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University of California.
GIFT OF
Class
I
I
APPLETOJ^S' MATHEMATICAL SERIES
NUMBERS UNIYERSALIZED
AN
ADVANCED ALGEBRA
BY
DAVID M. SENSENia, M.S.
PROFESSOR OF MATHEMATICS, STATE NORMAL SCHOOL, WEST CHESTER, PA.
PART SECOND
NEW YORK, BOSTON, AND CHICAGO
D. APPLETON AND COMPANY
1890
Copyright, 1890,
By D. APPLETON AND COMPANY.
PEEFAOE
NuMBEKS Universalized is believed to embrace all
algebraic subjects usually taught in the preparatory and
scientific schools and colleges of this country. For con-
venience, it is divided into two parts, which are bound sepa-
rately and together, to accommodate all kinds and grades
of schools sufficiently advanced to adopt its use.
Part Second is treated in five chapters, as follows : One
embracing serial functions, including development of func-
tions into series, convergency and divergency of infinite
series, the binomial formula, the binomial theorem, the
exponential and logarithmic series, summation of series,
reversion of series, recurring series, and decomposition of
rational fractional functions ; one treating of complex num-
bers, graphically and analytically, including fundamental
operations with complex numbers, general principles of
modulii and norms, and the development and representa-
tion of sine, cosine, and tangent ; one embodying a discus-
sion on the theory of functions, including graphical repre-
sentations of the meaning of the terms independent and
dependent variables, continuous and discontinuous func-
tions, increasing and decreasing functions, and turning
values and limits of functions, and also a treatment of
differentials and derivatives, and maxima and minima val-
ues of functions ; one treating of the theory of equations,
including a discussion of the properties of the roots, real
and imaginary, of an equation, methods of determining
the commensurable roots of a numerical equation, Sturm's
theorem for detecting the number and situation of real
roots, Horner's method of root extension. Cardan's for-
183681
iv PREFACE.
mnla for solving cubic equations, and a short treatment
of reciprocal and binomial equations ; one treating of de-
terminants and probabilities, so far as these subjects are of
interest and value to the general student. The volume
closes with a supplementary discussion of continued frac-
tions and theory of numbers.
The aim of the author in preparing this part of his work
has not been so much to give completeness to the various
subjects treated as to lead the student to a comprehension
of the fundamentals of a wider range of subjects, and to
cultivate in him a taste for mathematical investigation.
It is believed that the plan adopted will give the general
student a broader and more practical knowledge of algebra,
and will lead to better results in a preparatory course of
study for the university than would a completer treatment
of fewer subjects requiring an equal amount of space in
their development and more time in their mastery. While
a sufficient number of examples have been placed under
each head to offer opportunity for the application of the
principles and laws developed, there will not be found an
unnecessary multiplicity of them to retard the progress of
the pupil in his onward course.
In conclusion, the author desires to acknowledge his in-
debtedness to the English authors, Hall and Knight, Chr^s-
tal, Aldis, Whitworth, and 0. S. Smith, whose works he
frequently consulted, and from which he obtained many
new and valuable ideas.
David M. Sensei^ig.
Normal School, West Chester, Pa., \
December 2, 1889, \ •
OOIfTElsrTS,
CHAPTER IX.
PAGE
Serial Functions 315-353
Definitions, 315.
Development of Functions into Series 816-319
Theorem of indeterminate coefficients, 316. Expansion of rational
fractions, 316-318. Expansion of irrational functions, 318.
Convergency and Divergency of Infinite Series . . . 319-323
General definitions, 319. Fundamental principles, 320. Theorems,
321, 322. Exercises, 323.
The Binomial Formula 324-326
Development of, 324. Applications, 325, 336.
The Binomial Theorem 326-332
For positive exponents, 326, 327. For any rational exponents, 328-
) = nr«— I, 328. General demonstration of
binomial theorem, 328-330. Numerous corollaries and inferences,
327, 330-332. Exercises, 332.
The Exponential Theorem 333
The Logarithmic Series 334-337
Development of, 334, 335. Computation of logarithms, 335, 336.
Principles, 336, 337.
Summation of Series 337-344
Method by indeterminate coefficients, 338. Method by decompo-
sition, 339. The differential method, 339-342. To find the (n + l)th
term of a series, 340. To find the sum of n terms, 341. To inter-
polate terms, 342. Exercises, 343.
Reversion of Series . . 344
Recurring Series 345-349
Definitions, 345. To determine the scale of coefficients, 346. To find
the snm of n terms, 347. Exercises, 349.
Decomposition of Rational Fractional Functions . . . 350-353
Definition, 350. Principles, 350-352. Exercises, 353.
vi CONTENTS.
CHAPTER X.
PAGE
Complex Numbers 354-365
Graphical Treatment 354-361
Definitions, 354-358. To add complex numbers graphically, 359. To
multiply a complex number by a rational number, graphically, 360.
By a simple imaginary number, 360. By a complex number, 361.
General Principles of Complex Numbers ..... 361-362
Problem. To find the value of e^+y* 363
Graphical Representation of sin. y and cos. y , . . . 364-365
CHAPTER XI.
Theory of Functions 366-388
Definitions, 366-370.
Graphical Representation of Functions 370-374
Exercises, 374.
Differentials and Derivatives of Functions .... 374-382
Definitions, 374. Principles, 375-379. Exercises, 380-383. Concrete
applications, 380. Successive derivatives, 381.
Factorization of Polynomials containing Equal Factors . . 382-383
Maxima and Minima of Functions 385-388
CHAPTER XII.
Theory of Equations 389-422
Introduction, 389. Normal forms, 390, 391. Divisibility of equations,
393. Number of roots, 393. Relation of roots to coefllcients, 394.
Imaginary roots, 395. Fractional roots, 396. Relation of roots to
signs of equations, 396. Descartes' rule of signs, 398. Limits of ■
roots, 399. Equal roots, 400. Commensurable roots, 400-403. In-
commensurable roots, 403-409. Sturm's series of functions and
fundamental principles, 404-406. Sturm's theorem, 406-409. Hor-
ner's method of root extension, 409-415. Cubic equations, 415-419.
Cardan's formula, 416-419. Recurring equations, 419-422. Bino-
mial equations, 422.
CHAPTER XIII.
Determinants 423-441
Introduction, 423-425. Properties of, 425-430. Development of, 430-
432. Additional properties of, 432-435. Multiplication of deter-
minants, 435-437. Solution of simultaneous equations by determi-
nants, 438-440. Conditions of simultaneity, 440, 441. Sylvester's
method of elimination, 441.
CONTENTS. vii
PAGE
Probabilities 443-453
Definitions and fundamental principles, 442-444. Exclusive events,
444 - 44 6. Expectation, 446. Independent events, 447, 448. Inverse
probability, 448-450. Probability of testimony, 450, 451. Exer-
cises, 451-453.
SUPPLEMENT.
Continued Fractions 454-463
Definitions, 454, 455. Formative law of successive convergents, 455-
457. Properties of convergents, 457-460. Keduction of common
fractions to continued fractions, 460, 461. Reduction of quadratic
surds to continued fractions, 461, 462. Reduction of periodic con-
tinued fractions to simple fractions, 462. Approximation to the
ratio of two numbers, 462. Exercises, 463.
Theory of Numbers 464-482
Systems of notation, 464^i67. Divisibility of numbers and their
digits, 468-470. Even and odd numbers, 470-473. Prime and com-
posite numbers, 473-477. Perfect squares, 477^79. Perfect cubes,
479, 480. Exercises, 480-482
PAET SECOND.
CHAPTER IX.
SERIAL FUJVCTIOJTS.
I. Definitions.
662. Any expression containing a variable is called a
function of the variable.
Thus, ax -{-!?, a x~\ Va-\-a^, a", log. {a -\- x), and
a-{-l)x-\-cx^-{-doi?-\- etc. , are functions of x.
663. Any series containing variable terms is called a
serial function,
664. The expression f{x) represents any function of x,
and is read function x,
666. When two or more functions of the same variable
are used in a discussion, modified forms are used for dis-
tinction ; as,
1. f{x), F{x), <f>(x); read, / minor, f major, phi
functions of x.
^' f (^). /" i^\ f" (^) ; read, / prime, f second,
f third functions of x.
3. /i (x), /2 (x), fs (x) ; read, / one, f two, f three
functions of x.
316 ADVANCED ALGEBRA,
Development of Functions into Series.
Theorem of Indeterminate Coefficients.
566. If A-\-Bx-{-Ca^-\-D^etc. := A^-^- B^x-{-C^x^
-\- DiO^-\- etc., for any assigned value of x from — cx) to
+ 00, atid A, At, B, B^, C, C^, D, D^, etc., are independ-
ent of X, then will A = A^, B = B^, C=^Ci, D~ D^ etc.
Demonstration: Given
A->c Bx^- Cx^ -{-Dq? + etc. = A^ + B^x + CiX^ + D^a^ + etc., (A)
for any assigned value of x. Let a; = ; then A = Ai.
Therefore, A = Ai for every value of x. (1)
Subtract (1) from (A),
Bx + Gx^ ^Da?-^ etc. = BiX ^t CxX^ + D^a? -\- etc. (B)
Divide (B) by cc,
B ^Cx + Dxy^ ■\- etc. = ^1 + (7i a; + Di a;2 + etc. (C)
Let a; = 0, 5 = B^.
Therefore, B = Bi for every value of x. (2)
etc., etc., etc.
567. Corollary l. — If A + Bx -^ Cx^ + Dx^ -{- etc.
= 0, for any assigned value of x, then will A=0, B = 0,
(7=0, D = 0, etc.
568. Cor. 2. — A function of a single variaUe can he de-
veloped into a series of the ascending powers of the variable
in only one way.
For, if possible, let
f{x) = a-\-'hx-\-co? -\- etc. ; and
f{x) = «i + Jj a; + Ci x^ + etc. ; then will
a-^-lx-^cx^ -\- etc. = ax-^-lyX-^-c^o? -\- etc. ;
whence a = ai, J = Ji, c = c^, etc., and the two develop-
ments will be identical. -
2. Applications.
1. Expansion of Rational Fractions.
569. A rational fraction of a single variable may gen-
erally be developed into a series by dividing the numerator
EXPANSION OF RATIONAL FRACTIONS. 317
by the denominator, but a more expeditious method con-
sists in the application of the principle of indeterminate
coefficients.
1 — x
Ulustrations. — 1. Develop ^ into a series of the
ascending powers of x.
Let \^ = A + Bx + Cx^ + Da^ + etc. (A)
1. •{• X
Clear of fractions, and arrange the coefficients of the like powers
of X into columns.
1-x = A + B
+ A
x + C
+ B
x^ + D I x^+ etc. (B)
+ C
Equate the coefficients of the like powers of x [566],
A = l', A + B = -1, B + G = 0; C + D = 0, etc.
/. A = 1, B = -2, C = 2, D = -2, etc.
Substitute these values of the coefficients in (A),
2-^ = 1 - 2a; + 2a;2 - 2a;8 + etc.
1 + a;
Let the student divide the numerator by the denominator, and
show that the same result will follow.
570. The first term of the expansion may be obtained
by dividing the first term of the numerator by the first
term of the denominator, and the remaining terms by
indeterminate coefficients.
2. Develop , „ in the ascending powers of x.
X —J~ Xr
^^^ A 2 = »^~* + B^ + Cic + Dx^ + etc. (A)
X "T X
Clear of fractions and column coefficients,
x^+ D
+ Cb
ic« + etc. (B)
a = a + B I x+ C
+ ab \ + Bb
Equate coefficients,
(!) B + ab = 0. (2) G + Bb = 0. (3) i) + C& = 0, etc.
B = — ab, G = ab\ D = - ab\ etc.
Substitute these values in (A),
J— ^ = aa;-* — aft + a6'a; — a62a;' + etc.
a; + &a;*
318
ADVANCED ALGEBRA,
EXERCISE 87.
Develop to four terms :
2x — S
x-\-x^ + l
2.
3.
1-x
1
2x
1
l-x-\-x^
5.
9.
a-\-x
6x + 2x^
l-\-x-{-a^
2. Expansion of Irrational Functions.
Elnstrations. — 1. Expand to four terms Vl—x-\- x^,
^- 1 + Bx + Cx^ + Do? + etc.
Put Vl — a: + a;'
Square both members and column the coefficients,
Equate the coefficients,
+ 3(7
a;8+ 2i)
+ 3^(7
7? + etc.
(A)
(B)
(1)3^=-
B^-
(2) ^ + 3(7= 1.
3
(3) 32> + 3^C = 0.
^-8
D =
W
etc.
Substitute these values in (A)
1
2a; + |a;«+^a:«+etc.
-s/l - a; + ^2 = 1
2. Expand to three terms v8 — ^.
Put V8-a;2 = 2 + ^a; + Cx^ + Z>a:3 + ^a;* + etc.
Cube both members and column coefficients.
8-a;8 = 8 + 13^ I a; + 13 C
' + 6^
a;
2 + J53
+ 13i>
+ 13^(7
a:3 + 13^
+ 3J52(7
+ 6(7«
+ 13^i>
Equate the coefficients,
(1) 135 = 0. (3) 13 C + 6-B2 = - 1.
(3) 135C+13i> + 53 = 0.
. (4) 12jE^+3J52C7+6C8 + 135i> = 0.
... 5 = 0, C=-l, i>
Substitute these values in (A),
0, ^= -So5» etc.
V8
^ = ^-lV-38-8^-^^-
(A)
xl^ + etc.
CONVERGENCY OF INFINITE SERIES. 319
EXERCISE 88.
Expand to four terms :
1. a/4 — a; 4. Vl + a; 7. Va-{-x
2. ^l^X'-x' 5. a/27 + 0,-2 8. Vo^^
3. V9 + a;-3a;2 6. Vs + M^ 9. VoH^
Convergency and Divergency of Infinite Series.
General Definitions.
671. The limit of a series is the limit of the sum of n
terms of the series, when n is indefinitely increased ; that
is, when lim. n= co,
672. A series is convergent when its limit is a finite
constant, including zero,
673. A series is divergent when its limit is infinity.
674. A series is indeterminate when the sum of n terms
is finite hut does not approach any definite yalue as n is
indefinitely increased.
Thus, 1 — 1 + 1 — 1 + 1 — 1 + is indeterminate,
since, when n is even the sum is 0, and when n is odd
the sum is 1, however great n he taken.
676. For convenience of discussion, the following nota-
tion will be adopted :
1. The terms of a series will be represented in order
by Wi, Uz, Us w„, Un+i
2. The sum of n terms will be represented by Z7„, so
that Un = Ui-{-U2-\-Us-{- + w«. .
3. The limit of the series will be represented by U, so
that [7= wi + t^2 + % + .... + «^n + w«+i +
320 ADVANCED ALGEBRA.
Fundamental Principles.
576. X No series whose terms are all of the same sign
can he indeterminate.
For either the sum of n terms increases numerically
without limit as n is increased indefinitely, or else it can
never exceed some fixed value which it approaches as a
limit. Such a series is, therefore, either convergent or
divergent.
577. 2, A series of finite terfns whose signs are all
alike is divergent.
For, if we let a represent the numerical value of the
smallest term, then, numerically, U>na, whose limit
is c» , when lim. n = co and a is a finite quantity.
Thus, the series 1 + 2 + 44-8 + 16 + .... is divergent.
678. 3. If a series is convergent it will remain con-
vergent, and if divergent it will remain divergent, if any
finite number of terms he added to or subtracted from
the series.
For, the sum of any finite number of terms is finite,
and, therefore, can not change the nature of the limit of
the series when combined with the series by addition or
subtraction.
579. Ji.. If a series is convergent when its terms are
all positive, it is also convergent when its terms are all
negative, or some positive and some negative.
For its limit will have the same numerical value when
its terms are all negative as when they are all positive,
and will be numerically less when the terms do not all
have the same sign as when they do.
It must not be inferred from this principle that a series
THEOREMS. 321
is necessarily divergent when its terms are not all of the
same sign, if it is divergent wh«n they are alike in sign.
Such may or may not be the case.
Theorems.
580. I. In order that a series may he convergent, the
limit of the {n + l)th term, and the limit of the sum of
. any numder of terms beginning with the {n + ^)th term
must he zero, and conversely.
DemonBtration : If a series is convergent, then ultimately, if ti is
indefinitely increased,
(1) U-Un = o [4981 (2) U- Un+l = o
(3) U- Un+^ = o (4) U- Un+z = o
Subtract (1) from (2) ; (1) from (3) ; (1) from (4), etc. ; then,
(a) Un — Un + \ = o ; or, Un+\ = o ; whence, lim. w^^ i =
(6) Un — Un+i = o ; or, Un+\ + w„4.8 = o ; whence,
lim. (Wn+l + Wn+j) =
(c) Un — Un + 3 = o ; or, Un + 1 + Un + i + Un+z = o ; whencc,
lim. (Un + l + Un + 1 + Un + z) =
etc., etc., etc., etc.
581. II. If each term of a series whose terms are alter-
nately positive and negative is numerically greater than
the following term, the series is convergent.
Demonstration : Let U=ui — Ui + U9 — Ui+ ± w« T Un+i. . . .,
in which -Mi > 'Wa > Ws > '^4. . . ., be the given series.
(1) U — {Ui — Wa) + (Ws — Ui) + (We — w«) + etc.
(2) U = Ui — (Ui — Ws) — {Ui — -Mb) — etc.
From (1) it is evident that U is positive.
From (2) it is evident that, since U is positive, U<,Ui.
.'. U approaches Wi or some quantity less than Ui as a limit,
and the series is, therefore, convergent.
582. III. A series is convergent if after some particu-
lar term the ratio of each term to the preceding term is less
than unity.
322 ADVANCED ALGEBRA.
Demonstration: The most unfavorable case to convergency sup-
posable, under the conditions given, is evidently the one in which all
the terms have the same sign (say plus) and all the ratios described
are equal and each equal to the greatest of them. This is, therefore,
the only case that needs proof.
Let r be the greatest ratio after the nth term, but < 1 ; then,
Un + ^n + l + t^n + 2 + ^n + S + etC. = W„ + W» 7* + Wn r« + CtC. = q— ^?—
1 — r
[499, P.] = a finite quantity. Therefore, the whole series is con-
vergent [578J.
683. IV. A series of all positive or all negative terms
is divergent, if after some particidar term the ratio of
each term to the preceding term is equal to or greater
than unity.
Demonstration : The most unfavorable case to divergency, and the
only one that needs investigation, is the one in which all the ratios
described are equal and each equal to the least of them.
Let r be the least ratio after the nth term, but = or > 1 ; then,
«*n + w«+i + -Wn+s + 'Wn+s + ctc. is divergent [577] ; and hence the
whole series is divergent [578].
584. V. A series of positive terms is convergent if
each term is less than the corresponding term of a given
convergent series of positive terms.
Demonstration : Let CT = Wi + w, +....+ w„ + u^^i + be a
given convergent series ; and T = Vi + Va + + Vn + Vn+\ + . . . .
a series in which Vi < Wi , iJa < "Wa, . . . . t^n < w„ , VnJ^\ < w„+i ,
From the nature of addition, it is evident that Vn < Cn ; and
hence, too, lim. F„ < lim. C/« , or V <U', therefore, if U is converg-
ent V is convergent.
585. The foregoing principles and theorems will serve
to test the convergency and divergency of a very large
number of series, but are not of universal application,
inasmuch as they do not apply to all classes of series.
Note. — If the terms of a series are not all of the same sign no gen-
eral method can be obtained for testing their convergency or divergency.
586. The convergency or divergency of a series may
often be determined by grouping terms, as follows :
THEOREMS. 323
1 1 i i i
= 1 + (^3 + p) + (^3 + 5^3 + ^3 + fa) + (p +••••+ 1^3) + etc.
••• C^ < 1 + I + ^8 + I + etc.; or CA < J = 3 .
,'. The series is convergent.
EXERCISE 89.
1. Is the series T+o + o+T + "i' + ^^^' convergent ?
1 /C o 4
2. Test the series : l-\-^x-\-hoi?-{-^3?-\- for con-
vergency
1. When x<l. 2. When x>l. 3. When x = l.
3. Testtheseries:i-^ + ^-^+....
for convergency.
when a; < 1 ?
6. Testtheseries:i + ^ + ^ + ^ + etc.
for convergency.
6. Test the series :
. 1 rr^ 1 3 ^5 1 . 3 . 5 ^.7
^+2-3+2T4-5+2TT:^-T + '*'-
for convergency
1. When x<l. 2. When x>l. 3. When ic = 1.
Suggestion.— Lim. jj " = -j. Why!
7. Test the series : l + aj-f-a^^ + a^^H- etc. for con-
vergency
1. When x = l. 2. When x< 1. 3. When x>l,
8. Test 1 + 1 + ^ + ^^^+.... for con-
vergency.
324 ADVANCED ALGEBRA.
The Binomial Formula.
587. The binomial formula is used to find the prod-
uct of any number of binomial functions of the form of
Development.
{x + a){x + 'b) = x^ + (a+b)x+ab
Multiply both members hj x + c,
{x + a){x+b){x + c) = x^+(a + b)x^ + abx + cx^ + (ac + bc)x+abc
= x^ + {a+b + c)x^ + {ab + ac + bc)x +abc
Multiply both members hj x + d,
{x + a) (x + b) {x+ c) (x + d) =
a^+{a+b + c)x^ + {ab + ac + bc)x^ + abcx
+ dx^ + {ad + bd+cd)x'^ + {abd + acd + bcd)x+abcd
= a^+{a + b + c + d)a^ + {ab + ac + ad + bc + bd+cd)x^ +
{abc + abd + acd + bcd)x + abcd
Observe the following laws in these products :
i. The number of terms is one greater than the num-
ber of binomial factors,
2. The exponent of x in the first term equals the num-
ber of binomial factors, and decreases by unity in each
succeeding term,
S, The coefficient in the first term is unity ; in the
second term the sum of the second terms of the binomial
factors ; in the third term the sum of the products of the
second terms taken two together j in the fourth term the
sum of the products of the second terms taken three to-
gether, etc,
4' The last term equals the product of all the second
terms.
Are these laws true for any number of factors ?
Assume them true for r factors, so that
{x-{-a){x-\-b) {x-\-m) = x^-^pi of-^-\-pz af-*+ ....
Pr-\ x-\-p^, in which
THE BINOMIAL FORMULA. 325
p^ = a-\-h-{- +m
p^ = al)-\-ac-{- am-\-hc + bd-{- + Jm + etc.
^3 = abc-\-aI)d-\- + «5m + ....
p, = abc m. (A)
Multiply by {x + n), the (r + l)th factor, then
{x-\-a){x-{-b)..,.(x + n) =
X' + ^+PlX'-\- ^2^-^ + ....+ Pr^
-\- nx' + npi iC"- ^ + .... + npr-i x + npr
=:af+-'-^{pi-\-n)3f + (p2-{-np^)af-''-\-.... + npr
Laws 1 and 2 are evidently still true.
Pi-\-n = a -\- h -\- c -{- n,
Vi + ^ A = {d^ -\- d -\- am-\-'bc-\-hd-\-.,,.
•\-lm-\- etc.) -\-{an-\-ln-{- -^-mn),
which is still the product of the second terms taken two
and two.
• •••••
np, = alcd n. Therefore, all the laws still hold
true. Hence, if they are true for r factors, they are true
for r + 1 factors. But we found them true for four fac-
tors by multiplication ; hence, they are true for five fac-
tors ; and, if so, for six factors ; and so on. Therefore,
formula (A) is general.
Uote. — The number of products that enter into each coeflBcient
may be determined by the principles of combination.
Applications.
Ulnstrations. —
1. Expand {x + 1) (^ + 2) {x - 3) (a; + 4).
Solution :
pi = 1 + 2-3 + 4 = 4
i?a = (1 X 2) + (1 X -3) + (1 X 4) + (2 X -3) + (2 x 4) + (-3 x 4) = -7
i)3 = (lx2x-3) + (lx2x4) + (lx-3x4) + (2x-3x4) = -34
i>4 = lx2x— 3x4= —24
/. (x+l)(a;+2)(a;-3)(a;+4) = ic4 + 4ic»-7a;»-34a;-24.
326 ADVANCED ALGEBRA.
2. Factor a;* + 14 a;^ _|. ^^ ^2 _|_ I54 ^ _l_ ^go, if possible.
Let {x + a){x + h){x + c){x+d) = o^ + l^a? + 1l\x^ + l^^x + \20.
Then, 1. a-\-h + c + d = +14:
2. ab + ac + ad+bc + bd + cd = +71
3. a&c + a6c^ + «C(^ + 6ccZ = +154
4. aJc6? = +120
Resolve if possible +120 into four factors whose sum is +14.
These we find to be 2, 3, 4, 5.
.*. a = 2, b = d, c = 4, and d = 5.
Will these values satisfy 2 and 3 ?
ab + ac + ad+bc + bd + cd = 6 + 8 + 10 + 12 + 15 + 20 = 71, correct.
abc + abd + acd + bcd = 24:+S0 + 40 + 60 = 154, correct.
.-. x^ + Ux^ + 71x^ + 154:X + 120 = (x+2)(x+d)(x + 4)ix+n).
EXERCISE 90.
1. Expand (x -{-2){x-\- 3) (^ + 1)
2. Expand (x -j-3)(x — 2) (x — 3)
3. Expand {x + 2) (ir + 3) {x - 1) {x - 2)
4. Expand (:r + 3) (:r + 5) (a; - 2) (:c - 6)
5. Expand {x + 2) (rr + 2) {x -\-%){x-\- 2)
6. Expand {x — h){x — 5) (ic — 5) (a; — 5)
7. Expand (2a; + 1) (2a; + 3) (2:r - 5) (2 a;- 1)
Suggestion. — Put y for 2 a:.
8. Factor a;^ _j_ 9 ^^^s _{_ 26 a; + 24
9. Factor x^ -%x^ -%^x-\-m
10. Factor a:* + 5 a;^ + 5 a;^ — 5 a; — 6
11. Factor it-* - 2 a;^ - 25 a;^ + 36 a; + 120
12. Factor a;^ + 4 a;* - 13 a;^ - 52 ar^ + 36 a; + 144
The Binomial Theorem.
I . For Positive Exponents.
588. If, in the binomial formula [587, A], we assume
a = h=:c = dy etc., and r = w, then will
TEE BINOMIAL THEOREM. 327
1. {x^a){x + h){x-{-c).... ={x^ay.
2. a;'" = cc" ; x'-^ = aj~-^ ; a:'- ^ = x""-^ ; etc.
3. p^z=a-\-a-\-a-{- to n terms = na.
4. j92 = «^ + «^ + ^^ + = ^^ taken as many times
as there are combinations of 2 in 7i ; or,
n(n — l) 2
5. p^ = fl^aa + aaa + «a« + = a^ taken as many
times as there are combinations of 3 in w ; or,
Ps = -^ i ^'
• ••••••
6. p'=aXaXaXa to n factors = a\
589. Cor, 1, — If a and x le interchanged, {a + xy =
ar^na--^x-{-'^^^^a^-^x^-^...,-\-x\ (C)
From (B) and (C) it will be seen that the coefficients of
any two terms equidistant from the first and last terms
are numerically equal.
590. Cor, 2, — If X le made negative in (C), (a — xf =
a- -na''-^x-\- ^^^"^^ a'^-^x^ -,...±x\
591. Cor. 3, — The sum of the coefficients in (C) equals
zero.
For, put a = 1 and a: = 1 ; then
Li Li
ADVANCED ALGEBRA.
2. For any Rational Exponents.
— — — J = Tir*"^ for any
Demonstration : I. Let n = any positive integer.
Now, — ^-^ = ic"-^ + ra;«-2 + r^a;"-^ + + r«-J [134].
I =: lim. a:"— ^ + lim. roc:^—^ +
x-r Jx = r
+ lim. r»-i [401, 413] = r^-^ + r»-> + r«-i + . . . .
to n terms = nr^—'^.
If)
II. Ze^ w, = i- , a positive fraction.
-T a?^ — r"^ Xq — rq ..
Now, = — -. (1)
' a; — r x — r ^ '
Put Xq =y, or x = y9 '^ and r? = s, or r = s? ; then
a;» — r** yp — sp yp — sp y^ — s?
X — r ~ yt — si ~ y — s y — 8 '
Since x-=yi and r = s^, lim. y = s when lim. ic = r.
., Lim. (^^^:r.\ = ita. -S ^^^1^' ^ ^1^' i =
\aJ — r/x = r i y — s y — s)y = 8
pyP-i ^ qyq--^ [I, 416] = ^ yp-<i =
p J_ p—<i p P. _i
— (xq) = — xg =^na^—'^.
III. Lei n = —p, a negative integer or fraction.
__ x* — r* x—P — r-P /a;p — rP\
Now, = = — x-P r-P ( I .
' x — r x — r \ x — r J
VaJ — r/a: = r | \x — rj\x = r
— —r-^p,prP-^ [415] = — r'^«(— ?ir-»*-') = nr»-i.
General Dennonstratlon of BInonnial Theorem.
593. Let it be required to develop (a-\-xy, for any-
rational value of n, into a series of the descending powers
of X,
THE BINOMIAL THEOREM,
(a + ^)« = ja(l + f)["=a«(l+f)"
Put- = ^, or (1+ -y=(l + ^)«
"Put {\ + zY = I + Az Jr B z"^ + C z^ ■\- Dz/^ + . . , ,
Since z may have any finite value, put z — r\ then,
(1 + r)» = 1 + uir + i?r2 + Cr3 + i)r* + . . . .
Subtract (D) from (C),
(1 + zY - (1 + r)» = ^ (2; - r) + ^ (2« - r2) + C (2« - r3) + . . . .
Put ^ f or 1 + 2; and R f or 1 + r ; or Z — i2 f or 2 — r ; then,
Z^ - R^ = A{z - r) + B{z'^ - r«) + C{z^ - r^) + . . . .
Divide hj Z— R = z — r,
Z
329
(A)
(B)
(C)
(D)
(E)
(F)
(G)
Z — R \ z — r J \ z — r J
Let lim. 2; = r, then lim. Z=R, since Z=l + z, and i? = 1 + r ;
andlun.(§=|-")z=B =
.-. nR^-^ [592]
or, n(l + r)"-*
(H)
J. + 2 J5r + 3 02 + 4i)r3+ . . . .
(1)
(J)
Multiply by 1 + r, and column coefficients,
w (1 + r)*
+ 2B
r^ + dC
+ 4:D
r3 +
(K)
r-¥2B
+ 3(7
Multiply (D) by n,
n(l + r)» = 71 + J.nr + -e7ir2 + Cwr» + Z^nr4+ (L)
Equating the coefficients of the second members in (K) and (L),
we have
(l)A = n (2) A + 2B = An
{^)2B + dC = Bn (4) 3C + 4i> = i>w; etc.
A = n, B = — g— , C= ^
n{n
D= ''' ^''^ ^,etc.
Substituting the values in (C),
{1 + zY = l + nz+ -^ — '-z^ + r£ 2» +
n(n-l)(n-2)(n-S)^,_^^^^^ ^^^
330 ADVANCED ALGEBRA.
Substitute — for z (B), and multiply by o*» (A),
Cb
n(7i-l)(w — 2)a"-3 , w(7i — l)(w — 2)(n— 3) , ^ ' ^t.
—5^ '-j^ x'+— '^ '^ ^-a^-*x*+ (N)
594. Cor. 1. {X + i/)" = a- ^1 + ^=
= X" + «a?-' y + ^i^^af-^/ +
"('^-^>("-^> x-V+etc. (P)
This is the most general form of the binomial theorem, inasmuch
as X and y may be both variables.
595. Cor, 2, — By inspection it will le seen that
1, The rth term of the development of {x -\- yY =
n(n-l){n-^)..,.{n-r-^%) ^«_,+, ,_i
|r — 1 * ^
2, The {r + l)th term =
n{n^l){n-^2).,.,{n-r + 2)(n-r + l)
|r-l xr = |r '^ ^
3, The ratio of the (r -f- l)th term to the rth term =
n^r-\-l y
r
x' ' { r ) X
596. Cor, 3, — 1. If n is a positive integer equal to
r — 1, the coefficient of the {r + l)th term, which is also
the coefficient of the (n + 2)th term, will reduce to zero.
Therefore, the series will terminate with the {n-\-l)th
term, which will he y\
2. If n is negative or fractional no factor of the rth
THE BINOMIAL THEOREM. 331
term (r being a positive integer) will reduce to zero, how-
ever great r he taken. Therefore, the series will he infinite.
597. Cm-. 4,—
Since Urn, \ I —^ -'^ll' = —1 ,^ , it follows :
1. That the coefficients of all terms in the hinomial
theorem are finite hoioever far the theorem he expanded.
2. That if y <x the expanded form is convergent [583].
3. That if y> X, the literal part {not coefficient) of the
{r + Vjth term will increase indefinitely as r increases and
will ultimately become infinitely great, and as the coefficient
remains finite the whole term will become infinitely great.
Therefore the expanded form will be divergent [580].
Jf.. If y ■=. ±x the expansion will be indeterminate ;
hut (x + y)'' = (2 a:)" or (0)" = 2" a;* or 0.
5. The expansions of {x-\-yY and {y-\-xY can not
both be convergent for particular values of x and y ; only
the one that has the greater first term.
598. Cor, 5, — 1. The coefficient of the rth term will evi-
n "^ r I 1
dently be greatest when ^^— is first < 1 ; or when
n — r-\-l is first <r ; or when 2 r is first > w + 1, or
when r is first > T" .
2. The rth term when the expansion is convergent, or
n ~~ r I 1 1/
when xy y, is evidently greatest when ^^^— - . — is
r X
first < 1 ; or when {n — r-\-\)y is first <,rx ; or when
{n-\-l) y is first < {x-\-y) r ; or when r is first > ( -^— | y.
Illustration. — In the expansion of (8 + -^) , the
332 ADVANCED ALGEBRA.
greatest coefficient belongs to the term whose number is
— 4 + 1 1
first greater than — ^ — > ^^ TKy which is the first term.
The greatest term in the expansion of I 8 + -^ j is the
one which immediately follows in number, — \7~ x -5- ,
1 . . ^t ^
or — , which is again the first term.
00
EXERCISE 91.
Expand :
1. (a - 3 T'Y 3. (a; - 3 af 5. {x^ - 5)8
2. (2 + 5 xY 4. (2a^-\- 5y 6. (3 x^ + a^y
Expand to four terms :
7. (1 - x)i 9. (a^- 1)^ 11. («rr + b)-^
8.{a—x)^ 10. (a;^ + «)~^ 12. {x^ — a^~^
13. Extract the cube root of 126 to six decimal places.
Suggestion :
Vl26 = Vi25Tr = (125 + l)i = j53(l + ^)P = 5(l + ji^)*
-^r 3 125+ \2_ "" V125; + [3_
(^y+ etc. [ = 5(1 + 0026666 - -0000071 + -0000001) = 5-0132975.
14. Find to 5 decimal places : V65, VSO, V344, V3128
15. Find the 7th term of {2x-}-SY^
16. Find the 5th term of V4+^
a
17. Find the 6th term of
Va-\-x
18. Find the rth term of {a — x)~^
19. Find the greatest coefficient of (2 + x)^
20. Find the coefficient of the 5th term of {a — a^)~^.
21. Find the numerical value of the 10th term of
(7 — 5 3/^)", when y = 27 and n = S
TEE EXPONENTIAL THEOREM. 333
The Exponential Theorem.
