# Generating function of $a_{n}^2$ in terms of GF of $a_{n}$?

If we consider $A(x)$ as a generating function of a sequences $a_{n}$, is there any way to find the generating function of, say for example, the sequences : $v_{n}=a_{n}.a_{n+1}$ and $u_{n} = a_{n}^2$ in terms of $A(x)$?

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I don't expect an easy expression. For instance, you'd be hard pressed to express $\sum\limits_{k=0}^\infty \frac{x^k}{(k!)^2}$ in terms of the exponential function... –  Ｊ. Ｍ. Sep 20 '11 at 17:50
What you are looking for is the Hadamard product of generating functions. –  marty cohen Sep 20 '11 at 17:51

The problem of doing this is equivalent to the problem of finding the Hadamard product $\sum a_n b_n x^n$ of two generating functions $A = \sum a_n x^n$ and $B = \sum b_n x^n$. (The reason is that $a_n b_n = \frac{(a_n + b_n)^2 - a_n^2 - b_n^2}{2}$.)
$$\int_{0}^{2\pi} A(r e^{it}) B(r e^{-it}) \, dt = \int_0^{2\pi} \sum_{m,n} a_n b_m r^{n+m} e^{i(n-m)t} \, dt = 2\pi \sum_{n \ge 0} a_n b_n r^{2n}$$
provided that everything converges. If $A, B$ are sufficiently nice functions this integral may be computed in various ways, e.g. using complex analysis. See this blog post for details; there I consider a more general problem, that of computing the diagonal $\sum a_{n,n} x^n$ of a two-variable generating function $A = \sum_{m,n} a_{m,n} x^m y^n$.