$\arctan(\frac{x+1}{x-1})$ to power series

Question

EDIT: I just realize that I should start with the derivative of $\arctan(\frac{x+1}{x-1})$ , and keep going from there.

So $(\arctan(\frac{x+1}{x-1}))'=\frac{1}{1+x^2}$, does that mean this is the same series as $\arctan(x)$?

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I want to find an expression for $\arctan(\frac{x+1}{x-1})$ as a power series, with $x_0=0$, for every $x \ne 1$.

My initial thought was to use the known $\arctan(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}$, but I don't know how to keep going if I replace $x$ with $\frac{x+1}{x-1}$.

Thanks a lot!

Answer 1

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

Answer 2

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

Answer 3

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