??? D^2 tetration ??? tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 11/11/2009, 11:58 PM 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 Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 11/13/2009, 12:02 AM 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). « Next Oldest | Next Newest »