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correction to leasqr algo. : in dfdp.m

From: courtois
Subject: correction to leasqr algo. : in dfdp.m
Date: Fri, 12 Jan 2001 16:52:38 +0100 (CET)


I found that using leasqr.m there is a problem when matrix x has more
columns than rows. You'll find an easy solution below. I replace length
by size(x,1) which return the number of rows and not the max (row,

BTW I found that those routines are both in matcompat and optim which is
potentially a problem.

Anyway it is great ! I use it often and love it.

Thanx to Paul and other contributors.

function prt=dfdp(x,f,p,dp,func)
% numerical partial derivatives (Jacobian) df/dp for use with leasqr
% --------INPUT VARIABLES---------
% x=vec or matrix of indep var(used as arg to func) x=[x0 x1 ....]
% f=func(x,p) vector initialsed by user before each call to dfdp
% p= vec of current parameter values
% dp= fractional increment of p for numerical derivatives
%      dp(j)>0 central differences calculated
%      dp(j)<0 one sided differences calculated
%      dp(j)=0 sets corresponding partials to zero; i.e. holds p(j) fixed
% func=string naming the function (.m) file
%      e.g. to calc Jacobian for function expsum prt=dfdp(x,f,p,dp,'expsum')
%----------OUTPUT VARIABLES-------
% prt= Jacobian Matrix prt(i,j)=df(i)/dp(j)
n=length(p);      %dimensions
m=size(x,1); % Used to be : m=length(x);
ps=p; prt=zeros(m,n);del=zeros(n,1);       % initialise Jacobian to Zero
for j=1:n
      del(j)=dp(j) .*p(j);    %cal delx=fract(dp)*param value(p)
           if p(j)==0
           del(j)=dp(j);     %if param=0 delx=fraction
      p(j)=ps(j) + del(j);
      if del(j)~=0, f1=feval(func,x,p);
           if dp(j) < 0, prt(:,j)=(f1-f)./del(j);
           p(j)=ps(j)- del(j);
           prt(:,j)=(f1-feval(func,x,p))./(2 .*del(j));
      p(j)=ps(j);     %restore p(j)



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