599. The exponential theorem is the expansion of a" in
ascending powers of x, and is derived as follows :
T> ^ /^ lA"" - 1 nx(nx — l) 1
And (l . 1)'= 1 . 1 + ^ . ^-— I^^^ etc. (C)
Substitute (B) and (C) in (A),
x{x-l) x(x-^)(x-l)
1 + ^+ — \r^*- \r — -^'"'-
Suppose lim. n = ao, then
Put e for 1 + 1 + i-ft- + fo" + iT" + 6tc. ; then,
[£_ l£_ l±
e» = l+a: + j^ + -- + g-+ etc. (F)
Put ca; for a;, then e"" = 1 + ex + -rjr- + -r^- + etc. (G)
Let e" = a, and assume e as the base of a system of logarithms
[465], then c = loge a, read logarithm a to the base e. Substitute
these values in (G),
»- = l + ^log.a+^<!2|^ + ?!(l5|^ + eto.. (H)
which is convergent for all finite values of x [582].
This is the Exponential Theorem.
600. SchoUum. e = l + l+-i- + J- + ,-^ + etc. = 2-7182818. . . .
l^ l£- lL
is the base of the Napierian or natural system of logarithms, a system
universally used in theoretical work instead of the system based on 10,
which is used in practical work only.
334 ADVANCED ALGEBRA.
The Logarithmic Series.
601. The logarithmic series is the expansion of loge
(1 -f- ^) in the ascending powers of x, and is derived as
follows :
a> = l + ,log. a+ t^^ + t^^^ ,t,. p99,H]. (A)
Transpose 1 and divide by y,
-y- = logea + 2/f-^ + '-^"-^ + etcj. (B)
Let lim. y = 0, then
loge a = lim. I I
Put 1 + a; for a, then
loge (1 + a;) = lim. - ^1 + xy -l\-^^^
y r 12. |3_ ij/=o
= a; — -^r^ + -o — etc. Therefore,
lO o
logo (l+a:) = aj-^a;2+ -^x^— jX^ + etc. (C)
This is known as the Logarithmic Series.
602. The ratio of the (w + l)th term to the nth. term
x'^'^'^ x"" ^ n _ 1
IS I 7 ! — j — . X — r- . X .
n-{-l n n-\-l i_4__
Now, lim. / .x\ =x. Therefore, if a; < 1, numerically
li+i'7„=»
the series is convergent. It is, therefore, convergent for all values
of X between — 1 and + 1.
When x = l, loge 2 = 1— — + -^ — -j + etc., which is converg-
ent [5831.
When x= — l, log, ^=-'^('^+2+'o+T + ^^^-j' which is
divergent, since lim. ( -^-^ ) = lim.-! —1 ( r | >■ = — 1,
and all the terms have the same sign [583].
THE LOGARITHMIC SERIES. 335
603. Eesume
Put — X for oCf then
\og.{l-x)^-x-\x^-\7?-\^-\o?-.... (3)
Subtract (3) from (1), then
log. (1 + *) - log. (1 - X), or log, ([±|) [467, P. 3] =
2(.+ J + |'+....) (3)
171 — n ^ '
Put — I — for X, then.
m-{-n
Put m = n-{-l, then
iog,(„+i)-iog.«=2J^+i(^^y+
iog.(«+i)=iog.»+3J^-^+j(^^y+
+ ....[ (D)
5 \2 ^ + 1/
As this formula converges very rapidly for all values
of n, it may be used to find the Napierian logarithm of
any number from that of the next preceding number,
n being regarded an integer.
Computation of Logarithms.
604. The logarithms of composite numbers may be
readily found, when the logarithms of primes are known,
by Art. 467, P. 2.
/
OF THE
336 ADVANCED ALGEBRA.
The logarithms of prime numbers are found by formula
D, Art. 603.
niustrations. —
loge 1 = [466, P.]
loge 2 = loge (1 + 1) =
^ "*" ^ \ 3 + d^iW + 5l<^ + 7x~3^ + • • • • j
= 0-69314718 (by actual reduction).
loge 3 = loge (2 + 1) =
=1-09861228....
loge 4 = 2 loge 2 = 1 -38629436 ....
loge 5= loge (4 + 1) =
loge4 + 2(i + 3^3 + 5^ + ....)
= 1-60943791....
loge 6 = loge 3 + loge 2 = 1 '79175946 ....
loge 7 = loge (6 + 1) =
loge 6 + 2(1 + 3^ + ^^ + ....)
= 1-94591....
loge 8 = 31oge2 = 2-07944
loge 9 = 21oge3 = 2-19722
loge 10 = loge 5 + loge 3 = 2-30258509
etc., etc., etc.
605. Let a and i represent the bases of two systems
of logarithms and n any number.
Let logb n •= Xf then l' = 7i (1)
Let loga l — m, then oT =.1) (2)
dT" z=: If =: n, or log. n = mx (3)
loff. n mx
.*. , ^ = — = m : or
logb n X
log, n = m logb n. Therefore,
COMPUTATION OF LOGARITHMS. 337
Principle. — Multiplying the log^ of a number hy the
loQa of b gives the log^ of the number.
606. Loge n = logio n X loge 10 [605, P.]
.-. log.o^ = loge^ X j^ = log.^ X ^^30^
= loge /J X 0-4342944....
607. The number 0-4342944 is called the modulus
of the common system, and is represented by m.
Therefore,
JPrin, 2, — Log^Q n = m log, n.
By means of this principle the Briggean or common
logarithms may be derived from the Napierian or natural
logarithms.
I. Since loge (^ + 1) =
logio (» + !) =
]og.„»t + 3mj^^ + i(^^y+....| (H)
By means of this formula the common logarithms may
be computed directly.
Summation of Series.
609. No general method of summing series can be
given. Series of special types may sometimes be summed
by special methods. The student has already learned how-
to sum an arithmetical progression [486], a geometrical
progression [493], an infinite series of the geometrical
type [499], an arithmetico-geometrical progression [«500],
a series of square numbers [506], a series of cubic num-
338 ADVANCED ALGEBRA.
bers [507], and series dependent upon or resolvable into
these. A few additional methods will be given here.
610. I. Method by Indeterminate Coefficients.
This method is applicable when the nih. term is a
rational integi-al function of n.
niustration. — Find the sum Sn of n terms of the series :
lX22 + 2x32 + 3x4:2 + ..., + w(w + 1)2.
Solution :
Put Sn= 1 X22 + 2x32 + 3x42 + 4x52 +....+ 7i(W+l)2
== ^ + ^n + (7n2+i>w3+^7i4+
and, Sn+\ = 1x22 + 2x32 + 3x42 + 4x52 +....+ (?i + l)(n + 2)2
= ^ + ^(71 + 1) + (7(?i + l)2 +Z>(n + l)3 + ^(n + l)4+....
Then, by subtraction,
(w+l)(w + 2)2 = ^ + (2w + l)(7+(3n2 + 3n + l)i)
+ (4n3 + 6n2 + 4?i + l)^ + , or w3 + 5n2 + 8w + 4 =
{B+G+D+E) + {2G+^D + 4.E)n + {^D+QE)'n? + 4:En\
since all coefficients after E are zero, there being no more than four
terms in the expansion.
Equating the coefficients of the like powers of ti,
1.4^=1, 2. 3i> + 6^=5,
3. 2C+3i) + 4^=8, 4. 5 + C+Z) + ^ = 4;
17 7 5
whence, E = -^, 7) = -^, C = -r, and B = -^»
5 7 7 1
Sn= A+ -^n+ -^'n? + -^n^ + -jn^
To find A, put n = 1, then Sn = Si — the first term = 1 x 22.
...(lx2)2 = A+| + | + |- + i = ^ + 4;
whence A = 0; and
5 7 7 1
= i^(3n* + 14 n8 + 21^2 + lOn)
la
= ^(3n»+147i»+21»+10)=^(w+l)(n+2)(3w+5)
SUMMATION OF SERIES. 339
611. 2. Method by Decomposition.
This method is sometimes applicable when the wth
term is a rational fractional function of n, and is resolv-
able into the algebraic sum of the nth. terms of two or
more other series of the same nature.
niustration. —
Find the sum S,, of n terms of the series : ^r — ^ — 7 +
7 , 10 , , 37^ + l
3x4x5 ' 4X5X6 ' •**• ' (/i + 1) (/^ + 2) (/^ + 3) *
Solution :
3/1 + 1 _ A B C
^^^ (w + l)(w + 2)(n + 3) - n + 1 "^ n + 2 "^ 71 + 3' ^^®"
A = —ly B =z 5, and C = — 4 ; whence
3n + l _ / 1_ 5 4_\
(n + l){n + 2)(n + S)~ \ n + l'^n + 2 n + dj'
•• " ](7i + i)(n + 2)(w + 3)f
^V n + lj 2 3 4
\ n + 2)
4 n + 1
5 5 5
■o +^ +
3 4 n + 1 n + 2
2/ 4_\ _4_ 4 4 ^
V ^ + 3/ 4 71 + 1 n + 2 n + S
Adding the last three series, we have
"~V 2"^3"^w4-2 n + s)~6'^n + 2 n + 3
If lim. n = ao, then /Soo = 77 .
The Differential Method.
612. If the first term of any series be taken from the
second, the second from the third, the third from the
fourth, and so on, a new series will be formed which is
340 ADVANCED ALGEBRA.
called the first order of differences. If the first order of
differences be treated in the same manner as the original
series, a second order of differences will be formed, and
so on.
Thus, if we let a, I, c, d, e, , , . , be any series, then
h — a, c — h, d— Cy e — d, .... will be the first order
of differences ;
c — %'b-\-a, d — 2c-}-b, e — 2d-^c, the second
order of differences ;
d — dc-\-3b-a, e — Zd^^c-h, the third
order of differences, and so on.
613. If we let «!, Ji, Ci, fZi, represent the first
order of differences ;
«2? ^2? Cgj ^2* • • • • the second order of differences ;
^3 J ^3> ^3? ^3^ •••• the third order of differences;
and so on, we have the following scheme :
Series : a, l, c, dy e,
1st Differences: ^i, Jj, Ci, (?i,
2d Differences : ct^, hi c^y
3d Differences : ^3? ^3?
4th Differences : ^4, , and so on.
(hi ci2f (hy <^if are the^r^^ terms of the succes-
sive order of differences.
614. Problem 1. To find the {n + l)th term of a series.
Solution : Take the series a, 6, c, d, e, then from the above
scheme,
1. b —a =ai, whence b = a + ai (1)
bi — ai = a%, " bi = ai + a^ (2)
&a — aa = Os , " 63 = O9 + as (3)
&3 — as = a4 , " ba = a3 + at (4)
2. c =b +bi = a +2ai + tti, from (1 and 2) (5)
Ci = &i + 6a = ai + 2 as + as , from (2 and 3) (6)
Ca = Ja + 6s = as + 2 as + a4 , from (3 and 4) (7)
SUMMATION OF SERIES. 341
3. <? = c + Ci — a 4- 3 «! + 3 aa + aa , from (5 and 6) (8)
<?i = Ci + Ca = «! + 3 aa + 3 aa + a 4, from (6 and 7) (9)
4. e = 6^ + cZi = a + 4 «! + 6 CTa + 4 as + ^4 , from (8 and 9) (10)
Now, since
&r=a + ai, c = a + 2ai + aa,
(? = a + 3aj + 3aa + «3, e = a + 4ai + 6a3 + 4a8 + a4,
it will be observed that the coefficients of any term are the same as
the coefficients of a power of a binomial, whose index is one less than
the number of the term. Hence, the
/ H•.l^ X n(n—\) w(w— l)(w— 2) ^.,
(71+ l)th term = a+w Oi + -^ — - a^ + — j^^ ~ aa + [A]
615. Car, — The nth term =
;+in-l)a,+ ^''-^^^-'K + .... [B]
Illustrative Example. — Find the 7th term, also the nth
term, of the series : 1, 3, 6, 10,
Solution : 1st differences = 2, 3, 4, . . . .
2d " = 1, 1, . . . .
3d " =0, ....
.*. 1st. n = 7, a = 1, ai = 2, a<, = l, as =
Substitute these values in [B],
7th term = 1 + 6x2 + ^-^ x 1 = 28
2d. Put n = n, a = l, ai = 2, aa = 1, and aa = 0,
i-i, i-u i -. / ^^ n (n — l)(n — 2) . 7i(n—l)
then nth term = l + (w — l)x2+ ^^ '-^ x 1 = —^ — - .
« 2
616. Problem 2. To find the sum of n terms of a series.
Solution: Let it be required to find the sum of n terms of the
series a, b, c, d, e,
Assume the series 0, a, a + b, a + b + c, a + b + c + d, , then,
1st. The first order of differences of the assumed series is the
given series.
2d. The second, third, and ?ith orders of differences of the as-
sumed series are the first, second, and {n — l)th orders of differences
of the given series.
3d. The {n + l)th term of the assumed series is the sum of n
terms of the given series.
Hence, if in formula [A] we put for a, a for ai, ai for aa,
342 ADVANCED ALGEBRA.
etc., we shall have the sum of n terms of the given series. Doing so,
we shall have
c, n{n — l) w (n — 1) (n — 2)
Sn = na+ "^ ' ax+ -^ ^ U^+..,, [C]
Note. — This method is applicable only when some finite order of
differences will reduce to zero.
Example. — Find the sum of 10 terms of the series :
1 + 4 + 10 + 20 + 35+....
Solution : First Differences = 3, 6, 10, 15,
Second Differences = 3, 4, 5,
Third Differences = 1, 1,
Fourth Differences = 0, ....
.-. Put w = 10, a = 1, tti = 3, a, = 3, a^ — 1, and a^ = in
formula [CJ ; then,
^ _^10x9 „10x9x8 _ 10x9x8x7 ,
/S; = 10 + —5— X 3 + ^ ., X 3 + — s-^ — 7— X 1 = 715
3 2x3 2x3x4
617. Problem 3. To interpolate terms at regular inter-
vals between the terms of a given series.
Formula [B] may sometimes be used with advantage
to interpolate terms at regular intervals between the terms
of a given series.
Illustrations. — 1. Given
(651)3 = 423801, (653)3 = 426409,
(655)3 = 429025, and (657)3 = 431645,
to find the value of (652)3, (554)3^ ^^^ {'o^^f.
Solution: Series = 423801, 426409, 429025, 431649
First Differences = 2608, 2616, 2624
Second Differences = 8, 8
Third Differences =
Take formula a„ = a + (n — 1) «t + 10 «s + . . . .
L~_
Put a = 423801, a, = 2608, a, = 8, and 7i= l-^- , 2 g, and 3^
successively; then,
1. (652)8 _ 423801 + -|- x 2608 - | x 8 = 425104
2. (654)8 = 423801 + | x 2608 + | x 8 = 427716
3. (656)8 = 423801 + | x 2608 + ^ x 8 = 430336
SUMMATION OF SERIES. 343
V65T= 8-666831, V652'= 8-671266,
V653 = 8-675697, V654 = 8-680123, and
V655 = 8-684545, find V653-75.
Solution : Here a = 8-666831, «i = -004485, a, = - 0-000004,
tta = — 0-000001, 0^4 = 0, and tj- = 3 -j . Substitute these values in
n n^(. in-\){n-^) (n-l)(7i-2)(n-3)
a« = a + (71 — 1) «! H r^ tta H rK ^s ; then,
li /T'J' OQ1
V653-75 = 8-666831 + ^ x '004435- ^ x -000004- g^^ x -000001
= 8-666831 + -012196 - -000009 - -000001 = 8-679017.
EXERCISE 92.
Find the wth term and the sum of n terms of the
following series :
1. 2 + 6 + 12 + 20 + .... 4. 3 + 8 + 15 + 24 + ....
2. 1 + 9 + 25 + 49 + .... 5. 1 + 4 + 9 + 16 + ....
3. 1 + 3 + 6 + 10 + . . . . 6. 2 + 12 + 30 + 56 + . . . .
7. 6 + 24 + 60 + 120 + 210 + ....
8. 45 + 120 + 231 + 384 + 585 + ....
9. 1.3 .22 + 2.4.32 + 3 .5.42 + ....
10. 3. 5. 7 + 5. 7. 9 + 7. 9. 11 + ....
Sum to n terms and to infinity :
1.1. 1
344 ADVANCED ALGEBRA.
Find the value of :
16. %{n^{^n-2)} 17. :S j|(w + l)(^ + 2)
20. The log. 950 = 2-977724, log. 951 = 2-978181,
log. 952 = 2-978637, log. 953 = 2-979093,
find log. 952-375.
Reversion of Series.
618. If 2/ is a serial function of x, then may x be de-
veloped into a serial function of y, and the process is
called Reversion of Series.
Example. — Kevert y = a-\-lx-\-cx^+da?-{-ea^-\-,,,.
into a serial function oi y.
Solution : Put the series in the form
y—a = l)x-{-cx^-{-da^-\-ea^
Vvitx=zAiy-a)^B(jj-af + C{y-af + D{y-ay-\-....
Now, hx =bA(i/-a) + bB(i/-af + bC(y-af+bD{y-ay+....
cx^ = cA^ (y-a)^ + 2 c A B (y-af + {c B^ + 2c AC)(y-ay+
d a^ = d A^ (y-af + 3d A^ B {y-a)*+ . . . .
ere* = eA*{y—ay +
,\ y-a = bA(y-a) + (bB+cA^)(y-af
+ {b C+2c A B+d A^)(y-af
+ {b D + cB^ + 2c A C+2d A^ B+eA*)(y-a)*+ . . . .
Equating the coefficients,
1. bA = l; whence A = -j-
2. bB + cA^ = 0; whence B=- ^
3. bC+2cAB + dA' = 0; whence C= ^^^^^-^
4. bD + c& + 2cAC + SdA^B-\-eA* = 0;
¥e — 5bcd + 5c^
whence D = jji •
/. x = j{y-a)-y^{y-af+ ^ (y-af
bU — 5bcd + 5c*,
^i (y-a)*+....
REVERSION OF SERIES. 345
619. Cor.— It a = 0, then
1 c 3 , 2c^-bd , b^e-5bcd+6€^ . ,
EXERCISE 93.
Revert the following serial functions of x into serial
functions of y:
1. y=zx-\-x^-\-a^-\-xf^-\-
3. y = x-]-2x^-{-63^ + Ua^-\-.,J,
4. y = a; + a:^ + 2a;^ + 5a;^ +
Suggestion.— Let x = Ay + By^ + Cy^ + Dy'' +
5. y = l+a;+i^2+ia^+^a;* + ....
6. y = l + a;-2a:2_j_^
7. Find one value of x in the equation a^-{-43^ -]-6xz=l
Suggestion. —
Put 1 = 2/, and assume x = Ay + By^ + Cy^ + Dy*+...,
8. Find one valut of a; in ic^ + l^^ x = l
Recurring Series.
620. A series in the ascending powers of x, in which
each term, after one or more fixed terms, is px times the
preceding term, or px times the preceding -\-qa^ times
the next preceding term, or ^ a; times the preceding -{-qx^
times the next preceding -]-rx^ times the next preceding
term, or and so on, is a Recurring Series,
621. A recurring series is of the firsts second^ or nth
order, accordingly as each term, after the law begins, is
derived from one, two, or n preceding terms.
346 ADVANCED ALGEBRA.
622. The forms px, px-{-qx^, px-\- qx^ -\- rx^, and
so on, are called the order scales of the series ; and p,
p-\-Qy i? + 5' + ^^ and so on, the order scales of the co-
efficients.
Illustrations. — If we put p = 3, q = 4:, and r= —2,
then
1. 2-{-(ix-\~lSx^-\-64:X^ + .... isaseriesof the first
order.
2. 2-\-6x-^26x^-{-102x^-\- is a series of the
second order.
3. 2 + 6a; + 26a;2 + 98rz;-^ + 386i5^ + .... isaseriesof
the third order.
623. Prohlem 1. To determine the scale of coeflacients.
1. Let a-{-I)x-\-cx^-\-da^-\-..., be a recurring series
of the first order.
Then, bx = apx; cx^=p bx^ ; dx^ =pcx^ ; and so on.
b c d ^
i^ = - ; ^ = ^ 5 i? = - ; and so on.
2. Let a-\-hx-\-cx^-\-dx?-\- be a series of the
second order.
Then, a q x^ -{-h p x^ = cx^ (1); l)qx^-\-cpoi^ = da? (2);
whence, aq-\-lp =c {^)', hq -\-cp =: d (4).
Then, by elimination,
_ be — ad , _bd—(?
^ "" b^ — ac ^ ~ ¥ — ac
3. Let a-{-bx-\-cx^-\-da?-\-ex''-\-fx^-{-.... be a
recurring series of the third order. Then,
1. ar-\-bq-{-cp = d 2. br-}-cq-\-dp = e
3. cr-\-dq-\-ep =zf
By elimination the values of p, q and r may be found.
In the same manner the scale of coefficients of a recurring
series of any order may be found.
RECURRING SERIES, 347
624. SchoHum, — In order that the scale of coefficients
of any recurring series may he found, there must le given
at least twice as many terms of the series as there are
terms in the scale. In the exercises concluding this sub-
ject just twice as many terms of each series will he given
as are contained in the scale of the series. This will
enable the student to determine at a glance the order of
the series.
When this law of notation is not followed, as when the
nth term of a series only is given, it is usually hest to
expand the series and malce trial for a scale of two terms,
and, if the results thus obtained will not satisfy the series,
then make trial for a scale of three terms, and so on until
the proper scale is determined.
Example. — Find the scale of coefficients in the series
l-{-%x-\-^x^-\-^o(?-\' and expand the series.
Solution : This is a series of the second order, since four terms are
given. Hence,
1. g + 2i? = 3 2. 2 g + 3i? = 4
whence, p = 2 and q = —1, and the scale is 2 — 1. The series is
1 + 2a: + 3a;2 + 4a;3 + 5a;* + nos^-h
625. Problem 2. To find the sum of n terms of a
recurring series.
The method of finding the sum of a recurring series
is the same, whatever be the scale of the series. For the
sake of simplicity we will here assume a series of the
second order, whose scale is p-{-q, for illustration.
Let «o + «i ^ + «^2 ^ + • • • • «»-i ^^^ be a series of the
second order. Then,
>S„ = «o + aiX-{- a^x^ -\- + «„_! x'"'^
—pxSn= —pa^x—pa^T? — — ^«„_2a;"~*
— qx^S^ = —qa^x^ — — qa^^^af-^
— q a^-z^"" — q cin-\^'^'^
348 ADVANCED ALGEBRA.
Adding and remembering that the sum of the co-
efficients of each term from the third to the nth. in-
clusive is zero,
(1 —px — q 0^) /S'« = «o + («i — i? «o) ic
- (P «n_i + q a^-z) x^-q a^-i a;"+i ;
whence, S. = ^o + {a.-pa,)x
1 —px — qar
{pan-i-{-qan-2)x''-{-qan-i3f^'^
1 —px — qx^
626. Cor, — If X <1 and Urn. n = co, then
^ ___ ao-\-{ai—pao)x
1 ^px — qoir
Example.— Sum l^^x^bx^-\-l^ a;3+48 rr^+145 x^-{-
.... to 7 terms and to infinity, when x <1,
Solution: 1. 5jt? + Sg- + r = 18 3. 18^9 + 5g + 3r = 48
3. 48p + 18^ + 5r = 145
.*. p = 2, g = 3, and r = — 1.
S-, = l + 3a:+5a;2 + 18a;8 + 48a;4+1453:5^.4i6a;6
-2 a; /S't = -2a;-6a;2-10a;8-86a;4- 96a;«-290a;«-832a;''
-3a;2,S'T= -3a;2- 9a;8-15a;4_ 54a;*-144a;«-435a;''-1248ic8
+ x3>S'7= + x^+ 3a;4+ 5^^+ iga^e^ 48a;7+ 145a:«
+ 416a;9
.-. (l-2a;-3a;2 + a;3)>S'7 = l+a;-4a;2-1219a;''-1103a;8 + 416a:»
l + a;_4a:2_1219a;'-1103a:8 + 416a;9
/St
1+X — 4:X^
l-2a;-3a;2+a;»
627. If /S'ao is developed into a series by the method
of indeterminate coefficients, the original series may be
reproduced to any number of terms desired. Therefore,
/Soo is often called the generatrix of the series.
628. If the generatrix can be decomposed into partial
fractions, the general, or ni\i term, of the series may
easily be obtained, and hence, too, the sum of n terms.
RECURRING SERIES. 349
niustration. — Let it be required to find the nth term
and the sum of n terms of the series
Solution : It may readily be determined that i? = 2, g = 3, and the
l + 3a; 11 3 1
generatrix = ^_^^_.^^, = " 2 ' iT^ + 3 ' T^Ts^'
11 1.11,
^^^'-2 •n:T-=-2 + 2^-■2^+••••
r _ l^n ^ _ (-1)"^"-^ r493 b1 •
3 1 3 9 27 ,
^^^ 2T335=2 + 2^+-2^ +••••
_ (-l)na;n_l 3n+la;n_3
'^''~ 2a; + 2 "^ 6a;- 2 '
2^—1 3»i
nth term = (— 1)** . -g— + -g- . a?*-*.
EXERCISE 94.
Sum to infinity :
1. l + 3ic + 5a;2 + 13a;3 + ....
3. 2 + 2a; + 4a;2 + 14a;34-....
4. 3 + 2a;-7a;2- 38:^3-....
5. l + 2a; + 3a;2+ll:r3^35^_j_121a;5 + ....
6. l-3:zj + 5a;2_^5^^13^_|.61^5_|.__
7. Sum l + 2ir + 9a;2 + 33iz^+ to 6 terms.
8. Sum 1 — 2a;— 7a;2 — 8a;3 4- to 7 terms.
9. Sum 2 — a; + 6a;2 — 14a;3 + to 8 terms.
10. Sum l — ^x-{-^x^^Qo(?^x^ — ^x^-\- to 9 terms.
Find the nth. term and the sum of n terms of
11. l + 2a;-82;2 + 20a;3-
12. 1 + 5 a; -f 9 a;2 4- 13 a;3 _^
350 ADVANCED ALGEBRA.
Decomposition of Rational Fractional Functions.
629. To decompose a rational fraction is to find two
or more other fractions whose sum equals the rational
fraction.
630. It will be only necessary to show how to decom-
pose proper fractions, as all improper fractions may be
reduced to mixed numbers, which process will already
lead to a partial decomposition.
Principles.
631. 1. Any rational fraction of the form of
p
, ■ ., _.-,. j—j- — r may he decomposed so that
[X -f- a) \x -\- 0) .... {x-\-n)
(x-]-a){x-\-I)) (x-{-n) x-\-a x-\-b x-{-n
Illustration. — Put
3^24. 14:^-29 _ A . B . C .,.
{x-l){x-\-%)(x-^) x-1 ' x-{-2 ' x-3
Clear of fractions and arrange the terms according to the descend-
ing powers of a;,
3x^ + Ux -29 = (A + B + C)x^ - {A + 4:B - C)x -
{6A-SB + 2C).
Equating the coefficients [566], we have
(1) A +B+C=d (2) A + 4B-C= -U
(S) 6 A-BB + 2C= 29
Finding the values of A, B, and C by elimination, and substi-
tuting them in (A), we obtain
dx'^ + Ux-29 _ 2 d_ 4
{x — l)(x + 2){x-S)~ x-l x + 2'^ x-S'
632. In a similar manner it may be shown that
(a X -{- b) (c X -\- d) {mx-\-n)
ax-\-l) cx-\-d ' mx-\-n
RATIONAL FRACTIONAL FUNCTIONS. 351
633. 2, Any rational fraction of the form of
P
{p(^ -{- a X -{- h) {x^ -\- c X -\- d) {a^ -\- m x -\- n)
may he decomposed so that
P
{x" ^ ax ^h) {7^ -\- ex ^ d) . . . . {x^ -\- mx ^ n) ^
Ax + B Cx-\-D Mx-^JV
~r ^2 I ^ ^ I ,7 "T • • • • ~r
x^-\-ax-{-bx^-\-cx-\-d x^-\-mx-{-n'
„, , ^. T, i. 4:X^-8x^-63^-Ux-{-3
mustration.-Put ( ^^^^i)(^_^^i)(^.^^^^)
Ax + B Ox-^D Bx-\-F ..
^ x^-\-x-\-l ' x^ — x-\-l ' x^-\-x^%
Clear of fractions and arrange the terms according to the descend-
ing powers of ic,
4a;4 - 8a:3 - 5a;2 - 15a; + 3 =
{A + C + E)0!^ + {B + 2 C + D + F)a^
+ {2A-\-4.C + 2D-[-E)x'-^{-A + 2B + ^C+4:D + F)x^
+ (2A-^ + 2(7+3i) + ^)a; + (25 + 2i> + i^) (1)
Equating the coefficients [566],
(1) A + C+E =
(2) ^ + 2C+Z) + ^=4
(3) 2A + 4(7+2/> + J^= -8
(4)^-25-3(7-4i9-i^=5
(5)2A-^ + 2(7 + 3i; + ^=-15
(6) 2^ + 2Z) + ^=3
Finding by elimination the values of J., B, C, D, E, and F^ and
substituting them in (A), we have
4a4_8a^_5a;2_i5a;4.3 _
(a;2 + ic + 1) {x^-x+ 1) (ic* + a; + 2) ~
3 4 5
x^ + X + 1 x^ — X + 1 x^ + X + 2'
634. In a similar manner it may be shown that
P
{a x^ -^ b X -\- c) {dx^-\-ex-\-f) {m x^ -\- n x -{■ p)
Ax-\-B Cx-\-D Mx-\-N
ax^-\-bx-{-c d:x^ -\-ex +/ mx^ -\-nx -{-p *
352 ADVANCED ALGEBRA,
P
635. S, A rational fraction of the form of -, — ; — r-
-' J J J {x-\-aY
may he decomposed so that
p A B jsr
t /^ I ^\2 ~r • • • • "r
(x-^aY x-\-a (x-\-aY {x-\-ay
Illustration. — Put
i^-^r
_ A B G D
" x-%'^ {x-%)^^ {x-^f'^ {x- 2)* ^^^
Clear of fractions and arrange the terms according to the descend-
ing powers of x ; then,
3a:3 _ i4a;2 + 19a. _ 5 _ ^^3 _ (6 ^ _ ^)a;2 +
(12J-4^+ C)x-{%A-4:B + ^G-D).
Equating the coefficients, we have
1. ^ = 3 2. 6.4-5 = 14
3. 12A-4:B + C= 19 4. 8^ -45 + 2 C- 2) = 5
Solving these equations, and substituting the values of A, B, C,
and D in (A), we obtain
Sa^-Ux^ + 19x-5 3 4 1 1
{x-2)* x-2 (x- 2)2 (x - 2f {X — 2)4
636. In a similar manner it may be shown that
1 _p _^_+ -g + + ^
(ax-i-by ax + b ' (ax-^-bf ' ' (ax-^by
2 P ^ Ax-]-B
(x^ + ax-\-by a^-\-ax-\-b^
Cx + D Mx-^-N
{^j^ax^bf ' •••• • {x'-^-ax-irby
{aT^^bx-^cy a:^ + bx-^c'^
Cx^D Mx-{-JSr
{as^ + bx^cf ' •••* ' {a3^-\-bx-\-cy
4. Any rational fraction whose denominator may be
resolved into linear and quadratic factors may be decom-
posed by a combination of the above methods.
RATIONAL FRACTIONAL FUNCTIONS. 353
Ulustration. —
To decompose ^^^_ a)^^-iY i^^+p^ + ^y ' P^*
{x — a) {x — iy {x^ -\- p X -\- qY x — a x — b
, M Px^Q P'x-\-Q'
{x — df af-\-px-{-q {x^ -\- p x -{- q)*
EXERCISE 93.
Decompose into partial fractions :
5cc + 2 2a;-5 _ A B
^' A ^2 1 O ^ I -I I
x(x-{-l){x-\-2) X ' ic + 1 ' x-^2
3x-2 x^-x-{-l ^ 3x^-Ux-{-25
(a;4-lf (^ + 3)3 (a;-3) (a;2-a:+6)
aJ + (a — ^)a; — ic^ ■ {px-{-qy
^ ^ 10.
X^-{-X^ + l X(1-4:X^)
3a:3_8a:2 + 10 2a; + l
11< 7 TTT 12*
(a; - 1)* (a; - 1) (a;^ + 1)
13. -n r^ 14.
a^^6x-{-6 (x-l)(x^-i- ly
X5. .4t^+l^ 16. ^
(a; + 1) (x^ + 1) -"• a^ _|. :r7 _ 2^ _ ^
2a;g~lla; + 5 5 ^^3 _p g ^ _j_ 5 ^. ^
(^~3)(a;2_|.2a;-5) ' (ic^ _ j) (^ _|_ 1)3
1 x^-{-x-\-l
10 on ■
^^- a^^x'-x^-a^ (a;+l)(^' + l)
21. ^+^t^ . 22. ^' + ^
l_a;-a.'*-|-a:« "'•• {x-\-lf{x-2){x-{-3)
CHAPTER X.
COMPLEX J{ UMBERS,
Graphical Treatment.
637. If a straight line of any assumed length be taken
to represent the number one, then will a straight line
twice as long represent the number two, one three times
as long the number three, and so on. Thus, we see that
any number may be represented by a line.
638. A line representing a number is called a graph
number, or a vector. The point where a vector is sup-
posed to begin is called the origin, and the point where
it ends, the extremity.
639. A vector is fully determined when both its length
and direction are given. In a system of graphical repre-
sentation of numbers, a vector running rightward from
its origin represents a positive number and is positive, and
one running leftward from its origin represents a nega-
tive number and is negative,
640. If the vector +« be made to revolve about its
origin A, through an angle of 180°, or ir, it will become
the vector — a, or will be
multiplied by — 1 ; and if j^, ~ ^ i "'" " r^
the vector — a be revolved
about its origin A through an angle of 180°, or tt, it will
become the vector -|- a, or will be multiplied by — 1.
GRAPHICAL TREATMENT. 355
Therefore,
1. Revolving a vector through an angle of 180°, or tt, is
equivalent to multiplying it hy — 1.
2. Revolving a vector about its origin through an angle
of 360°, or 2 tt, is equivalent to multiplying it twice hy
— 1, or once ly + 1^ which does not affect its length or
direction.
3. Since — 1 = V— 1 X V— 1, revolving a vector
about its origin through an angle of 90°, or \ tt, is equiva-
lent to multiplying it by V— 1, or i [v. P. I. 299].
641. Motion about the origin of a vector in the direc-
tion the hands of a clock go is considered negative, and
counter-motion positive. The factor -\-i may, therefore,
be taken to represent circular motion about an origin
through an angle of 90°, or \ it, counter-clock-wise ; and
— i through an angle of 90°, or \ tt, cloclc-wise.
642. Since (+ i) X (+ i) X (+ i), or (+ if = - i, the
factor symbol — i may also denote circular motion about
an origin through an angle of 270°, or | tt, in a positive
direction.
643. Since {± i) X (± i), or (± if = - 1, the factor
symbol — 1 may denote
circular motion about an
origin through an angle of
180°, or TT, in either direc-
tion.
Illustrations. — 1. The B- ^
vector -\-a multiplied by
-{-i = AB revolved about
A in the positive direction
through an angle of 90° =
AB\
B'
+a
—at
356 ADVANCED ALGEBRA.
2. The vector + a multiplied by —l^AB revolved
about A in the positive or the negative direction through
an angle of 180° = A B",
3. The vector + a multiplied by —i = AB revolved
about A in the negative direction through an angle of 90°,
or in the positive direction through an angle of 270° =
A B'".
4. In a similar manner it may be shown that (— a) X
(+^) =AB"'', (-«)x(-l) =AB, and {- a) X {- i)
= AB\
644. From what has been explained thus far it will be
seen that, if a vector a units long running rigJitward from
its origin represents + «^ running leftward from its origin,
it will represent — a ; running upward from its origin,
-\-ai'y and running doiunward, — ai.
