Stationary point Code Error. Trying to find stationary points for the equation below. Was having a hard time doing it by hand so tried a code.. getting error for fsolve.

% Define the function f(x1, x2)
f = @(x) x(1)^2 + x(1)*x(2) + 3/2*x(2)^2 - 2*log(x(1)) - log(x(2));
% Define the gradient ∇f(x1, x2)
gradient = @(x) [2*x(1) + x(2) - 2/x(1); x(1) + 3*x(2) - 1/x(2)];
% Define a function that returns a vector for fsolve
stationary_points = fsolve(@(x) gradient(x), [0; 0], options);
% The variable stationary_points now contains the stationary points
disp('Stationary Points:');
disp(stationary_points);

Answer 1

[Star Strider]

One problem is using zero for any initial parameter estimate, and especially if the parameter is the only element in the denominator, since that becomes Inf and the solver immediately stops.

Start with different initial estimates instead —

% Define the function f(x1, x2)
f = @(x) x(1)^2 + x(1)*x(2) + 3/2*x(2)^2 - 2*log(x(1)) - log(x(2));
% Define the gradient ∇f(x1, x2)
gradient = @(x) [2*x(1) + x(2) - 2/x(1); x(1) + 3*x(2) - 1/x(2)];
% Define a function that returns a vector for fsolve
stationary_points = fsolve(@(x) gradient(x), rand(2,1));%, options);

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

stationary_points = 2×1

0.8944 0.4472

% The variable stationary_points now contains the stationary points
disp('Stationary Points:');

Stationary Points:

disp(stationary_points);

    0.8944
    0.4472

You apparently defined an options structure, however did not include it, so I changed the fsolve call to exclude it.

.

[Rian Sullivan]

Oh okay that makes sense. Thank you so much. I did have the options structure and I accidently took it out.. Thank you so much.

[Star Strider]

As always, my pleasure!

Answer 2

[Matt J]

The problem is strictly convex, so obviously the staitonary point is unique and lies at the global minimum. So why not just use fminunc?

f = @(x) x(1)^2 + x(1)*x(2) + 3/2*x(2)^2 - 2*log(x(1)) - log(x(2));

stationary_point = fminunc(f,[1;1])

Local minimum found.

Optimization completed because the size of the gradient is less than
the value of the optimality tolerance.

stationary_point = 2×1

0.8944 0.4472

[Rian Sullivan]

That makes complete sense to me. Sorry, learning coding as I go with this class haha thank you so much.

[Rian Sullivan]

So one more question.. We found the stationary point to be unique hence the global minimum, so there is no global max? or Even local min or local maxes right?

[Matt J]

No, a convex function with an open domain (like your problem where the domain is x1>0, x2>0) will not have a finite global max. On a closed domain, there will be a global max, but the gradient won't be zero there.

[Rian Sullivan]

Ah okay I understand now thank you so much!