Solving nonlinear function using fzero, Error Function values at the interval endpoints must differ in sign.

5 次查看(过去 30 天)
```
Imp=100;
t0=1e-6;
P=204000000;
Tf=2e-3;
x = fzero( @(x) myfunction(x, t0, Imp, P, Tf), [1.001, 10000]);
function [f] = myfunction( x, t0, Imp, P0, Tf)
f = Imp - (-(P0*t0*(x-1.0)*(x^(-Tf/(t0*(x-1.0)))-1.0))/(log(x)*(x^(-1.0/(x-1.0)) -x^(-x/(x-1.0))))+(P0*t0*(x^(-(Tf*x)/(t0*(x-1.0)))-1.0)*(x-1.0))/(x*log(x)*(x^(-1.0/(x-1.0))-x^(-x/(x-1.0)))));
end
```
x must bigger than 1.0
I don't think these input will make fzero suffer
thank you
  4 个评论
Walter Roberson
Walter Roberson 2022-6-4
编辑:Walter Roberson 2022-6-4
It is +100 at x=-1 but change x away from -1 and it goes complex, so at the moment I have no evidence that it has a real root.
Sam Chak
Sam Chak 2022-6-5
Once you have found the root of nonlinear function, can you verify if the solution really crosses 0?
Imp = 100;
t0 = 1e-6;
P0 = 204000000;
Tf = 2e-3;
f = @(x) Imp - (-(P0*t0*(x-1.0)*(x^(-Tf/(t0*(x-1.0)))-1.0))/(log(x)*(x^(-1.0/(x-1.0)) -x^(-x/(x-1.0))))+(P0*t0*(x^(-(Tf*x)/(t0*(x-1.0)))-1.0)*(x-1.0))/(x*log(x)*(x^(-1.0/(x-1.0))-x^(-x/(x-1.0)))));

请先登录,再进行评论。

回答(1 个)

Lateef Adewale Kareem
Imp=100;
t0=1e-6;
P=204000000;
Tf=2e-3;
x = nan;
options = optimset('Display','off'); % show iterations
x0 = 2;
while(isnan(x))
x = fzero( @(x) myfunction(x, t0, Imp, P, Tf), x0, options);
x0 = x0*1.2;
end
disp(['x = ', num2str(x)])
x = 1.762566874060497e+21
function [f] = myfunction( x, t0, Imp, P0, Tf)
f = Imp - (-(P0*t0*(x-1.0)*(x^(-Tf/(t0*(x-1.0)))-1.0))/(log(x)*(x^(-1.0/(x-1.0)) -x^(-x/(x-1.0))))+(P0*t0*(x^(-(Tf*x)/(t0*(x-1.0)))-1.0)*(x-1.0))/(x*log(x)*(x^(-1.0/(x-1.0))-x^(-x/(x-1.0)))));
end

类别

Help CenterFile Exchange 中查找有关 Optimization 的更多信息

标签

产品


版本

R2021b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by