646. One vector is added to another by placing its ori-
gin to the extremity of the other and giving it the direc-
tion indicated by its factor symbal. The vector of their
sum is the length and direction of the line joining the
origin of the vector to which addition is made with the
extremity of the vector
added. ^ +« ? +5 ^
Illustrations. —
1. The vector B C ^ ±^
(=+&) added to the G
vector A B {= -\-a) —h
■^^r-B
gives the vector A C ^ A ~^^
(= + (« + 5)).
2. The vector B C (= - b) added to the vector A B
(= -f a) gives the vector A C {= -{- {a — h)] , when a>h.
3. The vector BO (= — !)) added to the vector AB
(= -f a) gives the vector A C {= — (b — a)}, when a<b.
Note.— To subtracit a vector is to add the vector with contrary sign.
COMPLEX NUMBERS.
357
Representation of Complex Numbers.
646. Let it be required to represent graphically the
complex numbers a-^hi, a — hi, ^a-\-hi, and —a — hL
1. a^bi = +a+(+Ji).
.-. Vector {a-\-bi) = A C^ [645].
2. a — 'bi=-\-a-]-{—'bi).
.'. Vector {a-hi) = A C^ [645].
3. —a-\-hi=—a-\-{-\-hi).
.'. Vector {-a^hi) = AC^ [645].
4. — a — hi=— a-{-{— hi).
.-. Vector {--a-hi) = AC^ [645].
647. The vectors of the two terms of a complex num-
ber are called the components of the vector of the number.
Thus, A B and B C^ are the components of A C^.
648. The length of the vector of a complex number is
called the modulus of the vector, or the modulus of the
complex number, and is equal to the square root of the
sum of the squares of the lengths of the components.
Thus, mod. {a-\-hi) = mod. (a — hi) = mod. {—a-\-hi)
= mod. {—a — hi) = Va^ + h\
649. The direction of the vector of a complex number
is determined hy the angle which the vector makes with
358 ADVANCED ALGEBRA.
its horizontal component, which angle is called the ampli-
tude of the yector.
Thus, the amplitude of the vector A Ci is the angle
Ci A B, which for distinction will be represented by u)^ ;
the amplitude of A Cz is Cz A B', represented by Wg ; the
amplitude of A G^ is G^ A B\ represented by 6)3 ; and the
amplitude of A C4 is G^AB, represented by 0)4.
650. It is equally accurate, and sometimes more con-
venient, to define the amplitude of a vector as the angle
included between the vector and the vector + a, measured
in a positive direction from the vector +«.
Thus, the amplitude of A Gz is Gz A B, The ampli-
tude of A C3 is the reflex angle BAGs, described by re-
volving A B about A in Si positive direction until it coin-
cides with A C3. The amplitude of A C4 is the reflex
angle BAG4,.
651. A vector and its components may be constructed
from its modulus and amplitude as follows :
1. Draw an indefinite horizontal line, and select some
point in this line for the origin of the vector.
2. At the origin, deflect an angle with a protractor
equal to the amplitude of the vector, and in the proper
position.
3. Lay off from the origin, on the deflected side of the
angle, from a scale of equal parts, the modulus of the
vector. The vector is then determined.
^. At the extremity of the vector let fall a perpendicu-
lar to the horizontal line. This perpendicular and the
part of the horizontal line intercepted between the origin
and the foot of the perpendicular will he the components
of the vector. Their lengths may he ohtained hy actual
measurement on the scale of equal parts.
COMPLEX NUMBERS.
359
Problems.
652. To find the sum of two complex numbers, graphically.
the sum of a-\-hi and
+ a
Let it be required to find
— c-\-d i.
1. Add — c-\-di to a -{-hi,
graphically.
Solution : Construct a +1)i. Its
vector is A C. At its extremity, (7,
construct —c-^di. Its vector is C E.
Join A with E. AE is the vector of
the sum [645]. V(6 + df + (a- cf
is its modulus, and the angle EA B its
amplitude.
Proof : Add —c + di to a + bi
algebraically, and construct the sum.
Thus,
{a + bi) + {—c + di) =
{a — c) + {b + d) i.
Construct AB = +a, B C = —c ;
then AC = a — c. At C erect CE =
{b + d) i. Join A with E. The vector
A E will be identical with AE in. the
preceding diagram ; therefore, the solu-
tion is correct.
2. Add a-\-hi to —c-\-di.
Solution : Construct —c + di.
Its vector is A C. At its extremity,
C, construct + a + bi. Its vector
is CE. Join A with E. The vector
A E will be identical with the vector
AE in the first case, which proves
the commutative law of addition
graphically as applied to complex
numbers.
Exercise. — Find graphically the sum of :
1. a -{-hi and c-\- di 4. — a — hi and — c — di
2. a — hi and c — di 5. a -{-hi and a — hi
3. a -{-hi and c — di 6. 0-{-hi and — di
-\-di
+ bi
360
ADVANCUn ALGEBRA.
653. To multiply a complex number by a rational num-
ber, graphically.
Let it be required to multiply a-^-ii hy — c.
Solution :
Construct a-\-hi. Pro-
long its vector JL (7 to Z),
making AD z= G y. AG.
Revolve A D about A
through 180°, then AD
is the vector of — c times
a + & *.
Exercise. — Construct the vectors of :
1. (« + ^ t) X c 4. {—a — b i) X c
2. {a — bi)Xc b. {a — h i) X (— c)
3. {—a-\-bi)Xc 6. {— a -^ ii) X (— c)
664. To multiply a complex number by a simple im-
aginary number, graphically.
Let it be required to multiply
Solution : Construct — a + bi.
making AD = c x AG. AD
is the vector of c times — a
+ b i. Revolve A D about
A through an angle of 90°
clock-wise [641], or, which is
the same, draw AD — AD
and perpendicular to A D.
A D is the vector of — ci
times — a + bi. DAA! is
its amplitude.
Exercise. — Construct the vectors of
a -{-hi by — ci.
Prolong its vector AG to D^
d'
1. {a-{-li) Xci
2. (—a — bi) X ci
3. {a — bi) X (—ci)
4. (a + bi) X (—ci)
5. (—a + bi) Xci
6. (a — b i) Xci
7. (—a — bi)x (—ci)
8. (0 — bi) X (—ci)
9. (0 + b i) Xci
10. (O + bi) X (—ci)
COMPLEX NUMBERS.
361
^x
655. To multiply a complex number by a complex num-
ber, grraphically.
Let it be required to multiply — a-{-di by c — di.
Solution: The vector of
the sum of c times —a + bi
and — di times —a+bi
is required. Construct — a
+ b i. Its vector is A C.
Prolong AC to D, making
AD = c times AC; then
AD is the vector of c times
— a + bi. Prolong ^ Z> to
E, making DE = d times
A C, and revolve D E about
D through an angle of 90°
clock-wise, then is DE' the
vector of —di times —a
+ bi constructed at the extremity of A D. Join A and E'. A E' is
the vector of the sum of c times —a + bi and —di times —a + bi.
Exercise.— Multiply graphically :
1. a-\-bihyc-\-di 4. — a-^-bi hj c-\-di
2. a — hi hj c-\-di b, — a — lihj— c-\-di
3. a — bihjc — di 6. — a — bi hy — c — di
General Principles.
656. 1. The sum, the difference, the product, and the
quotient of two complex numbers are, in general, complex
numbers.
For, 1. {a^bi)-\r(c^-di) = {a^c)-\-{b-\-d)i.
2. {a-^bi)-(c + di) = {a-c)-{-{b-d)i.
3. {a-{-b i) {c + di) = ac-}-bci-{-adi — bd
= {ac — bd)-\-{bc-i-ad)i,
a-{-bi _ {a-\- bi) (c — di)
c-\-di ~ (c-\- d i) {c — d i)
4.
{a c -\-b d) -\- {b c — a d) i _ ac-\-bd
+
(be — ad\.
862 ADVANCED ALOEBRA.
667. 2. The sum and the product of two conjugate
complex numbers are real.
Eor, 1. {a-\-hi)-^{a — 'bi) z=%a.
2. {a-\-di)(a-bi) =a^ + bK
Scholium, a^ + b^ is the square of the modulus of
±a-\rbi and of ± a — bi, and is called the norm of
each. Therefore,
Cor, — The product of two conjugate complex numbers
equals their norm.
668. 8. The norm of the product of two complex num-
bers equals the product of their norms.
For, norm {a-\-b i) {c + di)
= norm {{ac — bd)-\- (ad-\-hc)i]
-{ac-bdf-\-{ad-\-b cf [657, Sch.]
= a^ c" -\-b^ d^ -^ a^ d^ -^-b^ d"
= norm {a-\-b i) multiplied by norm {c-\- d i).
Cor, — The modulus of the product of two complex num-
bers equals the product of their moduli.
669. Jf. If a-^bi = 0, then a = and b = 0.
For, it a -\- b i = 0, bi= —a and —b^ = a^;
whence, a^-\-P = 0, which is possible only when a =
and b = 0.
Cor, — If a complex number vanishes, its modulus van-
ishes ; and conversely, if the modulus vanishes, the complex
number vanishes.
660. S. Ifa-^bi = c-\-di, then a = c and b = d.
For, if a-\-bi = c-\-di, (a — c) -{- {b — d)i = ;
whence, a — c = and b •— d = [P. 4],
and a = c and b = d.
COMPLEX NUMBERS. 363
661. Problem. To find the value of e*+^*.
Solution : Assuming the exponential law of multiplication [275, P.],
and Formula (G), Art. 599, sufficiently general to include imaginary-
exponents; then
( 2/H"2 y^i^ y*i* ^ )
e«+y» = e' X ev' = e' <l+yi+ ^ + ^ + ^ + etc. j-
v^ v* y^
662. The expression 1~J2"I"|4~^ + ^*^* ^^ called
cosine y, and is written cos. y.
ti^ if* ip
The expression V ~ f^ ~\- h. vi "^ ^^^' ^^ called
sine y, and is written sin. y. Therefore,
e^' = cos. y-\-i sin. y. (B)
663. Resume the equations
ifi ^ -^/^
cos.y = l-| + |-|+etc. (1)
sin. 3/ = 2/-|- + |--|- + etc. (2)
e''* = COS. y -\-i sin. «/. (3)
Put —y tor y in (1), (2), and (3), then
COS. (- 2/) = 1 - || + 1^ - ^ + etc. = COS. y (4)
sin. {- y) = - y + y- -U- + y etc. = - sin. y (5)
e-** = COS. (— y) + i sin. (— y) — cos. y — i sin. y (6)
Multiply (3) by (6),
1 r= (cos. yf — i^ (sin. y)^ — cos.^ y + sin.' y (C)
Note, cos.*^ y denotes (cos. y)^ and sin,^ y denotes (sin. y)'.
Cor, sin. y = vl — cos.^ y ^
COS. y = a/1 — sm.^ y.
364 ADVANCED ALGEBRA.
664. Put ny ioT y in (B) ; then
e"*'* = COS. ny -\-i sin. n y,
Eaise (B) to the n'Cd power ; then
e""* =. (cos. y -\-i sin. yf.
.-. COS. ny -\-i sin. ny = (cos. y -\-i sin. «/)" (D)
665. Let n = % in (D) ; then
COS. 2 ?/ + ^ sin. %y •=■ (cos. «/ + *' sin. yY
= cos.^ «/ — sin.^ y -\-^ (sin. y cos. y) i.
.*. cos. 2«/ = cos.^ y — sin.^ ?/ [660] (E)
sin. 2y = 2 sin. y cos. «/ [660] (F)
666. Put x-\-y tor y in (B) ; then
g(x + y)i _. gxi ^ gyi _ gQg^ ^^ _|_ 1^) _|_ i; gin^ (^ _j_ ^)^
But e'* . y'^ = (cos. a; + ^ sin. ir) (cos. y ^i sin. ?/) (B)
COS. X COS. «/ — sin. x sin. «/ + * (sin. x cos. ly + cos. x sin. ?/)
'. COS. {x-\-y) = cos. a; cos. y — sin. a; sin. y [660] (G)
sin. (^ + ?/) = sin. x cos. ?/ + cos. x sin. ?/ [660] (H)
Graphical Representation of sin. y and cos. y.
667. It is evident that all the conditions expressed in
the equation sin.^ y + cos.^ «/ = 1 will be satisfied by as-
suming 1 as the modulus of a vector whose amplitude is
the variable angle y and whose components are sin. y and
COS. y. But, to make this expression conform to the nu-
merical values of sin. y and cos. y as expressed in Art. 662,
y must be taken to represent the number of vector units
in the arc which measures the amplitude, and sin. y as
the vertical and cos. y as the horizontal component of
the vector ; for in this way only would sin. y = and
cos. y = l when y = 0,
COMPLEX NUMBERS.
365
668. The ratio of sin. y to cos. y is called tangent y,
and is written tan. y. It may be expressed graphically
as follows :
Let BC = y, CE
= sin. y, and AE =
cos. y. At B draw an
indefinite tangent to
the circle. Prolong the
yector AC until it
meets the indefinite
tangent at i>. B D will
be tan. y. For, from
the similar triangles
DAB and CAB we
have BD'.CE '.'.AB'.AE', or,
B D : sin. y ::1 : cos. y ; whence,
sin. y
BD =
tan. y.
sin. y, and A L ^ cos. «/. The tri-
ces, y
If BF=zy, FL
angles A FL and B A K will be similar, and B K=^ tan. «/.
If B F G = y, then G L =. sin. y, and ^ Z = cos. ?/.
The triangles GAL and DAB wiU be similar, and D B =:^
tan. y.
If ^ i^ZT = y, then HE = sin. y, and AE = cos. «/.
The triangles KAB and -ST^ ^ will be similar, and KB
— tan. y.
Scholium, — So long as y < \ it [90°], sin. y and cos. y
are positive ; hence, tan. y (B D) is positive. When y >
i TT but < IT, sin. y is positive and cos. y negative ; he^ice,
tan. y (B K) is negative. When y > tt tut < f t, sin. y
is negative and cos. y negative ; hence, tan y {D B) is posi-
tive. When y > i TT but < 2 v, sin. y is negative and
cos. y is positive; hence, tan. y {KB) is negative.
CHAPTER XI.
THEORY OF FUJVCTIOJVS.
Definitions.
669. A quantity whose value changes, or is supposed to
change, according to a definable law, is a definite variable,
or simply a variable.
670. A variable whose law of change is not dependent
upon that of another variable is an independent variable,
671. A variable whose law of change is dependent upon
that of another variable is a dependent variable, and is
called a function of that variable.
Hence it is, that any expression containing a variable
is a function of that variable [562].
•
672. Any law of change may be imposed upon an inde-
pendent variable ; but, when it is once imposed, the law
of change of any function of the variable becomes de-
termined.
673. The simplest treatment of functions of a single
variable is that in which the variable is supposed to in-
crease or decrease uniformly by equal increments, finite or
infinitely small.
674. A function is said to be continuous so long as an
infinitely small change in the independent variable pro-
duces an infinitely small change in the function, and dis-
THEORY OF FUNCTIONS. 367
continuous when an infinitely small change in the inde-
pendent variable produces a finite or infinitely great change
in the function.
Ulustration. — Thus, the function - — — assumes all
values between + 1 and + oo as a; assumes all values be-
tween and + 1, and is, therefore, continuous from to
+ 00 ; but, as the value of x continues to increase from a
quantity infinitesimally less than + 1 to a quantity infini-
tesimally greater than + 1, or takes an infinitely small step
across + 1, the function takes a leap through the whole
gamut of numbers from + "^ to — oo , and is, therefore,
discontinuous between these values.
675. So long as a function increases in value as the
independent variable increases in value, and hence, too,
decreases in value as the independent variable decreases in
value, it is an iticreasing function ; but when it decreases
in value as the independent variable increases in value,
and, hence, increases in value as the independent variable
decreases in value, it is a decreasing function.
Ulustration. — Let y =f(x) = x^ — 4,x-\-3.
Assign values to x and calculate the corresponding
values of y by synthetic division [106], you will obtain
results as follows :
For a; = - 3, - 2, -1, 0, +1, +2, +3, +4, +5
y = +24, +15, +8, +3, 0, -1, 0, +3, +8
Here y decreases from + 24 to — 1 as a; increases from
— 3 to +2, and is, therefore, a decreasing function be-
tween these values of x ; and it increases from — 1 to +8
as X increases from +2 to +5, and is, therefore, an in-
creasing function between these values of x.
676. The maximum value of a function is the value at
which the function changes from an increasing to a de-
creasing function.
368 ADVANCUn ALGEBRA.
677. The minimum value of a function is the value at
which the function changes from a decreasing to an in-
creasing function.
678. The maxima and minima values of a function are
often called the turning values of the function.
679. A turning value of a function may be a finite
constant, zero, or infinity.
niustrations.— 1. Take ^^ = /(i») = 3 + (4 - xf.
As X increases from to 4, «/ decreases from 19 to 3 ;
and as x continues to increase from 4 to oo , y increases
from 3 to 00. Therefore, 3 is a turning value (a mini-
mum) of y.
2. Take y = {a- xf.
As X increases from to a, y decreases from a^ to ;
and as x continues to increase from a to co , y increases
from to 00 . Therefore, is a turning value of y.
As X increases in value from to + 1, (1 — xy decreases
from 1 to 0, and y increases from 1 to oo ; and as x con-
tinues to increase from 1 to oo , (1 — xY increases from
to 00 , and y decreases from oo to 0. Therefore, oo is a
turning value (a maximum) of y.
680. The limit of a function is the value of the func-
tion at which it ceases to be continuous.
Note. — Notice the distinction between the meaning of the word
limit as here used and as used in Art. 398. In the latter sense, <x>
would be the limit of y in illustration 3, Art. 679, instead" of a
maximum.
681. The limit of a function may be a finite constant,
zero, or infinity.
Illustrations. — 1. Take y =f(x) = 2 — —,
2
As X increases from to oo , ~ decreases from 2 to 0,
THEORY OF FUNCTIONS. 369
and y increases from to 2 ; and, as x can n(# be sup-
posed greater than oo , y can not become greater than 2,
neither can y begin to decrease at 2. Therefore, 2 is the
limit of y.
2. Take y z=z f{x) = x^{l^ x^).
As X decreases from 1 to 0, y decreases from 2 to ;
and as x can not be taken less than (negative) without
making y imaginary, y can not become less than 0, neither
can y change from an increasing to a decreasing function
at 0. Therefore, is the limit of y,
3. We have already seen [674] that y — f(x) = __
increases from 1 to oo as a; increases from to +1, and
thereafter becomes discontinuous. Therefore, (» is the
limit of y,
682. A function may have two sets of values approach-
ing the same or different limits for the same set of values
of the independent variable.
niustrations.— 1. Take y^ =/(a:) = 16 — a;^ ;
then y = ± a/16 - a;^.
Here are two values of y for each value of x, numeri-
cally equal but opposed in sign. As x increases from
to 4 one value of y decreases from 4 to 0, and the other
increases from — 4 to 0. It x becomes infinitesimally
greater than 4 both values of y become imaginary. There-
fore, is the limit of both values of y.
2. Take?/2 =/(^) = 4a:;
then «^ = ± 2 Vx.
Here, again, are two values of y for each value of x.
As X increases from to + oo , one value of y increases
from to -f °° and the other decreases from to — oo ;
and as x can not be supposed greater than -\- co , + °o ^s
the limit of one value of y and — oo the limit of the other
value.
683. The limit of an increasing function is a superior
370 ADVANCED ALGEBRA.
or maximum limit ; that of a decreasing function an iw-
ferior or minimum limit.
Graphical Representation of Functions of a
Single Variable.
684. Every function of a single variable may be ap-
proximately represented by a line, straight or curved,
called the graph of the function.
Method. — Let y = f{x). Assign successive values to x
and calculate the corresponding values of y. Construct
two indefinite straight lines intersecting each other at right
angles, one running right and left and the other up and
down from their intersection. These are the axes of refer-
ence. The first is the a;-axis and the second the i^-axis,
and their intersection the origin. Regard distance right-
ward from the y-axis positive^ and distance leftward nega-
tive ; distance upward from the a^-axis positive, and dis-
tance downward negative.
Assume a fixed length as a unit of scale, and lay off on
the a;-axis from the origin the successive values of x based
on this scale, and at the extremity of each x value, and on
a line parallel to the y-axis, lay off the corresponding values
of y. Thus will be located a series of successive points ;
draw a continuous line through these points ; it will be
the graph of the function, and its accuracy will depend
upon the nearness to each other of the successive values
of X taken, the relation of the unit of scale to that of x
and y, and the correctness of the instruments used in
plotting.
niustrations.— 1. Take y =f{x) = x^ — 20 x^ -\- QL
Assign special values to x and calculate the correspond-
ing values of y by synthetic division [106]. You will
readily derive the following table of values and make the
following plot :
FUNCTIONS OF A SINGLE VARIABLE. 371
X
y
64
1
44
2
3
-35
4
5
189
00.
00
- 1
44
- 2
- 3
-35
- 4
- 5
189
— 00
00
PLOT.
+ 2/
Observa-
tions. — 1. The
curved line
on the plot is
the graph of
the function.
2. The unit
of scale used in
plotting the
graph repre-
sents 20 units
of y to one
unit of X.
+
200
+
180
-f-
160
1
+
140
+
120
+
100
+
80
+^
v60
/
/.
.oN
\
/
+
20
\
/
4
-3
-2
A
A
+2
+3
u
\
/
20
\
y
V
/
40
\
/
__
60
-y
3. The graph exhibits three turning values of the function ; two
minima at the points (ic = 3, y = — 35) and (— 3, — 35), and one
maximum at the point (0, 64).
4. When a:=+oo, 2/r=+oo, and when a;=— oo, y = + co.
The graph, like the function, descends from (— oo , +oo) to (—3, —35),
then ascends from (— 3, — 35) to (0, 64), then descends from (0, 64) to
(3, — 35), then again ascends from (3, — 35) to (+ oo , + oo ). It is a
continuous graph from beginning to end.
5. At a? = 2, 4, — 2, and — 4, the graph crosses the ir-axis, exhibit-
ing the fact that for these values of x, y = f(x) = x^ — 20 x^ + 64t — 0.
The values of x that render f(x) = are, however, the roots of the
equation f(x) = 0; therefore, the values of the roots of f(x) = Q may-
be approximately found even if incommensurable, by plotting /(a;) = y
and determining with a scale of equal parts where the graph crosses
the a^^axis.
372
ADVANCED ALGEBRA.
2. Plot y = ± Va^-x = ± Vx(x+l){x -1).
The following table of values may readily be obtained
-^07
+
y
3
PLOT.
/
+
3.5
/
+
2
/
+
1.5
/
+
1
/
/
+
.5
/
-1
/d
■^
+.5
/
-fl.5
+2
+2.5
V
^
.5
\
1
\
1.5
\
y
2
\
2.5
\
_
3
\
\
+3
Observations. — 1. The graph consists of two
branches between the points (—1, 0) and (0, 0),
symmetrical with respect to the a;-axis. These
branches are confluent at the points mentioned.
2. The graph is discontinuous for all values
of X antecedent to — 1, counting from a; = — oo ,
and also for all values of x between and + 1.
3. The graph again consists of two branches,
symmetrical with respect to the a;-axis, for all
positive values of x greater than + 1.
4. The limits of the first branch are at (—1, 0)
and (0, 0) ; the limits of the second branch are also
at (—1, 0) and (0, 0) ; the limits of the third branch
are at (+ 1, 0) and (+ oo , + oo ) ; and the limits of
the fourth branch are at (+ 1, 0) and (+ oo , — oo ).
+ x
X
y
+ < + l
1
1-5
±1-4
2
±2-4
2-5
±3-6
3
±4-9
-1
-<-l
V-
-•2
±•44
-•4
±•58
-•5
±•61
-•6
± ^62
-•8
±•53
FUNCTIONS OF A SINGLE VARIABLE.
373
5. The first branch has a turning-point (maximum) somewhere
between (— '5, + -61) and (— '8, + 'SS). The second branch also has a
turning-point (minimum) between (— '5, — '61) and (— '8, — '53).
6. The branches of the graph meet the a;-axis when a: = 0, +1,
and — 1. These values are, therefore, the roots of f{x) = x^ — x=.0,
3. Plot 2/2 = a;3_9^^24a;^16^
or y = ± v^
y
X
y
<1
V-
1
1-5
± 1-76
2
±3
3
±1-41
4
5
±2
6
±4-47
7
±7-34
+ 00
±00
PLOT.
16.
Questions. —
1. How many
branches has
this graph ?
2. How many ^/
turning-points?
Locate and _
name them.
3. What are
the roots of the "~
function x^ —
9x2 -h 24a; -16 _
= 0!
4. Between
what limits are ""
the branches of
the graph con- _
tinuous ?
5. Where is
the function an ~~
increasingfunc-
tion and where
a decreasing?
6
/
5
/
4
/
/
3
/
2
/
1
f
\,
/
i
1
2
•\
/
5
6
1
\
/
\
2
V
/
\
3
\
4
\
5
\
\
6
\
\
y
\
374 ADVANCED ALGEBRA.
EXERCISE 96.
Plot and discuss the following functions :
(Use paper ruled in squares, called l)lotting-paper.)
1, y =^x-\-Q 6, 1/^ = a;^
2. y =Sx 1. y^ =:x^{x— 1)
3, y =81a;-3 8. y =a^ -'8x^-\-20x -10
4. y^ = 4:X 9. y =3x-\-lSx^ — 2a^
b. y^ = 16-x^ 10. y^ = 3^-{-ds^-5x-20
Differentials and Derivatives of Functions.
Definitions.
685. The limit of the ratio of the increment of a func-
tion to the increment of the independent variable pro-
ducing the increment of the function, when the limit of
the increment of the independent variable is zero, is called
the derivative of the function.
Thus, if we let y =/(a;), and represent the increment
of a; by A a; and the corresponding increment of y hj Ay,
then will lim. ( — - ] = the derivative of the function.
m
686. The limit of the increment of the independent
variable is called the differential of the independent vari-
alle, and is represented hy dx\ and the limit of the incre-
ment of the function is called the differential of the func-
tion, and is represented by dy.
dy
: -■ (fl)
Therefore, x^^x. , , _ ,
Q ax
Notice, dy and dx represent single quantities (differentials) and
are not equivalent to d x y and d y. x.
DIFFERENTIALS AND DERIVATIVES. 375
Ulustration. —
Let y = 7^ (1)
then, y-\- /\y = {x -^ /^xf = x^ -{-%x(Ax) -\-{^xf (2)
Subtract (1) from (2), ^y = '^xi^x) + {^xf (3)
Divide by a x, -^ = 2a; + A a; (4)
A X
.-. Lim.fA|) ^lim.(2a: + Aa;) [401,P.] (5)
.: -r^ = 2x = the derivative of a;',
dx
and dy = 2xdx = the differential of a;'.
687. The differential of a function equals the derivative
of the function multiplied ly the differential of the inde-
pendent variable,
688. The derivative of a function equals the differential
of the function divided by the differential of the independ-
ent variable.
689. // the differential of a function, and hence, too,
the derivative of a function, is positive, the function is an
increasing one; if negative, a decreasing one.
Principles.
690. Let y =f{x) = x% (1)
then, y-]- Ay = {x-\- Axy
= x''-\-nx''-'^ .Ax-{-A.{AxY (2)
in which A =
n{n-l) ^_, _^ n(n-l)in-2) ^_3 . ^ ^ + ^tc. [593].
Subtracting (1) from (2), Ay = nx^-^ . Ax + B .{Axf (3)
A tJ
Dividing by A x, — ~ = wa;*-i + -B . A a; (4)
A X
Lim. ( ^^ ) = lim. (na;«-i + 5. A a;)^ ^_o
.'. ~ = nx^-\ since Km. ^ = a finite constant [582], and
lim. A a; = 0. (5)
.'. dy = nx^-'^dx. Therefore,
376 ADVANCED ALGEBRA.
JPrin, 1. — The differential of a variable with a con-
stant exponent equals the continued product of the expo-
nent, the variable with its exponent diminished by unity,
and the differential of the independent variable.
Ulustrations. —
1. d{a^) = 4.a^dx 2. d{x)-^= -^x'^dx
3. dia + bxy^pia + bxy-'^dia + bx)
691. Let y = ax (1)
then, y -\- Ay = a(x-\- Ax) =zax-\-a{Ax) (2)
Subtracting, Ay = a{Ax) (3)
Dividing, — - = a
A X
Lim. — - = a
A X
whence, -j^ = a, and dy = adx. Therefore,
Prin, 2, — The differential of a constant times a vari-
able equals the constant times the differential of the vari-
able.
Thus, d{dx') = d . d(x') = 3 X 6x^ dx = 15a^dx.
692. Let y = ax-\-b (1)
then, y-\- Ay = a{x-{- Ax)-]-b =
ax-\-a(Ax)-\-b (2)
Subtracting, Ay = a(Ax)
Dividing, -^ = a; whence, ^ = «»
and dy = adx. Therefore,
I*rin, 3, — The diffetential of a constant term is zero.
693. Let V = f{x), w =/' (x), and z =/" {x) ; and
let y =z v-\-w — z (1)
then, y -\- Ay = v-Yav-^w-\-Aw — {z-]rAz)
■=V-\-W — Z-\-AV-[-AW— AZ (2)
DIFFERENTIALS AND DERIVATIVES. 377
Subtracting, Ay = Av + aw— a z (3)
T^. .,. , Ay A V AW AZ
DmdmgbyA^, — = — + — -— (4)
, ,. Ay T AV ,. AW ,. AZ
whence, lim. - — ~ = lim. h lim. lim. (5)
[413,P.]. ^"^ ^^ ^"^ AX '>
dy _dv dw d z
dx ~ dx dx dx ^ '
whence, dy = dv + dw — dz. Therefore,
Prin, 4, — The differential of a polynomial whose terms
are functions of the same independent variable equals the
algebraic sum of the differentials of its terms,
niustration. ^ (a;^ + 3 a;^ — 2 a; + 5)
= d{7^) + d{^x^) ^ d(-%x) -\- d{h)
= da^dx-{-6xdx^2dx= {Sa^ -\-6x — 2)dx.
694. Let V =f(x) and z = f {x),
and y = vz (1)
then, y-\- Ay = {v -\- /\v) {z -\- A z)
=iVZ-\-V./\Z-{-Z.l\V-\-llV.AZ (2)
Subtracting, 'Ay = v.Az + z.Av-\-Av.Az (3)
TV. .,. , Ay A z A V A V
Dividing by a x, — - = v . + z . + — - . a z
^ ^ 'ax ax ax ax
Lim. — ~ — lim. | v . ) + lim. ( z . )
AX \ AxJ \ AxJ
+ lim. (-^ . A zj
whence, dy = vdz + zdv. Therefore,
rrin, 5. — The differential of the product of two con-
tinuous functions of the same independent variable equals
the swn of the products obtained by multiplying each func-
tion by the differential of the other.
niustration. —
d{--^7^Xhx^) = -dar'xd(6x^)-\-6x^Xd{-Sa^)
= {-da^XiX6x-^-{-6xixSxi-3a^)}dx
= — 65x^ dx.
378 ADVANCED ALGEBRA.
Cor, d{vwz) = v.d{wz)-\-wz.dv [694, P.]
='Vwdz-\-vzdw-\-wzdv. [694, P.] ; etc.
696. Let V =: f{x), and z =f'(x); and
^ -1
y = — = vz ^ :
z
then dy = v , d {z-^) -\- z-"" d v [694, P.]
= —vz~^ dz-\-z~'^dv
_dv vdz_zdv — vdz
Therefore,
Prin» 6, — The differential of a fraction whose terms
are continuous functions of the same independent variable
equals the denominator into the differential of the numera^
tor minus the numerator into the differential of the de-
nominator, all divided by the square of the denominator,
\yy y'
_2xy^dx-'3x^y^dy
" ? •
696. Let y = log, x
then y-\- Ay = ]og,{x+Ax) = log, x(l + ^j
= log, X + loge (l + -^) [467, P. 2]
= log.a;+ — - ^ . L_^ + - . L_l - etc. [601, C]
— = - + 5. A2:; in which 5=_-.- + -.--^_ etc.;
which for very small values of A a; is convergent [582].
.-. Lhn. (^\ = lim. (^ +B. ax)
\AxJi,x = \X J^x-0
whence, -^ = —; and dy = — . Therefore,
CuX X X
DIFFERENTIALS AND DERIVATIVES, 379
Prin, 7. — The differential of the log, of a quantity
equals the differential of the quantity divided hy the quan-
tity itself
Cor, — Since logio x — m log, x [607, P.]
, ., . mdx
^.(logio^) = — ^.
Illustration. —
— ^ J ^■^^-\- (x-\-a)dx \ __ 1 /2x-\-a\ ,
"^ x-\-a{ x^ \'~ x-\-a\ X J
697. Let y = «*, in which « is a constant,
then, log. y =^ X log. a [468, P.]
d (log. y) = log. a . dx
or, -^ = log. a . t?a;
whence, dy = a" log. a .dx. Therefore,
JPrin, 8, — The differential of a constant with a vari-
able exponent equals the continued product of the original
quantity, the logarithm of the constant, and the differential
of the variable exponent.
Thus, d{a-\-l)'^=-d{a-^b)^={a^ bf^ log. {a + b)
X d (x^) = i x-^ (a + if^ log. (a + J) ^ x.
698. Problem. Find the differential of 05*.
Let y — ^
then, log. y = X log. a;
and d (log. y) = x . d (log. ic) + log. x , dx
dy dx . , J
or, — J = a; . 1- log. X . dx
X/ X
whence, dy — xf{\-\- log. x) dx,
G
380 ADVANCED ALGEBRA.
EXERCISE 97.
Differentiate :
1. y=.hao[^ —'^Ix^-^-'ilcx — d 2. y = 5x^z^ -\-z
^ ' 14. ?/ =
5' y— 2^ jg_ y = {x-\- ay (x - h)
= /^
7. 2^=V5aJ+6 19. 2/ = 3^'
z.f^%px 20. y = a;^^
9.f = Za^ 21. 2^ = ic*(a + ^^)-i
•^ ' , 22. «/ = lOge (a + Xf
^ ' ' 23. y = (f -^ d"
13. 2/ =
Vx-\-a 25. y = d'^-'
Applications.
EXERCISE 98.
1. At what rate is the area of a circle increasing when
the radius is 6 inches and is increasing at the rate of 3
inches per second ?
Solution : Let y = the area, and x = the radius ; then,
y z=z V x^
and dyz=2'irxdx.
This denotes that at any instant the rate of increase of the area is
2 rr X times as great as the rate of increase of the radius at the same
instant. But when the radius is 6 inches, it increases at the rate of 3
inches per second ; or, when x = 6 inches, dx = B inches.
.'. dy = 2ir X 6 inches x 3 inches = 36 ir square inches ; that is,
the area is increasing at such a rate that, if kept uniform for one sec-
ond, the increase would amount to 36 v square inches.
DIFFERENTIALS AND DERIVATIVES. 381
2. At what rate is the area of a square increasing when
the side of the square is 4 inches and is increasing at the
rate of 2 inches per second ?
3. The volume of a sphere increases how many times as
fast as its radius ? When its radius is 6 inches and in-
creases at the rate of 1 inch per second, at what rate is the
volume increasing ?
4. At what rate is the diagonal of a square increasing
when the side of the square is 8 inches and is increasing
at the rate of 2 inches per second ?
5. The radius of a circle is 4 inches and its circumfer-
ence is increasing at the rate of 2 tt inches per second. At
what rate is the radius increasing at the same instant ?
