04/09/2023, 07:59 PM
YEAH I KNOW TOMMY!
I WAS SUPER EXCITED LAST NIGHT AND DIDN'T PROVIDE ENOUGH DETAILS!
But essentially theres a level set \(\tau\) on kneser tetration that would be mapped to \(\gamma\) where \(|\beta'(\gamma(t))| = 1\) and \(|\text{tet}_K'(\tau(t))| = 1\). THIS LOOKS SUPER FUCKING PROMISING! The exact formula should also be unique. If we define a conjugate function WHICH MUST EXIST BY RIEMANN, such that \(\psi(\gamma(t)) = \tau(t)\). We can be sure that \(|\gamma'(t)| =1\) and \(|\tau'(t)| =1\). Then forcing \(\psi(0) = 0\) and \(\psi(-1) = -\infty\) we should have
\[
\psi(\text{tet}_K(z)) = \beta(\psi(z))\\
\]
I've already reduced the problem into the statement that:
\[
T(z) = \psi^{-1}(\beta(\psi(z)))\\
\]
And we must have:
\[
T'(z) = e^{ih(z)}\text{tet}_K'(z)\\
\]
Where \(h\) is real valued on \(\tau\)--this value will turn out to be precisely zero with some magic. Honestly, I'd considered this before, but I always thought it was a shot in the dark until I actually graphed the line \(|\beta'(z)| = 1\) across \(\beta(z)\); and it's an unbroken jordan curve from \(-\infty\) to \(\infty\). It also should be relatively easy to construct \(\psi\); and it should be possible to compute it rather effortlessly! Just need to do some contour integration.
The real trouble is that I only have a good idea of how to do this, with the assumption that Kneser is already constructed. I'm having trouble constructing \(\tau\) from scratch. But if you have any ideas on how to construct \(\tau\) from scratch, please share them! This would show that we can construct kneser from knowledge of a single unbroken level set. I'm focusing on implementing a fairly accurate rudimentary kneser function from Sheldon's program at the moment; that doesn't conflict with my code base. As great as Sheldon's program is; it's a logistic nightmare; and doesn't integrate well with my own code, which obeys the rules of pari much more. I should have some graphs up later tonight.
Take a look at this graph of \(\beta(z)\) with the level set \(|\beta'(z)| = 1\) for the moment! PERFECT CONTOUR JUST BEGGING TO BE FIDDLED WITH! I think the big hurtle will be showing that the left of this contour is a simply connected domain; which honestly, doesn't even look like much of a hurtle!!!! Then a standard riemann mapping should take care of the rest!!!!!!!!!!!!!!!
I WAS SUPER EXCITED LAST NIGHT AND DIDN'T PROVIDE ENOUGH DETAILS!
But essentially theres a level set \(\tau\) on kneser tetration that would be mapped to \(\gamma\) where \(|\beta'(\gamma(t))| = 1\) and \(|\text{tet}_K'(\tau(t))| = 1\). THIS LOOKS SUPER FUCKING PROMISING! The exact formula should also be unique. If we define a conjugate function WHICH MUST EXIST BY RIEMANN, such that \(\psi(\gamma(t)) = \tau(t)\). We can be sure that \(|\gamma'(t)| =1\) and \(|\tau'(t)| =1\). Then forcing \(\psi(0) = 0\) and \(\psi(-1) = -\infty\) we should have
\[
\psi(\text{tet}_K(z)) = \beta(\psi(z))\\
\]
I've already reduced the problem into the statement that:
\[
T(z) = \psi^{-1}(\beta(\psi(z)))\\
\]
And we must have:
\[
T'(z) = e^{ih(z)}\text{tet}_K'(z)\\
\]
Where \(h\) is real valued on \(\tau\)--this value will turn out to be precisely zero with some magic. Honestly, I'd considered this before, but I always thought it was a shot in the dark until I actually graphed the line \(|\beta'(z)| = 1\) across \(\beta(z)\); and it's an unbroken jordan curve from \(-\infty\) to \(\infty\). It also should be relatively easy to construct \(\psi\); and it should be possible to compute it rather effortlessly! Just need to do some contour integration.
The real trouble is that I only have a good idea of how to do this, with the assumption that Kneser is already constructed. I'm having trouble constructing \(\tau\) from scratch. But if you have any ideas on how to construct \(\tau\) from scratch, please share them! This would show that we can construct kneser from knowledge of a single unbroken level set. I'm focusing on implementing a fairly accurate rudimentary kneser function from Sheldon's program at the moment; that doesn't conflict with my code base. As great as Sheldon's program is; it's a logistic nightmare; and doesn't integrate well with my own code, which obeys the rules of pari much more. I should have some graphs up later tonight.
Take a look at this graph of \(\beta(z)\) with the level set \(|\beta'(z)| = 1\) for the moment! PERFECT CONTOUR JUST BEGGING TO BE FIDDLED WITH! I think the big hurtle will be showing that the left of this contour is a simply connected domain; which honestly, doesn't even look like much of a hurtle!!!! Then a standard riemann mapping should take care of the rest!!!!!!!!!!!!!!!
