Uniqueness

[Daniel]

Once again, thanks for your reply Henryk.

The technique I have developed for extending tetration is based on fixed points,

So what is your technique? As far as I remember you used always regular iteration at a real fixed point. And regular iteration at a complex fixed point leads to comlex values for real arguments. Which I would exclude from feasible solutions for tetrationals.

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. Rather, I consider the constraining of complex values for real arguments as a potential axiom that is worthy of exploration for extending tetration. Hopefully, different approaches to extending tetration that use the same axioms will produce consistent results.

Say you have an extension for tetration. What is the value of \( ^{-1} a \)? It is usually assumed to be 0, but it can actually be \( 2 \pi k \) where k is an integer; it has an infinite number of values.

…
Same with a tetrational. It has singularities at each negative integer number below or equal -2. So if you start with a powerseries development at 0 you can continue to any non-singular point along every path not hitting a singular point. So you have again infintely many values at every non-singular point depending how the path to that point revolves around the singularities.

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. This is a weakness of my technique in that it doesn’t provide a way to compute tetrationals that contain singularities. My guess is that these singularities have lead people to extend tetration using superlogarithms instead of superexponentiation.

I wonder if it isn’t true that any holomorphic function agreeing with the values of \( ^k a \), where k is a natural number, comes arbitrarily close to one of the infinite family of tetration solutions.

… (Note also that for a tetrational I would have the stronger condition \( f(x+1)=a^{f(x)} \) for all real/complex \( x \) instead of just for integers \( x=k \)).

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