I am using NonlinearModelFit to extract the gradient of what appears to be linear data in a logplot:

data = {{-0.010133333333333334`, 
    124.50000000000001`}, {-0.006933333333333333`, 
    610.`}, {-0.005600000000000001`, 
    75.58333333333333`}, {-0.004266666666666669`, 
    33.183673469387756`}, {-0.0029333333333333295`, 
    17.23926380368098`}, {-0.0015999999999999973`, 
    5.47681331747919`}, {-0.00026666666666666505`, 
    1.3872715510270095`}, {0.0010666666666666672`, 
    0.3445990722332671`}, {0.0023999999999999994`, 
    0.0713022981732469`}, {0.0037333333333333316`, 
    0.05300353356890459`}, {0.005066666666666671`, 
    0.021496815286624203`}, {0.006400000000000003`, 
    0.0028530670470756064`}, {0.008266666666666665`, 

expmodel = 
  NonlinearModelFit[data, a*Exp[b*x], {a, b}, x, MaxIterations -> 200];

  LogPlot[expmodel[x], {x, -0.01, 0.005}]}]

The function returned by NonlinearModelFit, however, does not match the data at all. I even tried to give realistic initial values to parameters a and b, but that didn't help.

  • 3
    $\begingroup$ Why not take a Log to Y-axis value and use LinearModelFit? $\endgroup$
    – Wjx
    Commented Jun 15, 2016 at 7:08
  • $\begingroup$ If this help, I can post a detailed answer :) $\endgroup$
    – Wjx
    Commented Jun 15, 2016 at 7:08
  • $\begingroup$ @Wjx That was a good idea :) $\endgroup$
    – BillyJean
    Commented Jun 15, 2016 at 7:12
  • $\begingroup$ So I will write an answer as soon as I get home. $\endgroup$
    – Wjx
    Commented Jun 15, 2016 at 7:32

2 Answers 2


Usually we'll tend to fit linear system, and in this situaiton, add Log to data's y-axis will help: Log[a E^(b x)]=Log[a]+b x

Truely, after this transformation, the points lay on a line~ great for fitting!


It seems that this method could really do the work, so naturally, here comes the following code for fitting using normal linear fitting method:

ldata = {#1, Log@#2} & @@@ data;
f[x_] = E^Normal@LinearModelFit[ldata, x, x]

Show[ListPlot[data, PlotStyle -> Darker@Green, PlotLegends -> {"Data"}], 
Plot[f[x], {x, data[[1, 1]], data[[-1, 1]]}, PlotLegends -> {"Fit"}]]
Show[ListLogPlot[data, PlotStyle -> Darker@Green,PlotLegends -> {"Data"}], 
Plot[Log@f[x], {x, data[[1, 1]], data[[-1, 1]]}, PlotLegends -> {"Fit"}]]

The resultant function is: E^(-0.0888526 - 717.01 x) and the fitting curve is shown below:


The result seems to be a bit wierd in normal scale, but in Log scale, it's quite a good fit!

Will this help?


FindFit seems to work

f[a_, b_, x_] = a*Exp[b*x];
expmodel[x_] = f[a, b, x] /. FindFit[data,f[a, b, x],{a, b},x]

50.3816 E^(-167.73 x)

I will compare it with

f1[x_]=E^(-0.08885264746563443` - 717.0097150284254` x)

which can be found from Wjx's method.

Show[{ListPlot[data, PlotRange->All], Plot[{expmodel[x],f1[x]}, {x, -0.01, 0.01}]}]
Show[{ListLogPlot[data], LogPlot[{expmodel[x], f1[x]}, {x, -0.01, 0.01}]}]

enter image description here

Definitely f1 appears much closer to data. Now if you calculate the RMS deviation

RootMeanSquare[(#[[2]] - expmodel[#[[1]]]) &/@ data]
RootMeanSquare[(#[[2]] - f1[#[[1]]]) &/@ data]



expmodel turns out to be the winner.

All these mess is due to your second data point. If you start from third data point Wjx's method will come pretty close to FindFit.

  • $\begingroup$ "FindFit seems to work" ??? With all of the data points very much below the fitted curve? $\endgroup$
    – JimB
    Commented Jun 15, 2016 at 14:58
  • $\begingroup$ check my modified answer. It will clear your doubt. $\endgroup$
    – Sumit
    Commented Jun 15, 2016 at 16:08
  • $\begingroup$ But f1 is simply the same structure as expmodel: f1 == Exp[Log[a] + b*x] and should end up with the same fit. So that's why I wonder about "FindFit" working. $\endgroup$
    – JimB
    Commented Jun 15, 2016 at 16:14
  • $\begingroup$ @JimBaldwin, the fitting algorithm is different in two cases. $\endgroup$
    – Sumit
    Commented Jun 15, 2016 at 17:18
  • $\begingroup$ Sorry. I see now. f1 is the antilog of the fit by @Wjx rather than a fit of f1 directly. $\endgroup$
    – JimB
    Commented Jun 15, 2016 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.