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
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