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

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 $$\arctan(x)$$?

--

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!

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

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

• Wow that is a cool answer! Jun 7 at 8:45

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.

• Great, exactly what I ended up with. Thanks again! Jun 7 at 8:45
• Always a pleasure ! :) Jun 7 at 8:49

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 $$\arctan\left(\frac{x+1}{x-1}\right)$$.

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