In the example Nonlinear Equations with Analytic Jacobian, the
function bananaobj evaluates F and computes the
Jacobian J. What if the code to compute the Jacobian is not
available? By default, if you do not indicate that the Jacobian can be computed in the
objective function (by setting the SpecifyObjectiveGradient option in
options to true), fsolve, lsqnonlin, and lsqcurvefit instead use finite differencing to approximate the Jacobian.
This is the default Jacobian option. You can select finite differencing by setting
SpecifyObjectiveGradient to false using
optimoptions.
This example uses bananaobj from the example Nonlinear Equations with Analytic Jacobian as the objective function, but
sets SpecifyObjectiveGradient to false so that
fsolve approximates the Jacobian and ignores the second
bananaobj output.
n = 64; x0(1:n,1) = -1.9; x0(2:2:n,1) = 2; options = optimoptions(@fsolve,'Display','iter','SpecifyObjectiveGradient',false); [x,F,exitflag,output,JAC] = fsolve(@bananaobj,x0,options);
The example produces the following output:
Norm of First-order Trust-region
Iteration Func-count f(x) step optimality radius
0 65 8563.84 615 1
1 130 3093.71 1 329 1
2 195 225.104 2.5 34.8 2.5
3 260 212.48 6.25 34.1 6.25
4 261 212.48 6.25 34.1 6.25
5 326 102.771 1.5625 6.39 1.56
6 327 102.771 3.90625 6.39 3.91
7 392 87.7443 0.976562 2.19 0.977
8 457 74.1426 2.44141 6.27 2.44
9 458 74.1426 2.44141 6.27 2.44
10 523 52.497 0.610352 1.52 0.61
11 588 41.3297 1.52588 4.63 1.53
12 653 34.5115 1.52588 6.97 1.53
13 718 16.9716 1.52588 4.69 1.53
14 783 8.16797 1.52588 3.77 1.53
15 848 3.55178 1.52588 3.56 1.53
16 913 1.38476 1.52588 3.31 1.53
17 978 0.219553 1.16206 1.66 1.53
18 1043 0 0.0468565 0 1.53
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.The finite-difference version of this example requires the same number of iterations to converge as the analytic Jacobian version in the preceding example. It is generally the case that both versions converge at about the same rate in terms of iterations. However, the finite-difference version requires many additional function evaluations. The cost of these extra evaluations might or might not be significant, depending on the particular problem.