I'm interested in the residue of $\exp(z-z^{-1})$ at $z=0$.

We have $(z-z^{-1})^n=\sum_{k=0}^n\binom{n}{k}z^{n-k}(-z)^{-k}=\sum_{k=0}^n \binom{n}{k}(-1)^k z^{n-2k}$ so

$\exp(z-z^{-1})=\sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^n \binom{n}{k}(-1)^k z^{n-2k}$

But as I see it, I don't really have a chance to get the coefficient $a_{-1}$ since for every odd $n$ there is a $k$ such that $n-2k=-1$.

Edit: A non-closed form of the residue would be $\sum_{n\ \text{odd}}\frac{(-1)^{\frac{n+1}{2}}}{\left(\frac{n+1}{2}\right)!\left(\frac{n+-1}{2}\right)!}$

  • $\begingroup$ It's probably easier to compute the residue by evaluating an integral around a circle. $\endgroup$ – Chappers Aug 3 '17 at 13:45
  • $\begingroup$ You mean $\int_0^{2\pi}i\exp(e^{it}-e^{-it})e^{it}dt$? $\endgroup$ – user424862 Aug 3 '17 at 13:47
  • $\begingroup$ Yes. That's $2\pi i$ times the residue at zero since the function's analytic elsewhere. $\endgroup$ – Chappers Aug 3 '17 at 13:50
  • $\begingroup$ Well, that's definitely an interesting approach but I don't see away to evaluate this integral without approximation. $\endgroup$ – user424862 Aug 3 '17 at 13:52
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    $\begingroup$ You can sum up the coefficients you get from each $(z - z^{-1})^n$ (write and odd $n$ as $2m-1$). An argument why that is legitimate should be given. If you want a "closed form", look at the Bessel functions. $\endgroup$ – Daniel Fischer Aug 3 '17 at 13:53

The residue is given by $$ \frac{1}{2\pi i}\int_{|z|=1} e^{z-1/z} \, dz, $$ which is probably easier to evaluate than a series. Putting $z=e^{i\theta}$ gives $$ \frac{1}{2\pi}\int_{-\pi}^{\pi} \exp{(e^{i\theta}-e^{-i\theta})} e^{i\theta} \, d\theta = \frac{1}{2\pi}\int_{0}^{2\pi} e^{2i\sin{\theta}} e^{i\theta} \, d\theta $$ This is a Fourier integral, of course. What is the Fourier expansion of $e^{2i\sin{\theta}}$? The answer is provided by the Jacobi–Anger expansion, $$ e^{iz\sin{\theta}} = \sum_{n = -\infty}^{\infty} J_n(z) e^{in\theta}, $$ from which we see that the integral pulls out the $n=-1$ coefficient, $J_{-1}(2)=-J_1(2)$, and the answer can't be expressed more simply.


$$ \begin{align} \left[z^{-1}\right]\sum_{k=0}^\infty\frac{\left(z-z^{-1}\right)^k}{k!} &=\left[z^{-1}\right]\sum_{k=0}^\infty(-1)^k\frac{\left(1-z^2\right)^k}{z^k\,k!}\\ &=\sum_{\substack{k=0\\k\text{ odd}}}^\infty\frac{(-1)^k}{k!}(-1)^{\frac{k-1}2}\binom{k}{\frac{k-1}2}\\ &=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(2k+1)!}\binom{2k+1}{k}\\ &=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{k!(k+1)!} \end{align} $$

  • $\begingroup$ Nice answer! (+1) $\endgroup$ – Markus Scheuer Aug 3 '17 at 14:19
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    $\begingroup$ Mathematica says that this sum is -BesselJ[1, 2] which looks to be the same as Chappers' answer. $\endgroup$ – robjohn Aug 3 '17 at 14:19

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