- integral needs the option 'ArrayValued',true to integrate vector valued functions.
- the term 1./r in your fun1 evaluates to [Inf ...]
- you redefine l (terrible choice of a variable name btw) as your loop variable, so by the time the loop is done (once you fixed it), l will have the value 2, and not be a 3-dim array.
Evaluate a function in a grid
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I want to plot an isosurface of a function*_ v_*.
Problem is that v doesn't accept matrix arguments (not vectorizable function as it contains Laguerre associated polynomials)
So if I create a meshgrid for the values of x,y,l
[r,th,l]=meshgrid(0:0.5:5,0:pi/2:2*pi,0:2);
x=r.*cos(th);
y=r.*sin(th);
Then I suppose to evaluate v using loops (counters) as I can not bypass x , y & l as arguments
But I'm doing something wrong, and as a consequence I'm not evaluating v in the points of the grid:
for l=0:2
k=k+1;
for r=0:0.5:5
i=i+1;
for th=0:pi/2:2*pi
j=j+1;
fun1=@(R)4*real(exp(-r.^2-R^2+2*1i*R*l./r).*(r+1i*R).^(2*l).*...
(mfun('L',n/2-l/2,l,r.^2+R^2)).^2);
v(i,j,k)=integral(fun1,-inf,inf);
end
end
end
does someone knows how to do the loops so I can obtain a v that matches in size with x, y, l so I can use:
isosurface(x,y,l,v) ??
or does someone knows how to obtain the mentioned isosurface v through an alternative way?
I actually need all the help I can get to do this.
Thanks, Pablo
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回答(1 个)
Jonathan Epperl
2013-4-24
Does that code run for you? I'm sure it doesn't, you should have mentioned that, along with the error messages...
Anyway, there are several problems:
Fix that, see if it works, if it doesn't come back here, this time with error messages, please.
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