Functional super-iteration and hierarchy of functional hyper-iterations

[bo198214]

For the function x -> x+1, you simply hyper-n-iterate the function f x0+1 times at x=x0, (eg. at x=2 you iterate 3 times, and at x=3.5 you iterate 4.5 times etc.) So I don't think that a non-constant function in the second argument would behave differently from a constant function in this respect.

If we take the equations as recursive definitions then \( \operatorname{It} \) is only defined for functions in the second argument that are constant. Thatswhy I proposed the different version.

Also \( I_0^m f = f \) would not be true, because at n=0, \( I^m_0 f = f \operatorname{It}_0 m = f(m), \mbox{ not }f. \)

Ok, I also made another mistake in the previous posting, so here again the recursive definition

\( (f \operatorname{It}_n g)(x) := (I_n^{g(x)} f)(x) \).
\( (I_0^m f)(x) := f(m) \)
\( I_n^1 f :=f \) and \( I_{n}^{m+1} f := f \operatorname{It}_{n-1} (I_n^m f). \)

From that follows
\( (f \operatorname{It}_0 g)(x) = (I_0^{g(x)} f)(x) = f(g(x)) \)

\( I_1^m f = f^{\circ m} \)
\( (f \operatorname{It}_1 g)(x) = (I_1^{g(x)} f)(x) = f^{\circ g(x)}(x) \)

\( I_2^m f = {^{\circ m} f} \)
\( (f \operatorname{It}_2 g)(x) = {^{\circ g(x)} f}(x) \)
....