How do I use fminunc with two variables?

I am trying to minize the follwing function q given x and muk where muk is each element of the array mu.

I first created the function q and dq, which is the gradient of q with two variables x and muk to be general.

Then, when I am work with the for loop over the size of mu, I was trying to plug in mu particularly and minize the function, q_, which is now a function of the variable x only. However, my code fails and if you guys can look at it. I'll very much appreciate it.

-------------------------------- Code is here -----------------------------------------------

mu = [1, 10, 100, 1000];
tau = 1 ./ mu;

q = @(x, muk) x(1) + x(2) + (muk / 2) * (x(1)^2 + x(2)^2 - 2)^2;
dq = @(x, muk) [1 + muk * 2 * x(1) * (x(1)^2 + x(2)^2 - 2); 1 + muk * 2 * x(2) * (x(1)^2 + x(2)^2 - 2)];

options = optimoptions('fminunc', 'Algorithm', 'trust-region', 'HessianApproximation', 'bfgs', 'SpecifyObjectiveGradient', true);

x0 = [1; 0];

xk = x0;

for k = 1:length(mu)
    
    a = mu(k);
    q_ = @(x)(q(x, a));
    dq_ = @(x)(dq(x, a));
    
    dq_2 = dq_(xk);
    
    
    while norm(dq_2) > tau(k)
        
        [x,fval] = fminunc(q_,xk,options)
        xk = x;
        dq_2 = dq_(xk);
    end
    
    
    xk
    
end

And I got these error at [x,fval] = fminunc(q_,xk,options) saying

Error using + 
Too many output arguments.

Error in HW4_3>@(x,muk)x(1)+x(2)+(muk/2)*(x(1)^2+x(2)^2-2)^2 (line 6)
q = @(x, muk) x(1) + x(2) + (muk / 2) * (x(1)^2 + x(2)^2 - 2)^2;

Error in HW4_3>@(x)(q(x,a)) (line 18)
q_ = @(x)(q(x, a));

Error in fminunc (line 345)
[f,GRAD] = feval(funfcn{3},x,varargin{:});

Caused by:
Failure in initial objective function evaluation. FMINUNC cannot continue.

Answer 1

[Torsten]

You tell "fminunc" that you supply the gradient of the objective, but you don't do it.

mu = [1, 10, 100, 1000];
tau = 1 ./ mu;

q = @(x, muk) x(1) + x(2) + (muk / 2) * (x(1)^2 + x(2)^2 - 2)^2;
dq = @(x, muk) [1 + muk * 2 * x(1) * (x(1)^2 + x(2)^2 - 2); 1 + muk * 2 * x(2) * (x(1)^2 + x(2)^2 - 2)];

options = optimoptions('fminunc', 'Algorithm', 'trust-region', 'HessianApproximation', 'bfgs', 'SpecifyObjectiveGradient', true);

x0 = [1; 0];

%xk = x0;

for k = 1:length(mu)
    
    a = mu(k);
    %q_ = @(x)(q(x, a));
    %dq_ = @(x)(dq(x, a));
    
    %dq_2 = dq_(xk);
    
    
    %while norm(dq_2) > tau(k)
        
        [x,fval] = fminunc(@(x)q_(x,a,q,dq),x0,options)
        %xk = x;
        %dq_2 = dq_(xk);
    %end
    
    
    x
    
end

Local minimum found.

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

x = 2x1

-1.1072 -1.1072

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

fval = -2.1123

x = 2x1

-1.1072 -1.1072

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

Local minimum possible.

fminunc stopped because the final change in function value relative to 
its initial value is less than the value of the function tolerance.

x = 2x1

-1.0123 -1.0123

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

fval = -2.0123

x = 2x1

-1.0123 -1.0123

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

Local minimum possible.

fminunc stopped because the final change in function value relative to 
its initial value is less than the value of the function tolerance.

x = 2x1

-1.0012 -1.0013

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

fval = -2.0012

x = 2x1

-1.0012 -1.0013

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

Local minimum possible.

fminunc stopped because the final change in function value relative to 
its initial value is less than the value of the function tolerance.

x = 2x1

-1.0001 -1.0001

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

fval = -2.0001

x = 2x1

-1.0001 -1.0001

<mw-icon class=""></mw-icon>

<mw-icon class=""></mw-icon>

function [z gz] = q_(x,a,q,dq)
  z = q(x,a);
  if nargout > 1
    gz = dq(x,a);
  end
end

[Sohyeon]

Thank you for your answer! It works perfectly!