Change of base formula for Tetration

[jaydfox]

I believe that you have a lot of good ideas, but actually I can not quite follow your explanations. Most of your formulas comes without any justifications nor proofs.

That your formula

\( {}^x a = \lim_{n \to \infty} \log_a^{\circ n}\left({}^{\left( n+x+\mu_b(a)\right)} b\right) \)

yields a proper tetration the resulting \( {}^xa \) must at least satisfy the identities

\( ^{x+1}a=a^{{}^xa} \) and \( {}^1 a=a \)

if \( {}^xb \) satisfies them. Can you verify this first?

Sorry, I thought it was implicit in the formula, but I've been looking at this for several days, so perhaps I just took it for granted:

\( \begin{eqnarray}
{}^x a & = & \lim_{n \to \infty} \log_a^{\circ n}\left({}^{\left( n+x+\mu_b(a)\right)} b\right) \\
\\[15pt]

\\
{}^{x+1} a & = & \lim_{n \to \infty} \log_a^{\circ n}\left({}^{\left( n+(x+1)+\mu_b(a)\right)} b\right) \\
& = & \lim_{n \to \infty} \log_a^{\circ n}\left({}^{\left( (n+1)+x+\mu_b(a)\right)} b\right) \\
& = & \lim_{n \to \infty} {\Large a}^{\left[ \log_a^{\circ (n+1)}\left({}^{\left( (n+1)+x+\mu_b(a)\right)} b \right) \right]} \\
& = & \lim_{n \to \infty} {\Large a}^{\left[ \log_a^{\circ (n-1)}\left( \log_a \left({}^{\left( (n+x+\mu_b(a)\right)} b \right)+\epsilon_1 \right) \right]} \\
& \approx & \lim_{n \to \infty} {\Large a}^{\left[ \log_a^{\circ n}\left({}^{\left( (n+x+\mu_b(a)\right)} b \right) \right]} \\
& \approx & \lim_{n \to \infty} {\Large a}^{\left[{}^{x} a\right]} \\
\end{eqnarray}
\)

The base case \( {}^1 a=a \) is guaranteed by the definition of \( \mu_b(a) \). If you run the formula and you don't get \( {}^1 a=a \), then your constant \( \mu_b(a) \) needs to be adjusted.