# How do you find a power series representation for  (x/(2-x))^3 and what is the radius of convergence?

Oct 21, 2016

${\left(\frac{x}{2 - x}\right)}^{3} \equiv {\sum}_{k = 2}^{\infty} \frac{k \left(k - 1\right)}{2} ^ \left(k + 2\right) {x}^{k + 1}$ which is convergent for $\left\mid x \right\mid < 2$

#### Explanation:

We have ${\left(\frac{x}{2 - x}\right)}^{3} = {\left(\frac{x}{2}\right)}^{3} {\left(\frac{1}{1 - \frac{x}{2}}\right)}^{3}$

Calling $y = \frac{x}{2}$ we have

${\left(\frac{x}{2 - x}\right)}^{3} \equiv {y}^{3} {\left(\frac{1}{1 - y}\right)}^{3}$

but

${d}^{2} / {\mathrm{dy}}^{2} \left(\frac{1}{1 - y}\right) = 2 {\left(\frac{1}{1 - y}\right)}^{3}$

Now, for $\left\mid y \right\mid < 1$ we have

$\frac{1}{1 - y} = {\sum}_{k = 0}^{\infty} {y}^{k}$ then

${y}^{3} {\left(\frac{1}{1 - y}\right)}^{3} = {y}^{3} / 2 {d}^{2} / {\mathrm{dy}}^{2} {\sum}_{k = 0}^{\infty} {y}^{k} = {y}^{3} / 2 {\sum}_{k = 2}^{\infty} k \left(k - 1\right) {y}^{k - 2}$ or

${y}^{3} {\left(\frac{1}{1 - y}\right)}^{3} = \frac{1}{2} {\sum}_{k = 2}^{\infty} k \left(k - 1\right) {y}^{k + 1}$

but $y = \frac{x}{2}$ so finally

${\left(\frac{x}{2 - x}\right)}^{3} \equiv {\sum}_{k = 2}^{\infty} \frac{k \left(k - 1\right)}{2} ^ \left(k + 2\right) {x}^{k + 1}$ which is convergent for $\left\mid \frac{x}{2} \right\mid < 1$ or $\left\mid x \right\mid < 2$