Has this tetration been considered before in the forum?

[JmsNxn]

If two functions \(F,G\) are holomorphic and bounded for \(\Re(s) > X\) and interpolate the same values, \(F = G\).

So that solution is Schroeder if holomorphic. Seems you are suggesting it's easy to see it is holomorphic aswell right?

Ya, it's not especially hard to see it's holomorphic. I mean, I'm not going to show it, so let's just stick to, if it is holomorphic it's Schroder, lol. But Simply Beautiful Art is a very reliable poster on Stack Exchange, and the argument makes a lot of sense. Essentially we are showing that:

\[
^n a - ^\infty a = \mathcal{O}(\lambda^n)\\
\]

For \(\lambda = \log(^\infty a)\) the multiplier, which means it has nice geometric decay. So as \(^\infty a + (^n a - ^\infty a) \lambda^x \to \,^\infty a\) like \(\lambda^n\) (geometrically with multiplier \(\lambda\)), the \(\log^{\circ n}\) part diverges like \(\lambda^{-n}\) (shoots us away from the fixed point geometrically like \(\lambda^{-n}\)).

This is essentially just an expression for the inverse schroder equation \(^z a = \Psi^{-1}(\lambda^z)\), nothing too fancy. It's just a really funky way of writing it.

It's definitely holomorphic though; especially if it's analytic on the real line, simply by the induced period, and the fact the Schroder function is unique in a neighborhood of the fixed point.


Just to strengthen the argument, remember that:

\[
\begin{align}

\Psi^{-1}(z) &=\, ^\infty a +z +...\\
\Psi^{-1}(\lambda^z) &= \,^\infty a + \lambda^z+...\\
\log\Psi^{-1}(\lambda^z) &= \log(^\infty a + \lambda^z + ...) =\, ^\infty a + \lambda^{z-1}+...\\
\Psi^{-1} (\lambda^{z+n}) &=\, ^\infty a + \lambda^{z+n}+...\\
&=\, ^\infty a + \left(^n a - ^\infty a\right) \lambda^{z}+...\,\,\text{per the asymptotic above}\\
\log^{\circ n}\left(^\infty a + \left(^n a - ^\infty a\right) \lambda^{z}\right) &\approx \Psi^{-1}(\lambda^z)\\
\end{align}
\]

SimplyBeautifulArt is basically just using/abusing this asymptotic relationship to create a very accurate heuristic. I see no reason it wouldn't work in the complex plane though. I may have miffed a couple of the convergence constants, but the basic idea is there, lol.