(08/08/2022, 07:56 PM)bo198214 Wrote: One can verify that also by direct calculation.
Assume we have these fixed points \(z_0\) and \(z_\infty\) where the Kœnigs/Schröder function f is 0 and \(\infty\) (has a pole) respectively.
Let \(\ell_c(z)=cz\) and \(\nu(z)=1/z\), then we take the derivative:
\begin{align}
(f^{-1}\circ \ell_c \circ f)' &= \frac{1}{f'\circ f^{-1}\circ \ell_c\circ f}\cdot (\ell_c'\circ f)\cdot f'\\
(f^{-1}\circ \ell_c \circ f)'(z_0) &= \frac{1}{f'(z_0)} c f'(z_0) = c
\end{align}
However for \(z_\infty\) we can not use this derivation because \(f'(z_0)=\infty\)? We need to find a safer way. But we found already the trick to take the inverse of \(f\) which is analytic at \(z_\infty\). So let \(f=\nu\circ g\) then
\[f^{-1}\circ \ell_c \circ f = g^{-1} \circ \nu \circ \ell_c \circ \nu\circ g = g^{-1}\circ \ell_{1/c} \circ g\]
and as above:
\[(f^{-1}\circ \ell_c \circ f )'(z_\infty) = (g^{-1}\circ \ell_{1/c} \circ g)'(z_\infty) = \frac{1}{c} \]
This is amazing in the sense that as long as the Kœnigs/Schröder function has an analytic 0 and a pole then the corresponding t-iterations have \(c^t\) and \(c^{-t}\) as fixed point derivations. So one could assume that this is the normal situation. However we consider it rather the exception, why?
Because the regular Kœnigs/Schröder function at one fixed point is often not analytic/a pole at the other fixed point.
Though we could have \(\lim_{z\to z_\infty} f^{-1}(c^t f(z)) = z_\infty \) for all t, the limit \(\lim_{z\to z_\infty} (f^{-1}\circ \ell_{c^t} \circ f)'(z) \) would be some other number for discrete t for those functions.
Quite mysterious I would say.
On the other hand if the fixed point derivations are not reciprocal, like with \(\exp_{\sqrt{2}}'(2) = \log(2)\) and \(\exp_{\sqrt{2}}'(4) = 2\log(2) \) then we can surely say that the regular iteration at one fixed point can not be the regular iteration at the other for all t (or just a continuous interval) because as Tommy pointed out: the superfunctions have different periods then.
This was largely my question. Can we find a Euclidean mapping (entire function) such that \(f(p_0) = p_0\) and \(f(p_1) = p_1\), such that \(f'(p_0) = 1/f'(p_1)\). And that additionally the orbits about a domain \(H\) of \(f\) tend to \(p_0\), and the orbits about a domain \(H\) of \(f^{-1}\) tend to \(p_1\).
I think this is possible, the trouble is, the fractional iterate will not be local about both fixed points. We can't fix a domain \(H\) which contains both fixed points. Even if \(H\) isn't simply connected, there should be a way to find a subdomain that is, that contains both fixed points--contradiction. This again would be my \(B(z)\) argument.
You're invariably getting \(p_1\) (\(p_0\)) to be on the boundary of the basin about \(p_0\) (\(p_1\)). The only way you have holomorphy at both is... if it's the Riemann sphere. Especially because we are making the coordinate change from \( f(z) \mapsto \lambda z\). And \(\lambda z\) technically has \(\infty\) (or \(0\)) at the boundary of the basin. But no holomorphic function on its basin is all of \(\mathbb{C}\), UNLESS it is \(\lambda z\).
This is deeply fascinating. We are definitely tapping the deep core of iteration right now, lmao.
EDIT: I've been devouring the Karlin Mcgregor paper, and I don't think we even need the extraneous paper you asked me to find. Karlin and Mcgregor prove a much more general result hidden as a result about real analysis. It specifically uses the term regular, and this should imply a domain about \([0,1]\) in the complex plane, but it includes \([0,1]\). I'll try to make a better write up of what I mean. But this is a much much better explanation of what I was trying to explain. MUCH MUCH better, lol.
