Fit Powerlaw to Data

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Tobias Bramminge
Tobias Bramminge on 6 Dec 2019
Commented: Star Strider on 8 May 2021
Hi all!
I need to fit following Power Law to some experimental data.
y = C(B+x)^n
The data I have is as the following:
STRESS = [0.574, 367.364, 449.112, 531.087, 596.241, 649.097, 695.038, 737.173, 815.008];
STRAIN = [2.8746e-04, 0.00063, 0.0459, 0.0901, 0.1320, 0.1725, 0.2132, 0.2557, 0.3579];
The variables are x and y are STRAIN and STRESS respectively, and I would like estimates for C, B and n.
I'm not sure how to use the fitting tool for this kind of specific model.
Any ideas or suggestions would be greatly appreciated!
Thank you very much!
  2 Comments
John D'Errico
John D'Errico on 6 Dec 2019
Edited: John D'Errico on 6 Dec 2019
One requirement is that you have the same number of values in each vector. You don't meet that basic requirement.
numel(STRESS)
ans =
8
numel(STRAIN)
ans =
9
Once you have actually told us the correct data, then you need to explain which variable is intended to be x, and which is y. If you want help, then make it possible, even easy, for someone to help you.
Tobias Bramminge
Tobias Bramminge on 6 Dec 2019
Thank you very much for your reply!
I am very sorry for the bad explanation, I am quite new at requesting help like this. I've updated the data and the variables are x and y.
Is this sufficient?

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Accepted Answer

Star Strider
Star Strider on 6 Dec 2019
Edited: Star Strider on 6 Dec 2019
Try this:
STRESS = [0.574, 367.364, 449.112, 531.087, 596.241, 649.097, 695.038, 737.173, 815.008];
STRAIN = [2.8746e-04, 0.00063, 0.0459, 0.0901, 0.1320, 0.1725, 0.2132, 0.2557, 0.3579];
yfcn = @(b,x) b(1).*(b(2)+x).^b(3);
B0 = [1E-6; 100; 2];
B = fminsearch(@(b) norm(STRAIN - yfcn(b,STRESS)), B0);
xv = linspace(min(STRESS), max(STRESS), 50);
yv = yfcn(B,xv);
figure
plot(STRESS, STRAIN, 'pg')
hold on
plot(xv, abs(yv), '-r')
hold off
grid
text(150, 0.27, sprintf('y = %.3E \\cdot (%.3f + x)^{%.3f}', B))
xlabel('STRESS')
ylabel('STRAIN')
Experiment to get different results.
EDIT (6 Dec 2019 at 17:50)
Added plot image:
1Fit Powerlaw to Data - 2019 12 06.png
  7 Comments
vedavathi
vedavathi on 8 May 2021
r square value?
Star Strider
Star Strider on 8 May 2021
vedavathi — It is a straightforward calculation, however it is easier to use fitnlm with ‘yfcn’ to get that and a number of other statistics.

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More Answers (2)

Stephan
Stephan on 6 Dec 2019

Image Analyst
Image Analyst on 6 Dec 2019
I'd use fitnlm() in the Statistics and Machine Learning Toolbox.
I suggest that you don't try to fit the first point. When I try to do that, it's impossible to get a fit. I get error messages that says the Jacobian is not well defined. "Warning: The Jacobian at the solution is ill-conditioned, and some model parameters may not be estimated well (they are not identifiable). Use caution in making predictions. " and it won't do the fit.
But if I try to fit from the second point onwards, I still get the warning but the fit seems reasonable. See plot below. If you want you could do a piece-wise fit where you fit everything to the left of the second point to a line.
0000 Screenshot.png

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