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David Goodmanson
David Goodmanson 2023년 12월 9일

0 개 추천

Hello Shreen,
Here are two ways although you will need the symbolics toolbox for the first one. I used benign values for z and nu and the two methods agree very closely. When either z or nu get excessively large there will be numerical issues, and you might explore to see which method gives up first.
nu = 1.2;
z = 0:.1:4;
% hypergeometic function 1F2
C = 2/(sqrt(pi)*gamma(nu+3/2));
L1 = C*(z/2).^(nu+1).*hypergeom(1,[3/2+nu,3/2],z.^2/4)
% numerical integration
D = 2/(sqrt(pi)*gamma(nu+1/2));
fun = @(th,z,nu) sinh(z*cos(th)).*sin(th).^(2*nu);
L2 = D*(z/2).^nu.*integral(@(th) fun(th,z,nu),0,pi/2,'arrayvalued',true);
checkdel = max(abs(L1-L2))
L1 =
Columns 1 through 9
0 0.0010 0.0046 0.0113 0.0214 0.0351 0.0528 0.0748 0.1012
Columns 10 through 18
0.1325 0.1691 0.2112 0.2594 0.3141 0.3758 0.4452 0.5228 0.6095
Columns 19 through 27
0.7059 0.8129 0.9316 1.0629 1.2080 1.3682 1.5449 1.7396 1.9540
Columns 28 through 36
2.1901 2.4498 2.7355 3.0496 3.3948 3.7742 4.1911 4.6491 5.1522
Columns 37 through 41
5.7048 6.3118 6.9783 7.7104 8.5143
checkdel =
3.5527e-15
ans = 3.5527e-15

댓글 수: 3

Thanks. I need to define the following
SturveL(0,k*r);
SturveL(1, k*r);
Because I used them inside other calculations to plot other functions. Are you understand me?
Pi = sym(pi);
syms z
nu0 = sym(0);
C0 = 2/(sqrt(Pi)*gamma(nu0+3/2));
L0 = C0*(z/2).^(nu0+1).*hypergeom(1,[3/2+nu0,3/2],z.^2/4)
L0 = 
nu1 = sym(1);
C1 = 2/(sqrt(Pi)*gamma(nu1+3/2));
L1 = C1*(z/2).^(nu1+1).*hypergeom(1,[3/2+nu1,3/2],z.^2/4)
L1 = 

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2023년 12월 9일

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