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 A relaxed zeta-extensions of the Recursive Hyperoperations MphLee Long Time Fellow Posts: 270 Threads: 23 Joined: May 2013 06/14/2013, 09:57 PM (This post was last modified: 06/03/2022, 02:12 PM by MphLee.) A relaxed $\zeta$-extensions of the Recursive Hyperoperations I want to show you an easy extension for hyperoperations. I don't want it to be the most natural, but I want to ask if someone already used this extension and if it can be usefull for something. Since is a bit different I want to use the plus-notation ($+_{\sigma}$) for the hyperoperators. I start with these basic definitions over the naturals $b,n,\sigma \in \mathbb{N}$: $o)\,\,\,S(n)=n+1$ $$o')\,\,\,B_b( \sigma+1):= \begin{cases} b, & \text{if} \sigma=0 \\ 0, & \text{if} \sigma=1 \\ 1, & \text{if} \sigma\gt 1 \\ \end{cases}$$ Then the recursive definitions of the operators $b,n,\sigma \in \mathbb{N}$ $i)\,\,\,b+_0 n=S(n)$ $ii)\,\,\,b+_{\sigma+1}0=B_b(\sigma+1)$ $iii)\,\,\,b+_{\sigma+1}S(n)=b+_{\sigma}(b+_{\sigma+1}n)$ Observation before the extension's definitons $b+_0 n=1+n$ $b+_1 n=b+n$ we can see that from rank zero to rank one we can define infinite functions $b+_\sigma n=\zeta_\sigma+n$ with  $1\lt\zeta_\sigma\lt b$ $b+_0 n=\zeta_{0}+n=1+n$ $b+_{0.5} n=\zeta_{0.5}+n$ $b+_1 n=\zeta_{1}+n=b+n$ Generalizing, now we can define $\zeta_b$ as a continous functions from the interval $0$ to $1$, to the interval $1$ to $b$: $Eiv)\,\,\,\zeta_b:[0,1]\rightarrow [1,b]$ $$Ev)\,\,\,\zeta_b(\varepsilon)=\begin{cases} 1, & \text{if \varepsilon=0} \\ b, & \text{if \varepsilon=1 } \\ \end{cases}$$ And we can define the operations with fractional rank starting from the interval $[0,1]$ $$Evi)\,\,\, b +_{\varepsilon}n=\zeta _b(\varepsilon)+_{1}n \,\, \text{ and} \,\, \varepsilon \in [0,1]$$ Other operations are these ( $\varepsilon \in ]0,1]$ and $k \in \mathbb{N}$ ): $b +_{k+\varepsilon}n=\zeta _b(\varepsilon)+_{k+1} n$ Example of $\zeta _b$ functions and the generated $\zeta _b$-hyperoperations: $\zeta _b(\varepsilon)=b^\varepsilon$ and $k \in \mathbb{N}$ for  $\varepsilon \in ]0,1]$  and $k \in \mathbb{N}$ $b +_{k+\varepsilon}n=b^\varepsilon+_{k+1} n$ MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ MphLee Long Time Fellow Posts: 270 Threads: 23 Joined: May 2013 06/03/2022, 02:08 PM Slightly related. I put this just to not forget. There are two way to explore: consider $$\Theta_q(z,y)$$ is a family of holomorphic solutions for every $$q\in [0,1]_\mathbb Q=[0,1]\cap \mathbb Q$$ to a0) $$\Theta_q(z,w+y)=w+\Theta_q(z,y)$$ a1) $$\Theta_q(z+w,y)=\Theta_q(z,\Theta_q(w,y))$$ b) $$\Theta_0(z,y)=\Theta_1(z,y)=z+y$$ In the above equation the goal is to consider only exotic solutions that are not constant when $$q\in(0,1)_\mathbb Q$$. We want to exclude the constant solutions. We have $$\Theta_q(0,w)=w+\Theta_q(0,0)$$... I hope we can have some periodic behavior or something like that. The main idea is that for rational ranks $$q\in [0,1]_\mathbb Q$$ we define a family of holomorphic $$a_q:\mathbb C\times \mathbb C\to\mathbb C$$ s.t. $$\mathbb C$$-equivariance holds. $$a_q(b,z+y)=\Theta_q(z,a_q(b,y))$$ With boundary conditions $$a_0(b,y)=y+1$$ and $$a_1(b,y)=b+y$$ and such that $$q\mapsto a_q$$ is smooth given a reasonable topological structure (manifold structure?) on $${\mathcal H}(\mathbb C\times \mathbb C)=\mathcal C^\infty (\mathbb C\times \mathbb C)$$ in a neighborhood of $$0$$ and $$1$$. