Generating Functions Interpretation - Expanding around other points? Generating functions are incredibly useful for solving all kinds of combinatorial problems. Whenever they are used, though, the generating function is always expanded around $x=0$ to obtain the series. Why is this the case? Is there a combinatorial use/interpretation of a 'generating function' expanded around other points on the line (or the complex plane)?
 A: Good question! For one thing, any power series  $$f(x) = \sum_n a_n (x-a)^n$$ may be converted into a power series based around $0$ by taking $$g(x) = f(x+a) = \sum_n a_n x^n.$$ and power series of this form are more convenient to work with.  For example, we frequently want to multiply generating functions, but how should we interpret a product of power series of the form
$$\left(\sum_n a_n (x-a)^n\right)\left( \sum_n b_n (x-b)^n\right)?$$  It's not clear whether we should expand it as a power series in $(x-a)$ or $(x-b)$, and it might not be possible to do either.
In combinatorics we usually use only formal power series, which means roughly that we do not want to consider issues of convergence.  For example, it's sometimes useful to consider a power series such as $$f(x) = \sum_{n=1}^\infty n!x^n$$ which only converges at $x=0$.  Here we don't think of $f(x)$ as a function.  It's just "a clothesline on which we hang up a sequence of numbers for display", to quote Wilf in the free downloadable book generatingfunctionology.  It doesn't make such to evaluate such a power series at a number such as $x=1$, but we can still add, multiply, and perform other operations with such power series and it's often meaningful combinatorially to do so. In this case it's not clear what it would mean to expand $f(x)$ as a power series around a point $a\neq0$.
Even in the case when $f(x)$ can be expanded around any point, you may not get integer coefficients.  For example, take the exponential function $$f(x) = e^x = \sum_{n=0}^\infty \frac{1}{n!}x^n$$ which is frequently useful as an exponential generating function.  If we expand $f(x)$ around $a=1$ we get  $$f(x) = \sum_{n=0}^\infty \frac{e}{n!} (x-1)^n$$ and the number $e = 2.718\cdots$ does not have any direct combinatorial significance.
