CHAPTER XXIII.

PARTIAL FRACTIONS. INDETERMINATE COEFFICIENTS.

291. IN Chapter VIII. it was shewn how to express as a single fraction the algebraic sum of any number of given fractions. It is often necessary to perform the converse operation, namely that of finding a number of fractions, called partial fractions, whose denominators are of lower dimensions than the denominator of a given fraction and whose algebraic sum is equal to the given fraction.

292. We may always suppose that the numerator of any fraction which is to be expressed in partial fractions is of lower dimensions in some chosen letter than the denominator. For, if this is not the case to begin with, the numerator can be divided by the denominator until the remainder is of lower dimensions: the given fraction will then be expressed as the sum of an integral expression and a fraction whose numerator is of lower dimensions than its denominator.

293. Any fraction whose denominator is expressed as the product of a number of different factors of the first degree can be reduced to a series of partial fractions whose denominators are those factors of the first degree.

For let the denominator be the product of the n factors x- a, x-b, x- c, ...; and let the numerator be represented by F(x), where F(x) is any expression which is not higher than the (n-1)th degree in a.

We have to find values of A, B, C,... which are independent of x and which will make

α

or, multiplying by (x − a) (x − b) (x — c)................,

C

+

с

In order that (i) may be an identity it is necessary and sufficient that the coefficients of like powers of a on the two sides should be equal. Now F(x) is of the (n-1)th degree at most, and the terms on the right of (i) are all of the (n − 1)th degree; hence, by equating the coefficients of xo, x1,... x2-1 on the two sides of (i), we have n equations which are sufficient to determine the n quantities A, B, C,......

The values of A, B, C,... can however be obtained separately in the following manner. Since (i) is to be true for all values of x, it must be true when x = a; and, putting a = a, we have F (a) = A (a - b) (a — c)......; and therefore A = F(a)/(a - b) (ac)...... Similarly we have B = F (b) / (b − a) (b − c)......; and so for C, D,....

...

We have thus found values of A, B, C,... which make the relation (i) true for the n values a, b, c, of x; and as the expressions on the two sides of (i) are of not higher degree than the (n - 1)th, it follows [Art. 91] that the relation (i) is true for all values of x.

In this identity put x=1; then 10-A. Now put x=2; then 13 B.

then (b-c) (c- a) (a - b) = A (x − b) (x − c) + B (x − c) (x − a)

+C (x − a) (x − b).

Putting xa, we have (b-c) (c − a) (a - b) = A (a - b) (a–c); therefore Ac-b; and the values of B and C can be written down from symmetry.

1=4, {(x+1)(x+2)... (x+n)} + A1 {x (x+2) (x+3)... (x+n) } + ... + A, {x(x+1)...(x+r−1)(x+r+1)... (x + n)} + ... + An {x (x+1)... (x + n − 1)}.

If we put x=0, all the terms on the right will vanish except the first, and we shall have 1=A,× |n, so that A=1/|n.

To find the general term, put x=-r; we then have

1=A, {(r)(-r+1)...(-1) (1) (2)...(n − r)},

that is 1 = ( − 1)” A,|r|n−r; hence A‚= ( − 1)” /|r [n—r.
Hence the required result is

x2+15

Ex. 5. Resolve

(x − 1)(x2+2x+5)

into partial fractions.

The factors of x2+2x+5 are the complex expressions x+1+2i and x+1-2i, where i is written for √ −1.

.. x2+15=4(x+1+2i) (x+1-2i) + B (x-1)(x+1-2i)

Put x=1; then 16=84, so that A=2.

+ C(x − 1)(x+1+2i).

Put x=-1-2i, then (1+2i)2 + 15 = B ( − 2 − 2i)(−4i), that is

12+4iB (8+8i); therefore B=

Change the sign of i

3+ i

2-2i'

294. We have in the last example resolved the given fraction into three partial fractions whose denominators are all of the first degree, two of the factors of the denominator being imaginary. Although it is for most purposes necessary to do this, the reduction into partial fractions, of a fraction whose denominator has imaginary factors, is often left in a more incomplete state. Take, for example the fraction just considered, and assume

[It is to be noticed that we must now assume for the numerator of the second fraction an expression containing x but of lower degree than the denominator.]

Then x2+15= A (x2+2x+5)+(Bx + C) (x-1).
Putting = 1, we have 168A, so that A = 2.

Put A = 2 in the above identity; then after transposition - a -4x+5= (Bx + C) (x − 1);

or, dividing by x-1, Bx+C=-x-5.

295. We have hitherto supposed that the factors of the denominator of the fraction which is to be expressed in partial fractions, were all different from one another. The method of procedure when this is not the case will be seen from the following examples.

2x+5= A (x − 3) + B (x − 1) (x − 3) + C (x − 1)2 (x − 3) + D (x − 1)3.

By equating the coefficients of xo, x1, x2, x3 on the two sides of the last equation, we shall have four equations to determine the four quantities A, B, C, D, so that the assumption made is a legitimate one. The actual values of A, B, C, D are not however generally best found from the equations obtained by equating the coefficients of the different powers of x. In the present case, the following method is more expeditious.

Put x-1=y; then we have

2+2y+5= A (y − 2) + By (y − 2) + Cy2 (y − 2) + Dy3.

Now equate coefficients of y°, y1, y2, y3, and we have 7 - 2A ; 2A-2B; 0=B-2C; and 0=D+C.

(1 + x)" = A + B (1 − 2x) + C (1 − 2x)2 + (1 − 2x)3 × integral expression.