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ JmsNxn Long Time Fellow Posts: 739 Threads: 104 Joined: Dec 2010 06/03/2022, 10:06 PM (This post was last modified: 06/03/2022, 10:08 PM by JmsNxn.) Hey, Mphlee This very much reminds me of Andy's slog method for Tetration. Specifically the initial iteration. By which, you would have analycity for $$\sigma \in \mathbb{R}^+/\mathbb{N}$$. The trick would be making it analytic at the natural numbers. These constructions appear more commonly in logic theory/computation theory. They don't really appear in analysis, because, sadly, I think it would be very difficult to massage this construction into analycity. Even if we could massage, it would definitely be very hard to prove convergence of the Taylor expansion which results. For example, what you could do to massage this is write: $$\zeta^1(\epsilon) = b^{\epsilon}\\$$ Your right, this creates a continuous solution at the endpoints $$\sigma =0,1$$. But it wouldn't be continuous in the first derivative. To make it continuous in the first derivative we need that: $$\frac{d}{d\epsilon}\Big{|}_{\epsilon = 0,1}\zeta_b^2(\epsilon) +_{k+1} n = \frac{d}{d\epsilon}\Big{|}_{\epsilon =0,1}b +_{k+\epsilon} n\\$$ Now, you'd have to make an iterative procedure to successively get this to go to the second derivative: $$\frac{d^2}{d\epsilon^2}\Big{|}_{\epsilon=0,1} \zeta_b^3(\epsilon) +_{k+1} n = \frac{d^2}{d\epsilon^2}\Big{|}_{\epsilon =0,1}b +_{k+\epsilon} n$$ Then, if you create a good enough sequence of functions $$\zeta^n$$, such that they converge uniformly (for analycity you're going to want this on a non trivial domain in $$\mathbb{C}$$, which would be very tricky), then $$\lim_n \zeta^n= \zeta$$ would give you an analytic solution (theoretically). But there'd be a lot of steps. This is very similar to the main principle of Andy's slog.  So you'd probably get a matrix formula for the various taylor coefficients about $$0$$ and $$1$$, and assuring they align (they satisfy the conjugate identity). Then your stuck with showing the taylor series converges, which will certainly be the hardest part. The idea you are suggesting does exist. I can only point to Joel David Hamkins talking about it, which is very similar to what you are doing here. https://mathoverflow.net/questions/14678...-the-reals, read the first answer, which is a very similar breakdown of what you have here, but he does it more generally. This though, largely subscribes itself to logic/computation stuff like that, as it isn't concerned necessarily with much more than continuity (or in some cases continuously differentiable). I don't quite understand what you are getting at with $$\Theta$$, could you elaborate? MphLee Long Time Fellow Posts: 270 Threads: 23 Joined: May 2013 06/06/2022, 07:37 PM Just few words, I'll reply in details asap. Yes, I was inspired by Andy's consecutive piecewise extensions method And I wanted to study the behavior of a couple of piece-wise extensions of goodstein. As of Hamkins' answer, it came in November, 4 months after my post here, and I was really excited at first when I red it. This concept is still valid. On an abstract level, everything is controlled by that first strip, like a boundary condition. It is in fact a kind of boundary condition and I can prove that it comes from the decomposition $$\mathbb R\simeq \mathbb N\times [0,1)$$. Interesting question is how to detect if the function on $$0\leq x <1$$ comes from something that globally is analytic/smooth. I'll add more. MathStackExchange account:MphLee Fundamental Law $(\sigma+1)0=\sigma (\sigma+1)$ « Next Oldest | Next Newest »

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