Assume we are working with
complex numbers . To reduce a
polynomial fraction to partial fractions is to rearrange it so it is the sum of terms of the form a/(x-b)
n
Suppose we have two
polynomials P(x) and Q(x), and we consider the fraction F(x)=P(x)/Q(x). The
Fundamental theorem of algebra tells us we can
factorise Q(x) completly, so
Q(x)=(x-p0)n0(x-p1)n1...(x-pk)nk
p
0,... p
k are the
poles of the fraction : they are where Q(x) is zero and thus where the fraction blows up. The aim of the game is to find numbers a
i,j and a
polynomial R(x) of degree less than that of Q such that :
F(x)=a
0,1/(x-p
0) + ... a
0,n0/(x-p
0)
n0+...+a
k,1/(x-p
k) + ... a
k,nk/(x-p
k)
nk + R(x)
R(x) is just the
remainder from the
division of P(x) by Q(x). Obtaining the a
i,ni is fairly easy. Just multiply the whole fraction by (x-p
i)
ni. All the terms in the
expression will now have a power of (x-p
i) in front of them, except the term the term with a
i,ni. Evaluating (x-p
i)
niF(x) at p
i allows you to get a
i,ni.
The trouble starts when you want to get the other coefficients. You can't just multiply by the appropriate polynomial and evaluate at the appropriate pole because some (x-p) terms will still be left in the denominator. One solution is to subtract ai,ni/(x-pi)ni from F(x). You can then multiply the resulting fraction G(x) by (x-pi)ni-1 and evaluate it at pi to obtain ai,ni-1. Repeat for all the multiplicities of all the poles and you will have F expressed in terms of partial fractions.
A simple example is F(x)=1/(x2-1). Its poles are 1 and -1.
F(x)(x-1)=1/(x+1)
F(x)(x+1)=1/(x-1)
Evaluating these 2 at 1 and -1 respectively gives 1/2 and -1/2.
Thus F(x) = 1/(2(x-1)) - 1/(2(x+1))
Although finding the coefficients is not always easy for large fractions, it makes many operations easier. In particular, integration is much easier, since we know the antiderivative of a/(x-b)^n.