[Scilab-users] find a zero of a system of nonlinear functions

[CRETE Denis]

Hello 
The first argument in fsolve is not the range of the variable in which the solution is looked for. Instead it is a guessed value:

// Start of code
function y = poly_cubic(x)
y = 3 - 10*x + 2*x^2 + 3*x^3;
endfunction
 
yr=fsolve(1.3,poly_cubic,%eps)
// yr  = 1.3027756  
yr=fsolve(-2,poly_cubic,%eps)
// yr  =  - 2.3027756  
yr=fsolve(0,poly_cubic,%eps)
// yr  =  0.3333333  
// End of code

By the way, for polynoms, it is more straightforward to do :
p=[3 2 -10 3]; roots(p)
// ans  =
  - 2.3027756  
    1.3027756  
    0.3333333  

HTH
Denis

-----Message d'origine-----
De : users [mailto:users-bounces at lists.scilab.org] De la part de Hermes
Envoyé : mardi 28 novembre 2017 11:13
À : users at lists.scilab.org
Objet : [Scilab-users] find a zero of a system of nonlinear functions

fsolve
I find a zero of a system of nonlinear functions:
y = 3 - 10 * x + 2 * x ^ 2 + 3 * x ^ 3 in the range x = -3: 0.01: 3

but fsolve returns as a solution a vector (in the scrip yr) of the same size as x.
I have varied the precision tolerance h to several values 1.d-3 to% eps. but I always get that infinity of solutions.
the only solution that I found is shown in the script.
Is it possible to improve this?

funcprot(0);
function y = poly_cubic(x)
 y = 3 - 10*x + 2*x^2 + 3*x^3;
endfunction

x=-3:0.01:3;
clf()
plot(x,poly_cubic);

disp("-2.3027756 0.3333333 1.3027756  ", "Roots to y = 3 - 10*x + 2*x^2 +
3*x^3")
h=%eps;
yr=fsolve(x,poly_cubic,h)
RR=strtod(unique(string(yr)));
RR=gsort(RR,'g','i');//sort the elements of the array RR in the increasing order.

disp(RR,"Roots values f(x)")
plot(RR,poly_cubic,"ro") 




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