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 pentation and hexation JmsNxn Long Time Fellow Posts: 568 Threads: 95 Joined: Dec 2010 09/03/2017, 11:52 PM (This post was last modified: 09/03/2017, 11:54 PM by JmsNxn.) There is a result posted on here about how the "eta constants" converge to 2 and the "euler constants" converge to 4. The n'th eta constant is the sup of the  x'th n'th hyperoperator root of x. And the "euler constants" are the actual values x_n such that x_n'th n'th hyperoperator root of x_n = n'th eta constant. I think from this we can show that anything less than 2 does not grow unbounded, (just observe the meaning of the n'th hyperoperator root). Any base less than the n'th eta constant has a bounded hyperoperator for N>n-1. We get something similar with the bounded hyperoperators as what you're talking about. It's a little messier but essentially there exists $\mathbb{Z}^n$  n-2'th hyperoperators. They satisfy the following really weird recursion $\alpha \uparrow^n_{a_1a_2...a_n} (\alpha \uparrow^{n+1}_{a_1a_2...a_n b} z) = \alpha \uparrow^{n+1}_{a_1a_2...a_n b} (z+1)$ for ANY arbitrary b, and $a_1,...,a_n \in \mathbb{Z}$ fixed. The first n numbers have to equal for this recursion to work though, but the last number can be anything. Essentially, everytime we go up a hyperoperator, we get another $\mathbb{Z}$ amount of iterates. I haven't actually proved this, but it's really straightforward. Firstly, there exists $\mathbb{Z}$ exponentiations. namely $f(z) = \,e^{2 \pi i a_1 z}\alpha^z$ for all integer a_1. Then we fix an a_1 and there exists a kind of fixed point that is unique (its uniqueness and what qualifies it is a little hard to explain, which is the weakest point of my argument). It's geometrically attracting and hence can be iterated. Now the iterate $g(z,\xi)$ is conjugate to SOME exponential $e^{2 \pi i a_2 z} \lambda^z$ for $\lambda$ the multiplier of the above fixed point. This generates $\mathbb{Z} \times \mathbb{Z}$ tetration functions g(z,1). (they interpolate the natural power towers, send to complex values and are bounded on SOME half plane of the complex plane that includes $\mathbb{R}^+$, and satisfy the recursion $f(g(z)) = g(z+1)$. This allows us to find another FIXED point and it's unique because of Schwarz' lemma. Rinse and repeat, and we get $\mathbb{Z} \times \mathbb{Z}\times \mathbb{Z}$ pentations. Then $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}\times\mathbb{Z}$ hexations, so on and so forth. This is all really roughly argued, more intuition than rigor. But I call it the "branches of hyper operators." « Next Oldest | Next Newest »

 Messages In This Thread pentation and hexation - by sheldonison - 08/07/2017, 07:33 PM RE: pentation and hexation - by JmsNxn - 08/21/2017, 08:05 PM RE: pentation and hexation - by sheldonison - 08/22/2017, 02:03 PM RE: pentation and hexation - by JmsNxn - 08/22/2017, 10:38 PM RE: pentation and hexation - by sheldonison - 09/03/2017, 10:11 PM RE: pentation and hexation - by JmsNxn - 09/03/2017, 11:52 PM RE: pentation and hexation - by sheldonison - 09/04/2017, 03:04 AM RE: pentation and hexation - by JmsNxn - 09/04/2017, 04:07 AM RE: pentation and hexation - by Ember Edison - 09/18/2019, 06:34 AM RE: pentation and hexation - by sheldonison - 09/18/2019, 02:34 PM

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