(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:
\( \alpha_u(v)=\lim_{n\to\infty}\frac{f^{[n]}(v) -
f^{[n]}(u)}{f^{[n+1]}(u)-f^{[n]}(u)}
\)
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
\( \alpha(z)=\lim_{n\to\infty} \tilde{\alpha}(f^{[n]}(z))-n \)
where \( \tilde{\alpha} \) is a sum of some negative powers (none in the hyperbolic case) and a logarithm for example for e^x-1 we get:
\( \tilde{\alpha}(x)={-2x^{-1}+\frac{1}{3}\ln(x) \).
Another formula (2.29 in the overview paper) that kinda combines hyperbolic and parabolic is:
\( \lim_{n\to\infty}
\frac{f^{[n]}(v)-f^{[n]}(z)}{f^{[n+1]}(z)-f^{[n]}(z)}=w\frac{1-\lambda^{w}}{1-\lambda}
\), \( f^{[w]}(z)=v \), \( \alpha_z(v)=w \)
where \( \lambda \) 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.