I also realize I haven't fully explained how to get \(|\Im(y)| < c\). This requires another trick. For \(s \approx 1\) we have that:
\[
1<s+1> y = 1 \oplus_{s+1,\theta,\mu} y \approx 1 \in \mathcal{W}\\
\]
This can be done for \(y \in \mathcal{W}\), but can be done outside of this domain, because \(1 \oplus_{s,\theta,\mu} y \in \mathcal{W}\) for all \(\Im(y) < c\). This just relies on \(1^y = 1\) regardless of \(y\).This is tricky, because now we have to talk about the repelling fixed point a bit. Let's take \(\sqrt{2}\) for a second. So we have a function:
\[
1<s> 2\\
\]
For \(s\approx 0,1,2\). We can also obtain a value for \(1<s>4\) following the exact procedure as before. It's just:
\[
1<s> 4 = \exp_{\sqrt{2}}^{\circ s + \theta}\left(\log_{\sqrt{2}}^{\circ s + \theta}(1) + 4\right)\\
\]
Where here \(\theta\) is an implicit function, constructed in the same manner as above. This allows us to do \(1<s>y\) for \(y \in \mathbb{R}^+\), because \(y^{1/y}\) has a fixed point at \(y\) (attracting or repelling or neutral), but for \(b = y^{1/y}\), we still have the above implicit functions. To extend further we'd get a maximal domain, for \(y \in \mathcal{Y} \subset\mathbb{C}\). But there'd be a restriction that \(y,y+1 \in \mathcal{Y}\)--which for a quick value gives us \(|\Im(y)| < c\). In this manner, we can always assign:
\[
\begin{align}
\mu(y) &= \log(y)/y\\
b(y) &= e^{\mu(y)}\\
\exp(\mu(y)y) &= y\\
1<s> y &= \exp_{b(y)}^{\circ s+\theta}\left(\log_{b(y)}^{\circ s+ \theta}(1) + y\right)\\
\end{align}
\]
This works identically as above, but you have to be careful now. We are going to keep \(s\approx 1\), but now we are solving:
\[
\begin{align}
1<s> \left(1<s+1>y\right) &= 1<s+1> y+1\\
\text{In a neighborhood of:}&\\
1<1> \left(1<2>y\right) &= 1<2>y+1\\
1\cdot 1^y &= 1^{y+1} = 1\\
\end{align}
\]
Which is always solvable through the implicit function--gives us a \(\theta(s)\) for \(|s-1| < \delta\). This is a more versatile form, because at \(s=1\) we are just writing an idempotent relation with \(1\). This will give us \(1<s+1> y\), to get \(1<s>y\) we just use the usual Goodstein operations in \(y\), and we get it for \(s\approx 1\). We may have trouble with getting \(1<s-1> y\), but I'm not sure exactly.
..........................Now that I think about it, this result should work where-ever \(y \in \mathbb{C}/\{0\}\) and \(b = y^{1/y}\) has an attracting or neutral fixed point... I believe every \(y\) should have one, but I may be wrong. (double checked, yes this will work for \(y \in \mathbb{C}/0\) up to branch cuts). Essentially the domain would be:
\[
\mathcal{H} = \{y \in \mathbb{C}/\{0\}\,|\, b = e^{\log(y)/y},\,\lim_{n\to\infty}|\exp_b^{\circ n}(0)| < \infty\} = \mathbb{C}/(-\infty,0)\\
\]
(This turns out to be all of \(\mathbb{C}\)--upto a branch cut of course...)
It may be easier to switch to the real positive line at this point... We'd definitely be able to construct:
\[
1<s> y\,\,\text{for}\,\,|s|,\,|s-1|,\,|s-2|<\delta(y)\,\,\text{and}\,\,y\in\mathbb{R}^+\\
\]
Shit... I really think this does more than I thought it'd do. Jesus this definitely works even better. This info dump just info dumped on me, and I think I see it better.
I strongly encourage you to info dump Mphlee. It doesn't have to be perfect. Just stretch the legs and get a good feel for it, and release it...
I can absolutely prove this for the real positive line, now that I think about it. It's super easy.
