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-Order                    Norm of

Iteration  Func-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]