I would like to know the procedure to do the power-series expansion of this function:
$$\LARGE{\mathrm{e}^{\frac{z}{2}(t-\frac{1}{t})}}$$
so that I would get this:
$$\sum_{r=0}^\infty \sum_{s=0}^\infty \frac{(-1)^s z^{r+s} t^{r-s}}{r!s!2^{r+s}}$$
I was unable to obtain this double series and the only thing I know is this basic formula:
$$\mathrm{e}^z=\sum_{k=0}^\infty\frac{z^k}{k!}$$
Basically, you need to use the exponential series and Newton's binomial theorem, before swapping the summation indexes, as follows : $$ \begin{array}{rcll} \displaystyle e^{\frac{z}{2}\left(t-\frac{1}{t}\right)} &=& \displaystyle \sum_{n=0}^\infty \frac{z^n}{2^nn!} \left(t-\frac{1}{t}\right)^n & (1) \\ &=& \displaystyle \sum_{n=0}^\infty \frac{z^n}{2^nn!} \sum_{k=0}^n \binom{n}{k} t^k (-t^{-1})^{n-k} & (2) \\ &=& \displaystyle \sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^{n-k} z^n t^{2k-n}}{2^n(n-k)!k!} & (*) \\ &=& \displaystyle \sum_{k=0}^\infty \sum_{n=k}^\infty \frac{(-1)^{n-k} z^n t^{2k-n}}{2^n(n-k)!k!} & (3) \\ &=& \displaystyle \sum_{k=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{(m+k)-k} z^{m+k} t^{2k-(m+k)}}{2^{m+k}((m+k)-k)!k!} & (4) \\ &=& \displaystyle \sum_{k=0}^\infty \sum_{m=0}^\infty \frac{(-1)^m z^{k+m} t^{k-m}}{2^{k+m}k!m!} \end{array} $$ where we used (1) the exponential series, (2) the binomial theorem, (3) a swap of the two sums and (4) an index shift such that $n := m+k$.
Alternatively, you may take advantage of Cauchy product formula, namely $$ \left(\sum_{i=0}^\infty a_i\right) \left(\sum_{j=0}^\infty b_j\right) = \sum_{n=0}^\infty \sum_{k=0}^n a_kb_{n-k}, $$ so that $$ e^{\frac{z}{2}\left(t-\frac{1}{t}\right)} = e^{zt/2}e^{-z/2t} = \left(\sum_{i=0}^\infty \frac{z^it^i}{2^ii!}\right) \left(\sum_{j=0}^\infty (-1)^j\frac{z^jt^{-j}}{2^jj!}\right) $$ and you will end up to the step $(*)$ above.
Addendum. The step going from $(*)$ to $(3)$ is made by inverting the order of summation. It cannot be done freely since the index $k$ depends on $n$ as an upper bound. It is to be noted that $k$ goes from $0$ to $\infty$, like $n$, but the index $n$ must increase for $k$ to reach higher values, because $k \le n$. This point of view can be reversed by saying that $k$ goes effectively from to $0$ to $\infty$ as a free variable and then $n$ takes all the values above $k$ (since $n \ge k$), namely from $k$ to $\infty$. This fact translates as $$ \sum_{n=0}^\infty \sum_{k=0}^n \longleftrightarrow \sum_{k=0}^\infty \sum_{n=k}^\infty $$