Andrew Robbins' Tetration Extension

[andydude]

Indeed, that would be strange. I'm not assuming convergence, but I am assuming that:

• convergence of the finite series (approximations with coefficient approximations as well) to the infinite series, or in other words: \( \lim_{n\rightarrow\infty} \text{slog}_b(z)_n \) where \( \text{slog}_b(z)_n = \sum_{k=0}^{n}s_{nk}z^k \) or in other words: \( \lim_{n\rightarrow\infty}\sum_{k=0}^{n} s_{n k}z^k \), and
  
• convergence of the infinite series (whose coefficients must also converge), or in other words: \( \lim_{n\rightarrow\infty}\sum_{k=0}^{n} s_{\infty k}z^k \)
  

are equivalent, but this may be an error as well.

What I can prove is that for the nth approximation of the super-log (the finite series):

• As \( b \rightarrow 1 \), \( s_{nk} \rightarrow (-1, 0, 0, \cdots, 0, 0, 1) \) with n+1 entries.
  
• As \( b \rightarrow \infty \), \( s_{nk} \rightarrow (-1)^{k-1} \left({n \atop k}\right) \) with k = 0 .. n.
  

From this it is easy to show \( \sum_{k=0}^{n} (-1)^{k-1} \left({n \atop k}\right) z^k = -(1-z)^n \)

Andrew Robbins