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

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$k)) when z is denominator? how many times should we diff ? Please don't down vote, really need this - Usually$Q$depends upon$m$, but you haven't indicated that. – Ross Millikan Jul 21 '12 at 15:04 You need to recover a sequence from its generating function. – Mhenni Benghorbal Jul 21 '12 at 15:05 So, you're asking how to find the coefficients of a Taylor series of a function? – Hurkyl Jul 21 '12 at 15:19 When you realize you've asked a poor question, the thing to do is to make your question better, rather than plead that people overlook the fact it's poor.... – Hurkyl Jul 21 '12 at 15:20 If$Q$is just a function of$x$, you can distribute it out as in did's answer from before. – Ross Millikan Jul 22 '12 at 4:55 show 2 more comments ## 2 Answers 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$k\geq0$. - how many times should we diff S? – M-Askman Jul 22 '12 at 1:50 @M-Askman$k$-times to obtain the$k$-th coefficient. – Peter Tamaroff Jul 22 '12 at 4:56 it sounds k-th coefficient of taylor series, not well understand how to do after get kth coefficient, the goal for Q is z-transform Q – M-Askman Jul 22 '12 at 7:57 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. – Andrea Mori Jul 22 '12 at 10:04 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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not fully understand, is it really S = Q*exp(z)? as Q is not a constant – M-Askman Jul 22 '12 at 1:59
@M-Askman: Ross has already asked you if Q is dependent on m. Without answering him, people can only assume what you wrote, and you wrote no dependence. That means you can simply use the distributive property, and get did's answer. – ex0du5 Jul 22 '12 at 4:42
sorry, if Q is constant, did is correct, usually classic generating function's Q is in terms of x – M-Askman Jul 22 '12 at 4:57