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

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

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Answer 1 · 2015-10-24T19:26:34

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

$\frac{x^{3}}{2 - x^{3}} = \infty \sum n = 0 2^{- n - 1} x^{3 n + 3}$ $x^3/(2-x^3) = sum_(n=0)^oo 2^(-n-1) x^(3n+3)$

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

Explanation:

The Maclaurin series for $\frac{1}{1 - t}$ $1/(1-t)$ is $\infty \sum n = 0 t^{n}$ $sum_(n=0)^oo t^n$

since $(1 - t) \infty \sum n = 0 t^{n} = \infty \sum n = 0 t^{n} - t \infty \sum n = 0 t^{n} = \infty \sum n = 0 t^{n} - \infty \sum n = 1 t^{n} = t^{0} = 1$ $(1-t) sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - t sum_(n=0)^oo t^n = sum_(n=0)^oo t^n - sum_(n=1)^oo t^n = t^0 = 1$

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

Then we find:

$\frac{2}{2 - x^{3}} = \frac{1}{1 - \frac{x^{3}}{2}} = \infty \sum n = 0 {(\frac{x^{3}}{2})}^{n} = \infty \sum n = 0 2^{- n} x^{3 n}$ $2/(2-x^3) = 1/(1-x^3/2) = sum_(n=0)^oo (x^3/2)^n = sum_(n=0)^oo 2^(-n) x^(3n)$

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

$\frac{x^{3}}{2 - x^{3}} = \frac{x^{3}}{2} \infty \sum n = 0 2^{- n} x^{3 n} = \infty \sum n = 0 2^{- n - 1} x^{3 n + 3}$ $x^3/(2-x^3) = x^3/2 sum_(n=0)^oo 2^(-n) x^(3n) = sum_(n=0)^oo 2^(-n-1) x^(3n+3)$

This is a geometric series with common ratio $\frac{x^{3}}{2}$ $x^3/2$ so converges when $∣ ∣ ∣ \frac{x^{3}}{2} ∣ ∣ ∣ < 1$ $abs(x^3/2) < 1$ which is when $∣ ∣ ∣ \frac{x}{\sqrt[3]{2}} ∣ ∣ ∣ < 1$ $abs(x/root(3)(2)) < 1$ , which is when $| x | < \sqrt[3]{2}$ $abs(x) < root(3)(2)$ . So the radius of convergence is $\sqrt[3]{2}$ $root(3)(2)$ .

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

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