Re: Functional equations for HermiteH[n,x]
- To: mathgroup at smc.vnet.net
- Subject: [mg59101] Re: Functional equations for HermiteH[n,x]
- From: Paul Abbott <paul at physics.uwa.edu.au>
- Date: Thu, 28 Jul 2005 02:28:36 -0400 (EDT)
- Organization: The University of Western Australia
- References: <dbie83$bt0$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
In article <dbie83$bt0$1 at smc.vnet.net>, janostothmeister at gmail.com
wrote:
> 1. I have found in the help that
> â??_z HermiteH[n, z]
> 2 n HermiteH[-1+n,z]
>
> Nice. I wanted to reproduce this myself.
D[HermiteH[n, z], z]
does exactly this.
> 2. I would also like to have H[n,-x]==-H[n,x],
> but even FunctionExpand does not produce this.
Because this is not true unless n is an integer.
> 3. This should be zero.
> FunctionExpand[HermiteH[n + 1,
> x] - 2x HermiteH[n, x] + 2n HermiteH[n -
> 1, x], n â?? Integers â?§ n > 0 â?§ x â?? Reals]
Series expansion is one possible approach here:
2 n HermiteH[n - 1, x] + HermiteH[n + 1, x] - 2 x HermiteH[n, x] +
O[x]^4 // FullSimplify
A better way to approach this is to use the generating function. Note
that the generating function and a wide range of properties can be
obtained at
http://functions.wolfram.com/Polynomials/HermiteH/
In particular, at
http://functions.wolfram.com/05.01.11.0001.01
one effectively has
E^(2 x t - t^2) == Sum[HermiteH[n, x] t^n/n!, {n,0,Infinity}]
> 4. This is known to be zero:
> Integrate[HermiteH[n, x] E^(-x^2, {x,-â??,â??},
> Assumptions ->(n â?? Integers â?§ n > 0)]
Use the generating function. Computing the integral
Integrate[E^(2 x t - t^2) Exp[-x^2], {x, -Infinity, Infinity}]
gives you the desired result.
> 5. This should be the KroneckerDelta[m,n]:
> Integrate[HermiteH[n, x]HermiteH[m, x]E^(-x^2), {x, -â??, â??},
> Assumptions -> (n â?? Integers â?§ m â?? Integers â?§ n > 0 â?§ m
> > 0)]
Series expansion of the integral,
Integrate[E^(2 x t - t^2) E^(2 x u - u^2) Exp[-x^2],
{x, -Infinity, Infinity}]
shows that the integral is not KroneckerDelta[m,n] but
Sqrt[Pi] KroneckerDelta[m,n] 2^n n!
Now try the integral of a product of 3 HermiteH functions ...
> I know, I know, mathematical program packages know everything except
> symbolic calculations, still...
Humans use the generating function (or other general techniques) to
compute such expressions. Using the same technique in Mathematica yields
the same results.
Cheers,
Paul
--
Paul Abbott Phone: +61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
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http://InternationalMathematicaSymposium.org/IMS2005/