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Topic: Legendre Polynomials normalised to Sin[theta]^m
Replies: 13   Last Post: Jul 1, 2013 12:13 PM

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Guido Walter Pettinari

Posts: 29
From: Italy & UK
Registered: 3/24/10
Re: Legendre Polynomials normalised to Sin[theta]^m
Posted: Jun 29, 2013 6:55 AM
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Dear Waldek,

Thank you for your reply. I use the definition of P_lm(x) in http://en.wikipedia.org/wiki/Associated_Legendre_polynomials#Reparameterization_in_terms_of_angles. This means that P_lm(x) vanishes for x=1 when m>1.

> For x really close to 1 you will get overflow or division
> by zero independently of way you comupute your function.


In my case, P_lm(x) is proportional to (1-x^2)^(m/2) by definition, thus dividing it by that same factor should give me a perfectly ok number.

Cheers,
Guido

Il giorno lunedì 24 giugno 2013 01:05:59 UTC+2, Waldek Hebisch ha scritto:
> Guido Walter Pettinari wrote:
>

> > I would like to add that, since I need to evaluate this function very close to theta=n*pi, I cannot just use the result from a Legendre polynomial routine and divide it by sin(theta)^m.
>
> >
>
>
>
> Why not? If I understood your notation P_lm(x) will be close to 1
>
> and the other term close to 0. As long as both terms have high
>
> relative accuracy division will produce accurate result.
>
> For x really close to 1 you will get overflow or division
>
> by zero independently of way you comupute your function.
>
>
>
> --
>
> Waldek Hebisch
>





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