Rank-Wise Approximations of Hyper-Operations

[JmsNxn]

This implicit solution exists uniquely across iterations. But if you ask for a TETRATION solution, it's not enough to declare uniqueness. Even while moving your base value, it's not enough.

There are infinite TETRATION solutions to these equations. There are infinite \(\text{Tet}(s)\) which satisfy \(\text{Tet}(0) = 1\). All you have to do is find countable \(z\) in \(\exp^{\circ s}(z)\) which orbits eventually hit \(1\). Honestly this is an artifact of the exponential function; and home to the iteration of transcendental functions.

This formula converges as your describing. But there's little to no general uniqueness. There are countably infinite solutions, and just because an algorithm evaluates to something, that doesn't qualify as a uniqueness condition. I can design a wrench for your algorithm which makes everything converge different. But it doesn't dissway the general idea.

I said tetration has to be unique.
You are saying using different fixed points to iterate off of would produce different tetrations. Then maybe it would also happen with higher hyper-operations. If tetrations of a converge, then a^^^x as x approaches infinity approaches a fixed point. And a^^^x as x approaches negative infinity approaches a different fixed point. Maybe you could do polynomial interpolation (Having points to interpolate on the left and right of the y-axis.) on the integer pentations, and higher hyper-operations with two fixed points approachable using real iteration.
For bases not in the Shell-Thron region, maybe you could analytically continue the hyper-operations there.

Oh it definitely would happen for all the hyperoperations. I'll do you one better. It's different for every function \(f\).

If \(f\) has a fixed point \(A\) and a fixed point \(B\) and assume there is a path connecting them \(\gamma\). Then \(f^{\circ s}\) can't be analytic on \(\gamma\). It has to pass the Julia set (of either \(f\) or \(f^{-1}\)), and it'll diverge.

So if \(f^{\circ s}(z)=\exp^{\circ s}(z)\) is holomorphic about \(z \approx L\), it cannot be holomorphic in the neighborhood of any other fixed point/periodic point. This throws a huge wrench in the gears--especially because the orbits of \(\exp\) are dense in \(\mathbb{C}\), and therefore arbitrarily close to a periodic point. The only solution to this, is to admit singularities/branchcuts and the such into the discussion.

So for example, when you write \(\exp^{\circ s}(z)\) using Kneser's construction we have the function \(\text{slog}(z)\) which is holomorphic for all \(z\) not a periodic point (incl fixed points), with singularities of some kind at these points. And Kneser's tetration \(\text{tet}_K(z)\) which is holomorphic for \(z \not \in (-\infty,-2]\). Then \(\exp^{\circ s}(z) = \text{tet}_K(s+\text{slog}(z))\), which has to account for the branching/singularities in both variables. And produces a whole bunch of nonsense.

One thing though, is that it is only analytic at the fixed point pair \(L,L^*\), and not analytic at any other fixed point/periodic point; but this is only true if \(\Im(s) > 0\) or \(\Im(s) < 0\). So again, you can't trace a path as I wrote above interweaving the two domains. In fact \(\exp^{\circ s}(L)\) will heavily depend on the value of \(s\), as \(\text{slog}(L)\) looks like \(-\infty\). I don't know too much about the dynamics, of this tbh, I might have mixed something up. But this is what I recall.