fitting a parabola giving unreasonable answer

7 vues (au cours des 30 derniers jours)
Michael Phillips
Michael Phillips le 11 Avr 2019
Hello,
I'm trying to fit a parabola to 4 data points using the following equation:
y = a.*exp(((-4.*pi.*b.*6.022e23)./(8.314.*1623)).*(((c./2.*(c-x).^2) - (1/3.*(c-x).^3))));
I'm getting an unreasonable result, which looks like this:
fit_to_sun1.svg
I think the equation is correct because I copy and pasted it from a function that employs it to create this graph, which models the same points:
lattice_strain_model_divalent.svg
  5 commentaires
Michael Phillips
Michael Phillips le 11 Avr 2019
The 4 data points:
r = [0.89;1.12;1.26;1.42].*1e-10;
D = [0.027322404;1.798850575;1.33;0.11];
r is x and D is y.
John D'Errico
John D'Errico le 12 Avr 2019
Modifié(e) : John D'Errico le 12 Avr 2019
But your model is not a parabola. It is a nasty to compute exponential thing. (Nasty in double precision arithmetic.)
Seems confusing. I'd suggest your problem is the huge dynamic range of the parameters. That gets the solver in trouble.
b = [2;-1.15e21;1.2e-10];

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Clay Swackhamer
Clay Swackhamer le 12 Avr 2019
Two things: I changed your independent values (r) to something that is not so small. Second, I made your equation more simple. I tried it with your original values but it didn't work for me. Hopefully this gets you off to a good start.
%Data
r = [0.89;1.12;1.26;1.42];
D = [0.027322404;1.798850575;1.33;0.11];
%Set up the fit
ft = fittype('a*r^2+b*r+c', 'independent', 'r');
opts = fitoptions('Method', 'NonlinearLeastSquares');
opts.Display = 'Off';
opts.StartPoint = [0.2, 0.2, 0.3];
%Conduct the fit
[fitresult, gof] = fit(r, D, ft, opts);
%Evaluate the function for plotting
a = fitresult.a;
b = fitresult.b;
c = fitresult.c;
r_model = linspace(min(r), max(r), 100); %create 100 points to evaluate the model on
D_model = a*r_model.^2+b*r_model+c;
%Make plots
plot(r, D, 'bo', 'markerSize', 6) %plot the data
hold on
plot(r_model, D_model, 'LineWidth', 2, 'Color', 'r') %plot the model
leg = legend('Data', 'Model');
leg.FontSize = 14;
model and data.png
  3 commentaires
Clay Swackhamer
Clay Swackhamer le 12 Avr 2019
No problem. If this was helpful would you mind accepting the answer? Thanks
Michael Phillips
Michael Phillips le 12 Avr 2019
yep no problem!

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