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## Re: exponential fitting

**From**: |
grundmann |

**Subject**: |
Re: exponential fitting |

**Date**: |
Thu, 5 Sep 2002 16:29:21 +0200 |

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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