6. A boy approaches a tree 90 feet high standing on a
level road at the rate of 3 miles an hour. At what rate
is he approaching the top of the tree when he is 220 feet
from the base ?
7. The diagonal of a cube is increasing at the rate of
36 inches per second, when the side of the cube is 5 inches
long. At what rate is the side increasing at the same
time ?
8. If X increases at the rate of '5 per instant, at what
rate is logio x increasing when a: = 42 ?
9. The logio 42 = 1-62325. What, then, would be the
logio 42*5, if the increase were uniform ? How does the
result compare with logio 4:2 '5 as found in the table ?
Successive Derivatives.
699. If the derivative of f{x) be treated as a new func-
tion of X [/i(^)], there may be found from it a second
derivative of f(x) [/g {x)"] in the same way as /i {x) was
derived from f{x), and so on, until a derivative is found
that is independent of x \_f{x^\
382 ADVANCED ALGEBRA.
Illustration. —
Let/ (a:) = x^ -]-4:a^ - '^a^ ^%7? - bx^^ -, then,
f^{x) - 5a:* + 16a,-3-9a;2 + 4^-5 [693, P. 688]
f^{x) = 20a;3^48^2_i3^_j_4
f^{x) = 60a:2_^96^,_i8
/4(a;) = 120a;4-96
f{x,) = 120
EXERCISE 99.
Find the first derivative of :
1. ic3_4ic2 + 7a; + 2 4. (a + a:)5 (a _ a:)3
2. (a; + 2)3 (:z; - 2)* 5. {a + a:^) (a - t?)
3. 2:(a; + 2) + a;2(:r + 3) 6. (« + a;)^ ^ (« - ic)*^
Factorization of Polynomials containing Equal
Factors.
700. Let f{x) = {x-\- tti) {x + a^) {x-\-a^) . . . . {x-^ a^)
= any polynomial composed of binomial factors of the
form of x-\-a; then
/i (x) = {x-\- ttz) (x + as) (x-\-a^) ....(x + «„) +
{x + «i) (a? + %) {x-{-a^) .... (x + aj +
{x + «i) (:r + «2) (:r + «4) . . . . (a; + o^J +
(^ + «i) (a; + ag) (^ + «3) i«^ + «» H-
[694, Cor. 688].
Observations. — 1. If no two factors of f(x) are alike, f{x) atid
fx (a;) have no common factor.
2. If two, three, or r factors of f{x) are equal, and all equal to
jc + a, then will a; + a, (a; + a)*, or (a; + a)'"^ be a common factor of
S{x) and /i (x).
POLYNOMIALS CONTAINING EQUAL FACTORS. 383
3. In general, if f{x) contains the factor x -^ a p times, {x + h)
q times, {x + c) r times then will {x + a)p-'^ {x + 6)?-* {x + cy-'^
.... be the H. C. D. of f(x) and /i (x).
4. The H. C. D. of f{x) and /i {x) contains one factor less of each
kind than does f{x).
701. Theorem, — Every polynomial composed of bino-
mial factors of the first degree, some of which are equal,
may he decomposed into factors containing no equal bino-
mial factors of the first degree.
For, let f{x) be a polynomial composed of binomial
factors of the first degree, some of which are equal, /i {x)
its first derivative, /' {x) the H. C. D. of f{x) and /i {x),
and <\>{x) the other factor oi f{x) ; then,
1. <^{x) will be devoid of equal factors of the first
degree [700, 4].
2. If /' {x) still contains equal factors of the first de-
gree it may be resolved into two factors, /" {x) and <^' {x),
in which <ii' {x) is devoid of equal factors [700, 4].
3. This process may be continued until no factor is
left that contains equal factors of the first degree, which
will be when the last H. C. D. found is unity.
Ulustration.— Eesolve x'' -\- x^ — 1% x^ — 1% 3^ -\- A:% x? -{-
48 a;^ — 64 ic — 64 into factors devoid of equal binomial fac-
tors of the first degree.
Solution :
f{x) = ic' + a;« - 12a;5 - 12a:* + ^oi? + 48a;2 - 64a; - 64
/a (a;) = 7a;6 + 62:6 - 60 a:* - 48a:3 + 144a;2 + 96a; - 64
f{x) = a:4 _ 8a;2 + 16 [158] = {x^ - 4)2 = {x +2)(a; + 2) (a; - 2) (a; - 2)
4>{x)=f{x)-^P{x) = x + l
.-. /(a;) = (a; + 2)(a; + 2)(a;-2)(a;-2)(a; + l).
EXERCISE 100.
Factor :
1. x^-^-'^Q^-llx^-VUx-m
2. a;«-5ic5 + a^ + 37tc^-86a;2_j_Y6a;-24
384
ADVANCED ALGEBRA.
3. o^-^x'-^Q x^-'^l ic5_|_2i6 2^+243 a^-^^Q x^-^^ a;-739
4. x^"" _ 30 a:8 + 345 x"^ - 1900 cc* + 5040 3? — 5184
5. a;^«-13a;8 4-42ic6-58ar* + 37a;2_9
Graphical Significance of /i (x).
702. Let m/i be the graph of y =f(x).
Let P be a point on the graph whose co-ordinates are x and y.
Let GE: = PR = Ax\ then will F'R = Ay.
Draw the secant line P'JPS, also the tangent line T'P T.
Take SB = 1, and draw BC = sin. /S and /SC = cos. S.
Now, the triangles P' PR and jB/SC are similar.
.-. ^ = §^=tan.>S'[668] (1)
.-. -^ =:tan. /S (2)
AX ^ '
Let the point P' approach the point P on the graph* so as to
make ax diminish uniformly; then will the secant line P' P S re-
volve about P and approach the tangent line T' P T &s its limit, and
the angle S will approach the angle T as its limit.
Lim. f -^"i = lim. (tan. S)i
or, :t-
dy^
dx
tan. T. Therefore,
The first derivative of a function is equivalent to the
tangent of the angle which a tangent line to the graph of
the function makes with the axis of abscissas.
MAXIMA AND MINIMA OF FUNCTIONS. 385
Maxima and Minima of Functions.
703. The maximum or minimum value of a quadratic
function may readily be found, as follows :
Example 1. — What is the maximum or minimum value
of x^-\-^x-\- 6, and what value of x will render it a maxi-
mum or minimum ?
Solution : Let f{x) = a;* + 8a: + 6 = w
Complete the square, ic* + 8 a: + 16 = w + 10
Extract the \/, x + 4 = ± ^/nl + l6
Transpose, x = —4± ^m + 10
Now, w < — 10, else would x be imaginary.
m = — 10 is the minimum value of / {x).
But when m — — 10, a; = —4; then, a; = — 4 renders f{x) =
a;* + 8 a; + 6 = — 10, a minimum.
Example 2. — What is the maximum or minimum value
of 8 a; — 3 2;^ + 9, and what value of x will render it a
maximum or a minimum ?
Solution : Let f{x) = 8a;-3a;2 + 9 = m
Complete the square, 9 a;* — 24 a; + 16 = 43 — 3 m
Extract the V» 3 a; - 4 = ± \/4S-dm
Transpose and divide, x = -^ ± -w \^4d — 3m
Now, 3 wi > 43, or m > 14 -o , else would x be imaginary.
1 ^
.'. w = 14-^ is the maximum value of f(x).
11 1
But, when m = 14-^ , x = 1-^; therefore, x = l-^ renders f(x)
= 8a;— 3a;* + 9 = 14-;3- , a maximum.
o
Example 3. — Divide 36 into two parts whose product
shall be the greatest possible.
Solution : Let x and 36 — a; = the two parts,
and X (36 — a;), or 36 a; — a;* = m.
Then, a; =i 18 ± \/^24: - m.
Now, m = 324 is a maximum ;
x = 18 and 36 - a; = 18.
386
ADVANCEB ALGEBRA.
704. General Method.—
Let mn he the graph ot y = f(x).
Y
/^
^^
— pVU
/a
A
X
/
/
P5^
J.
1
V
V V
Conceive a point, P, to move along the graph, carrying with it a
tangent Hne to the graph, in such a manner as to cause the abscissa
(ic) of the point to increase uniformly. Let v be the value of the vari-
able angle which the tangent line makes with the ic-axis. At P'
1? < 90 ; hence, tan. v, or /i {x\ is positive [668, Sch.]. This is true,
however near Pi is to pii. At P° the tangent line is parallel to
the rc-axis; hence, v = 0, and tan. v, or fx{x) =0. At P™, t;> 90;
hence, tan. v, or /i (a;), is negative [668, Sch.]. This is true, however
near P™ is to P°. Again, just before P arrives at P^, v > 90°, and
tan. V is negative ; when P is at P^,v = and tan. v = 0; when F
has just passed F^, v < 90 and tan. v is positive. Therefore,
706. Frin, 1» /i(a;) = at turning values of f(x),
Prin, 2, Immediately before a maximum value of
f{x), f\{x) is positive y and immediately after, negative,
Prin* 3, Immediately before a minimum value of
/(^)> /i(^) *^ negative, and immediately after, positive.
706. Caution 1.— A root of /i (x) = is
not necessarily the abscissa of a turning point.
For a tangent line to a graph may be parallel
to the a;-axis where there is no turning point,
as where two branches tangent to the same
line coalesce at the point of tangency. (See
diagram.)
It is only when Prin. 2 or Prin. 3 is satisfied, as well as /i (x) = 0,
that a turning point is established.
Caution 2. — There may be turning
points under peculiar conditions when
/i (x) 4= 0. For there may be turning
points where the tangent line to the
graph is not parallel to the a;-axis;
as where two branches coalesce and
cease. (See diagram.)
MAXIMA AND MINIMA OF FUNCTIONS. 387
707. Observations. — 1. So long as f{x) remains continuous, its
maxima and minima values succeed each other alternately.
2. If two successive turning values of f{x) have the same sign,
the graph of f{x) between these values can not cross the a;-axis, or
f{x) 4= between these values.
3. If two successive turning values of f(^) have opposite signs,
the graph of f{x) must cross the ic-axis between these values, or
f{x) = somewhere between these values.
4. \i x = a and x = h render f{x) = 0, and a^h, there must be
a turning value of f{x) between x = a and a; = 6.
Example. — Find the turning values of
f{x) = a;3 - 9 a;2 _^ 24 a; + 16.
Solution : f{x) = a:3 — 9a;2 + 24a;+16
/, {x) = 3a;« - 18a; + 24 = 0;
or, /,(a:) = a;2 — 6a; + 8 = 0;
whence, a; = 4 or 2, critical values.
/i(a;-Aa:). ^^4 , = (4 - A a;)* - 6(4- A a:) + 8 = -
( A a; = o f
/i(a;+ Aa;), ^^4 > = (4 + A a;)* - 6(4 + A a;) + 8 = +
■) A « = o )
.'. f{x) is a minimum when a; = 4
But f{x)x = 4 = 43 - 9 X 42 + 24 X 4 + 16 = 32.
.*. Minimum value of f{x) = 32
/i(a;-Aa;), ^^g . = (2 - a a;)»- 6(2 - A a;) + 8 = +
j A a; = o j
fi(x+ Ax)^ ^^2 I = (3 + A a;)« - 6(2 + A a;) + 8 = -
I A a; = o I
.*. f(x)x = 2 is a maximum
But f(x)x = 2 = 2» - 9 X 22 + 24 X 2 + 16 = 36
.*. Maximum value of f(x) = 36.
The value of f(x)x = a is best obtained by synthetic division, as in
Art. 106.
EXERCISE 101.
Find the maxima and minima values of :
1. 4:a^-15a^-}-12x-l 5. x^ -dx^ - 9x-\-6
2. 23^-21a^-{-36x-20 6. (a; - 1)* (a; + 3)^
3. x^-{-6x + 6 7. (re - af {x + bf
4.ic2_6a; + 5 B. x^ -3x^ -\-3x-\-H
388 ADVANCED ALGEBRA.
9. a^-^Qx-6 11. a^-^3(?-{-U
\0. a^ — Qx — h 12. a;* + a;3 + a;2 — 16
13. Show where a line a feet long must be divided so
that the rectangle of the two parts may be the greatest
possible.
14. Find the altitude of the maximum cylinder that can
be inscribed in a sphere whose radius is r.
Suggestion.— Let B C = Xy BD = r — x, and AB — y^
then, 2/^ = (^ + ^) (r — a;) = r* — ic*, and
/(a;) = F=ir2/2 X 2 a; = 3 ir a; (r^ — jr*)
/i(a;) = 2ira; X (-2 a;)
+ (r3_a;8)x2ir =
whence, x = -^ ^J~^
o
2
and, 2/2 _ ^3 _ 3.3 _ ,.2
o
2 /-
15. Find the altitude of the maximum cylinder that can
be inscribed in a cone whose altitude is a and whose radius
is J.
16. Find the volume of the maximum cone that can be
inscribed in a given sphere.
17. Find the area of the maximum rectangle that can
be inscribed in a square whose side is a,
18. What is the maximum convex surface of a cylinder
the sum of whose altitude and diameter is a constant a ?
19. Find the altitude of the maximum cylinder that
can be inscribed in a right cone whose altitude is a and
the radius of whose base is J.
20. Eequired the area of the maximum rectangle that
can be inscribed in a given circle.
21. Required the greatest right triangle which can be
constructed upon a given line as hypotenuse.
CHAPTER XII.
THEORY OF EQUATIOJ^S,
Introduction.
708. Equations of the first and second degree have
already been treated, and need no further attention here.
709. Jerome Cardan, an Italian mathematician (1501-
1576), published in 1545 a method of solving cubic equa-
tions, now known as *' Cardan's Formula." But, as this
formula is not finally reducible when the roots of an equa-
tion are real and unequal, it is not of much practical
value.
710. Eene Descartes, a French mathematician (1596-
1650), transformed the general bi-quadratic equation so as
to make its solution depend upon that of the cubic equa-
tion ; but, as he invented no new method of solving the
latter, the same difficulties are encountered in the applica-
tion of his rule as are met in Cardan's.
711. Nicholas Henry Abel, a Norwegian mathematician
(1802-1829), demonstrated, in 1825, the impossibility of a
general solution of an equation of a higher degree than
the fourth. Previous to that date many such solutions
were attempted.
712. The real roots of numerical equations of any de-
gree are, however, attainable through laws and principles
to be developed in this chapter.
390 ADVANCED ALGEBRA,
Normal Forms.
713. Theorem I. — Every equation of one unhnown
quantity with real and rational coefficients can he trans-
formed into an equation of the form of
Ax^-\-Bx^-'^-{-Cx^-^-i- .... +i; = 0,
in which A and all the exponents of x are positive in-
tegers, and each of the remaining coefficients, including L,
is either an integer or zero.
Note. L may be regarded the coefficient of a:P.
Demonstration. — 1. If the equation contains fractional terms, it
may be cleared of fractions.
2. If there are any terms in the second member, they may be
transposed to the first member.
3. All terms containing like exponents of x may be collected into
one term by addition.
4. If A is negative, both members may be divided by — 1.
5. If X contains negative exponents, both members may be multi-
plied by X with a positive exponent numerically equal to the greatest
negative exponent.
6. If X contains fractional exponents, x^ may be substituted for x,
in which m is the L. C. M. of the denominators of the fractional ex-
ponents.
The roots of the transformed equation will he the mth root of the
roots of the original equation.
1. The terms may now be arranged according to the descending
powers of x.
714. The equation
A af-i-Bx''--' + (7:r"-2-f . . . . + X = 0,
is known as the first normal form of an equation of one
unknown quantity, and will hereafter be represented by
Example. — Transform 3x^+-r—S-[-7 x~^ = - + -r-
x^ o xt
into the first normal form, and compare the corresponding
roots of the two equations.
NORMAL FORMS, 391
Solution: Given 3a;§ + -r — 8 + 7a;— f = — + -j. (A)
xi ^ xt
Clear of fractions, 9x + 12 — 24:X^ + 21 x—^ = 4xi + 9xi (B)
Transpose and collect terms,
9a; + 21a;-i-28a;i-9a;* + 12 = (C)
Multiply by xt ,
9a;t + 21-28a:f-9a;l + 12a;i = (D)
Puta; = a;«, 9a;9 + 21 - 282;* - 9^:^ + i2a:3 _ q (E)
Rearrange terms,
9a;9 + 0a;8 + 0a;' + 0a;«-28a;5-9a:4 + 12a:3 + 0a:2 + 0a: + 21 = (F)
The roots of (A) = Vof the roots of (F).
715. An equation that contains all the powers of x,
from the highest to the lowest, is called a Complete Equa-
tion, An incomplete equation may be written in the form
of a complete equation by supplying the wanting terms
with coefficients of zero.
Thus, a;^ — 4a;^ + 2a; — 5 = may he written 2:^ ± ar*
— ^7? ±0x^-\-2x — b = 0.
716. Theorem II, — The equation F^ {x) = ?nay be
transformed into an equation of the form of
x^-\-p,x''--' +p,x--'-\-., . .+^„ = 0,
in lohich the coefficient of a:" is unity ^ and each of the
remaining coefficients is either an integer or zero.
Demonstration.—
Take Fn{x) = Ax"" + Bxf^-'^ ^Cx!^-'^ ■¥ + L =
^ ^ X Ax"" Bx»-^ Cx»-^
Put X = -r, — r- + .„ , + -j^r^T + +L =
A A" A'*-^ A»-2
Multiply by .4„_i ,
a;« + Bx^-^ + A Cx'^-^ + + J.«-^ X =
Put pi for B, Pi iov AC, p„ for J.«-» L,
x"* + pi X^-^ + Pi X^-^ + + Pn=
This is the second normal form of an equation of one unknown
quantity, and will hereafter be represented by /« {x) — 0.
ADVANCED ALGEBRA.
717. Cor. 1, — Each root of /„ {x) = is A times as
great as the corresponding root of F^ {x) — 0.
718. Cor, 2, — The coefficient of the second term of
/„ {x) is the same as the coefficient of the second term of
F^ {x), and the succeeding coefficients of /„ {x) are obtained
ly multiplying the succeeding coefficients of F^ix), in
order, ly A, A^ , A^, ^"~^.
Note. — If terms are wanting, supply them with coeflScients of 0.
Example. — Transform the equation Ax^ — dx^-\-2a^ —
7 = into an equation of the form of f(x) = 0.
Solution :
Given Fix) = Aafi -Sx^ + Ox^ + 2x^ + Ox-7 = 0,
thenwm fix) = x^-dx^ + 4:x0a^ + Px2x^ + ¥x0x-4^x'7=z0 [718]
or, fix) = x^-dx^ + d2x^- 1792 = 0.
The roots of fix) = are 4 times as great as those of Fix) — 0.
EXERCISE 102.
Transform the following equations into equations of
the form of f{x) = 0. Compare the roots of the trans-
formed equation with the roots of the original equation.
1. Sa^-\-2x^-da^i-l!x^6 =
2. 2x' + 4:a^-x^-{-x^-1l =
3. 4:X^-\-3a^-5x^+llx-l=:0
4. Sx^-2x^ + 3xi-2x^-{-4:X^-2 =
5. x-^ 4_ 2 ic-^ + 3 x-^ - x-^ -^x-^-2x-^ + 2 =
6-lx+^x-i+lxi-^lx-^ + 3 =
7. x^'-dx^^2x^-\-dx-^-\-2 =
8. |^t+J^t_ 1^ + 1 =
DIVISIBILITY OF EQUATIONS.
Divisibility of Equations.
719. Theorem III. — If a is a root of F^ (x) = 0, then
X — a is a factor of F^ (x).
For, let F. {x) ^ {x - a) = F^_, {x) + ^
then, {i^„_i {x)\{x-a)-\-r = F^ {x) =
but, x — a = 0, since x = a.
r = 0;
whence, F„ (x) -^ (x — a) = i^„_i (x),
720. Cor. 1. — If a is an integral root of F^ {x) = 0,
it is a divisor of the absolute term of F^ {x) [163].
721. Cor. 2. — If X — a is a factor of F^ {x), then a is
a root of F^ {x) = 0.
For, F^{x) = {F,_, (x)} (a; - a) = ;
whence, x — a = 0, and x = a.
722. Cor. 3* — If x is a factor of F^ {x), then zero is a
root of Fr,{x)-0.
Number of Roots.
723. Theorem IV, F^ {x) = has at least one root.
The demonstration of this theorem may be found in
special treatises on the Theory of Equations. It is too
long and tedious to be introduced here.
724. Theorem, V. F^ {x) = has n roots and only n.
For, F^ {x) — has at least one root. [T. IV.J
Let a = one root of F^ {x) = ;
then, F„ (x) = { F„_t (x)] {x — a} =0 [T. IILl
.-. F,,_^{x) = 0,
Let b = one root of i^„_i (x) = ; [T. IV.]
then, F„_^ (x) = {F^^s (x)] {x-b}=0 [T. III.]
.-. F^_,{x) = 0.
394: ADVANCED ALGEBRA.
Now, as F^ {x) = is of the nth. degree, and each time
a root is removed by division the degree is lowered by
unity, it follows that n roots and only n can be removed
before F^{x) reduces to an absolute factor. Therefore,
Fn {x) = has n roots and only n.
725. Car, F^ (x) = may he written
A{x — a){x — b){x — c) (x — l) = Q;
or simply {x — a){x — b) (x — c) (x — l) = 0, in which
there are n factors of the form of x — r, the second terms
of which are the roots of F^ {x) = with their signs
changed, and may he positive or negative, fractional or
integral, rational, irrational, or imaginary, subject only
to restrictive conditions explained hereafter.
Relation of Roots to Coefficients.
726. OThearem Vl.—If F^ (x) = be put in the form
of x^ + B^ a;»-i + Ci a;*-^ + . . . . -|- Xi = 0, 5^^ dividing both
members of the equation by A, the coefficient of x"", then will
1. Bx = the sum of the roots vnth their signs changed.
2. Ci= the sum of the products of the roots taken two
together.
3. Di = the sum of the products of the roots with their
signs changed, taken three together.
4' El = the sum of the products of the roots taken four
together. And so on to
5. Li = the product of all the roots with their signs
changed.
Demonstration : Let the n roots of the equation he a, b, c, I;
then, Fn (x) = x'* + Bi x^-^ + Ci x^-^ + + Li
= {X - a){x - b){x - c) . . . . (X - I) [725].
After which the theorem is a direct inference from the binomial for-
mula [587], and the principle that " changing the signs of an even
IMAGINARY MOOTS. 395
number of factors does not change the sign of their product " [page
26, Ex. 3].
Cor. — Changing the signs of the alternate terms of
F^ {x) = changes the signs of its roots.
Imaginary Roots.
727. Theorem VII, — Imaginary roots can enter F^ {x)
= only in conjugate pairs.
For in this way only will their sum and the sum of
their products be real [657], as they must be [713].
728. Cor, 1, — The product of the imaginary roots of
Fn {x) = is positive.
For the product of each pair is positive.
Thus, (a -\-bi) {a '-bi) = a^-\- W.
729. Cor, 2, — When all the roots of F„ {x) = are
imaginary the absolute term is positive.
Suggestion. — For the equation is then of an even degree.
730. Cor, 3, F^ {x) = has at least one real root oppo-
site in sign to the absolute term, when n is odd.
731. Cor, 4, F„ (x) = has at least two real roots,
one positive and the other negative, if n is even and the
absolute term is negative.
732. Car. 5, — The sign of F^ {x) for any real value of
X depends on the real roots of F^ (x) = 0.
For the product of x — {a-\-b i) and x — {a — bi) =
{x — aY + b^, a positive quantity ; and this is true of every
pair of factors containing conjugate imaginary terms. '
733. Cor, 6, — Every entire function of x with real
and rational coefficients may be divided into real factors
of the first or second degree.
396 ^^ VANGED ALGEBRA.
Fractional Roots.
734. Theorem VIII, — JVo root of /„ {x) = can he a
rational fraction.
Take fn(x) = x^ + PiX^-''-{-p2X--^+ ....+jo^ =
[716]. If possible, let a; = t- , a rational fraction in its
lowest terms. Then, by substitution,
Jni- Jn-1 -1- Jn-2 "h • • • • "f ^« " ^.
Multiplying by 5**"^ and transposing terms, we have
an integer, which is impossible.
Scholium. — From this theorem it follows that the ra-
tional fractional roots of F^ {x) = may be obtained by
transforming F^ {x) = into /« (x) = and dividing the
roots of the latter equation by A, the coefficient of ic"~^ in
the former.
Relations of Roots to Signs of Equation.
735. Theorem. IX. — If F„ (x) = has no equal roots,
then Fn (x) will change sign-if x passes through a real root.
For, take F^{x) =\x — a){x — b){x — c) {x—l) =
[7^5] ; conceive x to start with a value less than the least
root and continually increase until it becomes greater than
the greatest root. At first, every factor of F^ (x) is nega-
tive, but, at the instant it becomes greater than the least
root, the sign of the factor containing that root will be-
come plus, while the others remain minus ; whence, F^ {x)
will change sign. It will, moreover, retain its new sign
until it passes over the next greater root, when it will
again change sign, and so on.
RELATIONS OF ROOTS TO SIGNS OF EQUATION. 397
736. Cor, 1, — If for any two assigned values of x,
Fr, {x) has different signs, one, or, if more than one, an
odd number of roots of F^ [x) = lie between these values,
737. Cor. 2, — If for any two assigned values of x,
F^ {x) has the same sign, either no root or an even number
of roots of Fr, {x) = lie between these values,
738. Some of the properties of F^ {x) = 0, already dis-
cussed, are beautifully illustrated by the following graph.
Y
1. It is seen that y = Fn{x)=:0 when a; = 1, 2, 3, and 5. There-
fore, these values of x are roots of i^„ {x) = 0.
2. Immediately before a; = 1, 2/ is positive, and immediately after
x = l, y is negative ; immediately before x = 2, y is negative, and
immediately after x = 2, y is positive, etc. ; illustrating that when x
passes over a real root, Fn (x) changes sign.
3. At a; = 3 two values of y become zero ; therefore, two roots
become identical, or, in other words, 3 is twice a root. Were the abso-
lute term of i^„ {x) so changed as to make y somewhat less, the a:-axis
would cross the graph twice between a; = 2 and a; = 4, once before
a: = 3, and once after, thus proving conclusively the duality of the
root 3, when y = 0.
4. Immediately before a; = — 2 and a; = 6, the graph approaches
398 ADVANCED ALGEBRA.
the ic-axis, but in each case makes a turn before reaching it, prevent-
ing, thereby, equal roots or unequal real roots. These turns locate the
position of imaginary roots. The truth of this statement becomes
manifest when we suppose the absolute term of Fn {x) to so change as
to cause y to gradually decrease, the a;-axis will gradually arise and
finally touch the graph at ic = — 2, thereby making two equal roots,
and, if y continues to decrease, the ic-axis will cross both branches
above the turn at x= —2, making two unequal real roots.
The student will be interested in observing the changes in the
roots if the absolute term of the equation so changes as to cause the
a;-axis to gradually move from the position Aix\ to the position A^x^-
5. It must not be assumed, however, that imaginary roots always
denote a turning point in the graph of the equation. Such may or
may not be the case.
739. If any two successive terms in a complete equa-
tion have like signs, there is a permanence of sign ; if
unlike signs, a variation of sign. Thus, in the equation
cc« - 5 a;5 + 8 a;* + 7 a;3 - 3 rc2 + 2 re - 5 =
there are five variations and one permanence.
740. Theorem X. — No complete equation has a greater
number of positive roots than there are variations of sign,
nor a greater numher of negative roots than there are per-
manences of sign.
Demonstration : Let the following be the successive signs of a com-
plete equation :
+ -- + + —
There are here two permanences and three variations. To intro-
duce another positive root, the equation must be multiplied by x — a.
The signs of the product will readily appear from the following
work :
+ — - + + —
-I- —
+ — — + + —
— + + -- +
+ —
The double sign denotes a doubt, growing out of an ignorance of
the relative numerical magnitudes of the terms added.
Now, a careful inspection will show that, whether we regard both
doubtful signs negative, both positive, or one negative and the other
RELATIONS OF ROOTS TO SIGNS OF EQUATION.
positive, the number of permanences will not be increased, but the
number of terms is increased by one ; therefore, the number of varia-
tions must be increased by at least one. Since the introduction of a
positive root introduces at least one variation, it follows that the num-
ber of positive roots can not exceed the number of variations.
In a similar manner, by introducing the factor x -\- a, it may be
shown that the number of negative roots can not exceed the number
of permanences of sign.
This is Descartes' celebrated rule of signs.
741. Car, 1, — If all the roots of an equation are real,
the number of variations equals the number of positive
roots, and the number of permanences equals the number
of negative roots.
742. Cm*. 2, — An equation whose terms are all positive
can have no positive roots,
743. Cor. 3. — An equation tvhose terms are alternately
positive and negative can have no negative roots.
Limits of Roots.
744. A nnmber known to be equal to or larger than
the largest root of an equation is called a superior limit
to the roots of the equation.
745. A number known to be equal to or smaller than
the smallest root of an equation is called an inferior limit
to the roots of the equation.
746. Theorem XI. — If the first h coefficients of F^ (x)
are positive, and P is the smallest of them, then, if Q is
numerically the largest subsequent coefficient, \/ ^ + 1 is
a superior limit to the roots of F^ {x) = 0.
Demonstration : It is evident that the case in which x must have
the greatest value to make Fn (x) = when the first h coefficients are
positive, is the one in which these coefficients are all equal to the least
one of them (P), and the remaining n + 1 — h coefficients are all
400 ADVANCED ALGEBRA.
negative and each- equal to the greatest among them {Q). Therefore,
the value of a; is a superior limit to the roots of Fn {x) — 0, if
Pa^ + i — Paari + i-k _ Qx'^ + ^-^—Q
or, z — z
X— 1 X— 1
or, Pa?' + i-A(a;*-l) = ^(a;» + ^-*-l)
or if, P(x^ - 1) = ^, since 1 > (l - ^^^^^
+ 1.
= l/|
747. Cor, — If the signs of the alter^iate terms of an
equation he changed, then loill the superior limit to the
roots of the transformed equation, with its sign changed,
be the inferior limit to the roots of the original equation
[726, Cor.].
Equal Roots.
748. Theorem XII, — If F^ {x) = has equal roots, it
may he separated into two or more equations with unequal
roots.
This is a direct inference from Art. 701.
Commensurable Roots.
749. The integral and rational fractional roots of F^ (x)
= are called its commensurable roots.
750. Problem 1. To find the commensurable roots of
Fn {X) = 0.
Solution : Pursue the following line of investigation :
1. Determine the number of roots the equation has [724].
2. Determine how many roots may be positive and how many
negative [739].
3. Determine the limit to the positive and the negative roots
[746, 747].
COMMENSURABLE ROOTS. 401
4. Determine what integral numbers may be roots [720].
5. Find and remove the integral roots by synthetic division [719,
105].
6. Determine whether there are any equal roots [701], and if so,
remove them by synthetic division.
7. Find the rational fractional roots from the equation resulting
from the removal of the integral roots, and according to Theorem
VIII, Scholium.
niustrations. — 1. Find the commensurable roots of
F^ (x) = 24 cc* + 122 a;3 + 5 a;2 - 26 a; - 5 = 0.
Solation :
1. This equation has four roots, all real, or two real [724, 731].
2. There are one variation and three permanences of sign ; there-
fore, there can not be more than one positive nor more than three
negative roots [739].
3. The only integral roots possible are +1, — 1, +5, and — 5
[720].
4. The largest positive root < 4/ — + 1, or < 2 [746].
5. Neither + 1 nor — 1 is a root, since F^ (x) is not divisible by
either a; — 1 or a; + 1, as witness :
-1)24 + 122+ 5- 26- 5
- 24-
98 +
93-
67
+
98-
1)24 + 122 +
24 +
93+ 67-
5-26-
146 + 151 +
72
5
125
146 + 151 + 125 + 120*
Note. — It is evident that when + 1 is a root of Fn ix) = 0, the
sum of the positive coefficients must equal the sum of the negative
coefficients ; and, if — 1 is a root, + 1 is a root if the signs of the
alternate terms are changed. These facts determine a more expe-
ditious method of testing whether either + 1 or — 1 is a root.
6. — 5 is a root, as witness :
-5)24 + 122+ 5-26-5
-120-10 + 25 + 5
+ 2-5-1
7.' The resulting equation, after removing the root — 5, is F^ (x)
= Mx^ + 2 x^ — 5 X — 1 = 0, which has no integral roots.
Transform Fa (x) = into an equation of the form of /a {x) = 0,
f3(:x) = afi + 2x^- 120 a: - 576 = [733, Sch.].
402 ADVANCED ALGEBRA.
8. /s {x) = has three roots [724], only one of which can be posi-
tive, and the largest positive root possible is ^^77 = 24.
9. The divisors of 576 not exceeding 24 are 1, 2, 3, 4, 6, 8, 9, 12,
16, 18, 24. From the relative values of the positive and negative co-
efficients it will be seen at a glance that a; > 6.
10. + 8 and + 9 are not roots, but + 12 is a root, as witness :
4-8)1+ 2-120-576
+ 8 +
80-
-320
+ 10-
40-
-896*
+ 9)1
.+ 2-
120-
-576
+ 9 +
99-
-189
+ 11-
21-
-765*
hl2)l
.+ 2-
120-
-576
+ 12 + 168 + 576
14+ 48
11. The resulting equation, after removing the root + 12 from
/a ix) =z 0, is /a (a:) =: a;2 + 14 a; + 48 = ; whose roots are found to be
- 8 and - 6 [3311.
12 8
12. The four roots of Fiix) = Q are, therefore, —5, +04.' "oi*
and - ^ [733, Sch.], or -5, + -^, - g , and - ^.
2. Find the commensurable roots of
/^ (a;) = a;^ -f 3 a;« - 1 2 a;5 - 3 6 a:* -f 48 a:^ ^
144 a;2_g4^_ 192 = 0.
Solution : It may readily be found by synthetic division that + 2,
— 2, and — 3 are the only integral roots of this equation.
The resulting equation, after removing these roots, is fi (a;) = a:* —
8 a:* + 16 = 0. If any of the roots of this equation are integral, they
must be equal to one or more of the roots already found. They may,
however, all be incommensurable or imaginary.
Factoring ft (a;), we have {x^ — 4) (a;* — 4) = ;
whence, a; = + 2, — 2, +2, — 2.
Therefore, all the roots of /t (x) = are ±2, ±2, ± 2, and — 3.
EXERCISE 103.
Find the commensurable roots of :
1. a^-^x^-\-'ix-b = ^ 2. a:3_6a;2-f-10a;-8 =
3. a^ - 11 2:2 -f 41 2; - 55 =
COMMENSURABLE ROOTS. 403
4. a:3 + 6a:2_j_i4^_{.12 =
5. :r*-3a;3_2a;2 + 12a:-8 =
6. 12:^3 _|_3a;2_3^._2 = o
7. a:* + 2a:3-7a;2-8a; + 12 =
8. 2r* - 8 a;3 + 10 a;2 + 24 a; + 5 =
9. a;5 + 3a;4-3a:3-9a:2-4a;-12 =
10. a;*-5ic3 + 3a;2 + 2iC + 8 =
11. 8a:3_i6^8_3^_|_2i = o
12. 16a;^-48a:3_^ 32^:2 _^12a;-9 =
13. 3a;5 + 2a:*-21a:3-14a:2 + 36a; + 24 =
14. 9 a;5 + 81 :r* + 203 a:3 + 99 a,-2 - 92 a; - 60 =
15. 18a;5 + 9a;* + 22a;3 + lla;2-96a;-48 =
16. a:* + 4ic3_i3 2;3_28a; + 60 =
17. ar* + 2a:3_ii^2_12a; + 36 =
18. ic5 + 4a;* + a:3_iQ^_4^_|.3^0
19. 3 a;6 + 22 a;5 + 8 a;* - 42 :z;3 _ Y9 ^2 _ 52 ^ _ 12 =
Incommensurable Roots.
