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I found this problem in the book Complex Analysis by Christian Berg, and I got stuck at this point.

I have shown the first part, which is, that the Laurent series of the function $$f(w)=\exp\left(\frac{z}{2} \left(w-\frac{1}{w}\right)\right) =\sum_{n=-\infty}^{\infty} J_n(z)w^n,\quad w \in \mathbb{C}\backslash\{0\},$$ where $$J_n(z)=\frac{1}{2\pi}\int_{-\pi}^\pi e^{i(z\sin t-nt)}dt=\frac{1}{\pi}\int_{0}^\pi \cos(z\sin t-nt)dt,\quad n \in \mathbb{Z}$$ denotes the Bessel function of order $n$.

Now in the second part I have to show that $z(J_{n-1}(z)+J_{n+1}(z))=2nJ_n (z)$.

I don't even have a real approach to this. Any help, and other ideas would be appreciated!

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  • $\begingroup$ Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. $\endgroup$ Mar 9, 2023 at 9:59
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    $\begingroup$ Show that $\cos (z\sin t - (n - 1)t) + \cos (z\sin t - (n + 1)t) = 2\cos (z\sin t - nt)\cos t$. Then use this to show that the problem is equivalent to showing that $$ \frac{2}{\pi }\int_0^\pi {(z\cos t - n)\cos (z\sin t - nt){\rm d}t} = \frac{2}{\pi }\int_0^\pi {\frac{{\rm d}}{{{\rm d}t}}\sin (z\sin t - nt){\rm d}t} $$ is $0$. $\endgroup$
    – Gary
    Mar 9, 2023 at 11:33

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We want to prove $$ z(J_{n-1}(z)+J_{n+1}(z))=2nJ_n (z) . \tag1$$ We know $$ \exp\left(\frac{z}{2} \left(w-\frac{1}{w}\right)\right) =\sum_{n=-\infty}^{\infty} J_n(z)w^n. \tag2$$ Fix a complex number $z$. Write $$ f(w) = \exp\left(\frac{z}{2} \left(w-\frac{1}{w}\right)\right) . \tag3$$

Work with that generating function.
$$ \sum_{n=-\infty}^{\infty} J_n(z)nw^{n-1}= f'(w) \\ \sum_{n=-\infty}^{\infty} 2nJ_n(z)w^{n}=2wf'(w) \\ \sum_{n=-\infty}^{\infty} J_{n-1}(z)w^{n} =\sum_{n=-\infty}^{\infty} J_{n}(z)w^{n+1} = wf(w) \\ \sum_{n=-\infty}^{\infty} zJ_{n-1}(z)w^{n} = zwf(w) \\ \sum_{n=-\infty}^{\infty} J_{n+1}(z)w^{n} =\sum_{n=-\infty}^{\infty} J_{n}(z)w^{n-1} = \frac{f(w)}{w}. \\ \sum_{n=-\infty}^{\infty} zJ_{n+1}(z)w^{n} = \frac{zf(w)}{w}. $$ So $(1)$ is equivalent to $$ zwf(w) + \frac{zf(w)}{w} = 2wf'(w) . $$ Now recall $(3)$. It suffices to prove $$ zw \exp\left(\frac{z}{2} \left(w-\frac{1}{w}\right)\right) +\frac{z}{w}\exp\left(\frac{z}{2} \left(w-\frac{1}{w}\right)\right) = 2\frac{d}{dw}\exp\left(\frac{z}{2} \left(w-\frac{1}{w}\right)\right) $$ This is merely an elementary calculus exercise.

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  • $\begingroup$ Thanks. This really helped:) $\endgroup$
    – Morten1305
    Mar 9, 2023 at 21:17

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