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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

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    If the "variables" $x$ and $z$ are unrelated, you can just treat $Q(x)$ as constant (which is, in the sense that $dQ/dz=0$ identically) and the problem is absolutely trivial.2012-07-22

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

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    sorry, if Q is constant, did is correct, usually classic generating function's Q is in terms of x2012-07-22
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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 $k\geq0$.

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    Well, if the coefficient depends on $x$ different from $z$ the procedure remains the same: if there is a dependance from $m$ you get it differentiating as I said, if not the problem it is even more trivial and see did's answer.2012-07-22