Thread Subject:

Hermite polynomials with complex parameters.

Hello,
       Can anybody guide me how to simulate hermite polynomials with complex numbers as given below...

H_n(x) is the given hermite polynomial with 'n' being a complex number...x is real number in this case


Thanks and kind regards,
Xiang

On 3/1/2012 3:07 AM, Xiang wrote:
> Hello,
> Can anybody guide me how to simulate hermite polynomials with
>complex numbers as given below...
>
> H_n(x) is the given hermite polynomial with 'n' being a complex number

How can 'n' be complex? n is the order of the Hermite polynomial.
It must be non-negative integer.

--Nasser

"Nasser M. Abbasi" <nma@12000.org> wrote in message <jinhmi$5qa$1@speranza.aioe.org>...
> On 3/1/2012 3:07 AM, Xiang wrote:
> > Hello,
> > Can anybody guide me how to simulate hermite polynomials with
> >complex numbers as given below...
> >
> > H_n(x) is the given hermite polynomial with 'n' being a complex number
>
> How can 'n' be complex? n is the order of the Hermite polynomial.
> It must be non-negative integer.
>
> --Nasser
- - - - - - - - -
  The Hermite differential equation

 u" - 2*x*u' = -lambda*u

can have solutions with complex values of lambda rather than just non-negative integers. According to Wikipedia

 http://en.wikipedia.org/wiki/Hermite_polynomials

an explicit formula is given in terms of a contour integral in Courant and Hilbert 1953. (My own copy of Courant and Hilbert is written in German and my German is very rusty so I'm afraid I can't understand their explanation.)

  I don't know of any matlab routines that are designed to generate these contour integrals.

Roger Stafford

Hello Roger,
                Thanks for your reply. For general 'n', its actually hermite function. I couldnot find it in MATLAB. The same is available as HermiteH in mathematica, which saved my day.

Xiang


"Roger Stafford" wrote in message <jioqjh$dgl$1@newscl01ah.mathworks.com>...
> "Nasser M. Abbasi" <nma@12000.org> wrote in message <jinhmi$5qa$1@speranza.aioe.org>...
> > On 3/1/2012 3:07 AM, Xiang wrote:
> > > Hello,
> > > Can anybody guide me how to simulate hermite polynomials with
> > >complex numbers as given below...
> > >
> > > H_n(x) is the given hermite polynomial with 'n' being a complex number
> >
> > How can 'n' be complex? n is the order of the Hermite polynomial.
> > It must be non-negative integer.
> >
> > --Nasser
> - - - - - - - - -
> The Hermite differential equation
>
> u" - 2*x*u' = -lambda*u
>
> can have solutions with complex values of lambda rather than just non-negative integers. According to Wikipedia
>
> http://en.wikipedia.org/wiki/Hermite_polynomials
>
> an explicit formula is given in terms of a contour integral in Courant and Hilbert 1953. (My own copy of Courant and Hilbert is written in German and my German is very rusty so I'm afraid I can't understand their explanation.)
>
> I don't know of any matlab routines that are designed to generate these contour integrals.
>
> Roger Stafford

On 3/1/2012 10:53 PM, Xiang wrote:
> Hello Roger,
> Thanks for your reply. For general 'n', its actually hermite
>function. I couldnot find it in MATLAB. The same is available as HermiteH in
>mathematica, which saved my day.
>
> Xiang
>

How did you manage to use Mathematica's HermiteH[n,x] function for
complex n? Can you show an example?

>H_n(x) is the given hermite polynomial with 'n' being a complex number...
>x is real number in this case

--Nasser

"Xiang " <shaik.nancy@gmail.com> wrote in message <jipjno$r89$1@newscl01ah.mathworks.com>...
> Hello Roger,
> Thanks for your reply. For general 'n', its actually hermite function. I couldnot find it in MATLAB. The same is available as HermiteH in mathematica, which saved my day.
- - - - - - - - -
  In case it is of interest to you, Xiang, I believe now that matlab can compute the Hermite function with a complex n parameter in terms of the Symbolic Toolbox's 'hypergeom' function. The formula for it would be:

 H_n(z) = 2^n*sqrt(pi)*(1/gamma((1-n)/2)*hypergeom(-n/2,1/2,z^2) ...
                    - 2*z/gamma(-n/2)*hypergeom((1-n)/2,3/2,z^2));

  You might try comparing matlab's results with those of mathematica.

