# solving a triple sigma equation (or triple integral)

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형준 이 on 14 Jul 2022
Commented: 형준 이 on 18 Jul 2022
above equation is what i want to express by mathlab code
for definite the A(x,y,z)
A(x,y,z) : random size 3-D vector 00x00x00 (but example is 6x61x17)
below is my try, but i think that is not correct
it is hard to make code... please somebody help me
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
clear
% for A = 6 x 61 x 17 (3-d vector)
A = ones(6,61,17);
k = linspace(10.4, 31.4, 6);
pii = linspace(0, pi, 61); % pii = 𝝓
E = linspace(-2, 2, 17);
for x = 1:6
for y = 1:61
for z = 17
for f = 1:6
for p = 1:61
for e = 1:17
A(x,y,z) = A(x,y,z) + (k(f)).^2 * (1/(norm([x y z]-[cos(pii(p)) sin(pii(p)) E(e)]))) * (1-cos(pii(p))-sin(pii(p))) * exp(1i*k(f)*(((norm([x y z]-[cos(pii(p)) sin(pii(p)) E(e)])))-1));
end
end
end
end
end
end
Torsten on 15 Jul 2022
When you approximate an integral by a sum, there must appear terms of the form
(k(i+1)-k(i))*(phi(j+1)-phi(j))*(E(n+1)-E(n))
for the volume element in which you evaluate the function.
I don't see this volume element in your finite sum.

Chunru on 15 Jul 2022
% for A = 6 x 61 x 17 (3-d array)
A = ones(6,61,17);
k = linspace(10.4, 31.4, 6);
pii = linspace(0, pi, 61); % pii = 𝝓
E = linspace(-2, 2, 17);
% reshape to 3d array
k3d = reshape(k, 6, 1, 1); % along 1st dimension
p3d = reshape(pii, 1, 61, 1); % along 2nd dimension
e3d = reshape(E, 1, 1, 17); % along 3rd dimension
for x = 1:6
for y = 1:61
for z = 1:17
R = sqrt((x-cos(p3d)).^2 +(y-sin(p3d)).^2 + (z-e3d).^2);
tmp = k3d.^2 ./ R .* (1-x*cos(p3d)-y*sin(p3d)) .* exp(1i * k3d .*(R-1));
A(x,y,z) = sum(tmp, 'all');
end
end
end
imagesc(20*log10(abs(A(:, :, 8)))); % slice for z=8
형준 이 on 18 Jul 2022

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