[Scilab-users] use of fsolve

[Paul Carrico]

Dear All

 

Additional « silly » question : in the system herebellow, the solver finds 4
values  (1 per variable), but we know there’s 4 couples of results due to
the 2nd order ; of course the values depend on the starting point

 

I feel there’s a strategy in order to find all the solution but I don’t know
(or I don’t remember) how to proceed : any advice or reading in order to
solve such kind of system ?

 

Thanks

 

Paul

 

De : users-bounces at lists.scilab.org [mailto:users-bounces at lists.scilab.org]
De la part de Radovan Omorjan
Envoyé : lundi 2 septembre 2013 17:32
À : International users mailing list for Scilab.
Objet : Re: [Scilab-users] use of fsolve

 

Hello,

If you have some additional parameters in your function like this 

f(x,a,b..)=0
 
searching x, then use list() in fsolve() calling with the function name as
the first argument and parameters as subsequent arguments - like this f.e.

clc, clear
function f=foo(x, cst, R_)
    f(1) = (x(3) - cst(3))*(x(3) - x(1)) + (x(4) - cst(4))*(x(4) - x(2)) ;
    f(2) = (x(1) - cst(1))*(x(3) - x(1)) + (x(2) - cst(2))*(x(4) - x(2)) ;
    f(3) = (x(1) - cst(1))^2 + (x(2) - cst(2))^2 - R_(1)^2 ;
    f(4) = (x(3) - cst(3))^2 + (x(4) - cst(4))^2 - R_(2)^2 ;
endfunction
 
// constants
a = [106 ; 30 ; 54 ; 15] //a would be cst
b = [9 ; 9]              //b would be R_
 
//Initial guess 
 x1 = zeros(4,1)
 
// calculation using fsolve
// ind - termination indicator
// x - solution
// f - function values at the solution
[x,f,ind] = fsolve(x1,list(foo,a,b))

Regards,
Radovan

On 9/2/2013 3:32 PM, Carrico, Paul wrote:

Dear All
A basic question : what is the correct synthax is order to call the
constants cst and R_ in the foo function ? do I need to use "global"
variable ?
Thanks
Paul
function [f, ind]=foo(x, cst, R_, ind)
    f(1) = (x(3) - cst(3))*(x(3) - x(1)) + (x(4) - cst(4))*(x(4) - x(2)) ;
    f(2) = (x(1) - cst(1))*(x(3) - x(1)) + (x(2) - cst(2))*(x(4) - x(2)) ;
    f(3) = (x(1) - cst(1))^2 + (x(2) - cst(2))^2 - R_(1)^2 ;
    f(4) = (x(3) - cst(3))^2 + (x(4) - cst(4))^2 - R_(2)^2 ;
endfunction
 
function [f, ind]=foo2(x, ind)
    cst = [106 ; 30 ; 54 ; 15]
    R_ = [9 ; 9]
    f(1) = (x(3) - cst(3))*(x(3) - x(1)) + (x(4) - cst(4))*(x(4) - x(2)) ;
    f(2) = (x(1) - cst(1))*(x(3) - x(1)) + (x(2) - cst(2))*(x(4) - x(2)) ;
    f(3) = (x(1) - cst(1))^2 + (x(2) - cst(2))^2 - R_(1)^2 ;
    f(4) = (x(3) - cst(3))^2 + (x(4) - cst(4))^2 - R_(2)^2 ;
endfunction
 
// constants
cst = [106 ; 30 ; 54 ; 15]
R_ = [9 ; 9]
 
x0 = cst
x1 = zeros(4,1)
 
// calculationusing fsolve
T1 = fsolve(x0,foo)
T2 = fsolve(x1,foo2)

 

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