08/12/2007, 06:41 AM
Now to back up my claim with some dense math. The conversions are all basic, so high school level calculus should be sufficient to follow (if you take your time), except for the new notation for tetration, which we should all be familiar with if we're visiting this forum.
To see that the constant \( \mu_b(a) \) exists, and furthermore that its value converges superexpontially, i.e., \( \frac{1}{O\left({}^n e\right)} \), consider the following. Note that this is simply extending my previous calculations to generic bases a and b.
First, from the definition of the conversion formula I gave above, for a fixed large n, we get:
\( \begin{eqnarray}
{\Large ^{\normalsize x} a} & = &
{\Large log_a^{\circ n}\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}
\end{eqnarray} \)
Now, let's see what happens if we increase n by 1.
\( \begin{eqnarray}
{\Large ^{\normalsize x} a} & = &
{\Large log_a^{\circ \normalsize (n+1)}\left({}^{\left(\normalsize (n+1)+x+\mu_b(a)\right)} b\right)} \\
& = & {\Large log_a^{\circ \normalsize (n+1)}\left(b^{\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)} \\
& = & {\Large log_a^{\circ n}\left(log_a \left(a^{\left( log_a(b) \times \left({}^{\left(\normalsize n+x+\mu_b(a)\right)}\right) b\right)}\right)\right)} \\
& = & {\Large log_a^{\circ n}\left({log_a(b) + \left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)} \\
& = & {\Large log_a^{\circ n}\left(\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)\ \times\ \left(1\ +\ \frac{log_a(b)}{{}^{\left(\normalsize n+x+\mu_b(a)\right)} b}\right)\right)} \\
& = & {\Large log_a^{\circ n} \left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)\ +\ \epsilon_{\small 1}},\text{ given }\epsilon_{\small 1}\ \approx\ \frac{log_a(b)}{{}^{\left(\normalsize n+x+\mu_b(a)\right)} b} \to 0
\end{eqnarray}
\)
However, if you're not convinced as I am, consider taking it to n+2:
\( \begin{eqnarray}
{\Large ^{\normalsize x} a} & = &
{\Large log_a^{\circ \normalsize (n+2)}\left({}^{\left(\normalsize (n+2)+x+\mu_b(a)\right)} b\right)} \\
& = & {\Large log_a^{\circ \normalsize (n+2)}\left(b^{\left(b^{\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)}\right)} \\
& = & {\Large log_a^{\circ \normalsize (n+2)}\left(a^{log_a(b)\times \left(a^{log_a(b) \times \left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)}\right)} \\
& = & {\Large log_a^{\circ \normalsize (n+1)}\left(log_a(b)\ +\ a^{\left(log_a(b) \times a^{\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)}\right)} \\
& = & {\Large log_a^{\circ \normalsize (n+1)}\left(\left(a^{\left(log_a(b) \times a^{\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)}\right)\ \times\ (1+\epsilon_2)\right)} \\
& = & {\Large log_a^{\circ n}\left(\left(log_a(b)\ +\ a^{\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)\ +\ log_a(1+\epsilon_2)\right)} \\
& = & {\Large log_a^{\circ n}\left(\left(\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)\ \times (1+\epsilon_1)\right)\ +\ \epsilon_{2'}\right)} \\
& = & {\Large log_a^{\circ n} \left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)\ +\ \epsilon_{\small 1'}},\text{ where }\epsilon_{\small 1'} \approx \epsilon_1 (1+log_b(a)\epsilon_{2'})
\end{eqnarray}
\)
Notice that \( \epsilon_{2'} \ll \epsilon_1 \), thus supporting the claim that this superlogarithmic constant converges superexponentially. For base e, once you've found it within a factor of 1 part in 1000, the next iteration will get you accuracy of 1 part in e^1000. The next will get you within e^(e^1000). Of course, machine precision will necessarily cut your ascent off pretty early.
To see that the constant \( \mu_b(a) \) exists, and furthermore that its value converges superexpontially, i.e., \( \frac{1}{O\left({}^n e\right)} \), consider the following. Note that this is simply extending my previous calculations to generic bases a and b.
First, from the definition of the conversion formula I gave above, for a fixed large n, we get:
\( \begin{eqnarray}
{\Large ^{\normalsize x} a} & = &
{\Large log_a^{\circ n}\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}
\end{eqnarray} \)
Now, let's see what happens if we increase n by 1.
\( \begin{eqnarray}
{\Large ^{\normalsize x} a} & = &
{\Large log_a^{\circ \normalsize (n+1)}\left({}^{\left(\normalsize (n+1)+x+\mu_b(a)\right)} b\right)} \\
& = & {\Large log_a^{\circ \normalsize (n+1)}\left(b^{\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)} \\
& = & {\Large log_a^{\circ n}\left(log_a \left(a^{\left( log_a(b) \times \left({}^{\left(\normalsize n+x+\mu_b(a)\right)}\right) b\right)}\right)\right)} \\
& = & {\Large log_a^{\circ n}\left({log_a(b) + \left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)} \\
& = & {\Large log_a^{\circ n}\left(\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)\ \times\ \left(1\ +\ \frac{log_a(b)}{{}^{\left(\normalsize n+x+\mu_b(a)\right)} b}\right)\right)} \\
& = & {\Large log_a^{\circ n} \left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)\ +\ \epsilon_{\small 1}},\text{ given }\epsilon_{\small 1}\ \approx\ \frac{log_a(b)}{{}^{\left(\normalsize n+x+\mu_b(a)\right)} b} \to 0
\end{eqnarray}
\)
However, if you're not convinced as I am, consider taking it to n+2:
\( \begin{eqnarray}
{\Large ^{\normalsize x} a} & = &
{\Large log_a^{\circ \normalsize (n+2)}\left({}^{\left(\normalsize (n+2)+x+\mu_b(a)\right)} b\right)} \\
& = & {\Large log_a^{\circ \normalsize (n+2)}\left(b^{\left(b^{\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)}\right)} \\
& = & {\Large log_a^{\circ \normalsize (n+2)}\left(a^{log_a(b)\times \left(a^{log_a(b) \times \left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)}\right)} \\
& = & {\Large log_a^{\circ \normalsize (n+1)}\left(log_a(b)\ +\ a^{\left(log_a(b) \times a^{\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)}\right)} \\
& = & {\Large log_a^{\circ \normalsize (n+1)}\left(\left(a^{\left(log_a(b) \times a^{\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)}\right)\ \times\ (1+\epsilon_2)\right)} \\
& = & {\Large log_a^{\circ n}\left(\left(log_a(b)\ +\ a^{\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)}\right)\ +\ log_a(1+\epsilon_2)\right)} \\
& = & {\Large log_a^{\circ n}\left(\left(\left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)\ \times (1+\epsilon_1)\right)\ +\ \epsilon_{2'}\right)} \\
& = & {\Large log_a^{\circ n} \left({}^{\left(\normalsize n+x+\mu_b(a)\right)} b\right)\ +\ \epsilon_{\small 1'}},\text{ where }\epsilon_{\small 1'} \approx \epsilon_1 (1+log_b(a)\epsilon_{2'})
\end{eqnarray}
\)
Notice that \( \epsilon_{2'} \ll \epsilon_1 \), thus supporting the claim that this superlogarithmic constant converges superexponentially. For base e, once you've found it within a factor of 1 part in 1000, the next iteration will get you accuracy of 1 part in e^1000. The next will get you within e^(e^1000). Of course, machine precision will necessarily cut your ascent off pretty early.
