(07/19/2022, 10:05 AM)Catullus Wrote: What would happen if you used the analytic continuation of the Kneser method to do that?
For a fixed n greater than one, then using the analytic continuation of the Kneser method how fst would n[1.5]x grow?
Would it grow half-exponentially?
Hmm, so this is a weird question.
First of all, you can't use Kneser to construct this, as Kneser is a tetration solution, this is an iterated exponential--outside of the purview of Kneser. This construction is specifically about the repelling iteration, and not the attracting iteration, nor is it based off of any tetration per se, just the iterated exponential. Additionally it is within the Shell-Thron region, and so, the Kneser fixed point pair algorithm wouldn't make sense for the repelling case. I don't think you'd even be able to run a Kneser like iteration about the repelling point, but I could be mistaken.
Either way, to answer your question ignoring the Kneser part; it kind of grows half exponentially, kinda doesn't. It would look something like this:
\[
n [1.5] x = \exp^{\circ 0.5}_{1+\log(x)/x+O(\log(x)/x)^2}\left( A x\right)
\]
For some constant \(A > n\), but not much bigger than \(n\).
So this means it grows half exponentially, but not technically, because the base of the half exponential \(b \to 1\).
If I took:
\[
\text{slog}\left(n[1.5]x\right) \approx \text{slog}(x) + C(x)\\
\]
And \(C(x)\) would be about a half maybe, as \(x \to \infty\), it would be around there. It would probably be a tad chaotic. So ya, I guess. Close enough to half exponential. Then, I guess technically you could write a function:
\[
\text{tet}_K(\text{slog}(x) + C(x)) = n[1.5]x\\
\]
Which would be a Kneser form of this function; but in and of itself; the construction would be tedious with Kneser. Best I could say is \(C(x) \approx 1/2\) as \(x \to \infty\). So again, about a half exponential.
Regards, James