11/26/2007, 11:25 AM
You can just solve for n explicitly as \( n = \lfloor \text{slog}_b(e)\rfloor - 1 \), since you can write your inequality that defines n as
\( \exp_b^{-2}(e) < \exp_b^{n}(1) \le \exp_b^{-1}(e) \)
\( \exp_b^{-1}(e) < \exp_b^{(n+1)}(1) \le \exp_b^{0}(e) \)
or \( {}^{(n+1)}b = \exp_b^{(n+1)}(1) = \exp_b^a(e) \) where \( -1 < a \le 0 \). Taking the super-logarithm we get \( n+1 = \text{slog}_b(\exp_b^a(e)) \) which is a decreasing function of b, so if we take the floor: \( n+1 = \lfloor \text{slog}_b(\exp_b^0(e)) \rfloor = = \lfloor \text{slog}_b(e) \rfloor \) so \( n = \lfloor \text{slog}_b(e)\rfloor - 1 \).
It is interesting to note that both the floor of slog and the ceiling of slog can be defined independently of the function itself. This means if you do use this to define n then there is no chance of circular definitions
Andrew Robbins
\( \exp_b^{-2}(e) < \exp_b^{n}(1) \le \exp_b^{-1}(e) \)
\( \exp_b^{-1}(e) < \exp_b^{(n+1)}(1) \le \exp_b^{0}(e) \)
or \( {}^{(n+1)}b = \exp_b^{(n+1)}(1) = \exp_b^a(e) \) where \( -1 < a \le 0 \). Taking the super-logarithm we get \( n+1 = \text{slog}_b(\exp_b^a(e)) \) which is a decreasing function of b, so if we take the floor: \( n+1 = \lfloor \text{slog}_b(\exp_b^0(e)) \rfloor = = \lfloor \text{slog}_b(e) \rfloor \) so \( n = \lfloor \text{slog}_b(e)\rfloor - 1 \).
It is interesting to note that both the floor of slog and the ceiling of slog can be defined independently of the function itself. This means if you do use this to define n then there is no chance of circular definitions
Andrew Robbins
