# [seqfan] BIPRODUCT of Two Sequences (Hadamard)

Marc LeBrun mlb at fxpt.com
Sun Nov 7 22:44:48 CET 2004

>=Paul D. Hanna
> Given two sequences with known g.f., what is the g.f. of the sequence
> resulting from the term-by-term product of the sequences?

It can be expressed as a contour integral "convolving" the two GFs.  The
integrand "kernel" is basically like f(z)g(1/z)/z (with some scale factor
involving pi).  As z goes around the unit circle (say) the individual cross
terms are like f[p] exp(ipt) g[q] exp(-iqt).  The orthogonality of exp
causes all of these to be zero except when p=q (around contours, GFs and
Fourier series are "morally equivalent").

Of course, as with integration generally, the value of the resulting
integral may or may not be expressible in "closed form" as easily as the f
and g were.

> Suppose we call this the "BIPRODUCT" of the two sequences
> (is there a better term for this?).

I learned this from vol. 1 of Z. A. Melzak's (fascinating but typo-marred)
"Companion to Conrete Mathematics".  Unfortunately I can't seem to find my
copy right now to give you a better reference...

> Q:  Can someone extend this to higher orders --
> i.e., what is biproduct( 1/(1-ax-bx^2-cx^3), 1/(1-dx-ex^2-fx^3) ) = ?

Since they're both rational you can probably push it through a computer
algebra system (although last time I tried the contour integrals tended to
give them indigestion).

You might also be able to compactly derive it using the matrix
representation.  A linear recurrence is a matrix analog of a geometric
series, with the scalar expression for the sum in terms of the ratio,
1/(1-r), going to (I-R)^(-1), where R is the recurrence matrix.

> Would this interest anyone to come up with a more general formula?

It'd be interesting, but probably messy...alas I don't have time to try
right now, but some other computist might...
"Enjoy"!

PS: Let g be the GF for n-th powers.  "The Hadamard product of g and g^2 is
zero for n>2" is equivalent to FLT.