Show that

\begin{equation} e^{\frac{c}{2}(z-1/z)}=\sum_{n=-\infty}^{\infty}a_nz^n \end{equation}


\begin{equation} a_n:=\frac{1}{2\pi}\int_0^{2\pi}\cos(n\theta-c\sin(\theta))d\theta \end{equation}

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    $\begingroup$ This is an incredibly inconvenient - but known and understood - reference to the $J$-Bessel function. From that, my answer here also answers this question. $\endgroup$ – davidlowryduda May 17 '14 at 5:49
  • $\begingroup$ @mixedmath In the link that you sent me to, I believe you have a mistake/typo, you prove that $$e^{c/2(z-z^{-1})}=\sum J_n(x)z^n$$ which is contrary to your claim, however, thanks, I have been able to prove it $\endgroup$ – user115850 May 17 '14 at 6:39
  • $\begingroup$ Since you know markup properly, can you please edit $z-1/z$ as $z-\frac{1}{z}$ or $\frac{z-1}{z}$ as appropriate? Thanks! $\endgroup$ – MattAllegro May 17 '14 at 7:04
  • $\begingroup$ @MattAlegro, it can't mean anything but the former $\endgroup$ – vonbrand May 17 '14 at 10:16

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