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I am trying to fit a Lorentzian distribution to my data, and I was trying the solution provided by blochwave in this post.

The model I am using is $f(x) = \frac{a}{(\frac{b-x}{c})^{2} +1}-1$, which I attempt to use like so:

LorentzianMODEL = (a/(((b - x)/c)^2 + 1)) - 1;
LorentzianFIT = NonlinearModelFit[Data,LorentzianMODEL, {a, b, c}, x, MaxIterations -> 500, 
Method -> {NMinimize}]

However when I plot LorentzianFIT I get the following: enter image description here

Which I do not understand, as my data looks like: enter image description here

I add my data here:

{{9.58*10^7,-78.4119},{9.581*10^7,-78.2809},{9.582*10^7,-77.6955},{9.583*10^7,-77.9017},{9.584*10^7,-77.6219},{9.585*10^7,-77.7346},{9.586*10^7,-77.3264},{9.587*10^7,-77.0228},{9.588*10^7,-77.0116},{9.589*10^7,-76.9978},{9.59*10^7,-77.1691},{9.591*10^7,-76.3959},{9.592*10^7,-76.5368},{9.593*10^7,-76.5986},{9.594*10^7,-75.9251},{9.595*10^7,-76.4683},{9.596*10^7,-76.1168},{9.597*10^7,-75.986},{9.598*10^7,-75.9121},{9.599*10^7,-75.198},{9.6*10^7,-75.3556},{9.601*10^7,-74.9559},{9.602*10^7,-74.8481},{9.603*10^7,-75.0825},{9.604*10^7,-74.8915},{9.605*10^7,-74.7229},{9.606*10^7,-74.3703},{9.607*10^7,-74.5724},{9.608*10^7,-74.0785},{9.609*10^7,-73.9684},{9.61*10^7,-74.1571},{9.611*10^7,-73.7761},{9.612*10^7,-73.608},{9.613*10^7,-73.4015},{9.614*10^7,-73.2046},{9.615*10^7,-73.0437},{9.616*10^7,-72.7368},{9.617*10^7,-72.671},{9.618*10^7,-72.6937},{9.619*10^7,-72.201},{9.62*10^7,-72.3267},{9.621*10^7,-71.8846},{9.622*10^7,-71.8481},{9.623*10^7,-71.6784},{9.624*10^7,-71.4149},{9.625*10^7,-71.0938},{9.626*10^7,-70.9062},{9.627*10^7,-70.787},{9.628*10^7,-70.4523},{9.629*10^7,-70.3844},{9.63*10^7,-70.093},{9.631*10^7,-69.936},{9.632*10^7,-69.9041},{9.633*10^7,-69.5365},{9.634*10^7,-69.2365},{9.635*10^7,-69.2724},{9.636*10^7,-68.9441},{9.637*10^7,-68.4834},{9.638*10^7,-68.377},{9.639*10^7,-68.2777},{9.64*10^7,-67.9335},{9.641*10^7,-67.8276},{9.642*10^7,-67.3677},{9.643*10^7,-67.2121},{9.644*10^7,-66.9543},{9.645*10^7,-66.5628},{9.646*10^7,-66.3919},{9.647*10^7,-66.2225},{9.648*10^7,-65.8333},{9.649*10^7,-65.4605},{9.65*10^7,-65.2045},{9.651*10^7,-64.86},{9.652*10^7,-64.6347},{9.653*10^7,-64.3055},{9.654*10^7,-63.921},{9.655*10^7,-63.6229},{9.656*10^7,-63.3132},{9.657*10^7,-62.9083},{9.658*10^7,-62.4926},{9.659*10^7,-62.1753},{9.66*10^7,-61.7799},{9.661*10^7,-61.3998},{9.662*10^7,-60.9837},{9.663*10^7,-60.6005},{9.664*10^7,-60.13},{9.665*10^7,-59.7679},{9.666*10^7,-59.362},{9.667*10^7,-58.8551},{9.668*10^7,-58.3979},{9.669*10^7,-57.9813},{9.67*10^7,-57.4714},{9.671*10^7,-57.0503},{9.672*10^7,-56.6426},{9.673*10^7,-56.2068},{9.674*10^7,-55.8046},{9.675*10^7,-55.4163},{9.676*10^7,-55.1658},{9.677*10^7,-54.9431},{9.678*10^7,-54.8101},{9.679*10^7,-54.7577},{9.68*10^7,-54.7391},{9.681*10^7,-54.8365},{9.682*10^7,-55.0157},{9.683*10^7,-55.203},{9.684*10^7,-55.4826},{9.685*10^7,-55.8251},{9.686*10^7,-56.1373},{9.687*10^7,-56.549},{9.688*10^7,-56.8983},{9.689*10^7,-57.2847},{9.69*10^7,-57.6948},{9.691*10^7,-58.0864},{9.692*10^7,-58.5025},{9.693*10^7,-58.8397},{9.694*10^7,-59.1333},{9.695*10^7,-59.6324},{9.696*10^7,-59.9333},{9.697*10^7,-60.2584},{9.698*10^7,-60.6089},{9.699*10^7,-60.8731},{9.7*10^7,-61.2167},{9.701*10^7,-61.5615},{9.702*10^7,-61.7549},{9.703*10^7,-62.0251},{9.704*10^7,-62.3054},{9.705*10^7,-62.5325},{9.706*10^7,-62.7754},{9.707*10^7,-63.2065},{9.708*10^7,-63.3878},{9.709*10^7,-63.5167},{9.71*10^7,-63.8702},{9.711*10^7,-64.0812},{9.712*10^7,-64.2383},{9.713*10^7,-64.4088},{9.714*10^7,-64.5798},{9.715*10^7,-64.8458},{9.716*10^7,-65.0902},{9.717*10^7,-65.2676},{9.718*10^7,-65.4861},{9.719*10^7,-65.5697},{9.72*10^7,-65.8855},{9.721*10^7,-65.9254},{9.722*10^7,-66.1197},{9.723*10^7,-66.3695},{9.724*10^7,-66.4352},{9.725*10^7,-66.6378},{9.726*10^7,-66.7653},{9.727*10^7,-66.9068},{9.728*10^7,-67.0008},{9.729*10^7,-67.1592},{9.73*10^7,-67.2369},{9.731*10^7,-67.3674},{9.732*10^7,-67.5434},{9.733*10^7,-67.8334},{9.734*10^7,-67.8152},{9.735*10^7,-67.8429},{9.736*10^7,-68.0408},{9.737*10^7,-68.0338},{9.738*10^7,-68.4292},{9.739*10^7,-68.4913},{9.74*10^7,-68.4621},{9.741*10^7,-68.6884},{9.742*10^7,-68.5909},{9.743*10^7,-68.9225},{9.744*10^7,-68.9119},{9.745*10^7,-69.1127},{9.746*10^7,-69.2131},{9.747*10^7,-69.3178},{9.748*10^7,-69.4601},{9.749*10^7,-69.3414},{9.75*10^7,-69.4669},{9.751*10^7,-69.7638},{9.752*10^7,-69.7311},{9.753*10^7,-69.8488},{9.754*10^7,-69.8445},{9.755*10^7,-69.9179},{9.756*10^7,-70.2205},{9.757*10^7,-70.3532},{9.758*10^7,-70.5463},{9.759*10^7,-70.3361},{9.76*10^7,-70.4333},{9.761*10^7,-70.6723},{9.762*10^7,-70.4073},{9.763*10^7,-70.5563},{9.764*10^7,-70.8198},{9.765*10^7,-71.0367},{9.766*10^7,-70.9699},{9.767*10^7,-71.0749},{9.768*10^7,-71.1531},{9.769*10^7,-71.1948},{9.77*10^7,-71.2866},{9.771*10^7,-71.2072},{9.772*10^7,-71.5042},{9.773*10^7,-71.4413},{9.774*10^7,-71.4487},{9.775*10^7,-71.495},{9.776*10^7,-71.6052},{9.777*10^7,-71.9859},{9.778*10^7,-71.7211},{9.779*10^7,-71.9483},{9.78*10^7,-71.7974}}
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12
  • $\begingroup$ Consider: 1) rescaling the values of your abscissae; 2) providing better starting values for the parameters. $\endgroup$
    – MarcoB
    Commented Jul 5, 2016 at 21:53
  • $\begingroup$ @MarcoB I am unsure what you mean by re-scaling my axis, do mean just have a scaling factor on the $x$ variable? For point 2 I tried this, I put in values for a,b,c based on just looking at the plot - no joy! Also from what I understand the point of the Method -> {NMinimize} is that you don't need to make any guesses at all. $\endgroup$
    – user27119
    Commented Jul 5, 2016 at 22:01
  • 1
    $\begingroup$ Try this: {a -> 25, b -> 9.68, c -> 0.04}, with the rescaled data and model without the final "-1". Better? (One does not even need Mathematica to find the parameters - a little trial and error and a plotting routine will do). $\endgroup$
    – Peltio
    Commented Jul 5, 2016 at 22:41
  • 2
    $\begingroup$ Rescaling usually helps in keeping numerical error at bay and tidying the expressions. The -1 does nothing to the peak when you are 55 units under the axis. Making that a variable will just set the offset, but then you are using a slightly more elaborate model - that's the refinement I suggested above. More importantly you can't change the concavity of the curve with that -1; that's why I suggested to shift all the data above the axis. To see how to select the parameters 'by hand' I suggest you read the second answer in the question you linked. $\endgroup$
    – Peltio
    Commented Jul 5, 2016 at 23:09
  • 1
    $\begingroup$ @Peltio, "One does not even need Mathematica to find the parameters" - sure, the initial guess can be eyeballed, but you need to use something to get the least squares fit, and I sure as peas ain't doing Levenberg-Marquardt by hand. ;) $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Jul 6, 2016 at 2:00

