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 Does tetration take the right half plane to itself? JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/10/2017, 07:46 PM (This post was last modified: 05/10/2017, 07:49 PM by JmsNxn.) This question is rather straightforward. Let $1 < \alpha < \eta$. We know that there exists a unique bounded tetration function $F(z) = \alpha \uparrow^2 z$ that is holomorphic for $\Re(z) > 0$. But is $\Re (F(z)) > 0$? Does the bounded tetration function take the right half plane to itself? If this is true, this manages to prove a lot of things. First of all, it follows that tetration has only one fixed point $\omega$ such that $F(\omega) = \omega$, that is geometrically attracting $0, which follows by the Schwarz lemma. Secondly $F^{\circ n}(z) \to \omega$ for all $z \in \mathbb{C}_{\Re(z) > 0}$.  All of the orbits of $F$ tend to this fixed point. This implies we can construct a complex iteration $F^{\circ z}(\zeta):\mathbb{C}_{\Re(z) > 0} \times \mathbb{C}_{\Re(\zeta)>0} \to \mathbb{C}_{\Re(\zeta)>0}$. Therefore giving us pentation $F^{\circ z}(1) = \alpha \uparrow^{3} z$ that ALSO takes the right half plane into itself. Therefore it ALSO has a unique fixed point, this fixed point is ALSO geometrically attracting, all of the orbits of pentation tend to this fixed point, and now we can rinse and repeat to construct hexation, so on and so forth.  The great part about this is how it qualifies the sequence of bounded analytic hyper-operators. It gives a lot of great properties.  We get the following things for free. $\alpha \uparrow^n z$ for $n\ge 2$ is holomorphic in $z$ and analytic for $1 < \alpha < \eta$  (I still haven't really managed to show the hyper-operators are analytic in the base argument, only continuous; but with this lemma, it follows trivially).  $\alpha \uparrow^n z : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}_{\Re(z) > 0}$ (before I simply wrote they send to the complex plane and focused on their behaviour in $\mathbb{R}^+$). $\alpha \uparrow^n z$ has a purely imaginary period (something that gives a lot of cool things, like a Fourier series representation for example). if $\omega_{n-1}$ is the fixed point of $\uparrow^{n-1}$ or the limit at infinity of $\uparrow^n$, then $|\alpha \uparrow^n z - \omega_{n-1}|< C q^x$ for $x = \Re(z)$ for some $0 < q < 1$. (Exponential decay is always nice.) ...(I could go on)... And most importantly, these functions satisfy the holy grail of functional equations $\alpha \uparrow^n (\alpha \uparrow^{n+1} z) = \alpha \uparrow^{n+1} (z+1)$ The functional equation is something I could never truly get perfect, because I only managed to show it on the real positive line, without sending the right half plane to itself, the composition in the complex plane is technically ill defined. So all in all, I've boiled a whollllllllleeeee swash of questions into one question. Does $\alpha \uparrow^2 z : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}_{\Re(z) > 0}$? If anyone's curious, I can explain how I've approached the question. It's a little convoluted, so I'll leave it out till someone asks me. « Next Oldest | Next Newest »

 Messages In This Thread Does tetration take the right half plane to itself? - by JmsNxn - 05/10/2017, 07:46 PM RE: Does tetration take the right half plane to itself? - by JmsNxn - 05/10/2017, 08:22 PM RE: Does tetration take the right half plane to itself? - by sheldonison - 05/15/2017, 08:16 PM RE: Does tetration take the right half plane to itself? - by sheldonison - 05/15/2017, 09:00 PM RE: Does tetration take the right half plane to itself? - by JmsNxn - 05/16/2017, 04:09 AM RE: Does tetration take the right half plane to itself? - by sheldonison - 05/16/2017, 03:34 PM RE: Does tetration take the right half plane to itself? - by JmsNxn - 05/16/2017, 08:46 PM RE: Does tetration take the right half plane to itself? - by JmsNxn - 05/16/2017, 04:46 AM

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