(09/16/2021, 10:54 PM)tommy1729 Wrote: Just commenting on this small thing.
You bring it up often.
And I have mixed feeling about it :^)
The thing is , it is correct and relevant to many tetration related ideas. It has its upsides and downsides.
However notice for MOST \( z \) , the orbits \( \exp^{\circ R}(z) \) USUALLY DO NOT get arbitrarily close to the orbit \( \exp^{\circ R}(0) \) where \( R \) is a positive real that is not an integer.
tommy1729
I know it looks that way tommy, that's not the truth. Which is what I'm saying.
The points where \( \exp^{\circ n}(z) \) do not aggregate towards \( \exp^{\circ n}(0) \) are of zero measure in the lebesgue measure. This means they are isolated points, and are precisely the cyclic points of exp which are not dense, but are sparse. Despite that they look numerous is nothing to us seasoned mathematicians--they are sparse. So if you pick a point \( z \) at random in \( \mathbb{C} \); then almost surely does its orbit aggregate towards \( \exp^{\circ n}(0) \). This means, that pretty much all the points do aggregate.
Now I only use this to say that,
\(
\beta_\lambda(s) = q\,\,\text{for}\,\,q\,\,\text{a cyclic point}\\
\text{then}\\
\beta_\lambda(s+1) = \exp(q)/(1+e^{-\lambda s})\,\,\text{is not a cyclic point}\\
\)
This allows us, to reiterate again, that the beta function \( \beta_\lambda(s) \to \infty\,\,\text{as}\,\,\Re(s) \to \infty \); because there are no cyclic points of beta due to it's functional equation. This is a central fact. This does not mean that tetration tends to infinity necessarily; but almost everywhere it will.
The trouble is there are A LOT of cyclic points (they're more like neutral points as \( q+2\pi i \) produces a different orbit than \( q \) but they still bounce around and not hit infinity). Like a lot a lot of them. But if you deleted them, you'd still have the complex plane almost everywhere.
It's important to remember if this weren't the case, then there would be no holomorphic slog. As slog has a singularity at each cyclic point. If what you were saying is true, that these points are dense in \( \mathbb{C} \) or "most of the points", then slog would be a holomorphic function with a dense amount of singularities. No such holomorphic function can exist then.