Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $A(x)$?

share|cite|improve this question
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
up vote 6 down vote accepted

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 $a_n b_n = \frac{(a_n + b_n)^2 - a_n^2 - b_n^2}{2}$.)

Hadamard products are difficult to compute in general. However, many classes of generating functions (e.g. the rational ones) are closed under Hadamard product. One way to compute Hadamard products is to consider the integral

$$\int_{0}^{2\pi} A(r e^{it}) B(r e^{-it}) \, dt = \int_0^{2\pi} \sum_{m,n} a_n b_m r^{n+m} e^{i(n-m)t} \, dt = 2\pi \sum_{n \ge 0} a_n b_n r^{2n}$$

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 $A = \sum_{m,n} a_{m,n} x^m y^n$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.