Okay, Mphlee,
I'll go very slow. We are going to set, universally \(b = e^{\mu}\), and \(e^{\mu \omega} = \omega\) and \(0 < |\mu\omega|<1\)--we're going to always assume these relations, but it can get tricky. This is intended to mean that \(b\) is in the Shell-Thron region, and \(\omega\) is its fixed point. We are going to call the domain \(\mathcal{W}\) the domain of said fixed points. So, for example, \(\mathcal{W}\cap \mathbb{R} = (1/e,e)\) coinciding with when \(b \in (e^{-e},e^{1/e})\). I'm just going to investigate \(1<s> \omega\) for \(\omega, \omega + 1 \in \mathcal{W}\); just to set the scene.
This is going to be a long post, and I'll skip a few steps, but we're going to solve the problem in stages. Please be aware, that the actual mechanics I'd be using don't appear until the very end of this post.
This is the first step, which gets us a third of the way....
Let us define a modified version of Bennet's commutative hyper operators:
$$
x \oplus_{s,\theta,\mu} y = \exp_b^{s+\theta}\left(\log_b^{\circ s + \theta}(x) + \log_b^{\circ s + \theta}(y)\right)\\
$$
Here, \(\theta\) is a holomorphic 1-periodic function which satisfies \(\theta(0) = 0\). We know now that,
$$
x\oplus_{s,\theta,\mu} \omega = \exp_b^{s+\theta}\left(\log_b^{\circ s + \theta}(x) + \omega\right)\\
$$
Is a holomorphic function in \(s\), but also interpolates \(x+\omega,\,x\omega,\,x^{\omega}\) seamlessly. It does so regardless of \(\theta\), so long as \(\mu\) is associated with \(\omega\) in the manner above. We are going to let \(\mu\) move freely though.
Now, let us define the function:
$$
G(s,\theta,\mu) = \left(1\oplus_{s-1,\theta,\mu} \left(1\oplus_{s,\theta,\mu} \omega\right) \right) - 1 \oplus_{s,\theta,\mu(\omega+1)}(\omega+1)\\
$$
For \(s=1\) we get that \(G = 0 \), no matter what \(\theta\) or \(\mu\) do. We are trying to find an implicit function in \(\mu\) and \(\theta\), such that this result still holds locally. Implicit function theorem, literally does all the work.
$$
G(s,\theta(s,\omega),\mu) = 0\\
$$
Such that \(\theta(1,\omega) = 0\).
And voila, we've found a solution for \(|s-1| < \delta\), in which we can call:
$$
1 <s> \omega = 1 \oplus_{s,\theta(s,\omega),\mu(s,\omega)} \omega\\
$$
Which satisfies the Goodstein equation for \(\omega \in \mathcal{W}\cap\mathcal{W}-1\).
This absolutely gets us holomorphic in a neighborhood of multiplication/addition/exponentiation, for \(\omega \in \mathcal{W}\cap\mathcal{W}-1\). To extend further so that we get \(\mathcal{W} \pm k\), would require repeating the above argument for each iteration. The domains would be very sensitive here. This would be a bitch of a proof by induction. But should follow similarly.
This requires a bit more finesse than what I've written here, because we'd need to consider \(\theta\) as a parameter which is 1-periodic and zero at naturals, but locally this is not a problem; it's just \(\theta(1) = 0\) and we're only talking about \(|s-1| < \delta\); the period never pops up.
I can't imagine what it's like in Europe right now. You guys have my greatest sympathies. The worst we have here is increased food prices and ridiculous gas prices--bad inflation all around. But there's zero threat of a looming war; and for that my heart goes out to you. Heart is with you Mphlee. Although, historically if war breeds something; it breeds deep intellectual breakthroughs (especially in Europe); just get arrested like Weil for draft dodging and write your best work in prison
. Jokes aside, keep well and stay strong Mphlee.
Regards, James
This is the second step; paying attention to \(\mu\)...
