Re: Functional equations for HermiteH[n,x]

*To*: mathgroup at smc.vnet.net*Subject*: [mg58827] Re: Functional equations for HermiteH[n,x]*From*: dh <dh at metrohm.ch>*Date*: Wed, 20 Jul 2005 00:29:24 -0400 (EDT)*References*: <dbie83$bt0$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Hi Janos, Mathematica does not know too man features about Hermite polynomials. But needed features can be added. See below sincerely, Daniel janostothmeister at gmail.com wrote: > Hi, All, > > 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. > > FullForm[Hold[â??_z HermiteH[n, z]]] > Out[31]//FullForm= > Hold[D[HermiteH[n,z],z]] > > Then, it should also work for me: > D[Hermite[n,z],z] > > \!\(\* > RowBox[{ > SuperscriptBox["Hermite", > TagBox[\((0, 1)\), > Derivative], > MultilineFunction->None], "[", \(n, z\), "]"}]\) > > But it does not. Yes it does if you write HermiteH instead of Hermite > > 2. I would also like to have H[n,-x]==-H[n,x], > but even FunctionExpand does not produce this. This can be done by adding a rule to HermiteH. First you must unprotect HermiteH: Unprotect[HermiteH] then you can define: HermiteH[n_, -x] := -HermiteH[n, x] > > 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] This can be achieved by defining a rule for H[n+1,x] as above: HermiteH[n_Integer+1,x_]:= 2 x HermiteH[n,x]-2 n HermiteH[n-1,x] > > 4. This is known to be zero: > Integrate[HermiteH[n, x] E^(-x^2, {x,-â??,â??}, > Assumptions ->(n â?? Integers â?§ n > 0)] A further rule is needed here. You would assigne this rule not to Integrate, but to HermiteH by: HermiteH/; ..... > > 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)] Same thing as 4) > > > I know, I know, mathematical program packages know everything except > symbolic calculations, still... > > Can anybody help me? > > Thanks, > > JÃ¡nos >