# Use the power series To determine a power series centered at 0, for the function.Identify the interval of convergence. f(x)=ln(x^2+1)

Use the power series
To determine a power series centered at 0, for the function.Identify the interval of convergence.
$f\left(x\right)=\mathrm{ln}\left({x}^{2}+1\right)$
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Jaycee Figueroa
$g\left(x\right)=\frac{1}{1+x}=1-x+{x}^{2}-{x}^{3}+{x}^{4}$
We must recognize a connection between 1/(1+x) and the naturallog. well if we integrate what we have now we'd end up with $\mathrm{ln}\left(1+x\right)$, so that's a start, but we want to end up with $\mathrm{ln}\left({x}^{2}+1\right)$. So we'd need $2x/\left[1+{x}^{2}\right]$. So we have
$g\left({x}^{2}\right)=\frac{1}{1+{x}^{2}}=1-{x}^{2}+{x}^{4}-{x}^{6}+{x}^{8}$
Multiplying everything by x
$xg\left({x}^{2}\right)=\frac{x}{1+{x}^{2}}=x-{x}^{3}+{x}^{5}-{x}^{7}+{x}^{9}$
And by 2
$2xg\left({x}^{2}\right)=\frac{2x}{1+{x}^{2}}=2x-2{x}^{3}+2{x}^{5}-2{x}^{7}+2{x}^{9}$
Integrating letting $u=1+{x}^{2}du=2xdx$, we find
$\mathrm{ln}\left(1+{x}^{2}\right)={x}^{2}-\frac{{x}^{4}}{2}+\frac{{x}^{6}}{3}-\frac{{x}^{8}}{4}+\frac{{x}^{10}}{5}$
Hence
$\mathrm{ln}\left(1+{x}^{2}\right)=\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n+1}\frac{{x}^{2n}}{n}$
$\underset{n\to \mathrm{\infty }}{lim}|\frac{{x}^{2n+1}}{n+1}\ast \frac{n}{{x}^{2n}}|<1$
$|x|\underset{n\to \mathrm{\infty }}{lim}|\frac{n}{n+1}|<1$
|x|<1
Now we must check the endpoints individually. For x =1, we have a alternating harmonic series, which isconvergent. Same for x = -1. So the intervalis [-1,1]
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Greyson Landry
$1-x+{x}^{2}-{x}^{3}+{x}^{4}-{x}^{5}+{x}^{6}-{x}^{7}+{x}^{8}-{x}^{9}+{x}^{10}-{x}^{11}+{x}^{12}-{x}^{13}+{x}^{14}-{x}^{15}+....................$
$f\left(x\right)=\mathrm{ln}\left({x}^{2}+1\right)={x}^{2}-{x}^{4/2}+{x}^{6/3}-{x}^{8/4}+{x}^{10/5}-{x}^{12/6}+............................$