HOLY FUCKING SHIT! THIS DOES \(\alpha <s> y\) for \(\alpha \in (1,e^{1/e})\) and \(y \in \mathbb{R}^+\)!!!! WE JUST HAVE TO RELATE TO THE FRACTIONAL CALCULUS!!!!!! OMG!!!!!!!!!!!!!!!!! Okay, I need to write a quick write up. This goes deeper than I thought. I always thought this was a half solution so I didn't give a shit. YES!!!
\[
1<s+1> y = 1 \oplus_{s+1,\theta,\mu} y \approx 1 \in \mathcal{W}\\
\]
This can be done for \(y \in \mathcal{W}\), but can be done outside of this domain, because \(1 \oplus_{s,\theta,\mu} y \in \mathcal{W}\) for all \(\Im(y) < c\). This just relies on \(1^y = 1\) regardless of \(y\).This is tricky, because now we have to talk about the repelling fixed point a bit. Let's take \(\sqrt{2}\) for a second. So we have a function:
\[
1<s> 2\\
\]
For \(s\approx 0,1,2\). We can also obtain a value for \(1<s>4\) following the exact procedure as before. It's just:
\[
1<s> 4 = \exp_{\sqrt{2}}^{\circ s + \theta}\left(\log_{\sqrt{2}}^{\circ s + \theta}(1) + 4\right)\\
\]
Where here \(\theta\) is an implicit function, constructed in the same manner as above. This allows us to do \(1<s>y\) for \(y \in \mathbb{R}^+\), because \(y^{1/y}\) has a fixed point at \(y\) (attracting or repelling or neutral), but for \(b = y^{1/y}\), we still have the above implicit functions. To extend further we'd get a maximal domain, for \(y \in \mathcal{Y} \subset\mathbb{C}\). But there'd be a restriction that \(y,y+1 \in \mathcal{Y}\)--which for a quick value gives us \(|\Im(y)| < c\). In this manner, we can always assign:
\[
\begin{align}
\mu(y) &= \log(y)/y\\
b(y) &= e^{\mu(y)}\\
\exp(\mu(y)y) &= y\\
1<s> y &= \exp_{b(y)}^{\circ s+\theta}\left(\log_{b(y)}^{\circ s+ \theta}(1) + y\right)\\
\end{align}
\]
This works identically as above, but you have to be careful now. We are going to keep \(s\approx 1\), but now we are solving:
\[
\begin{align}
1<s> \left(1<s+1>y\right) &= 1<s+1> y+1\\
\text{In a neighborhood of:}&\\
1<1> \left(1<2>y\right) &= 1<2>y+1\\
1\cdot 1^y &= 1^{y+1} = 1\\
\end{align}
\]
Which is always solvable through the implicit function--gives us a \(\theta(s)\) for \(|s-1| < \delta\). This is a more versatile form, because at \(s=1\) we are just writing an idempotent relation with \(1\). This will give us \(1<s+1> y\), to get \(1<s>y\) we just use the usual Goodstein operations in \(y\), and we get it for \(s\approx 1\). We may have trouble with getting \(1<s-1> y\), but I'm not sure exactly.
..........................Now that I think about it, this result should work where-ever \(y \in \mathbb{C}/\{0\}\) and \(b = y^{1/y}\) has an attracting or neutral fixed point... I believe every \(y\) should have one, but I may be wrong. (double checked, yes this will work for \(y \in \mathbb{C}/0\) up to branch cuts). Essentially the domain would be:
\[
\mathcal{H} = \{y \in \mathbb{C}/\{0\}\,|\, b = e^{\log(y)/y},\,\lim_{n\to\infty}|\exp_b^{\circ n}(0)| < \infty\} = \mathbb{C}/(-\infty,0)\\
\]
(This turns out to be all of \(\mathbb{C}\)--upto a branch cut of course...)
It may be easier to switch to the real positive line at this point... We'd definitely be able to construct:
\[
1<s> y\,\,\text{for}\,\,|s|,\,|s-1|,\,|s-2|<\delta(y)\,\,\text{and}\,\,y\in\mathbb{R}^+\\
\]
Shit... I really think this does more than I thought it'd do. Jesus this definitely works even better. This info dump just info dumped on me, and I think I see it better.
I strongly encourage you to info dump Mphlee. It doesn't have to be perfect. Just stretch the legs and get a good feel for it, and release it...
I can absolutely prove this for the real positive line, now that I think about it. It's super easy.
HOLY FUCKING SHIT! THIS DOES \(\alpha <s> y\) for \(\alpha \in (1,e^{1/e})\) and \(y \in \mathbb{R}^+\)!!!! WE JUST HAVE TO RELATE TO THE FRACTIONAL CALCULUS!!!!!! OMG!!!!!!!!!!!!!!!!! Okay, I need to write a quick write up. This goes deeper than I thought. I always thought this was a half solution so I didn't give a shit. YES!!!