until the year 1935, when it will require 42 digits.
47. 4.251.
48. (1) 100,100; more accurate value 100,081 ; (2) 86,460; more accurate
value, 85,442.
49. 1547 miles.
50. 146,700 sq. kilometers.
Article 120. Page 170
2. 1.3862. 3. 0.6825. 4. 2.7302. 5. 3.822.
6. 1.661. 7. 2.096. 8. 1.431. 9. 3.980.
10. 0.856. 11. 1.625. 12. 2.393.
260 ANSWERS
Article 122. Pages 173-175
4. x = 0.8115. 5. x = 0.1853. 6. * = ± 2.390.
7. x=lor2. 8. 3 = 3 or — 1.
9 . n = logrl-loga 1Q log [(r - 1> + a]- loga
log r log r
12. x = 1.61, y = 2.56. 13. * = 6.84, y = 10.84.
14. (1) 11.90, (2) 11.68, (3) 11.55.
With a seven-place table of logarithms, the following more accurate result
may be obtained:
(1) 11.89, (2) 11.64, (3) 11.55.
15. 0, ± 1.32. 16. 3.96. 17. 0.00003776. 18. 18,360.
19. A; = 0.126. 30. 5.5 minutes. 21. x=^-=J.
3
22. x = 25 and - 4. 28. s = ^(e* - e~*).
Article 128. Pages 181-182
1. -J !_. 8 . 2 + 5 1
x-3 x + 2 x 8(1 - x) 3(x + 2)
3. 3 +, * • 4. -^ + 2 2
4(x - 3) 4(x + 1) x-1 (x-l)» (x-l)» x+2'
5. -? L.. 6. -1 — *-+ 5
x-1 x— 2 2x x-1 2(x — 2)
-2,3 * 1 , 2
7. H • 8. - +
X + 1 X - 1 X (X - 1)2
9. -ZJ-+ 1 +l£±J. 10. x + i+l+-i_+ *
X — 1 (x — l)' 2 x 2 + 1 X X — 1 X + 1
11. —1 6 -+— I 12. 2 * + 2 !
2(x-l) x-1 2(x-3) x-3 x2+l z + 1
1Q 2 , 5 5 1A 1 , 2 3
13. 1 • 14. - +
x— 1 x— 3 x— 2 x x— 3 x+2
15. 1 §_ + _J? 16. I- *
2(x-l) x-2 2(x-3) x x 2 + 4
17. _i 4 *+ 8 . 18. -?-!-!+ 2
3(x-l) 3(x 2 + x + l) x x2 x* x — 1
19. 1 + ! 1
20.
1-lOx 3(1 +3x) 3(l+3x)2
11 7 11 11
8(x-3) 2(x-l)3 4(x-l)2 8(x-l)
ANSWERS 261
«. ., 8 „ ,. + _" + 21 „ , +
2(1 - 8x)« 8(1 - 3x)* 82(1 - 3as) 32(1 + x)
1 + 2 ~* ■ 23.-1-4 X
a; + 2 x 2 — 2a; + 5 x-1 (x 2 — x + 1)
2
24. _2_+_2_+2jMl2. 26 . _J 2_ + _3x-4_
x-fl x — 1 x 2 +l »+l x — 1 x 2 + x+l
. gg _^ 27 27 27 6__1 5 4_
* " 2(2x-3) 2(2x-3)3 (2x-3)4* x x 2 x+1 (x + 1)
28.-2 3__x^ip > 29 . x + 2 + „3, + 4
2
x — 2 x -f 2 x 2 + 4 x + 1 x — 5
80. -i * + * 2x + 3. . # 31 J__2 _8 5_
x — 3 x 2 -x+l (x 2 -x+l) 2 x 2 x x— 1 (x — l) 2
82. 3 + -5 ^ ™-. 33. -L + - 2 — i^+i
x + 2 (x + 2)2 (x + 2)' x + 3 x - 3 x* + 3
84. , * ±+ *-*
(X — l) 2 X — 1 X 2 — X + 1
85. 1 1 + * . 86. x + 2 x
x 2 + x + 2 x 2 + x + 1 (x 2 + x + 2) 2 x 2 + 1 (x 2 + l) 2
87. -2- + _I_ + -l*+i_. 38. _J_ + * 3 + 2x 2 + 2 <
1-x (l-x)3 Hx + x2 3x + 2 (x 2 +l) 2
Article 132. Pages 185-186
1. 720. 2. 5040. 3. 60. 4. 1,860,480.
