Finding Zero of Sum of Functions by Iteration

I am trying to sum a function and then attempting to find the root of said function. That is, for example, take:

Consider that I have a matrix, X, and vector, t, of values: X(2*n+1,n+1), t(n+1)

for j = 1:n+1
        sum = 0;
    for i = 1:2*j+1
        f = @(g)exp[-exp[X(i,j)+g]*(t(j+1)-t(j))];
        sum = sum + f;
    end
        fzero(sum,0)
end

That is,

I want to evaluate at

j = 1

f = @(g)exp[-exp[X(1,1)+g]*(t(j+1)-t(j))]
fzero(f,0)

j = 2

f = @(g)exp[-exp[X(1,2)+g]*(t(j+1)-t(j))] + exp[-exp[X(2,2)+g]*(t(j+1)-t(j))] + exp[-exp[X(3,2)+g]*(t(j+1)-t(j))]
fzero(f,0)

j = 3

etc...

However, I have no idea how to actually implement this in practice.

Any help is appreciated!

PS - I do not have the symbolic toolbox in Matlab.

Answer 1

[jgg]

I think the solution is to write a function:

 function [ sum ] = func(j,g,t,X)

    sum = 0;
    for i = 1:2*j+1
        f = exp(-exp(X(i,j)+g)*(t(j+1)-t(j)));
        sum = sum + f;
    end
 end

Then loop your solver

 for j=1:n
  fun = @(g)func(j,g,t,X);
  fzero(fun,0)
end

You'll have to debug the function inside( f = stuff ) (since I don't know what it's supposed to do) so that it returns what you want, but this should solve your problem.

I can get this to solve, but I'm not sure if your f function does what you want it to do.

[Gregory McArthur]

This is really impressive, actually, and probably what I would do if I was better at MATLAB.

Thanks!