Uniqueness

[bo198214]

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.

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.

Yes, but you know if you consider the global behaviour of holomorphic functions, they can be many valued, actually infinitely many valued. Best example is the logarithm. It has one powerseries development at 1, but you can continue it along any path not crossing 0 and you get infinitely many values for one point, depending on how often the path to the point revolves around 0.

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.

Then \( ^{-2} a \) also has an infinite number of values \( \omega^2 = \omega \). It seems to me that this leads to \( ^{-\omega} a \) having as many values as the continuum has points.

I am reluctant to discuss this before there is not a proper definition and meaning of what is meant by \( ^{-\omega} a \). Either we work with complex numbers or we work with cardinalities, or we work with ordinalities. But I never heard about a number system that unites complex numbers with cardinalities or ordinalities. The only thing about infinity that is allowed with the complex numbers is to add an infinity which makes the complex plane into a complex sphere. But this infinity is not a cardinality and not an ordinality.

This results in \( ^n a \) consisting on an \( \aleph _1 \) family of solutions.

Yes and no. Seeing this in the context of global holomorphic functions: For any point you have a path to this point and this path winds finitely many times around each singularity. For (countably) infinitely many singularities we get \( \aleph_0^{\aleph_0}=\aleph_1 \) many possibilities for the path to wind around the singularities. And hence we can have possibly so many values for each point.

But dont forget that it is one global holomorphic function (many-valued). The whole function can be given by one power-series development for example at 0. And probably there is only one out of the many at 0, that is real analytic.

So what you are speaking about can be summarized as many-valuedness of one holomorphic function. Imho it makes no sense to diversify this into different holomorphic solutions on cutted domains. (If you see the logarithm as a global function you dont need the cut at the negative real axis. But if you put a cut somewhere than it looks whether you have infinitely many logarithms.)

but there are an infinite number of fixed points.

And this is a completely different uniqueness concern. This is about whether there are indeed different global functions. I.e. two real analytic powerseries developments at 0 which have somewhere different coefficients.

It is my understanding that there are \( \aleph _2 \) number of curves, but holomorphic functions are much more restricted.

Yes, I would agree; You get all real anlytic global functions (defined at 0) by going through all real powerseries developments at 0. And this is \( |\mathbb{R}|^{\aleph_0}=\aleph_1 \)

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.

This question perhaps becomes obsolete after distinguishing many-valuedness of one holomorphic function from different holomorphic tetrationals. (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 \( x=k \)).