# Exponential least-squares fitting and initial parameters

1. Dec 13, 2013

### rkaminski

Dear All,

I would like to do an exponential function least-squares fitting, but having two or more exponents. For example the function looks like this:

$y (x) = A \exp (-x/a) + B \exp (-x/b)$

where $A$, $a$, $B$ and $b$ are the least-squares fitted parameters. My question is how to obtain the initial parameters? I could assume that that my initial function follows the one-exponential function and then get initial estimates of $a$ and $A$, but then how to get initial values of both $a$ and $b$, and other parameters? I cannot assume for example $b = 0$ obviously. Thanks in advance.

Best wishes,

2. Dec 14, 2013

### Staff: Mentor

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. Dec 14, 2013

### JJacquelin

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 non-linear regression is transformed to a linear regression thanks to some convenient preliminary numerical intégrations.
See pages 73-75 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 non-linear regression.

4. Dec 14, 2013

### rbj

like least-squares. (i didn't even think that the least-squares method had iterations or initial conditions.)

5. Dec 14, 2013

### Stephen Tashi

In general least squares fitting amounts to trying to find values of variables that minimize a (non-linear) function. There may be different combinations of values of the variables that do equally well. Are you applying some minimization method that requires "initial" values to get it started?

6. Dec 14, 2013

### JJacquelin

Of course, the least-squares method doesn't require iterations or initial conditions.
But the least-squares 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 least-squares method only works for A and B.
So you cannot use the least-squares method to compute a, b, A and B.
You have to search a more sophisticated method.

7. Dec 14, 2013

### rkaminski

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. Non-linear least-squares 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 least-squares then there is no problem like this obviously.

8. Dec 14, 2013

### JJacquelin

Thay is exactly what I intended to say. Sorry if my writting was not clear enough when I was talking of linear least-squares fitting.

9. Dec 14, 2013

### Staff: Mentor

You can, you just don't get analytic solutions. That's why initial values and a fitting procedure are used.

10. Dec 14, 2013

### JJacquelin

OK. You are right. That is what I call "sophisticated methods".
The method I propose avoids initial values and itterative procedure, because the preliminary numerical integrations leads to a linear mean squares fitting.

11. Dec 15, 2013

### AlephZero

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/research_reports/CSRSR2009-04.pdf

"General purpose" optimization methods tend not to work well, because the slowest decaying exponential tends to dominate the others.