Normalization of JacobiP function

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Luis Isaac
Luis Isaac 2016-1-12
编辑： Torsten 2016-1-12
Dear;
I am new in MuPAD I would like to know if the matlab function JacobiP (which computes the Jacobi polynomial) is normalized to 1. To prove this I computed the following code:
syms x
a = 3.5;
b = 7.2;
P3 = jacobiP(3, a, b, x);
w = (1-x)^a*(1+x)^b;
int(P3*P3*w, x, -1, 1)
The answer is the symbolic expresion of the integral:
int((1 - x)^(7/2)*(x + 1)^(36/5)*(- (1284731*x^3)/16000 + (853923*x^2)/16000 + (44247*x)/16000 - 42439/16000)^2, x, -1, 1)
If the Jacobi polynomials are orthonormalized, the result should be 1 How can I evaluate this integral numerically? I tried (<http://es.mathworks.com/help/symbolic/mupad_ref/numeric-int.html>)
numeric::int(P3^2*w,x,-1,1)
or
numeric::int(P3^2*w,x,-1..1)
but Matlab return an error.
Many thanks in advance;
  2 个评论
Luis Isaac
Luis Isaac 2016-1-12
Thank you very much;
As can be seen the Matlab definition of JacoviP are not normalized.
So what is the normalization of the JacobiP, or how can I get orthonormalized Jacobi polynomilas whitout need to calculate the integral.
Thanks again;
Torsten
Torsten 2016-1-12
编辑：Torsten 2016-1-12
n=3;
a=3.5;
b=7.2;
normfactor=1/(2^(a+b+1)/(2*n+a+b+1)*gamma(n+a+1)*gamma(n+b+1)/(gamma(n+a+b+1)*gamma(n+1)));
fun=@(x) normfactor*jacobiP(n, a, b, x).^2.*(1-x).^a.*(1+x).^b;
value=integral(fun,-1,1);
Best wishes
Torsten.

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采纳的回答

Torsten
Torsten 2016-1-12
a = 3.5;
b = 7.2;
fun=@(x) jacobiP(3, a, b, x).^2.*(1-x).^a.*(1+x).^b;
value=integral(fun,-1,1);
Best wishes
Torsten.

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