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/