1 Answer 1

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5
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Your model function has no chance of reproducing your data (you have negative values, a linear skew, etc). Here is a different model, which is still a Lorentz peak, but with an added linearly varying baseline, and much better starting values for the parameters.

nlm = NonlinearModelFit[
        data, scale 1/(b (1 + (-a + x)^2/b^2)) + linear x + offset, 
        {{a, 96800000}, {b, 400000}, {scale, 10000000}, {offset, -80}, linear}, 
        x
      ];

Show[
 ListPlot[data, PlotRange -> All],
 Plot[nlm[x], Evaluate@Flatten@{x, MinMax[data[[All, 1]]]}, PlotStyle -> Red]
]

Mathematica graphics

Mathematica graphics

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2
  • $\begingroup$ Thanks! This looks great! If I can ask though, how do you choose the adapted model as well as the starting values, is this just from experience and looking at the data? $\endgroup$
    – user27119
    Commented Jul 5, 2016 at 22:46
  • $\begingroup$ @QuantumPenguin It was mostly trial and error. I added the linear skew because of the shape of your data, then used a simple Plot of the new model to see how the model responded to the parameters, and to find the ballpark values that would somewhat reproduce what you have. I then fed those values as starting points to the fitting function. $\endgroup$
    – MarcoB
    Commented Jul 7, 2016 at 14:11

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