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I was trying to find this identity of Bessel function

$$e^{-2i\gamma t} J_{\left|n\right|}(2\gamma t) = e^{\large \frac{\pi i}{2}} \sum_{k=|n|}^{\infty} \frac{(-i\gamma t)^k}{k!}\binom{2k}{k-n}$$

on some books like "Watson: theory of Bessel function" or "Abramowitz and Stegun" but I can not find it. I have also tried to derived it using the Frobenius method for the Bessel equation but there are some coefficients that don't match. Does anyone know where to find the identity or how i can prove/verify it? Any advice is welcome. Thank you in advanced

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It will be assumed that $n$ is a positive integer.

By definition of the functions $e^z$ and $J_n(z)$ $$\begin{align} e^zJ_n(i z)&=\sum_{k=0}^\infty\frac{z^k}{k!} \sum_{l=0}^\infty\frac{(-1)^l}{l!(n+l)!}\left(\frac {iz}2\right)^{n+2l}\\ &=\left(\frac{iz}2\right)^n\sum_{m=0}^\infty z^{m} \sum_{l=0}^{\left\lfloor\frac m2\right\rfloor}\frac1{(m-2l)!\,l!\,(n+l)!\,4^l}\\ &\stackrel*=i^n\sum_{m=0}^\infty \frac1{(n+m)!}\binom{2n+2m}m\left(\frac z2\right)^{n+m}.\tag1 \end{align} $$ Substituting in $(1)$ $z=-2i\gamma t$, $m=k-n$ one obtains your identity with prefactor $i$ replaced by $i^n$ (I assume it was a typo).

Appendix:

In $(\stackrel*=)$ we used the combinatorial identity: $$ \sum_{l=0}^{\left\lfloor\frac m2\right\rfloor}\frac1{l!(n+l)!(m-2l)!4^l} =\binom{2n+2m}m\frac1{(n+m)!2^m}\\ \iff \sum_{l=0}^{\left\lfloor\frac m2\right\rfloor} \binom {n+m}l\binom{n+m-l}{m-2l}2^{m-2l} =\binom{2n+2m}m, $$ a proof of which can be found here.

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  • $\begingroup$ Thank you very much! Yes there is a typo in the formula, so sorry for that! $\endgroup$ – mik Apr 12 at 8:52
  • $\begingroup$ Hi, please I cannot get the proof of the combinatorial identity. I mean, for getting $|S\cap E_i|=2$ which $i$ should I choose? I'am sorry for this stupid question but I would like to understand well :) $\endgroup$ – mik Apr 13 at 14:59
  • $\begingroup$ @mik You can choose either of two $i$'s. And it is the choice which gives rise to the factor $2^{m-2l}$. $\endgroup$ – user Apr 13 at 15:33
  • $\begingroup$ I think I'am not getting the problem. We should count the number of subsets $S$ of size $m$, right? So for instance, consider the set $\{1,2,3,4\}$ so $n=2$ and $i=1,2$. If we have $m=2$, then the only possible set of $i$ for which $|S\cap E_i|=2$ is $\{1,2\}$. Then I have two possible waya to have $|S\cap E_i|=1$. Now what does $2^{m-2k}$ represents? Then is the $k$ the parameter that count the intersection $|S\cap E_i|=k$? Sorry but I think I cannot visualize the problem. Thank you in advance $\endgroup$ – mik Apr 14 at 16:13
  • $\begingroup$ @mik Let consider the set $\{1,2,3,4\}$ as an example. It has 6 subsets of two elements. Translated in the language of $E_i$ they are: $E_1,E_1^-E_2^-,E_1^-E_2^+,E_1^+E_2^-,E_1^+E_2^+,E_2$, where $E_i^-$ means an odd element (1 or 3), and $E_i^+$ means an even element (2 or 4). $\endgroup$ – user Apr 14 at 20:51
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There's a typo in the exponential factor on the right-hand side -- it should read $\exp(i\,\pi n/2).$ Mathematica can just about do the identity. Using properties of the Bessel functions for integer $n,$ I'll write (the corrected form) as

$$ (*) \quad \quad \sum_{k=n}^\infty \frac{(-1)^{n-k}}{k!} \Big(\frac{x}{2}\Big)^k \binom{2k}{k-n}=e^{-x}\,I_n{(x)} $$ where $I_n(x)$ is the modified Bessel function. Mathematica sums the left-hand side as $$ \sum_{k=n}^\infty \frac{(-1)^{n-k}}{k!} \Big(\frac{x}{2}\Big)^k \binom{2k}{k-n} = \Big(\frac{x}{2}\Big)^n \frac{1}{n!} \, {_1}F_1(n+1/2, 2n+1, -2x)$$ where ${_1}F_1(a,b,z)$ is the confluent hypergeometric function and is equivalent to the Kummer M-function $M(a,b,z).$ The numbers in the next steps refer to formulas in the Digital Libray of Mathematical Functions. Use formula 13.2.39 to determine that $M(n+1/2, 2n+1, -2x)=\exp{(-2x)}M(n+1/2, 2n+1, 2x).$ Next use 13.6.9 to determine $M(n+1/2, 2n+1, -2x)=\exp{(-x)}n!(x/2)^{-n}I_n(x).$ Formula (*) is thus true.

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  • $\begingroup$ Thank you! Yes there is a typo, I am so sorry for that! $\endgroup$ – mik Apr 12 at 8:51

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