Proving Bessel equation $J_{0}(u+v) = J_0(u)\cdot J_0(v)+2\sum_{s=1}^{\infty}J_s(u)\cdot J_{-s}(v)$

Question

I've already found

$$J_n(u+v) = \sum_{s=-\infty}^{\infty}J_s(u)\cdot J_{n-s}(v)$$

here.

Now I'm tryint to arrive at

$$J_{0}(u+v) = J_0(u)\cdot J_0(v)+2\sum_{s=1}^{\infty}J_s(u)\cdot J_{-s}(v)$$

If I just plug $m=0$ we have: $$j_0(u+v) = \lim_{q\to -\infty}\lim_{p\to\infty}\sum_{s=q}^p J_s(u)\cdot J_{-s}(v)$$

right? As the definition of a sum from $-\infty$ to $\infty$ are $2$ limits. How do I get at what I need?

Answer 1

For $s$ an integer, $J_{-s}(x) = (-1)^s J_s(x)$ (this is not trivial, and requires writing out the sums and throwing away the terms that have Gamma-functions with nonpositive integral arguments). Then you have \begin{align} J_0(u+v) &= \sum_{s=-\infty}^{\infty} J_s(u)J_{-s}(v) \\ &= \sum_{s=-\infty}^{-1} J_s(u) J_{-s}(v) + J_0(u)J_0(v) + \sum_{s=1}^{\infty} J_s(u) J_{-s}(v) \\ &= \sum_{s=-\infty}^{-1} (-1)^s J_{-s}(u) J_{-s}(v) + J_0(u)J_0(v) + \sum_{s=1}^{\infty} J_s(u) J_{-s}(v) \\ &= J_0(u)J_0(v) + 2\sum_{s=1}^{\infty} J_s(u) J_{-s}(v). \end{align} since $J_s(u)J_{-s}(v) = (-1)^s J_{-s}(u)J_{-s}(v) = J_{-s}(u)J_s(v) $.