The \(\varphi\) method of semi operators, the first half of my research

[tommy1729]

So, I've shifted my research into calculating:

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

On a more maximal domain. By this we approach errors at around \(|\log(y)|=1\), but otherwise the functions behave cleanly.

This is a large domain in \(y\), where the errors, white outs and glitches, are precisely near \(|\log(y)|=1\). This goes hand in hand with the manner I have programmed these constructions. So many of the errors near here are human error on the manner of code, and not a mathematical error. Many more detailed tests have taught me that \(x [s] y\) is holomorphic for \(\Re(y) > 1\) when \(x > e\) and \(0 \le s \le 2\).

To exemplify this, I will use some graphs, they are not proof--just descriptions of the way it's working.



This is a graph over \(\Re(y) > 0\), done largely.

Zooming in on that error which represents the values \(|\log(y)| = 1\) we get a closer picture. Here \(\Re(y)>1\) and the white of this graph is error code. It does not represent the analytic function. It represents an error on my part within the coding.




By this I want to say that \(x[s]y\) is holomorphic on much larger domains than originally thought.

what exactly are those pictures ??

There are 3 complex variables x,s,y , so I do not see how you can map those into  1 picture.

***

another thing is to add a restriction on the idea of superfunctions.

that could really make the ultimate (extra) uniqueness criterion.

I mean the condition f^[s+t](z) = f^[s](f^[t](z)) = f^[t](f^[s](z)) as you might have guessed.

regards

tommy1729