751. The incommensuraUe roots of an equation are
best sought for after all the commensurable roots have
been removed by division and the resulting equation trans-
formed into an equation of the form of /« {x) = 0.
752. The first step necessary in the search for the
values of the incommensurable roots of an equation is to
find the number and situation of such roots.
Jacques Charles Frangois Sturm, a Swiss mathemati-
cian (1803-1855), discovered a method of doing this in
1829, known as Sturm's method.
404 ADVANCED ALGEBRA.
753. Sturm's Series of Functions. — Assuming that
f^ (x) = has no equal roots, this eminent mathematician
formed a series of functions, as follows :
The first two terms of the series are /„ (x), and its first
derivative, which we will now represent by /«_i (x).
The other functions, and which are called Sturmian
functions, are derived as follows : Divide /„ (x) by /„_i (x),
and represent the remainder with its sign changed by
/«_2 {x). Divide /„_i {x) by /„_2 {x), and represent the re-
mainder with its sign changed by /„_3 (x) ; continue this
process until the last remainder with its sign changed is
an absolute term. Eepresent this remainder /o {x). There
will then be n-\-l of these functions, as follows :
/n (^), /«-l {X), fn-2 {x) ....fo [x).
Caution. — Care must be taken in the operation of successive divis-
ion not to reject any negative factors except in the remainders.
764. Relation of the terms of Sturm's series of func-
tions. — If we put qi, qz, q^ as the successive quo-
tients obtained in finding the Sturmian functions, it is
evident that
/»(^)=/«-l(^)^l-/n-2W (1)
/„_! {X) = /._2 (X) q2 - fn-3 (^) (^)
fn-2 (X) = fn-3 (x) §'3 " /«-4 (^) (3)
/„_3 (x) = ./;_4 (x) q, - /„_5 (x) (4)
/„_4 (x) = /«_5 (x) q, - /„_6 (x) (5)
etc., etc., etc.
755. Fundamental Principles.
1. No two consecutive functions can vanish, i. e., be-
come 0, for the same value of x.
For, if possible, \Qtx = a make /„_2 = and /„_3 = ;
then will /„_4 = [754, 3], and hence, too, /n_6 =
INCOMMENSURABLE ROOTS. 405
[754, 4], and so on until lastly /« (:?;) = ; but /o {x) is
the absolute term and can not be zero. Therefore, etc.
2, If any one of the functions intervening hetween
/„ {x) and /o {x) vanishes for any value of x, the two ad-
jacent functions have opposite signs for this value.
Thus, \i x=:a causes /„_3 {x) to vanish, /„_2 (x) —
-f.-,{x) [754, 3].
3. If any value of x, as x = a, causes any intervening
function to vanish, then will the number of variations and
the nuniber of permanences in the signs of the functions
he the same for the immediately preceding and the imme-
diately succeeding values of x, i. e., for x = a — <z> and
x = a-\- <^ .
For the two adjacent functions will have opposite signs
when x-=a [755, 2], and will not change their signs for
any value of x from x = a — c> and a; = « -|- o , since no
root of either can lie between these values [755, 1]. But
the function in question does change its sign, since x passes
over a root of the function in going from a; = a — o
to ^ = « + o . If the signs of the three functions for
x = a — <^ are +? +> — > for x = a-\-o they will be
+ , — , — , which in either case form one permanence
and one variation. Similarly, +, — , — will change to
+ ,+,—; — , +, + will change to —,—,+; and
— , — , + will change to — , +, +.
4' If any value of x causes /„ {x) to vanish, then will
one variation in the signs of the functions be lost in pass-
ing from the immediately preceding value of x to the im-
mediately succeeding value.
{X - «„) +
(1st term)
(*-««) +
(2d term)
{x-a,) +
(3d term)
406 ADVANCED ALGEBRA.
Let f,{x) =
{x — «i) {x - ttg) {x — a^) (x — a,) ;
then /,_! (a;) =
{x — ttz) {x — as) {x — a^) .
(x — tti) {x — Us) (x — a^ .
{x — «i) {x — a^ {x — a^ .
{x — «i) {x — ttz) {x — a^) (x — «„) + (4th term)
(x — tti) (x — ttz) {x — a^) x—a,^_i^ (nth term)
Now, if x equals, say as, then will fn(^) and all the
terms of /„_i (x), except the third, vanish.
Now, the third term of /„_i (x) contains all the factors
of fn (x) except x — a^. Therefore, if x is infinitesimally
less than %, x — % will be negative and /„ (x) and /„_i {x)
will have opposite signs or will form a variation ; but, if x
is infinitesimally greater than as, x — as will be positive
and /„ (x) and /„_i (x) will have like signs or will form a
permanence. Therefore, a variation is lost in passing from
tts — O to «3 -j- O .
766. These principles are true if /„ (x) contains imagi-
nary roots as well as when all the roots are real, since the
signs of the functions depend wholly upon the real factors
they contain [732].
Sturm's Theorem.
757. The number of variations of sign lost in the terms
of the Sturmian series, as the value of x continuously
changes from a to b, a being less than h, equals the num-
ber of real roots of f„ (x) = lying between a and b.
Demonstration. — For each time the value of x, in ascending from
a to b, passes over a root of /» {x) = 0, there is lost one variation of
sign [755, 4] and only one [755, 3].
INCOMMENSURABLE ROOTS. 407
758. Cor, 1. — The theorem is equally true for F„ {x)
= 0, there being nothing in the demonstration of it to
restrict its application to /« {x) = 0.
769. Cor, 2, — The difference between the number of
variations when + oo and — cc are substituted for x in
the series is the number of real roots in the equation.
760. Cor. 3, — The difference between the number of
variations when and + oo are substituted for x is the
number of positive roots, and, when and — oo are sub-
stituted for X, the number of negative roots.
761. Eemark 1.— It is evident that the sign of the absolute term
of a function is the sign of the value of the function, when a; = 0.
762. Remark 2. — The sign of the first term of a function is the
sign of the value of the function, when a; = ± oo .
For, Ax^ = Acc^-^ .x> Bx:^-'^ + Ca;"-* + Dai^-^ +
+ Lx""-^ > Bx^-^ + Cx:^-^ + Dx?"-^ + + L, when a; = ± oo .
763. Remark 3. — The sign of the value of a function for any
integral or decimal value of x is best determined by the method ex-
plained in Art. 106.
Illustration. — Find the sign of i^4 (a:) = 3 a:* — 2 a;^ +
7a;2 — 3a;-8 when x=l'2.
Solution : The value of i^4 {x) when a: = 1-2 is + -3808, as witness:
1.3)3 _ 2 +7-3-8
3-6 + 1-92 + 9-984 + 8-:
1-6 + 8-92 + 6-984 + -3808*
.'. The sign of F^{x) is +.'
Note. — In practice it is usually not necessary to make the last
multiplication and addition to determine the sign of the value.
764. Remark 4. — Though it is not usually best to apply Sturm's
method of solution to equations before the commensurable roots have
been removed by division, on account of the great labor involved in
deriving and evaluating the different functions when the equation is
of a high degree, yet such a course may be pursued. If there are
equal roots, the fact will appear in deriving the functions, and if there
are integral or fractional roots they will be discovered in evaluating
the functions to determine their signs.
408 ADVANCJSn ALGEBRA.
Example. — Determine the number and situation of the
real roots in /s (a:) = o^ — 12 a;^ + 57 a; — 94 = 0.
Solution : /a {x) = a;^ — 13 a;2 + 57 a; - 94
/2(a;) = 3a;2-24a; + 57
fi{x) = —x + ^
foix)=-
Substituting in these functions as follows, we shall have :
For x=+Go, + + — — one variation.
For X = 0, _ 4- 4. _ two variations.
For x= —cOy _ + + _ two variations.
There is, therefore, one real root between and 4- oo . There is
no negative root. Therefore, there are two imaginary roots.
To find the situation of the real root, we proceed as follows :
For re = 1, we have — + + _ two variations.
For a; = 2, we have — + + _ two variations.
For ic = 3, we have — + ± _ two variations.
For aj = 4, we have + + — — one variation.
Therefore, there is one real root between 3 and 4, or the first
figure of the real root is 3.
To find the next figure, we proceed as follows :
For X — 3-1, we have — + _ — two variations.
For a; = 3*2, we have — + _ _ two variations.
For X = 3-3, we have — + _ _ two variations.
For X = 3-4, we have + + — — one variation.
Therefore, the root lies between 3*3 and 3'4, or the first two fig-
ures of the root are 3-3.
By a continuation of this process the root might be extended to
any number of figures. A more expeditious method, however, is
known, and will be explained hereafter, for extending a root after a
sufficient number of figures have been found to distinguish the root
from any other root lying near it. Thus, if an equation had the two
roots 3*1256. . and 3*1234. . ., the first four figures of each root only
would be found by Sturm's theorem.
When it is known, as in the above example, that only one real
root lies between two numbers, it becomes necessary only to study
the signs of /„ (x), since passing over the roots of the intermediate
functions does not cause a change in the number of variations.
The same conclusion will be reached by the simple application of
Art. 735, since /a (x) changes sign between a; = 3 and a; = 4.
INCOMMENSURABLE ROOTS. 409
EXERCISE 104.
Find the number and situation of the real roots in the
following equations :
1. a:3_4a;2-6a; + 8 = 4. x^-10a^-[-Qx-\-l =
2. x^ + 6x^'-Sx + 9 = 5. 2:r*-lla;2^82,_16_0
3. a^-{-3s^-'6x-[-2 = 6. xf" - Ux^ + Ux -3 =
Horner's Method of Root Extension.
765. In 1819, W. G. Horner, an English mathema-
tician, published an elegant method of extending a root
of an equation to any desired number of places, after a
sufiBcient number of initial figures have been found by-
other methods to distinguish the root from other roots of
the equation. This method is based upon the following
principle :
766. Principle, — If F^ {x) he continuously divided hy
X — a, the successive remainders will he the coefficients in
inverse order of an equation whose roots are a less than
the roots F^ {x) = 0.
Demonstration :
Take En (x) = A a?— ^ + 5a:«-2 +
+ Jx^ + Kx + L = (A)
Put Xi + a = X, or Xi=x — a;
En {Xi + a) = A(xi + «)«-! + i? (a;, + a)«-2 +
+ J{Xi + af 4- K{xi + a) + i = (B)
Expand terms, bracket coefBcients of like powers of Xx , and rep-
resent the coefficients of the transformed equation by -4j , ^i , . . . .
«/i , Kx, Li', then,
En {Xx) = Ax a;i»-i + Bx Xx^-^ + . . . .
Jx Xx^ -{■ KxXx-\- Lx = (C)
Now, the roots of (C) are evidently a less than those of (A).
Substitute Xi = x — a m. (C),
En{x -a) = Ax{x- ay + Bx(x- a)"-! +
+ Ji{x- af + Kx{x-a) + Lx = (D)
Of 1 HE
-NIVERSITY \
410 ADVANCED ALGEBRA.
Now, Fn (x — a) is evidently equivalent to Fn (x), and will leave
the same remainder when divided by a; — a as will Fn (x) -i-{x — a).
But, if Fn {x — a) is continuously divided by x — a, the successive
remainders will he Li, Ki, Ji, £i, and Ai, or the coefficients
of Fn (xi) in inverse order. Therefore the theorem.
Applications.
1. Transform aH^ -\- x^ -}- x^ -^ 3 x — 100 = into an
equation whose roots are 2 less than those of the given
equation.
Foniii
( + 3
Famii
+ 1
+ 1
+ 3
-100
+ 2
+ 6
+ 14
+ 34
+ 3
+ 7
+ 17
- 66*
+ 2
+ 10
+ 34
+ 5
+ 17
+ 51*
+ 2
+ 14
+ 7
+ 31*
+ 2
The transformed equation is x^ + 9 x^ + 31 x^ + 51x — 66 = 0.
Explanation. — Dividing by x — 2, by synthetic division, the co-
efficients of the first quotient are 1 + 3 + 7 + 17, and the first re-
mainder is — 66, the absolute term of the transformed equation.
Dividing 1 + 3 + 7 + 17 again by + 2, the second quotient is
1 + 5 + 17, and the second remainder, or the coefficient of x in the
transformed equation, is + 51.
Dividing 1 + 5 + 17 again by + 2, the third quotient is 1 + 7,
and the third remainder, or the coefficient of x^ in the transformed
equation, is + 31.
Dividing 1 + 7 again by + 2, the fourth quotient is 1, and the
fourth remainder, or the coefficient of a;' in the transformed equation,
is +9.
Therefore, the transformed equation is x* + 9x^ + dlx^ + 51x —
66=0.
Query. — Could you tell by inspection that the roots of the trans-
formed equation are less than those of the original equation ?
Query.— Since 1 + 9 + 31 + 51 > 66, can a;* ^. 9 ^^a + 31 ^s + 51 a;
— 66 = have a positive root equal to or greater than unity ? Why,
or why not f
INCOMMENSURABLE ROOTS. 411
2. Transform rz;* + 9 a;^ + 31 a;^ + 51 a; - 66 = into an
equation whose roots are '8 less than those of the given
equation.
1+9 +31 +51 -66 (-8
-8 7-84 31-072 656576
-0-3424*
9-8
-8
38-84
8-48
82-072
37-856
10-6
-8
47-32
9-12
+ 119-928*
11-4
-8
+ 56-44^
12-2*
The coefficients of the first quotient are 1 + 9-8 + 38-84 + 82-072,
and the first remainder is — 0-3424, which is the absolute term of the
transformed equation. The second remainder, or the coefficient of rr,
is 119-928. The third remainder, or the coefficient of x^, is 56-44.
The fourth remainder, or the coefficient of ofi, is 12-2.
The transformed equation is
Q^ + 12-2 a;3 + 56-44 a;^ + 119-928 a: - -3424 = 0.
3. Transform
ir* + 12-2 0^3 _|_ 56-44 a;2 + 119*928 a; - -3424 =
into an equation whose roots are '002 less than those of
the given equation.
1 12-2 56-44 119-928 --3434 (-003
•002 -024404 -112928808 + -240081857616
12-202 56-464404 120-040928808 -'102318142384*
•002 -024408 -112977624
12-204 56-488812 120-153906432*
•002 -024412
12-206 56-513224*
•003
12-208*
The coefficients of the first quotient are 1 + 12-202 + 56-464404 +
120-040928808, and the first remainder, or the absolute term of the trans-
formed equation, is — -102318142384. The second, third, and fourth
remainders, or the coefficients of x, x^, and x^, are 120-153906432,
56-513224, and 12-208.
The transformed equation is
ic* + 12-208x3 + 56-513224 a;2 + 120-153906432 a; - -102318142384 = 0.
412
ADVANCED ALGEBRA.
4. The integral part of one of the roots of a^ -f x^ + ^^
+ 3 a; — 100 = is 2. Extend the root.
Form.
1 +1
+ 2
+ 3
+ 1
+ 6
+ 7
+ 10
+ 3
+ 14 H
+ 17
+ 34
+ 51*
- 100 . 1 2-802
- 34
- 66*
+ 2
65-6576
+ 5
+ 17
+ 14
+ 31*
-0-3424*
+ 2
31-072
82-072
37-856
119-928*
+ 7
+ 2
+ 9*
7-84
. 38-84
8-48
47-32
9-12
56-44 *
•240081857616
- -102318142384*
•8
•112928808
120-040928808
•112977624
120-153906432 *
+ 9-8
•8
10-6
•8
•024404
56-464404
•024408
56-488812
•024412
56-513224*
11-4
•8
12-2*
•002
12-202
•002
12-204
•002
12-206
•002
12-208*
Explanation. — 1. Transform the equation into one whose roots are
less by 2. The new equation is ocf^ + Q x^ -{- ^Ix"^ -^^ bl x — QQ = 0.
The roots corresponding to the one we are considering will now be a
decimal.
2. Since (-1)2 = -01 ; (-1)3 = -001 ; and (-l)* = •OOOl, the first three
terms are small in comparison to 51a:; therefore, 51 ic = 66 nearly,
whence 51 may be taken as a trial divisor to find the next figure of
the root, considerable allowance being made for the omitted terms.
At first we would be tempted to try "9 for the value of x. But, upon
transforming the equation into one whose roots are less by -9, we shall
find that the absolute term will become positive, which shows that "9
INCOMMENSURABLE ROOTS,
418
is a superior limit. We therefore use -8 for the next term of the
root, and transform the equation into one whose roots are '8 less. The
transformed equation is x^ + 12'2a;3 + 56-44 a;^ + 119-9282; - -3424 = 0.
The root of this equation is now less than -1.
3. Omitting the first three terms of the equation on account of
their smallness, and using the coefficient of x as a trial divisor, we see
that the root is less than '01 and is about -002. The next figure of
the root is therefore 0, and the following one 2. Transform the equa-
tion into one whose roots are less by '002 ; the resulting equation is
Q^ + 12-208 a;3 + 56-513224 a;^ + 120-153906432 a; - •102318142384 = 0.
The work may be extended as far as we please.
76 7o Bemark 1. — When the number of decimal places in the
absolute term becomes equal to the number of such places desired in
the root, we may begin to drop one figure in the preceding term (trial
divisor), two in the next preceding term, and so on toward the left.
When all the figures of the first term are exhausted, the remaining fig-
ures of the root may be found by simply dividing by the trial divisor.
768. Bemark 2.— The absolute term after each transformation
must be negative, else would the last figure of the root used be too
large (a superior limit).
769. Bemark 3. — The method may be applied with equal facility
to extending an integral root after a sufficient number of initial fig-
ures have been obtained by trial or by Sturm's Theorem to distinguish
the root from others of the equation. It may be used with exactness
whenever there is an exact root ; hence, the incorrectness of the title
"Horner's Method of Approximation" given the method by most
authors.
770. Bemark 4. — The negative roots are the numerical equiva-
lents of the positive roots of the equation resulting from changing the
signs of the alternate terms, and may be found accordingly.
5. Solve a? - 1728 = 0, or find the Vi728,
Solution.
(12
+
10
10
+
100
100
200
300*
-1728
1000
-728*
10
20
+ 728
10
30*
64
"364
32
414
ADVANCED ALGEBRA.
6. Extract the 5tli root of 4312345 to thousandths,
e., solve approximately x^ — 4312345 = 0.
Solution.
1
-4312345 1 21-229
20
400
8000
160000 3200000
20
400
800
8000
24000
82000
48000
80000 *
160000
-1112845*
20
640000
40
1200
1200
800000 *
20
84101
884101
884101
- 228244 *
60
2400
1600
4000*
20
4101
88304
972405 *
80
84101
4203
88304
4806
198220-84882
20
101
100*
- 30028-15168 *
4101
102
18699-2416
991104-2416
1
101
4208
103
4306
92610*
18877-3264
1
886-208
1009981-5680 *
102
98496-208
1
103
104
4410*
890-424
94886-682
894-648
1
21-04
104
1
4481-04
21-08
95281-280*
105*
4452-12
21-12
•2
105-2
•2
4473-24
21-16
105-4
4494-40 *
-2
105-6
•2
105-8
•2
106-0*
The number of decimal places in the second remainder is greater
than the number required in the root; therefore, the remaining fig-
ures may be found by dividing the remainder by 1009981-568 [767,
Rem. 1].
INCOMMENSURABLE BOOTS. 4-15
EXERCISE 108.
Solve :
1. x^ - 704 a; - 58425 = 2. a? - 15348907 =
3. x^-\-Za?'-^x-'i =
4. iC*-4a;3-6a:2 + 32a;-26 =
5. ir*-19a;3 + 24a;2 + 712a;-40 =
6. 2^5 + 12 a;* + 59a;3 + 150a:2 + 201a; + 94 =
7. 3a;* + 24a:3 + 68a:3_^g2^_964 =
8. Find the cube root of 2 9. Find the fifth root of 5
10. a:3 4-ii<?;2_io2a;+181 =
11. a:* + 9ar^ + 31a;2 + 51rr-66 =
12. a;5 + 2 2;* + 3 i?^ + 4ic3 + 5 a; - 54321 =
13. One root of the equation o? -\-2x^ -\-^x — 13089030 is
235. Find a cubic equation whose root is 225.
Cubic Equations.
771. A cubic equation containing an integral root may
be readily factored.
Let — a be a root of a cubic equation, then x-\-a \^
a factor of the equation [719]. Let x^-{-mx-\-n be the
other factor ; then,
(x-\-a){ci^-\-mx-{'n) = (A)
or, x^ -\- {a -\- m) 7? -\- {a m -\- n) X -\- a n = (B)
We now observe that if we subtract the factor a of the
absolute term from the coefficient of x^, and the factor n
from the coefficient of .t, the latter remainder divided by
the former will give a, the root with the sign changed.
This, then, is the condition under which a factor of the
absolute term is a root with the sign changed.
416 ADVANCED ALGEBRA.
Illustration. — Solve a? — x^ — 4:X-\-4: = 0,
Solution : 1. The factors of + 4 are +3 and + 2 ; — 2 and — 2 ;
+ 4 and + 1 ; and — 4 and — 1
Try whether + 4 is a root with the sign changed.
Take — 1 and — 4, the coeflScients of x^ and x.
Subtract, + 4 and + 1, the factors of the absolute term.
Divide, —5) — 5 ( which =}= 4.
.'. 4 is not a root with the sign changed.
2. Try whether — 2 is a root.
Take - 1 and - 4
Subtract, + 2 and + 2
Divide, -3) -6( + 2
.'. — 2 is a root.
Now, x^ — x^ — 4tx + 4: = {x + 2){x^ — ^x ■\- 2) = 0;
whence, a; = — 2, 2, and 1.
Cardan's Formula.
772. I. The general cubic equation a^ -\- a oi? -{- h x -\- c
= may be transformed into an equation of the form of
y''-\-py^q = 0, by putting x = ij- -a.
Demonstration : Take a^ + ax"^ + hx + c = Q. (A)
Put
x = y-
g-a; then.
a^ = y^-
-ay^-\- -a^y-—a^
ix^ =
2 1
ayi--a^y+—a^
hx-
by -g-ttft
c =
c
.'. a^ + ax^ + bx + c = y^ + (*- -g-^M^ + (27^' "s""* + ^)
1 2 1
Put p tor b— -^ a*, and q tor ^a^ — -^ah + c; then,
a:^ + ax'' + bx + c = y^+py + q = 0. (A)
CUBIC EQUATIONS. 417
773. II. The equation y^-{-py-\-q=^0 may be trans-
formed into a quadratic by putting y — z— —
o z
Demonstration: Take y^ -^-py + {
1=0.
Put
y = z-
P .
~ 3z'
then.
yZ^2%
P'
p^
27 £3
py =
pz
P'
dz
,*,
y^ +py + q = z^
—
pz
27 z^
+ Q =
whence,
27z'^ + 27 qz^-
pz
= 0;
or,
2« + qz^-^
^1^'
= 0.
(B)
(C)
774. III. The roots of z^ -\- q z^ — ^;;: p^ = are
/c7
p
-(-l^\/j + ff'^'''^'''
^ 3z
(-f+vT^+(-iVS+# <»'
This is Cardan's formula. ^
The values of x may be obtained by subtracting —a
from the values of y.
775. IV. Cardan's formula fails when all the roots are
real and unequal.
For, let the roots be 0^ + V^, a — V~b, and c. Then,
since the coefiBcient of y^ is 0,
(^ a^ ^/b) -^ (^ a-j- VJ) - c = [726, 1] ;
whence, c = — 2 «. _
The equation whose roots are a-j-V^, a—Vb, and
-2a, is y^-{Sa''-\-b)y-2{a^-ab) = 0.
.-. p = -{da^-\-b), md q = 2{a^-ah),
whence, A/^ + |! = L2 _ ^ j j V- 3 &, which
is im-
418 ADVANCED ALGEBRA.
aginary, and, therefore, irreducible when h is positive, or
when the roots are all real and unequal.
Illustrations.— 1. Solve ic^ — 3 a;^ -}- 4 = 0.
Here a = — 3 ; hence, x = y — I —^- j = ^ + 1.
Substitute, (2/ + l)^ - 3 (y + 1)^ + 4 = 0.
Reduce, 2/3 — 3^ + 2 = 0. (1)
P 1
Here « = — 3 ; hence, y = z — :^ = z + — .
oz z
Substitute, (^0 + ^V - 3 T^ + i) + 2 = 0.
Reduce, z^ + 2 z^ = — 1.
Complete the square, z^ + 2z^ + 1 = 0.
Extract the V* z^ + 1 = 0.
Factor, (z + l)iz^ - z + 1) = ;
1 1 /— T
whence, z = — i or-^ i-^V — 3*
x = y + 1 = z + — +1= — 1, or 2, or 2.
776. Sometimes an integral root can only he approxi-
mately found.
2. Solve x^-\-dx^-{-9x^l3 = 0. (A)
Here a = 3 ; hence, x = y — l^)=y — l.
Substitute,
(2/ - 1 f + 3 (2/ - 1)2 + 9 (2/ - 1) - 1 3 = 0.
Reduce, y^ + 6y-20 = 0. (1)
p 2
Here p = Q; hence, y = z — ^ z= z .
^ ' ' ^ dz z
Substitute in (1),
(._|)%e(._|)_.o = o.
Reduce, z^-20z^ = 8. (2)
Complete the square, 2« -20z^ + 100 = 108.
Extract the \/, z^ - 10 = ± 10-392304.
Transpose, ^« = 20-392304 or - -392304+
Extract the V» '^ = 2*73+ or - -73 +
x=:y-l = z-- r\-l = 2-73 --73 + 1 = 3, or --73 + 2-73 + 1 = 3.
z
These two values are identical. The other two roots are found by
dividing equation (A) by x — d.
CUBIC EQUATIONS. 419
EXERCISE 106.
Solve :
1. a;3-3a:2 + 7a;-5 = 8. iC^H- a;^ _ 8a; — 12 =
2. x^ — Q7^-\-l()x — S = 9. a:3_a;2 — 8a;+12 =
3. rc3 - 11 0:2 + 41 2; — 55 = 10. ar^ - 11 2; - 20 =
4. a:3 + 6a:2^i4^_^12 = 11. a:^ - 26^:+ 60 =
5. a:3_4a;8 + 5a;-6 = 12. a:^ _ 4 ^jS _|. 3 _. q
6. a?-\-^x'-\-Qx-\-S = 13. a;3_4a;2_Ya;_^10 =
7. a:3_|.7^_j_i6a;4-12 = 14. a:3_|_4a;2_ 7^;- 10 =
Recurring Equations.
777. A Recurring Equation is one in which the co-
efficients of the first and last terms, and of those equi-
distant from the first and last terms, are numerically
equal, and the signs of the corresponding terms are either
alike throughout or unlike throughout ; as,
1. a;5-4a;* + 5a:3 + 5a:2~4ic + l=:0.
2. a;^ + 3a:*-2a;3 + 2ic2_3^_l_0.
3. a;« + 4a;5-5a,-* + 3ic3-5a;2 + 4ir + l = 0.
778. In a recurring equation of an even degree in
which the corresponding terms have unlike signs, the
middle term is wanting.
For, according to definition, it is both positive and
negative.
779. A Reciprocal Equation is one such that, if a is a
root, — is also a root.
a
780. Thetyrem I, — A recurring equation is also a re-
ciprocal equation.
+ ;;^ + ;;^+----±;72±-:;±l = o (C)
420 ADVANCED ALGEBRA.
Demonstration : Let a be a root of
f^{x) = x^ + Ax^-'^ + Bx^-'^ + . . . .± Bx^ ± Ax ± 1 = (A)
then, a« + J.a«-» + J5a«-2 + ±Ba^ ±Aa±\ = (B)
Substitute — for x in /„ {x) = 0,
whence, 1 + Aa + Ba^+ ±B w^-^ ± A a»-' ± a* = (D)
or, a" + ^a«-i + 5a«-2+ ± ^a^ j|. ;ia^ ^ 1 _ (E)
Now, (E) is identical with (B) ; therefore, if a is a root of (A),
— is also a root.
a
781. Theorem II. — A recurring equation of an odd
degree has -\- 1 for a root when the signs of the correspond-
ing terms are unlike.
Demonstration :
Let a;2» + i + ^a;2« + ^a;2«-» + -Bx^-Ax-1=0' (A)
then, (a;2»+i - 1) + Axix^''-^ - 1) + Bx^ix^''-^ - 1) + =0 (B)
Now, each term of (B) is divisible hj x — 1 [134, P.] ;
.'. a; — 1 = 0, or x=l.
782. Theorem III, — A recurring equation of an odd
degree has — 1 for a root when the signs of the correspond-
ing terms are alike.
Demonstration :
Let ir2« + i + J.a;2« + 5a;2«-i +....+ 5a;2 + Aaj + 1 = (A)
then, (a;2« + i + 1) + Ax(x^''-'^ + 1) + Bx^x^-""-^ + 1) + =0 (B)
Now, each term of (B) is divisible by a; + 1 [135, P.] ;
.*. a; + 1 = 0, or a; = — 1.
783. Theorem, IV. — A recurring equation of an even
degree has + 1 and — 1 for roots when 4he signs of the
corresponding terms are unlike.
Demonstration :
Let a;2» + ^a:'"- ^ + 5a:2«-2+ — Bx^ — Ax — \ = (i (A)
then, (x2 «- 1) + ^ a: (a;2«-2 - 1) + 5 a;2(a;2 "-4-1)+ =0 (B)
Now, (B) is divisible by both a; - 1 and a; + 1 [134, 136, P.] ;
.*. a; — 1 = and x + 1 = 0; whence, a; = ± 1.
RECURRING EQUATIONS, 421
784. Thetyrem V, — A recurring equation of an even
degree may le transformed into an equation of one half
the degree when the signs of the corresponding terms are
alike.
Demonstration :
Let a;2»+^a;2»»-J + 5a;2«-2+ + ^a;^ + ^a; + 1 = (A)
Divide by x*, and collect terms,
(^»+l) + A(^-.+ ji^)+£(x«-«+jjL) + ....+P=0 (B)
Put X ^ = z ; then will
X '
x^ + -. = z^-2
x^
and, in general, each term of (B) may be transformed into a term of
only half the degree.
Illustration. —
Take a;^ + ^o;''^ - 3 a:* + 2 a;^ - 3 2;3 + 4a; + 1 = 0. (A)
Divide by a:8, jc« + 4a:2-3a; + 2- - + -^ + -^ = (B)
•' X x^ x^ ^ '
Rearrange the terms and factor,
Put x->r - = y\ (1)
then, a;2 + 2 + -2 = 2/^ ; (2)
or, a;2 + |, = 2/2-2;
JO
3 1
and, a;3 + 3a;+ - + ^ = 2/«;
^'^'+^+K'^-'^)=^'
a;3+ J-=y3_3y. (3)
Substitute (1), (2), and (3) in (C),
(2/3 - 32/) + 4(2/2 - 2) - 3y + 2 = 0;
whence, j/^ + 4 2/* — 6 2/ — 6 = 0.
422 ADVANCED ALGEBRA.
EXERCISE 107.
Solve :
Z. QC^ — dx^-\-^x — l = 4. ic* + 3a;=^ — 3a;— 1 =
5. 2a;*-5a;3 + 4a;2~5a; + 2 =
6. a;s + 5a;* + 10a;3 + 10:z;2_j_5^_^l_0
7. 6a;5-ir*-432;-'' + 43a;2 + ir-6 =
8. bx'>-\-ll7^-^%x?-^^x^-^llx-{-b =
Reduction of Binomial Equations.
785. A Binomial Equation is an equation of two terms,
one of which is absolute ; as, a;" ± « = 0.
786. Every Mnomial equation can te reduced to the
form 2/** ± 1 = 0.
Demonstration : Take the general binomial equation a;* ± a = 0.
Put ^"^ for X,
y
a
r
+ a = ; whence, 2/" ± 1 = 0.
787. r ± 1
=
is a recurring equation, and may be so
solved.
Illustrative Solutions.— 1. Solve a:* + 1 = 0. (A)
Divide by x^,
^^ + ^ = (1)
Put
^^\ = y (3)
Square
x^ + 2+l, = y^
Transpose,
^' + ^, = y'-^
Substitute in
(1),
y^-2 =
Factor, (y + ^2 ), (y - a/s") = ;
whence, y — ± \/2.
Substitute in (2), x + -= ± \/2 (3)
X
whence, x = ^ {\/2 ± ^/'^), or - ^ {^ T ^-2)-
chapter xiii.
determij^a:n'ts akb probabilities.
Introduction.
788. In the polynomial
«! 1)2 Ci — «i ^3 C2 + «8 ^3 ^1 — «2 ^1 ^3 + ^3 ^1 ^2 — ^3 ^2 ^1 ^ (-^)
it will be seen :
1. That the letters a, h, and c of each term are ar-
ranged in natural order.
2. That the subscript figures, 1, 2, and 3, are dis-
tributed among the letters in the six different terms in
as many ways as possible, using all in each term and
making no repetitions.
3. That the first term contains no inversions of sub-
script figures, they advancing in natural order from left to
right ; the second term contains one inversion, 3 standing
before 2 ; the third term contains two inversions, 2 and 3
both standing before 1 ; the fourth term contains one in-
version, 2 standing before 1 ; the fifth term contains two
inversions, 3 standing before 1 and 2 ; and the sixth term
contains three inversions, 3 and 2 both standing before 1,
and 3 standing before 2.
4. That in the positive terms there is an even number
of inversions (zero being regarded an even number), and
in the negative terms there is an odd number of inver-
sions.
424
ADVANCED ALGEBRA.
(B)
789. If we now arrange the nine different quantities
found in (A) in a square, as follows :
«i a2 a^
h h h ;
Ci Cg C3
form all the possible products of them taken three to-
gether, using in each product one and only one from each
row, and one and only one from each column ; arrange
the factors of the products in the natural literal order ;
consider those products positive which have an even num-
ber of inversions of subscript figures, and those negative
which have an odd number ; and take the algebraic sum
of these products, we will have :
«i ^2 C3 — a^ J3 (?2 + «3 ^3 Ci — «3 li C3 + a^ bi <?2 — % ^2 Ci . (A)
Therefore, form (B) may be taken as the representative
of form (A), and when so taken it is called a determinant,
and (A) its development.
790. Definition. — A Determinant is any n^ quantities
arranged in a square, as follows :
at
«2 «3
h
\ h
Cx
C% C3
(C)
Pi P2 PZ Pn
and interpreted to denote the algebraic sum of all the
products that may be formed by taking one and only one
quantity from each row, and one and only one from each
column ; arranging the letters of each product in the natu-
ral literal order, and regarding all products positive that
have an even number of inversions of subscript figures,
and all negative that have an odd number of such inver-
sions.
DETERMINANTS. 425
791. The quantities contained in a determinant are
called the elements of th^ determinant.
792. Determinants are divided into orders, named
second, third, nila., accordingly as they contain 2^, 3^,
n^ elements.
Thus,
is a determinant of the second
h h
order. Form (B) is a determinant of the tMrd order, and
form (C) a determinant of the ntli order.
793. The diagonal joining the upper left-hand element
with the lower right-hand element is called the 'principal
diagonal ; and the one joining the upper right-hand ele-
ment with the lower left-hand element the secondary
diagonal.
794. The product of all the elements along the principal
diagonal is called the principal term of the development.
