This code computes the coefficient of the leading order term, not the value
of the polynomial.

Thanks,

Jiahao Chen
Staff Research Scientist
MIT Computer Science and Artificial Intelligence Laboratory

On Mon, Feb 2, 2015 at 11:43 AM, Andrei Berceanu <[email protected]>
wrote:

> I came up with this, so far
>
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> *function compute_hermite_polynomial(n)    P = Poly([1])    const x =
> Poly([0;
> 1])
>     for i = 1:n        P = 2x*P - polyder(P)    end    Pend*
>
> On Monday, February 2, 2015 at 5:24:21 PM UTC+1, Andrei Berceanu wrote:
>>
>> Andras, no worries :) Now I understand why I couldn't find the
>> polynomials in your gist!
>>
>> //A
>>
>> On Monday, February 2, 2015 at 5:19:49 PM UTC+1, Andras Niedermayer wrote:
>>>
>>> Sorry, I meant Cubic Hermite Interpolation. Now I see you're looking for
>>> Hermite polynomials.
>>>
>>> On Monday, February 2, 2015 at 4:50:00 PM UTC+1, Andras Niedermayer
>>> wrote:
>>>>
>>>> I was looking for Hermite polynomials and haven't found any code. I
>>>> have some (very unpolished) code.
>>>>
>>>> I haven't made a public package yet, since it needs to be improved
>>>> (especially in terms of efficiency, also documentation). Unfortunately, I'm
>>>> unlikely to have time for this in the near future, so I'll just post a link
>>>> to a gist:
>>>> https://gist.github.com/afniedermayer/57873094430e8ddb201c
>>>>
>>>> I mainly used it with the output of the ODE.jl.
>>>>
>>>> I hope this is a useful starting point...
>>>>
>>>> Best,
>>>> Andras
>>>>
>>>> On Monday, February 2, 2015 at 4:38:57 PM UTC+1, Andrei Berceanu wrote:
>>>>>
>>>>> Yes, exactly, in order to generate plots like
>>>>> http://en.wikipedia.org/wiki/Hermite_polynomials#
>>>>> mediaviewer/File:Hermite_poly_phys.svg
>>>>>
>>>>> //A
>>>>>
>>>>> On Monday, February 2, 2015 at 4:36:55 PM UTC+1, Jiahao Chen wrote:
>>>>>>
>>>>>>
>>>>>> >  Is there an easy way to compute Hn(x)?
>>>>>>
>>>>>> Do you mean to evaluate a given Hermite polynomial of order n at a
>>>>>> value x?
>>>>>>
>>>>>

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