Logan Wyatt

2022-07-01

Maclaurin series of $\mathrm{ln}\left(2+{x}^{2}\right)$
I know that $\mathrm{ln}\left(1+x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n-1}}{n}{x}^{n}$
So is $\mathrm{ln}\left(1+{x}^{2}\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n-1}}{n}{x}^{2n}$?
I don't know what to do next then.

Oliver Shepherd

Expert

We are looking at $\mathrm{ln}2+\mathrm{ln}\left(1+\frac{{x}^{2}}{2}\right)$. Use the series for $\mathrm{ln}\left(1+t\right)$, and replace $t$ everywhere by $\frac{{x}^{2}}{2}$

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