Error using trustnleqn (line 28)

Question

Amirul Hezry 2020-2-24

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/507314-error-using-trustnleqn-line-28

评论： Amirul Hezry 2020-2-25

Hello everyone,

I'm trying to solve non linear equations system using fsolve. I'm beginner in matlab I just have no idea what's going on with my program when I tried to run it, there's an error message in the command window:

function F = BEB(x)
n = 50; %number of stations
F(1) = x(1) - (2) / (1-x(2))*((1+x(2))+ 16*(1+2*x(2))+ ((x(2)^3)/(1-2*x(2)))*((x(2)/1-x(2))-((x(2)/1-x(2))^6)/((x(2)/1-x(2))^2))*((64*x(2)^3)/1-3*x(2))*(2*(x(2)/1-x(2))-((2*(x(2)/1-x(2)))^6))/(2*(x(2)/1-x(2))^2)); 
F(2) = (1-x(1))^n + x(2) -1;

i run it:

fun = @BEB
x0 = [0 0]
x = fsolve (fun, x0)

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Walter Roberson 2020-2-24

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/507314-error-using-trustnleqn-line-28#answer_417159

在 MATLAB Online 中打开

Part of your expression is

((x(2)/1-x(2))^6)/((x(2)/1-x(2))^2))

The right hand side of that is (x(2)/1 - x(2))^2 which is (x(2) - x(2))^2 which is 0^0 which is 0. You are dividing by that zero. Division by 0 is going to get you one of -inf, +inf, or nan, depending what the numerator works out as. Your numerator happens to work out as 0 because it also has the X(2)/1-x(2) basic unit.

Therefore your expression always includes 0 / 0 which is NaN, so BEB always returns NaN as the first value. Most routines will give up when they see a nan result.