How to write a function and create 3 nested loop then evaluate an integral?
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Fatin
el 6 de Oct. de 2014
Comentada: Fatin
el 7 de Oct. de 2014
I have a matrix of Density values (rho) of size 15x121 , I have lambda is a row of size 1x20 and the values of lambda between 0 and 2 .I want to evaluate the following function : f(t)= [exp(-lambda*t)*rho] then I want the integral of f with respect to t and the integral from t_j-1 to t_j The values of t is between 0 and 1 , and those are 15 intervals (( i.e 15 intervals [t_j-1,t_j] ))
Thanks for any help
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Respuesta aceptada
Mike Hosea
el 6 de Oct. de 2014
Editada: Mike Hosea
el 6 de Oct. de 2014
I'm not sure if there is supposed to be a relationship between the t intervals and rho, or whether the fact that there are 15 rows and 15 intervals is a coincidence. I'm assuming it's a coincidence, but hopefully you will see how to proceed from the principle of the following.
I've made up some data just so it would run.
lambda = linspace(0,2,17);
rho = rand(15,121);
t = linspace(0,1,16);
Here's brute force and ignorance, so to speak, with nested loops and collecting each (scalar) integral in a 4-D array.
f = @(t,lambda,rho)exp(-lambda*t)*rho;
Q = zeros([size(rho),numel(lambda),numel(t)-1]);
for i1 = 1:size(rho,1)
for i2 = 1:size(rho,2)
for i3 = 1:numel(lambda)
for j = 1:numel(t)-1
Q(i1,i2,i3,j) = integral(@(x)f(x,lambda(i3),rho(i1,i2)),t(j),t(j)+1);
end
end
end
end
We can improve on this by using the 'ArrayValued' option of INTEGRAL, treating rho as a matrix input to the qs function rather than a scalar.
f = @(t,lambda,rho)exp(-lambda*t)*rho;
Q = zeros([size(rho),numel(lambda),numel(t)-1]);
for i3 = 1:numel(lambda)
for j = 1:numel(t)-1
Q(:,:,i3,j) = integral(@(x)f(x,lambda(i3),rho),t(j),t(j)+1,'ArrayValued',true);
end
end
But why not let lambda be an array in qs as well? We just need to be clever so that we do every combination of rho elements and lambda elements. Here I've used an outer product of vectors and reshaped it.
f = @(t,lambda,rho)reshape(rho(:)*exp(-t*lambda(:).'),[size(rho),numel(lambda)]);
Q = zeros([size(rho),numel(lambda),numel(t)-1]);
for j = 1:numel(t)-1
Q(:,:,:,j) = integral(@(x)f(x,lambda,rho),t(j),t(j)+1,'ArrayValued',true);
end
Now if each interval of t needs only one row of rho, it's a slightly different problem. But we can do that, too.
Más respuestas (1)
yonatan gerufi
el 6 de Oct. de 2014
Editada: yonatan gerufi
el 6 de Oct. de 2014
for 3 dim integral use:
integral3
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