because NaN or Inf function value encountered during search while using fzero

2022 6 2
1 个回答
回答已采纳
更新时间：2022 6 2
2 次查看（30 天）
Ran in:

0 个投票

Ran in:
I have a nonlinear equation---5e6-5e6*(exp(-aa*xt)-exp(-bb*xt))/(exp(-aa*1e-6)-exp(-bb*1e-6))=0
I am thinking is that due to the function is too nonlinear, and fzero cannot find the result?
For reference, the results should be 1e-6
aa=810930;
bb=1216395;
func = @(xt) 5e6-5e6*(exp(-aa*xt)-exp(-bb*xt))/(exp(-aa*1e-6)-exp(-bb*1e-6));
x0 = 0.00001;
roots = fzero(func,x0)
Exiting fzero: aborting search for an interval containing a sign change because NaN or Inf function value encountered during search. (Function value at -0.0008092 is Inf.) Check function or try again with a different starting value.
roots = NaN

请先登录，再回答此问题。

 采纳的回答

Matt J
Matt J 2022-6-2
编辑：Matt J 2022-6-2
Ran in:

0 个投票

Ran in:
The problem occurs because func() has no axis crossing.
aa=810930;
bb=1216395;
func = @(xt) 5e6-5e6*(exp(-aa*xt)-exp(-bb*xt))/(exp(-aa*1e-6)-exp(-bb*1e-6));
x0 = 0.00001;
fplot(func,[0,10e-6])
Based on the plot, it appears you should be looking for the minimum of func() rather than a root.
xmin=fminsearch(func,x0)
xmin = 1.0000e-06

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Problem-Based Optimization Setup 的更多信息

产品

版本

R2021b

标签

Community Treasure Hunt

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

Start Hunting!

Translated by