[seqfan] BIPRODUCT of Two Sequences (Hadamard)

[Karol A. PENSON]

   Dear Seqfans, an interesting version of the BIPRODUCT
     exists when applied to egf rather than to gf:
     in Maple notation

      if F(x)= sum(f(n)*x^n/n!,n=0..infinity) and
         G(x)= sum(g(n)*x^n/n!,n=0..infinity)   then
      
     subs(y=0, F(x*d/dy)*G(y))= subs(y=0, G(x*d/dy)*F(y))=
     sum(f(n)*g(n)*x^n/n!,n=0..infinity);

    Many examples of this relation ( which we call the product formula)
    in the Quantum Field Theory and in Combinatorics were subject
    of our two recent papers, which can be found in the ArXives under:

       http://fr.arxiv.org/pdf/quant-ph/0405103
       http://fr.arxiv.org/pdf/quant-ph/0409152
        
     Comment are welcome.

                     Karol A. Penson     


   

On Mon, 8 Nov 2004, Brendan McKay wrote:

> Date: Mon, 8 Nov 2004 09:20:11 +1100
> From: Brendan McKay <bdm at cs.anu.edu.au>
> To: seqfan at ext.jussieu.fr
> Subject: Re: [seqfan] BIPRODUCT of Two Sequences (Hadamard)
> 
> 
> 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.
> > 
> 

-- 
_________________________________________________________________________
Karol A. PENSON
Universite Paris 6              |  Internet : penson at lptl.jussieu.fr.
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