Taylor Series of $ \frac{1}{1-x^2} $ about x=2

Question

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?

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}

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

Answer 2

Hint: Write $1 + x^2 = (x - 2)^2 + 4(x - 2) + 5 $

Answer 3

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 $$f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots$$

Answer 4

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.

Answer 5

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.