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

2 次查看（过去 30 天）

Miraboreasu ，2022-6-4

0
链接

Direct link to this question

https://ww2.mathworks.cn/matlabcentral/answers/1733485-solving-nonlinear-function-using-fzero-error-function-values-at-the-interval-endpoints-must-differ

评论： Sam Chak ，2022-6-5

```

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
显示隐藏 3更早的评论

Sam Chak 2022-6-5

@Miraboreasu

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。

请先登录，再回答此问题。

回答（1 个）

Lateef Adewale Kareem 2022-6-4

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1733485-solving-nonlinear-function-using-fzero-error-function-values-at-the-interval-endpoints-must-differ#answer_979060

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