On Fri, Jun 26, 2015 at 8:14 AM, mimatopo <address@hidden> wrote:
Hi,
I tested different input and it really works very nice, thanks a lot for
your help! But there is on more thing:
For further data processing I need the 'lambda' as described in my initial
posting. For the posted example data, this value in matlab is -0.1686 which
I did not manage to reproduce so far...
Do you have any idea how to try this?
Best regards, Tom
Yes try 1/p(2)
or try
F=@(x, p) p(1)*exp(x*p(2));