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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

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