03/31/2018, 04:19 AM
(This post was last modified: 03/31/2018, 04:29 PM by sheldonison.)
(03/30/2018, 07:37 PM)JmsNxn Wrote: It seems upsetting that the only tetrations that could satisfy this are non-analytic. I'm not prone to believe this, only because it doesn't look nice...
There's an old thread http://math.eretrandre.org/tetrationforu...hp?tid=236
I started the thread in 2009, before I had written generic programs for analytic tetration for any base, and I was using an excel spreadsheet to approximate analytic tetration. I estimated that as x gets arbitrarily large
\( \text{slog}_2(x)-\text{slog}_e(x)\approx1.1282 \)
The 2009 thread continues on to discuss what I called "the wobble"...
Credit needs to go to William Paulsen and Samuel Cowgill in their upcoming paper which discusses these issues more rigorously than I can. But it was quickly clear in the 2009 thread that there is an inherent wobble when comparing tetration bases; this was apparent for bases a little bit bigger than eta=exp(1/e) using straightforward techniques on an excel spreadsheet. The limit as x get arbitrarily large does not converge to a simple number like the 1.1282 estimate, but instead converges to a 1-cyclic function near that value. For base(2) and for base(e), if you use Kneser's construction; then the 1-cyclic limit is graphed below.
\( \text{slog}_e(\text{tet}_2(x))-x \)
On this forum, other ideas like "the base change function" were discussed, where you use Peter Walker's idea to define tetration base(a) from tetration base(b). For example, you could define tetration base(2) from Kneser's tetration base(e). The relevant equations might look something like this. But the "h" function below is conjectured to be nowhere analytic, even though Walker proved it is \( C^{\infty} \) for the case in his paper. Walker defined the base(e) slog from the Abel function for iterating \( x\mapsto\exp(x)-1 \). This is mathematically conjugate (or exactly equivalent) to iterating base eta. \( y\mapsto\eta^y\;\;\eta=\exp(1/e)\;\;x=\frac{y}{e}-1 \)
\( h_n (x)=\ln_b^{[n]}(\exp_a^{[n]}(x)) \)
\( h(x)=\lim_{n\to\infty}h_n (x) \)
\( \text{slog}_b(x)=\text{slog}_a(h(x))-\text{slog}_a(h(1))); \) /* constant to guarantee slog_b(1)=0 */
- Sheldon