Defining an anonymous function with 3*(N+1) variables

2021 4 8
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更新时间：2021 4 10
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I need to define a function
f = @(X)...
where X has 3*(N+1) variables in the form: X(:,1), X(:,2), ... X(:,3*(N+1)) . The problem is that the expression for the function is very long. However, there is a pattern which I want to take advantage of.
The function is
Notice that the function can be expressed in terms of summations and products. I want to take advantage of this fact so that I don't have to manually type out the products and sums. Is there any way of doing this using for loops?

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David Hill
David Hill 2021-4-8
What is the X matrix or array? X(i+N+1) appears to exceed the array and X(i+2N+2) is much bigger than the array. Therefore without understanding what X and how you are indexing into it with i and j, I cannot help.
Anthony Gurunian
Anthony Gurunian 2021-4-8
That's my mistake, there is 3*(N+1) variables
Cris LaPierre
Cris LaPierre 2021-4-8
Perhaps I'm getting hung up on symantics here, but it appears you have a single variable, X. Is it correct to say that your array X has 3*(N+1) columns?
Anthony Gurunian
Anthony Gurunian 2021-4-8
Yes that is correct. I'm trying to use the syntax required here: https://www.mathworks.com/matlabcentral/fileexchange/53477-monte-carlo-integration
" f is a vectorized function a la f=@(X)sin(X(:,1))+cos(X(:,2))"

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David Hill
David Hill 2021-4-8

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f=prod(exp(-beta*(diff(x(1:n+1)).^2+diff(x(n+2:2*n+2)).^2+diff(x(2*n+3:3*n+3)).^2)))*sum((x(1:n+1)-x(1:n+1)').^2+(x(n+2:2*n+2)-x(n+2:2*n+2)').^2+(x(2*n+3:3*n+3)-x(2*n+3:3*n+3)).^2,'all');

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David Hill
David Hill 2021-4-8
Even simplier
x=reshape(x,[],3);
f=prod(exp(-beta*(sum(diff(x,1).^2),2)))*sum((x(:,1)-x(:,1)').^2+(x(:,2)-x(:,2)').^2+(x(:,3)-x(:,3)').^2,'all');
Anthony Gurunian
Anthony Gurunian 2021-4-8
Thanks for your response. Sorry for not being clear, but the 'variables' are the columns of X. So you can't say X_i - X_j is X(1:n+1)-X(1:n+1)' for example. I am trying to use the syntax required in https://www.mathworks.com/matlabcentral/fileexchange/53477-monte-carlo-integration
" f is a vectorized function a la f=@(X)sin(X(:,1))+cos(X(:,2))"
David Hill
David Hill 2021-4-8
It should be very close to the solution. Is X a 1x(3n+3) array? If so, it should work.
Anthony Gurunian
Anthony Gurunian 2021-4-8
编辑：Anthony Gurunian 2021-4-9
You don't know the number of rows in X when defining this function. Each variable is a collumn vector. The number of rows is defined later in integralN_mc(). It depends on the number of times the function f is called per itteration, but this number is also adapted within integralN_mc() for efficiency.
Rik
Rik 2021-4-9
I don't see an indication that his code depends on a specific number of rows. The number of columns is a different matter, but that is also possible.
David Hill
David Hill 2021-4-9
编辑：David Hill 2021-4-9
Is there a different f for each row? If not, I don't understand your equation. If so, then just do a loop with my equation. How can you not know the number of rows? X must be defined before this equation is processed.
for k=1:size(x,1)
f(k)=prod(exp(-beta*(diff(x(k,1:n+1)).^2+diff(x(k,n+2:2*n+2)).^2+diff(k,x(2*n+3:3*n+3)).^2)))*...
sum((x(k,1:n+1)-x(k,1:n+1)').^2+(x(k,n+2:2*n+2)-x(k,n+2:2*n+2)').^2+...
(x(k,2*n+3:3*n+3)-x(k,2*n+3:3*n+3)).^2,'all');
end
Or,
for k=1:size(x,1)
X=reshape(x(k,:),[],3);
f(k)=prod(exp(-beta*(sum(diff(X,1).^2),2)))*sum((X(:,1)-X(:,1)').^2+...
(X(:,2)-X(:,2)').^2+(X(:,3)-X(:,3)').^2,'all');
end
Anthony Gurunian
Anthony Gurunian 2021-4-10
I ended up writing a custom function f = Rg(X) and called it using integralN_mc(@(X) Rg(X), ...)

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