How to find back from generating function $\sum(Q(x)*z^m/m!)$ to $Q(x)$?

Question

How to find back from generating function $\sum_{m=0}^{\infty}(Q(x)\frac{z^m}{m!})$ to $Q$?

In other words, find $Q$ from $\sum_{m=0}^{\infty}(Q(x)\frac{z^m}{m!})$.

finally is generating function depends on $x$ and $z$

Update The reason for this backward action, is forward action from dsolve differential equation solved as special function such as kummer, even convert to ratpoly also can not be success

for example

$\sum_{m=0}^{\infty}(Q(x)\frac{z^m}{m!})$ = 2/(Pi*(exp(x/z)+exp(-x/z)))

how subs(z=0, diff(f, z$k)) when z is denominator? how many times should we diff ?

Please don't down vote, really need this

Answer 1

Since $S=\sum\limits_{m=0}^{+\infty}Q\frac{z^m}{m!}$ is such that $S=Q\cdot\mathrm e^z$, one has $Q=S\cdot\mathrm e^{-z}=S\cdot\sum\limits_{m=0}^{+\infty}(-1)^m\frac{z^m}{m!}$.

Answer 2

If $$ S(z)=\sum_{m=0}^\infty Q_m\frac{z^m}{m!} $$ (note that the coefficient $Q_m$ does depend on $m$) then, rather obviously, $$ Q_k=\frac{d^kS}{dz^k}(0) $$ for all $k\geq0$.