795. If the elements on the principal diagonal are
known in order, the entire determinant may be written ;
hence it is that a determinant is often expressed by a
modified form of the principal term of its development ;
as, \_a^lzc^ ^J, or S(±ai^2^3 i?«)-
796. It is evident that there are as many terms in the
development of a determinant of the wth order as there
are permutations of n things taken all together, or \n.
Properties of Determinants.
797. If we rearrange the factors of the terms in form
(A) so as to place the subscripts in natural order, we shall
have
«! ^8 ^3 — «i Cg ^3 + Ci ttg ^3 — 5i ^2 <^3 + ^1 ^2 <^3 " ^1 ^3 «3' (^i)
4:26 ADVANCED ALGEBRA.
It will be observed that in form (Aj),
1. The value of each term and of the entire polyno-
mial is the same as in form (A).
2. The first term contains no literal inversion ; the
second term contains one, c standing before h ; the third
term contains two, c standing before both a and b ; the
fourth term contains one, h standing before a; the fifth
term contains two, h and c both standing before a ; and
the sixth term contains three, c and h both standing before
a, and c before h
3. The terms which contain an even number of literal
inversions are positive, and those which contain an odd
number negative,
798. If we now interchange the rows and columns in
(B), giving us the form
ttx hi Ci
«2 h ^2 ; (D)
% h ^3
make all the possible products of three elements, using,
each time, one and only one from each row, and one and
only one from each column ; arrange the factors of the
products so that the subscripts stand in natural order ;
consider those products positive which have an even num-
ber of literal inversions, and those negative which have an
odd number ; and take the algebraic sum of these prod-
ucts, we shall have
«1 ^2 ^3 "~ ^1 ^2 ^3 + ^1 ^2 ^3 — ^1 ^2 ^3 + ^1 <^2 ^3 " ^1 ^2 ^3 • (-^-i)
This shows that in a determinate of the third order an in-
terchange of rows and columns does not change the value.
Is this law true for a determinant of the nth order ?
1. It is evident that the number of terms in the de-
velopment of both forms is the same, each being [w.
PROPERTIES OF DETERMINANTS. 427
2. Each term in the development of either form has a
corresponding term of equal numerical yalue in the devel-
opment of the other form, because both developments con-
tain all the possible products of n elements that can be
formed from the n^ elements by taking one and only one
from each row and one and only one from each column.
3. The signs of the corresponding terms toill be the
same. For the number of literal inversions in a term of
the second development is equal to the number of sub-
script inversions in the corresponding term of the first
development, as will readily appear from the fact that, if
a subscript in any term of the first development follows r
subscripts greater than itself, then, in the second develop-
ment, the letter containing this subscript must precede r
letters antecedent to it in the natural order. Therefore,
JPrin. 1. — An interchange of rows and columns in a
determinant of any order does not change the value of the
determinant.
799. In form (A) and in form (Ai) the second term
equals minus the first term with the subscripts of h and c
interchanged ; the third term equals minus the second
term with the subscripts of a and c interchanged ; and so
on, showing that any term in the development of a deter-
minant of the third order equals minus some other term in
the development with the subscripts of two factors inter-
changed.
Is this law true for the development of a determinant
of the nth order ?
1. It is evident that, if Pcr^^ be any term in the de-
velopment of a determinant of the nth order, then will
Fcrnhhe numerically another term of the development;
because P in both instances is the product oi n — 2 ele-
4:28 ADVANCED ALGEBRA.
ments, none of which are taken from rows c and Jc, and
none from columns r and w; and c^h^ and c^h, are
different elements taken from these rows and columns
and combined with P. Therefore, the products are not
identical.
2. The signs of the original and the derived terms are
always opposite. For,
(1) Suppose the two subscripts interchanged to be con-
secutive. Let the original term be P c^ h^ Q, and the de-
rived term P c„ k^ Q. Since m and n follow all the sub-
scripts contained in P and precede all contained in Q, an
interchange of them can not affect the number of inver-
sions they make with the subscripts of either P or Q ; but
such an interchange will either change a natural into an
inversion or an inversion into a natural, either of which
will evidently cause a change of sign,
(2) Suppose the two subscripts interchanged to be non-
consecutive. Let the original term be Pc^Q Jc^ R, and
the derived term P c^Qhn R. Suppose Q to contain q
subscripts. Let m, in the original term, interchange con-
secutively with each of the subscripts in Q and with the
subscript of c, then will it make q-{-l interchanges before
it becomes the subscript of c, n will now be the subscript
of the first element in Q. Let it now interchange con-
secutively with each of the remaining subscripts in Q and
with the subscript of h ; then will it make q interchanges
before it becomes the subscript of Ic. Therefore, for the
two subscripts of c and Jc in the original term to inter-
change there must be made 2q-{-l, or an odd number of
consecutive interchanges, each one of which will cause a
change of sign (1) in the entire term. Therefore, the sign
of the term will be changed. Therefore,
PROPERTIES OF DETERMINANTS, 429
Prin. 2. — If tioo subscripts he interchanged in any
term of the development of a deter miiianty another term
of the development will be obtained whose sign is opposite
to that of the original term.
800. If we let Ph,^QKR be a term in the develop-
ment of a determinant, then will Ph^QKR be the term
formed by the elements which occupy the same places, if
rows h and h be interchanged, and will have the same
sign. But Ph„,QKR is also a term of the development
of the original determinant, and has there an opposite sign
to Ph„,Q KB [P. 2]. Therefore,
Prin, 3, — Interchanging two rows in a determinant
changes the sign of the determinant.
Cor. 1, — Interchanging two columns in a determinant
changes the sign of the determinant.
801. It is evident that if two columns or two rows of a
determinant are in every respect alike, an interchange of
them would not affect either the form or value of the de-
terminant. But, according to Principle 3, the sign of the
value would be changed. Now, both these statements can
be true only when the value of the determinant is zero.
Therefore,
Prin, 4, — A determinant that has two rows or ttvo col-
umns identical equals zero.
802. Since every term in the development of a deter-
minant contains one factor and only one from each row
and one and only one from each column, it follows that,
Prin, 5. — Multiplying or dividing all the elements of
one row or one column of a determinant by any quantity
multiplies or divides the determinant by that quantity.
430
ADVANCED ALGEBRA.
Cor, 1, — Changing the signs of all the elements in any
row or column changes the sign of the determinant.
Car, 2, — If two rows or tiuo columns of a determinant
differ only hy a common factor, the value of the deter-
minant is zero,
803. Definitions. — If any number of rows and the same
number of columns be deleted (stricken out) of a deter-
minant, the remaining elements, taken in order, form a
determinant called a minor, and the elements common to
the deleted rows and columns form another minor. These
minors are said to be complementary.
Thus, in the following determinant of the fourth order.
h
-do
^T
d,
-dr
the complementary minors are
h di
h ds
and
«2
t?4
804. If a single row and a single column be deleted,
the remaining minor is called the principal minor, and it,
together with its complementary minor, which in this in-
stance is a single element, are called cofactors.
805. Problem. To develop a determinant.
Let it be required to develop
«i «8 ^3 a^
h h h h
C\ Cg ^3 C4
d\ d^ dz di^
into a series of determinants of a lower order.
Let ^1, Ai, At^ and A^ represent respectively the cofactors of
«i , «a , as , and a^ . Then, it is readily seen that ail the terms in the
PROPERTIES OF DETERMINANTS.
431
development containing the factor a^ are formed from ai and its co-
factor Ai , and the sum of these terms is ai Ai . Similarly, the sum
of all the terms containing a^ is a^ At, the sum of all the terms con-
taining aa is tta Aa, and the sum of all the terms containing a* is
at At . Now, in each term of a^ A^ there occurs one more inversion
than in each term of ai Ai , since a subscript 2 will precede a subscript
1 ; similarly, in each term of as -4s there occur two more inversions
of subscripts than in ai Ai, and in each term of a* A^ there occur
three more inversions of subscripts than in Ui Ai.
Therefore,
«! tti az at
b\ bi bz bi
C\ Ci Cz C4
di di dz d^
= ai
bi bz
Ci Cz
di dz
&4
C4
— «2
b\ bz
Ci Cz •«
di dz
74 +
^4
bi bi &4
bi bi bz
Oz
c
d
1 Ci C4
I di d.
— ai
ci Ci Cz
di di dz
Therefore,
Mtde, — Multiply each element of the first row hy its
cof actor, making the products alternately plus and minus,
and talce the algebraic sum of the results,
806. Scholium, — The successive application of this rule
will eventually make the full development of any deter-
minant depend upon the development of a determinant of
the second order. Thus,
h h h
Cz C^ Ci
dz ds t?4
= h
C3 64
ds di
-K
Cz Ci
dz di
+ *4
Cz C3
do d%
h (^3 di — d ds) — bs (cz di — C4 dz) + h (Cz d^ — c^ dz)
Ulnstrative Example.— Find the value of
Solution :
2 3 4
3 2 4
4 3 2
2(2x2
-3
+ 4
4 X 3) - 3 (3 X 2
4x4) + 4(3 x3-2 x4) =
- 16 + 30 + 4 = 18.
432
ADVANCED ALGEBRA.
EXERCISE lOa
Find the value of :
10.
13.
3 15
2.
4 3 2
3.
4 2 7
5 14
16 4
2 4 5
a 5
5.
m n
6.
-a d
m p
a b (
p n
2 3 2 3
8.
3 2 3 2
2 3 3 2
3 2 2 3
2 3 12
3 2 14
2 3 12
14 3 2
11.
3 4 12
14.
6 8 2 4
12 3 4
2 4 6 8
1
2 3 4
9.
2
-13 4
3
2 13
1
2 10
2 3 4 2
12.
2 3 4 2
2 3 4 2
2 3 4 2
3-1 4
2 1-3
4 2 7
h-\-c c h
c a-\-c a
h a a-\-h
2 3 4
2 3 4
3 2 2
4 12
a h c d
& f 9 ^
a I c d
i k I m
12 1
15.
2 4 1
3 6 1
4 8 1
a
h
c
d
a
b
c
d
n
n
n
n
a
c
h
d
c
a
d
I
Additional Properties of Determinants.
807. JPHn, 6, — If every element of one row, or column,
of a determinant is a binomial, the determinant can be
expressed as the sum of two other determinants, one of
which is derived from the original determinant by drop-
ping the second terms of the binomials and the other by
dropping the first terms.
Demonstration. — Each term of the development contains one and
only one of the binomial elements as a factor. Therefore, each term
of the development can be separated into two terms, one of which is
the first term of the binomial factor times the remaining factors of
PROPERTIES OF DETERMINANTS.
433
a
b + c
b
a
b
b
a
c
b
b
a + c
a
=
b
a
a
+
b
c
a
c
a + b
b
c
a
h
c
b
b
the term, and the other the second term of the binomial factor times
the remaining factors of the term. The sum of the component parts
that contain the first terms of the binomial elements will form a de-
terminant which is independent of the second terms of the binomial
elements, and the sum of the component parts that contain the second
terms of the binomial elements will form a second determinant which
is independent of the first terms of the binomial elements.
Thus,
808. Cor. 1, — If every element in any row, or column,
consists of m terms, the determinant can be expressed as
the sum of m qther determinants.
809. Oor. 2. — If the elements of r rows, or columns,
consist of a,h,Cf m terms respectively, the determi-
nant can le expressed as the sum of abc m determi-
nants.
810. Scholium,, — These truths are of value in reducing
a determinant when one or more of the derived determi-
nants reduce to zero. Thus,
a h^
•a-\-c c
abc
a a
c
ace
d e+d^f f
=
d e f
+
d df
+
df f
g h-\-g-\-h h
g h k
9 9^
g k k
a h c
ale
d e f
+ + [801, P.] =
d e f
g h h
9
h I
;
811. Prin, 7* — If to all the elements of any row, or
column, he added equimultiples of the corresponding ele-
ments of any other row, or column, the value of the deter-
minant will remain unchanged.
Demonstration. — Consider a determinant of the third order. Thus,
Prove
«1
«2
^3
h
h
h
Ci
Cz
Cz
434
ADVANCED ALGEBRA.
Demonstration.
a\ + paz aa (H
ai ai
3^8
paz
as oz 1
h +i?&3 &2 ^3
=
b\ &2 bz
+
p bz bi bz
Ci +pcz Ci Cs
C\ Ci Cz
pcz Ci Cz
a\ a<i, as
dX di Clz
bx h h
+ [802, Cor. 2] =
b\ bi bz
C\ Cn Cs
C\ C2
Cz
The method of proof employed in this example is general, and is,
therefore, applicable to a determinant of any order.
812. Car, — It may also be shown that
ai-{-pa2-\-qas a^
Ci -\rpc2-\-q Cg -cg
«3
Oz
«3
^3
; etc.
«i a^
Ci C2
That is.
To all the elements of any row, or column, may be added
equimultiples of the corresponding terms of a second row,
or column, and again equimultiples of the corresponding
terms of a third row, or column, etc.
813. Scholium, — This principle is of practical value
in the reduction of a determinant, if, by its application,
two rows, or columns, can be made identical, or one of
them a multiple of the other. Thus,
5 + 8 + 11 8 11 24 8 11
6+9 + 12 9 12 = 27 9 12 =0
7 + 10 + 13 10 13 30 10 13
[802, Oor. 2]. It may also be used to simplify a complex
determinant.
5 8 11
6 9 12
=
7 10 13
814. Prin, 8. — If the elements of one column of a de-
terminant be multiplied by the cof actors of {he correspond-
ing elements of another column, the sum of the products,
taTcen alternately plus and minus, is zero.
Demonstration. — Take two determinants, alike in every respect,
except that, in the second, the gth column is identical with the ^th
column. Now, in the first, the sum of the products of the elements
PROPERTIES OF DETERMINANTS.
435
in the pih. column and the cofactors of the corresponding elements of
the ^th column, when the products are taken alternately plus and
minus, is apAq — bpBg +
The value of the second determinant is
agAg-bgBq + ..,. [805, R.] = [801, P.J.
But it is evident that aq , bq , are identical with Up, bp, ....
Therefore,
apAq — bpBq + =0.
EXERCISE 109.
Find the value of
3 2
5 3
7 4
8
6
-8
3
4
5
-1
5
9
4
6.
a+p
b-hp
o+p
7. Prove
8.
Prove
-2
2
8
-4
6
-9
2
2 2
3 -3
b + q
c-^q
a-\-q
h-\-c
q-\-r
aJ^ — be
1
1
1
1
c -\-r
a-\-r
b-^-r
c-\-a a-\-b
r^-p p-\-q
z-^x x-{-y
(? — ab
+
y-z
z — y
a —p
b — p
c —p
= 2
3.
12 -5
-3 4
7
X —
z —
-z x-y
• X y — X
X x-\-y
b-q
c-q
a — q
a b
p q
X y
(^ — ab b^ — ac
ciSabc-a^-b^-c")
Multiplication of Determinants.
B15. Lemma,—
ai bi Ci I
m
n
«2 ^2 02 p
% bs c-3 s
a^i
a^g
t
yi
y2
r
u
^2
=
«2 ^2 ^2
«3 ^3 ^3
a;3
^3
2^3
X
Xi yi Zi
^2 y2 %
^^ yz z^
436
ADVANCED ALGEBRA,
Demonstration.—
ax h\ ci I m n
ai h Ci p q r
as bs Cz s t u
x\ y\ Zi
a:2 2/2 ^2
() xz yz Zz
+ az
= ai
bi Ci p q r
5i
c\ I m n
bz Cz 8 t u
&3
Cz s t u
a;i yi zi
— as
xi yx zx
Xi yi Zi
Xi 2/2 ^2
xz yz zz
Xz yz Zz
bx cx I m n
bi Ci p q ■ r
a;i 2/1 Zx
= ai &2
iCs 2/2 ^2
^ Xz yz Zz
Cz s t u
a;i 2/1 ^1
Xi 2/2 2^2
— ai&8
Xz yz Zz
Ci
p q r
xx yx Zx
Xi 2/2 Zi
Xz yz Zz
— tti bx
— az bi
Cz s t u
Xx yx Zx
Xi yi Zi
+ a2&3
Xz yz Zz
Cx I m n
ici 2/1 2;i
Xi yi Zi
+ aa&i
Xz yz Zz
Ci
p q r
Xx yx Zx
Xi yi Zi
Xz yz Zz
Cl
I m n
Xx yx Zx
Xi yi Zi
Xz yz Zz
Xx yx Zx
ax bi Cz
Xi yi Zi
Xz yz Zz
— ax bz Ci
Xx
2/1
Zx
Xi
2/2
Zi
Xz
2/3
Zz
Xx yx Zx
Xi
2/1 Zx
Xx yx Zx
ai bx Cz Xi yi Zi
+ ai bz Cx
Xi yi Zi
+ az bx Ci
Xi yi Zi
Xz yz Zz
Xz yz Zz
Xz yz Zz
Xx yx Zx
ax bi C\
Xx yx Zx
az bi Cx
Xi yi Zi
=
ai bi Ci
X
Xi yi Zi
[789, (A)].
Xz yz Zz
az bz
Cz
Xz y» Zz
Scholinm. — It will be seen that the elements ?, m, n, p, q, r, s, t, u
of the first member do not appear in the second member. This is due
to the fact that in the development of the first member all the terms
containing these elements eventually vanish.
816. Problem. To find the product of two determinants
of any order in terms of a determinant of the
same order.
Example. — Eind the product of :
«1 ^1 Ct Xi j/i Zx,
«2 hz Cz X Xz yz ^2
«3 ^3 ^3 ^3 Vz %
MULTIPLICATION OF DETERMINANTS. 437
Explanation :
a\ hi C\
xi yi zx
tti bi Ci
X
Xi 2/2 22
az h Cz
Xi ys zz
a\ hx C\ —\
0-1
-1
Xx yx Zx
Xi 2/2 Zi
xz yz zz
-1
0-
[815]
0-100
0-10
0-1
aiXi + a%yi + aaZi biXi + b^yi + baZi CiXi + dyi + dZi Xi yt Zi
aiXi-\:aiyi + aaZi biXt + btyi + baZi CiXt + c^yi + CaZa Xi yi z^
axXz-\-aiy%-^aaZa biXa + b^ya + bsZa CiXa + C^ya + CaZa Xa ya Za
(This last form is derived from the preceding by adding to the
first column ax times the 4th column + a^ times the 5th column +
az times the 6th column ; to the second column, bx times the 4th +
&2 times the 5th ^ bz times the 6th ; and to the third column, Cx times
the 4th + c% times the 5th + Cz times the 6th [812])
ax Xx + as 2/1 + (H Zx Ix Xx + &2 yi + bz zx Cx Xx + 62^1 + Cz Z\
ax Xi + a^yi + az z% bx Xi + b^y^ + bz z^ Cx Xi + Co 2/2 + Cz 22
ax xz + a^yz + az Zz bx xz + b^yz + bz Zz C\ X3 + c^yz + cz Zz
[805, R.].
Let the student observe how the elements of the first column of
the product are derived from the elements of the multiplicand and
multiplied ; then, how the elements of the second and third columns
are found. The laws which he will observe are general for determi-
nants of any order.
Example. — Sliow that, according to the laws above ob-
served.
X
^1 yi
^2 Vz
ax Xz + ttz 1/2 h ^% + ^2 ^2
EXERCISE 110.
Find the value of :
2 3
1 2
X
3 1
2 4
3.
12 3
3 1
2 3 1
X
2 13
13 2
12 3
1 3
2 -2
X
3
1 2
2 3
10
3 10
X
2 2 2
2 3
13 2
438 ADVANCED ALGEBRA.
Applications.
817. I. Solution of Simultaneous Equations of the
First Degree,
1. Solve aiX-\-'biy = ri (A)
«2^+^2 2/ = ^2 (B)
Solution : Multiply (A) by ^i and (B) by —A^ , in which ^i and
Ai are the eofactors of ai and as in the determinant [ ai h ], and
take the sum,
{ai Ai — tti Aii)x + {hi Ai — h Ai)y = ri Ai — r2 A^ (C)
Now, hi Ai-hA.i = [814, P.]. Therefore,
(ai A\ — tti Ai) X = ri Ai — n An ; whence,
aiAi — a^Ai [ ai os J ^ ' -■
Again, multiply (A) by Bi and (B) by — -Bs , in which Bi and
Bi are the eofactors of h\ and B^ , and take the sum,
(ai Bi - tti Bci)x + (bi Bi -h Bi)y = n Bi - n Bi (D)
Now, ai Bi-a2Bi = [814, P.]. Therefore,
2. Solve aiX-\-biy-\-CiZ = ri (A)
«2^+^2«/+^2^=^2 (B)
a3^ + hy + CsZ = rs (C)
Solution : Multiply (A) by J.i , (B) by —Ai, (C) by J3 , in which
Ai , Ai , and ^Is are respectively the eofactors of ai , as , and as in
the determinant [ ai fts Cs ], and add the resulting equations,
(ai Ai — tti Ai + as A3 )x + (bi Ai — bi At + bzAz)!/
+ (ci Ai — Ci Ai + C3 Az)z — n Ax — Ti Ai + r% Az
Now, the coefficients of y and z vanish [814, P.]';
^ ^ nA,-r,A,^nA, ^ [n t, ..]
ai J.1 — as J-s + as -43 [ ai 02 C3 J
Then, by symmetry,
^ [ ai 62 C3 ] [ ai 62 cz ]
In a similar manner it may be shown that, if we take n equations
of the first degree of the form of ai ic + 61 2/ + Ci 2 + . . . . = ri , mul-
MULTIPLICATION OF DETERMINANTS. 439
tiply the first by A\ , the second by —A2 , the third by Az , , in
which Ai , A2, A3, , are the cofactors ot ai , a^ , as , , and
take the sum, the coefficients of all the unknown quantities, except x,
will vanish, and we shall have
X = F — ^-^ — ^ j ; and, by symmetry,
[ai h cz ]' •"
[a, r, ^3....] ^^^^ Therefore,
^ [ai bi cz ]'
Principle, — Any unknown quantity in a complete sys-
tem of simultaneous equations of the first degree equals a
fraction whose denominator is a determinant formed from
the coefficients of the terms of the equations taken in order,
and whose numerator is formed from the denominator hy
replacing the coefficients of the unknown quantity hy the
corresponding right members of the equations,
EXERCISE 111.
Solve :
I. 3a;-|-2y = 16 2.6x — ^y= 6
'Zx — ^yz^ 2 2x-\-5y = 21
3. ax-]-by = c 4. (a-\-b)x — (c-\-d)y=: m
mx-\-ny — d (a-^b)x-{- (c — d)y = n
5. 2x-\-dy — 2z=5 6. x--y-\-z=6
dx-2y-{-4:Z=16 Sx-{-6y — Sz=14:
4:X — 3y— z=—6 2x-{-4:y + Sz = 20
1. ax-\-by-\- cz = d s, {a-\-b)x-\-by-\-az = m
cx-\-by-\-az = e ax-\-{a-{-b)y-{-bz = n
bx-\-cy-\-az = h bx-\-ay-\-{a-\-b)z = r
9. x-\-y-{-z-\-u = 14: 10. ax-{-by+cz=m
X — y~^z — u= — 2 bx-^ cy-\-au = n
x-\-y — z — u= — 4: cx-\-az-\-bu=p
X — y — z-\-u= ay-{-bz-\-cu = q
11, 2x^Sy + 2z-{- u = -12
3x-[-2y-dz-{-2u = 12 -
x — Sy-\-4:Z + 3u= — 24:
2x-]-2y-dz-4cU=z2S
440 ADVANCED ALGEBRA.
12. ax-\-'by-{-cz-\-du=p
ax — dy-\-cz — du = q
ax-\-'by ^ cz — du = r
ax^by — cz-\-du = s
13. 2x + 3y — 4:Z + 2u + 3v = 19
dx-2y-\-2z — du + 4:V = 13
2x — 4:y-\-3z — 2u + 2v= 5
ic+ y — dz-\-2u-\- v= 7
x-\-2y-{-dz — 4:U-\-5v = 2d
818. II. To determine under what condition (n + 1)
equations of n unknown quantities may he simultaneously
true.
Assume the equations
«i^ + ^iy + ^i = 0, (A)
«2^' + ^22/ + ^2 = 0, (B)
and «3 a^ + ^3 y + ^3 = (C)
to be simultaneously true.
Multiply (A) by d , (B) by - ^ , and (C) by Cs , in which C^ ,
Ci , and Oa are the cofactors of ci , ca , and Cs in the determinant
[ ai 62 C3 ], and add the results. Then,
(ai C — aid + tti Cs) X + (bi Ci — h Ci + h Cs) y
+ {ci Ci - Ci d + cs Cz) = [Ax. 2].
But, ai C — aid + «3 G = &i (7i — &2 Ci + bsC9 =
ci Ci -Cid + cz Cz = [ ai bi Cz] [814, P. ; 805, R.].
.'. [ «! bi C3 ] = is the condition under which (A), (B), and
(C) are simultaneously true.
Note. — The equation [ ai bi Ca ] = is called the eliminant of
the group.
In a similar manner it may be shown that n + 1 equations of n
unknown quantities, of the form of
ai a; + 61 y + . . . . + ri = 0,
are simultaneously true, when
[ai 62 C8 rn4i ] = 0.
Note.— Equations that are simultaneously true are said to consist^
that is, they are consistent.
MULTIPLICATION OF DETERMINANTS, 44I
EXERCISE 112.
Test the consistency of :
1. 2a; + 3y-13 =
2. 3a;-f-2y-17 =
bx — '^y— 3 =
2a;-5«/ + 15 =
819. III. To eliminate x from any two rational inte-
gral equations in x.
Ulustrations. — 1. Eliminate x from the equations
aoi?-\- bx-\-c =
ma^-\-nx-{-r =
Solution : It is evident that
ax^ + bx^ + ex =
ax^ + bx + c =
mx^ + nx^ + rx =0
and
mx'^ + nx + r =
are simultaneously true.
Therefore,
a b c
a b c
m n r
=
m n r
a
b
c
d
e
a
b
c
d
e
m
n
P
<1
m
n
P
?
m
n
P
?
(A)
(B)
2. Eliminate x from the equations
ax^-{-l)3^ -\- cx^ -\-dx-\- e =0
mQ?-\-noi? -\-p x-{-q=zO
Solution : It is evident that
ax^ + bx!^ + cx^ + dx^ -{■ ex =0
ax!^ + bx^ + cx^ + dx + e =
ma^ + nx* +px^ + qx^ =0
mx^ + nx^ + px^ + qx =0
ma^ + nx^ -\- px + q =
are simultaneously true. Therefore,
=
This method is known as Sylvester's Method of Elimination.
442 ADVANCED ALGEBRA,
Probabilities.
Definitions and Fundannental Principles.
820. When the number of ways in which an event may
occur is greater than the number of ways in which it may
fail, and the ways are equally likely to happen, we say :
1, The event is 'probable.
2, The event is likely to happen,
3, The chance is in favor of the event,
U* The odds are in favor of the event,
821. When the number of ways in which an event may
fail is greater than the number of ways in which it may
occur, and the ways are equally likely to happen, we say :
1, The event is improbable,
2, The event is not likely to happen,
3, The chance is against the event,
U, The odds are against the event,
822. When the number of ways in which an event may
occur is equal to the number of ways in which it may fail,
and the ways are equally likely to happen, we say :
1, The occurrence and failure of the event are equally
probable,
2, The event is as liJcely to happen as to fail.
3, There is an even chance for and against the event,
U' The odds are even for and against the event,
823. If an event can occur in a ways and fail in b
ways, and the ways are equally likely to happen, we need
more definite language to express the exact probability or
chance of the event. Thus, we say :
1, The odds are as a to b in favor of the event,
2, The odds are as b to a against the event.
PROBABILITIES. 443
824. If we let k represent the probability of any par-
ticular way happening, a the number of ways favorable to
the event, and h the number of unfavorable ways, then
will ah represent the probability of the event happening,
and h k the probability of its failing, and ak-\-hk, or
{a + h) k, certainty^ which is taken as the unit of measure,
then (a-\-V)k = \\ whence, k = , ;
and ak — -, , the probability or chance of the
event, and Ik — , , the probability or chance against
the event. Therefore,
Prin, 1, — The prolaMUty or chance of an event hap-
pening equals the number of favorable ways divided by the
whole number of ways,
Prin. 2. — The probability or chance of an event fail-
ing equals the number of unfavorable ways divided by the
ivhole number of ways,
825. Since an event is certain to happen or fail, and
certainty is expressed by unity, it follows that,
Prin, 3, — The probability of an event happening equals
unity minus the probability that it will fail j and the prob-
ability that it will fail equals unity minus the probability
that it will happen.
Illustration. — If there are 3 black and 2 white balls in
a bag containing only 5 balls, what is the chance,
1. That a black ball will be drawn on the first trial ?
2. That a black ball will not be drawn on the first trial ?
Solution : 1. There are 3 favorable ways out of 5 to draw a black
Q
ball ; therefore, the chance is -=• (Prin. 1).
3. There are 2 unfavorable ways out of 5 to draw a black ball,
namely, the two favorable ways for drawing a white ball ; therefore,
2
the chance of failing to draw a black ball is -g . Or,
444 ADVANCED ALGEBRA,
That a black ball will be drawn or not drawn on the first trial
Q
is certainty. The chance for drawing a black ball is ■^; therefore,
3 2
the chance of failure is 1 — -^ = -=■ .
826. Exclusive Events. — Two or more events are mu-
tually exclusiye when the happening of one of them pre-
cludes the possibility of any other one happening. Thus,
if a coin be thrown up, it may fall either head or tail. If
it fall head, or is supposed to fall head, it can not fall tail,
or be supposed to fall tail, in the same throw. Falling
head and falling tail are, therefore, mutually exclusive
events.
827. In a bag are d balls ; a of them are white, h blue,
c red, and the remaining ones yellow. What is the chance
of drawing, on the first trial,
1. Either a red or a white ball ?
2. A red, a white, or a blue ball ^
Solution : 1. The chance of drawing a red ball is -r , and the
chance of drawing a white ball is -^ ; and the chance of drawing
either a red or a white ball is — -j- = -j + -^ .
2. The chance of drawing a red ball is -^ ; of drawing a white
h
ball, -^ ; of drawing a blue ball, -^ ; and of drawing a red, a white, or
a blue ball, ^ ; which equals "^ + ^ + "^ • Therefore,
JPrin. 4, — The chance that one of several mutually ex-
elusive events will happen equals the sum of their separate
chances of happening.
EXERCISE 118.
1. "What is the chance of throwing 4 with a single die ?
Suggestion. — A die has six faces, which are equally liable to turn up,
but only one of these contains four dots. Therefore, the chance is -^ .
PROBABILITIES. 445
2. What is the chance of throwing an even number
with a single die ?
, — Three of the faces have an even number of dots;
3 1
therefore, the chance is ^ , or ^.
3. If the odds be 4 to 3 in favor of an event, what are
the respective chances of the success and failure of the
event ?
Suggestion. — There are 4 points out of 7 favorable and 3 out of 7
unfavorable to the happening of the event ; therefore, the respective
chances of success and failure are -y and -=- .
4. If 4 coppers are tossed, what are the odds against
exactly 2 turning up head ?
Suggestion. — Each coin may fall in two ways; hence, the four
coins may fall in 2* = 16 ways [550, Cor.]. The two coins that may
4x3
turn up head can be selected from the four coins in .^ , or 6 ways.
fi ^ —
Therefore, the chance of success is ir^ , or -5- ? and the chance of fail-
3 5 16 8
ure is 1 — -^ = ^ . Therefore, the odds are as 5 to 3 against the
event.
5. In a bag are 7 white and 5 red balls ; if two are
drawn, find the chance that 1 is red and 1 white.
12 X 11
Solution : Two balls can be selected from 12 balls in — r^ — = 66
if.
ways. One white ball can be selected from 7 white balls in 7 ways, and
1 red ball from 5 red balls in 5 ways. Hence, 1 white ball and 1 red
ball can be selected from 7 white and 5 red balls in 7 x 5, or 35 ways.
Therefore, 35 out of 66 ways are favorable to drawing 1 white and 1
35
red ball. Therefore, the chance is p^s .
66
6. Twenty persons take their seats at a round table.
What are the odds against two persons thought of sitting
together ?
Solution: Let the two persons be A and B. Besides the place
where A may sit, there are 19 places, two of which are adjacent to
him, and the remaining 17 not adjacent. Any of these B may select.
Therefore, the odds are as 17 to 2 against A and B sitting together.
446 ADVANCED ALGEBRA.
828. Expectation, — The value of any probability of
prize or property depending upon the occurrence of some
uncertain event is called an Expectation,
7. A person holds a tickets in a lottery in which the
whole number of tickets issued is n. There is only one
prize offered, and this is worth %p. What is the person's
expectation ?
Solution : It is evident that the n tickets are worth ^p, and that
the tickets are of equal value before the drawing; therefore, the a
tickets are worth — of $ jo, which is $ « x — . Therefore,
829. Prin, 5. — The expectation of an event equals the
product of the sum to he realized and the chance of the
8. A person is allowed to draw two bank-notes from a
bag containing 8 ten-dollar bills and 20 two-dollar bills.
What is his expectation ?
28 X 27
Solution : The two notes can be drawn from 28 notes in — r^ —
\jL
= 378 ways. Two ten-dollar notes can be drawn from 8 ten-dollar
8x7
notes in . = 28 ways. Two two-dollar notes can be drawn from
— 20 X 19
20 two-dollar notes in — ^ — = 190 ways.
\A
One ten-dollar note and one two-dollar note can be drawn from
8 ten-dollar notes and 20 two-dollar notes in 8 x 20 = 160 ways.
Therefore,
28
The chance of drawing $20 is 5^^, and the expectation is $ 1.482V*
190
The chance of drawing $4 is -^^^ , and the expectation is $2.01-iV5.
1 fiO
The chance of drawing |12 is 5^=5 , and the expectation is $5.07ff .
o7o
.'. The entire expectation is $1.48^ + $2.01 ^ + $5.07 1 = $8.57 y.
9. A bag contains a £5 note, a £10 note, and six pieces
of blank paper of the same size and texture as a bank-note.
Show that the expectation of a man who is allowed to draw
out one piece of paper is £1 17^. 6d,
PROBABILITIES. 447
830. Independent Events, — Two or more events are in-
dependent of each other when the happening of one of
them does not affect the probability of any other one's
happening.
10. There are h balls in one bag, a of which are white ;
d in another, c of which are white ; and / in another, e of
which are white. Show that the chance of drawing one
white ball from each bag in a single trial is -r X ^ X ^ .
Solution : One ball can be drawn from each bag in b x d x f ways
[550]. One white ball can be drawn from each bag in a x c x e ways
[550]. Therefore, the chance of drawing a white ball from each bag
. a X c X e ^^.. T^ ^^ a c e _,, .
■« hlTdlTf t^^*' P- 1] = 6 "^ d >< 7 • T'^^'^'o^^'
831. JPrin. 6, — The chance of two or more independent
events happening simultaneously is the product of their
several chances of happening,
832. Car. 1, — The chance of two or more independent
events failing simultaneously is the product of their several
chances of failing,
833. Cor, 2. — The chance of one of two independent
events failing and the other happening is the product of
the chance that one fails and the chance that the other
happens.
11. A can solve 3 problems out of 4, B 5 out of 6,
and C 7 out of 8. What is the chance that a certain
problem will be solved, if all try ?
Solution : Unless all fail, the problem will be solved. The chance
that A will fail is i , that B wUl fail -^ , that C wiU fail g , that aU
will fail -J X -^ X -^ = Yq^ • Therefore, the chance of success is -^ ,
834. Dependent Events. — In a series of events, any
assumed event is said to be dependent upon a preceding
4AS ADVANCED ALGEBRA.
event, if the happening of the preceding event changes the
probability of the happening of the assumed event.
