how to use fsolve function
16 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
%I have 3 nonlinear equation. I need to find unknowns I tried fsolve function but I couldn't use it properly, could anyone explain for me how can I use fsolve function?
(-4.71777*10^21)+(2.10263*(10^15)*x0)-(3.49196*(10^8)*(x0^2))+(28*x0^3)+(1.56321*(10^14)*y0)-(3.76035*(10^7)*x0*y0)-(1.16399*(10^8)*y0^2)+(28*x0*y0^2)+(1.11156*(10^15)*z0)-(2.6739*(10^8)*x0*z0)-(1.16399*(10^8)*z0^2)+(28*x0*z0^2)==0
(-7.62057*10^20)+(1.56321*(10^14)*x0)-(1.88017*(10^7)*x0^2)+(1.16012*(10^15)*y0)-(2.32797*(10^8)*x0*y0)+(28*(x0^2)*y0)-(5.64052*(10^7)*y0^2)+(28*y0^3)+(1.7955*(10^14)*z0)-(2.6739*(10^8)*y0*z0)-(1.88017*(10^7)*z0^2)+(28*y0*z0^2)==0
(-5.41882*(10^21)+(1.11156*(10^15)*x0)-(1.33695*(10^8)*x0^2)+(1.7955*(10^14)*y0)-(1.33695*(10^8)*y0^2)+(2.41161*(10^15)*z0)-(2.32797*(10^8)*x0*z0)+(28*(x0^2)*z0)-(3.76035*(10^7)*y0*z0)+(28*(y0^2)*z0)-(4.01085*(10^8)*z0^2)+(28*z0^3)==0
%x0,y0,z0 are the unknowns.
2 Commenti
Risposta accettata
Walter Roberson
il 24 Feb 2014
eqn1 = @(x0, y0, z0) (-4.71777*10^21)+(2.10263*(10^15)*x0)-(3.49196*(10^8)*(x0^2))+(28*x0^3)+(1.56321*(10^14)*y0)-(3.76035*(10^7)*x0*y0)-(1.16399*(10^8)*y0^2)+(28*x0*y0^2)+(1.11156*(10^15)*z0)-(2.6739*(10^8)*x0*z0)-(1.16399*(10^8)*z0^2)+(28*x0*z0^2);
eqn2 = @(x0, y0, z0) (-7.62057*10^20)+(1.56321*(10^14)*x0)-(1.88017*(10^7)*x0^2)+(1.16012*(10^15)*y0)-(2.32797*(10^8)*x0*y0)+(28*(x0^2)*y0)-(5.64052*(10^7)*y0^2)+(28*y0^3)+(1.7955*(10^14)*z0)-(2.6739*(10^8)*y0*z0)-(1.88017*(10^7)*z0^2)+(28*y0*z0^2);
eqn3 = @(x0, y0, z0) (-5.41882*(10^21)+(1.11156*(10^15)*x0)-(1.33695*(10^8)*x0^2)+(1.7955*(10^14)*y0)-(1.33695*(10^8)*y0^2)+(2.41161*(10^15)*z0)-(2.32797*(10^8)*x0*z0)+(28*(x0^2)*z0)-(3.76035*(10^7)*y0*z0)+(28*(y0^2)*z0)-(4.01085*(10^8)*z0^2)+(28*z0^3);
fun = @(x) [eqn1(x(1), x(2), x(3)); eqn2(x(1), x(2), x(3)); eqn3(x(1), x(2), x(3))];
now you can fsolve(fun, x0)
0 Commenti
Più risposte (1)
Matt J
il 24 Feb 2014
Modificato: Matt J
il 24 Feb 2014
Your initial guesses x0, y0, z0 should not be passed in separate scalar arguments to myfun (notice your myfun also accepts only a single input argument vector, as it should!). Also, fsolve will not return the solution vector components as separate scalar arguments.
Vedere anche
Categorie
Scopri di più su Symbolic Math Toolbox in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!