(05/04/2009, 01:06 AM)Tetratophile Wrote: 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.
But anyway its no *definition*.
See have your original equation:
\( f \operatorname{It}_n x+1 = f \operatorname{It}_{n-1} (f \operatorname{It}_n x). \)
and lets add the initial condition:
\( f \operatorname{It}_n 1 = f \)
Interpreting x as constant function, with the above lines we can derive the function
\( f \operatorname{It}_n x \) for any constant function \( x \).
And that is all! We can no derive what \( f \operatorname{It}_n g \) means for any non-constant function \( g \). Just because on the left side there are only constant functions in the second argument.
Well if we allow any function \( g \) for \( x \), not only constant functions, then it is still no definition, because we have no initial condition, which stops the recursion.
For a constant function \( m \), the right side needs to evaluate \( m-1 \) in the second argument, to evaluate this it must be evaluated at \( m-2 \) and so on until one derives at m=1. This the initial condition and the recursion is finished.
If you put however any function \( g \) there then on the right side in the second argument \( g-1 \) needs to be evaluated then \( g-2 \) then \( g-3 \) and so on but this recursion does never stop because \( g-k \) never becomes a constant function for which we know how to evaluate.
Think of writing a computer program to evaluate your operator. It can not guess what your intention was for non-constant second arguments. But I can guess and thatswhy I proposed the definition I gave.