  Note: The above formula comes from Mathematica's website at:

 http://functions.wolfram.com/HypergeometricFunctions/HermiteHGeneral/02/

Roger Stafford

@Roger, Thanks for your reply, I will try it in matlab also...
@Nasser, You can simply try this in mathematica
HermiteH[1 + I, 0.5]

Nevertheless you will get answer for valid n,x only..where n is a complex number

--Xiang


"Roger Stafford" wrote in message <jipr4c$ia9$1@newscl01ah.mathworks.com>...
> "Xiang " <shaik.nancy@gmail.com> wrote in message <jipjno$r89$1@newscl01ah.mathworks.com>...
> > Hello Roger,
> > Thanks for your reply. For general 'n', its actually hermite function. I couldnot find it in MATLAB. The same is available as HermiteH in mathematica, which saved my day.
> - - - - - - - - -
> In case it is of interest to you, Xiang, I believe now that matlab can compute the Hermite function with a complex n parameter in terms of the Symbolic Toolbox's 'hypergeom' function. The formula for it would be:
>
> H_n(z) = 2^n*sqrt(pi)*(1/gamma((1-n)/2)*hypergeom(-n/2,1/2,z^2) ...
> - 2*z/gamma(-n/2)*hypergeom((1-n)/2,3/2,z^2));
>
> You might try comparing matlab's results with those of mathematica.
>
> Note: The above formula comes from Mathematica's website at:
>
> http://functions.wolfram.com/HypergeometricFunctions/HermiteHGeneral/02/
>
> Roger Stafford

On 3/2/2012 1:29 AM, Xiang wrote:

> @Nasser, You can simply try this in mathematica
> HermiteH[1 + I, 0.5]
>
> Nevertheless you will get answer for valid n,x only..where n is a complex number

May be I do not understand something.

The above only gives you one numerical value. Not a polynomial,
which is what you asked for:

In[7]:= HermiteH[1 + I, 0.5]
Out[7]= 1.9951895074974817 - 0.29935560945022155*I

I thought you wanted a Hermite polynomial, as in

In[8]:= HermiteH[5, x]
Out[8]= 120*x - 160*x^3 + 32*x^5

which only works for non-negative integers.

If I type

HermiteH[1 + I, x]

then Mathematica does not do anything with it. It can't
evaluate it.

anyway, if what you did worked for you, all is good.

--Nasser

"Nasser M. Abbasi" <nma@12000.org> wrote in message <jipvcj$ro$1@speranza.aioe.org>...
> The above only gives you one numerical value. Not a polynomial,
> which is what you asked for:
> In[7]:= HermiteH[1 + I, 0.5]
> Out[7]= 1.9951895074974817 - 0.29935560945022155*I
> I thought you wanted a Hermite polynomial, as in
> In[8]:= HermiteH[5, x]
> Out[8]= 120*x - 160*x^3 + 32*x^5
> which only works for non-negative integers.
> If I type
> HermiteH[1 + I, x]
> then Mathematica does not do anything with it.
- - - - - - - - - -
  Hi Nasser. I see nothing contradictory in the various results you get with HermiteH. HermiteH(n,z) when n is a positive integer is in fact a polynomial function of z which can be given a specific polynomial coefficient format rather than simply a value. However when n is otherwise, there is no way that Mathematica can express what function of z it is, so as you say, it "does not do anything with it". On the other hand if you give a specific value for z, then HermiteH can return with the corresponding function value for that z which it apparently does in every case, hence your single result for HermiteH[1+I,0.5].

  When Xiang originally asked "how to simulate hermite polynomials" it was presumably a minor misnomer which was later corrected by saying "For general 'n', its actually hermite function."

Roger Stafford