05/03/2009, 01:08 PM
EDIT 5/2/09: For the hierarchy of iterations, or hyper-iterations, we can use the operator:
\( \operatorname{It}_k: \lbrace f: \mathbb{R} \rightarrow \mathbb{R} \rbrace \times \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace \rightarrow \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace, \) where It_0 is function composition (g It_0 f = g(f)). All of the operators has the property \( f \operatorname{It}_n x+1 = f \operatorname{It}_{n-1} (f \operatorname{It}_n x). \) So unless otherwise specified by the use of the down arrow like this, \( f {\operatorname{It}\downarrow}_k , \) super-iteration, like other hyper-iterations, is evaluated from the top down, like ordinary tetration, not from the bottom up as I said originally. I am sorry for any confusion the mistake in the original post may have caused.
ANOTHER EDIT 5/2/09: Ictually I am not even sure whether iteration is not associative, but i'll assume it is not unless proven otherwise.
\( \operatorname{It}_k: \lbrace f: \mathbb{R} \rightarrow \mathbb{R} \rbrace \times \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace \rightarrow \lbrace f:\mathbb{R} \rightarrow \mathbb{R} \rbrace, \) where It_0 is function composition (g It_0 f = g(f)). All of the operators has the property \( f \operatorname{It}_n x+1 = f \operatorname{It}_{n-1} (f \operatorname{It}_n x). \) So unless otherwise specified by the use of the down arrow like this, \( f {\operatorname{It}\downarrow}_k , \) super-iteration, like other hyper-iterations, is evaluated from the top down, like ordinary tetration, not from the bottom up as I said originally. I am sorry for any confusion the mistake in the original post may have caused.
ANOTHER EDIT 5/2/09: Ictually I am not even sure whether iteration is not associative, but i'll assume it is not unless proven otherwise.
