Nonlinear equation numerical solution

Question

Tadej Cigler 2020-7-21

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/568326-nonlinear-equation-numerical-solution

编辑： Matt J 2020-7-21

Hi,

I have a following relation

C = 1.2;                         
D = 9420;                  
E = -2.0;                        
B = 0.1973;
f = D*sin(C*atan(B*x-E*(B*x-atan(B*x))));

and I would like to X if Y = 9250.

When I use below code:

syms x0 y0
y0 = 9250;
eqn = y0 == StiffnessMagicFormula(D,C,B,E,x0);
slip_angle = vpasolve(eqn, x0)

I obtained following result:

slip_angle =

-4.1564971021120118282594483668007e191020734

But actually result should be around 6.8 .

I tryed also with functions: fsolve, roots and find but were all unsuccessful.

BR, Tadej

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

Matt J 2020-7-21

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/568326-nonlinear-equation-numerical-solution#answer_468759

编辑：Matt J 2020-7-21

在 MATLAB Online 中打开

I tryed also with functions: fsolve, roots and find but were all unsuccessful.

fsolve is overkill for such a simple 1D problem. Use fzero instead:

C = 1.2;                         
D = 9420;                  
E = -2.0;                        
B = 0.1973;
y0 = 9250;
f = @(x) D*sin(C*atan(B*x-E*(B*x-atan(B*x))))-y0;
x0=fzero(f,6.8)

For me, this results in

x0 =
    6.9606

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。

Answer 2

Fabio Freschi 2020-7-21

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/568326-nonlinear-equation-numerical-solution#answer_468756

在 MATLAB Online 中打开

% params
C = 1.2;                         
D = 9420;                  
E = -2.0;                        
B = 0.1973;
y0 = 9250;
% function
fun = @(x)D*sin(C*atan(B*x-E*(B*x-atan(B*x))))-y0;
% solution
x0 = fsolve(fun,0)