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Good Morning, I have a set of experimental data which i want to fit with a formula that I know it works. I tried to write a code in mathematica notebook that I put it here

data = {{1.45045*10^8, 1.}, {7.44768*10^8, 0.8787}, {1.81148*10^9,0.7013}, {3.34183*10^9, 0.5414}, {5.33711*10^9,0.3959}, {7.80159*10^9, 0.2703}, {1.07275*10^10,0.1857}, {1.41183*10^10, 0.1399}, {1.79806*10^10,0.1022}, {2.2302*10^10, 0.07084}, {2.70884*10^10,0.04772}, {3.23398*10^10,0.03562}, {3.80655*10^10,0.02567}, {4.42474*10^10, 0.01419}, {5.08944*10^10,0.01431}, {5.80178*10^10, 0.01108}};

model = 1/(1 + σ^2/DD*b)^(DD/σ)^2;
nml = NonlinearModelFit[data, model, {DD, σ}, b]

Show[
  Plot[nlm[b], {b, 0, 10^11}],
  ListPlot[data, PlotStyle -> {Darker@Green, PointSize[0.03]}]
]

It gives some errors. I want to show the superposition of original data and the fitting function on semiLog (in the ordinate axis). Can you help me Please? Thank you. Have a nice day.

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  • $\begingroup$ It's because it's trying out negative values. Add constraints so it doesn't try to explore negative values for DD and σ and it works: nml = NonlinearModelFit[data, {model, σ > 0, DD > 0}, {DD, σ}, b]. You can get the parameters with nml["ParameterTable"]. Also watch out, you are plotting nlm[b] not nml[b] - must be a typo? $\endgroup$
    – flinty
    Commented Sep 11, 2020 at 16:34
  • 2
    $\begingroup$ The fit itself is terrible though. Maybe you should consider a better model. $\endgroup$
    – flinty
    Commented Sep 11, 2020 at 16:40
  • $\begingroup$ Your problem is the wildly different scalings of the two variables. Try data[[All, 1]] = data[[All, 1]]/10^10 and you'll get a near perfect fit with no need for constraints. $\endgroup$
    – JimB
    Commented Sep 11, 2020 at 17:03
  • $\begingroup$ Alternatively, if you use model = 1/(1 + \[Sigma]^2/DD*(b/10^10))^(DD/\[Sigma])^2;, then the fit will also work with no problems. $\endgroup$
    – JimB
    Commented Sep 11, 2020 at 21:11
  • $\begingroup$ @flinty How did you judge the fit of the model to be terrible? Once one accounts for the scaling of the predictor variable, then it looks fine. $\endgroup$
    – JimB
    Commented Sep 11, 2020 at 21:13

2 Answers 2

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Changing the model definition from

model = 1/(1 + σ^2/DD*b)^(DD/σ)^2;

to

model = 1/(1 + σ^2/DD*(b/10^10))^(DD/σ)^2;

will fix the problem.

data = {{1.45045*10^8, 1.}, {7.44768*10^8, 0.8787}, {1.81148*10^9, 0.7013},
  {3.34183*10^9, 0.5414}, {5.33711*10^9, 0.3959}, {7.80159*10^9, 0.2703},
  {1.07275*10^10, 0.1857}, {1.41183*10^10, 0.1399}, {1.79806*10^10, 0.1022}, 
  {2.2302*10^10, 0.07084}, {2.70884*10^10, 0.04772}, {3.23398*10^10, 0.03562},
  {3.80655*10^10, 0.02567}, {4.42474*10^10, 0.01419}, {5.08944*10^10, 0.01431},
  {5.80178*10^10, 0.01108}};
model = 1/(1 + σ^2/DD*(b/10^10))^(DD/σ)^2;
nlm = NonlinearModelFit[data, model, {DD, σ}, b];
nlm["BestFitParameters"]
(* {DD -> 2.05829, σ -> 1.22768} *)
Show[ListPlot[data], Plot[nlm[x], {x, 0, 6 10^10}, PlotRange -> All]]

Data and fit

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Consider not using a pre-conceived model:

Quantile Regression

qrFunc = ResourceFunction["QuantileRegression"][data, 6, 0.5][[1]];

Simplify[qrFunc[t]]

ListPlot[{data, {#, qrFunc[#]} & /@ data[[All, 1]]}, 
 Joined -> {False, True}, 
 PlotLegends -> {"Original", "QuantileRegression"}]

enter image description here

FindFormula

ffFunc = FindFormula[data];

(* ffFunc = Piecewise[{{1.3784372096038422 - 
       0.018976070116910493*Log[#1] - 1.504938184291008*^-10*#1, 
      Inequality[1.45045*^8, LessEqual, #1, Less, 
              2.664504676051075*^9]}, {9.462060930568198 - 
       0.4091705591204782*Log[#1] + 1.6776712526675216*^-11*#1, 
            
      Inequality[2.664504676051075*^9, LessEqual, #1, Less, 
       2.177107509602669*^10]}, {2.3369081717676714 - 
       0.09516385999471486*Log[#1], 
            
      Inequality[2.177107509602669*^10, LessEqual, #1, Less, 
       3.5441247654932556*^10]}, {1.5828759881621233 - 
       0.0634289715131976*Log[#1], 
            
      Inequality[3.5441247654932556*^10, LessEqual, #1, Less, 
       4.756828177139195*^10]}}, 0] & ; *)

ListPlot[{data, {#, ffFunc[#]} & /@ data[[All, 1]]}, 
 PlotLegends -> {"Original", "FindFormula"}]

enter image description here

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  • $\begingroup$ Thank you for your answer. However, if I try to fit a different set of data (from the same experiment but at different temperature) it does not work. $\endgroup$
    – Michaels
    Commented Sep 11, 2020 at 19:00
  • $\begingroup$ @Michaels FindFormula is a little bit of "wild card" -- try it at least a few times with the same data, you should get different fit functions. Also, I used FindFormula to illustrate my point. In general, I would use Quantile Regression. $\endgroup$
    – Anton Antonov
    Commented Sep 11, 2020 at 20:01

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