05/04/2009, 01:06 AM
(This post was last modified: 05/04/2009, 01:51 AM by Base-Acid Tetration.)

(05/03/2009, 10:27 PM)bo198214 Wrote:(05/03/2009, 09:45 PM)Tetratophile Wrote: If one defined the It operator for functions in the second argument the definition would not be a "recursive" definition?What do you mean by "true for all functions"? That x is an arbitrary function?

then say that the functional equations are true for the It operator for all functions, but the equations are not "recursive". ok.

Yes. x can, in principle, be any function*; interpreting the "iterants" (the n in f [It_k] n) as constant functions rather than as simply numbers is what makes the hierarchy possible (since a function outputs numbers we can use as iterant). I should have written the iterant as g(x) to emphasize that.

The order of the operations is that It_k for bigger k are performed first, and, if the same operator is nested, perform the innermost one first; top down, inside out.

*as long as everything is defined: f(x), g(x), some real x, some natural n, f(x) It_(n+1) g(x), f(x) It_(n) [f(x) It_(n+1) g(x)-1]