(07/05/2022, 12:17 AM)Daniel Wrote:(07/05/2022, 12:06 AM)JmsNxn Wrote: I understand that, Daniel. I apologize if my response seemed hostile. Wasn't my intention, lol.
All I'm saying is that the Taylor series approach, is inherently Schroder's construction. I'm not doubting that it works in any way shape or form. But I suggest you observe your iteration \(f^{\circ 1/2}(z)\) of \(f = \sqrt{2}^z\), and trace \(z\) from \(2 \to 4\). Somewhere along that path there is a singularity (probably a tiny discontinuity/jump at about 1E-10 height). If not, you've done something wrong.
Hey, it's all good JmsNxn. I'm always happy to get thoughtful feedback, even if it is not what I have hoped for.
My construction not only encompasses the cases of Abel and Schroeder's functional equations. See my page on generating flows based on the Abel's case. I'd run your test but unfortunately I'm now without Mathematica for the first time in thirty years, so I guess I need to get good at GP-Pari.
Lol, ya!
Pari-gp is the ultimate tool. I only use mathematica to plot complicated graphs. Even that is difficult.
You can always sail the high seas and yarr me matey, acquire Mathematica by more nefarious means.
All I'm saying, is that mathematically, you cannot have an iteration holomorphic at \(2\) and at \(4\). This is detailed beautifully in Trappmann's and Kouznetsov's paper on iterated exponentials. It deals specifically with \(\sqrt{2}\). They define exactly 4 iteration types. and sadly, those are the only ones about \(2,4\). And all of them cannot be holomorphic at both fixed points. It's just a cruel joke that nature plays on us.
https://www.researchgate.net/profile/Hen...ion_detail
The only solution to this is to do something extravagant and fancy with super functions, then you can get holomorphy on a larger domain, but still, it is not holomorphic at both fixed points.