# a map between two expansions of a function

Is there something remarkable about the following map between two kinds of expansion $f(x)=\sum_n f_n x^n$ and $\tilde{f}(x)=\sum_n \frac{f_n x^n}{n!}$ ?

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@Eugene: your question seems vague. It is possible to say something about the passage from an ordinary generating function to an exponential generating function, so the answer is yes. But what is it that you really want to know? –  Pete L. Clark Jun 6 '11 at 12:06
It takes "division by x and ignoring the remainder" to differentiation; this is certainly a noteworthy algebraic property. –  James Cranch Jun 6 '11 at 13:21
It can be modeled by a variation of the Laplace transform (or possibly its inverse). The details are left as an exercise... –  Qiaochu Yuan Jun 6 '11 at 18:10
Motivation came from the star-product (Moyal-Weyl) algebras. It turned out that certain star-product can be represented just as a pointwise multiplication of functions if one passes from one kind of expansion to another one. The map is not local, but who cares if it allows to transform complicated star-product equations to algebraic ones for functions of commuting variables. –  Eugene Starling Jun 7 '11 at 6:24

I don't know your motivation, but the passage between the two expansions is compatible with the theory of "$q$-integers". Recall that an integer $n$ can be written as the sum $1+1+\cdots+1$ ($n$ times). Its "$q$-analog" is the number $[n]_q = 1 + q + \cdots + q^{n-1}$. This is $n$ when $q=1$ and $1$ when $q=0$.
Similarly, $n!$ has a $q$-analog given by $[n]_q! = $1$_q \cdot [2]_q \cdots [n]_q = (1) \cdot (1+q) \cdot (1+q+q^2) \cdots (1+q+\cdots +q^n)$. If $q=1$ this is the usual factorial, and if $q=0$ it is simply $1$.
Thus if you consider $$f_q(x) = \sum \frac{f_n}{[n]_q!} x^n,$$ then $q=0$ corresponds to your $f$ and $q=1$ to your $\tilde{f}$.
There are some surprising extensions of a number of combinatorial objects to $q\neq 1$, see for example the Wikipedia page on q-analogs. For example, if $\tilde{f}$ satisfies a differential equation, then $f$ satisfies an appropriate $q$-analog of it (this was effectively mentioned in one of the comments, as the $q$-derivative at $q=0$ is simply multiplication by $1/x$).