Nonlinear Curve Fitting Problem
in [Matlab]
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From: Anton on 21 Apr 2010 16:24 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
From: John D'Errico on 21 Apr 2010 17:41 "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? Sigh. You need to fit this curve very accurately. Of course, the model that you pose does not look even remotely like the data that you have. You cannot fit a model simply by wanting very much for it to fit. Prayer won't help either. Since you apparently have no idea of a good model for this data, we cannot help you. The best I can offer is the simple polyfit. P = polyfit(x,n,2); At least it fits quite nicely. And anything of a higher order than a quadratic polynomial is pretty much a waste of cpu cycles. John
From: James Phillips on 23 May 2010 09:04 This data set fits extremely well to the standard Steinhart-Hart equation: n = coeff_A + coeff_B * ln(x) + coeff_C * ln(x)^3 and appears to both interpolate and extrapolate smoothly. I found this by using the 'Function Finder' at http://zunzun.com Wikipedia link to Steinhart-Hart equation: http://en.wikipedia.org/wiki/Steinhart%E2%80%93Hart_equation James Phillips zunzun(a)zunzun.com http://zunzun.com "Anton " <anton.mazurenko(a)gmail.com> wrote in message <hqnmt4$1bj$1(a)fred.mathworks.com>... > Hello, I need to fit a curve very accurately. The function is > > n = ( k * sqrt(1 + x^2 / h^2)+phi ) / pi > > where k, h, phi are fitting parameters > and the data is > n = > 2 > 3 > 4 > 5 > 6 > 7 > > x = > 3.7910 > 4.0640 > 4.2850 > 4.4930 > 4.6750 > 4.8440 > > I tried doing a nonlinear least squares, but it comes up a fitted curve that is essentially straight, so the residuals go up and then down, and are not randomly distributed. > > Can you suggest a better method? > > Thank you, > > Antonmaz
From: John D'Errico on 23 May 2010 09:53 "James Phillips" <zunzun(a)zunzun.com> wrote in message <htb944$fa7$1(a)fred.mathworks.com>... > This data set fits extremely well to the standard Steinhart-Hart equation: > > n = coeff_A + coeff_B * ln(x) + coeff_C * ln(x)^3 > > and appears to both interpolate and extrapolate smoothly. I found this by using the 'Function Finder' at http://zunzun.com > > Wikipedia link to Steinhart-Hart equation: http://en.wikipedia.org/wiki/Steinhart%E2%80%93Hart_equation > > James Phillips > zunzun(a)zunzun.com > http://zunzun.com Sorry, but this is just silly. Ridiculous in the extreme. Note that the model you pose results in a fit that while accurate, is in fact poorer than the fit that you get from a simple quadratic polynomial. The model you pose has a residual standard deviation of 0.012987. The R^2 is 0.99996. While these are reasonably good numbers for a curve fit, note that had you tried a simple quadratic polynomial, you would have found that it yields a lower residual error and a higher r^2. The quadratic model has a residual standard deviation of 0.011293, with R^2 = 0.99996. And the last that I checked, quadratic polynomials also extrapolate fairly smoothly. Note that the quadratic has the same number of coefficients, (three) as the model that you pulled out of a hat. My point is that throwing a canned routine at your data to automatically test a variety of models until you find one that fits is a remarkably fatuous strategy, especially if you can't bother to compare the results to a simple quadratic polynomial fit on the same data. Yes, you are proud of your code. But next time try showing some common sense when you post a response like this. John
From: James Phillips on 23 May 2010 13:57
"John D'Errico" <woodchips(a)rochester.rr.com> wrote in message <htbbvv$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 |