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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!

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    $\begingroup$ Really, start with the derivative and simplify it. $\endgroup$ Commented Jun 7, 2021 at 8:28
  • $\begingroup$ Thank you @IvanNeretin, so it has the same derivative as $\arctan(x)$, is that mean that they share the same series? $\endgroup$ Commented Jun 7, 2021 at 8:30
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    $\begingroup$ Plug 0 in both $\arctan(x)$ and $\arctan(\frac{x+1}{x-1})$. You should be able to see the difference. $\endgroup$
    – Jujustum
    Commented Jun 7, 2021 at 8:38
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    $\begingroup$ @CalculusLover Do not forget that there could be a non-zero constant of integration when returning to the original function. $\endgroup$
    – Gary
    Commented Jun 7, 2021 at 8:41
  • $\begingroup$ @Gary, yes I just figured out that is $\frac{\pi}{4}$. Thanks! $\endgroup$ Commented Jun 7, 2021 at 8:44

3 Answers 3

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

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  • $\begingroup$ Wow that is a cool answer! $\endgroup$ Commented Jun 7, 2021 at 8:45
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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 $\arctan\left(\frac{x+1}{x-1}\right)$.

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  • $\begingroup$ Thank you for that, the derivative has a $+$ in this case. $\endgroup$ Commented Jun 7, 2021 at 8:46
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    $\begingroup$ I'd double check that if I were you $\endgroup$
    – Stuck
    Commented Jun 7, 2021 at 8:48
  • $\begingroup$ check it again my friend :) $\endgroup$ Commented Jun 7, 2021 at 9:30
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    $\begingroup$ wolframalpha.com/input/… $\endgroup$
    – Stuck
    Commented Jun 7, 2021 at 9:32
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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.

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    $\begingroup$ Great, exactly what I ended up with. Thanks again! $\endgroup$ Commented Jun 7, 2021 at 8:45
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    $\begingroup$ Always a pleasure ! :) $\endgroup$
    – GuyPago
    Commented Jun 7, 2021 at 8:49

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