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NLS fit for exponential distribution

On Jun 12, 2011, at 18:57 , Diviya Smith wrote:

            
If a==0, then a*exp(-m*x) does not depend on m. So don't use a=0 as initial value.
That's not a sum of exponentials. Did you mean a*(exp(-m1*x) + exp(-m2*x)) + c? Anyways, same procedure with more parameters. Just beware the fundamental exchangeability of m1 and m2, so don't initialize them to the same value.