An identity about Bessel functions

Question

How can I prove

$\frac { 2n}{\rho}J_n (\rho)=J_{n-1}(\rho)+J_{n+1}(\rho)$

?

When $J_n$ is n'th order Bessel function.

I tried a lot, but I don't know how to construct $"n"$ in the LHS.

Is there any hint?

Thanks for your help.

Answer 1

Bessel functions appear in a particular generating function : $$ \exp\left(\frac{\rho}{2}\left(t-\frac{1}{t}\right)\right)=\sum_{n\in\mathbb{Z}} J_n(\rho)\,t^n. \tag{1}$$ By differentiating both sides of the previous identity with respect to $t$ we get: $$ \frac{\rho}{2}\left(1+\frac{1}{t^2}\right)\exp\left(\frac{\rho}{2}\left(t-\frac{1}{t}\right)\right) = \sum_{n\in\mathbb{Z}} J_n(\rho)\, n t^{n-1}\tag{2} $$ hence: $$ \frac{\rho}{2}\left(t+\frac{1}{t}\right)\sum_{n\in\mathbb{Z}}J_n(\rho)\, t^n = \sum_{n\in\mathbb{Z}} J_n(\rho)\, n t^n\tag{3} $$ and by equating the coefficients of $t^n$ in the LHS and RHS of $(3)$ your identity follows.