Associated Topics || Dr. Math Home || Search Dr. Math

### General Expression for Partial Fractions

```
Date: 02/09/99 at 04:01:38
From: TOMMY
Subject: Partial fractions

Could you please tell me the general algorithm to convert an expression
into partial fraction form?

Thank you!
```

```
Date: 02/09/99 at 10:56:56
From: Doctor Rob
Subject: Re: Partial fractions

Thanks for writing to Ask Dr. Math!

This applies to the quotient of two polynomials, g(x)/f(x). If g(x) has
degree at least as large as that of f(x), then divide:

g(x) = q(x)*f(x) + r(x),  deg(r) < deg(f) or r = 0
g(x)/f(x) = q(x) + r(x)/f(x)

Then you want to express r(x)/f(x) as a sum of fractions, and the
degree of r(x) is smaller than that of f(x). This means that you can
assume without loss of generality that g(x) has degree less than that
of f(x). We let n = degree(f).

Now if f(x) has real coefficients, we know that we can factor it as
follows:

t
f(x) = K * PRODUCT (x - a[i])^e[i] *
i=1
u
PRODUCT (x^2 - b[j]*x + c[j])^d[j]
j=1

where the a's, b's, c's, and K are all real numbers, and the d's and
e's are positive integers. The a's are all the real roots of f(x), and
the quadratic factors have as roots all the complex conjugate pairs of
roots of f(x).

Then the partial fraction theorem says that there are constants A[i,r],
B[j,s], and C[j,s] such that

t  e[i]
g(x)/f(x) = SUM SUM  A[i,r]/(x-a[i])^r +
i=1 r=1
u  d[j]
SUM SUM  (B[j,s]*x+C[j,s])/(x^2-b[j]*x+c[j])^s
j=1 s=1

This may be the "general expression" to which you were referring in
your question. Next, multiply through by f(x),

g(x) = SUM SUM A[i,r]*[f(x)/(x-a[i])^r] +
SUM (B[j,s]*x+C[j,s])*[f(x)/(x^2-b[j]*x+c[j])^s]

and do the indicated divisions, which will leave no remainders. Now
there are two ways to find the A's, B's, and C's. The first is to
expand everything in sight into a polynomial, and gather together all
terms involving the same powers of x. Then you will have an equation
stating the equality of g(x) on the left and a polynomial in x on the
right side. For them to be equal, the coefficients of each power of x
on each side must be equal, and this gives you n real linear equations
in the n real unknown A's, B's, and C's. These you can solve by various
methods, for example, Gaussian elimination.

The second way is to substitute for x in the resulting equation n
different values. Each will give you a real linear equation relating
the values of the real unknown A's, B's, and C's. Again you can solve
these by various methods, for example, Gaussian elimination.

Example: Write (x+1)/x^4 using partial fractions. Then

f(x) = x^4

so t = 1, u = 0, a[1] = 0, e[1] = 4, and the theorem says that there
exist constants A[1,1], A[1,2], A[1,3], and A[1,4] such that

(x+1)/x^4 = A[1,1]/x + A[1,2]/x^2 + A[1,3]/x^3 + A[1,4]/x^4

Multiplying through by x^4 and dividing out, you get

x + 1 = A[1,1]*x^3 + A[1,2]*x^2 + A[1,3]*x + A[1,4]

Now equating coefficients on both sides of the equation, we get

A[1,1] = 0
A[1,2] = 0
A[1,3] = 1
A[1,4] = 1

This system of linear equations is already solved for us, so we are
done, and

(x+1)/x^4 = 1/x^3 + 1/x^4

If instead we substitute x = -1, 0, 1, and 2, we would get the system
of equations

0 =  -A[1,1] +   A[1,2] -   A[1,3] + A[1,4]
1 =                                  A[1,4]
2 =   A[1,1] +   A[1,2] +   A[1,3] + A[1,4]
3 = 8*A[1,1] + 4*A[1,2] + 2*A[1,2] + A[1,4]

These have, of course, the same solutions as before.

If the coefficients of f(x) are complex numbers, then factor f(x)
completely over the complex numbers, obtaining

t
f(x) = K * PRODUCT (x-a[i])^e[i]
i=1

where the a's are complex roots of f(x), the e's are positive integers,
and K is a complex number. Then the partial fraction theorem says that
you can find constants A[i,r] such that

t  e[i]
g(x)/f(x) = SUM SUM  A[i,r]/(x-a[i])^r
i=1 r=1

This may be the "general expression" to which you were referring in
your question. Now proceed as before, with u = 0. Now the n A's will
be complex numbers, in general, that are found by solving n linear
equations with complex coefficients.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Polynomials

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search