Holomorphic semi operators, using the beta method

[JmsNxn]

Note: Shouldn't the domain \(\mathcal H\) have \(0\) removed?

Yes you are absolutely right. The domain \(\mathcal{H}\) is essentially a domain of holomorphy of \(f(z) = z^{1/z}\) so it would exclude zero and some kind of branch cut to infinity. I definitely made a typo saying \(\mathcal{H} = \mathbb{C}/(-\infty,0)\), it should be \(\mathcal{H} = \mathbb{C}/(-\infty,0]\). You won't have holomorphy at \(0\), but you will have a continuous limit. I'm sticking to \(1<s> y\) at the moment, but this looks like it should at least work for \(\alpha<s> y\) for \(\alpha \approx 1\).



I'm currently trying to see if I can resurrect my old notes on Fourier analysis applied to semi-operators, so that we can take \(s \approx 0,1,2\) and turn it into \(0 \le \Re(s) \le 2\)--this is the real tricky part. But for the moment I have \(1<s> y\) for \(y \in \mathcal{H}\) and \(s \approx 0,1,2\). I'm trying to think of a feasible way of constructing \(\theta\) at the moment. I know it exists, thanks to the implicit function theorem, but I have no idea how to get it. I'll need to think of a limit formula, but I'm scratching my head about this.



Getting to \(\alpha \in (1,e^{1/e})\) shouldn't be hard once you have it for \(1\)...



I'm definitely going to try writing this up; but this is going to be a long project. Probably the summer...



I think the key relation we are using from Bennet, is that:



\[

\begin{align}

\exp_b^{\circ s}(\log_b^{\circ s}(1) + y + \delta) &= \exp_b^{\circ s+\epsilon}(\log_b^{\circ s+\epsilon}(1) + y)\\

&= \exp_{b+\kappa}^{\circ s}(\log_{b+\kappa}^{\circ s}(1) + y)\\

\end{align}

\]



This allows us to locally change the base of \(\exp,\log\), without affecting their values at \(s=0,1,2\), and affect the value of \(y\) (at least locally). Now we're just looking for a geodesic (lol, a path through all the values) which manages to solve Goodstein.



Call \(F(z) =\log(z)/z\), and \(Y = 1 \oplus_{s,\theta,F(y)} y\), then:



\[

\begin{align}

G(s,\theta) = 1\oplus_{s-1,\theta,F(Y)} Y - 1 \oplus_{s,\theta,F(y+1)} y+1\\

G(s,\theta(s)) = 0\\

\end{align}

\]



We would construct \(\theta\) like this for \(s\approx 2\) and \(s-1 \approx 1\). Then, we want to find the value of \(\theta\) that satisfies this as we move \(s\) (beyond a local scenario, this is the tricky part), since it's satisfied for \(s=0,1,2\) prima facie. Then our solution becomes:



\[

1 <s> y = 1\oplus_{s,\theta(s),F(y)} y\\

\]



Now we have to talk about the types of iterations we use to construct \(\theta\). Thank god for the beta method thesis. We can construct functions \(\theta\)'s such that \(\theta(s+1) = \theta(s)\) and \(\theta(s+2\pi i/\lambda) = \theta(s) - 2\pi i/\lambda\).



This means you should think of \(\exp_b^{\circ s + \theta}(1)\) as a different tetration that \(\exp_b^{\circ s}(1)\), they are essentially both tetration... The difference will be they have different periods. That is all.  So we can think of moving \(\theta\), as really just stretching and shrinking the period of our Bennet hyperoperation. That is, we can instead of talking about \(\theta\), just talk about a value \(\Re \lambda > 0\)...



That's the craziest part about all this.  This result draws a lot on using the \(\beta\) method; where we can get \(\exp_b^{\circ s}\) for arbitrary period in \(s\), and still retain large areas of holomorphy. This is to say, quite literally, think of the \(\theta\) in \(\oplus_{s,\theta,\mu}\) as choosing "what tetration we use". I don't mean what base of tetration, that's handled by \(b = e^\mu\). I mean, what tetration with base \(b = e^{\mu}\) we are using. And we can uniquely identify and compare these things based solely on one parameter, \(\lambda\); which forms the period \(2 \pi i / \lambda\).

These form very special \(\theta\) functions, such that:

\[
\begin{align}
\theta(s+1) &= \theta(s)\\
\theta(s+2\pi i / \lambda) &= \theta(s) - 2\pi i / \lambda\\
\end{align}
\]

We can classify these functions very well; the beta method produces the "best" possible class essentially.

So instead of writing, \(\theta\), in all of these posts, we can instead talk about \(\lambda\); which is implicitly choosing which tetration is the right tetration to turn bennet to goodstein.

EDIT:

I'm waiting for your info dump Mphlee. It doesn't have to be perfect. Just infodump. Who cares. It's all love here. I can't believe I was about to just post this one thread and drop everything and never touch it again. But after your replies it started making more and more sense, and now I defs have \(1<s>y\)...