Parabolic Formal Powerseries

[tommy1729]

this is completely logical , almost trivial :

near fixpoint 0 , for small x and n > 2 ,  we have

f(x) = a x + b x^n + O(x^(n+1))

Notice the other fixed points are always a nonzero distance away from 0.

so if say a is 2 we are nearly solving

y = 2 x.

and hence the hyperbolic method works by the approximation y = 2x near 0.
AND we then get a unique solution.

...

Dear Tommy,
I think you heavily confusing stuff here.
We are not talking about taking inverses,
we talk about fractional iterates.
Let me give you a quick overview of the problem:
The regular iteration in the hyperbolic case has the form
\begin{align}
f(x) &= bx + O(x^2)\\
f^{\circ t}(x) &= b^tx +O(x^2)
\end{align}
So for t=1/q we have q fractional roots (with multiplicity) corresponding to \(b^{\frac{1}{q}}\).
In the case b=1 however the regular iteration look like this:
\begin{align}
f(x) &= x + cx^n + O(x^{n+1})\\
f^{\circ t}(x) &= x + t c x^n + O(x^{n+1})
\end{align}
So one would have only one q-th iterative root if 1 is the first coefficient.


So, then Leo came up with the example of \(f(x)=-x+x^2\) and could not find a
second iterative root for that and showed it can not exist.
Which then lead to the general question for which \(b=e^{2\pi i k/q}\) there would exist a
Q-th iterative root of 
\[f(x)=e^{\frac{2\pi i p}{q}}x + c x^n + O(x^{n+1})\]
and the answer is iff:

\[ {\rm valit}[f^{\circ q}] = Qqm \]

for some integer \(m\ge 1\). Then there would exist Q iterative Qth roots of f of the form 
\[ f^{\circ \frac{1}{Q}|k}(x)=e^{\frac{\frac{2\pi i p}{q}+2\pi i k}{Q}}x + O(x^2) , \quad 0\le k < Q\]

I am aware we are not talking About inverses.

But they matter.

Because when values are repeated in a neighbourhood ( we get multiple “ flows “ ) of a fix , we get multiple superfunctions adressing all those Locally multiple univalent neighbourhoods.
And thus we get also multiple fractional iterates. 

Just like exp(x) - 1 has 2 solutions that fundamentally disagree ( no 1 periodic function unites them ) 

Regards 

tommy1729