(03/19/2015, 10:22 AM)sheldonison Wrote:(03/18/2015, 09:49 PM)tommy1729 Wrote: Since jan 2015 I am very intrested in parabolic fixpoints again.
I take the fixpoint at 0 for simplicity.
Maybe I havent read the right books, but I feel a large gap in understanding of the subject.
Here is the asymptotic series for the Abel function for the parabolic case; for example for \( f(z)=\exp(z)-1 \)
We have \( \alpha(f(x) ) = \alpha(x)+1 \)
\( \alpha(x) \sim \frac{-2}{z} +
\frac{\ln(z)}{3}
- \frac{z}{36}
+ \frac{z^2}{540}
+ \frac{z^3}{7776}
- \frac{71\cdot z^4}{43556}
+ \frac{8759\cdot z^5}{163296000}
+ \frac{31 \cdot z^6}{20995200}
- \frac{183311 \cdot z^7}{16460236800} + ... \)
My understanding is that the series is asymptotic because there are actually two different superfunctions and Abel functions, depending on whether you are working with the repelling sector, for z>0, or the attracting sector, for z<0. From a practical point of view, for \( \Re(z)<0 \) one can iterate \( z \mapsto \exp(z)-1 \) a few times or for \( \Re(z)>0 \) one can iterate \( z \mapsto \ln(z+1) \) a few times, before calculating \( \alpha(z) \), so that z is closer to zero. Then the series convergence is remarkably good, allowing results of arbitrarily high precision without any difficulties. Also, for the Abel function we can add any arbitrary additive constant to \( \alpha(z)+k \). For \( \Re(z)<0 \) it is customary to use \( k=\frac{-\pi i}{3} \), or equivalently to replace the \( \frac{\ln(z)}{3} \) term with \( \frac{\ln(-z)}{3} \). Again, this is because there are really two completely different superfunctions for f(z), with different Abel functions, that have the same series expansions.
What you have is not a Taylor series, so how did you find this expansion.
Also this does not answer the op.
I know ways to get the Abel , but i forgot why getting to z=0 helps.
Probably perturbation theory.
Btw for x + x^N the situation is different for Every N.
Lots to do.
Regards
Tommy1729
