One way to define Bessel functions is
$$ e^{\frac{x}{2}(t-\frac{1}{t})}=\sum_{n=-\infty}^{+\infty}J_{n}(x)t^n. $$
How do I prove that?
I can't see a way of writing the L.H.S. as a geometrical (or Laurent) series on $t$ to compare the coefficients with the definition of $J_n(x)$. Searching for a more deep relation for the comparison, I arrived at
$$ \sum_{n,k=0}^{\infty}\frac{(-1)^{n-k}}{2^{n}k!(n-k)!}x^{n}t^{2k-n}=J_0(x)+\sum_{l=1}^{\infty}J_{l}(x)\left[t^{l}+\frac{(-1)^{l}}{t^{l}}\right] $$
Where the L.H.S. comes from the expansion of $e^{\frac{x}{2}\left(t-\frac{1}{t}\right)}$ about $x=0$ and of $\left(t^2-1\right)^{n}$ about $t^2=0$.
Of course, when analysing the coefficients I would make use of the series form of the Bessel functions:
$$ J_{n}(x)=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k! \Gamma(n+k+1)}\left(\frac{x}{2}\right)^{2k+n} $$
Let me know if there are mistakes or if what I did is just not useful here.
Thanks!