Find the Taylor Series of a function

Question

My goal is to find the Taylor Series (around zero) for the following function:

$$f\left(x\right)\:=\:\frac{x}{\:\left(1-x\right)\left(1+x\right)^2}$$

I figured out that I should use partial fractions and make it a sum of known series.

Unfortunately, I can't get rid of the square at the denominator, therefore I am struggling to get a simple form that I can work with.

I tried Ax+B, A+B+C and so on. Couldn't solve it yet.

Is it really the way to solve it? If so, what should I do with the square at the denominator?

Thanks for helping,

Answer 1

You can't get rid of the square. You can write your function as $$f(x)=\frac{A}{1-x}+\frac{B}{1+x}+\frac{C}{(1+x)^2},$$ and use $$\frac{1}{(1+x)^2}=-\frac{d}{dx}\frac{1}{1+x}.$$ Does that help?

Answer 2

The typical approach to partial fractions in a case like yours is $$\frac{A}{1-x} + \frac{Bx+C}{(1+x)^2}$$ or, equivalently, $$\frac{A}{1-x} + \frac{B}{1+x} + \frac{C}{(1+x)^2}$$(where by "equivalent" I don't mean that you get the same $B$ and $C$, only that the two different approaches ultimately cover exactly the same cases).

Answer 3

You need the model $$ \frac{x}{(x-1)(x+1)^2}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x+1)^2}. $$ This yields $$ x=A(x+1)^2+B(x-1)(x+1)+C(x-1)=(A+B)x^2+(2A+C)x+(-B-C) $$ for all $x\in\mathbb R$. Therefore we get the system \begin{align} A+B&=0\\ 2A+C&=1\\ B+C&=0 \end{align} whose solution is $A=\frac13$, $B=-\frac13$ and $C=\frac13$ hence $$ \frac{x}{(x-1)(x+1)^2}=\frac13\cdot\frac1{x-1}-\frac13\cdot\frac1{x+1}+\frac13\cdot\frac1{(x+1)^2} $$