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

Mar 17, 2017

${x}^{2} \arctan \left({x}^{3}\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(2 n + 1\right) {x}^{6 n + 5} = {x}^{5} - {x}^{11} / 3 + {x}^{17} / 5 + \ldots$

The radius of convergence is $R = 1$.

#### Explanation:

$\frac{1}{1 - x} = {\sum}_{n = 0}^{\infty} {x}^{n}$

Replacing $x$ with $- {x}^{2}$:

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

Integrating to get $\arctan \left(x\right)$:

$\arctan \left(x\right) = \int \frac{\mathrm{dx}}{1 + {x}^{2}} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} \int {x}^{2 n} \mathrm{dx}$

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

Letting $x = 0$ shows $C = 0$:

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

Replacing $x$ with ${x}^{3}$:

$\arctan \left({x}^{3}\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(2 n + 1\right) {\left({x}^{3}\right)}^{2 n + 1} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(2 n + 1\right) {x}^{6 n + 3}$

Multiplying by ${x}^{2}$, which can be brought into the series:

x^2arctan(x^3)=x^2sum_(n=0)^oo(-1)^n/(2n+1)x^(6n+3)=color(blue)(sum_(n=0)^oo(-1)^n/(2n+1)x^(6n+5)

This is the Maclaurin series for ${x}^{2} \arctan \left({x}^{3}\right)$. To find its radius of convergence, find the ratio $\left\mid {a}_{n + 1} / {a}_{n} \right\mid$ where ${a}_{n} = {\left(- 1\right)}^{n} / \left(2 n + 1\right) {x}^{6 n + 5}$:

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = \left\mid \frac{{\left(- 1\right)}^{n + 1} / \left(2 \left(n + 1\right) + 1\right) {x}^{6 \left(n + 1\right) + 5}}{{\left(- 1\right)}^{n} / \left(2 n + 1\right) {x}^{6 n + 5}} \right\mid = \left\mid \frac{{\left(- 1\right)}^{n + 1} / \left(2 n + 3\right) {x}^{6 n + 11}}{{\left(- 1\right)}^{n} / \left(2 n + 1\right) {x}^{6 n + 5}} \right\mid$

$\left\mid {a}_{n + 1} / {a}_{n} \right\mid = \left\mid {\left(- 1\right)}^{n + 1} / {\left(- 1\right)}^{n} \left(\frac{2 n + 1}{2 n + 3}\right) {x}^{6 n + 11} / {x}^{6 n + 5} \right\mid = \left\mid {x}^{6} \left(\frac{2 n + 1}{2 n + 3}\right) \right\mid$

Through the ratio test, we know the series converges when ${\lim}_{n \rightarrow \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid < 1$. Finding the limit:

${\lim}_{n \rightarrow \infty} \left\mid {a}_{n + 1} / {a}_{n} \right\mid = {\lim}_{n \rightarrow \infty} \left\mid {x}^{6} \left(\frac{2 n + 1}{2 n + 3}\right) \right\mid = {\left\mid x \right\mid}^{6} {\lim}_{n \rightarrow \infty} \left\mid \frac{2 n + 1}{2 n + 3} \right\mid$

The internal limit is $1$. We want this when it's less than $1$:

${\left\mid x \right\mid}^{6} < 1 \text{ "=>" } \left\mid x \right\mid < 1$

Thus the radius of convergence is color(blue)(R=1.