# Thread: Deriving Bessel generating function

1. ## Deriving Bessel generating function

So I am trying to prove that

$exp\left[\frac{x}{2}(t-\frac{1}{t}\right]=\sum^{\infty}_{n=-\infty} J_{n}(x)t^{n}$

What I did so far is to set $t=e^{i \theta}$ and I came down to something like that using fourier series
$exp\left[\frac{x}{2}(t-\frac{1}{t}\right]=\frac{2i\sin(\pi\sin \theta)}{\pi}\left[\frac{1}{2i\sin \theta} + \sum^{\infty}_{n=1} \frac{(-1)^{n}}{n^{2}-\sin^{2} \theta} (i\sin\theta\cos nx - n\sin nx)\right]$

I am not even sure it is correct or if it is the right way, I am just confused as to where to go from here. Any suggestions??? I really need help this thing is due soon. Any help is welcome!!!

2. ## One hint

We have:
$\exp{\frac{zt}{2}} = \sum_{m=0}^\infty \frac{z^m t^m}{2^m m!}$

and

$\exp\left(-\frac{z}{2t}\right) = \sum_{k=0}^\infty (-1)^k\frac{z^k t^{-k}}{2^k k!}$

Because the above series are uniformly convergent we have:
$\exp\frac{z}{2}\left(t - \frac{1}{t}\right) = \sum_{k=0}^\infty \sum_{m=0}^\infty \frac{(-1)^k}{k!m!} \left( \frac{z}{2} \right)^{k+m} t^{m-k}$

From this point, you can easily see what to do

3. Thanks, I can't believe I didn't think of that!!!

4. Now using that generating function formula I am supposed to prove the following,

$J_{n}(x+y)=\sum^{\infty}_{k=-\infty} J_{k}(x)J_{n-k}(x)$

I came down to something like

$exp\left[\frac{(x+y)}{2}(t-\frac{1}{t})\right]= \sum^{\infty}_{n=-\infty} J_{n}(x)t^{n}\sum^{\infty}_{m=-\infty} J_{m}(y)t^{m}$

Again, I have no idea if I am going in the right direction or where to go from here in order to prove this thing. Any suggestion???

5. It is a good start. You should get to the point where you have two power series: $\sum \alpha_n t^n$ and $\sum \beta_n t^n$ which are equal for t in some open set... Then we can say, that their coefficients are equal.

The coefficients of this series shoud be (of course) $J_{n}(x+y)$ and $\sum^{\infty}_{k=-\infty} J_{k}(x)J_{n-k}(x)$