MATLAB: 1 iteration in fsolve: new x and no more

fsolveiterationOptimization Toolbox

I want to use fsolve to only use one iteration: get the Jacobian, get x1 and done. However when test the code below, which tries to find the zero of x^2, i see the following:
-iteration 0, 2 function evaluations: it seems f(x0) and 1 for finite difference
-iteration 1, 2 function evaluations: it seems f(x1) and 1 for finite difference
I do not want the last two evaluations. Is this possible?
I mean, this doubles the computation time, but it adds nothing for me. Especially since the Jacobian that is reported is the jacobian at x0 (not at x1). If it would at least give me the jacobian at x1 as output i could calculate the next iteration myself.
The solution would to program a similar routine myself, but I rather use fsolve (i also want to use JacobPattern for example).
clear all;
clc;
fun_res = @(xx)xx.^2;
x0 = 1;
ex_fl = -10;
iter = 0;
while ex_fl < 1 && iter < 100
    iter = iter+1;
    %[xx_it(iter,1),res(iter,1),ex_fl,~,jac(iter,1)] = fsolve(fun_res,x0,optimoptions(@fsolve,'Display','iter','MaxIterations',1));
    [xx_it(iter,1),~,ex_fl] = fsolve(fun_res,x0,optimoptions(@fsolve,'Display','iter','MaxIterations',1));
    x0 = xx_it(iter);   
    
end

Best Answer

One option is to use fsolve just to get the Jacobian and then generate x1 yourself. This gives you access to JacobPattern and all the other Jacobian calculation features that fsolve offers.
To do so, replace your objective function f(x) with f(x)-f(x0), where x0 is thte initial point. Note that this does not change the Jacobian. Because x0 is a root of the modified objective, fsolve will return x0 and its Jacobian as output in zero iterations. Example:
fun=@(x) x.^2/2;
options=optimoptions(@fsolve,'MaxIter',1);
x0=rand(5,1);
f0=fun(x0);
[x,fval,ef,stats,JacobianNumerical]=fsolve(   @(x)fun(x) - f0 ,x0,options);
Equation solved at initial point.

fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
JacobianNumerical,
JacobianNumerical = 5×5
    0.3167         0         0         0         0
         0    0.5539         0         0         0
         0         0    0.7692         0         0
         0         0         0    0.0906         0
         0         0         0         0    0.5526

JacobianAnalytical=diag(x0)
JacobianAnalytical = 5×5
    0.3167         0         0         0         0
         0    0.5539         0         0         0
         0         0    0.7692         0         0
         0         0         0    0.0906         0
         0         0         0         0    0.5526
Also, it will do this with no extra Jacobian calculations, as seen from,
stats
stats = struct with fields:
       iterations: 0
        funcCount: 6
        algorithm: 'trust-region-dogleg'
    firstorderopt: 0
          message: '↵Equation solved at initial point.↵↵fsolve completed because the vector of function values at the initial point↵is near zero as measured by the value of the function tolerance, and↵the problem appears regular as measured by the gradient.↵↵<stopping criteria details>↵↵Equation solved. The final point is the initial point.↵The sum of squared function values, r = 0.000000e+00, is less than sqrt(options.FunctionTolerance) = 1.000000e-03.↵The relative norm of the gradient of r, 0.000000e+00, is less than options.OptimalityTolerance = 1.000000e-06.↵↵'
Related Question