24小时热门版块排行榜

返回列表

zidingxiang

金虫 (小有名气)

应助: 0 (幼儿园)
金币: 774.8
散金: 69
红花: 1
帖子: 114
在线: 31.7小时
虫号: 958621
注册: 2010-03-01
性别: MM
专业: 玻璃材料

[求助] 运行结果与设置精度不符，结果应与初值接近才对已有1人参与

1.首先建立函数myfun,储存方程组编程如下将myfun.m保存到工作路径中:
function F = myfun( x )
%this function designed for solving nonlinear equations
a = x(1);
b = x(2);
c = x(3);
d = x(4);
e = x(5);
f = x(6);

F = [a+b*10+c*10^2+(d+e*10)/(0.6438^2-f^2)-2*1.620015/(1.620015^2-1)*(-0.87+2.6);
a+b*10+c*10^2+(d+e*10)/(0.6328^2-f^2)-2*1.620507/(1.620507^2-1)*(-0.87+2.7);
a+b*10+c*10^2+(d+e*10)/(0.5893^2-f^2)-2*1.622702/(1.622702^2-1)*(-0.87+2.8);
a+b*10+c*10^2+(d+e*10)/(0.5461^2-f^2)-2*1.625401/(1.625401^2-1)*(-0.87+2.9);
a+b*10+c*10^2+(d+e*10)/(0.4800^2-f^2)-2*1.631029/(1.631029^2-1)*(-0.88+3.3);
a+b*10+c*10^2+(d+e*10)/(0.4358^2-f^2)-2*1.636369/(1.636369^2-1)*(-0.88+3.6)];

end

2.运行程序
x0=[-3e-005,-1.5e-008,-5.0e-011,3.0e-007,2.0e-010,0];
options=optimset('Display','iter','TolFun',1e-020,'TolX',1e-020,'MaxIter',1e100);
format long;
[x,fval]=fsolve(@myfun,x0,options)
运行结果显示如下：

                                       Norm of    First-order Trust-region
Iteration  Func-count    f(x)       step       optimality radius
   0       7       106.837                   2.5e+003             1
   1       14       1.58981    0.0393664          28.2             1
   2       21    0.0425606    0.0513838          36.8             1
   3       28    0.0198142 0.000578732       0.414             1
   4       35    0.0194798 0.000755402       0.541             1
   5       42    0.0194749 8.50804e-006       0.00609             1
   6       49    0.0194748 1.11053e-005       0.00795             1
   7       56    0.0194748 1.25077e-007    8.95e-005             1
   8       63    0.0194748 1.6325e-007    0.000117             1
   9       70    0.0194748 1.84156e-009    1.32e-006             1
10       77    0.0194748 2.41651e-009    1.58e-006             1
11       84    0.0194748 2.49273e-011    1.22e-008             1
Optimizer appears to be converging to a point which is not a root.
Relative function value changing by less than max(options.TolFun^2,eps) but
sum-of-squares of function values is greater than or equal to sqrt(options.TolFun)
Try again with a new starting guess.

x =

0.000167225746880 0.001972242468804 0.019722574638040 0.006232656757062 0.062323567700574 0.000000000137224

fval =

0.060290300323624
  -0.083434220551602
  -0.030535103096213
0.083729176350573
  -0.030583681232919
0.000533528265654

注：函数精度远远小于设定精度值；参数值应接近于设定值，但相差很远。

回复此楼