While doing a calculation on frequency modulation I am encountering the sum $$ \sum_{m=-\infty}^{+\infty} J_{n+m}(z)J_m(z) $$ which looks awfully similar to the identity: $$ J_n(t+r) = \sum_{m=-\infty}^{+\infty} J_k(t) J_{n-k}(r) $$ Except the former sum is over indices with fixed difference, while the latter is with fixed sum.
For the case of $n=0$, I see my expression has the simplification $$ J_0^2(z) + 2\sum_{m=1}^{\infty}J_m^2(z)=1 $$ Does a simplification exist for $n\neq 0$?
Ok, after a bit more searching I have found on DLMF: $$ \mathscr{C}_{\nu}\left(u\pm v\right)=\sum_{k=-\infty}^{\infty}\mathscr{C}_{\nu% \mp k}\left(u\right)J_{k}\left(v\right) $$ Where $\mathscr{C}_{\nu}$ is a "cylinder function". The Bessel functions $J_v$ are cylinder functions, so taking the bottom row of the $\pm$ we get: $$ \sum_{k=-\infty}^{\infty}J_{\nu + k}\left(z\right)J_{k}\left(z\right) = J_\nu(z-z) = J_\nu(0)=\delta_{\nu 0} $$ Which is what I had been finding to be the case numerically.