I went to the examples of leasqr and copied example #2, then adjusted it for my data.
% Define functions
leasqrfunc = @(x, p) p(1) * exp (-p(2) * x ) +p(3);
leasqrdfdp = @(x, f, p, dp, func) [exp(-p(2)*x), -p(1)*x.*exp(-p(2)*x),0*x+1];
what I did was added +p(3) to the original function. I then calculated the partial diff with respect to p(3). this is " 1 ".
when I put a 1 in the 3rd position in leasqrdfdp, then the program does not work. but when I make the 3rd term to by 0*x + 1 then the program runs and gives the correct answer.
should just +1 work???
here is the program that works
% Define functions
leasqrfunc = @(x, p) p(1) * exp (-p(2) * x ) +p(3);
leasqrdfdp = @(x, f, p, dp, func) [exp(-p(2)*x), -p(1)*x.*exp(-p(2)*x),0*x+1];
% generate test data
t = [0 2 4 6 13]';
#p = [1; 0.1];
#data = "" (t, p);
data="" 1457 1426 1408 1387]';
#rnd = [0.352509; -0.040607; -1.867061; -1.561283; 1.473191; ...
# 0.580767; 0.841805; 1.632203; -0.179254; 0.345208];
% add noise
% wt1 = 1 /sqrt of variances of data
% 1 / wt1 = sqrt of var = standard deviation
wt1 = (1 + 0.01 * t) ./ sqrt (data);
#data = "" + 0.05 * rnd ./ wt1;
#wt1=[1 1 1 1 1]' *1;
% Note by Thomas Walter :
%
% Using a step size of 1 to calculate the derivative is WRONG !!!!
% See numerical mathbooks why.
% A derivative calculated from central differences need: s
% step = 0.001...1.0e-8
% And onesided derivative needs:
% step = 1.0e-5...1.0e-8 and may be still wrong
F = leasqrfunc;
dFdp = leasqrdfdp; % exact derivative
% dFdp = dfdp; % estimated derivative
dp = [01; 0.1;1];
pin = [160; 1.0; 1150];
stol=0.000001; niter=50;
minstep = [0.01; 0.00001;.1];
maxstep = [010; 0.2; 1000];
options = [minstep, maxstep];
global verbose;
verbose = 1;
[f1, p1, kvg1, iter1, corp1, covp1, covr1, stdresid1, Z1, r21] = ...
leasqr (t, data, pin, F, stol, niter, wt1, [], dFdp, options);
f1
p1
xx=0:.1:30;
yy=F(xx,p1);
hold on
plot(xx,yy)
hold off