03/24/2022, 12:00 PM
(03/24/2022, 11:13 AM)MphLee Wrote: It's hard to follow for me. It is my fault. I'm not even remotely familiar with perturbation methods and those theta mappings. Some points are really obscure: again my fault. Let's see if you can drop some candies for me.
Quote:I should've clarified, that first of all, this is just intended for \(0 \le \Re(s) \le 2\). This would not make Tetration, or the job of finding inbetween tetration in any meaningful way. This is why I don't even like this solution, But it is doable. It's essentially just run Bennet's commutative hyperoperations, but paste them together in a meaningful way to give a hyper-operator structure.
Ok lets start: how far this is from this
\[\begin{align}x<s>y&=x\odot_s y &&0\le \Re(s)\le 2\\
x<s+1>y+1&=x<s>(x<s+1>y)&&{\rm otherwise}\end{align},\\\]
modulo some perturbation business you use to force the Goodstein equation over that domain?
The following point is particularly obscure. We can say that \(x<s>\omega\) are a family of functions \({\mathbb C}/{\mathcal E}\times \mathcal{W}\to \mathbb C\), as the rank varies, where \(\mathcal{W}\) contains all the fixed points associated to \(\mu\) s.t. \(e^\mu\) is in the ST-region, i.e. if I remember well, when it's infinite tower converges (to the fixed point). You tell me to compute em by \(F(x,s,\mu)\), a function that we know how to compute using a tetration function with base \(b=e^\mu\).
Then what do you mean by
Quote:We can delineate an equivalence class for the Goodstein functional equation so we have a bunch of functions that \(x<s>\omega \pm k\) must equal for \(1 \le \Re(s) \le 2\). Now we play the implicit function game...
Also is \(F(s+1)\) intended to be \(F(x,s+1,\mu\)?
Quote:For brevity's sake's, let's assume we can find where \(F(s+1) \in \mathcal{W}\) (the domain of fixed points). Then \(x <s> F\) is a valid operation--it can now be assigned the value \(x<s+1>\omega +1 = x<s> \omega'\).
The starting point, if I'm following you, is to extend \({\mathbb C}/{\mathcal E}\times \mathcal{W}\to \mathbb C\) outside \(W\), somewhere in \(\mathcal{W}+\mathbb Z\), whenever \(F(x,s+1,\mu)\) still lands in \(\mathcal{W}\). But what if the new fixed point is associated with another ST-base? The holomorphic semi operators are expected to agree across all bases \(b\) only for rank 0 (addition) and rank 1 (multiplication) but to "ramify" for the other ranks. So I don't understand how all the pieces can fit.
I stop here because I don't have time to parse the theta mapping part atm.
Yes! You are right in all of your questions. Everything is exactly as you described. You're just forgetting:
\[
\log^{\circ s + \theta}(x) + \omega = \log^{\circ s}(x) + \theta_2 + \omega
\]
So as we move \(\omega\) we move \(\theta\); we can move \(\mu\) just as well here. And they'll satisfy a similar small change which can respect Goodstein.
By the part of equivalence classes, I meant, we can take all the values \(x<s+1>\omega+1\) could equal based on the fact it equals \(x <s> (x<s+1>\omega)\). Now we are going to try to glue the domains together for \(\mathcal{W}\) and \(\mathcal{W}-1\); so their intersection is holomorphic. Consider \(s \approx 1\), we are just checking that:
\[
\begin{align}
1<s-1>1<s>\omega = 1<s> \omega+1\\
1<0>1<1>\omega = \omega + 1\\
\end{align}
\]
Essentially we check that this function can be holomorphic for \(\omega \in \mathcal{W} \cap \left(\mathcal{W} - 1\right)\) for \(s \approx 1\). Once you can show that you can extend indefinitely in either direction up to a value \(c\) for \(|\Im(\omega)|\le c\).You do that in a neighborhood of \(s \approx 1\), you do it for \(s \approx,0,2\) as well. Then doing it in between would definitely be: hope for the best using your theta mappings

I'm gonna stick for the moment with \(1<s> \omega\) for \(s\approx 1\). I think I could explain this better. Without going off the deepend which you'd have to do for \(x <s> y\); Lmao!