5. 40,320. 6. 8. 7. 3360 ; 34.650. 8. 5040.
9. 336. 10. 360. 11. 30. 12. 180.
Article 136. Pages 187-189
3. 31.
6. 1140 ; 220.
2. 9. 31.
12. 462.
16. 327,600.
19. 20.
22. 4095.
25. 3,303,300.
1.
210.
4.
502,601.
7.
10.
10.
63.
13.
37T
17.
45.
80.
30,240.
23.
1820; 456.
26.
100!
60 ! 20 ! 20 !
2.
40.
5.
70.
8.
n = 17, r
11.
25.
15.
18.
21.
4950.
52! 4!
(13 !)4
3360.
24.
10,080.
262 ANSWERS
Article 139. Pages 191-192
1. |f. 8. Jf. 4. tV 5. $6. 6. 0.60804. 7. (1) fc (2) £;
(3) A- 8. $ 3.
Article 140. Page 193
i. a. a. a- 3. ^. 4. a- *• f •• «• 7. A- •• A-
9. (1) 0.03; (2) 0.68; (3) 0.12; (4) 0.17.
Article 141. Pages 194-195
1. (1)A ; (2) |J. 2. (1) 0.2886; (2) 0.9446.
3- iHH- «• «•
5. a 10. 6. (1) 0.108 ; (2) 0.994.
7. The latter. 8. (1) 0.087 ; (2) 0.503.
10. «h. 11. H
12. (1) I; (2) flfc. 18. A's expectation is $36; B'sis$30.
14. (i) ._64i™?_- 0.000852; (2) 89,827 = 0.00117 ; (8)0.00000100;
y 75,994,575 v J 75,994,575 W
(4) 76.
Article 144. Page 202
2. - 25. 8. 5. 4. 2. 5. — 99.
Article 149. Pages 211-212
1. 2, 1. 2. 1,0, — 1. 8. J, oo, 1.
4. 2, 4, —1,-3. 5. A; = 4. 6. Z :y : Z=— 1 : 1 :2.
7. 6x«-2x 4 — 9x 2 + 6x- 1 = 0. 8. a* + 6* + c* - Sofcc = 0.
9. Yes. 10. 2, or - if ±.
Article 156. Page 220
1. 3. 2. -J. 8. 2V2. 4. J.
5. _§«. e. 4. 7. i. 8. 1.
2 a
9. Increases without limit when x-> — 1.
10. 1. 11. 0. 12. i.
Article 163. Page 231
1. Convergent. 2. Convergent. 3. Convergent.
4. Divergent. 5. Divergent. 6. Convergent.
7. Convergent. 8. Divergent. 9. Convergent.
ANSWERS
263
Article 167. Pages 235-236
1. Convergent.
4. Convergent.
7. Divergent.
10. Divergent.
13. 0.7834.
16. Convergent.
2. Convergent.
5. Convergent.
8. Divergent.
11. 0.8965.
14. 0.2877.
17. Divergent.
3. Convergent.
6. Convergent.
9. Convergent.
12. 0.4794.
15. Convergent.
18. Convergent.
19. Divergent for all values of x.
20. Convergent for | x | < 1, also for x = — 1.
21. Convergent for x > 0, and x < — 1.
28. Convergent for x > — 1 and x < — 3.
22. Convergent for \x\ < 1 .
Article 168. Pages 237-238
2. Convergent for | x | < \. 3. Convergent for | x | < 3.
4. Divergent for all values of x, for which the series is denned.
5. Convergent for — oo < x < oo .
7. Convergent for | x | < 1.
9. Convergent f or | x \ < \.
11. Convergent for | x | < 1.
18. Convergent for — oo < x < oo .
14. Convergent for — 1 < x < 1 and for x = — 1.
15. Convergent for — 1 < x < 1. 16. 0.0872-.
6. Convergent for - oo < x < oo.
8. Convergent for | x \ < 1.
10. Convergent for | x \ < J.
12. Convergent for — oo < x < oo.
4.
