lsqnonlin and stretched exponential function

7 次查看（过去 30 天）
Alessandro
Alessandro 2020-3-29
评论： Torsten 2020-3-29
Hi all,
I am currrently trying to fit my data with lsqnonlin instead lsqcurvefit for a comparative reason and I am facing some troubles.
Starting with the easiest example in https://se.mathworks.com/help/optim/ug/lsqnonlin.html, I modify the function as:
fun = @(r) r(1).*exp(-d.*r(2)).^r(3)-y
This function yields x = [1.0376 3.1962 0.4104].
From here, I calculate my errors in the way:
[x,resnorm,residual,exitflag,output,lambda,J] = lsqnonlin(fun,x0);
N = length(y(:,1));
[Q,R] = qr(J,0);
mse = sum(abs(residual).^2)/(size(J,1)-size(J,2));
Rinv = inv(R);
Sigma_var = Rinv*Rinv'*mse;
x_er = full(sqrt(diag(Sigma_var)));
Here, the values I get are not making any sense to me x_er = [0.02701 6458034.8422 829226.8633]; especially for x(2) and x(3).
Fit and errors are totally fine if I fit similar data in the form: (i) r(1).*exp(-d.*r(2)) or (ii) r(1).*exp(-d.*r(2))+r(3).*exp(-d.*r(4));
Could you please help me? Am I missing something?
Thanks a lot.
Best wishes
Alessandro
  4 个评论
显示 2更早的评论
Alessandro
Alessandro 2020-3-29
Yes, indeed but I need the errors as well.
When I insert the exponent in the function, i.e., r(3), something goes extremely wrong.
Any idea?

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Torsten
Torsten 2020-3-29
编辑：Torsten 2020-3-29
error_bounds = nlparci(x,residual,'jacobian',J)
after the call to lsqnonlin.
If you don't have a toolbox with nlparci, search the web for nlparci.m.
  8 个评论
显示 6更早的评论
Alessandro
Alessandro 2020-3-29
Thanks. Right but I still don't get why the errors I got are making no sense to me
Torsten
Torsten 2020-3-29
There is no such thing as confidence intervals for dependent parameters.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Solver Outputs and Iterative Display 的更多信息

标签

产品


版本

R2019b

Community Treasure Hunt

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

Start Hunting!

Translated by