7楼: Originally posted by crazyiyy at 2014-07-14 17:06:19
您好！ 我还有最后一个问题，就是matlab种一定要给出初值， 请问初值的调试有没有什么方法或者经验啊，或者说如何判断当前初值得到的解是正解。...

function jie_fun
u0=[0.5 0.5 0.5];
u=fsolve(@myfun,u0)
f_val=myfun(u)



function f=myfun(u)
x=u(1);
y=u(2);
z=u(3);
f(1)=1/((4.5+(8.5)*(x+y-x*y)+(33/2)*(x+y-x*y)^2+(33/2)*(x+y-x*y)^3+(33/2)*...
    (x+y-x*y)^4)*(1-(x+y-x*y))/(1-(x+y-x*y)^5)+(1-(12/(48*2^3-4)))*(1-x)+6*...
    (12/(48*2^3-4))+(1-(12/(48*2^3-4)))*(1-x)*(12*(1-y)+1))-z;
f(2)= (13*(1-(1-z)^8)+9*z*(1-z)^8)*(1-x)*(1-y)-x;
f(3)=(1-(1-z)^8)/(2-(1-z)^9)-y;

function jie_fun
clear a;clc
format long
u0=[0.5 0.5 0.5];
u=fsolve(@myfun,u0)
f_val=myfun(u)



function f=myfun(u)
x=u(1);
y=u(2);
z=u(3);
f(1)=1/((4.5+(8.5)*(x+y-x*y)+(33/2)*(x+y-x*y)^2+(33/2)*(x+y-x*y)^3+(33/2)*...
    (x+y-x*y)^4)*(1-(x+y-x*y))/(1-(x+y-x*y)^5)+(1-(12/(48*2^3-4)))*(1-x)+6*...
    (12/(48*2^3-4))+(1-(12/(48*2^3-4)))*(1-x)*(12*(1-y)+1))-z;
f(2)= (13*(1-(1-z)^8)+9*z*(1-z)^8)*(1-x)*(1-y)-x;
f(3)=(1-(1-z)^8)/(2-(1-z)^9)-y;

3楼: Originally posted by 月只蓝 at 2014-07-14 16:14:10
以下程序第对2楼中程序的改进，提高解的小数点后的位数。


function jie_fun
clear a;clc
format long
u0=;
u=fsolve(@myfun,u0)
f_val=myfun(u)



function f=myfun(u)
x=u(1);
y=u(2);
z=u(3);
...