08/13/2007, 07:31 PM
andydude Wrote:Let me make a short unification. The slog is simply a solution to the Abel equation for \( F(x)=b^x \):\( {}^{x + slog_b(a)}{b} = \exp_b^{[x]}(a) \)
\( f(F(x))=f(x)+1 \)
which is merely our initial condition written for \( f=\text{slog}_b \) instead of \( f^{-1}(x)={}^xb \).
It is well-known that this corresponds to a fractional iteration via
\( F^{\circ t}(x)=f^{-1}(f(x)+t) \)
Moreover that any fractional iteration must be of this form.
