Coefficients of a certain Laurent series

Question

If $$e^{t(z - 1/z)/2} = \sum_{n \in \mathbb{Z}} J_n(t)z^n$$

is the Laurent expansion in $\mathbb{C}^*$, how can I show that

$$ J_{-n}(t) = (-1)^n J_n(t)$$

Any help would be appreciated.

Answer 1

HINT:

$$e^{t(z-z^{-1})/2}=\sum_{n=-\infty}^{\infty}J_n(t)z^n \tag 1$$

Now, enforce the substitution $z\to -z^{-1}$ followed by $n\to -n$.

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SPOILER ALERT: Scroll over the highlighted area to reveal the solution.

Starting from $(1)$, we enforce the substitution $z\to -1/z$ to obtain $$\begin{align}e^{t(-z^{-1}+z)/2}&=e^{t(z-z^{-1})/2}\\\\&=\sum_{n=-\infty}^{\infty}J_n(t)(-1)^n z^{-n}\tag 2\end{align}$$Then, in $(2)$, transforming the summation index by letting $n\to -n$ reveals $$e^{t(z-z^{-1})/2}=\sum_{n=-\infty}^\infty (-1)^n\,J_{-n}(t)\,z^n \tag 3$$Comparing $(1)$ and $(3)$ and exploiting the uniqueness of the Laurent expansion, yields the coveted relationship $$J_n(t)=(-1)^nJ_{-n}(t)$$