No, no, no James, you can not wind that easily out of the situation with statements like "can you even call it regular iteration" ...
Regular iteration is not something designed for entire function or so (OMG its meromorphic - I die).
It is a general method for a function analytic at a fixed point (i.e. in a (possibly small) vicinity of the fixed point) and this is what I provided.
The given constructions are as proper regular iterations as it can get.
\[\frac{2^t z}{1+(2^t-1)z} = 2^t z -(2^t - 1)2^t z^2 + (2^t - 1)^22^t z^3 -(2^t - 1)^32^t z^4 + \dots\]
\[\frac{z}{1+tz}=z -tz^2 + t^2z^3 -t^3z^4 + \dots\]
Remember regular iteration was always a local property. We take the powerseries expansion at the fixed point, then there can only be one formal powerseries parametrized by t hat satisfies \(f^{\circ s}\circ f^{\circ t} = f^{\circ s+t}\). Then we prove that this powerseries has a radius of convergence for all t (hyperbolic case), which it has i.e. my hyperbolic series converges for \(|z|<\left|\frac{1}{2^t-1}\right|\).
This is the standard setup of regular iteration I remember a theorem of Écalle that states - in the parabolic case - that this formal iteration powerseries has a positive radius of convergence for all t if and only if the iterative logarithm series has a positive radius of convergence.
The iterative logarithm or Julia function \(j\) is a function that satisfies \(j\circ f = f' \cdot j\) and it has properties similar to the normal logarithm, i.e. \(\text{logit}[f^{\circ t}] = t\cdot\text{logit}[f]\) (and one can construct the Abel function via \(\int \frac{1}{j} \)).
We can verify Écalles theorem with our parabolic example:
\[\text{logit}\left[\frac{z}{1+z}\right] = - z^2\] (do the math yourself)
And voilá this "series" has infinite radius of convergence.
This brings me to an idea of reverse engineering, just looking at simple expression for the logit and then obtaining a function with parabolic fixed point that also has positive radius of convergence for all t.
But alas I am anyways already above my time budget for tetration - not good for my health *sigh*
But honestly James this is a lot of explaining why a counter example does not work. Better would be that these cases come up while developing the proof, because I can not see the role of this demand that we have a fixed \(D\times N\) in a proof, on the other hand I see your point that this would be a nice property to have.
Regular iteration is not something designed for entire function or so (OMG its meromorphic - I die).
It is a general method for a function analytic at a fixed point (i.e. in a (possibly small) vicinity of the fixed point) and this is what I provided.
The given constructions are as proper regular iterations as it can get.
(08/03/2022, 02:52 AM)JmsNxn Wrote: It's also important to note that this still isn't a counter example to the proposition. Because this iteration does not have the local expansion:Only because there are poles in t does not mean that it lacks the given expansion. Put it in your favourite computer algebra tool and look at the coefficients, these are:
\[
\begin{align}
f^{\circ t}(z) &= \lambda^t z + O(z^2)\,\,\text{for}\,\,|z|<\delta\,\,\Re(t) > 0\\
f^{\circ t}(z) &= 1 + \mu^t (z-1) + O(z^2)\,\,\text{for}\,\,|z-1| < \delta\,\,\Re(t) > 0\\
\end{align}
\]
Because there are poles in \(f^{\circ t}\) at precisely:
\[
\begin{align}
(2^t-1)z &= -1\\
2^t - 1 &= -1/z\\
t &= \log(1-1/z)/\log(2) + k\pi i/\log(2)\,\,\text{for}\,\,k\in\mathbb{Z}\\
\end{align}
\]
\[\frac{2^t z}{1+(2^t-1)z} = 2^t z -(2^t - 1)2^t z^2 + (2^t - 1)^22^t z^3 -(2^t - 1)^32^t z^4 + \dots\]
\[\frac{z}{1+tz}=z -tz^2 + t^2z^3 -t^3z^4 + \dots\]
Remember regular iteration was always a local property. We take the powerseries expansion at the fixed point, then there can only be one formal powerseries parametrized by t hat satisfies \(f^{\circ s}\circ f^{\circ t} = f^{\circ s+t}\). Then we prove that this powerseries has a radius of convergence for all t (hyperbolic case), which it has i.e. my hyperbolic series converges for \(|z|<\left|\frac{1}{2^t-1}\right|\).
This is the standard setup of regular iteration I remember a theorem of Écalle that states - in the parabolic case - that this formal iteration powerseries has a positive radius of convergence for all t if and only if the iterative logarithm series has a positive radius of convergence.
The iterative logarithm or Julia function \(j\) is a function that satisfies \(j\circ f = f' \cdot j\) and it has properties similar to the normal logarithm, i.e. \(\text{logit}[f^{\circ t}] = t\cdot\text{logit}[f]\) (and one can construct the Abel function via \(\int \frac{1}{j} \)).
We can verify Écalles theorem with our parabolic example:
\[\text{logit}\left[\frac{z}{1+z}\right] = - z^2\] (do the math yourself)
And voilá this "series" has infinite radius of convergence.
This brings me to an idea of reverse engineering, just looking at simple expression for the logit and then obtaining a function with parabolic fixed point that also has positive radius of convergence for all t.
But alas I am anyways already above my time budget for tetration - not good for my health *sigh*
Quote:To me, iteration about a fixed point, means that we have a fixed neighborhood \(N\) and a fixed domain \(D\) in \(\mathbb{C}\) such that this object is holomorphic; and D is closed under addition of elements; \(\{D,+\}\) is a semigroup. Notably, this parabolic iteration is \(\mathbb{R}\) analytic in \(t\), but isn't holomorphic for any strip \(|\Im(t)| < \delta\) and \(|z| < \delta\). You have to shrink \(\delta\) as \(t\) grows, and thereby you are making a more dynamic kind of iteration.
...
But honestly James this is a lot of explaining why a counter example does not work. Better would be that these cases come up while developing the proof, because I can not see the role of this demand that we have a fixed \(D\times N\) in a proof, on the other hand I see your point that this would be a nice property to have.
