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## Re: regarding leasqr

**From**: |
Martijn |

**Subject**: |
Re: regarding leasqr |

**Date**: |
Mon, 04 Jul 2011 15:43:29 +0200 |

Hi Preeti,
I did not look into your formula's but I have the impression that a
number of parameters are 100% correlated and therefore impossible to
determine. A simple example: fitting f(x,p) = p(1)*exp(p(2)*x + p(3)) to
a set of data will fail because p(1) and p(3) are correlated: increasing
p(3) with 1 has the same effect as multiplying p(1) with e. It is
therefore impossible to determine them from a fit.
Your problem might be similar.
Good luck,
Martijn
On Mon, 2011-07-04 at 14:48 +0200, preeti gaikwad wrote:
>* *
>* *
>* Hello all, *
>* *
>* I am the new user in fitting please help me to solve my*
>* query. I am using leasqr to estimate parameters from the experimental*
>* data. *
>* *
>* my fuction is *
>* *
>* *
>* s=A.*exp(-x./tabs).*(D./D0).^2.*exp(-(pi*pi).*(D.*x)./L.^2);*
>* *
>* where D is *
>* *
>* D=(D0.*tloc.^(a))./(tloc.^(m).+x.^(m)).^(a./m);*
>* *
>* I would like to calculate *
>* *
>* *
>* *
>* A = p(6);% constant value *
>* *
>* D0=p(2); *
>* *
>* tloc=p(3);*
>* *
>* a=p(4);*
>* *
>* m=p(5);*
>* *
>* Using the leasqr *
>* *
>* *
>* [f, p] = leasqr(t, y, pin,'conv_fnc', tolerance, max_iterations,*
>* weights, dp, dFdp, options);*
>* *
>* unfortunately i am getting value whatever is my pin *
>* *
>* *
>* like my pin is *
>* *
>* pin = [0.5 25 0.000000005 1 7 1000 0]*
>* *
>* *
>* and for all value of x I have straight line like but my experimental*
>* data is exponential decreasing function. *
>* *
>* *
>* *
>* and the defined fuction is s also exponential decay function. My quest*
>* is why leasqr is not giving the new calculated value from the fun*
>* s???? THANKS FOR YOUR HELP in advance. *
>* *
>* *
>* *
>* regards P*
>* _______________________________________________*
>* Help-octave mailing list*
>* address@hidden*
>* https://mailman.cae.wisc.edu/listinfo/help-octave*