Fitting Experimental data knowing the possible fitting formula

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.

• 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? Sep 11 '20 at 16:34
• The fit itself is terrible though. Maybe you should consider a better model. Sep 11 '20 at 16:40
• 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.
– JimB
Sep 11 '20 at 17:03
• 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.
– JimB
Sep 11 '20 at 21:11
• @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.
– JimB
Sep 11 '20 at 21:13

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]] Consider not using a pre-conceived model:

Quantile Regression

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

Simplify[qrFunc[t]]

ListPlot[{data, {#, qrFunc[#]} & /@ data[[All, 1]]},
Joined -> {False, True},
PlotLegends -> {"Original", "QuantileRegression"}] 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"}] • 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. Sep 11 '20 at 19:00
• @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. Sep 11 '20 at 20:01