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Scilab Help >> Optimization and Simulation > fsolve

fsolve

find a zero of a system of n nonlinear functions

Syntax

```[x, v, info] = fsolve(x0, fct)
[x, v, info] = fsolve(x0, fct, fjac)
[x, v, info] = fsolve(x0, fct, fjac, tol)
[x, v, info] = fsolve(x0, fct, tol)```

Arguments

x0

real vector (initial value of function argument).

fct

external (i.e function or list or string).

fjac

external (i.e function or list or string).

tol

real scalar. precision tolerance: termination occurs when the algorithm estimates that the relative error between x and the solution is at most tol. (`tol=1.d-10` is the default value).

x :

real vector (final value of function argument, estimated zero).

v :

optional real vector: value of function at x.

info

optional termination indicator:

0 improper input parameters. algorithm estimates that the relative error between x and the solution is at most tol. number of calls to fcn reached tol is too small. No further improvement in the approximate solution x is possible. iteration is not making good progress.

Description

find a zero of a system of n nonlinear functions in n variables by a modification of the powell hybrid method. Jacobian may be provided.

`0 = fct(x) w.r.t x.`

`fct` is an "external". This external returns `v=fct(x)` given `x`.

The simplest syntax for `fct` is:

`[v]=fct(x).`

If `fct` is a character string, it refers to a C or Fortran routine which must be linked to Scilab. Fortran calling sequence must be

```fct(n,x,v,iflag)
integer n,iflag
double precision x(n),v(n)```

and C Syntax must be

`fct(int *n, double x[],double v[],int *iflag)`

Incremental link is possible (help `link`).

`jac` is an "external". This external returns `v=d(fct)/dx (x)` given `x`.

The simplest syntax for `jac` is:

`[v]=jac(x).`

If `jac` is a character string, it refers to a to a C or Fortran routine which must be linked to Scilab calling sequences are the same as those for fct. Note however that v must be a nxn array.

Examples

```// A simple example with fsolve
a=[1,7;2,8];
b=[10;11];

function y=fsol1(x)
y=a*x+b
endfunction
function y=fsolj1(x)
y=a
endfunction

[xres]=fsolve([100;100],fsol1);
a*xres+b

[xres]=fsolve([100;100],fsol1,fsolj1);
a*xres+b

// See SCI/modules/optimization/sci_gateway/fortran/Ex-fsolve.f
[xres]=fsolve([100;100],'fsol1','fsolj1',1.e-7);
a*xres+b```

For some starting points and some equations system, the fsolve method can fail. The fsolve method is a local search method. So, to have a good chance to find a solution to your equations system, you must ship, a good starting point to fsolve.

Here is an example on which fsolve can fail:

```// Another example with fsolve
function F=feuler(x, r)
F=x-r-dt*(x.^2-x.^3);
endfunction
function J=dFdx(x)  //Definition of the Jacobian
J=1-dt*(2*x-3*x^2);
endfunction

r = 0.04257794928862307 ;
dt = 10;

[x,v,info]=fsolve(r,list(feuler,r),dFdx); // fsolve do not find the solution
disp(v); // The residual
disp(info); // The termination indicator

[x,v,info]=fsolve(1,list(feuler,r),dFdx); // fsolve find the solution
disp(v); // The residual
disp(info); // The termination indicator

clf();
x=linspace(0,1,1000);
plot(x,feuler(x))
a=gca();
a.grid=[5 5];```

So, each time you use fsolve, be sure to check the termination indicator and the residual value to see if fsolve has converged.

• external — Scilab Object, external function or routine
• qpsolve — linear quadratic programming solver
• optim — non-linear optimization routine