Andrew Robbins' Tetration Extension

[andydude]

So what do you actually prove?

That:
\( \lim_{b \rightarrow 1} \text{slog}_b(z)_n = z^n - 1 \) (expanded about z=0) and:
\( \lim_{b \rightarrow \infty} \text{slog}_b(z)_n = -(1-z)^n \) (expanded about z=0),
which means in the limit:
\( \text{slog}_1(z) = \lim_{n \rightarrow \infty} z^n - 1 = \begin{cases}
-1 & \text{ if } 0 \le z < 1 \\
0 & \text{ if } z = 1
\end{cases} \)
and:
\( \text{slog}_{\infty}(z) = \lim_{n \rightarrow \infty} -(1-z)^n = \begin{cases}
-1 & \text{ if } z = 0 \\
0 & \text{ if } 0 < z \le 1
\end{cases} \)

One of the nice things about having an exact form for all approximations is that the approximations are invertible (so we can find tetration as well) but the limit to infinity makes it discontinuous, which is not invertible.

Andrew Robbins