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least square curve fitting on a nonlinear equation , set of data available

9 vues (au cours des 30 derniers jours)
Hi all, I want to use least square curve fitting on a nonlinear equation , set of data available , I did the analytics on paper and want to be sure using pure code.
to summrize the method;
we first partial derivative with respect to a and equate it to 0
then partial derivative with respect to b and equate it to 0 as well.
we end up with 2 equations ( nonlinear).
how to solve and find a and b using code? I am new to matlab, any help would be appriciated.
the equation is where y and x are the data set and it goes 27 iterations.
Thanks.

Réponse acceptée

Star Strider
Star Strider le 17 Fév 2023
Modifié(e) : Star Strider le 17 Fév 2023
What you appear to be describing is the derivation of a linear least square regression. In that context, ‘linear’ implies ‘linear in the parameters’, such the partial derivatives of the objective function with respect to each parameter are not functions of that parameter or of any other parameters.
Your data are best determined by fminsearch. There are other options, however those require the Optimization Toolbox, the Statistics and Machine Learning Toolbox or Curve Fitting Toolbox.
Example —
xv = 1:10;
yv = sqrt(xv);
objfcn = @(b,x) b(1).*(1-exp(-b(2).*x));
B0 = rand(2,1);
[B,fval] = fminsearch(@(b)norm(yv-objfcn(b,xv)), B0)
B = 2×1
3.2928 0.2495
fval = 0.3841
figure
plot(xv,yv,'x', 'DisplayName','Data')
hold on
plot(xv, objfcn(B,xv), '-r', 'DisplayName','Regression Fit')
hold off
grid
xlabel('x')
ylabel('y')
legend('Location','best')
text(1,3, sprintf('$y = %.2f\\ (1-e^{-%.2f \\ x})$',B), 'Interpreter','latex')
EDIT — (17 Feb 2023 at 16:58)
Corrected typographical errors.
.

Plus de réponses (1)

Matt J
Matt J le 17 Fév 2023
Modifié(e) : Matt J le 17 Fév 2023
I would recommend this FEX download,
fun={@(b,x) (1-exp(-b*x))};
[b,a]=fminspleas(fun,b0); %b0 is initial guess of b

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