MathGroup Archive 2007

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: rather complicated NonlinearFit

On Jan 30, 9:53 pm, fonfas... at 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:

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

  • Prev by Date: Re: loglogplot
  • Next by Date: reverse polish notation
  • Previous by thread: Re: Re: rather complicated NonlinearFit
  • Next by thread: Re: rather complicated NonlinearFit