Taylor Series of $ \frac{1}{1-x^2} $ about x=2 I am trying to form a taylor series of the following:
$ \frac{1}{1-x^2} $ about $x=2$ 
I tried factoring the equation such that it becomes the following:
$ \frac{1}{{(1+x)}{(1-x)}} $
I tried to substitute $ x = h + 2 $ into the equation and obtained the following after using partial fractions to decompose the result:
$ \frac{1}{2(h+3)} - \frac{1}{2(h+1)} $
I do not know how to proceed from here.
I know I can just compute all the derivatives of the expression and evaluate them. But this would be non-trivial. Could someone please advise me on how I could solve this question?
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\begin{align}
{1 \over 1 - x^{2}}&=\half\,{1 \over 1 - x} + \half\,{1 \over 1 + x}
=-\,\half\,{1 \over 1 + \pars{x - 2}} + {1 \over 6}\,{1 \over 1 + \pars{x - 2}/3}
\\[3mm]&=-\half\sum_{n = 0}^{\infty}\pars{-1}^{n}\pars{x - 2}^{n}
+ {1 \over 6}\,\sum_{n = 0}^{\infty}\pars{-1}^{n}\pars{x - 2 \over 3}^{n}
\\[3mm]&=\sum_{n = 0}^{\infty}\pars{-1}^{n}
\bracks{-\,\half + {1 \over 6}\,\pars{1 \over 3}^{n}}\pars{x - 2}^{n}
\end{align}

$$\color{#00f}{\large%
{1 \over 1 - x^{2}}
=\sum_{n = 0}^{\infty}\pars{-1}^{n}\,
\half\,\pars{{1 \over 3^{n + 1}} - 1}\pars{x - 2}^{n}}\,,\qquad
\color{#000}{\large\verts{x - 2} < 1}
$$

A: Hint: Write $1 + x^2 = (x - 2)^2 + 4(x - 2) + 5 $
A: Notice that
$$\frac{1}{1 - x^2} = 1 + x^2 + x^4 + x^6 + \cdots = \sum_{k = 0}^\infty x^{2k}$$
provided $|x| < 1$. In general, however,
$$\frac{1}{1 - x} = \frac{1}{1 - a} + \frac{x - a}{(1 - a)^2} + \frac{(x - a)^2}{(1 - a)^3} + \cdots \quad \text{if} \quad |x - a| < 1.$$
Now, simply replace $x$ with $x^2$ and set $a = 2$.
The equality above follows from the formula
$$f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots$$
A: Hint: Since $x = (x-2) + 2$, then $x^2 = [(x-2) + 2]^2 = (x-2)^2 + 4(x-2) + 4$ so
\begin{align*}
\frac{1}{1 - x^2} &= \frac{1}{1 - [(x-2)^2 + 4(x-2) + 4]} = \frac{1}{-(x-2)^2 -4(x-2)-3}\\
&= -\frac{1}{[(x-2) + 3][(x-2) + 1]} \, .
\end{align*}
Now use partial fractions to finish it off.
A: If we have a function that has been expanded in a power series about the point $x = x_{0}$
\begin{equation}
f(x) = \sum\limits_{n=0}^{\infty} a_{n} (x - x_{0})^{n}
\end{equation}
and if the radius of convergence of this series is non-zero, then we can generate a power series expansion about a new point $x = x_{1}$ if this point is in the radius of convergence of the original power series. The new series is
\begin{equation}
f(x) = \sum\limits_{k=0}^{\infty} b_{k} (x - x_{1})^{k}
\end{equation}
\begin{equation}
b_{k} = \sum\limits_{n=0}^{\infty}
\begin{pmatrix}
n+k \\
k
\end{pmatrix}
a_{n+k} (x_{1} - x_{0})^{n}
\end{equation}
It is interesting that there exists a procedure to do this even if it is not practical to do so without a computer.
