Holomorphic semi operators, using the beta method

[JmsNxn]

is x [s] f(x) = x 

a degree of freedom ?

can we freely choose f(x) and work from there ??

regards

tommy1729

It is a degree of freedom yes, but I would write it using \(\varphi\). So that:

\[
x\,[s]_{\varphi} f(x) = x\\
\]

Then there is some \(\varphi\) where this statement is true (given that \(f\) is reasonably well behaved).

proof ??

I mean such that x [s] y is analytic in x , y , s to be clear.

and yes f(x) analytic ofcourse.

regards

tommy1729

Sure no problem:

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

Take \(\log^{\circ s}_{f(x)^{1/f(x)}}\) on both sides, and cancel out the \(\log^{\circ s}_{f(x)^{1/f(x)}}(x)\): so that:

\[
f(x) + \varphi = 0\\
\]

And then writing

\[
\varphi(x) = -f(x)\\
\]

Gives you your answer.

So by construction:

\[
x [s]_{-f(x)} f(x) = x\\
\]

Not sure how that helps with anything, or what you are going to use it for. This also doesn't really help because you'd be asking for \(f(x) > e\), and I doubt that would really happen. There's no value \(f(x) > e\) such that \(x \langle s\rangle f(x) = x\). By nature of this looks like \(x + f(x) > x + e\) or in between \(x \cdot f(x) > x \cdot e\) which obviously can't equal \(x\). So these values would be beyond our purview.