lsqnonlin and stretched exponential function

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Alessandro 29 Mar 2020
댓글: Torsten 29 Mar 2020
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));
Thanks a lot.
Best wishes
Alessandro

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Alessandro 29 Mar 2020
I used this:
rng default
d = linspace(0,3);
y = exp(-1.3*d) + 0.05*randn(size(d));
I get the same issues with my experimental data.
I also attach everything else on my test script:
fun = @(r) r(1).*exp(-d.*r(2)) .^r(3)-y;
x0 = [1 4 0.2];
[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)))';
figure(1)
hold off
plot(d,y,'ko',d,x(1).*exp(-x(2).*d).^x(3),'b-')
Thanks
A
Ameer Hamza 29 Mar 2020
You code seems to give correct output
Alessandro 29 Mar 2020
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?

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채택된 답변

Torsten 29 Mar 2020
편집: Torsten 29 Mar 2020
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.

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Torsten 29 Mar 2020
Your function to be fitted is
y= r1*exp(-d*r2*r3).
with three fitting parameters.
Say r1 = 1, r2 = 3 and r3 = 4 is what the solver returns.
But why should this result be better than
r1 = 1, r2 = 2 and r3 = 6 because both give the same fitted data.
So only the product r2*r3 is relevant, and the result returned by the solver was only arbitrary.
Mathematically speaking, parameters r1 and r2 do not contain individual information -only their product does.
Test it by choosing different initial values for r2. My guess is that the final r2 and r3 in each run will be different, but their product will remain constant.
If you search for "streched exponential function" in Wikipedia, it's of the form
exp(-r2*d^r3), not
exp(-r2*d)^r3.
Alessandro 29 Mar 2020
Thanks. Right but I still don't get why the errors I got are making no sense to me
Torsten 29 Mar 2020
There is no such thing as confidence intervals for dependent parameters.

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