how to use fsolve function

%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 个评论

Post the code you used.
function F = myfun(x)
F = [(-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;(-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);(-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)]
x0 = [-5; -5]; y0=[1 1] z0=[1 1]
options=optimset('Display','iter');
[x0,y0,z0,fval] = fsolve(@myfun,x0,y0,z0,options) % Call solver

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 采纳的回答

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)

更多回答(1 个)

Matt J
Matt J 2014-2-24
编辑:Matt J 2014-2-24

0 个投票

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.

2 个评论

could you write the codes I need please
[solution,fval] = fsolve(@myfun,[x0,y0,z0],options) % Call solver

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