Tetration fixed points Daniel Long Time Fellow Posts: 281 Threads: 95 Joined: Aug 2007 07/20/2022, 07:51 AM (This post was last modified: 07/22/2022, 04:45 AM by Daniel.) Edit: Changed from Pentation Fixed Points to Tetration Fixed Points. Thanks to JmsNxn for pointing this out The identity $z \uparrow \uparrow \infty=\frac{\mathrm{W}(-\ln{z})}{-\ln{z}}$ with some constraints gives a tetration fixed point. To extend this scheme to pentation I believe I need slog, which is fine, but I would also need a generalization of the Lambert W function. Haven't folks on this forum explored hyper or super Lambert W functions. Daniel Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022 07/20/2022, 08:22 AM (This post was last modified: 07/20/2022, 08:25 AM by Catullus.) The super Lambert W function was talked about here. Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ Daniel Long Time Fellow Posts: 281 Threads: 95 Joined: Aug 2007 07/20/2022, 08:33 AM (This post was last modified: 07/20/2022, 08:35 AM by Daniel.) (07/20/2022, 08:22 AM)Catullus Wrote: The super Lambert W function was talked about here. Thanks but not what I was looking for. What is discussed is the inverse of $x^x$. Maybe I was thinking of some of Galidakis' work. Daniel Catullus Fellow Posts: 213 Threads: 47 Joined: Jun 2022 07/20/2022, 08:35 AM (This post was last modified: 12/15/2022, 05:20 AM by Catullus.) The also discussed the inverse of x^superE^^x. Please remember to stay hydrated. ฅ(ﾐ⚈ ﻌ ⚈ﾐ)ฅ Sincerely: Catullus /ᐠ_ ꞈ _ᐟ\ MphLee Long Time Fellow Posts: 376 Threads: 30 Joined: May 2013 07/20/2022, 09:22 AM (This post was last modified: 07/20/2022, 03:41 PM by MphLee.) Yes, forum members have started the study of this matter. I'd name first Rubtosv and Romerio's 2006 report (2006) Notes on Hyper-operations Progress Report NKS Forum III I have not time to brush of the rust, but If I recall correctly everything starts with this heuristics. Let $H_n(x)=b[n]x$ be n-th rank hyperexponentiation in base $b$ then we have $V_n(b)=\lim_{x\to \infty} b[n]x$ then $H_n(V_{n+1}(b))=V_{n+1}(b)$ thus if ${\rm hrt}_n(d,x)$ is the $d$-th degree n-hyperroot function the equation $b[n]p=p$ is solved by $p=V_{n+1}(b)$ or by the inverse of $b={\rm hrt}_n(p,p)$: call that $W_n(b)=p$. So we use the next rank... and roots. All of this obviously depends on attracting or repelling fixedpoints. Separate analysis needs to be carried for odd rank (exp, pentation, heptation,...) and even ranks (tetration, hexation, octation,...). I suggest you these threads: (2007) Generalized recursive operators (2007) Exploring Pentation - Base e (2015) Mizugadro, pentation, Book (2017) pentation and hexation Note that $V_{n+1}(b)$ and $W_n(b)$ must be related somehow. Also we could go also make the distinction $V_n^-(b)=\lim_{x\to -\infty} b[n] x$. It is important to highlight that $W_3$ is not Lambert W function, so I apologize for choosing the letter W for it. Lambert W is the inverse of $b\mapsto be^b$ because $W(b)e^{W(b)}=b$; while $W_3$ inverts $b\mapsto b^{1/b}$ because $b^{W_3 (b)}=W_3(b)$; the function $V^+_4(b)={}^{\infty}b$ is the infinite tower, so it somehow inverts $b\mapsto {\rm srt}(\infty,b)$; From this heuristics Rubtsov and Romerio derive the concept that self hyperrots are deeply related to infinite degree hyperroots of the next ranks. Because $\boxed{{\rm hrt}_{n+1}(\infty,b)\simeq {\rm hrt}_{n}(b,b)}$ MSE MphLee Mother Law $(\sigma+1)0=\sigma (\sigma+1)$ S Law $\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$ JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 07/20/2022, 10:20 PM (This post was last modified: 07/20/2022, 10:24 PM by JmsNxn.) I can answer in the bounded case, and the bounded case only. If you construct the bounded analytic hyper-operators you get the following