12. Find the chance of drawing 3 white balls in suc-
cession from a bag containing 5 white and 3 red balls.
Solution : The chance of drawing a white ball on the first trial
K
is -5- . Having drawn a white ball, there remain in the bag 7 balls,
o
4 of which are white. The chance of drawing a white ball on the
4
second trial is therefore -=- . Similarly, the chance of drawing a white
Q
ball on the third trial is -k . Therefore, the chance of drawing three
5 4 3 5
white balls in succession is q- x -=- x -^ [831] = -^r^ . Therefore,
835. Prin, 7. — The chance that a series of events
should happen is the continued product of the chance that
the first should happen, the chance that the second should
then happen, the chance that the third should follow, and
so on,
13. In one of two bags are 3 red and 4 white balls, and
in the other 5 red and 3 white balls, and a ball is to be
drawn from one or other of the bags. Find the chance
that the ball drawn will be white.
Solution : The chance that the first bag will be chosen is -^ . Then,
a
4
the chance of drawing a white ball from the first bag is -y ; hence, the
14 2
real chance of drawing a white ball from the first bag is -^ of -=r = -=^ .
Similarly, the chance of drawing a white ball from the second bag is
13 3
— of -Q = T5 • These events are mutually exclusive ; therefore, the
\ ^ ^ . 3 3 53
chance required ^^ -y + jg = htr •
836. Inverse Probability. — When an event is known
to have happened from one of two or more known causes,
the determination of the chance that it has happened from
PROBABILITIES. 449
any particular one of these causes is a problem of inverse
prohalility,
14. It is known that a black ball has been drawn from
one of two bags. The first of these bags contained in balls,
a of which were black, and the second n balls, b of which
were black. What is the chance that the ball was drawn
from the first bag ?
Solution : Suppose that 2 N drawings were made. The chance is
that N were made from each bag. In the N drawings from the first
bag the chance is that ~ x N were black balls. In the drawings
from the second bag the chance is that — x iV were black balls.
Therefore, in 2 JV drawings, the chance is that ( — H jN were
black balls. Therefore, the chance that a black ball was drawn from
the first bag is (— xiV") -^(— + —) N = — "'\ .
837. Theorem, — If an event is believed to have been
produced by some one of the causes Pi, Pg? A^ P»>
which are mutually exclusive, and Pi, Pzy Pzy Pn rep-
resent the respective probabilities of these causes when no
other causes exist, then the probability that P, produced
the event is -^ t—^ "-p^ i-^ — .
Demonstration. — Let N be the number of trials made in produc-
ing the event. The first cause operated N x Pi times ; therefore,
on the supposition that no other causes operated than those named,
the probability that the event was produced by the first cause is
iV X Pi X pi. Under similar restrictions, the probability that the
event was produced by the second cause is iV x Pg x jsa ; by the third
cause, N X Pz x ps; by the rth cause, N x Pr x pr ; by any one of
the causes, N{Pipi + P^p^ + Psps + + P«i?«). Therefore, the
real chance of its having been caused by the rth cause, or P, , is
N X Pr X Pr _ Pr X Pr
N{Pipi + Ps^s +.... + PnPn) ~ P\ P\ + P^Pi + + PnPn'
15. Four bags were known to contain 3 red and 4 white,
4 red and 3 white, 5 red and 1 white, and 4 red and 4
450 ADVANCED ALGEBRA.
white balls respectively. A white ball was drawn at ran-
dom from one of the bags. Find the chance that it was
drawn from the second bag.
Solution: Pi =P2 = Ps = P4 = i, i?i = y, i>2 = |, Pi = \,
and i?4 = o" . Therefore, the required probability is
1 ^ i
4^7 T 9
2_/4 i 1 1\ :5
4V7 "^ 7 "*■ 6 ■*■ 2/ 3
35
838. Probability of Testimony.— HhQ following exam-
ples illustrate how to deal with questions relating to the
credibility of testimony :
16. A speaks the truth a times in W2, B J times in n,
and c times in r. What is the chance that a statement
is true which all affirm ? Which A and B affirm and
C denies ?
Solution : 1. The statement is either true or false. If true, all have
spoken the truth ; the probability of which is — x — x — = .
^ ' ^ •' m n r mnr
If false, all have lied ; the probability of which is
\ m)\ n)\ r) mnr
Hence, the probability of the truth of the statement is,
ahc ^ {abc (m—a){n—b)(r—c))_ abc
mnr ' {mnr mrir ) ~ abc + {m—a){n—b)(r—c)'
2. If the statement is true, A and B have told the truth and C
has lied ; the probability of which is — x — xfl ) = — ^^ ,
^ ■' m n \ r J mnr
If the statement is false, A and B have lied and C has told the truth ;
the probability of which is fl- ») (l- * ) (^) = '"'-"»"-^>' .
^ J \ mj\ nj\rj mnr
Hence, the probability of the truth of the statement is,
ab{r—c) ^ \ab{r—c) {m—a){n—b)c\_ a b (r—c)
mnr ' { mnr mnr ) ~ ab{r—c) + {m—a){n—b)c
17. A, B, and tell the truth to the best of their
knowledge and belief. A observes correctly 4 times out
of 5, B 3 times out of 5, and C 5 times out of 7. What
is the probability that a phenomenon occurred (which was
PROBABILITIES. 451
just as likely to fail as to occur), provided all had equal
opportunity of observing, and all report its occurrence ?
What if A and B report its occurrence and its failure ?
Solution : 1. The phenomenon either occurred or failed. If it oc-
curred, A, B, and C observed correctly ; the probability of which is
-^ X -^ X -=-. The inherent probability that it would occur is -^ .
o o 7 -©
Hence, the probability that the assumption that it occurred is correct
• 1 1 1 A -A
is 2 X g X g X ^ _gg.
If it did not occur, all observed falsely ; the probability of which
12 2
is -^ X -=■ X ■=■ ; and the probability of the correctness of the assump-
11222
tion that the phenomenon failed is7rX-=-x-=-x-=- = r-— , Hence,
2 5 7 175
the chance that the phenomenon occurred is 5^ -^ ( ^^ + 7^ ) = -- .
do \35 175/ 16
2. The probability of the correctness of the assumption that the
, . 1 4 3 2 12
phenomenon occurred iS7rX-=-x-=-x-7=- = 77==.
^ 2 5 5 7 175
The probability of the correctness of the assumption that the
, ..,,.1125 1
phenomenon failed \s -z -k -^ y. -^ y. -p^ = ^.
Hence, the chance of the event is ^^ -«- f r^ "^ s^) ~ 17*
Note. — For a fuller treatment of Choice and Chance than space will
permit to give in this book, see Whitworth's " Choice and Chance."
EXERCISE 114.
1. If A's chance of winning a race is — and B's chance
1 17
■Z-. show that the chance that both will fail is rr.
o 24
2. If the odds be m to w in favor of an event, show
that the chance of the event is — ; — , and the chance
m-{-n
against the event is — ; — .
7n-\-n
3. If the letters e, t, s, n be arranged in a row at ran-
dom, show that the chance of having an English word is — .
M O
452 ADVANCED ALGEBRA.
4. Show that the chance that the year 1900 + 2?, in
which X < 100, is a leap-year, is ^ .
5. A draws 3 balls from a bag containing 3 white and
6 black balls ; B draws 1 ball from another bag containing
1 white and 2 black balls. Show that A's chance of draw-
ing a white ball is to B's chance as 16 to 7.
6. Show that when two dice are thrown the chance that
the throw will amount to more than 8 is -5 .
lo
7. Show that the chance of throwing exactly 11 in one
throw with two dice is :7^ .
lo
8. One purse contains 5 sovereigns and 4 shillings ;
another contains 5 sovereigns and 3 shillings. Show that
the chance of drawing a sovereign is —rj , if a purse is
selected at random and a coin drawn from it at random.
Show that the expectation of the privilege is 125. 2 Yig d,
9. There are three independent events whose several
chances are ^, -r-, and ■^. Show that the chance that
one of them will happen and only one is j^ .
10. If two letters are taken at random out of esteemed,
show that the odds against both being e are the same as
the odds in favor of one at least being e,
y. 11. A letter is taken at random out of each of the
words choice and chance. Show that the chance that they
are the same letter is — .
b
12. A bag contains 6 black and 1 red ball. Show that
the expectation of a person who is to receive a shilling for
every ball he draws out before drawing the red one is 3
shillings.
PROBABILITIES. 453
13. Two numbers are chosen at random. Show that
the chance is ^ that their sum is even.
14. An archer hits his target on an average 3 times out
of 4. Show that the chance that he will hit it exactly
27
3 times m 4 successive trials is 77-7 .
15. A box contains 10 pairs of gloves. A draws out a
single glove ; then B draws one ; then A draws a second ;
then B draws a second. Show that A's chance of drawing
a pair is the same as B's ; and that the chance of neither
^ . . . 290
drawmg a pair is ^^ .
16. Show that with two dice the chance of throwing
more than 7 is equal to the chance of throwing less than 7.
17. Two persons throw a die alternately, with the under-
standing that the first who throws 6 is to receive 11 cents.
Show that the expectation of the first is to that of the
second as 6 to 5.
18. A's chance of winning a single game against B is ^ .
Show that his chance of winning at least 2 games out of 3
. 81
^^-25-
19. A party of n persons take their seats at random at
a round table. Show that it is w — 3 to 2 against two
specified persons sitting together.
20. Show that the chance that a person with 2 dice
will throw double aces exactly 3 times in 5 trials is
\m) ^ \36J
X 10.
21. There are 10 tickets, five of which are numbered
1, 2, 3, 4, 5, and the rest are blank. Show that the prob-
ability of drawing a total of ten in three trials, one ticket
33
being drawn each time and replaced, is r^r^ .
SUPPLEMEl^T.
COJ^TIJfUED FBACTIOJfS,
I. Definitions.
839. An expression in the form of
a
h-{-c
d-\-e
/+ etc.,
is a Continued Fraction,
840. The discussion in this section will be limited to
continued fractions in the form of
1
«4-i
J+i
c+ etc.,
and -, -7-, -, etc., will be called Partial Fractions,
a' h c
841. A continued fraction may be written in a more
convenient form, as follows :
111 ill
842. When the number of partial fractions in a con-
tinued fraction is finite, it is a terminating continued frac-
tion ; when infinite, an interminate continued fraction.
843. If at some stage in an interminate continued frac-
tion one or more partial fractions begin to repeat in the
same order, it is called a periodic continued fraction.
CONTINUED FRACTIONS. 455
844. A periodic continued fraction is pure when it
contains no other than repeating partial fractions, and
mixed when it contains one or more partial fractions be-
fore the repeating ones.
rrt. 1 1 1 1
1 1
is a pure periodic fraction ;
11111
1 i i
is a mixed periodic fraction.
845. The fraction resulting from stopping at any stage
is called a convergent,
2. The Formative Law of Successive Convergents.
846. In the continued fraction
111 1 1 1 i
— = the first convergent.
- , -T = r , -. = the second convergent.
- , -T , - = . , , ..x — i — = the third convergent.
a -{■ b -{- c {ab-\-l)c-{-a ^
It will be seen that
t The numerator of the third convergent is the numer-
ator of the second convergent multiplied by the denomina-
tor of the third partial fraction, plus the numerator of the
first convergent; and
2. The denominator of the third convergent is the de-
nominator of the second convergent multiplied by the de-
nominator of the third partial fraction, plus the denomi-
nator of the first convergent.
456 ADVANCED ALGEBRA.
Will these laws hold true in the formation of any con-
vergent from the two preceding convergents ?
P O Tf ^
Let p-, -~f -^, and -^ be respectively the (/i — 2)th,
^1 Vi -^1 ^1
{n — l)th, nth, and {n + l)th convergents ; and p, q, r,
and s the denominators of the (n — 2)th, {n — l)th, nth,
and (?i + l)th partial fractions.
Suppose the laws to hold true in the formation of the
convergent ^^ , then will ^ = 7Q^, ' (^)
Now, from the nature of the continued fraction, -^ may
1 R ^1
be formed by putting r-\ — f or r in -^ . Therefore,
S 2il
^_ (r+l)8+i-
(«r + l)e +^
?P
(sr + l)e. + ^
sA
sB +Q
_ ^(re+p)+e
Therefore, if the laws are applicable in the formation
of the nth. convergent, they are also applicable in the
formation of the {n + l)th convergent. But we have
seen that they do apply in the formation of the third
convergent, and, hence, apply in the formation of the
fourth convergent, and so on. Therefore, in general,
1. The numerator of the nth convergent equals the
numerator of the (n — l)th convergent multiplied by the
denominator of the nth partial fraction, plus the numera-
tor of the {n — 2)th convergent ; and
2. The denominator of the nth convergent equals the
denominator of the (n — l)th convergent multiplied by the
denominator of the nth partial fraction, plus the denomi-
nator of the (n — %)th convergent.
PROPERTIES OF CONVEROENTS, 457
Example. — Find the first 8 convergents of the con-
tmued fraction ^^3_^i^4^5^^ + 4 + 3-
Solution :
R _ rQ -\-P _\ i i 1? ^ ?1? ^ 3118
Ri~ rQx-^Px~ 2' 7' 9' 43' 324' 491' 2188' 7055"
Properties of Convergents.
847. Take the continued fraction
_1 1 1 1
«zx.ll .1>..11
whence, i^i<2/
I, .1 1^.111
whence, «+ - _^ - < « + -_^-^-_^.. . . ;
and - , -T , - > Vi etc. Therefore,
Prin. 1, — The successive convergents are alternately
greater and less than the continued fraction {the odd orders
being too great and the even orders too small).
848. The difference between the first two conyergents
= 5— 7-T = , , . ^x = unity divided by the prod-
a aJ + 1 a (a 5 + 1) ^ ^ ^
uct of their denominators. Is this a general law ?
P Q J?
Let -^f -TT, and ^- be the (w— l)th, wth, and {n-\-l)th.
convergents, and p, q, and r the denominators of the
458 ADVANCED ALGEBRA.
{n — l)th, nth, and {n -f l)th partial fractions, and let '^
denote difference hetioeen.
Assume -— r^ -^ = — „ ^ — - = „ .. ; then will
Ri'^ Qi QiRi QiRi
_ PQ.^QPr _ _j_
" Gi^i "ft A* ^^
Therefore, if the law holds good for the difference be-
tween the {n — l)th and wth convergents, it will also for
the difference between the {n + l)th and the wth conver-
gents. But we have seen that it does hold good for the
difference between the first and second convergents, and,
hence, it will for the difference between the next higher
pair, and so on. Therefore,
JPrin. 2, — The difference between any two consecutive
convergents equals unity divided by the product of their
denominators.
849. Since PQ^'^QP^ = 1 [848, A], P and P^ can
not have a common factor, neither can Q and Q^.
Therefore,
Prin, 3, — Bvery convergent is in its lowest terms.
850. If we let ^=- represent the true value of the con-
tinned fraction ; then will
P U ^P Cmn 1 ^ Q ^ P Q
''' p[-u.<p;Q.^^'''^''''''urQ;^p;Qr'
Hence, if either -^ or ^ be used for -^, the error will
be less than p ^ , or less than -^ .
-^1 Vi VI
PROPERTIES OF CONVERGENTS, 459
P Q R
851. Let -^, -jr, and -^ be three consecutive con-
-t 1 vi ^1 -^ ^
versrents whose terminal partial fractions are -, -, and
1 U , P ^
— : and -77- , the true value of the continued fraction.
r Ui
TT 7? 1
Then, y^ differs from ^- only in the use of r H — — etc.
t/i ^ III s-\- ,
for r. Put r H r etc. = a;.
s-\r
1
ftC^ft + i'i)
, P U _P_ xQ +P _ x{PQ^-P,Q)
a;
A (a: ft + Pi)
Now, a; > 1, and Pi < ft ,
1 ^ a;
< p /^ /. . D X ; or.
•• ft(^ft + i^i) ^A(^ft + A)
-^ is nearer -^ than is ^-. Therefore,
vi ^1 -^1
2*Hn. 4. — The MgJier the order of a convergent the
nearer does it approach to the true value of the continued
fraction.
852. Cor. — A continued fraction is the limit of its con-
vergent s ; or, if y be a continued fraction and x its vari-
able convergent, y = lim. x.
853. The denominators of successive convergents in-
crease more rapidly than their numerators [846 ; 1, 2] ;
therefore, of any two convergents, that is the greater
which has the greater denominator. But may there not
460 ADVANCED ALGEBRA.
be some other fraction, not a convergent, with smaller de-
denominator, that is a nearer approximation to a continued
fraction than a given convergent ?
Suppose -Yjr not a convergent, and nearer to -^ than ■—,
^^ U M Q U ^' ^'
and ifi < §1 ; then — ^ -^ < ^ ^ -^ ;
M P ^ Q P MP^^M^P
/^
< -^ -- D- ; or, TT-B <
But MtPi < QiPi, since M^ < ft.
,'. MPi '^ MiP < 1; which is impossible, since M,
Ml, P, and Pi are integral. Therefore,
Brin* 5, — Any convergent is nearer the true value of
a continued fraction than any fraction with smaller de-
nominator.
Problems.
851. 1. To reduce a common fraction to a terminating
continued fraction.
Since an improper fraction is equivalent to an integer
and a proper fraction, it will be necessary only to investi-
gate a method for extending a proper fraction.
Let - = a proper fraction in its lowest terms.
Divide both terms by h, and put for the improper frac-
a c
tion -T , the mixed number p-\- -r] then.
h
a
1
c
1
Divide both terms of -r
by c, and
put
h _
c
= ^ +
d
n, J
a
1
1
q^d
c
PROPERTIES OF CONVERGENTS. 461
d c p
Again, divide both terms of - by d, and put -7 = r + 77 ;
then, _5 ^ 1 1 1
a ~ p -\- q -\- r -\- e_
d
It will now be seen that the denominators of the suc-
cessive partial fractions have been obtained as follows :
b) a {p
bp
11.
d) c (r
rd
e etc.
Since a and I are integral, they have a highest com-
mon divisor, and the division will eventually terminate.
Therefore, the continued fraction will be a terminat-
ing one,
Rule. — To reduce a proper fraction to a terminating
continued fraction, find the highest common divisor of its
terms by successive division, and use the quotients in regu-
lar order for the denominators of the partial fractions,
855. 2. To reduce a quadratic surd to a continued
fraction.
Ulnstrations. — 1. Eeduce ^26 to a continued fraction.
Solution : V36 = 5 + —
= 10 + :4— . = 10 + "^
10 + 1 ^ 10 + 1
10 + Jl_
X
■• '»^ = «+fo + K + ^+--=5+^
462 ADVANCED ALGEBRA.
2. Eeduce a/iQ^ to a continued fraction.
Solution : ^19 = 4 + —
X
Vl9-4 3 ccj
3 Vl9 + 2 , 1
iCi = —y= = - — = = 1 + —
5 ^19"+ 3 „ 1
ars = -7= = - — s = 3 + -
VlQ - 3 2 iCs
Scholiwm, — A quadratic surd may always le reduced
to a periodic continued fraction if the expansion is carried
sufficiently far,
856. 3. To reduce a periodic continued fraction to a
simple fraction.
The periodic continued fraction
ili_lll __
p + q -h r ~ p-\- q -i- r -\-x~
qr + qx-^1 _^
pqr-\-pqx-\-p-{-r-\-x
whence, {pq-{-l)x^-\-ipqr — q -\-p + r)x = qr-^-l
The value of x found from this equation is the value
of the continued fraction.
857. 4. To approximate the ratio of two numbers.
Example. — When the diameter of a circle is 1, the cir-
cumference is 3 •1415926 + . Approximate the ratio of the
diameter to the circumference.
Solution :
-I oi^iKno« 10000000 1111 rp^ , .,
1 : 3.1415926 = 3^jj^= 3 ^ y ^ j^ ^ J ^.... [Prob. 1].
The successive convergents, which are also the successive approxi-
^ ^, ^. 17 106 113 ,
mationsof the ratio, are : -g, ^, ^, g^g, etc.
PROPERTIES OF CONVERGENTS, 46_3
EXERCISE 1 18.
Reduce to continued fractions :
125
^' 317
140 100 106
^' 213 ^" 999 ^' 729
6. ViO
6. Vl2 7. a/30 8. V57
9. -3183
10. 3-1416 11. 67°, 20', 30"
Find the successive convergents of :
"•UJ +
^ ^ 13 ^ ^ ^ ^
4+5 3+1+9+1
i i
"■2 + 3
15 ^ ^ ^ 16 ^ ^
Find the true value of :
17. -r 18.
4
i i _Q 1 i „ i 1 i
2 + 4 *^" 2 + 4 ^"-1 + 2 + 1
21. Find a series of common fractions converging to
1: y/J.
22. Express approximately the ratio of a liquid quart
(57*75 cu. in.) to a dry quart (67*2 cu. in.).
23. The square root of 600 is 24*494897, and the cube
root of 600 is 8*434327. Find a series of four common
fractions approximating nearer and nearer to the ratio of
the latter to the former.
24. The imperial bushel of Great Britain contains
2218*192 cu. in., and the Winchester bushel 2150*42 cu.
in. Find the nearest approximation, that can be expressed
by a common fraction whose denominator is less than 100,
of the ratio of the latter to the former.
25. Two scales of equal length having their zero points
coinciding also have the 27th gradation of the one to coin-
cide with the 85th gradation of the other. Show that the
7th and 22d more nearly coincide than any other two gra-
dations.
464 ADVANCED ALGEBRA.
THEORY OF JfUMBEBS.
Systems of Notation.
I. Definitions.
858. Notation is the art of expressing numbers by
means of characters.
869. A system of notation is a method of expressing
numbers in a series of powers of some fixed number.
860. The order of progression on which any system of
notation is founded is called the scale of the system, and
the fixed number on which the scale is based is called the
radix,
861. Any integral number, except unity, may be taken
as the radix. When the radix is two, the scale and system
are called binary ; when three, ternary ; when four, qua-
ternary ; when five, quinary ; when six, senary ; when
seven, septenary ; when eight, octary ; when nine, nonary ;
when ten, denary or decimal ; when eleven, undenary ;
when twelve, duodenary ; etc.
862. In the decimal or denary system,
56342 = 5x10,000 + 6x1000 + 3x100 + 4x10 + 2 =
5Xl0* + 6xl03 + 3Xl02 + 4xl0 + 2; ,
or, in inverse order,
2 + 4x10 + 3x102 + 6x10^ + 5x10*.
In the octary system,
34725 = 3 X8* + 4X83 + 7x88 + 2x8 + 5;
or, in inverse order,
5 + 2x8 + 7x82 + 4x83 + 3x8*.
SYSTEMS OF NOTATION. 465
.*. In general, if r be taken as the radix, and
ttQ, a-i, «3, «3 . . . . ««_!
as the n digits of a number, reckoning in order from right
to left, the number is represented by
«•-! y-""^ + a^«-2 ^""^ + ««-3 ^""^ + .... + «2 r^ -}- «i r + «o
863. Theorem, — Any integral number may he expressed
in the form of
ar^'-^-h r"~^ + ^ ^*~^ + -\-p r^-\-qr + s,
in which the coefficients are each less than r.
Demonstration : Let N equal the number of units in any number,
and r" the highest power of the radix less than N.
Divide N by r", and let the quotient be a and the remainder N',
Then N=ar'' + N'.
Now, a is less than r, else r" would not be the highest power of
r less than N; and JV' is less than r".
Divide N' by r«-^ and let the quotient be b and the remainder
iV". Then N' = Ir^-"^ + iV", in which 6 < r and N" < r«-i.
In like manner, divide N" by r*-^, and let the quotient be c, and
the remainder N'". Then iV" = cr'-^ + N"\ in which c <r and
N'" <r*-8.
If this process be continued, a remainder, 5, will eventually be
reached less than r. Therefore, JV= ar* + 6r*— * + cj*—''^+
+ pr^ + qr + s, in which the coeflacients are each less than r.
864. Car, — In any system of notation, the number of
digits including is equal to the radix,
866. Problem. To express a given number in any pro-
posed scale.
Solution : Let N be the number and r the radix of the proposed
scale.
Suppose JV= ar" + &r«-» + cr«-2 + + pr^ + qr + s, it is
required to find the values of a, b, c, p, q, s.
N 8
— = ar^-^ + br^-^ + ci-*^—^ + + pr + q ■{ — .
r T
Therefore, the remainder, after dividing N by r, is the last digit.
Suppose N' = ar^-i + br^—^ + cr*-^ + + ^r + q.
— = ar»— 2 + 6r"— 3 + cr^—^ + +«? + —.
r r
Therefore, the remainder, after dividing N' by r, is the next digit.
4:6Q ADVANCED ALGEBRA.
Suppose iV" = ar"-2 + 6r"-' + cr*-^ + + ^,
N" n
= ar»-3 + 6r«— * + cr«-6 + + ^.
r r
Therefore, the remainder, after dividing N" by r, is the next digit.
Etc., etc., etc.
Therefore,
Hvle, — Divide the numher ly the radix, then the quo-
tient hy the radix, and so on until the quotient becomes less
than the radix ; the successive remainders will he the digits
of the number, beginning with the units.
Illustrative Examples. — 1. Express 35432 (denary scale)
in the senary scale ; also, 35432 (senary scale) in the octary
scale.
(1) 6 )35432 (2) 8 )35432
6 )5905 -2 8 )2545 -4
6 ) 984 -1 8 )212 -1
6 ) 164 -0 8) 14-0
6)_2_7-2 1-2
4-3 .-. 35432,. = 6 = 12014r=8
.-. 35432r = io = 432012^ = 6
Explanation of (2) :
35 -T- 8 = (3 X 6 + 5) -5- 8 = 23
8 = 2, and 7 over ;
74-^8 = (7x 6 + 4) -^8 = 46^8 = 5, and 6 over;
63 -f- 8 = (6 X 6 + 3) -^ 8 = 39 -^ 8 = 4, and 7 over;
72-f"8 = (7x6 + 2)-s-8=44-^8 = 5, and 4 over.
2. Express 35439 (denary scale) in duodenary scale ;
also, 34439 (nonary scale) in un denary scale.
Kote-^The undenary scale needs a character to represent ten, and
the duodenary scale two characters to represent ten and eleven. We
will represent ten by t and eleven by e.
(1) 12 )34439 (2) 11 )35439
12 ) 2 8 6 9 - e 11 ) 2852 -5
12 )239 -- 1 11 )236 -8
12}J^-e 11)18-8
1-7 1-6
.-. 34439r = io = 17eler=i2 .-. 35439^ = 9 = 16885^ = 11
SYSTEMS OF NOTATION. 467
EXERCISE 116.
1. Find the sum in senary scale of 4532^ = 6,
3452r = 6, 5423^ = 6, and 3251^ = 6
2. Find the difference (octary scale) of 3574^ = 8 and
2756r = 8
3. Multiply 36425r = 7 by 8; also 26436,. = 8 by 10
4. Divide 4765^54^ = 11 by 9; also 2e58if3r = i2 by 11
5. Express 43250^ = 5 in the denary scale.
6. Express 38472r = 9 in the septenary scale.
7. Express 35243^ = e in the duodenary scale.
8. Express Set 950r = 12 in the quaternary scale.
9. Find the sum (denary scale) of 3472^ = s and 5842^ = 10
10. Find the difference (nonary scale) of 5 ^34^ = 11 and
6432r = 7
11. What is the radix of the scale in which 476^ = 10
= 2112?
Suggestion.— Let r = the radix ; then will 2r' + r^ + r + 2 = 476.
12. In what scale is 3 times 134 = 450?
13. In what scale is 135^ = 6 = 43 ?
14. What is the H. C. D. of 36^ = 8, 48^ = 8, and 60r = 8 ?
15. Multiply 28r = 9 by 45r = 9; also square 25^ = 6
16. In what scale is 1552 the square of 34 ?
17. Show that 35, 44, and 53 are in arithmetical progres-
sion in any scale of notation.
18. Show that 1331 is a perfect cube in any system of no-
tation.
19. Show that 14641 is a perfect fourth power in any sys-
tem of notation.
20. Show that 11, 220, and 4400 are in geometrical pro-
gression in any system of notation.
468 ADVANCED ALGEBRA.
Divisibility of Numbers and their Digits.
866. Theorem I. — If a number ^ Ny he divided hy any
factor of r, r^, r^ etc., respectively {r being the radix), it
will leave the same remainder as when the number ex-
pressed by the last term, the last two terms, the last three
terms, etc., is divided by the same factor.
Dem(mstration : Suppose x a factor oi r, y a factor of r^, and z a
factor of r^, etc.
iV= ar«-* + Jr~-2 + + pr^ + qr + s.
Now, X is certainly a factor of every term of N, except s; y, a,
factor of every term, except qr + s; and z, a factor of every term,
except pr^ + qr + s, etc. Therefore,
N s
1. — = an integer h — .
rt iV . , qr + 8
2. — = an integer + .
N^ t)r^ + <7 7* 4- s
3. — = an integer + ; which was to be proved.
z z
867. Car, — In the decimal system of notation,
1. A number is divisible by any factor of 10, if the
units^ digit is divisible by that factor,
2. A number is divisible by any factor of 100, if the
number expressed by the last two figures is divisible by
that factor.
S. A number is divisible by any factor of 1000, if the
number expressed by the last three figures is divisible by
that factor.
868. Theorem II. — The difference between a number
and the sum of its digits is divisible by the radix less one.
Bemonstration :
Let N =z ar"— > + &r»-2 + + pr^ -^ qr ■{• s = any number;
then, a + 6+....4-jp + g' + 5 = the sum of the digits.
Now, a (r«-» —1) + b (r»-9 -1) + + p (r^— 1) + q (r— 1) = the
difference between the number and the sum of its digits, and every
term is divisible by r — 1.
DIVISIBILITY OF NUMBERS, 469
869. Cor. — In the decimal system of notation,
The difference letween a number and the sum of its
digits is divisible by 9 or S.
870. Theorem III, — A number, N, divided by r—1,
leaves the same remainder as the sum of its digits divided
by r — 1, r being the radix.
Demonstration. — Put s for the sum of the digits; q and q' for
the quotients; and c and c' for the remainders.
1. N= q{r-l) + c
2. s = q' (r — l) + c'
.*. N—s = (q — q') (r — 1) + (c — c').
Now, iV — s is divisible by r — 1 [T. II], and (q — q') (r—1) is
evidently divisible by r — 1; therefore, c — c' is divisible by r — 1.
But c and c' are each less than r — 1 ; hence, c — c' = 0, or c = c',
871. Car, — In the decimal system,
A number is divisible by 9, if the sum of its digits is
divisible by 9.
872. Theorem IV, — If from a number, N, we sub-
tract the digits of the even powers of r, and add those of
the odd powers, the result will be divisible by r + 1.
Demonstration. — Let iV= ar* + br^ + cr^ + dr + e.
Add —a + b — c + d — e^
then, a(r* — 1) + b (r^ + 1) + c{r^ — 1) + d{r + 1), the result, is divis-
ible by r + 1, since every term is divisible by r + 1.
873. Theorem, V, — If a number, N, be divided by
r + 1, the remainder will be the same as when the differ-
ence between the sums of the digits of the even and odd
powers of r is divided by r-\-l.
Demonstration.— Put d for the difference between the sums of the
digits of the even and odd powers of r ; g and q' for the quotients ;
and c and c' for the remainders ; then will
iV= g(r + 1) + c,
and d = q* {r + \) + c'.
.-. N-d = {q-q'){r + l) + c-c'.
Now, N — d is, divisible by r + 1 [T. IV], and {q — q') (r + 1) is
evidently divisible by r + 1. Therefore, c — c' is divisible by r + 1.
But c and c' are each less than r + 1 ; hence, c — c' = 0, or c = c'.
470 ADVANCED ALGEBRA.
874. Cor, — In the decimal system of notation,
A number is divisible by 11, if the difference between
the sums of the digits in the even and odd places is divis-
ible by 11,
Even and Odd Numbers.
875. An even number is a number that is exactly di-
yisible by 2.
876. An odd number is a number that is not exactly
divisible by 2.
877. If we let x represent any integral number in-
cluding zero, and regard zero as an even number, it be-
comes evident that the general formula for an even num
ber is 2 x, and for an odd number 2 ic + 1.
878. Theorem, I, — The sum, of any number of even
numbers is even.
Demonstration. — Let 2 a:i , 2 iCa , 2 X8 , 2xn represent n even
numbers ; then will their sum be
2a;i + 2a:a + 2a:8 + + 2xn = ^{xi + x^ -^ Xt + + a^n)
an even number.
879. Jlieorem II. — The sum of an even number of
odd numbers is even.
Demonstration.— Let 2a;i + 1, 2a;a + 1, 2a;8 + 1 + + 2a;an + 1
represent 2» odd numbers ; then will their sum be
(2a;x + 1) + {2xu + 1) + (2xs + 1) +.... + (2x2n + t) =
2xi +2Xi + 20^8 + + 2a;an+ 2n =
2 (a^i + a:a + iCa + . . . . + x^n + n), an even number.
880. Theorem III, — The sum of an odd number of
odd numbers is odd.
Demonstration. — Let 2xi + 1, 2ira + 1, 2a;8 + 1, 2a;9«+i + 1
represent 271 + 1 odd numbers ; then will their sum be
(2a;i + 1) + (2a;a + 1) + (2a:8 + 1) + . . . . + 2a;3«+i + 1) =
2xi + 2a;a + 2a;8 + + Sa^sn+i + 27i + 1 =
2{xi-\-Xi + Xf\- + x%n+\ + n) + 1, an odd number.
EVEN AND ODD NUMBERS. 471
881. Theorem IV, — The sum of an equal even number
of even and odd numbers is even.
Demonstration.— Let (2xi + 1) + (2a;a + 1) + . . . . + (3a:an + 1) =
the sum ot %n odd numbers ;
and 2a;'i + 2a;'a + +2ic'a« =
the sum of 2 n even numbers ; then will their sum be
{2(a;i + a:'x) + 1} + 1 2 (a^a + ic'a) + If + + \'i{Xin + x'^n) + 1},
which is even [T. II].
882. Thefyrem V. — The sum of an equal odd number
of even and odd numbers is odd.
Demonstration. — Let (2a;i + 1) + (2a;a + 1) + + (2a'9» + i +1) =
the sum of 2 ?i + 1 odd numbers ;
and 2a;'i + 2a;'a + + 2a;'2» + i =
the sum of 2n + 1 even numbers ; then will their sum be
{2{Xi + x\) + 1\ + {2(2:2 + a;'a) + 1} + + {2{Xin+i + x'^n + i) +1},
which is odd [T. IIIj.
883. Theorem VI. — The difference bettoeen two num-
bers, if both are odd or both even, is even.
Demonstration. — 1. Let 2 x and 2 x' be two even numbers.
Their difference is 2 a; — 2 a;' = 2 (a; — a;'), which is even.
2. Let 2 a; + 1 and 2 a;' + 1 be two odd numbers.
Their difference is (2 a; + 1) — (2 a;' + 1) = 2 a; — 2 a;' = 2 (a; — x%
which is even.
884. Theorem; VII, — The difference between an odd
and an even number is odd.
Demonstration. — Let 2 a; + 1 be any odd number, and 2 x' any
even number.
Their difference is (2 a; + 1) — 2aJ' = 2 (a; — a;') + 1, which is odd.
885. Theorem VIII,— The product of any number of
even numbers is even.
Demonstration. — Let 2 a:, , 2 a^a , 2 a^s , . . . 2 a:„ be n even numbers.
Their product is 2(2*-ia;i , Xi, Xz, a;„), which is even.
