From: rwinther@my-dejanews.com Subject: Re: Products of Hermite Polynomials Date: Sun, 21 Feb 1999 21:52:05 GMT Newsgroups: sci.math Keywords: Products of Hermite Polynomials as sums of Hermite Polynomials In article , Garry Smith wrote: > > I need to expand the product of two Hermite polynomials as a sum of Hermite > polynomials. I can't find anything close enough in tables. Can anyone give > any advice? > It's hard to believe this isn't published *somewhere*. Did you try Gradshteyn and Ryzhik's "Table of Integrals Series and Products" or Erdelyi, Magnus, Oberhettinger and Tricomi's "Higher Transcendental Functions" (the Bateman Manuscript)? For the G&R book I would look also for the integral from -infinity to infinity of exp(-x^2/2)*H_m(x)*H_n(x)*H_p(x) dx ; then the answer given is (2^p)*p!*sqrt(pi) times the coefficient of H_p(x) in the expansion of H_m(x)*H_n(x) I have a result which I came up with empirically (I computed H_m*H_n for m=1, 2, and 3 and for enough values of n that I could see a pattern). I am all but certain that it's correct, but I haven't managed to prove it yet. Any takers? The n=1 case follows immediately from the recurrence relation H_{m+1}(x) = 2x*H_m(x) - 2m*H_{m-1}(x) once H_1(x) is substituted for the 2x in the first term on the right. The n=2 case can also be derived from the recurrence relation with a bit more effort, and I abandoned trying to prove my conjecture from the recurrence relation. Clearly the expression must be symmetric in m and n, and it is that property (plus the fact that it works for all the cases I computed by hand) that gives me the confidence that the conjecture is correct. I get: H_m(x)*H_n(x) = \sum_{i=0}^{min(m,n)} [(2^i/i!)*(m!/(m-i)!)*(n!/(n-i)!)*H_{m+n-2i}(x)] A similar result then follows for the so-called alternative Hermite polynomials, usually denoted He_m(x), which satisfy the recurrence relation He_{m+1}(x) = x*He_m(x) - m*He_{m-1}(x) and which are related to the H_m's by He_m(x) = 2^(-m/2)*H_m(x/sqrt(2)) He_m(x)*He_n(x) = \sum_{i=0}^{min(m,n)} [(1/i!)*(m!/(m-i)!)*(n!/(n-i)!)*He_{m+n-2i}(x)] Ron Winther -----------== Posted via Deja News, The Discussion Network ==---------- http://www.dejanews.com/ Search, Read, Discuss, or Start Your Own