Article 169. Pages 239-240
1. 1 - 6x + 27 x 2 - 108x3 + 405x 4 -.
2. 1 + x— Jx 2 + |x3 — fx 4 + ....
X
8. 1 +
9 81
5 X3-^X4 +
243
V2
(1 + f x + § £x* + Hi* 3 + Hfl & +
12. 2.236.
17. 5.066.
13. 3.873.
18. 9.008.
14. 1.010.
19. 2.002.
x|<l.
■)• |»l<*.
15. 9.993.
20. 0.8016.
16. 2.005
2. 0.2877.
Article 171. Page 242
3. log e 16 = 2.772. log, 17 = 2.833.
INDEX
[Numbers refer to pages.]
Abscissa, 33.
Absolute inequality, 86.
Absolute value, 87.
Addition, 3.
associative law of, 4.
commutative law of, 3.
of radicals, 26.
Algebraic expressions, 18.
Algebraic reductions, 18-30.
Algebraic solution, 148-149.
Alternating series, 234.
Amplitude, 112.
Argument, 112.
Arithmetical means, 102.
Arithmetical progression, 101.
common difference of, 101.
elements of, 101.
nth term of, 101.
sum of, 102.
Associative law,
of addition, 4.
of multiplication, 4.
Axis,
of imaginaries, 111.
of reals, 111.
Base of system of logarithms, 156, 169.
Binomial expansion, 94.
general term of, 94.
Binomial series, 238.
Binomial theorem, positive integral
exponent, 93.
Biquadratic equation, 151.
Cardan's formula for cubic, 150.
Circle, 77.
Clearing an equation of fractions, 47.
Coefficients,
denned, 24.
in terms of roots, 153.
variable, 153.
Combinations, 186.
Combined variation, 98.
Common difference, 101.
Commutative law,
of addition, 3.
of multiplication, 4,
Complex numbers, 110.
absolute value of, 112.
addition of, 113.
amplitude of, 112.
argument of, 112, 115.
conjugate, 115.
" division of, 119.
equal, 112.
extraction of roots of, 117.
modulus of, 112.
multiplication of, 114.
subtraction of, 113.
Complex roots of equations, 129.
Conditional inequality, 86, 88.
Conjugate complex numbers, 115.
Constant, 31.
of variation, 97.
Continuous graphs, 36, 125.
Convergence of infinite series, 221.
comparison test of, 224.
ratio test of, 229, 232.
Coordinate axes, 33.
Coordinates, system of, 33.
Cubic equation, 149.
Decimals, repeating, 106.
Defective equation, 45, 46.
Degree of an equation, 43.
De Moivre's theorem, 116.
Descartes' s rule of signs, 135.
Determinant,
defined, 50.
elements of, 50, 53.
evaluation of, 202.
minors of, 201-203.
of nth order, 198.
of second order, 50.
of third order, 53.
principal diagonal of, 50, 53,
197.
properties of, 198.
solution of equations by, 50,
54.
Direct variation, 97.
Discriminant, 68.
Distributive law, 4,
265
266
INDEX
[Numbers refer to pages.]
Divergence of infinite series, 221.
comparison test of, 227.
ratio test of, 229, 232.
Division, 5.
Division by zero, 6.
Eliminant, 211.
Elimination, 209-212.
Ellipse, 77.
Equal complex numbers, 112.
Equality, 41.
conditional, 41.
identical, 41.
Equations,
algebraic solution of, 148.
biquadratic, 151.
cubic, 149.
defective, 45.
degree of, 43.
equivalent, 44.
equivalent systems of, 44.
exponential, 172.
fractional, 47.
general, of degree n, 126.
graphical solution of systems of, 52.
incompatible, 52.
inconsistent, 52.
in p-form, 139.
in quadratic form, 63.
linear, 49.
number of roots of, 66, 126.
quadratic, 43, 59-74.
quartic, 43.
quintic, 43.
rational integral, 43.
roots of, 43.
simultaneous, 75.
solution by radicals, 148.
symmetrical, 81.
transformations of, 131.
Equivalence of equations, 44-47.
