# How do you find the power series for f(x)=int(ln(1+2t^2)dt from [0,x] and determine its radius of convergence?

Feb 28, 2017

Series is:

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

Converges absolutely for: $x \in \left(- \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$

#### Explanation:

For $f \left(x\right) = {\int}_{0}^{x} \ln \left(1 + 2 {t}^{2}\right) \setminus \mathrm{dt}$, we have from the FTC:

$f ' \left(x\right) = \ln \left(1 + 2 {x}^{2}\right)$

We note the following power series:

$\ln \left(1 + z\right) = {\sum}_{n = 1}^{\setminus \infty} {\left(- 1\right)}^{n + 1} \frac{{x}^{n}}{n} = x - {x}^{2} / 2 + {x}^{3} / 3 - \ldots$ for $| x | < 1$

So:

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

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

Integrating this:

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

We know from the integral that $f \left(0\right) = 0 \implies C = 0$, so:

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

For convergence we apply the ratio test:

${\left\mid \frac{{\left(- 1\right)}^{n + 2} {2}^{n + 1} \frac{{x}^{2 n + 3}}{\left(n + 1\right) \left(2 n + 3\right)}}{{\left(- 1\right)}^{n + 1} {2}^{n} \frac{{x}^{2 n + 1}}{n \left(2 n + 1\right)}} \right\mid}_{n \to \infty}$

$= {\left\mid - 2 {x}^{2} \cdot \frac{n \left(2 n + 1\right)}{\left(n + 1\right) \left(2 n + 3\right)} \right\mid}_{n \to \infty}$

$= {\left\mid - 2 {x}^{2} \cdot \frac{2 + \frac{1}{n}}{2 + \frac{5}{n} + \frac{3}{n} ^ 2} \right\mid}_{n \to \infty}$

$\implies \left\mid 2 {x}^{2} \right\mid < 1$

This will definitely converge for: $x \in \left(- \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$