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

Oct 24, 2015

Use the Maclaurin series for $\frac{1}{1 - t}$ and substitution to find:

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

with radius of convergence $\sqrt[3]{2}$

#### Explanation:

The Maclaurin series for $\frac{1}{1 - t}$ is ${\sum}_{n = 0}^{\infty} {t}^{n}$

since $\left(1 - t\right) {\sum}_{n = 0}^{\infty} {t}^{n} = {\sum}_{n = 0}^{\infty} {t}^{n} - t {\sum}_{n = 0}^{\infty} {t}^{n} = {\sum}_{n = 0}^{\infty} {t}^{n} - {\sum}_{n = 1}^{\infty} {t}^{n} = {t}^{0} = 1$

Substitute $t = {x}^{3} / 2$

Then we find:

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

Multiply by ${x}^{3} / 2$ to find:

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

This is a geometric series with common ratio ${x}^{3} / 2$ so converges when $\left\mid {x}^{3} / 2 \right\mid < 1$ which is when $\left\mid \frac{x}{\sqrt[3]{2}} \right\mid < 1$, which is when $\left\mid x \right\mid < \sqrt[3]{2}$. So the radius of convergence is $\sqrt[3]{2}$.