Re: Partial fraction decomposition

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Posted by Soroban on September 30, 2002 at 15:17:31:

In Reply to: Partial fraction decomposition posted by sash on September 30, 2002 at 02:46:37:

: please help me with this problem which I've attempted some of it.
: A/(x-1)+ B/(x-1)^2+ C/(x-1)^3+ D/(x+7)+ (Ex+F)/(x^2+4)+ (Gx+H)/(x^2+4).

Hello, Sash!
Does that last fraction have (x^2+4)^2 in the denominator?
(Obviously, we shouldn't have two identical denominators.)

Many purists will complain, but the "Forbidden Zeros" method is much faster.

Multiply through (both sides) by the LCD: (x-1)^3 (x+7)(x^2+4)^2

On the left side, we have the numerator, f(x). On the right, we get:
A(x-1)^2 (x+7)(x^2+4)^2 + B(x-1)^2(x+7_(x^2+4)^2 + C(x+7)(x^2+4)^2
+ D(x-1)^3 (x^2+r)^2 + Ex(x-1)^3 (x+7)(x^2+4) + F(x-1)^3 (x+7)(x^2+4)
+ Gx(x-1)^3 (x+7) + H(x-1)^3 (x+7)

Now, select any value of x which makes a factor equal to 0.

Let x = 1. Everything vanishes except 200C ~ and the left side: f(1).
We have already found C.

Let x = -7, and we have -1438208D = f(-7).

Because of the repeated factors, this technique will NOT pop out all the values
as quickly. But it eliminates the need for expanding a fifth-degree
polynomial and collecting terms.

The objection most mathematicians/teachers have to this method is that
the values we are using (x = 1, -7) are the very values which are NOT in
the domain of the original function. This bothered me for some time,
but considering the work I avoid, I said, "I can live with it!"

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