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f1 = ROOT.TF1("f1","([2]/pow(x,[3]))/(1+exp(([0]-x)/[4]))+[1]",300,2000)
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Results
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The following plots show the evolution of the fitted parameter *\[0\]* as a function of time for both FLE and FHE. This parameter is likely to described the mean FLE (FHE) threshold. |
- The red cross are the obtained values and the error bars are the errors given by MINUIT
- First plot is the fitted FLE threshold as a function of time and the second plot is the same but for FHE threshold
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