Nonlinear Curve Fitting Problem
in [Matlab]
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From: John D'Errico on 23 May 2010 15:56 "James Phillips" <zunzun(a)zunzun.com> wrote in message <htbq9f$as6$1(a)fred.mathworks.com>... > "John D'Errico" <woodchips(a)rochester.rr.com> wrote in message <htbbvv$jtc$1(a)fred.mathworks.com>... > > > > Sorry, but this is just silly. Ridiculous in the extreme. > > I apologize for posing an well-understood alternative physical model that might help explain the underlying physics of the data set, even though you did not. Better? > > Note the number of digits of precision in the data set. No. You picked out an arbitrary, random model, that happens to fit well. Now you claim that it will help to explain the underlying physics. There is no physical justification for choosing this model from the OP's query. That it happens to fit well does not suggest that it explains anything. The quadratic fits just as well. Does it explain anything? Of course not. But I'm not trying to claim that it will. Both models extrapolate smoothly. So what? The number of digits of precision provided tells you nothing here. Sorry, but this is just an abuse of modeling. John
From: Mark Shore on 23 May 2010 16:31 "James Phillips" <zunzun(a)zunzun.com> wrote in message <htbq9f$as6$1(a)fred.mathworks.com>... > "John D'Errico" <woodchips(a)rochester.rr.com> wrote in message <htbbvv$jtc$1(a)fred.mathworks.com>... > > > > Sorry, but this is just silly. Ridiculous in the extreme. > > I apologize for posing an well-understood alternative physical model that might help explain the underlying physics of the data set, even though you did not. Better? > > Note the number of digits of precision in the data set. > > James One usually uses a more complex model where there is either an understood physical basis for doing so, or where trial and error over a range of parameters has eliminated simpler methods. Here we are given a grand total of six points, with no explanation of what they might represent physically - if anything. With a little digging, I find that the Steinhart–Hart approximation is used to model thermistor resistance vs. temperature. OK, good, but is there any indication this is relevant for the set of numbers shown? After all, a fifth-order polynomial would fit the data both perfectly, and pointlessly.
From: James Phillips on 23 May 2010 17:41 Perhaps this is a good place for the original poster to chime in. Anton, any thoughts or remarks on this thread? James
From: Star Strider on 30 May 2010 21:54
"Anton " <anton.mazurenko(a)gmail.com> wrote in message <hqnmt4$1bj$1(a)fred.mathworks.com>... > Hello, I need to fit a curve very accurately. The function is > > n = ( k * sqrt(1 + x^2 / h^2)+phi ) / pi > > where k, h, phi are fitting parameters > and the data is > n = > 2 > 3 > 4 > 5 > 6 > 7 > > x = > 3.7910 > 4.0640 > 4.2850 > 4.4930 > 4.6750 > 4.8440 > > I tried doing a nonlinear least squares, but it comes up a fitted curve that is essentially straight, so the residuals go up and then down, and are not randomly distributed. > > Can you suggest a better method? > > Thank you, > > Antonmaz With due note to the fact that I may be missing something, I didn't have any trouble fitting it: nfit = inline('( B(1) .* sqrt(1 + x.^2 ./ (B(2).^2) ) + B(3)) ./ pi', 'B', 'x'); % B(1) = k, B(2) = h, B(3) = phi B0 = rand(3,1); [Bfit, nrsd, nJ, covB, mse] = nlinfit(x, n, nfit, B0); This estimates: Parameters: k = 604.376 h = 12.857 phi = -624.226 MSE = 0.017 If I wrote your equation correctly in my "inline" function, this works. If I didn't, rewrite it and post your rewrite. Run it to see if it fits your data, and post your parameters. Frank |