07/11/2022, 08:45 PM
So, this is actually a defining property of the standard Schroder iteration. But it's a little difficult to fully flesh out why. Now, to begin I'll construct an arbitrary iteration which has a constant noodle, and show there are many of them.
If you iterate locally about a fixed point, and your solution satisfies \(f^t(p) = p\), then the iteration is expressible via Shroder iteration (provided that \(|f'(p)| \neq 0,1\)). For convenience, assume that \(|f'(p)| < 1\).
You can actually prove this pretty fast. Assume that \(f^t(x)\) is a super function in \(t\) about a fixed point \(p\), and \(x\) is in the neighborhood of \(p\). Assume that \(f^t(p) = p\). Well then:
\[
\Psi(f^{t+1}(x)) = \lambda \Psi(f^t(x))\\
\]
So that:
\[
\theta(t) = \frac{\Psi(f^t(x))}{\lambda^t}\\
\]
And we know that there must be some 1-periodic function \(\theta(t)\), such that:
\[
f^t(x) = \Psi^{-1}\left(\lambda^{t}\theta(t) \Psi(x)\right)\\
\]
In fact, any periodic function will work fine here, and will have a constant noodle, will be a super function, but will not be Schroder iteration.
Now let's add one more constraint, let's say that \(f^{t}(f^{s}(x)) = f^{t+s}(x)\). Well then, we have that \(\theta\) must be constant. By which, we are guaranteed that it's a Schroder iteration...
So to clarify. Any fractional iteration with a constant noodle is a Schroder iteration. But there are plenty of superfunctions of \(f\) which have a constant noodle, but in turn, they aren't fractional iterations then (don't satisfy the semi-group law).
Now to return to Gottfried's question a bit more robustly. It depends on how we qualify fractional iterations, as to whether \(f^t(p)\) is constant or not. But I'll take the classic approach, and say that \(f^t(x) : \mathbb{R}^+ \to \mathbb{R}^+\) and that it satisfies the semigroup property.
Every element \(q \in \mathbb{Q}^+\) has some value \(m\) such that \(f^{mq}(p) = p\). And therefore:
\[
f^{mq}(x) = \Psi^{-1}(\lambda^{mq} \Psi(x))\\
\]
This is where it gets fun; take an \(|x-p| < \delta\) so that \(x\) is in the basin about \(p\). Well then:
\[
\lim_{m\to\infty} \frac{f^{mq}(x) - p}{\lambda^{mq}} \to \Psi(x)\\
\]
Or rather:
\[
\lim_{t \to \infty} \frac{f^t(x) - p}{\lambda^t} \to \Psi(x)\\
\]
Hmmmmmm.... Well this just means that \(f^t\) is the Schroder iteration... Any non Schroder iteration, this limit will not converge and will continue to oscillate (there'll be a 1-periodic \(\theta\) somewhere in there).
This shows two things, any iteration about a fixed point, which has the semi-group property is Schroder's iteration, and by result has a constant noodle.
BUTTT
Superfunctions, don't have to have a constant noodle--but they don't have the semi-group property then.
Now you may be wondering how Kneser works then... Kneser as a fractional iteration is not holomorphic about any fixed point or periodic point, it has a singularity there. So Kneser sneaks past the above proof by not being a local iteration.
If you iterate locally about a fixed point, and your solution satisfies \(f^t(p) = p\), then the iteration is expressible via Shroder iteration (provided that \(|f'(p)| \neq 0,1\)). For convenience, assume that \(|f'(p)| < 1\).
You can actually prove this pretty fast. Assume that \(f^t(x)\) is a super function in \(t\) about a fixed point \(p\), and \(x\) is in the neighborhood of \(p\). Assume that \(f^t(p) = p\). Well then:
\[
\Psi(f^{t+1}(x)) = \lambda \Psi(f^t(x))\\
\]
So that:
\[
\theta(t) = \frac{\Psi(f^t(x))}{\lambda^t}\\
\]
And we know that there must be some 1-periodic function \(\theta(t)\), such that:
\[
f^t(x) = \Psi^{-1}\left(\lambda^{t}\theta(t) \Psi(x)\right)\\
\]
In fact, any periodic function will work fine here, and will have a constant noodle, will be a super function, but will not be Schroder iteration.
Now let's add one more constraint, let's say that \(f^{t}(f^{s}(x)) = f^{t+s}(x)\). Well then, we have that \(\theta\) must be constant. By which, we are guaranteed that it's a Schroder iteration...
So to clarify. Any fractional iteration with a constant noodle is a Schroder iteration. But there are plenty of superfunctions of \(f\) which have a constant noodle, but in turn, they aren't fractional iterations then (don't satisfy the semi-group law).
Now to return to Gottfried's question a bit more robustly. It depends on how we qualify fractional iterations, as to whether \(f^t(p)\) is constant or not. But I'll take the classic approach, and say that \(f^t(x) : \mathbb{R}^+ \to \mathbb{R}^+\) and that it satisfies the semigroup property.
Every element \(q \in \mathbb{Q}^+\) has some value \(m\) such that \(f^{mq}(p) = p\). And therefore:
\[
f^{mq}(x) = \Psi^{-1}(\lambda^{mq} \Psi(x))\\
\]
This is where it gets fun; take an \(|x-p| < \delta\) so that \(x\) is in the basin about \(p\). Well then:
\[
\lim_{m\to\infty} \frac{f^{mq}(x) - p}{\lambda^{mq}} \to \Psi(x)\\
\]
Or rather:
\[
\lim_{t \to \infty} \frac{f^t(x) - p}{\lambda^t} \to \Psi(x)\\
\]
Hmmmmmm.... Well this just means that \(f^t\) is the Schroder iteration... Any non Schroder iteration, this limit will not converge and will continue to oscillate (there'll be a 1-periodic \(\theta\) somewhere in there).
This shows two things, any iteration about a fixed point, which has the semi-group property is Schroder's iteration, and by result has a constant noodle.
BUTTT
Superfunctions, don't have to have a constant noodle--but they don't have the semi-group property then.
Now you may be wondering how Kneser works then... Kneser as a fractional iteration is not holomorphic about any fixed point or periodic point, it has a singularity there. So Kneser sneaks past the above proof by not being a local iteration.