Bessel Equations Addition Formula So, I'm considering yet another tricky proof involving Bessel Functions.  Basically, I'm trying to figure out how the following is true:
$$J_n(\alpha + \beta) = \sum_{m = -\infty}^\infty J_m(\alpha)J_{n - m}(\beta)$$
Right now, I'm speculating that it has something to do with the Hankel Representation of Bessel functions, but I'm having a hard time making any headway.  Any suggestions or directions I could take for this?
 A: There are many different ways to define the Bessel function. I will assume we use the first series given in Wikipedia:
$$ J_\alpha(x) = \sum_{m \geq 0} \frac{(-1)^m}{m!\Gamma(m+\alpha+1)}\left(\frac{x}{2}\right)^{2m + \alpha}.$$
Claim: We have the generating function relation
$$e^{\frac{x}{2}(z + z^{-1})} = \sum_{n \in \mathbb{Z}}J_n(x)z^n.$$
Proof: Expand out the power series, and then collect terms of the same power - it will be exactly this series definition for the Bessel function. Explicitly:
$$\begin{align}
e^{\frac{x}{2}z}e^{-\frac{x}{2}\frac{1}{z}} &= \sum_m \frac{(x/2)^m}{m!}z^m\sum_k\frac{(-1)^k(x/2)^k}{k!}z^{-k} \\
&=\sum_{n \in \mathbb{Z}}\left(\sum_{\substack{m-k = n \\ m,k \geq 0}} \frac{(-1)^k}{m!k!}\left(\frac{x}{2}\right)^{2k}\left(\frac{x}{2}\right)^n\right)z^n \\
&= \sum J_n(x)z^n 
\end{align}$$
A bit slick, but not really that bad. $\diamondsuit$
Many many things can be proven with the generating function, which is why we so often use generating functions.
Claim: Your claim is true.
$$ J_n(x+y) = \sum_{k \in \mathbb{Z}}J_k(x)J_{n-k}(y)$$
Proof: The idea of the proof is to use the generating function, and let $t$ be a formal variable. We will match up coefficients of $t^n$ after expanding a series in two different ways. 
$$\begin{align}
\sum_n J_n(x+y)t^n &= e^{\frac{1}{2}(x + y)(t + t^{-1})} \\
&=e^{\frac{1}{2}x(t+t^{-1})}e^{\frac{1}{2}y(t+t^{-1})} \\
&= \sum_k J_k(x)t^k \sum_m J_m(y)t^m \\
&= \sum_{n \in \mathbb{Z}}\left(\sum_{k \in \mathbb{Z}}J_k(x)J_{n - k}(y)\right)t^n,
\end{align}$$
and equating coefficients gives your result. $\diamondsuit$
