(07/09/2010, 12:27 PM)tommy1729 Wrote: why is the radius zero again ?

cant find your paper about parabolic iteration ...

Actually I dont know exactly the proof, but in most cases the convergence radius is 0 for parabolic iteration, particularly for the iteration of e^x-1 which is in turn equivalent to iteration of e^(x/e).

For parabolic Abel function there is a very old formula by Lévy (2.20 in the overview paper), which is:

which however is not usable for numeric calculation as it is too slow.

A formula given by Ecalle (2.22 in the overview paper) is much more usable and works for both cases hyperbolic and parabolic. It is

where is a sum of some negative powers (none in the hyperbolic case) and a logarithm for example for e^x-1 we get:

.

Another formula (2.29 in the overview paper) that kinda combines hyperbolic and parabolic is:

, ,

where is the derivative at the fixed point 0, which is 1 in the parabolic case and you take the limit of lambda->1. I am in a hurry a bit. So perhaps more detailed later.