Re: exponential fitting

[grundmann]

On Wed, 4 Sep 2002 16:47:49 +0200
Gert Van den Eynde <address@hidden> wrote:

> expfit returns the root mean square (rms): 
> sqrt(sum((data(i)-fitting_evaluation(i))**2)), you would want to square this, 
> no ?
> 
Yes, thank you. I didn't know that expfit returns the root mean square. 


> expfit will return a term where b = 0 and a is the constant 'c' you're 
> looking for (exp(0*x) = 1).

Sorry, I don't understand what you've written. AFAIK expfit uses y=a*exp(b*x) 
fitting the data and returns three values: a, b, and the rms. Actually I don't 
need the value of 'c', but to get the right 'b' I have to fit it with 
y=a*exp(b*x)+c. That's the problem. 

And there is another problem: the fit is really bad. 
Fitting the same data with expfit and leasqr, I get:

fit-function: y=a*exp(b*x)+1  
expfit: y=9.7*exp(-0.0029*x)+1   R^2=54.8
leasqr: y=9.5*exp(-0.0018*x)+1   R^2=97

I used a konstant c=1, otherwise I can't compare the results.

But expfit seems to be very stable and easy to use, while leasqr is very 
sensitive to the starting values.


Jens Grundmann



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