Taylor-expansion through change of variables

Question

I need to make a Taylor series expansion about $x = 2$ of the function: $$f(x)=\frac{1}{1-x^2}$$ The general formula for Taylor series is: $$f(x)=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...$$

I know I can make this expansion by directly computing successive derivatives of $f(x)$ and evaluating them at $x=2$ but that would be burdensome. I know there is more elegant way to do this through change of variables. But I can't get my head around that concept. I guess the starting point should be the series expansion about $x = 0$: $$f(x)=1+x^2+x^4+x^6+...$$ But I'm not sure where to go from there.

Answer 1

$f(x)=\frac 1 2 (\frac 1 {1-x} +\frac 1{1+x})$ so it is enough to write down the Taylor expansion for the two terms separately. Now $\frac 1 {1-x}=-\frac 1 {1+(x-2)}=-\sum\limits_{n=0}^{\infty}(-1)^{n} (x-2)^{n}$. You can handle the second term in a similar fashion.

Answer 2

$$\frac1{1-x^2}=\frac{1/2}{1-x}+\frac{1/2}{1+x}.$$ To find the Taylor series at $x=2$, write $x=2+y$. Then $$\frac1{1-x^2}=\frac{1/2}{-1-y}+\frac{1/2}{3+y} =-\frac12\sum_{n=0}^\infty (-y)^n+\frac16\sum_{n=0}^\infty\frac{(-y)^n}{3^n}$$ etc.

Answer 3

HINT:$\frac{1}{1-x^2}=\frac{1}{2}[\frac{1}{1-x}+\frac{1}{1+x}]=\frac{1}{2}[\frac{-1}{1+(x-2)}+\frac{1}{3+(x-2)}]=\frac{-1}{2}[1+(x-2)]^{-1}+\frac{1}{6}[1+(x-2)/3]^{-1}$, etc.

Answer 4

Let $x=2+y$, then $$f(x)=\frac{-1}{(y+1)(y+3)}=\frac{1}{2}\left( \frac{1}{y+3} -\frac{1}{1+y}\right)=\frac{1}{2} \left( \frac{1}{3}\frac{1}{1+y/3} -\frac{1}{1+y}\right).$$ So when $|y|<1$ you can expand both the terms as infinite GP as $$f(x)=\frac{1}{2} \left( \frac{1}{3} [1+(y/3)+(y/3)^2+(y/3)^3+....]- [1+y+y^2+y^3+...] \right), ~\mbox{if}~ |y|<1.$$