(06/10/2021, 12:41 PM)MphLee Wrote: Graphically they're obviously two different functions... Also the limit at imaginary infinity seems a good argument because in that way the two solutions have a substantial "topological" difference.
About the "stretching the cylinder" I suspect that that are many ways to do that. Is it possible that we have no way to compare those way to stretch the cylinder?
I mean... something like measuring where the purely imaginary tetrations go... idk... like tet(i)... (secretly I'm thinking about an Euler identity for tetration)
The best way I could think of this was measuring the period in \( z \); this is difficult to explain though.
In many ways,
\(
\text{tet}_\beta(z) = \lim_{n\to\infty} \lim_{\lambda \to 0} \beta_\lambda(z+n+x_0)\,\,\text{for}\,\,\lambda = \mathcal{O}(n^{-\epsilon})\\
\)
So the sequence has a period \( \ell_n \) in \( s \) of about,
\(
\ell_n = 2 \pi i \mathcal{O}(n^{\epsilon})\\
\)
So what we are doing is limiting \( \ell_n \to \infty \). Really, we're mapping \( \ell_n/2 = \infty \) and \( -\ell_n/2 = -\infty \).
I can't think of any non-obvious way to do this. The furthest I really got was using the implicit function theorem.
Let
\(
F_\lambda(z) - \text{tet}_\beta(z) = 0
\)
Which can define an implicit local solution. This function will be (as far as I can tell?) split into \( \lambda^+ \) and \( \lambda^- \) for the upper (resp. lower) half plane. Then,
\(
\lambda(z) = \sum_{k=1}^\infty c_k e^{2 \pi i k z}\\
\lambda(z) = \sum_{k=1}^\infty \overline{c_k} e^{-2\pi i k z}\\
\)
But this is about as far as I got; because using the implicit function theorem seems like a cop out. I'm still trying to think of a better argument for constructing the actual "stretching" function \( \lambda(z) \); other than the limit definition.