I have a minimisation problem at hand using lsqnonlin. According to the output, there are 8 iterations required to get to the final answer. Is it possible to find the values the lsqnonlin function calculates during those 8 iterations. The way I did it now is by including the line
(where x0 is the calculated values). This however gives me much more than 8 times the intermediate x0 values.
Also, if anyone has an idea to speed up the code, that would be very helpfull too.
The full function looks like this:
F = permute(squeeze(sum(IVW.*x0,2))-EVW(:,1),[2,1]) ;