How do you find the taylor series for (1/(1-x^2)) centered around a = 3?

Aug 2, 2016

$\frac{1}{1 - {x}^{2}} \equiv - \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / {2}^{n + 1} \left(1 - \frac{1}{2} ^ \left\{n + 1\right\}\right) {\left(x - 3\right)}^{n}$

for $1 < x < 5$

Explanation:

Making now $y = x - 3$

$f \left(x\right) = \frac{1}{1 - {x}^{2}} \equiv \frac{1}{1 - {\left(y + 3\right)}^{2}} = {C}_{1} / \left(1 + y + 3\right) + {C}_{2} / \left(1 - y - 3\right)$
$= {C}_{1} / \left(y + 4\right) - {C}_{2} / \left(y + 2\right)$

Solving for ${C}_{1} , {C}_{2}$ the condition

${C}_{1} \left(y + 2\right) - {C}_{2} \left(y + 4\right) = 1 , \forall y \in \mathbb{R}$ or equivalently

$\left({C}_{1} - {C}_{2}\right) y + 2 {C}_{1} - 4 {C}_{2} = 1 , \forall y \in \mathbb{R}$ we obtain

${C}_{1} = {C}_{2} = - \frac{1}{2}$ so

$\frac{1}{1 - {x}^{2}} \equiv \frac{1}{2} \left(\frac{1}{y + 2} - \frac{1}{y + 4}\right) = \frac{1}{2} \left(\frac{1}{2 \left(\frac{y}{2} + 1\right)} - \frac{1}{4 \left(\frac{y}{4} + 1\right)}\right)$

Now considering that for $\left\mid z \right\mid < 1$

${\lim}_{n \to \infty} \frac{{z}^{n} - 1}{z - 1} = - \frac{1}{z - 1} = {\sum}_{k = 0}^{\infty} {z}^{n}$

restraining $\left\mid \frac{y}{2} \right\mid < 1$ then

$\frac{1}{\frac{y}{2} + 1} \equiv - {\sum}_{n = 0}^{\infty} {\left(- \frac{y}{2}\right)}^{2}$ and
$\frac{1}{\frac{y}{4} + 1} \equiv - {\sum}_{n = 0}^{\infty} {\left(- \frac{y}{4}\right)}^{2}$

Putting all together

$\frac{1}{2} \left(\frac{1}{y + 2} - \frac{1}{y + 4}\right) \equiv s \left(y\right)$

$s \left(y\right) = - \frac{1}{2} \left(\frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(- \frac{y}{2}\right)}^{n} - \frac{1}{4} {\sum}_{n = 0}^{\infty} {\left(- \frac{y}{4}\right)}^{n}\right)$

This serie in convergent for $\left\mid y \right\mid < 2$ or equivalently $1 < x < 5$

Finally, for $1 < x < 5$ we have $f \left(x\right) \equiv s \left(x - 3\right)$

$s \left(x - 3\right) = - \frac{1}{2} {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / {2}^{n + 1} \left(1 - \frac{1}{2} ^ \left\{n + 1\right\}\right) {\left(x - 3\right)}^{n}$