05/02/2022, 06:49 PM
Woooh, very interesting! Those are very nice looking identities. The trouble I'm having with it, is that none of my computations will work outside of the region \(0 \le \Re(s) \le 2\). So any of the iteration formulas that go beyond that, I wouldn't be able to compute and verify. To do that you would first have to solve for: \(f(z) = 2 <s+1> z\), and take the iteration \(f^{\circ z}\) to get \(<s+2>\), and that would be far too computationally exhausting.
I can test for \(2 <s> y\) for all \(y > 1\), and verify the functional equation. The trouble is, I don't know how to glue these solutions together. The monodromy theorem guarantees there is a function, but I can't for the life of me find a way to compute it. All the trouble now is describing the surface \(\boldsymbol{\varphi} = (\varphi_1,\varphi_2,\varphi_3)\). I can generate all the solutions (which produces a surface in \(\mathbb{C}^3\) (so it has complex dimension 2 for \(s,y\) fixed)), but I don't know which is the correct solution we want. Well, I mean, I do know the correct solution, but I don't know how to test for finding the correct solution. So for the moment I've been trying to derive a differential equation.
I'm trying to see if there's an obvious formula for the first term in \(w\):
\[
2<s>(2<s+1> y+w) = 2 <s+1> (y+1+w)\\
\]
I think there might be a first term relation which can help us determine the solution we want. That's mostly what I'm trying to track by what I wrote above.I'm just scratching my head trying to figure out how to ensure:
\[
\varphi_2(2<s+1>y+w,s) = \varphi_1(y+w,s)\\
\]
Because, we require this identity to paste all the solutions together. The trouble is, this isn't a local equation, so it doesn't help us for \(s,w \approx 0\). If we had a local equation it'd be easy to test in pari-gp; which is why I'm looking for a differential equation test instead. Jesus this is frustrating... There's gotta be something I'm missing to test this better.
I am very sure that this method is not equivalent to the Ramanujan method for a simple reason. For \(\Re(s) \ge 2\), we are given \(\alpha \uparrow^s 1 = \alpha\), which, due to the identity theorem, we must have it for \(0 \le \Re(s) \le 2\). This doesn't happen for \(<s>\). So instantly they are in disagreement. This is again, why I didn't use the uparrow notation, to make sure there was no confusion.
As to the iterated/infinite composition, I'm not sure where this would stand. As far as I'm concerned, zoologically speaking, these are two very different animals--from two entirely different genuses, they just both happen to live in tetrationland, lol.
I can test for \(2 <s> y\) for all \(y > 1\), and verify the functional equation. The trouble is, I don't know how to glue these solutions together. The monodromy theorem guarantees there is a function, but I can't for the life of me find a way to compute it. All the trouble now is describing the surface \(\boldsymbol{\varphi} = (\varphi_1,\varphi_2,\varphi_3)\). I can generate all the solutions (which produces a surface in \(\mathbb{C}^3\) (so it has complex dimension 2 for \(s,y\) fixed)), but I don't know which is the correct solution we want. Well, I mean, I do know the correct solution, but I don't know how to test for finding the correct solution. So for the moment I've been trying to derive a differential equation.
I'm trying to see if there's an obvious formula for the first term in \(w\):
\[
2<s>(2<s+1> y+w) = 2 <s+1> (y+1+w)\\
\]
I think there might be a first term relation which can help us determine the solution we want. That's mostly what I'm trying to track by what I wrote above.I'm just scratching my head trying to figure out how to ensure:
\[
\varphi_2(2<s+1>y+w,s) = \varphi_1(y+w,s)\\
\]
Because, we require this identity to paste all the solutions together. The trouble is, this isn't a local equation, so it doesn't help us for \(s,w \approx 0\). If we had a local equation it'd be easy to test in pari-gp; which is why I'm looking for a differential equation test instead. Jesus this is frustrating... There's gotta be something I'm missing to test this better.
I am very sure that this method is not equivalent to the Ramanujan method for a simple reason. For \(\Re(s) \ge 2\), we are given \(\alpha \uparrow^s 1 = \alpha\), which, due to the identity theorem, we must have it for \(0 \le \Re(s) \le 2\). This doesn't happen for \(<s>\). So instantly they are in disagreement. This is again, why I didn't use the uparrow notation, to make sure there was no confusion.
As to the iterated/infinite composition, I'm not sure where this would stand. As far as I'm concerned, zoologically speaking, these are two very different animals--from two entirely different genuses, they just both happen to live in tetrationland, lol.