Matlab: Nonlinear equation Optimization - matlab

Question

This question is related to the post below:
Matlab: Nonlinear equation solver
With 8 variables x0-x8, I got great results. However, when I increase to solving 10 variables, the results aren't so good. Even if my "guess" is close to the actual value and change the max iteration to 100000, the results are still poor. Is there anything else I can do?
Here is the code:
function F = fcn(x)
F=[x(6) + x(7) + x(8) + x(9) + x(10)-2 ;
x(6)*x(1) + x(7)*x(2) + x(8)*x(3) + x(9)*x(4) + x(10)*x(5) ;
x(6)*x(1)^2 + x(7)*x(2)^2 + x(8)*x(3)^2 + x(9)*x(4)^2 + x(10)*x(5)-2/3 ;
x(6)*x(1)^3 + x(7)*x(2)^3 + x(8)*x(3)^3 + x(9)*x(4)^3 + x(10)*x(5) ;
x(6)*x(1)^4 + x(7)*x(2)^4 + x(8)*x(3)^4 + x(9)*x(4)^4 + x(10)*x(5)-2/5 ;
x(6)*x(1)^5 + x(7)*x(2)^5 + x(8)*x(3)^5 + x(9)*x(4)^5 + x(10)*x(5) ;
x(6)*x(1)^6 + x(7)*x(2)^6 + x(8)*x(3)^6 + x(9)*x(4)^6 + x(10)*x(5)-2/7 ;
x(6)*x(1)^7 + x(7)*x(2)^7 + x(8)*x(3)^7 + x(9)*x(4)^7 + x(10)*x(5) ;
x(6)*x(1)^8 + x(7)*x(2)^8 + x(8)*x(3)^8 + x(9)*x(4)^8 + x(10)*x(5)-2/9 ;
x(6)*x(1)^9 + x(7)*x(2)^9 + x(8)*x(3)^9 + x(9)*x(4)^9 + x(10)*x(5)
];
end
clc
clear all;
format long
x0 = [0.90; 0.53; 0; -0.53; -0.90; 0.23; 0.47; 0.56; 0.47; 0.23]; %Guess
F0 = fcn(x0);
[x,fval]=fsolve(#fcn, x0) %solve without optimization
options = optimset('MaxFunEvals',100000, 'MaxIter', 100000); %optimization criteria
[x,fval]=fsolve(#fcn, x0, options) %solve with optimization
Here are the actual values I'm trying to get:
x1 = 0.906179
x2 = 0.538469
x3 = 0.000000
x4 = -0.53846
x5 = -0.906179
x6 = 0.236926
x7 = 0.478628
x8 = 0.568888
x9 = 0.478628
x10 = 0.236926

Answer 1

The result of such optimization functions like fsolve depends on the initial point very much. A non-linear function like yours can have a lot of local minima and your option is to randomly dice the initial point and hope it will lead the optimization to a better minimum than before.
You can do like this:
clear;
options = optimset('MaxFunEvals',2000, 'MaxIter', 1000, 'Display', 'off');
n = 200; %how many times calculate f with different initial points
z_min = 10000; %the current minimum Euclidian distance between fval and zeros
for i=1:n
x0 = rand(10, 1);
[x,fval]=fsolve(#fcn, x0, options);
z = norm(fval);
if (z < z_min)
z_min = z;
x_best = x;
f_best = fval;
display(['i = ', num2str(i), '; z_min = ', num2str(z_min)]);
display(['x = ', num2str(x_best')]);
display(['f = ', num2str(f_best')]);
fprintf('\n')
end
end
Change the maximum number of the optimization loops and look at the z value. It shows how close your function is to a zero vector.
The best solution I've got so far:
x_best =
0.9062
-0.9062
-0.5385
0.5385
0.0000
0.2369
0.2369
0.4786
0.4786
0.5689
f_best =
1.0e-08 * %these are very small numbers :)
0
0.9722
0.9170
0.8740
0.8416
0.8183
0.8025
0.7929
0.7883
0.7878
For this solution z_min is 2.5382e-08.