(07/07/2011, 05:02 AM)sheldonison Wrote: So, what you have is a way to evaluate the tower of exponents and logarithms. I have to study it more. I know the short-cut for \( \ln(\ln(b^b)) \), but I don't know any short-cuts for taller towers, like \( \ln(\ln(\ln(b^{b^b})) \). That tower sequence is very similar to the base change constant. If you have an alternative equation to evaluate higher towers, that would be very interesting. Could you write out the equation for \( a_3 \)? As you noted, in the limit, this sequence from your earlier post converges.
\( a_3 = (\ln(b) + \ln^{\circ 2}(b)) \,\,\bigtriangleup_{-1}^e\,\,\ln^{\circ 3}(b) \)
\( a_4 = ((\ln(b) + \ln^{\circ 2}(b)) \,\,\bigtriangleup_{-1}^e\,\,\ln^{\circ 3}(b))\,\,\bigtriangleup_{-2}^e \,\,\ln^{\circ 4}(b) \)
where
\( p \le 0 \)
\( a\,\,\bigtriangleup_p^e\,\,b = \exp_e^{\circ p}(\exp_e^{\circ -p}(a) + \exp_e^{\circ -p}(b)) \)
Quote:Let us call the value the sequence conveges to \( a_n \). If you take the equation, \( k=\text{slog}_e(a_n) \), then it is the same as this limit, for the constant "k" for the base change equation,
\( k=\lim_{n \to \infty} {\text{slog}_e(\exp_b^{[n]}(1))-n=\lim_{n \to \infty} {\text{slog}_e(\log_e^{[n]} (\exp_b^{[n]}(1))) \).
This is the "k" constant for:
\( \text{sexpBaseChange}_b(z) = \lim_{n \to\infty}\log_b^{[n]}\text{sexp}_e(z+k+n) \)
For large enough numbers, k is an approximation for how the delta for how the two different sexp bases would represent large numbers. You might expect that the "k" constant represented how to convert from one sexp base to another, but the value is only approximate, and in fact has a 1-cyclic period. For base e, and base eta, the 1-cyclic function k(z) function is approximately 0.5835+/-0.0004
\( k(z)=\lim_{n \to \infty} {\text{slog}_\eta(\text{sexp}_e(z+n))-n \)
- Sheldon
that's really quite interesting
