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Partial Fractions

Date: 01/29/98 at 07:51:00
From: yanzhen
Subject: Partial fractions

Dear Dr. Maths,

How do I express the following in partial fractions?

(i)     ---------

(ii)    ---------
Thank you for your time!

Date: 01/29/98 at 14:00:48
From: Doctor Rob
Subject: Re: Partial fractions

First you have to factor the denominators into linear or quadratic 
factors. In this case

   1 - x^3 = (1 - x)*(1 + x + x^2)

   1 + x^3 = (1 + x)*(1 - x + x^2)

Those factors will be the denominators of the partial fractions.  
The numerators will be of lower degree, with unknown constant 
coefficients, so the numerator of a fraction with a degree-1 
denominator will just be an unknown constant, and the numerator of a 
fraction with a degree-2 denominator will be a degree-1 polynomial 
with unknown constant coefficients.  Thus

   3/(1-x^3) = A/(1-x) + (C+B*x)/(1+x+x^2)

where A, B, and C are constants, to be determined.  This must be an
identity, that is, true for all values of x.  Now clear fractions, and 
you get

   3 = A*(1+x+x^2) + (C+B*x)*(1-x)

Expand and collect terms, bringing everything to one side:

   0 = -3 + A + A*x + A*x^2 + C - C*x + B*x - B*x^2

   0 = (A - B)*x^2 + (A + B - C)*x + (A + C - 3)

This must still be an identity. Now there are two ways to proceed.

First, if this is true for all x values, we can pick a few and 
substitute them in, and get a system of linear equations in A, B, 
and C to solve:

   x = 0:   A + C - 3 = 0.
   x = 1:   3*A - 3 = 0.
   x = -1:  A - 2*B + 2*C - 3 = 0.

From the second equation A = 1, so from the first equation, C = 2, 
and then from the last equation, B = 1.

The second method is observe that if this is an identity, then the
coefficients of each power of x must all be zero. This gives the 
system of equations

   A - B = 0,
   A + B - C = 0,
   A + C - 3 = 0.

The solution is again A = B = 1, C = 2.

Either way, you have the identity

   3/(1-x^3) = 1/(1-x) + (x+2)/(x^2+x+1)

The second problem is similar. Using the above as a model, you should 
be able to solve it yourself.

-Doctor Rob,  The Math Forum
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Associated Topics:
High School Basic Algebra
High School Polynomials

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