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The power law function for gamma-ray burst is defined for amplitude ($A$), energy ($E$), pivot energy ($E_{piv}$) (fixed) and index $\lambda$ as

$ f(E; A, \lambda) = N(E) = A*\left(\dfrac{E}{E_{piv}}\right)^\lambda $

Where we use a range of energy for model fitting of GRB. The fitting gives us a single value for all variables except for energy (which is the input), and error propagation for the said formula can be calculated by partial differentiating it with respect to it's variables. What I'm confused at is that what value will I give to $E$ when I'll evaluate the error expression

$ err_{pl} = \displaystyle\sqrt(\left(\frac{E}{E_{piv}}\right)^{2\lambda}s_A^2 + \frac{A^2\lambda^2\left(E/E_{piv}\right)^{2\lambda-2}s_E^2}{E_{piv}^2} + A^2\left(\frac{E}{E_{piv}}\right)^{2\lambda}\log\left(\frac{E}{E_{piv}}\right)^2s_{\lambda}^2) $

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  • $\begingroup$ Have you tried plotting the error versus E? $\endgroup$ – Ed V Jul 8 at 13:34
  • $\begingroup$ okay, If I got your point correct yes I can plot errors versus E as an envelope over the fitted model since the error is symmetrical so it doesn't change over the entire energy range $\endgroup$ – Syed Ali Jul 8 at 14:24
  • $\begingroup$ Just to make sure I understand the situation: when you do the nonlinear curve fitting, of your real data to your model (top equation), I assume you know and enter the pivot energy and hold it constant in the fit. The fit returns estimates of A and lambda and their standard errors. Then your error equation gives you the error as a function of E. If the error was independent of E, then you could use any reasonable E. If my understanding is correct, no need to respond! Contiinued in next comment. $\endgroup$ – Ed V Jul 8 at 16:54
  • $\begingroup$ You may find these papers helpful: "A simple, all-purpose nonlinear algorithm for univariate calibration," by J. Tellinghuisen, Analyst 125, 1045-1048 (2000), and "Statistical Error Propagation," by J. Tellinghuisen, J. Phys. Chem. A 105, 3917-3921 (2001). $\endgroup$ – Ed V Jul 8 at 16:56

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