Parabolic Formal Powerseries

[bo198214]

\[
\text{The half iterate of}\,\,-z + \lambda z^2\,\,\text{has no power series at 0 for all}\,\,\lambda \neq 0\\
\]

Yes, this follows from \(g_1^{{\rm valit}[f]} = 1\), the reasoning is like that:
First we integer iterate to make the first coefficient 1 (because this is much easier to handle than any primary roots).
\begin{align}
a(z)&=-z+\lambda z^2\\
f(z)&=a(a(z)) = z - \lambda z^2 + \lambda (-z + \lambda z^2) = z - 2\lambda z^3 + \lambda^2 z^4\\
{\rm valit}[f] &= 2
\end{align}
If you now have an iterational square root \(h\circ h = a\) then follows  \({h_1}^2 = -1\). But surely also \(h\circ f=f\circ h\) and hence \({h_1}^{2} = 1\) which is a contradiction.

But whom did you quote above?

Which is absolutely provable. And for higher powers of \(z\), it's reducible to this.

It can quite literally just become a question of \(y = z^n\) and there are \(n\) branches--in no world is \(y\) holomorphic at \(0\).

It is not generally true for higher powers, it really has to do with the valit. For example if you have
\begin{align}
a(z)&=-z+z^5\\
f(z)&=a(a(z))=z-2z^5 + O(z^6)\\
{\rm valit}[f]=4
\end{align}

Then you can take 4 4th iterational roots of f with first coefficients 1,i,-1,-1. Two of these roots are square roots of \(a\) corresponding to \(h_1=i,-i\):
\begin{align}
h^{|1}(x) & = x - \frac{1}{2} x^{5} - \frac{5}{8} x^{9} - \frac{25}{16} x^{13} - \frac{605}{128} x^{17} + O(x^{21})\\
h^{|2}(x) & = i x - \frac{1}{2} i x^{5} - \frac{5}{8} i x^{9} - \frac{25}{16} i x^{13} - \frac{605}{128} i x^{17}+ O(x^{21})\\
h^{|3}(x) & = - x + \frac{1}{2} x^{5} + \frac{5}{8} x^{9} + \frac{25}{16} x^{13} + \frac{605}{128} x^{17}+ O(x^{21})\\
h^{|4}(x) & = -i x + \frac{1}{2} i x^{5} + \frac{5}{8} i x^{9} + \frac{25}{16} i x^{13} + \frac{605}{128} i x^{17}+ O(x^{21})\\
\end{align}
(Unfortunately I use the indices already to address the coefficients, so I could not call it \(h_1,h_2,h_3,h_4\), so called it in that strange way)
In this simple case they are just the same function (i.e. the default regular iteration \(f^{\circ 1/4}\)) applied to x,ix,-x,-ix.
But this is different when you have more complicated \(a\).

For even powers we have: if \(a(z)=-z+z^{2n}\), (i.e. \({\rm valit}[a]=2n-1\)) then \({\rm valit}[a\circ a]=4n-2\).
Means we never get \({\rm valit}[a\circ a] = 4n\) so we never have those \(\pm i\) roots.

I love how you reduce everything into taylor coefficients. I reduce everything into integrals or something like that. I love the taylor coefficient approach

I think this is the most philosophical thing about complex function theory I learned, 
that the whole global behaviour of a holomorphic function is just given by the Taylor series at one point (because from there you just continue it anywhere you want).
But really I spend some time with formal powerseries - and built a sage-python library for all those iteration stuff (Though not with those Brent-optimizations, just the naive approach with recurrence formulas)
Up to now I only considered the standard case \(f_1=1\), and always wondered why the unit roots are not considered.
It really costed me some grey hairs to fiddle myself through the unit root case.
But now I am close to finish that case too.