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

4 ビュー (過去 30 日間)
```
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 件のコメント
Walter Roberson
Walter Roberson 2022 年 6 月 4 日
編集済み: Walter Roberson 2022 年 6 月 4 日
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 年 6 月 5 日
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)))));

サインインしてコメントする。

回答 (1 件)

Lateef Adewale Kareem
Lateef Adewale Kareem 2022 年 6 月 4 日
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

カテゴリ

Help Center および File ExchangeOptimization についてさらに検索

タグ

製品


リリース

R2021b

Community Treasure Hunt

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

Start Hunting!

Translated by