I'm looking for a closed form for the coefficients of the series expansion of $\arctan(x)$ at some real $x_0$. If $x_0 = 0$, the expansion is of course the well-known

$$\sum_{n\geq 0} \frac{(-1)^n z^{2n+1}}{2n+1}\ .$$

In general, due to analyticity, the expansion is $$\sum_{n\geq 0} \arctan^{(n)}(x_0) \cdot (x - x_0)^n = \\\arctan(x_0) + \sum_{n\geq 0}\left(\frac{d^n}{dx^n}\frac{1}{1+x^2}\right)_{\!|\, x=x_0}\! \cdot (x - x_0)^{n+1}$$

but I'm struggling to find a closed form for these coefficients $$\left(\frac{d^n}{dx^n}\frac{1}{1+x^2}\right)_{\!|\, x=x_0}$$

In fact, I don't necessarily need a closed form as such; a recursive formula would also be all right.

UPDATE: Mathematica says the $n$-th coefficient in the series expansion of $\arctan(x)$ at $x = x_0$ is: $$\frac{-i}{2}\left((-i-x_0)^{-(n+1)} - (i-x_0)^{-(n+1)}\right)$$ One can probably derive this using a partial fraction expansion like Daniel Fischer suggested in his comment and then the geometric series. But I'm struggling to simplify this result to a ‘purely real’ formula.

UPDATE: After some rewriting, I found the following formula: $$\arctan(x) = \arctan(0) + \sum_{n=1}^\infty \frac{\sin(n\arg(i-x_0))}{n(x_0^2+1)^{\frac{n}{2}}} (x - x_0)^n$$

I guess one can simplify the $\arg(i-x_0)$ using $\arctan$. Not sure if the result is the easiest form one can get though.

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    $\begingroup$ If you're not afraid of complex numbers, the partial fraction decomposition $$\frac{1}{1+x^2} = \frac{1}{2i}\biggl(\frac{1}{x-i} - \frac{1}{x+i}\biggr)$$ helps finding a closed form for the derivatives. $\endgroup$ – Daniel Fischer Jan 13 '17 at 22:14
  • $\begingroup$ Yes, Mathematica gave me something like that as well when I asked it for the coefficients, but I'd very much prefer a ‘purely real’ solution if possible. $\endgroup$ – Manuel Eberl Jan 13 '17 at 22:24
  • $\begingroup$ Actually, I thought about it a bit more and I think I know what to do now. Thanks! $\endgroup$ – Manuel Eberl Jan 14 '17 at 9:25
  • $\begingroup$ At second thought, no, I still get stuck. $\endgroup$ – Manuel Eberl Jan 14 '17 at 9:52
  • $\begingroup$ Have you seen this? $\endgroup$ – J. M. is a poor mathematician Jan 14 '17 at 15:27

As shown in the updates of my question, I did find some rather ugly closed-form solutions for the coefficients, but I wasn't very happy with them.

I then realised that, given the partial fraction decomposition pointed out by Daniel Fischer, I could simply find a linear recurrence equation for them:

$$\arctan(x) = \arctan(x_0) + \sum_{n\geq 1} \frac{b_n}{n}(x-x_0)^n$$ $$\quad\text{where}\quad b_n = \begin{cases} 0 & \text{if}\ n = 0\\ \frac{1}{1+x_0^2} & \text{if}\ n = 1\\ -\frac{1}{1+x_0^2}\left(2x_0 b_{n-1} + b_{n-2}\right) & \text{otherwise}\end{cases}$$

That's pretty much exactly what I wanted – purely real and easy to compute. Thanks for all the suggestions!


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