计算结果感觉误差好大，是由于初值取值造成的影响吗，还是要设置option来控制精度，第一次使用matlab一窍不通，请大家帮忙看下。
First-OrderNorm of
IterationFunc-count Residual optimality Lambda step
0 8 39.2026 1.13e+003 0.01
1 16 0.00103045 0.705 0.001 0.294456
2 24 3.55079e-005 0.0727 0.0001 0.237055
3 32 7.28141e-006 0.19 1e-005 0.144512
4 40 2.91808e-006 0.0123 1e-006 0.0248491
5 48 2.90763e-006 3.58e-007 1e-007 0.000338087
6 56 2.90763e-006 1.02e-009 1e-008 1.90914e-006
Equation solved, fsolve stalled.
fsolve stopped because the relative size of the current step is less than the
default value of the step size tolerance and the vector of function values
is near zero as measured by the selected value of the function tolerance.
y =
-2.2413 2.2083 -0.1310 0.1318 1.0596 0.0211 -0.0272
fval =
0.0000
0.0008
-0.0008
0.0003
0.0000
0.0000
-0.0011
0.0006
下面是我要解的方程组
function F=myfun(x)
b=1;a=1;
F=[-0.0575*x(3)/a-x(7)*a*0.5-x(6);
x(3)/a^2+x(4)+17.39*x(7)*log(a)+0.8695652174*b*log(a);
-x(1)/(x(5)^2)-x(2)+x(3)/(x(5)^2)+x(4)-92.49880554*x(7)*log(x(5))+0.8695652174*b*log(x(5));
0.0065*x(1)/x(5)-0.0026*x(2)*x(5)-0.0575*x(3)/x(5)+x(7)*x(5)*((0.7-0.4*log(x(5)))/1.4-0.9347826087);
-2*x(1)/(x(5)^2)+109.8901099*x(7)-b;
2*x(1)/(1.5^2)-109.8901099*x(7)-b;
0.444444444444444*x(1)/a^2+x(2)+109.8901099*x(7)*log(1.5*a);
-x(1)*log(1.5*a/x(5))+0.5*(1.5^2*a^2-x(5)^2)*x(2)-x(3)*log(x(5)/a)+0.5*(x(5)^2-a^2)*x(4)+17.39130435*x(7)*(0.5*x(5)^2*(log(x(5))+0.5)-0.5*a^2*(log(a)+0.5))+109.8901099*x(7)*(0.5*1.5*1.5*a^2*(0.5+log(1.5*a))-0.5*x(5)^2*(log(x(5))+0.5))+0.0209742*1.5^2*a^2*b+0.8695652174*b*(0.5*x(5)^2*(log(x(5))+0.5)-0.5*a^2*(log(a)+0.5))];
选取的初值是x0=[-2,2,-0.5,0.5,1.05,0.01,-0.01]
