Holomorphic semi operators, using the beta method

[JmsNxn]

Hey everyone! Some more info dumps!

I haven't talked too much about holomorphic semi-operators for a long time. For this brief exposition I'm going to denote the following:

\[
\begin{align}
x\,<0>\,y &= x+y\\
x\,<1>\,y &= x\cdot y\\
x\,<2>\,y &= x^y\\
\end{align}
\]

Where we have the identity: \(x<k>(x<k+1>y) = x<k+1> y+1\). Good ol fashioned hyper-operators.

...

First note : 

In my notebook - and maybe posted here too - i found the identity: \((x<k+1>y) <k>x= x<k+1> y+1\) is consistant with
  \[
\begin{align}
x\,<0>\,y &= x+y\\
x\,<1>\,y &= x\cdot y\\
x\,<2>\,y &= x^y\\
\end{align}
\]

This seems a nicer choice or not ?

Why not this ? because it is slower ?

Second note : Im going to ignore holomorphic for now because I do not believe that. Might explain later ( more time ) and maybe already did in the past.

Third note :

Which fractional iteration for exp and log ?? There are many and they do not agree on the 2 real fixpoints or ( in case of base e^(1/e) a single real fixpoint that is not analytic ! )
These problems and choices are not simultaniously adressed , picked and motivated.

 Fourth note : why non-commutative ? 

Fifth note : 

you basicly are looking for a function f_1(a,b,s) and " find " the solution f_2(a ,b ,s , f_3(a,b,s) ) where f_3(a,b,s ) is unknown , undefined and unproven analytic.

that feels like solving the quintic polynomial as exp(a_0) + f(a_0,a_2,a_3,a_4,a_5) for some unknown f ...

forgive my parody.   

I could continue but I respect you

regards

tommy1729

1) I don't want left associative, who wants left associative....

...

Like I commented , your original equation mentioned IS left associative.

Haha, sorry late at night. Meant to say do not want right associative. I apologize, was really tired when I wrote that.

What I meant to say, is your equation will not generate tetration, it'll generate the lower tetration. I don't want the lower tetration. I still want \(x <3> y\) to be tetration, but I'm not going to go that far out. And in order for that to happen you must satisfy goodstein's equation and not the equation you wrote.

Sorry, I apologize. Sometimes I stay up too late and mix up my words.

To clarify, YES! We are looking for solutions to this equation:

\[
x<s> \left(x <s+1> y\right) = x <s+1> (y+1)\\
\]

For natural numbers you'll note that this will generate tetration for \(x<3>y\) (upto a normalization constant) and \(x <4> y\) will be pentation, so on and so forth. I don't want to go that far out though, I only care for \(0 \le \Re(s) \le 2\).

Also, since you don't like the idea of local holomorphy; we can instead refer to \(x,y > e\) and \(0 \le s \le 2\) and write:

\[
x <s>_\varphi y = \exp^{\circ s}_{y^{1/y}}\left(\log^{\circ s}_{y^{1/y}}(x) + y + \varphi\right)
\]

Which is analytic so long as \(\varphi > - e\) (at least). And the iteration we choose is the Schroder iteration about the repelling fixed point of \(b = y^{1/y}\) (so about \(y\) since \(y > e\)). Now all the discussion of surfaces is reduced to a discussion of surfaces in \(\mathbb{R}^3\), and we don't have to worry about Riemann surface stuff. This makes the problem much more manageable.