Functional super-iteration and hierarchy of functional hyper-iterations

[Base-Acid Tetration]

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.

Your operator was defined with my operator. Here is the definition that you gave:

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

From that follows
\( (f \operatorname{It}_0 g)(x) = (I_0^{g(x)} f)(x) = f(g(x)) \)
\( I_1^m f = f^{\circ m} \)
\( (f \operatorname{It}_1 g)(x) = (I_1^{g(x)} f)(x) = f^{\circ g(x)}(x) \)

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

Your I operator is basically the same as my It operator; the only differnce is the notation. Can you tell me exactly what is different about your definition of the hyper-iteration than mine?

[f It_n g](x) := [f It_n g(x)](x) - Let me clarify what it means. For any n, theoretically the procedure is
1. Evaluate g(x) at c first.
2. Hyper-n-iterate f to the OUTPUT of Step 1.
3. Evaluate the resulting function at c.

For n=1 (iteration), to evaluate this expression at any given natural x=c:
1. Evaluate g(x) at c first.
2. Iterate f to the OUTPUT of Step 1.
3. Evaluate the resulting function at c.

If you substitute g for f, it becomes: evaluate f(x) at c, then iterate f to the result, evaluate the resulting function at c. This is what (f^f)[x] f It_2 2 = \( (I_n ^2) f(x) \) is.

For It_2, at any natural x=c, the procedure is:
1. Evaluate g(x) at c.
2. Evaluate f(x) at c.
3. Perform the steps to calculate [f It_n f©]© g© (the OUTPUT of g(x) at c) times:
3a. Iterate f to the OUTPUT of 2.
3b. Evaluate the resulting function, at x. Make the result the input for 3a.
4. Evaluate the result of Step 2 at x.
At least for integer values of g(x), your computer program could evaluate g(x), and then iterate f to the OUTPUT of the function g(x).

ps How do you do © so that it comes out as ( c ) without spaces instead of a copyright symbol?

This can be extended to the higher hyper-operators by repeating the steps at It_n-1 g(x) times for It n.
So we both agree that it is not the FUNCTION per se, but the OUTPUT to which f is hyperiterated.