Fitting log-linear data

Question

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`, 
    0.004938271604938271`}};

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

Show[{
  ListLogPlot[data],
  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.

Answer 1

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;
ListPlot@ldata
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?

Answer 2

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}]}]

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

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

137.858

354.308

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.