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fsolve — find a zero of a system of n nonlinear functions
[x [,v [,info]]]=fsolve(x0,fct [,fjac] [,tol])
| 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 : | real vector (value of function at x). | ||||||||||
| info | : termination indicator
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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 calling sequence 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 Calling sequence 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 calling sequence 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.
// A simple example with fsolve
a=[1,7;2,8];b=[10;11];
deff('[y]=fsol1(x)','y=a*x+b');
deff('[y]=fsolj1(x)','y=a');
[xres]=fsolve([100;100],fsol1);
a*xres+b
[xres]=fsolve([100;100],fsol1,fsolj1);
a*xres+b
// See routines/default/Ex-fsolve.f
[xres]=fsolve([100;100],'fsol1','fsolj1',1.e-7);
a*xres+b
| << fit_dat | impl >> |