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 Has this tetration been considered before in the forum? MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 11/04/2021, 10:53 AM (This post was last modified: 11/04/2021, 10:53 AM by MphLee.) I hope you all can forgive my naivety but I found this on MathSE. It somehow seems familiar but at the same time, since I missed a lot of the nitty gritty details of the tetration extensions in the past discussions (not the beta method ones), it somehow look new to me. In this question How to evaluate fractional tetrations? (March 2020) the user Simply Beautiful Art cites a chain of previous question of his/her, and in that the author claim the method is probably equivalent to Kneser. The formula is derived from the assumption of asymptotic behavior analogous to that of Gamma function. The chain of questions where this is laid out are: Dec 26, 2019 Numerical instability of an extended tetration Dec 26, 2019 Verifying tetration properties Dec 29, 2019 Verifying uniqueness of my tetration Of the three only the first received attention by forum users (Gottfried). There only the question if computation was unstable of the formula non-convergent was treated. The other posts maybe were not noticed by the forum experts. So my question: is it something new? Was already discussed here? MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 11/05/2021, 12:25 AM (11/04/2021, 10:53 AM)MphLee Wrote: I hope you all can forgive my naivety but I found this on MathSE. It somehow seems familiar but at the same time, since I missed a lot of the nitty gritty details of the tetration extensions in the past discussions (not the beta method ones), it somehow look new to me. In this question How to evaluate fractional tetrations? (March 2020) the user Simply Beautiful Art cites a chain of previous question of his/her, and in that the author claim the method is probably equivalent to Kneser. The formula is derived from the assumption of asymptotic behavior analogous to that of Gamma function. The chain of questions where this is laid out are: Dec 26, 2019 Numerical instability of an extended tetration Dec 26, 2019 Verifying tetration properties Dec 29, 2019 Verifying uniqueness of my tetration Of the three only the first received attention by forum users (Gottfried). There only the question if computation was unstable of the formula non-convergent was treated. The other posts maybe were not noticed by the forum experts. So my question: is it something new? Was already discussed here? You can prove it's the standard Schroder iteration pretty simply. First of all, it converges for $$\Re(x) > X$$ for large enough $$X$$. Then notice that it is periodic with the same period as Schroder's iteration; which is $$2 \pi i / \log \log (^\infty a)$$. Then notice that it's bounded in the right half plane $$\Re(x) > X$$. Then notice that it interpolates the values $$^na$$ for $$n > X$$. Now pull out your Ramanujan identity theorem: If two functions $$F,G$$ are holomorphic and bounded for $$\Re(s) > X$$ and interpolate the same values, $$F = G$$. Easy peasy. I have never in my life seen that expression for tetration though. Especially odd looking... MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 11/05/2021, 01:34 AM (11/05/2021, 12:25 AM)JmsNxn Wrote: 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? MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 11/05/2021, 02:06 AM (11/05/2021, 01:34 AM)MphLee Wrote: (11/05/2021, 12:25 AM)JmsNxn Wrote: 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. MphLee Fellow Posts: 184 Threads: 19 Joined: May 2013 11/05/2021, 02:33 AM Thank you for dumbing this down a bit for me! It helped MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 11/05/2021, 05:12 AM (This post was last modified: 11/05/2021, 05:13 AM by JmsNxn.) No, problem! This is a really clever result. I had never noticed that before. Simply Beautiful Art always has crazy solutions to problems. They're a really smart cookie! « Next Oldest | Next Newest »

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