search minimum of a parameter function with the constrain that one parameter must be integer

Hi,

I'm looking for the minimum of sum((f(x)-data(x))^2), where f(x)= (sum(exp(1i .* x .* jz))).^2 where jz = 1,2,...nz and nz is my integer parameter and data(x). So f is function of x while (sum(f(x)-data(x)))^2 is function of nz.

I think I have to use fmincon, but I don't find the syntax to tell MatLab that the constrain is nz = integer.

I'm using this approach beacuse I wasn't able to use lsqcurvefit because when I run the summation to evaluate f(x) matlab doesn't understand tha f is function of x and gives an indexes error.

Many thanks

Gianluca

Answer 1

[Andrew Newell]

This is an integer programming problem, and MATLAB does not have a function for that (although the FEX does have code for solving linear or quadratic mixed integer problems). However, you could cheat. Since your sum is equal to (a^(p+1)-a)/(a-1), where a = exp(1i*x), you could treat p as a real variable, fit it, then round off the answer.

Given the way you formulated the minimum (your f(x) being the square of a function g(x), and minimizing sum(g(x)^2-data^2), you are fitting the modulus of g(x) to the data, where g(x) is your sum of exponentials. I will generate some sample data assuming the integer is 6:

xdata = rand(100,1);  n = 6;
a = cos(xdata);
ydata = abs((a.^(n+1)-a)./(a-1)) + 0.1*randn(size(xdata)); %Add some noise

Now define the function to fit

fun = @(p,x) abs((x.^(p+1)-x)./(x-1));

Fit to the data:

p0 = 3;
p = lsqcurvefit(fun,p0,cos(xdata),ydata);
p = round(p)

p =

     6

If you want to put some constraints on your parameters, see the fields lb and ub in the structure options in the documentation for lsqcurvefit.

Note: revised to take into account new information.

Answer 2

[gianluca messina]

You are right. The only problem now is that I have a second parameter inside the exponential and matlab find it negative, that hasn't any physical meaning. Using lsqcurvefit I don't see a method to superimpose the constrain: parameter > 0.

ps my function is real because i'm taking the square modulus of the summation.

Thank you Gianluca