Hey everyone! Some more info dumps!

I haven't talked too much about holomorphic semi-operators for a long time. For this brief exposition I'm going to denote the following:

\[

\begin{align}

x\,<0>\,y &= x+y\\

x\,<1>\,y &= x\cdot y\\

x\,<2>\,y &= x^y\\

\end{align}

\]

Where we have the identity: \(x<k>(x<k+1>y) = x<k+1> y+1\). Good ol fashioned hyper-operators.

Now there exists a really old thread on here, where using fatou.gp you could get really close to a solution of semi-operators. Let's let \(b \in \mathfrak{S}\), be in the interior of the Shell Thron region. Let \(\exp/\log\) be base \(b\). Let \(\omega\) be the fixed point assigned.

Then:

\[

x <s> \omega = \exp^{\circ s}(\log^{\circ s}(x) + \omega)\\

\]

Which is holomorphic and allows us to solve for all \(\omega \pm k\) for all \(s\). Now, the idea is that we have to solve implicit equations in log. I've never had a familiarity with this since I've investigated \(\beta\), but it should be doable on the following domain. If you take all forward and backwards iterates of \(\omega \pm k\) for \(\omega \in \mathcal{W}\); which is the domain of the fixed points. You should be able to construct an implicit solution to the equation:

\[

x <s> (x<s+1>y) = x <s+1> y+1\\

\]

For all \(x \in \mathbb{C}/\mathcal{E}\) and \(y \in \mathcal{W} + \mathbb{Z}\)--where \(\mathcal{E}\) is measure zero in \(\mathbb{R}^2\).

I mean, this problem is really solved if you think of it implicitly. We are just varying \(\mu,\lambda\) until we find a solution to the above equation while we freely move \(s\). This is very fucking difficult to do. I have not done it, as this would require a good 20 pages of work, but it is definitely possible. I may come back to this, but for the moment my brain is switching to PDE/ODE territory, and this type of research is secondary.

Regards, James

I haven't talked too much about holomorphic semi-operators for a long time. For this brief exposition I'm going to denote the following:

\[

\begin{align}

x\,<0>\,y &= x+y\\

x\,<1>\,y &= x\cdot y\\

x\,<2>\,y &= x^y\\

\end{align}

\]

Where we have the identity: \(x<k>(x<k+1>y) = x<k+1> y+1\). Good ol fashioned hyper-operators.

Now there exists a really old thread on here, where using fatou.gp you could get really close to a solution of semi-operators. Let's let \(b \in \mathfrak{S}\), be in the interior of the Shell Thron region. Let \(\exp/\log\) be base \(b\). Let \(\omega\) be the fixed point assigned.

Then:

\[

x <s> \omega = \exp^{\circ s}(\log^{\circ s}(x) + \omega)\\

\]

Which is holomorphic and allows us to solve for all \(\omega \pm k\) for all \(s\). Now, the idea is that we have to solve implicit equations in log. I've never had a familiarity with this since I've investigated \(\beta\), but it should be doable on the following domain. If you take all forward and backwards iterates of \(\omega \pm k\) for \(\omega \in \mathcal{W}\); which is the domain of the fixed points. You should be able to construct an implicit solution to the equation:

\[

x <s> (x<s+1>y) = x <s+1> y+1\\

\]

For all \(x \in \mathbb{C}/\mathcal{E}\) and \(y \in \mathcal{W} + \mathbb{Z}\)--where \(\mathcal{E}\) is measure zero in \(\mathbb{R}^2\).

I mean, this problem is really solved if you think of it implicitly. We are just varying \(\mu,\lambda\) until we find a solution to the above equation while we freely move \(s\). This is very fucking difficult to do. I have not done it, as this would require a good 20 pages of work, but it is definitely possible. I may come back to this, but for the moment my brain is switching to PDE/ODE territory, and this type of research is secondary.

Regards, James