Holomorphic semi operators, using the beta method

[JmsNxn]

Hi, had literally zero time in the last three months to write or study, but I continue to read all the new exciting posts.
Said that, I think I have a hole in my memory but I remember something of your 2010/2011 work on semi operators, i.e. non-integer hos. One of those,  the cheta one, was originally a piecewise fix of Bennet's hyperoperations (aka distributive Hyperoperations) but extended to the real.
Using the superfunction of exp via beta method to build a continuous spectrum of homomorphic abelian groups extending Bennet's sequence seems really interesting.

But here you are doing something different.Since my mind is a bit foggy lately I can't follow it with ease. How exactly can you obtain the Goodstein equation while holding true for \(\omega\) an exp-fix-point the following equation \[x\,<s>\,\omega=\exp^s(\log^s(x)+\omega)?\quad\quad (*)\]

I know you can't write the details right now but even abstractly I don't see the intended direction of the implication.

Let \(\odot_s\) be the standard Bennet hyperoperations for a fixed base \(b\) and extended to the real ranks \(s\) using a tetration function. If we have a sequence of operators \(<s>\) satisfying the property \( (*) \) then why don't we have

\[
\begin{align}
x\,<s>\,\omega&=\exp^s(\log^s(x)+\omega)\\
&=\exp^s(\log^s(x)+\log^s(\omega))\\
&=x \odot_s \omega\\
\end{align}\]?

If the above is true how can \(<s>\) also satisfy the Goodstein equation, i.e. \(<s+1>S=<s><s+1>\)? Isn't \(<3>\) different from tetration?

Note: we have however a chain of holomorphic Goodstein equations. Observe that
\[
\begin{align}
x\odot_{s+1} ({}^{s}b\odot_s y)&=(x\odot_{s+1}{}^{s}b) \odot_s(x \odot_{s+1} y)\\
&=\exp^{s+1}(\log^{s+1}(x)+\log^{s+1}({}^{s}b)) \odot_s(x \odot_{s+1} y)\\
&=x \odot_s(x \odot_{s+1} y)\\
\end{align}\]

Since \({\bf 1}_{0}=0\); \({\bf 1}_{1}=1\); \({\bf 1}_{2}=b\); and \({\bf 1}_{s}=\exp_b^s(0)={}^{s-1}b\); we can define the homomorphic image of the successor as \({\rm succ}_{s+1}(y):={\bf 1}_{s+1}\odot_s y\) and write the previous as \[x\odot_{s+1} {\rm succ}_{s+1}(y)=x \odot_s(x \odot_{s+1} y)\]

Hey, Mphlee! Great to hear from you again.

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.

Essentially you would be pasting together a large swath of functions starting from the holomorphic functions:

\[
x \odot_{s,\mu} \omega(\mu) = F(x,s,\mu)\\
\]

This initially creates a holomorphic function for \(b=e^\mu\) in the Shell-Thron region. Then we describe \(x<s>\omega = F(x,s,\mu)\). 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...

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 \(x<s+1>\omega +1 = x<s> \omega'\).


Notice I haven't talked about \(\lambda\), or what could be equivalently done with a \(\theta\) mapping. The goal now is to make this whole mess holomorphic. We are going to essentially, continuously perturb the iteration of \(\log^{\circ s}\). As every type of iteration of \(\log\) works fine for \(0,1,2\), we don't break the initial Goodstein functional equation.

Again, I haven't worked out the details, but the final expression should look something like this (remember, \(b = e^\mu\)):

\[
x<s>\omega = \exp_b^{\circ s + \theta(s,x,\omega)}(\log_b^{\circ s + \theta(s,x,\omega)}(x) + \omega)
\]

Where \(\theta(0,x,y) = 0\) and \(\theta(s+1,x,y) = \theta(s,x,y)\). This expression is only viable for \(\omega\) a fixed point though, we then have to nest solutions for \(\omega+\pm k\), but and we are choosing \(\theta\) in just the right way so that "the ends line up," essentially.

We have so much god damn freedom moving \(\theta\), that this should be very doable. I can't fill in the cracks. Again, I'm switching gears from iteration theory; so this is just an info dump on what I've been mulling over lately.

Hopefully this makes sense, I'm kind of just shooting at the wall and seeing what sticks, lol. But I do believe this holds some weight. Think of it as taking all the Bennet hyperoperations, and finding a path along them where the typical Goodstein functional equation works.

Regards, James

As an important point, it's helpful to think about \(1 <s> \omega\) for \(s\approx 1\), and how this relationship plays out; when you try to add a theta mapping. Honestly, the implicit function theorem should take care of everything... I mean, we're just trying to find \(1<s-1> 1<s>\omega = 1<s> \omega+1\) which has a point of solution at \(s=1\), and we're just trying to find the geodesic which continues to satisfy this. On the real line this would constitute looking inbetween \(-e,e\), and checking we're glued together properly for \(-e,e-1\).