How to Decompose into Partial Fractions

Question

Why is it useful to write:

$$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x+1}+\frac{}{x-1}$$

and not:

$$\frac{}{(x+1)^2(x-1)}=\frac{}{(x+1)^2}+\frac{}{x-1}$$

when decomposing into partial fractions?

Answer 1

For example, try decomposing $\frac{1}{(x+1)^2(x-1)}$ using the second form: \begin{align} \frac{1}{(x+1)^2(x-1)} &= \frac{A}{(x+1)^2} + \frac{B}{x-1} \\ \frac{1}{(x+1)^2(x-1)} &= \frac{A(x-1) + B(x+1)^2}{(x+1)^2(x-1)}. \end{align} Then $B$ must be zero by looking at the $x^2$ term, and then $A$ must be zero as well. So as the commenter pointed out, it just doesn't work. But using the first form we get $$\frac{1}{(x+1)^2(x-1)} = -\frac{1}{2 (x+1)^2}-\frac{1}{4 (x+1)}+\frac{1}{4 (x-1)}.$$

Answer 2

Suppose you have a rational function $r(x) = \frac{p(x)}{q(x)}$ with $p(x)$ and $q(x)$ polynomials and the degree of $q(x)$ is larger than the degree of $p(x)$. You can examine its behavior near the zeroes of $q(x)$. If $q(y) = 0$ then you can expand $r(x)$ around $x= y$ as:

$$r(x) = \frac{A_n}{(x-y)^n} + \frac{A_{n-1}}{(x-y)^{n-1}} +\cdots$$

where $n$ is the multiplicity of the zero at $x=y$ of $q(x)$. Suppose that to approximate $r(x)$ globally, you add up the expansion for each zero of $r(x)$ and you keep only the singular terms. Then that would not just be an approximation to $r(x)$, it would be the same as $r(x)$, because the difference between $r(x)$ and the sum of all the singular parts opf all the expansions obviously has no singularities left. Therefore that difference is a polynomial. However, this polynomial tends to zero at infinity because $r(x)$ tends to zero at infinity and all the singular terms we have subracted also tend to zero. This means that the polynomial is in fact identical to zero.

So, you see that each term of the partial fraction expansion comes from the expansion around the singularities of the rational function and you need to keep all the singular terms, not just the most dominant ones with the largest negative power.