Here the factors of the denominators x -22 being x and 1-x, 1+x A A' 1+x A+ (Ã'—A)x

we shall have

X

+ ; or

x 1 -Xx

X

And consequently, by equating the coefficients of the like terms of the two numerators,

A-1 and A'-A-1, or A'=1+A=1+1=2;

fractions.

Here the factors of the denominator x-3 being x, 1-x, and A A' A" 1+x, we shall have

1+x2

X

+

x 1-x 1+x the fractions to the same denominator, 1+x2___A(1—x2)+A′(x+x2)+A′(x—x2)

;; or, by reducing

; whence 1+x=A+

(A'+A′′)x+(—A+A'—A”) x2. And, by equating the coefficients of the like terms, A=1, A'+A"=0, and -A+A'—A”=1. From which equations we have A=1, A'=1, and A"-1. 1+x2 1 1

1

+
X 1-x 1+x

the simple fractions.

3. It is required to convert the rational fraction

Here, by putting the denominator 2-2x+4=0, the two roots of the equation will be found to be 1+2, and 1—1√2, and consequently the factors are x-(1+1√√2), and x—(1 — {√2) ;

—(1+}√2)Ä—(1—}√2)A'+(A+A')z, where, putting

—(1+}√/2)A—(1—{√2)A'—†, and A + A'=-, we shall have, from these equations, A, and A'-; therefore

In cases of this kind, where the highest term of the equation, representing the denominator of the given fraction, has any coefficient prefixed to it, the numerator and denominator must be each

RULE II. When the factors of the denominator of the given fraction are all equal, or some of them equal and others unequal, the decomposition of it into simple factors cannot be effected in the way before pointed out, but must now be be taken of the form

B'

(x_pjait.

A'

B[] A •+ + + &c. x-p X-T X-T

where p represents the root of one of the equal factors, contained in D, and r, r', &c. those of the unequal factors.*

In which case, as in the former, if all the partial fractions, so formed, be brought to a common denominator, and the resulting fraction be reduced, when necessary, so as to have the same denominator D as the proposed fraction, the comparison of the coefficients of the like terms, as before, will give the values of the numerators sought.

1. Required to convert the rational fraction

1+x

(1-x)

-2 into its equivalent simple fractions. Here, according to the rule, we have

(1—x)2

B
(1-x)2 x

B'

=

B+B′(1−x)
(1-x)2

B+B'-B'x

; and by

(1-x)2

equating the coefficients of the homologous terms, B'-1, and B+B'=1, or B=2; whence

1-5x

2. It is required to convert the rational fraction

(1−x) (1+x)® into its equivalent simple fractions. Here, according to the above

rule,

1-5x

B'
+ +
(1—x)(1+x)2 ̄ ̄ ̄(1+x)2 1+x 1-x
B{(1+x)(1−x)}+B′{(1+x)2(1—x)}+A(1+x)3
(1—x)(1+x)3

divided by this coefficient, as above, in order to obtain the true factors; which equal division, it is evident, will not alter the value of the fraction.

*That the former rule cannot hold, in the case where the denominator of the given fraction has equal factors, is obvious; for a+bx B B' if we take, for instance, + the two partial [x-p]2 x-p x-p fractions, on the right hand side of the equation, would form in

B(1—x)+B′(1—x2)+A(1+2x+x2)
(1-x)(1+x)

B+B+A+(2A-B)x+(A-B')
(1-x)(1+x)'

; whence, by putting B+B'+

A=1, 2A—B——5, and A—B' — 0, we shall have, from these equations, B3, B'1, and A-1; therefore

1-5x (1−x)(1+x)3

3

1

1

are the simple fractions.

