 # Find the Taylor series centered at zero for the function f(x)=ln(2 + x^2). Determine the radius of convergence of this series. BenoguigoliB 2020-12-15 Answered
Find the Taylor series centered at zero for the function $f\left(x\right)=\mathrm{ln}\left(2+{x}^{2}\right)$. Determine the radius of convergence of this series.
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Find the Taylor series centered at zero for the function $f\left(x\right)=\mathrm{ln}\left(2+{x}^{2}\right)$. Determine the radius of convergence of this series.
Consider a given function is
$f\left(x\right)=\mathrm{ln}\left(2+{x}^{2}\right)$.
Now,
$f\left(x\right)=\mathrm{ln}\left[2\left(1+\frac{{x}^{2}}{2}\right)\right]$
$=\mathrm{ln}\left(2\right)+\mathrm{ln}\left(1+\frac{{x}^{2}}{2}\right)$
$=\mathrm{ln}\left(2\right)+\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n+1}\left(\frac{{x}^{2}}{2}{\right)}^{n}}{n}...\left[\mathrm{ln}\left(1+x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n+1}{x}^{n}}{n}\right]$
$=\mathrm{ln}\left(2\right)+\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n+1}{x}^{2n}}{n{2}^{n}}$
Hence, the required Taylor's series of $f\left(x\right)=\mathrm{ln}\left(2+{x}^{2}\right)$ is
$\mathrm{ln}\left(2\right)+\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n+1}{x}^{2n}}{n{2}^{n}}$
Consider a Taylor's series is $\mathrm{ln}\left(2\right)+\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n+1}{x}^{2n}}{n{2}^{n}}$
To find a Taylor's series, consider a sequence $<\frac{\left(-1{\right)}^{n+1}{x}^{2n}}{n{2}^{n}}>$
Now, by ratio test,
$L=\underset{n\to \mathrm{\infty }}{lim}|\frac{{a}_{n+1}}{{a}_{n}}|$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{\frac{\left(-1{\right)}^{n+1}{x}^{2n+2}}{\left(n+1\right){2}^{n+1}}}{\frac{\left(-1{\right)}^{n+1}{x}^{2n}}{n{2}^{n}}}|$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{n{2}^{n}{x}^{2n+2}}{\left(n+1\right){2}^{n+1}{x}^{2n}}|$
$=\underset{n\to \mathrm{\infty }}{lim}|\frac{n{x}^{2}}{\left(n+1\right)2}|$
$=|\frac{{x}^{2}}{2}|$

The series is convergent if $L<1$. Hence, $|\frac{{x}^{2}}{2}|<1$

It implies that 0

Check at the end point $x=\sqrt{2}$. Then,

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

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

It is convergent series.

Hence, the radius of convergence is $0$

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