2
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ 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
0
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.