# 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.

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Rian Sullivan on 11 Feb 2024
Commented: Star Strider on 11 Feb 2024
% 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);

Star Strider on 11 Feb 2024
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.
% 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.
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##### 2 CommentsShow NoneHide None
Rian Sullivan on 11 Feb 2024
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 on 11 Feb 2024
As always, my pleasure!

### More Answers (1)

Matt J on 11 Feb 2024
Edited: Matt J on 11 Feb 2024
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
Matt J on 11 Feb 2024
Edited: Matt J on 11 Feb 2024
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 on 11 Feb 2024
Ah okay I understand now thank you so much!