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For integral $m$, we have

$$J_m(x) = \sum_{j=0}^\infty \frac{(-1)^j}{j!\ \Gamma(j+m+1)}\left(\frac{x}{2}\right)^{2j+m} = \sum_{j=0}^\infty \frac{(-1)^j}{j!(j+m)!}\left(\frac{x}{2}\right)^{2j+m}$$

I'm curious if $J_m(x) = i^m$ for some limit of $x$.

If we send $x\to 2i$, for example, we get

$$J_m(2i) = i^m\sum_{j=0}^\infty \frac{1}{j!(j+m)!}$$

which doesn't quite do it.

zahbaz
  • 10,441
  • Plug the value $x$ that does this into the recurrence, and see what it is... Here is the recurrence: $$J_{n-1}(x)+J_{n+1}(x) = \frac{2n}{x};J_n(x)$$ – GEdgar Apr 05 '17 at 23:20

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