Re: rather complicated NonlinearFit

*To*: mathgroup at smc.vnet.net*Subject*: [mg73098] Re: rather complicated NonlinearFit*From*: "Robert Dodier" <robert.dodier at gmail.com>*Date*: Fri, 2 Feb 2007 05:19:27 -0500 (EST)*References*: <epp7cl$dvl$1@smc.vnet.net>

On Jan 30, 9:53 pm, fonfas... at interia.pl wrote: > here is the formula of the equation: > 1-Exp[-0.5 (A1 Exp[-E1/(8.31 x)] (1-(Sqrt[((A1 Exp[-E1/(8.31 x)])/(A2 > Exp[-E2/(8.31 x)]+A3 Exp[-E3/(8.31 x)])) 0.05])/(1+Sqrt[((A1 Exp[-E1/ > (8.31 x)])/(A2 Exp[-E2/(8.31 x)]+A3 Exp[-E3/(8.31 x)])) 0.05]))+A3 > Exp[-E3/(8.31 x)] ((Sqrt[((A1 Exp[-E1/(8.31 x)])/(A2 Exp[-E2/(8.31 x)] > +A3 Exp[-E3/(8.31 x)])) 0.05])/(1+Sqrt[((A1 Exp[-E1/(8.31 x)])/(A2 > Exp[-E2/(8.31 x)]+A3 Exp[-E3/(8.31 x)])) 0.05])))], > {x},{A1, A2, A3,E1, E2, E3} > should anyone want to give it a try, here you can find a csv file with > data:http://www.chemia.uj.edu.pl/~stelmach/fitfile.csv A couple of general comments -- (1) I see the values of x in the data file are in the range 300--800. Try fitting the data with new variables (for both the dependent and independent variables) such that the mean and standard deviation of each new variable are 0 and 1, respectively. This is just to make the numbers less likely to overflow or underflow -- there is no statistical meaning to this. (2) Given the form of the equation (with sums of exponentials), it's very likely that the 6 parameters are not well-determined by the data you have, i.e., there are wide ranges of values which give almost the same goodness of fit. That's not necessarily a bad thing -- for one thing it means you can afford to tolerate inaccuracies in the numerical analysis -- but it does mean that if you are trying to draw conclusions based on specific values of the parameters, your conclusions will be spurious. For what it's worth Robert Dodier