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Exponential leastsquares fitting and initial parameters 
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#1
Dec1313, 09:38 PM

P: 3

Dear All,
I would like to do an exponential function leastsquares fitting, but having two or more exponents. For example the function looks like this: [itex]y (x) = A \exp (x/a) + B \exp (x/b) [/itex] where [itex]A[/itex], [itex]a[/itex], [itex]B[/itex] and [itex]b[/itex] are the leastsquares fitted parameters. My question is how to obtain the initial parameters? I could assume that that my initial function follows the oneexponential function and then get initial estimates of [itex]a[/itex] and [itex]A[/itex], but then how to get initial values of both [itex]a[/itex] and [itex]b[/itex], and other parameters? I cannot assume for example [itex]b = 0[/itex] obviously. Thanks in advance. Best wishes, Radek 


#2
Dec1413, 10:48 AM

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P: 11,576

Where does your data come from? In particular, can you look at the data to choose initial parameters?
If yes: Draw it on a logarithmic scale, see if you can identify the two exponentials there, estimate the parameters. If no, or the method above does not help: Use arbitrary starting values (with different values for a and b), hope that the fit converges  it should do that if the model is not completely wrong. 


#3
Dec1413, 12:38 PM

P: 756

Hi !
I think that the best way is to use a method which do not requires guessed initial values of the parameters. There is a straightforward method for doing that : the nonlinear regression is transformed to a linear regression thanks to some convenient preliminary numerical intégrations. See pages 7375 of the paper "Régressions et équations intégrales" published on Scribd : http://www.scribd.com/JJacquelin/documents The theory is written in French, but you don't need it. The practical application for fitting your function is written in English (The notations are not the same of yours. It is not difficult to change the symbols). With this method, the approximates of the parameters are generaly sufficient for the direct practical use. But, if you need a specific fitting according to some particular criteria, it's up to you to use the good values already obtained as initial values for an iterative process using a software for nonlinear regression. 


#4
Dec1413, 12:58 PM

P: 2,251

Exponential leastsquares fitting and initial parameters



#5
Dec1413, 01:23 PM

Sci Advisor
P: 3,248




#6
Dec1413, 02:07 PM

P: 756

But the leastsquares method doesn't work in the case of the of the equation : y = A*exp(x/a)+B*exp(x/b) because the parameters a and b are in the argument of the exponential function. The leastsquares method only works for A and B. So you cannot use the leastsquares method to compute a, b, A and B. You have to search a more sophisticated method. 


#7
Dec1413, 02:12 PM

P: 3

Hi JJacquelin,
I will have a look at the French papers you send. These are very interesting. No problem for me to understand them:) However, I don't understand your last post. Nonlinear leastsquares method does require initial values of the parameters. That is why the procedure needs to be cycled and in each case we obtain better estimates of the refined parameters. If it were linear leastsquares then there is no problem like this obviously. Radek 


#8
Dec1413, 02:27 PM

P: 756




#9
Dec1413, 04:05 PM

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P: 11,576




#10
Dec1413, 05:58 PM

P: 756

The method I propose avoids initial values and itterative procedure, because the preliminary numerical integrations leads to a linear mean squares fitting. 


#11
Dec1513, 03:36 PM

Engineering
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P: 6,931

This is an important issue in some fields of engineering, and specialized methods have been developed for it. See here for a summary and bibliography:
http://www.csrc.sdsu.edu/csrc/resear...RSR200904.pdf "General purpose" optimization methods tend not to work well, because the slowest decaying exponential tends to dominate the others. 


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