Mathematical Methods for Physics and Engineering : A Comprehensive Guide

1.4 PARTIAL FRACTIONS

in partial fractions, i.e. to write it as

f(x)=

g(x)

h(x)

=

4 x+2

x^2 +3x+2

=

A 1

(x−α 1 )n^1

+

A 2

(x−α 2 )n^2

+···.

(1.43)

The first question that arises is that of how many terms there should be on

the right-hand side (RHS). Although some complications occur whenh(x)has

repeated roots (these are considered below) it is clear thatf(x) only becomes

infinite at thetwovalues ofx,α 1 andα 2 , that makeh(x) = 0. Consequently the

RHS can only become infinite at the same two values ofxand therefore contains

only two partial fractions – these are the ones shown explicitly. This argument

can be trivially extended (again temporarily ignoring the possibility of repeated

roots ofh(x)) to show that ifh(x) is a polynomial of degreenthen there should be

nterms on the RHS, each containing a different rootαiof the equationh(αi)=0.

A second general question concerns the appropriate values of theni.Thisis

answered by putting the RHS over a common denominator, which will clearly

have to be the product (x−α 1 )n^1 (x−α 2 )n^2 ···. Comparison of the highest power

ofxin this new RHS with the same power inh(x)showsthatn 1 +n 2 +···=n.

This result holds whether or noth(x) = 0 has repeated roots and, although we

do not give a rigorous proof, strongly suggests the following correct conclusions.

• The number of terms on the RHS is equal to the number ofdistinctroots of

h(x) = 0, each term having a different rootαiin its denominator (x−αi)ni.

• Ifαiis a multiple root ofh(x) = 0 then the value to be assigned toniin (1.43) is

that ofmiwhenh(x) is written in the product form (1.9). Further, as discussed

on p. 23,Aihas to be replaced by a polynomial of degreemi−1. This is also

formally true for non-repeated roots, since then bothmiandniare equal to

unity.

Returning to our specific example we note that the denominatorh(x) has zeros

atx=α 1 =−1andx=α 2 =−2; thesex-values are the simple (non-repeated)

roots ofh(x) = 0. Thus the partial fraction expansion will be of the form

4 x+2

x^2 +3x+2

=

A 1

x+1

+

A 2

x+2

. (1.44)

We now list several methods available for determining the coefficientsA 1 and

A 2. We also remind the reader that, as with all the explicit examples and techniques

described, these methods are to be considered as models for the handling of any

ratio of polynomials, with or without characteristics that make it a special case.

(i) The RHS can be put over a common denominator, in this case (x+1)(x+2),

and then the coefficients of the various powers ofxcan be equated in the