I realize I've let a good amount of freedom with \(\mu\), but it's for good reason. We can instead write our function like this \(\mu: \omega \mapsto \mu(\omega)\). So instead we write our \(G\) like:
$$
G(s,\theta,\omega) = \left(1\oplus_{s-1,\theta,\mu(\omega)} \left(1\oplus_{s,\theta,\mu(\omega)} \omega\right) \right) - 1 \oplus_{s,\theta,\mu(\omega+1)}(\omega+1)\\
$$
This version is asking solely for a function \(\theta(s,\omega)\); and is a slightly clearer manner of writing what I wrote above. The above has a bit too many free variables. But in this case. All we are implicitly finding is a function:
$$
\theta(s,\omega) : \{s \in \mathbb{C}\,:\,|s-1| < \delta\}\times \left(\mathcal{W}\cap\mathcal{W}-1\right) \to \{\theta \in \mathbb{C}\,:\,|\theta| < \delta'\}\\
$$
Such that \(G(s,\theta(s,\omega),\omega) = 0\). Nothing more, nothing less. This is exactly what I wrote above, but it may be a tad difficult to follow without explicitly describing \(\mu\)'s dependence to \(\omega\). In short, we can think of moving \(\mu\) and moving \(\omega\) synonymously. I just moved \(\mu\) instead of \(\omega\) in the above.
The third step is setting up a similar equation which respects Goodstein's equation inductively...
It's also important to note, that this isn't the exact answer, this would just be proof that this method of thought can lead us to the answer. The correct function you would actually want is:
$$
G(s,\theta,\omega) = \left(1\oplus_{s-1,\theta,\mu\left(1\oplus_{s,\theta,\mu(\omega)} \omega\right)} \left(1\oplus_{s,\theta,\mu(\omega)} \omega\right) \right) - 1 \oplus_{s,\theta,\mu(\omega+1)}(\omega+1)\\
$$
Which has a second recursive call in \(\mu\), which assures us a greater deal of convenience when dealing with proving \(\mathcal{W} + \mathbb{Z}\). Nonetheless the mathematics is emboldened by what was written above; it's just not gonna be perfect. This \(G\) is the \(G\) you really want, which still satisfies:
$$
G(1,\theta,\omega) = 0\\
$$
So finding, \(\theta(s,\omega)\) as above which satisfy:
$$
G(s,\theta(s,\omega),\omega) = 0\\
$$
Is not far off at all.
This would mean, as I first wrote, that:
$$
1 < s> \omega = 1 \oplus_{s,\theta,\mu(\omega)} \omega\\
$$
And through the above implicit function theorem, for \(|s-1| < \delta\) and \(\omega \in \mathcal{W}\cap\mathcal{W}-1\):
$$
1<s-1>1<s>\omega = 1<s> \omega+1\\
$$
It satisfies this so long as \(1<s>\omega \in \mathcal{W}\) and \(\omega + 1 \in \mathcal{W}\). And the exact formula is the above implicit functions.
The ENTIRETY of this argument is based on the fact \(\log^{\circ s + \theta}(x) = \log^{\circ s}(x) + \theta'\), which allows us to move the perfect line between \(x + \omega, x\omega, x^{\omega}\), in an implicit manner to allow us to compare operations between \(\omega\) and another \(\omega'\).
I should also disclaim. I do not know how to extend this to \(0 \le \Re(s) \le 2\), I am just sure of \(|s-1| < \delta\) and by comparison \(|s|,|s-2| < \delta\). My argument to extend it further would involve fourier analysis, and I see how I might do it, but I'm not too sure yet...
The exact domain that this would produce holomorphy for \(1<s>\omega\) as we vary \(\omega\); would be for \(|\Im(s)| \le |c|\), where \(c = \sup \Im (\mathcal{W}\cap\mathcal{W}-1)\). This means this would only work on a strip of the complex plane. The height of the strip would be exactly \(c = \Im(a)\) where \(a\) is on the border of the Shell Thron region and \(a+1\) is on the border of the Shell Thron region. This is a constant I cannot for the life of me identify with anything else. This would definitely be a new constant.
$$
\begin{align}
a &= \lim_{n\to\infty} \exp_b^{\circ n}(0)\\
|\log(b)a| &= 1\\
a+1 &= \lim_{n\to\infty} \exp_{d}^{\circ n}(0)\\
|\log(d)(a+1)| &= 1\\
c &= |\Im(a)| = |\Im(a+1)|\\
\end{align}
$$
The nested recursion allows us to talk about \(\mathcal{W}\pm k\) pretty clearly now. Literally, all we've done, Mphlee; is add theta mappings to Bennet's commutative operations; and we're playing fast and loose in a neighborhood of multiplication. To extend further would definitely be more difficult.
Nothing but love; I hope this makes a bit more sense...
To sum everything up, we can solve the equation:
$$
1<s> y\\
$$
When \(|\Im(y)| \le c\) and \(|s-1| < \delta = \delta(y)\).