1-x

(1+x)2 1+x

Note. When the factors of the denominator of the given fractions are few in number, we can always find the partial fractions into which it is convertible by the preceding methods; but as the calculation, in other cases, becomes more laborious, it will be here proper to show, that any one of the numerators of these fractions 'may be deduced immediately from N and D, independently of the rest, as follows

RULE III. 1. When the factors of the denominator of the given fraction are all unequal, or of the form

N A

A'

A"

A"
X-r Xmp!!!

D + + +

&c. take that which consti

N

tutes the denominator of the simple fraction that is to be determined, and let S denote the product of all the rest of the factors; then if the root or value of x (found by putting the factor thus taken 0) be substited for x in the formula merator of that fraction; and the same rule will hold for all the rest.

S' it will give the nu

2. If the factors of the denominator are equal, or some equal and others unequal, as in the form

A'

B

+ -+ &c......+ +

-B'

N A D X-r x -r (x—p)2 + (x—p)o_1+ &c. let S denote the product of all the factors in the denominator of the given fraction, except one, as before; then find the simple fractions, due to the unequal factors, by the first part of the rule, and for the unequal factors proceed as follows:

taking x-p=0, or x=p, &c. &c., observing when

tors being x, 1.

2

1+22

and 1 + 2

Rule 1,

x(1—x)(1+x) X

the factor x-p, in any case of this kind, becomes xp, to take x. in these formulæ, p, instead of +p; and if it be simply z, or x0, take x=0; which operations being performed, the sum of the fractions thus obtained, together with the former, will give all the simple fractions into which the given fraction is resolvable.

1. It is required to convert the rational fraction into its equivalent simple fractions. Here, the factors of the denomina

x, we shall have, as in Example 2,

N

1+22

x-0, the first factor, A =

1, x be

N 1+x2

ing=0; and for 1-x, the second factor, A'

1, x being = 1; also, for 1+x, the third factor, A"

1+22

N

A"=

S

the simple factors, as before.

2. Convert the rational fraction

1

x3(1—x)2(1+x)

into its equiva

lent simple fractions. Here, according to the rule, we have A B B' B"

++ +=+ + where 23 x2 x (1-x)2 -X numerators of the first set of factors, and C, C', those of the latter. Hence, for the unequal factor 1+x, we have

B, B', B", are the

== x being = -1. And x3(1—x)2 x3-2x++x5 4'

for x3, x2, and x, or (x—0)3, (x—0)2, and x

`ing=0. Again, for (1-x)2, and 1-x, in the second set of equal

-x

2. Let

1

;

X 1. -x 1 -3x1+x 1- -X 1+2x 1-2x

and

- 1. Therefore

7

4(142) 4(1+x)

each be redu

1—5x2+4x

1

6

2

16

+ +

1+5x+3x2
(1+x)2(1+2x) x(1+x)(2—x)(1—2x)”

duced to its equivalent simple fractions.

(1+x)2 1+x (1+2x)2+1+2x

1 1

Ans. +

+

1

x1+x 2- 1

Of Recurring Series.

152 A Recurring Series is a rank, or progression of quantities, so constituted that each succeeding term is formed from the sums, or differences of some multiples of a certain number of the preceding terms, taken continually in the same way. Thus, if 1+6x+12 +48x3+120x2+&c. be the given series, we shall have the third term 12x=xX2d term +62 × 1st term; the 4th term 48xxx 3d term +62 × 2d term; and so on. In which case the compound expression z, 622, or simply 1, 6, is called the scale of relation of the several terms.

And if 1+4x+6x2+11x+28x+63x+&c. be the given series, we shall have the 4th term 11x3 2x X 3d term x2X2d term +33× 1st term; the 5th term 28x1 = 2x × 4th term x2X3d term +32X2d term; and so on.

Where 2x, x, +3x3, or 2, -1, +3, is the scale of relation, the numbers composing it being the multipliers by which the seyeral terms of the series, or their coefficients are produced.* These

*This branch of the science appears to have been first treated on by Demoivre, in his Miscellanea Analytica, and his Doctrine of Chances and has since been considerably improved by Euler,