sequence of analytic functions: \begin{align} \alpha \in (1,\eta)\,\, & n \ge 1\,\,\Re(z) > 0\\ \alpha \uparrow^n z &: \mathbb{C}_{\Re(z)> 0} \to \mathbb{C}_{\Re(z) > 0}\\ \alpha \uparrow^1 z &= \alpha^z\\ \alpha \uparrow^n 1 &= \alpha\\ \alpha \uparrow^n ( \alpha \uparrow^{n+1} z) &= \alpha \uparrow^{n+1} (z+1)\\ \end{align} Then, I believe you are asking for the fixed points of tetration not pentation, because the formula you wrote is for the fixed points of exponentiation, or the infinite power tower. To begin, in this schema: $\lim_{k\to\infty} \alpha \uparrow^2 ...(k\,\text{times})...\uparrow^2 \alpha = \omega_2\\$ And additionally: $\lim_{x \to \infty} \alpha \uparrow^3 x \to \omega_2\\$ There are ways to evaluate the $\omega_2$ constant more effectively, but they're probably not what you are looking for. The Lambert W identity for the infinite tetration is more of a coincidence than a rule of law for hyper operators. But: $\alpha \uparrow^2 \omega_2(\alpha) = \omega_2(\alpha)\\$ This is a geometrically attracting fixed point for $\alpha \in (1,\eta)$, and this is enough to construct an implicit solution. Which honestly would be the fastest route, as you'd just be solving for the taylor expansion in $\alpha$ about a point of your desire. There is no real quick way, personally I would use the iterative limit, as it's fast and simple. But if that's not up your alley, then I'd suggest you expand the Schroder function: $\alpha \uparrow^2 z = \Psi_\alpha^{-1}\left(\Psi_\alpha(1) (\log(\alpha)\omega_1)^z\right)\\$ Where $\omega_1$ is your Lambert W expression above. Now you are asking for a point: $\Psi^{-1}_\alpha\left(\Psi_\alpha(1) (\log(\alpha)\omega_1(\alpha))^{\omega_2(\alpha)}\right) = \omega_2(\alpha)\\$ Discovering a taylor expansion from here is difficult, but should be doable. This reduces into, $\lambda(\alpha) = \log(\alpha)\omega_1(\alpha)$ and $a_k(\alpha) > 0$: $\alpha \uparrow^2 z = \omega_1 - \sum_{k=1}^\infty a_k(\alpha) \lambda(\alpha)^{kz}\\$ Which then isn't too hard to run a an algo to get: $\omega_1 - \sum_{k=0}^\infty a_k(\alpha) \lambda(\alpha)^{k\omega_2(\alpha)} = \omega_2(\alpha)$ It's especially fast to construct this function too because: $\alpha < .... <\omega_n(\alpha) <....<\omega_2(\alpha)< \omega_{1}(\alpha) < e$ MphLee Long Time Fellow Posts: 376 Threads: 30 Joined: May 2013 07/21/2022, 11:39 AM How the previous rank' omega constant pops out in this forumla? Quote:$\alpha \uparrow^2 z = \Psi_\alpha^{-1}\left(\Psi_\alpha(1) (\log(\alpha)\omega_1)^z\right)\\$ Can be this extended to Quote:$\alpha \uparrow^{n+1} z = \Psi_{n,\alpha}^{-1}\left(\Psi_{n,\alpha}(1) (\log(\alpha)\omega_n)^z\right)\\$ Where the $\Psi_{n,\alpha}$ is the Schroeder of $\alpha \uparrow^n-$. MSE MphLee Mother Law $(\sigma+1)0=\sigma (\sigma+1)$ S Law $\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$ JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 07/22/2022, 12:57 AM (This post was last modified: 07/22/2022, 01:00 AM by JmsNxn.) (07/21/2022, 11:39 AM)MphLee Wrote: How the previous rank' omega constant pops out in this forumla? Quote:$\alpha \uparrow^2 z = \Psi_\alpha^{-1}\left(\Psi_\alpha(1) (\log(\alpha)\omega_1)^z\right)\\$ Can be this extended to Quote:$\alpha \uparrow^{n+1} z = \Psi_{n,\alpha}^{-1}\left(\Psi_{n,\alpha}(1) (\log(\alpha)\omega_n)^z\right)\\$ Where the $\Psi_{n,\alpha}$ is the Schroeder of $\alpha \uparrow^n-$.Hmm, no not exactly Where you have $\log(\alpha)\omega_n$ you should have $\lambda(\alpha) = \frac{d}{dz}\Big{|}_{z=\omega_n} \alpha \uparrow^n z$. We just get lucky with the exponential fixed point, the derivative formula satisfies $\frac{d}{dz}\Big{|}_{z=\omega_1} \alpha^z = \log(\alpha) \omega_1$. « Next Oldest | Next Newest »

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