The free variables in your definition of psi are causing
problems. There are other ways to deal with free variables, but I
think you are better off defining psi as a function of six
variables. Try this (no more free variables):
psi(i,j,a,b,r,s):=sqrt((2*i+1)*(i+j+1)/2)*jacobi_p(i,0,0,a)*((1-s)/2)^i*jacobi_p(j,2*i+1,0,b)$
Also, I'd guess that you need to integrate s from -1 to 1, not -1 to -1:
(%i21) integrate(integrate(psi(1,1,2*(1+r)/(1-s)-1,s,r,s)^2, r, -1, -s),
s, -1, 1);
(%o21) 1
maxima-bounces at math.utexas.edu wrote on 10/07/2010 10:46:31 AM:
> [image removed]
>
> [Maxima] integrate returns 0
>
> Bart Vandewoestyne
>
> to:
>
> maxima
>
> 10/07/2010 10:46 AM
>
> Sent by:
>
> maxima-bounces at math.utexas.edu
>
> Hello all,
>
> I'm trying to check a property of so-called 'Koornwinder-Dubiner'
> polynomials. According to
>
> http://www.springerlink.com/content/1l4h576r307l62n6/
>
> they are defined over the reference triangle {r,s>=-1 and r+s <= 0} as
>
> a:2*(1+r)/(1-s)-1;
> b:s;
> psi(r,s):=sqrt((2*i+1)*(i+j+1)/2)*jacobi_p(i,0,0,a)*((1-s)/2)
> ^i*jacobi_p(j,2*i+1,0,b);
>
> for certain i and j. According to the literature, these
> polynomials are orthonormal with respect to the L2-norm over
> the reference triangle, so I compute (i choose i=j=1):
>
> i:1;
> j:1;
> integrate(integrate((psi(r,s))^2, r, -1, -s), s, -1, -1);
>
> and maxima returns 0 for the result of this integral, which is
> not what I expect for a polynomial that is orthonormal. I would
> expect 1 there.
>
> Is maxima giving me a wrong answer here, or am I misunderstanding
> something?
>
> Regards,
> Bart
>
> --
> "Share what you know. Learn what you don't."
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