Evaluation of formulas, 27.
Expansion,
binomial, 94.
Expectation, 191.
Exponent,
fractional, 12.
negative, 13.
positive integral, 10.
zero, 13.
Exponential equations, 172.
Exponential series, 240.
Extremes, 102, 104.
Factorial, 92.
Factoring, 21.
solution of quadratic by, 62.
Factor theorem, 126.
Ferrari's solution of the biquadratic,
151.
Fraction,
complex, 19.
definition of, 5.
rational, 110.
Fractional equation, 47.
Function,
denned, 31.
defined at isolated points, 36.
graph of a, 35.
rational integral, 42.
zeros of, 39.
Functional notation, 32.
Fundamental theorem of algebra, 126.
General equation of degree n, 126.
Geometrical progression, 103.
elements of, 103.
infinite, 105.
ratio of, 103.
nth term of, 103.
sum of, 104.
Graphical representation of complex
numbers, 111.
Graphical solution of systems of equa-
tions, 52.
Graphs,
continuous, 36, 125.
discontinuous, 37.
Graphs,
of functions, 35.
of functions with imaginary
factors, 130.
of polynomials, 125, 128-130.
of quadratic functions, 70.
Groups, substitution, 148.
Harmonica! progressions, 107.
means, 107.
Horner's method, 141.
Identical equality, 41.
Identity, 18, 41.
Imaginary numbers, 30, 110.
conjugate, 115.
Imaginary roots, 68, 130.
Incompatible equations, 52.
Inconsistent equations, 52.
Indeterminate forms, 218.
INDEX
267
[Numbers refer to paces.]
Index laws, 10.
fractional exponents, 12.
negative exponents, 13.
positive integral exponents, 10.
zero exponents, 13.
Induction, mathematical, 90.
Inequalities, 86.
absolute, 86.
conditional, 86, 88.
sense of, 86.
transposition of terms of, 86, 87.
Infinite geometrical progressions, 105.
Infinite roots, 154.
Infinite series, 105, 221.
alternating, 234.
binomial, 238.
convergence and divergence of,
221.
exponential, 240.
logarithmic, 241.
power, 236.
Infinitesimal, 214.
Infinity, 217.
Interpolation, 164.
Inverse variation, 97.
Irrational numbers, 23, 110.
Irrational roots, 141.
Joint variation, 98.
Knowns, 42.
Limit of a variable definition, 213.
theorems concerning, 215.
Limiting value of a function, 218.
Linear equations, 49.
simultaneous, 50.
Location of roots of an equation, 137.
Locus of an equation, 75.
Logarithmic series, 241.
Logarithms, 156.
calculation of, 175.
characteristic of, 160.
common, 159.
mantissa of, 160.
modulus of, 170.
natural or Naperian, 159.
properties of, 157.
table of, 162, 163.
Mathematical induction, 90.
Means,
arithmetical, 102.
geometrical, 104.
harmonical, 107.
Multiple roots, 127.
Multiplication, 4.
associative law of, 4.
commutative law of, 4.
distributive law of, 4.
of radicals, 26.
Multiplying equations, 44, 45.
Nature of roots of quadratic, 68.
Negative numbers defined, 5.
Negative roots, 145.
Number, of roots of an equation, 66,
126.
reciprocal of a, 5.
Number systems, 110.
Numbers,
complex, 110.
graphical representation of com-
plex, 111.
graphical representation of real, 2.
imaginary, 30, 110.
integral, 110.
irrational, 23, 110.
positive, 110.
rational, 23.
real, 1.
Numerical equations, 126.
Ordinate, 33.
Origin, 33.
Parabola, 70.
Parentheses, rules for removal of, 19.
Partial fractions, 177-182.
Permutations, 183.
Polar form of complex number, 112.
Polynomial,
denned, 36.
degree of, 121.
graphs of, 35, 125, 128, 130.
of the nth degree, 121.
zeros of, 126.
Positive roots, 135.
Power series, 236.
convergence of, 236.
Probability, 190.
derived from observation, 191.
Product of roots of quadratic, 69.
Progressions,
arithmetical, 101.
elements of, 101.
geometrical, 103.
harmonical, 107.
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