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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.

Good luck,

Martijn

On Mon, 2011-07-04 at 14:48 +0200, preeti gaikwad wrote:
>
>
> Hello all,
>
> 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
>
>
>
> regards P
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