??? D^2 tetration ???

[tommy1729]

could it be true that by andrew robbins method :

f(x) = 2^f(x-1) f(0) = 1 ( coo base 2 tetration thus )

g(x) = df^2/dx^2 f(x)

g(0) = 0 ???


regards

tommy1729

[andydude]

By nth apprimation to intuitive tetration (base 2):

Code:

n     sexp_2'(0) = -slog_2''(1)/slog_2'(1)^3
===============
5     0.0108519
10    0.0171036
15    0.0173285
20    0.0173507
25    0.0173537
30    0.0173539

So it seems as though this sequence converges to something other than zero.
If you're looking for a base with "nice" properties, I would recommend base (3.0885325).

What makes base (3.0885325) special is that it is the maximizer of \( ([4]x)^{-1}(x) \), and maximum of \( x[5]\infty \) over the interval (?, 1.6353245). I think this is analogous to how base (e) is the maximizer of \( ([3]x)^{-1}(x)=\sqrt[x]{x} \) and the maximum of \( x[4]\infty \) over the interval (0.065988, 1.44467).