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I have a problem while trying to make a nonlinear fit of the following data:

qn = NSolve[Tan[q] == 3*q/(3 + 1870715*q^2) && 0 < q <= 65, q]

Qn = {3.141593164052022`, 6.283185562410763`, 9.424778130923505`, 
      12.566370741974769`, 15.707963370041444`, 18.849556006615824`, 
      21.99114864805175`, 25.132741292526145`, 28.27433393902618`, 
      31.41592658694417`, 34.557519235893395`, 37.69911188561605`, 
      40.840704535933654`, 43.982297186718704`, 47.12388983787773`, 
      50.26548248934059`, 53.40707514105368`, 56.5486677929753`, 
      59.69026044507251`, 62.831853097318984`}

eq1 = 1 - Sum[6*alpha*(1 + alpha)/(9 + 9*alpha + 
 (alpha*Qn[[i]])^2)*Exp[-(Qn[[i]])^2*d*t/r^2], {i, 1, 10}] /. alpha -> 1870715 
 /. r -> 4*10^(-4)

NonlinearModelFit[{{0.17*60, 0.0301}, {0.5*60, 0.02408}, {0.83*60, 0.01806}, 
  {1*60, 0.02408}, {1.5*60, 0.07224}, {2*60, 0.16856}, {3*60, 0.13244}, 
  {5*60, 0.13244}, {8*60, 0.1505}, {9*60, 0.14448}, {10*60, 0.13244}}, eq1, d, t]

General::nomem: The current computation was aborted because there was insufficient memory available to complete the computation.

Throw::sysexc: Uncaught SystemException returned to top level. Can be caught with Catch[[Ellipsis], _SystemException]. >>

SystemException["MemoryAllocationFailure"]

Thank you!

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    – Michael E2
    Commented May 16, 2016 at 15:57
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    – Michael E2
    Commented May 16, 2016 at 15:57

1 Answer 1

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You need to scale down the parameter d. One way is to modify the equation by dividing d by 10^12:

Qn = {3.141593164052022, 6.283185562410763, 9.424778130923505, 
   12.566370741974769, 15.707963370041444, 18.849556006615824, 
   21.99114864805175, 25.132741292526145, 28.27433393902618, 
   31.41592658694417, 34.557519235893395, 37.69911188561605, 
   40.840704535933654, 43.982297186718704, 47.12388983787773, 
   50.26548248934059, 53.40707514105368, 56.5486677929753, 
   59.69026044507251, 62.831853097318984};

eq1 = 1 - Sum[6*alpha*(1 + alpha)/(9 + 9*alpha + (alpha*Qn[[i]])^2)*
       Exp[-(Qn[[i]])^2*(d/10^12)*t/r^2], {i, 1, 10}] /. 
    alpha -> 1870715 /. r -> 4*10^(-4);

data = {{0.17*60, 0.0301}, {0.5*60, 0.02408}, {0.83*60, 0.01806},
 {1*60, 0.02408}, {1.5*60, 0.07224}, {2*60, 0.16856}, {3*60, 0.13244},
 {5*60, 0.13244}, {8*60, 0.1505}, {9*60, 0.14448}, {10*60, 0.13244}};

nlm = NonlinearModelFit[data, eq1, d, t] ;
nlm["BestFitParameters"]
(* {d -> 0.6517443755400871`} *)
nlm["ParameterErrors"]
(* {0.1906294119832146`} *)

That gets you a value of d being 6.517443755400871*10^11 on the original scale with the corresponding standard error of 1.906294119832146*10^11.

But the fit looks pretty horrible:

Show[ListPlot[data, Frame -> True, PlotRange -> All], Plot[nlm[t], {t, 0, 600}]]

Data and model fit

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  • $\begingroup$ Thank you, Jim! It works better for other data sets. $\endgroup$
    – Dmitry
    Commented May 20, 2016 at 4:04
  • $\begingroup$ Now I have one more question. What kind of exponential term can be added to the current fitting in order to have decay for large values of t? $\endgroup$
    – Dmitry
    Commented May 20, 2016 at 4:07
  • $\begingroup$ I know nothing about the subject matter but it sure looks like the model is constructed with some theory in mind. Are you looking for a new theory or just want to predict the observations well? Will you be needing to compare estimated parameters among different datasets? In any event, I think that's a new and legitimate question. $\endgroup$
    – JimB
    Commented May 20, 2016 at 4:12
  • $\begingroup$ I need to predict observations well. My idea is that smth like 1- (1/a)*exp(-x)-(1/(1-a))*exp(x) should lead to competition of exponents. Here a is one more parameter for fitting. $\endgroup$
    – Dmitry
    Commented May 20, 2016 at 4:21
  • $\begingroup$ There is no need to compare parameters for different data sets. Parameter d is determined from each data set separately $\endgroup$
    – Dmitry
    Commented May 20, 2016 at 4:27

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