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Octave can solve sets of nonlinear equations of the form
F (x) = 0
using the function fsolve
, which is based on the Minpack
subroutine hybrd
. This is an iterative technique so a starting
point will have to be provided. This also has the consequence that
convergence is not guarantied even if a solution exists.
Given fcn, the name of a function of the form
f (
x)
and an initial starting point x0,fsolve
solves the set of equations such thatf(
x) == 0
.If fcn is a two-element string array, or a two element cell array containing either the function name or inline or function handle. The first element names the function f described above, and the second element names a function of the form
j (
x)
to compute the Jacobian matrix with elementsdf_i jac(i,j) = ---- dx_jYou can use the function
fsolve_options
to set optional parameters forfsolve
.
When called with two arguments, this function allows you set options parameters for the function
fsolve
. Given one argument,fsolve_options
returns the value of the corresponding option. If no arguments are supplied, the names of all the available options and their current values are displayed.Options include
"tolerance"
- Nonnegative relative tolerance.
Here is a complete example. To solve the set of equations
-2x^2 + 3xy + 4 sin(y) = 6 3x^2 - 2xy^2 + 3 cos(x) = -4
you first need to write a function to compute the value of the given function. For example:
function y = f (x) y(1) = -2*x(1)^2 + 3*x(1)*x(2) + 4*sin(x(2)) - 6; y(2) = 3*x(1)^2 - 2*x(1)*x(2)^2 + 3*cos(x(1)) + 4; endfunction
Then, call fsolve
with a specified initial condition to find the
roots of the system of equations. For example, given the function
f
defined above,
[x, info] = fsolve (@f, [1; 2])
results in the solution
x = 0.57983 2.54621 info = 1
A value of info = 1
indicates that the solution has converged.
The function perror
may be used to print English messages
corresponding to the numeric error codes. For example,
perror ("fsolve", 1) -| solution converged to requested tolerance
When no Jacobian is supplied (as in the example above) it is approximated numerically. This requires more function evaluations, and hence is less efficient. In the example above we could compute the Jacobian analytically as
function J = jacobian(x) J(1,1) = 3*x(2) - 4*x(1); J(1,2) = 4*cos(x(2)) + 3*x(1); J(2,1) = -2*x(2)^2 - 3*sin(x(1)) + 6*x(1); J(2,2) = -4*x(1)*x(2); endfunction
Using this Jacobian is done with the following code
[x, info] = fsolve ({@f, @jacobian}, [1; 2]);
which gives the same solution as before.