# Documentation

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## Nonlinear Equations with Analytic Jacobian

This example demonstrates the use of the default trust-region-dogleg fsolve algorithm (see Large-Scale vs. Medium-Scale Algorithms). It is intended for problems where

• The system of nonlinear equations is square, i.e., the number of equations equals the number of unknowns.

• There exists a solution x such that F(x) = 0.

The example uses fsolve to obtain the minimum of the banana (or Rosenbrock) function by deriving and then solving an equivalent system of nonlinear equations. The Rosenbrock function, which has a minimum of F(x) = 0, is a common test problem in optimization. It has a high degree of nonlinearity and converges extremely slowly if you try to use steepest descent type methods. It is given by

$f\left(x\right)=100{\left({x}_{2}-{x}_{1}^{2}\right)}^{2}+{\left(1-{x}_{1}\right)}^{2}.$

First generalize this function to an n-dimensional function, for any positive, even value of n:

$f\left(x\right)=\sum _{i=1}^{n/2}100{\left({x}_{2i}-{x}_{2i-1}^{2}\right)}^{2}+{\left(1-{x}_{2i-1}\right)}^{2}.$

This function is referred to as the generalized Rosenbrock function. It consists of n squared terms involving n unknowns.

Before you can use fsolve to find the values of x such that F(x) = 0, i.e., obtain the minimum of the generalized Rosenbrock function, you must rewrite the function as the following equivalent system of nonlinear equations:

$\begin{array}{c}F\left(1\right)=1-{x}_{1}\\ F\left(2\right)=10\left({x}_{2}-{x}_{1}^{2}\right)\\ F\left(3\right)=1-{x}_{3}\\ F\left(4\right)=10\left({x}_{4}-{x}_{3}^{2}\right)\\ ⋮\\ F\left(n-1\right)=1-{x}_{n-1}\\ F\left(n\right)=10\left({x}_{n}-{x}_{n-1}^{2}\right).\end{array}$

This system is square, and you can use fsolve to solve it. As the example demonstrates, this system has a unique solution given by xi = 1, i = 1,...,n.

### Step 1: Write a file bananaobj.m to compute the objective function values and the Jacobian.

function [F,J] = bananaobj(x)
% Evaluate the vector function and the Jacobian matrix for
% the system of nonlinear equations derived from the general
% n-dimensional Rosenbrock function.
% Get the problem size
n = length(x);
if n == 0, error('Input vector, x, is empty.'); end
if mod(n,2) ~= 0,
error('Input vector, x ,must have an even number of components.');
end
% Evaluate the vector function
odds  = 1:2:n;
evens = 2:2:n;
F = zeros(n,1);
F(odds,1)  = 1-x(odds);
F(evens,1) = 10.*(x(evens)-x(odds).^2);
% Evaluate the Jacobian matrix if nargout > 1
if nargout > 1
c = -ones(n/2,1);    C = sparse(odds,odds,c,n,n);
d = 10*ones(n/2,1);  D = sparse(evens,evens,d,n,n);
e = -20.*x(odds);    E = sparse(evens,odds,e,n,n);
J = C + D + E;
end

### Step 2: Call the solve routine for the system of equations.

n = 64;
x0(1:n,1) = -1.9;
x0(2:2:n,1) = 2;
[x,F,exitflag,output,JAC] = fsolve(@bananaobj,x0,options);

Use the starting point x(i) = –1.9 for the odd indices, and x(i) = 2 for the even indices. Set Display to 'iter' to see the solver's progress. Set SpecifyObjectiveGradient to true to use the Jacobian defined in bananaobj.m. The fsolve function generates the following output:

Norm of    First-order   Trust-region
Iteration  Func-count   f(x)       step       optimality    radius
0          1       8563.84                      615               1
1          2       3093.71           1          329               1
2          3       225.104         2.5         34.8             2.5
3          4        212.48        6.25         34.1            6.25
4          5        212.48        6.25         34.1            6.25
5          6       102.771      1.5625         6.39            1.56
6          7       102.771     3.90625         6.39            3.91
7          8       87.7443    0.976563         2.19           0.977
8          9       74.1426     2.44141         6.27            2.44
9         10       74.1426     2.44141         6.27            2.44
10         11        52.497    0.610352         1.52            0.61
11         12       41.3297     1.52588         4.63            1.53
12         13       34.5115     1.52588         6.97            1.53
13         14       16.9716     1.52588         4.69            1.53
14         15       8.16797     1.52588         3.77            1.53
15         16       3.55178     1.52588         3.56            1.53
16         17       1.38476     1.52588         3.31            1.53
17         18      0.219553     1.16206         1.66            1.53
18         19             0   0.0468565            0            1.53

Equation solved.

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

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