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Questions on the use of Mathematica to construct models for approximating empirical data.

I have mentioned this in a comment already, but this seems like a good opportunity to provide some related discussion in the form of a full-fledged answer. In Mathematica 8, we can take advantage of …
answered Feb 25 '12 by Oleksandr R.
One way to address this issue, and one that I think works well, is to fit to an explicitly complex model. While such an approach initially seems unnecessary for this real-valued problem, the crucial p …
answered Aug 26 '12 by Oleksandr R.
receive a unique and hopefully meaningful result. If this doesn't occur, you will need to experiment with the Method option. Any method supported by NonlinearModelFit can be used here, since ComplexFit merely transforms how the problem is stated before calling the usual fitting functions. …
answered Jul 6 '13 by Oleksandr R.
I just add that fitting under the assumptions applied in Jim Baldwin's answer was a large part of the motivation for me to write the TransformedFit package (but please get the updated code from …
answered Mar 25 '17 by Oleksandr R.
I think as much discussion as can reasonably be had on this issue has already taken place on this site, although the solution might not be readily apparent without the benefit of experience. This is n …
answered Mar 2 '14 by Oleksandr R.
below), it seems good enough for most applications that require fitting complex data. BeginPackage["TransformedFit`"]; ClearAll[TransformedParameter]; SetAttributes[TransformedParameter, HoldRest … ] -> 53.249, Re[b] -> 1.897, Im[b] -> 0.632} *) This is also useful for fitting real-valued data where the model may become erroneously complex-valued for certain values of the parameters. For example, from …
answered Jan 21 '13 by Oleksandr R.
described in the documentation.) {bins, counts} = HistogramList@RandomVariate[dist, 10000]; For fitting we need a list of bin centres and probabilities, rather than boundaries and counts: data …
answered Jul 11 '13 by Oleksandr R.
specify "CoordinateSystem" -> "Cartesian" (or specify it implicitly by omitting that option entirely), and you will get the best-fitting complex values for the parameters. …
answered May 1 '14 by Oleksandr R.
I think OP could have done this by themselves, or at least indicated whether or not they had tried it or anything else. We are not just here to do the leg-work for your Ph.D. while you feign helplessn …
answered May 21 '14 by Oleksandr R.
actual fitting), and so its requirements are even further removed from what one can provide in LibraryLink libraries. This situation exists for many Mathematica functions, and the requirement for the code … implementation involving not much more than simple arithmetic. For these reasons I already implemented it in compiled code, and in fact specifically for a fitting application. The same code may help you, if FindFit is really the only thing you need to reproduce in C. …
answered Jan 25 '15 by Oleksandr R.
is A, purple is B, and gold is X: The key to the exercise, of course, is the simultaneous fitting of all three datasets in order for the rate constants to be determined self-consistently. To … ] := Through[sol[k1, k2, k3][t], List][[i]] /; And @@ NumericQ /@ {k1, k2, k3, i, t}; The fitting is now straightforward. Although it will help if reasonable initial values are given, this is not …
answered Jul 13 '13 by Oleksandr R.