[seqfan] BIPRODUCT of Two Sequences (Hadamard)

[Brendan McKay]

People with recent editions of Maple can do this using the
'gfun' package.  Maybe Mathematica has something similar.

Brendan.


* Marc LeBrun <mlb at fxpt.com> [041108 08:47]:
> >=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?).
> 
> Hadamard product.
> 
> 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.
>