Alan Smith

2021-12-26

$\mathrm{arctan}\left(\frac{x+1}{x-1}\right)$ to power series
I want to find an expression for $\mathrm{arctan}\left(\frac{x+1}{x-1}\right)$ as a power series, with ${x}_{0}=0$, for every $x\ne 1$
My initial thought was to use the known $\mathrm{arctan}\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}{x}^{2n+1}}{2n+1}$ but I don't know how to keep going if I replace x with $\frac{x+1}{x-1}$

zesponderyd

Let
It can be easily showed that:
${f}^{\prime }\left(x\right)=\frac{1}{1+{x}^{2}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n}{x}^{2n}$
Integrating both sides yields that $\mathrm{\exists }C\in \mathbb{R}$ such that:
$\mathrm{arctan}\left(\frac{1+x}{1-x}\right)+C=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-1\right)}^{n}{x}^{2n+1}}{2n+1}$
Check for f(0) to conclude C and you're done.

Annie Levasseur

Use the identity $\mathrm{arctan}\left(\frac{x+1}{x-1}\right)=-\mathrm{arctan}\left(\frac{x+1}{1-x}\right)=-\left(\frac{\pi }{4}+\mathrm{arctan}\left(x\right)\right)$

karton

Since
$\frac{d}{dx}\mathrm{arctan}\left(\frac{x+1}{x-1}\right)=-\frac{1}{1+{x}^{2}}$
you can express the RHS as a power series, and then integrate the result to get the desired series for your original function $\mathrm{arctan}\left(\frac{x+1}{x-1}\right)$

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