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This operation is similar to discrete convolution and cross-correlation, but has binomial coefficients:

$$f(n)\star g(n)=\sum_{k=0}^n \binom{n}{k}f(n-k)g(k) $$

Particularly,

$$a^n\star b^n=(a+b)^n$$

following binomial theorem.

I just wonder if there is a name for such operation and where I can read about its properties.

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To be precise, one should write $(f\star g)(b)$ rather than $f(n)\star g(n)$. For example, what if $n=5$ and $f(5)=3$ and $g(5)7$. Would $f(n)\star g(n)$ then be $3\star 7$? Obviously that's nonsense. – Michael Hardy Jan 30 '12 at 19:28
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Composition is distributive against any binary functions. That is $(f\circ w) \star (g \circ w) = (f \star g) \circ w$ for any f,g,w and binary operation $\star$ – Anixx Jan 30 '12 at 19:48

3 Answers

up vote 15 down vote accepted

It's called a binomial convolution in Graham, Knuth, and Patashnik's Concrete Mathematics. I don't have that text in front of me (but I bet someone here can give you a page number), but here's a reference on the Fermat's Last Theorem blog.

It would also be worth checking out Section 2.3 of Wilf's Generatingfunctionology. This is on exponential generating functions. The property of interest is that if $F(x)$ and $G(x)$ are the exponential generating functions of $f(n)$ and $g(n)$, respectively, then $F(x)G(x)$ is the exponential generating function of $f(n) \star g(n)$.

(FYI: You can download the second edition of Generatingfunctionology from Wilf's web site.)

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Its page 366 in the second edition. – J. M. Nov 3 '10 at 2:33
@J.M.: Thanks. I knew someone would find it quickly. :) – Mike Spivey Nov 3 '10 at 2:37

Remark that one can employ the powerful umbral calculus to compute closed forms for many such binomial convolutions of special functions, e.g. from p. 161 of Roman: Umbral Calculus:
alt text

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Note for casual readers: page 17 of the book defines the concept of an "Appell sequence". – J. M. Nov 3 '10 at 2:56
I wonder where can I find a closed form expression of binomial Bell's numbers of higher order? I.e. I need $\sum_{k=0}^{x-1} B_n^x \star B_n^k$ – Anixx Nov 3 '10 at 3:02
@MathFacts: You might try posting that as a separate question. – Mike Spivey Nov 3 '10 at 20:48

It corresponds to multiplying exponential generating functions (check generatingfunctionology).

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