Proof of the Bessel $ J_n $ using the generating function

Question

Given that $\sum\limits_{n=-\infty}^{+\infty}t^nJ_n(x)=e^{x(t-t^{-1})/2}$, prove that $J_n(x)=\sum\limits_{k=0}^{+\infty}\frac{(-1)^k}{k!(n+k)! }(x/2)^{n+2k}$ by expanding $$e^{x(t-t^{-1})/2}=\sum_{m=0}^{+\infty} \frac{1}{m!}(t-t^{-1})^m (x/2)^m$$ and that $$ (t-t^{-1})^m = \sum_{k= 0}^ {m} (-1)^k \frac {m!}{k!(m-k)!} t^{m-2k} $$

It also gives a hint to set $ n=m-2k $ but when i make this change i can't quite understand how should the boundaries of the dummy variables be.

edit: I should clarify a bit better what is my problem. The above gives $$e^{x(t-t^{-1})/2}=\sum_{m=0}^{+\infty} \frac{1}{m!}(\sum_{k= 0}^ {m} (-1)^k \frac {m!}{k!(m-k)!} t^{m-2k}) (x/2)^m$$

$$e^{x(t-t^{-1})/2}=\sum_{m=0}^{+\infty} \sum_{k= 0}^ {m} (-1)^k \frac {(x/2)^m}{k!(m-k)!} t^{m-2k} \qquad (1) $$

But we have to get from that above, into $$e^{x(t-t^{-1})/2}=\sum_{n= -\infty}^{+\infty} (\sum\limits_{k=0}^{+\infty}\frac{(-1)^k}{k!(n+k)! }(x/2)^{n+2k} )t^{n} \qquad (2) $$ by setting $ n = m-2k $ so that we can say $J_n(x)=\sum\limits_{k=0}^{+\infty}\frac{(-1)^k}{k!(n+k)! }(x/2)^{n+2k}$. What i can't understand is, how do we get from $ 0\leq m < +\infty $ and $ 0\leq k < m $ of eq.(1) to $ -\infty < n < +\infty $ and $ 0\leq k < +\infty $ of eq.(2).

Answer 1

A useful technique when dealing with factorials in the denominator is to let the indices be unrestricted and then use $1/n! = 0$ for $n < 0$.

Answer 2

$J_n(x)$ is the coefficient of $t^n$ in the expansion. So you just want to know when $ t^n$ appears in the expansion and add the coefficients. This happens when $m-2k=n$, with the constraint $0\leq k\leq m$. For $n\geq 0$, this is the same as saying that $m=2k+n$ with $k\geq 0$ (because the other constraint $k\leq m$ is automatically satisfied), which means that the sum is over $k\geq 0$.

So, you get the $J_n(x)$ for $n\geq 0$.

When $n<0$ actually the one expansion they are telling you to prove doesn't make too much sense since you have that one of the summands is $\frac{1}{n!}(x/2)^{n}$, which doesn't make sense, unless you extend the factorials to the negative numbers (which can be done, but in my opinion it's abuse of notation.)

Now, to deal with $J_n(x)$ for negative numbers, you might just use $J_n(x)=(-1)^nJ_{-n}(x)$.