We know that the exponential map is infinitely differentiable; is there a tetration map $TET$ defined on $(0,\infty)$ such that

1) $TET(x) > TET(y)$ when $x > y$

2) $TET(e^x) = TET(x) +1$

3) $TET(x) $ is infinitely differentiable on $(0,\infty)$

4) $TET(1) = 0$

5) (optional) $TET$ has a local taylor series at each point

Note here that $TET(e) = 1, TET(e^e) = 2, TET(e^{e^e}) = 3...$


A real analytic inverse to tetration was constructed in

H. Kneser. Reelle analytische Lösungen der Gleichung $\phi(\phi(x)) =\mathrm{e}^x$ und verwandter Funktionalgleichungen. J. Reine Angew. Math. 187 (1949), 56–67.

Holomorphic analytic continuation of inverse tetration is discussed in

H. Trappmann and D. Kouznetsov. Uniqueness of holomorphic Abel functions at a complex fixed point pair. Aequationes mathematicae 81 (2011), 65–76.

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  • $\begingroup$ A nice approach (in my view) is the essay of Andrew Robbins "OnAnalyticIteration.pdf" (or so) of (I think 2009) but whose url must be searched since it changed over the years. It describes how he came from an amateurish hobby fiddling to a seemingly valid (asymptotic) solution under the guiding requirement of being infinitely differentiable. (If you don't find the article I could send you a copy) $\endgroup$ – Gottfried Helms Apr 18 '19 at 0:32

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