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If we iterate + (addition), we get * (multiplication).
If we iterate * (multiplication), we get ^ (exponentiation).
If we iterate ^ (exponentiation), we get ^^ (tetration).
... etc.
If we iterate o (composition), we get ☉ f(x) ☥^N (steinix-ankh).

BUT
The uniteration of ^^ (tetration) is ^ (exponentiation).
The uniteration of ^ (exponentition) is * (multiplication).
The uniteration of * (multiplication) is + (addition).
... etc.
If we uniterate ☉ f(x) ☥^N (steinix-ankh), we get o (composition).

The question is that: What is the uniteration of the composition?
What ??

If you are asking about the fractional superfunctions , that is a known open problem.

Otherwise I do not understand what you said.

Regards

Tommy1729
I get what are you saying.
It is a very mysterious problem but at the same time is a bad defined problem (as you pose it): it means that it can be formalized in different ways leading to different difficulties.

Let's analyze your question:
"It is possible to "uniterate" the composition operator?"
In order to make this clear we have to define what "uniterate" means.
And to do this we need to understand what "iterate" means.

But first let see that the analogy between hyperoperators and composition can be misleading in the sense that composition doesn't fit well in that analogy scheme
See:
Quote:-We iterate ADDITION $+:{\mathbb N}\times{\mathbb N}\to{\mathbb N}$ (a function over the naturals) and we obtain MULTIPLICATION $\cdot:{\mathbb N}\times{\mathbb N}\to{\mathbb N}$
-We iterate MULTIPLICATION $\cdot:{\mathbb N}\times{\mathbb N}\to{\mathbb N}$ (a function over the naturals) and we obtain EXPONENTIATION $\wedge:{\mathbb N}\times{\mathbb N}\to{\mathbb N}$
So the simplest general scheme we can find is that we start with a function over the natural numbers and we get another over the natural numbers.

Quote:iteration's definition attempt 1: Given a function $*:{\mathbb N}\times{\mathbb N}\to{\mathbb N}$ we obtain its iterate as a new binary function $*'$ over the natural numbers
$m*'1:=m$
$m*'(n+1):=m* (m*'n)$
So we have that $+\mapsto (+'=\cdot) \mapsto (\cdot'=\wedge)\mapsto \uparrow^2\mapsto ...\mapsto \uparrow^n\mapsto \uparrow^{n+1}\mapsto ...$

NOTE: with this definition we have that $m*'2:=m* m$, $m*'3:=m* (m*m)$ and so on, and we also have (up some fine details) all the hyperoperation sequence starting from the addition or the successor.

BUT

composition doesn't fit into this perfectly because composition is not an operator that takes argument and values on the naturals. Composition takes functions and gives functions so we need a broader concept of iteration if we want to apply it on the composition:
Quote:iteration's definition attempt 2: Given a generic function $f: X \times X\to X$ over a set $X$ we obtain its iterate as a new binary function, we call it $F$, that takes two different arguments: a value in $x\in X$ and a natural number so it is a function $F:X\times {\mathbb N}\to X$ over $X$
$F(x,1):=x$
$F(x,n+1):=f(x,F(x,n))$
NOTE: note that an infix operator a*b is the same as a binary function f(a,b). Seeing this makes the analogy between the first and second definition striking.

At this point we are free to iterate composition and get what is called "function iteration" or just iteration: just take $X$ to be a set of function that is "composable" (a monoid of functions or a category) and take as binary operator $\circ(f,g)=f\circ g$ that is the well known function composition. Then its iteration is the function $F(f,n)=f^{\circ n}$, that is, a binary function $F:F:X\times {\mathbb N}\to X$ defined as
$f^{\circ 0}:={\rm id}$
$f^{\circ n+1}:=f\circ f^{\circ n}$
NOTE: this is the classical recursive definition of iteration.

NOW WE COME AT YOUR QUESTION! It is possible to reverse this process?

Iteration: ${\rm function}\rightarrow {\rm iterate}$
Uniteration: ${\rm ???}\leftarrow {\rm function}$

We notice something important: with iteration we start with a generic function and we end with another function (its iterate) that has a natural number argument.

$f(x,y)\rightarrow {\rm iteration} \rightarrow F(x,n)$

So iteration goes from the set $X^{X\times X}$ (the set of binary function over X) to the set $X^{X\times {\mathbb N}}$ (of functions over X that takes a value in X and one in N)

${\rm iteration} :X^{X\times X}\to X^{X\times {\mathbb N}}$

It is reasonable that "uniteration", whenever we define it, should go in the opposite direction
${\rm uniteration} :X^{X\times {\mathbb N}}\to X^{X\times X}$

Partial conclusion
The story here gets quite long and tricky so i'll go for the simplest and shortest answer and avoid all the technicalities: the composition is not in the domain of the "uniteration" because it has not natural numbers as argument, in other words it is not the iteration of nothing, at least not in the sense of my two definitions above.

to understand this better just try to imagine what this function should satisfy: it need to be a hypothetical function $f \diamond g$ such that

$f\diamond g\rightarrow {\rm iteration} \rightarrow f\circ g$

but we already know that the iteration of such operation MUST take arguments in the natural numbers too! But COMPOSITION DOES NOT!

---
Said that alot of things remain obscure.
First of all is not sure that the pair "iteration-uniteration" forms a pair of inverse functions and the details to fix are really many. In some interpretation the one is the inverse of the other in the sense that going Back and forth doesn't generally takes us to the same place but to the same place up to some notion of equivalence that I will not make precise.

In many context iteration fails to be uniquely defined and is multivalued, in many other context also the "uniteration" fails to be single valued too!!! SO both can be multivalued and so they just are relation and not functions anymore...so it gets weird to talk about iterating these an integer number of times (let alone talking about fractional-complex ranks).

The story is very long and i could talk of alot more but there are many IF out there.