Cot, — Any power of an even number is even,
886. Theorem IX, — The product of any number of
odd numbers is odd.
472 ADVANCED ALGEBRA.
Demonstration. — Let SiCi + 1, 2a;a + 1, 2a;n + 1 be n odd num-
bers. It is evident, from the nature of multiplication, that the prod-
uct of these numbers will contain the factor 2 in every term, except
the last, which will be 1. That is, the product will have the form of
2 a;' + 1, which is odd.
887. Cm; — Any power of an odd number is odd.
888. Theorem X, — The product of any number of odd
and even numbers is even.
Demonstration. — The product of the odd numbers is odd [T. IX],
and may be represented by 2 a; + 1.
The product of the even numbers is even [T. VIII], and may be
represented by 2 x'.
.-. The entire product is 2x' {2x+l) = 2(xx'+x'), which is even.
Example. — It is required to divide one dollar among 15
boys, giving to each boy an odd number of cents. Is this
question possible ?
Prime, Composite, Square, and Cubic Numbers.
I. Definitions.
889. A Prime Number is a number that can not be
produced by multiplying together factors other than itself
and unity.
A prime number is divisible only by itself and unity.
890. A Composite Number is a number that may be
produced by multiplying together other factors than itself
and unity.
A composite number is divisible by other factors than
itself and unity.
891. A Square Number is one that may be resolved
into two equal factors.
892. A Cubic Number is one that may be resolved into
three equal factors.
PRIMES. 473
893. Two or more numbers are prime to each other
when they have no common factor, except unity.
2. Primes.
894. Theorem I, — The number of primes is unlimited.
For, let n be the number of primes, and, if n is not
unlimited, let p be the greatest prime number. Then will
2x3x5x7xllX .... XphQ divisible by all primes not
greater than jo ; and (2 X 3 X 5 X 7 X 11 X X i?) -f 1
not be divisible by any prime not greater than j9. There-
fore, (2X3X5X7X11X Xi?) + 1 is itself a prime
greater than p, or is divisible by a prime greater than p.
In either case, p is not the greatest prime. Therefore, n
is unlimited.
895. Theorem II, — Every prime number, except 2 and
3, belongs to the form 6 ;^ ± 1.
For, every number evidently belongs to one of the
forms Qn, 6 w + 1, 6 w + 2, 6 ?^ + 3, 6 tj + 4, or 6 w + 5,
in which n may be any integer including 0. Now, 6w,
6 ?i + 2, and 6 w + 4, are each divisible by 2, and 6 tj + 3
by 3 ; hence, these forms are composite, except when ^ =
in 6 ?^ + 2 and 6^ + 3, in which case we have the primes
2 and 3.
The only forms remaining to contain primes are Qn-\-l
and 6^ + 5. But 6 w + 5 = (67^ + 6) - 1 = 6 (?i + 1)-1
= 6 7^' ■— 1. Therefore, the general form 6 7^ ± 1 contains
all primes, except 2 and 3.
Scholium, — It must not be inferred from this propo-
sition that all numbers expressed by 6 7i ± 1 are prime.
Thus, when w = 4, 6^ + 1 = 25; and when n = 11,
6 71 — 1 = 65.
474 ADVANCED ALGEBRA.
Cor, — Every prime above S, increased or diminished
hy unity, is divisible by 6,
896. Theorem III, — iVo rational formula can repre-
sent primes only.
For, if possible, let a-{-bx-\-cc(^-}-dcc^-\- be
prime for all values of x.
When X = m, let a-\-bx-^cx^-\-da?-\- =i?;
then, p = a-\-bm-\-cm^-{-dm^-\-
When X = m-{-np, leta-\-bx-\-cx^-\-da^-\- =g ;
then, q = a-{-b{m-{- np) -{- c{m-\- npf +
d(m-\- npY + . . . .
= a-\-bm-\-cm^-\-dm^-}- -\-rp
=z p -{- r p = p (1 -\- r)f a composite number.
897. Scholium, — The form n^-\-n-\-4:l is prime for all
values of n from to 39 inclusive, and the form 2 ^^ + 29
for all values of n from to 28 inclusive. These forms
have been discovered by trial, and are not demonstrable.
898. Theorem, IV, — If a number is not divisible by a
factor equal to or less than its square root, it is a prime.
For, let N=xxy be any number not prime. Then,
if x = y, W= /, and y = VW, But, if ic > VW, then
y < ViV, since x X y = J^- But JV is divisible by y.
Therefore, if iV is not prime, it is divisible by a factor
equal to or less than VJSf, Hence, too, if a number is not
divisible by a factor equal to or less than its square root,
it is prime.
3. Composites.
899. Theorem I. — If a number is a factor of the prod-
uct of two numbers and is not a factor of one of them, it
is a factor of the other.
COMPOSITES. 475
Thus, let a; be a factor of a h, and not a factor of a ;
then will it be a factor of 5.
For, — may be reduced to a terminating continued
fraction [854]. Let — be the conyergent next in yalue
to -. Then, ^ ^ — = — [848, P.] ; whence,
X q X qx ^ -^^
px'^aq = l; and hpx ^ abq =: h.
Now, hp X and abq ?iie each divisible by x ; therefore,
their difference, h, is divisible by x,
900. Car, — If a number is prime to each of two or more
other numbers, it is prime to their product.
901. Theorem II. — Every composite number may be
resolved into one set of prime factors and into only one
set.
1. Any composite number {N) is the product of two
or more factors each less than N, which are all composite,
all prime, or some composite and some prime. As many
of these as are composite are again resolvable into other
factors less than themselves, and so on, until no factor is
further resolvable into factors less than itself and greater
than unity, at which stage all the factors are prime.
2. Let one set of prime factors oi N he a, b, c, ,
and, if possible, let another set be ^, g', r, ; then will
axbxcx =pXqXrX
Now, suppose a different from q, r, , then it is not
contained in q X r X [900] ; it must, therefore, be
contained in p, but this can only be when a=p, since p
is a prime. But, it a=p, bXcX =qXrX ;
from which it follows as before that b is identical with one
of the factors in qXr X ; etc.
476 ADVANCED ALGEBRA.
902. Theorem III, — The product of any r consecutive
numbers is divisible hy [r.
^ n(n — l)(n — 2) (n — r-\-l) . , , , ,
For, — ^ — 1 ^^ ' — - IS the product
of r consecutive numbers divided by [r, and it is also the
number of combinations of n things taken r together,
which is evidently a whole number.
903. Cor, J.— The coefficient of the {r + l)th term of
^, , . .... . n{n-l)(n-2)....{n-r-\-l)
the bmomial theorem is -^ — ,
[595] ; therefore,
The coefficient of every term of the binomial theorem is
integral when n is a positive integer,
904. Cor, 2, — If we represent
n(n-X)(n-2) (n-r±V) ^^ ^^ .^ ^^jj^^^ ^^^^^
All factors of the numerator that are prime and are
greater than r are divisors of q.
905. Theorem IV, — Fermafs Theorem. If p be any
prime number, and a be a number prime to p, then
a^~^ — 1 will be divisible by p.
Demonstration : a^ = [1 + (a — 1)]p
=^ 1 + p{a-l) + ^^^~^\ a-lf + . . . .+ {a- V)p (A)
.-. aP — {a - \)P -1 = p{a -1) + --^g + ^*^^-
= a multiple of p [901. 140, P.]. "~ (B)
Let a = 3, then
ap ~.{a — l)P — \ = 2/' — 2 = a multiple of p.
Let a = 3, then
aP - (a-\)P -1 = ^p -2p -I = {^P - 3) - {2P - 2)
= a multiple of p.
... Si* — 3 is a multiple of p [157, P.].
PERFECT SQUARES, 477
By continuing this process, it may be shown by induction that
aP — a is a multiple of p.
But aP — a = a(aP—^ — 1) and a is prime to p; therefore,
aP-^ — 1 is divisible by p.
Perfect Squares.
906. Theorem I. — Uvery square number is of the form
3 m or 3 m + 1.
For, every number is of the form of 3x or 3 a: ± 1.
Now, {3xY = dx^ = d{3x^)=z3m; and
{3x±lY=(9af±6x + l) = 3{3x±2)-\-l = 3m-{-l.
907. Theorem II. — Every square number is of the
form 4^m or 4 m + !•
For, every number is of the form of ^x, 4cX-\-l,
4a: + 2, or 4a- + 3.
Now, {4.xf = 16a^ = 4.(4:3^) = 4:m;
{4:X + iy =16a^-^ 82; + l = 4(4a;2 + 2:r) + l
= 4m + l;
{4:X + 2Y = 16a^-\-lGx + 4. = 4: (4: a^ + 4: x -\- 1)
= 4:m;
and {4:X-{-3y = 16 a^ -\- 24: x -{- 9 = 4: {4: x^ -{- 6 x -\- 2) + 1
=4/7i + l.
908. Theorem, III, — Uvery square number is of the
form 5 m or 5 m ± 1.
For, every number is of the form 6x, 5 ic ± 1, or
5 a: ± 2.
Now, (5 xY =z 25x^ = 5 {6a^) = 5m;
(5a; ±1)2 = 25a;2±10a; + l = 5(5a;2±2^) + l
= 5m-\-l ;
and (5.'c±2)2 = 25a;2±20a; + 4 = 5 (5 3^ ± 4:X + 1) -1
= 5 m — 1.
478 ADVANCED ALGEBRA,
909. Theorem IV, — If a^-\-l^ = (^ when a, h, and c
are integers, then luill ale he a multiple of 60,
For, 1. a^ and h^ can not both be of the form 3 m + 1,
else would c^ be of the form 3m + 2, which is not a
square. Therefore, either a or J is a multiple of 3 [906].
2. a^ and h^ can not both be of the form of 4^ + 1,
else would c^ be of the form 4m + 2, which is not a
square. Therefore, either a ov h must be a multiple of 4,
or each of them a multiple of 2 [907]. In either case,
ahc is a multiple of 4.
3. a^ and ¥ can not both be of the form 5 m + 1 or
5 m — 1, else would (^ be of the form 5 m ± 2, which is
not a square. Therefore, either a^ or h^ must be of the
form 5 m, or one of the form bm-\-l and the other of
6 m — 1 [908]. In the former case, either a or 5 is a
multiple of 5, and in the latter, c is a multiple of 5, and
in either case, ahc is a multiple of 5.
4. Since ale is a multiple of 3, 4, and 5, and these
numbers are prime to each other, ale is a multiple of 60.
Scholium, — By means of this theorem and the formula
a = V{c -{-l){c — I), rational values of a, I, and c may
he determined ly inspection that will satisfy the equation
a^^¥=:(^.
910. Problem. To determine the rational value of x
that will render x^ +px + q a perfect square.
Solution : Let x^ +px + q = (x + mf
then, x^ + px + q = x^ + ^mx + m^
whence, x = ^ , in which m may have any
rational value from -- oo to + oo .
Illustration. — What value of x will render a;^ — 7 a; + 2
a perfect square ?
PERFECT SQUARES. 479
Solution : Here p = — 7, g = 2, and let m = 5,
25-2 23 ,6
then, X = _^_-^Q = 1117 = - 1 17
. r, o 529 161 „ 3844 /62\2
911. Cor, 1, — For m > Vq and 2m <p, or m < Vq
and 2m > p, m being positive, x will be positive,
912. Cor. 2. — Put m^ — q = n{p — 2m); then
q = m^ — n {p — 2 m) ;
X = n, an integer ; and
oi?-\-px-\-q — x^ -\- p X -\- m^ — n (p — 2 m)
= {n-\- my, an integer.
7f? —
913. Cor. 3.— Put X = -^ — — m; or
p — 2m
m=^ ± a/^ - q, then,
x^-\-px-\-q = 0, and a; = — | :f a/ ^
which conforms to Art SSI.
Perfect Cubes.
914. Theorem I. — Every cube is of the form 4om or
4m ±1.
For, every number is of the form 4:X, 4:X-{-l, 4 a; + 2,
or 4 a; + 3.
Now, {4:xY = 64:X^ = 4.(16x^) = 4:m;
{4:X-\-iy = 64:a^-\-A83^-\-12x-{-l
= 4 (16 a:^ + 12 rc2 + 3 a;) + 1 = 4 m + 1
(4a;4-2)^ = 64a:3_^96^2_j_43^_|_8
= 4(16x3 + 24a;2 + 12a; + 2)=4m
(4a; + 3)^ = 64ar^ + 144a;2 + 108a; + 27
= 4:{Ux^-\-36x^-i-27x+'7)-l = 4:m-l
480 ADVANCED ALGEBRA.
915. Problem. To determine rational values of x that
will render a? + px^ ■\- qx + r a perfect cube.
Solution : Put x^ -{■ px"^ + qx ^^ r = {x + mf \ then,
(^ — 3m)a;2 + (5'-3m2)a; + (r — m3) = 0. (A)
1. Put bm—p, or m = -^p\ then
_ m^ — r _ jp«-27r
^- q-'dm^- 27q-9p' ' ^^"^
(pS 27 r « \'
~ V 27g-9i?2 ) ' ■
Cor.—If9pq-27r-'2p^ = 0, or r = ^^^Il^^\
3 Q '^*
x^-\-p^ + qx-\-r = 0, and x = ^^—^^.
2. Put m* = r, and suppose r = ri% then m = ri ; and (A) will
become (i? — 3 ri) x^ + {q — B Ti^) x = 0; whence,
x= ^o ; and x^+px^ + qx + r = x^+px^ + qx + ri^;
and (re + m)^ = ( ^^3^^ ) . Therefore,
^ — 3ri
_o )
Cor, — If q =p n, x= —ri, and
7? -\-p :i^ -\-qx-\-rx ^^'
, Scholium, — Other values under particular suppositions
may he obtained hy putting 3m^ = q = 3qi^.
EXERCISE 117.
1. Find which of the following numbers are prime :
19r, 251, 313, 281, 461, 829, 957.
2. Find the least multiplier that will render 3174 a
perfect square.
3. Find the least multiplier that will render 13168 a
perfect cube.
PERFECT CUBES. 481
4. Find which of the following numbers are divisible
by 9, which by 11, and which by both 9 and 11 : 11205,
24530, 342738, 25916, 558657.
5. Show that, if ^ + 3' is an even number, then is j9 — g
also an even number, provided p and q are integral.
6. Show that every cube number is of the form 7 w or
7/^±l.
7. Find such a value of x as will render the wax
rational.
Suggestion. — Put ^/a x = p.
8. Find such values of a; as will render vax-\-h
rational.
9. Prove that 2*" — 1 is a multiple of 15.
' 10. Show that no square number is of the form 3 » — 1.
11. Show that n{n-\-l){^n-\-l) is divisible by 6.
12. Show that {n^ -f 3) (w*+ 7) is divisible by 32, when
n is odd.
13. Show that n^ — n is a multiple of 30.
14. Show that the fourth power of any number is of
the form 5 m or 5 m + 1.
15. Every even power of every odd number is of the
form 8 w + 1'
16. Show that every square can be expressed as the
difference between two squares.
17. Show that a' -\-a and a' ^a are even numbers.
18. Show that every number and its cube leave the
same remainder when divided by 6.
19. If ?^ > 2, show that n^ — 5n^-{-4:n is divisible by
120.
20. It n is a, prime number greater than 3, show that
n^ — 1 is divisible by 24.
482 ADVANCED ALGEBRA.
21. Find such a value of x as will render Vox rational.
Suggestion. —
Let ^/a x = p, any rational quantity, then will re = — .
22. Find such a value of a; as will rationalize vax-{-h.
Suggestion.— Put ^/ax + b = p, and prove x = — .
23. Find such a value of x as will rationalize
Suggestion. — Put \/ax^ + bx = px, and prove x = g_ .
24. Find such a value of x as will rationalize
Suggestion. —
Put '\/ax^ + bx + c^ = px + c, and prove x = ^_ ^ .
25. Find such a value of a; as will rationalize
Va^a^-\-bX'^c.
Suggestion. —
Put ^/a^ x^ + bx + c = ax + p, and prove a; = , _ ^ — .
26. Find such a value of a; as will render Vax^-{-bx-\-c
rational when J^ — 4 « c is a perfect square.
Suggestion.— Put \/ax^ + bx + c = 0, and b^ — 4ac=z q^, and
-b ±q
prove a; = -2^.
27. Find such a value of x as will rationalize
, , p^-b
Suggestion.— Put ^a^ ->cbx^ = px, and prove x = ^ .
28. Find such a value of a: as will rationalize
Vaa^-^ba^-\-cx-\-dK
Suggestion.- ^,_^^^
Put Va^ + 6a;« + ca; + d« = 2^ a:+d, and prove x = -^
d«
ANSWEES.
1. l+x+x'+x'
Exercise
87.
2.
1
3 +
2 4
'^'U
4. 1+x-
-x^-
-7?
6.
-d + 5x-
-2x^-
-Saf
81
3. 1+x-x^-x*
5. l-x-x^ + 2a^
a a' a*
9. l+a: + a;2— a;8
Exercise 88.
1- 2- ^^+ 4^^+ 3^6^' 2. 1+ i^- |^«- Ix^
^ , 1 109 , 109 3 ^ . 1 1 5 5 ,
3' ^+ 6 ^- 216^ + 3888^ ^' ^+ 3 ^- 9 ^ "^ 81^
27 3.27« "3.27*
^ a 1 23 25 „ I . X x^ ^ x^
^'^^r2^^27T2^-2A2^ '-"^^^-^^I^l
_ia; a;' 7 ^ a:*ic«5a;'
3a§ 9a^ 81al ^a^ ga*^ 81a8
Exercise 89.
1. Divergent. 2. Convergent; divergent; divergent.
3. Convergent. 4. Convergent. 5. Divergent.
6. Convergent; divergent; divergent.
7. Divergent; convergent; divergent. 8. Convergent.
Exercise 90.
1. x^ + Qx^ + llx + Q 2. a;3_2a;2-9a; + 18
3. a;4+2a;3-7a;«-8a; + 12 4. a:4-37a;2-24a;+180
484 ADVANCED ALGEBRA.
5, a;4 + 8a:3 + 24a:2 + 32a: + 16 6. a;4_20a;3 + 150a:2-500a;+625
7. 16 a;4-16 2:3-64.1:2 + 4 a; +15 8. (a: + 2) (a; +3) (a: + 4)
9. (a:-3)(a:-4)(a:+5) 10. (a: + 2)(a:+3)(a: + l)(a;-l)
11. (a; + 2)(a:-3)(a: + 4)(a:-5) 12. (a; + 2)(a;-2)(a;+3)(a;-3)(a;+4)
Exercise 91.
1. a4_i2a3a;2 + 54a2a:4_i08aa;« + 81a:4
2. 32 + 400 a: + 2000 x^ + 5000 x^ + 6250 a:* + 3125 a:*
3. a:6-18aa:^ + 135a2a:4_540a3a;3+i2i5a4a:2_i458a6a:+729a«
4. 128 a:!^ + 2240 x^^ + 16800 a;io + 70000 a? + 175000 a:« + 262500 a:^ +
218750 a:2 + 78125
5. a;4-40 xl + 700 3:^-7000 a;t + 43750 a:2-175000a:l +437500 x
- 625000 a:i + 390625
6. 2187 a;¥ + 5103 at a:^ ^. 5103 ^3 ^^.^ + 2835 ai a:l + 945 a« a:^ +
189 a¥a:f + 21 a9a:§+a¥
7. l__a:-ga:2-_a:3
8. at — -5- a~i ^ — 77 ^~ ^ ^^ — 01 ^~~ ^ ^
o 9 81
9.xt-|x-i-|x-5-j|g^-|
10. a;— 4 — — aa;~4 + —a^x—i — r-^a^a:~4
-o o Id
11. a~%x~% — — a~V Ja:~ (T + — a~V h^x~^ —
12. a:— ^ + -^ aix—^ + -^aia;— V + — a^a;"^
14. 8-06225; 8*94427; 7*006796; 5-000980
15. 2449440a.-* 16. - -^^a^ 17. - ^a""^^'
,3, ^r(r+l)(r+2)(r + 3)^_,,^,,^_.
Li
19. 2^ 20. iJ^ 21.
343
Exercise 92.
1. n(n + l); ■|n(n + l)(n+2) 2. (2n-l)«; ■|n(4w«-l)
AKSWURS. 485
3. ^n(n + l)', ^n(n + l)(n + 2) 4. w(n + 3); ^n(n + l)(2n + 7)
5. w2; ^nin+l){2n+l) 6. 2n{2n-l); -n{n + l){4n-l)
o o
1. n{n + l)in+2); -^n(n + l){n + 2)(n+S)
8. n(n + 4)(n+S); -jn(n + l){n + 8)in + d)
9. n(n + 2)(n + lf; ^n(n + l)(n + 2)(n + d)i2n + d)
10. {2n + l){2n + 3){2n + 5); -i(2w + l)(2n + 3)(2w + 5)(2w + 7) -IS^
o o
n 1 J./ 1 1^ 1 1 1_\ 11
3?i + l' 3 3 V "^ 2 "^ 3 w + l w + 2 n + dj' 18
13 iri____i___V 1
8 V4 2(?i + l)(n + 2)y' 32
^ 6 n + ll ^ 21
180 12(2n + l)(2n + 3)(2w + 5)' 180
15. i _ -_1^+^; 4 16. ^(9n^ + 10n^-Sn-4)
4 2(ri + l)(w + 2)' 4 12^ ^
^^ n(n + l)(n + 2)(n+3) ^g _w_
[4 ■ n + 1
19. n + 1 K; 20. 2-978809
Exercise 93.
1. x = y-y^ + y^-y*+ 2. x = y+ -^y^+ -^y^+ 24 2/*+
3. x = y—2y^ + Sy^—4:y^+ 4. x = y—y^+y^—y'' +
b.x = (y-1) - 1 (i/-l)2 + 1 (^-1)3 _ ^ (y_l)4 + . . . .
6.x = (y-1) + 2iy-lf + liy-lf + 30(2/-l)*+ ....
7. a; = i-|2 + |^-^ = '17590144 8. a; = -00999999
Exercise 94.
1.^ = 3, ^=-1; ., i~^ , 2. » = 2, q = 2; , ^^"*"^^ ^
^ ^ 1— 3a; + a;2 ^ ' ^ 1— 2aj— 2a;2
o *, ;t Q 2 + 8a; ^ . ^ 3-lOa;
3.i? = 5, 5= -3; , r:^.Q^2 4.i? = 4, ^=-5;
l_5a; + 3«2 -./---,«- "' i_4a;+5a;2
486 ADVANCED ALGEBRA.
TOO o 1—x—^x^
^ o o A l-5x + Sx^
„ ^ ^ l-x-1809x^-Udlx-'
'J'P = ^^9 = ^; 1_3^_3^.
8. p = 2, ^ = — 3 ;
9.p=-2, q = 2;
l-2ic + 3a;2
2 + 3a:-2208a;« + 1616ic9
10. i)=l, q=-l, r=l;
1— a; + a;2— ic*
Exercise 95.
, 2 3 „ 3 2
1. — TR + 7i 2.
a:+2 a;-2 2a; + l 2a;-l
,12 3 ^3 5
3. - + -— r + — ^ 4.
X x + 1 x + 2 x + 1 (x + l)^
15 7 - 3 4
°' ^J_.q /'^_La\2 "'" /^j..q\3 ^' <rJ_9. "^
a:+3 (a;+3)2 (a; + 3)3 iC+2 ' (x-df
P , Q
7. + 8. — ^- — +
a—x o+x px + q (px+q)^
^2 3 ,^222
x^ + x + 1 x^—x + 1 X l + 2a; 1— 2a;
5 7_ 1 3
(rc-l)* (a;-l)3 "^ (ar-l)^ "^ a:-l
12 ^ 4. ^-^^ 13 1 3 4. »
1 a;+l a;+5 ,, 1 x + l_
1)
^^' 4(a;-l) "^ 4(a;« + l) "*" 2(a;« + l)« ^^" 2(a;+l) "^ 2(a;2 +
,^111 1 9 1 x+1
16. - - + -^ . + jt; TT +
X ' x^ x^ ' 8(a;-l) ' 8(a; + l) 4(a; + l)« 4(a;2 + l)
3a: 1 ,„ 1 1 3 3 2
17. -o-t; = K 18. r -: + ^—-^ - . ■ ..o +
a:« + 2ic-5 a;-3 ic-l ic + 1 (a; + l)« (a; + l)2 (a:+l)*
,^111 1 9 1 a; + l
19. + -J 8 +
a; a:» a:» - 8(2;-l) 8(2; + l) 4(a; + l)» 4(a;» + l)
1 x + 1
^"" 3(a; + l) "^2(0:2 + 1) ^^'
22.
^iV^6'Tf^i2;S^.
487
3
3
"•"SCaj + l)
1 "^-^
• 4:{X-lf
8(a;-l)
4(0:2+1)
8
13
(x + 1) 6 (a; + 1)2 " 45(a:-2) 20 (a; + 3)
Exercise 97.
1. dy = {15ax^-6bx + 2c)dx
2. dy = 15z^x^dx + 10x^zdz + dz 3. <?y = (3a;2 + 6a; + 4)fZa;
Q
4. f?3/ = 5m2(a+Z>a;)"»''-icZa; 6. (?2/ = (a:+ -^ — a;-2)f?a;
^ , ad^a; 1 „ , 5dx 1
6,dy — — J- . r 1, dy=-^- . ^
2ci (aa; + 6)t 3 (5aj + 6)f
8. <Zy = /i/— . <Za; 9, dyz=z—i^a^) . x—^dx
10. dy= . — , 11. dy = (— ax^+bx—i]dx
a ^a^-x^ ^ V2 ;
,^ , 2a;3(4a+a;) , ,_ , i^x+a)dx
12. dy= — -T^ — ~- . dx 13. dy = - -tt-
^ («+^)* ^ (a;+a)l
14. cZyzz:- ^^^^^ 15. dy = {x+a){x+})f{^x+Za + h)dx
(6'+a;2)
16. <?2/ = (a; + a)»-»(a;-6)i»-»(2a;+a-&)(?a; 17. dy-"^^^ . da;
o ^ ^
18. 6?2/ = (logea:)2 . 19. dy = 9iQ(?^{^ + \og^Z + \oz^x)dx
X
20. dy = 2a»*.a;logea<ia; 21. dy=z- ^(2a + 3a;2)
(a+a;2)t.
22. dy = ^-j-| 23. dy- (^^Y. (loge c-loge d)dx
«^ 7 2<?a; «,^ , , , dx
(l_a;)(l-a;2)i ^ ^' a;
Exercise 98.
2. 16 sq. in. per second. 3. 4'irr2; 144 ir cu. in. per second.
4. 2 V' 2 in. 5. 1 in. per second. 6. About 6 mi. an hr.
7. 12 V^ in. per sec. 8. -00517 9. 1-62842 ; -00003
Exercise 99.
1. 3a;2-8a; + 7 2. (a;+2)2(a;-2)3(7a; + 2)
3. 3a:2 + 8a;+2 4. (a+a;)4(a-a:)2(8a-2a;)
5. -8a:' + 5aa:*-3aa;2 6. -10a; . ^^^
(a—xf
p
488 ADVANCED ALGEBRA.
Exercise 100.
1. {x + Zf{x-2f 2. (x-2)^{x-iy(x+d)
3. {x + Sf(x-3f(x^+x+l) 4. (x-2)\x + 2f{x-df(x+Sy
5. {x-lY(x+iy{x+d)(x-d)
Exercise 101.
Q
1. Max. 1 -J ; min. —5 2. Min. —3 ; max. —128
3. Min. -4 4. Min. -4 5. Max. 10 ; min. -22
6. Min. 0; max. 18+ 7. Max. (^Y
8. No turning values. 9. Min. —14 10. Min. —14
11. Min. 12. Min. —16 13. At the middle point.
1 32 8
15. o-a 16. ^ V r*, or ^r^ of the vol. of the sphere.
O ol at
17. -^a^ or an inscribed square. 18. -j <ir a^ 19. -^ a
20. Square = 2r' 21. An isosceles triangle.
Exercise 102.
1.x* + 2 a^-9 x^ + 63 a;-135 = ; roots of fn {x) = dx roots of Fn (x)
2. x^i-4: x*—2 x^+4 a;2— 112 = ; roots of /« {x) = 2x roots of Fn (x)
3. x^ + 12 a;*-320 x^ + 1792 a;- 1024 = ; roots of /„ (x) =
4x roots of ^n(a;)
4. a;'^-2a;* + 9a;3-18a;2 + 108a;-162 = 0; roots of /„ (a:) =
3 V^oots of (A)
5. a:«-2 a;5 + 2 x*-4: a:* + 24 a:^ + 32 a; + 32 = ; roots = 2 Vroots of {A)
6. a;3« + 8 . 9^x^+36 . 9»a;24 + 10 . 9^9a;'«-4 .934 = 0; roots =
9 Vroots of {A)
7. a;''-a;6 + 18a:*-162a;9-2187 = 0; roots = 3 Vroots of {A)
8. a;«-10a;*-36a:4_i2288 = 0; roots = 4 Vroots o f (^)
9. a;4 + 6800a;-9000 = ; roots = 10 Vroots of (A)
Exercise 103.
1. a; = l 2. a; =
5. a; = 1, 2, 2, - 2
: 4 3. a; = 5
6.a; = i, ■
—
4. a; =
1 2
2' 3
7. a; = l, 2, -2, -3
8. a; = 5, -
■1
9. a;= +2, -2, -3
10. a: = 2, 4
-2
ANSWUES. 489
13. ic = 3, -2, -1 14. a:=-3, -5, -1, ±|
15. a: = -l, ll, -li 16. ic = 2, -3, -5, 2
17. re = 2, 2, -3, -3 18. a; = 1, 1, -2, -2, -2
19. a; = -1, -1, -1, 2, -3, -|
Exercise 104.
1. -1 + , 4 + , -9+ 2. -6-6+ 3. 0-3 +
4. 0-8 + , 3 + , -3 + , -0-1 + , -0-6 +
5. 2 + , -2+ 6. 2 + , 0-6 + , 0-4 + , -3 +
Exercise 105.
1. 775 2. 240 3. -87938, -2*53208, -1-34729
4. 3-85808, 1-60601, 1-44327, -2-90737
5. -5, 4-05607, 11-15306, 12-79085
6. -1-12579 7. 2-34244 8. 1-25992 9. 1-3797
10. 3-2131, 3-2295, -17-4426 11. -80285, -5-4335
12. 8-41445 13. a;3 4.32a;2 + 343a;-13087800 =
Exercise 106.
1. a; = l, 1±2\/^ 2. a; = 4, 1±\/^
3.a; = 5, 3±\/^ 4,x=-2±^/^
6,x = d, i(l±V^ 6,x=-4, _i(l±v^
1, x= -2, -2, -3 8. x = 3, -2, -2
9. a: = 2, 2, -3 10. a; = 4, -2±'v/-^
1
2
13. a; = 5, 1, -2 14. a: =-5, -1, 2
11. a;=r-6, 3±'\/^ 12. a; = 1, ^ (3± V^l)
Exercise 107.
1. a; = l, i(l±^Z3") 2. a^ = -l, 2±\^
Z.x = \, -1, i(3±V5) 4. a: = l, -1, \{-^±V~^)
5. a; = i(5±V2T, ±a/^ 6. a; = -l, -1, -1, -1, -1
7. a; = l, 2, i, -3, -i 8. a; = -l, -5, -1, 2±V3
490
ADVANCED ALGEBRA,
Exercise 108.
1. -9
2. -59 3. 65
4. o&c— 2a6*
5. ^mnp
6. 4:abc 7. 28
8. -82
9. -108
10. 11.
12.
13.
14. 15.
Exercise 109.
1.
2. 3.
4.
5. Sxyz
6. 6a&c-2a3_2 63_2c3
1. 10
Exercise 110.
-48 3. -199
4.
1. X
3. X
4. X-
6. X:
7. X ■
Exercise 111.
2. a; = 3, y = S
y
z
9. X:
4, y = 2
cn—hd _ ad—c m
an—bm^ ^~an—bm
cm—dm + cn + dn _ an + bn—am + bm
2{ac-bd) ' ^~ 2iac-bd)
2, 2/ = 3, 2 = 4 6. x = 5, y = l, 2 = '.
abd—acd—abe + ec^ + abh—bc h
a^ b—a^ c—a b c + c^ + a b^—b^ c
a^ e—a^ h—a cd+c^h + ab d—b c e
aH—a^c—abc + c^ + ab^—bU
abh—ace—bch + c^d + b^e—b^d
a^ b—a^ c—a bc + c^+a b^—¥ c
_ a^(m + n—r) +ab(m—n—r) +b^(7n—n + r)
"" 2aF+2¥
_ a^(n—m + r) +ab{2an—r—m—n) +b^(n—r+m)
— 2a3 + 2 63
a^{r—n-k-m) +ab(2r—n—r—m) +b^(r+n—m)
■ 2aF+2¥
2, 3/ = 3, 2 = 4, u = 5
y =
z=.
10. x =
c a
b c
b c
b c
m
a b
—n
ab
+p
c a
-Q
c a
a b c
a b c
a b c
ab
c a
b c
b c
a
ab
-b
ab
+ c
c a
a b c
a b c
a b c
ANSWURS.
491
y = -{a
11. a; = 2, _
_p + q—r—s
n a
m c
m c
pah
-b
p a b
+ c
71 a
q b c
q b c
q b c
c n a
b m
b m
Op b
-b
Op b
+ c
c n a
a q c
a q c
a q c
c n
b c m
b c m
a p
-b
a p
+ c
c n
a b q
a b q
a b q
= 3, Z--
= -3,
w=-3
p—q + r—8
4&
p—q—r+8
Ac ' '"" 4d
13. a; = 1, J/ = 2, 2 = 3, u = 4, v = 5
u =
\-
-^c.d
-i- c . d
1. Consistent.
Exercise 112.
2. Inconsistent.
i 1 1 i i i
2+1+1+6+2+4
111
Exercise 115.
1 1
2.
_ _ i 1 i
1 + 1 + 1+11 + 6
3. ^ .-=^
6. 3 +
1
+ 1 + 99
i i
4 i i 1 i i
6+1+7+6+2
5. 3 +
2 + 6
1 1
7. 5 +
1^ ]_
2 + 10
8. 7+ ^ . 4
!_ 1^
1 + 4
1 1
3 + 7 + 17 + 2 + 1 + 8
11-67+^^^^^^^^^^deg.
10. 3 +
12.
i J_ 1
7+16 + 11
3 13 68
2' 7' 80' 157
1^ 1^ 10 11
3' 4' 39' 43
1 2 7 16
23
13.
15.
17.
21.
23. 4, 4
14.^,
1 ' 3' 10'
\/5-2
2 5 J^
3' 9' 12
1 1 15
2 ' 3 ' 29
55
79'
etc.
18. V6-2
16. 2 , ^,
19. VS
16'
10
23
24 55 189
55' 126' 433
33 109
' 76' 251
etc.
etc.
20. i(VlO-l)
19 26
33' 45
21
61
123
etc.
22.
24.
492 ADVANCED ALGEBRA.
Esercise 116.
1. 25542 2. 616
3. 434005,
327454
4. 58072 1, 32853:^
5. 2950
6. 125344
7. 2e23 8. 202033030
9. 7692
10. 14860
11. r = 6 12. r=6
13. r = 14
14. 2
15. 1414, 1201
16. r = 7
Exercise 117.
1. 197, 251, 313, 281, 461, 829 2. 6 3. 1646
4. By 9: 11205, 342738, 558657
By 11 : 24530, 342738, 25916, 558657
By 9 and 11 : 342738, 558657
1. x=^ 8. a: = ^
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