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Re: weighted residuals are not real


From: Ben Abbott
Subject: Re: weighted residuals are not real
Date: Fri, 18 Nov 2011 08:46:30 -0500

On Nov 18, 2011, at 8:27 AM, preeti gaikwad wrote:

> On 18 November 2011 14:21, Ben Abbott <address@hidden> wrote:
> 
>> On Nov 18, 2011, at 7:10 AM, preeti gaikwad wrote:
>> >
>> > On 18 November 2011 12:55, Ben Abbott <address@hidden> wrote:
>> >
>> >> On Nov 18, 2011, at 6:30 AM, preeti gaikwad <address@hidden> wrote:
>> >>
>> >>> Hello all,
>> >>>
>> >>>     I am thee new user of octave could u please help me?
>> >>> I am getting this error using "
>> >>> weighted residuals are not real
>> >>> error: called from:
>> >>> error:   /usr/share/octave/packages/3.2/optim-1.0.12/leasqr.m at line 
>> >>> 394, colu
>> >>> mn 4"
>> >>>
>> >>> do not understand why using leasqr to fit my data.....when I hv 20 
>> >>> points i dont hv this problem but still it is not fitting well but if i 
>> >>> reduce some points i m getting this error....could u please tell me why 
>> >>> it is so?
>> >>
>> >> Can you show us what you did to cause this error?
>> >>
>> >> Perhaps a simple example we can try ourselves?
>> >>
>> >> Ben
>> >
>> > Function body:::::::::::::::::::::
>> >
>> >
>> > function F = trans_fnc(x,p) %
>> >
>> > if nargin~=2
>> >    error('This function is waiting for 2 arguments')
>> > end
>> >
>> > ls = p(1); % in um
>> > la = p(2);% in um
>> > offset = 0.0;
>> > c=0.3; %speed of light in um/fs
>> > n1 = 1.05;                       % average refractive index
>> > n2 = 1;
>> > R =((n1-n2)/(n1+n2))^2;
>> > z0 = (2/3)*ls*(1+R)/(1-R);
>> > ze = (la/2)*log((1+(1/la)*z0)/(1-(1/la)*z0));
>> > zp = ze;
>> > %% model function:
>> > F =1.0./( 
>> > (la/ze)*sinh((zp+ze)/la)*sinh(ze/la)./sinh(((x+offset)+2*ze)/la)); % 
>> > inverse transmittance
>> >  This is my function body and then I have to find the value of ls and la I 
>> > have x value and y but for particular x value it is fitting but for some x 
>> > its not fitting.....
>> >
>> > where using leasqr as follows
>> > for i=1:nr  % i is the LAMBDA VARIABLE
>> >       y=1./T(i,:);
>> >       %% start leasqr, be sure that 'verbose' is not set
>> >       global verbose; verbose = false;
>> >       [f, p] = leasqr (x, y, pin,'trans_fnc', tolerance, max_iterations, 
>> > weights, dp, dFdp, options);
>> >       ls(i) = p(1);
>> >       la(i) = p(2);
>> >       fit(i,:) = f;
>> >
>> > end
>> >
>> > importantly i do not understand how to fixed the weighted residual in my 
>> > case it is
>> > weights = ones (1, nc);
>> >
>> >  where nc is column value for y
>> > thanks for ur help in advance
>> 
>> Please reply at the bottom of the email so that those arrive later can 
>> follow along. Also, these emails are archived in several places on the web. 
>> Using plain text ensures they'll be reproduced accurately. So, please don't 
>> post in html or rtf.
>> 
>> Regarding you problem can you provide all the inputs to leasqr?
>> 
>> Ben
> 
> thanks ben....yes i m giving the guess value
> 
> pin = [0.1 3];% ls;la
> options.bounds=[0.0001 10;0.001 100];
> and this guess was almost near to the calculated value but dont know why its 
> not fitting 

Ok, what what of the other inputs?

        tolerance
        max_iterations
        dp
        dFdp

Ben


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