Uniqueness

[bo198214]

Yes, you are correct in how my technique works. I do disagree with your a priori exclusion of my approach as not being a feasible solution.

Yes of course not a priori. If we had no methods for real analytic tetrationals we would would of course fall back to other methods that are still good, that is they have complex values on the real axis, or are not analytic on the real axis.

But we have even 5 methods for a real analytic tetrational, i.e. a function that hopefully satisfies:
* real on the real axis (>2) (on one branch)
* strictly increasing on the real axis
* holomorphic on the whole complex plane with exception of \( z\le -2 \).
* satisifying \( f(z+1)=\exp_b(f(z)) \) and \( f(0)=1 \) for complex all \( z \) (though one has to be careful with branching, so this maybe true only when considering \( f \) as a global holomorphic function).

The \( log_a(log_a(1)) \) used to define a tetrational for –2 is infinitely multi-valued. Yes, if you restrict complex values from real arguments, then you just obtain singularities.

Well this possible non-singularity at -2 is an interesting phenomenon when considering tetration as a global function. It means that only on one (or some) branch(es) of the global function there is a singularity at -2. If the path to the point -2 winds differently around the other singularities, then there may be a continuation to -2 (yielding no singularity there). Thats really an intriguing topological manifold where the singularities change depending on which branch we reside. Where on the other hand the branches depend on the singularities, as they come to existence by different winding around the (other) singularities.

My guess is that these singularities have lead people to extend tetration using superlogarithms instead of superexponentiation.

Not that I would know of. The superlogarithm to base \( b \) corresponds to the Abel function of \( b^x \). Its just in the theory of regular fractional iterates that the Abel function plays a bigger role and is easier usable and constructable than its inverse, the superexponentiation in this case.

Yes, I agree. I show this by demonstrating \( f^a(f^b(z))-f^{a+b}(z)=O(n) \) in Mathematica.

Ya but that is already done without Mathematica. Which is necessary as it is no proof for arbitrary \( n \), but you can only verify it for some given big \( n \). However there is a thorough mathematical base of regular iteration, this is for example presented in the book:

Jean Écalle, Théorie des invariants holomorphes, Publ. Math. Orsay No. 67–7409, 1974

and also to some extent in the book:

Kuczma, Iterative functional equations, 1990

And it is the definition of fractional iterates that they satisfy
\( f^1=f \)
\( f^{s+t}=f^{s}\circ f^{t} \)

and regular fractional iterates are shown to be fractional iterates.