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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
>
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