What are the criteria for partial fractions?

Question

Consider the example of

$$1\over (u^2 - 1)(u+1)$$.

$${Au + B\over u^2 - 1} + {C \over u + 1} = {Au^2 + Au + Bu + B + Cu^2 - C\over (u^2 - 1)(u+1)} = {1\over (u^2 - 1)(u+1)}$$

$$\begin {cases}A + C =0 \\ A + B = 0\\ B - C = 1\end{cases}$$

Adding $3$ and $1$,

$$\begin {cases}A + B =1 \\ A + B = 0\end{cases}$$

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The degree of denominator is $3$ and that of numerator is $0$, so I should not have got inconsistent group of equations.

What are the criteria for partial fractions ? what am I missing in this example ?

Also Wolfram Alpha gives proper partial fraction of the following.

Answer 1

Let's change variables to make things clearer (won't affect the theory, of course). If we put $1+u=x$, the original fraction becomes $$ \frac{1}{((x-1)^2-1)x} = \frac{1}{(x^2-2x)x} = \frac{1}{x^2(x-2)}. $$ Now, your question boils down to what happens when we add two fractions $$ \frac{A}{x}+\frac{Bx+C}{x(x-2)}. $$ But we can put them over the common denominator $x(x-2)$, so $$ \frac{A}{x}+\frac{Bx+C}{x(x-2)} = \frac{Ax-2A+Bx+C}{x(x-2)} = \frac{(A+B)x-2A+C}{x(x-2)}! $$ So this is the problem: no matter what polynomial we put on top of these fractions, they will never produce a denominator that is $x^2(x-2)$!

Okay, what about $$ \frac{A}{x^2} + \frac{Bx+C}{x(x-2)}? $$ This looks better, since we have to put it over a common denominator $x^2(x-2)$. Then we get $$ \frac{Ax-2A+Bx^2+Cx}{x^2(x-2)}. $$ Therefore we're forced to take $B=0$. We also have to get rid of the $x$ term, so $A=-C$. Then you can choose $A$ in order to get $-2A=1$, so $$ \frac{-1/2}{x^2} + \frac{1/2}{x(x-2)} $$ is a valid partial fraction decomposition for $1/x^2(x-2)$ (and equally, $$ \frac{-1/2}{(1+u)^2} + \frac{1/2}{u^2-1} = \frac{1}{(u+1)(u^2-1)}). $$

Answer 2

sunce we have $$\frac{1}{(u^2-1)(u+1)}=\frac{1}{(u-1)(u+1)^2}$$ you must make the ansatz $$\frac{A}{u-1}+\frac{B}{u+1}+\frac{C}{(u+1)^2}$$

Answer 3

The function $$ f(u)=\frac{1}{(u^2-1)(u+1)}=\frac{1}{(u-1)(u+1)^2} $$ has a simple pole at $u=1$ and a double pole at $u=-1$. It follows that its partial fraction decomposition has the following form: $$ f(u) =\frac{A}{u-1}+\frac{B}{u+1}+\frac{C}{(u+1)^2} $$ where $A$ can be found through $$ A = \lim_{u\to 1}(u-1)\,f(u) = \lim_{u\to 1}\frac{1}{(u+1)^2} = \frac{1}{4} $$ and $C$ can be found through $$ C = \lim_{u\to -1}(u+1)^2\,f(u)=\lim_{u\to -1}\frac{1}{u-1}=-\frac{1}{2}. $$ We have $A+B=0$ because $f(u)$ decays faster than $\frac{1}{u^2}$ at infinity, hence $B=-\frac{1}{4}$ and finally

$$ f(u) = \frac{1/4}{u-1}-\frac{1/4}{u+1}-\frac{1/2}{(u+1)^2}. $$