08/15/2007, 11:29 PM
I've been reviewing my superlogarithmic constants, and realized I forgot a factor of log_b(a) when I was calculating, so, for example, \( \mu_2^{\small -1}(1) \) is closer to 2.574062876128519046336549796971138669.
What this means is that if you tetrate base 2 n times, and tetrate base 2.574062876128519046336549796971138669... only {n-1} times, then as n goes to infinity, these answers will converge.
Here are the first few terms of each, using a=2, b=2.574062876128519046336549796971138669..., with \( \mu_b(a)=-1 \):
\(
\begin{array}{c|ccccc}
n & {}^{(n-1)} b && {}^{n} a && \frac{{}^{(n-1)} b}{{}^{n} a} \\
\hline
0 & 0.00000000000000000000 && 1.00000000000000000000 && --- \\
1 & 1.00000000000000000000 && 2.00000000000000000000 && 2.00000000000000000000 \\
2 & 2.57406287612851904634 && 4.00000000000000000000 && 1.55396359471068572326 \\
3 & 11.4014349335617688931 && 16.0000000000000000000 && 1.40333213259865135794 \\
4 & 48044.9265758905624669 && 65536.0000000000000000 && 1.36405661681012200348 \\
5 & 1.46881265672123\times 10^{19728} && 2.0035299304068\times 10^{19728} && 1.36404729441755212824 \\
6 & 10^{10^{6.03122606263029537156455}} && 10^{10^{6.03122606263029537156455}} && 1.36404729441755212824 \\
\end{array}
\)
As you can see, the ratio goes to log_a(b) with increasing n. As it is, for n=6, the best I could do was to compare their logarithms. For row n=7, the discrepency in the logarithms would require about 19750 digits of accuracy, so for all practical intents and purposes, you would work with something in the neighborhood of n=6 when doing a base transformation between a and b (assuming you had an exact solution for either a or b).
What this means is that if you tetrate base 2 n times, and tetrate base 2.574062876128519046336549796971138669... only {n-1} times, then as n goes to infinity, these answers will converge.
Here are the first few terms of each, using a=2, b=2.574062876128519046336549796971138669..., with \( \mu_b(a)=-1 \):
\(
\begin{array}{c|ccccc}
n & {}^{(n-1)} b && {}^{n} a && \frac{{}^{(n-1)} b}{{}^{n} a} \\
\hline
0 & 0.00000000000000000000 && 1.00000000000000000000 && --- \\
1 & 1.00000000000000000000 && 2.00000000000000000000 && 2.00000000000000000000 \\
2 & 2.57406287612851904634 && 4.00000000000000000000 && 1.55396359471068572326 \\
3 & 11.4014349335617688931 && 16.0000000000000000000 && 1.40333213259865135794 \\
4 & 48044.9265758905624669 && 65536.0000000000000000 && 1.36405661681012200348 \\
5 & 1.46881265672123\times 10^{19728} && 2.0035299304068\times 10^{19728} && 1.36404729441755212824 \\
6 & 10^{10^{6.03122606263029537156455}} && 10^{10^{6.03122606263029537156455}} && 1.36404729441755212824 \\
\end{array}
\)
As you can see, the ratio goes to log_a(b) with increasing n. As it is, for n=6, the best I could do was to compare their logarithms. For row n=7, the discrepency in the logarithms would require about 19750 digits of accuracy, so for all practical intents and purposes, you would work with something in the neighborhood of n=6 when doing a base transformation between a and b (assuming you had an exact solution for either a or b).
~ Jay Daniel Fox
