(08/21/2018, 02:33 AM)Chenjesu Wrote: Well I appreciate you taking the time to right about it. When I look at the formula you posted again though, I can understand how it applies to complex bases, but I still don't see how it addresses complex heights. You're still only picking an integer 2 to define the tetration, but I don't see how it tells me what something like \( ^{i/3}(z) \) outside of maybe the sexp/slog formula that also doesn't seem to have an explicit representation.
I never worked too much on complex bases. My formula WILL NOT work for \( z \) complex in your scenario. If you want to work with these bases, I developed a formal solution, but it is god awful, and it only works in the domain
\( \Omega = \{z\in\mathbb{C}\,|\,\lim_{n\to\infty} z^{z^{...(n\,times)...^z}} \to A \neq \infty\} \)
There is no nice way to calculate the function \( ^{i/3}z \), and I never proved the limiting expressions I had converged. Essentially I can prove that there is a dense subset of \( \Omega \) call this \( \mathcal{K} \) where \( ^{s} z \) is a holomorphic function in \( s \) (on wildly different domains for each \( z\in \mathcal{K} \)). Then through some rough limiting arguments we can take \( z_n \to z \) where \( z_n \in \mathcal{K} \) and \( z\in\Omega \) and prove that \( ^sz_n\to\, ^sz \) (which is kind of where I wasn't sure how to continue). These are highly non unique though and branch cuts pop up wildly and in a crazy chaotic way.
Again though, this is in the hemisphere of knowledge, it's not proven per se in a modern sense.
Sadly, something like \( ^{i/3}z \) is a nightmare. If you insist, I can write out the formula, but it's a nightmare, a god awful mess of a nightmare.