How do you find a power series representation for $\frac{x}{1 - x^{2}}$ $x/(1-x^2)$ and what is the radius of convergence?

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How do you find a power series representation for $\frac{x}{1 - x^{2}}$ $x/(1-x^2)$ and what is the radius of convergence?

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Answer 1 · 2015-10-24T19:51:38

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

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

with radius of convergence $1$ $1$ .

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 $t = x^{2}$ $t = x^2$ to get:

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

Multiply by $x$ $x$ to get:

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

This is a geometric series with common ratio $x^{2}$ $x^2$ , so it will converge if $∣ ∣ x^{2} ∣ ∣ < 1$ $abs(x^2) < 1$ , which is when $| x | < 1$ $abs(x) < 1$ . That is, the radius of convergence is $1$ $1$ .

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How do you find a power series representation for $\frac{x}{1 - x^{2}}$ $x/(1-x^2)$ and what is the radius of convergence?

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