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MathGroup Archive 2005

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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
The University of Western Australia         (CRICOS Provider No 00126G)    
AUSTRALIA                               http://physics.uwa.edu.au/~paul
        http://InternationalMathematicaSymposium.org/IMS2005/


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