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 Zeration = inconsistant ? MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 10/01/2014, 11:27 AM (This post was last modified: 10/01/2014, 03:39 PM by MphLee.) Everything depend on the choice of the recursive definition used to define the Hyperoperations sequence. Using a classic definition over the naturals (starting from the addition) leaves an opening for a non-trivial solution for zeration below the addition because if we try to define it starting from the higher Hos. (the addition) we have to use the cutoff subtraction ${-}^{*}$ (not defined if $a\lt n$ ). $a[0]n=(a+((a{-}^{*}n)+1))$ This means that for the recursive definition of the hyperoperation sequence on the naturals we have an infinite ammount of functions that "models" Zeration (some are commutative!) and each of those contains the successor function: in other worlds, those models should behave as the successor function over a restricted subset of $\mathbb{N} \times \mathbb{N}$). Rubtsov and Romerio tried to reach the uniqueness of the solution by introducing a different recursive picewise definition of the Hyperoperations sequence. Inside their operation family (Rubtsov-Romerio Hyperoperation Family) Zeration happens to unique and coincides with the classic Zeration (the one defined by Rubtsov). Anyways taking the Goodstein definition of the Hyperoperations, Zeration is defined as the successor. $G(0,a,n):=n+1$ The problems arise when we want to extend the Hyperoperations to the reals, and we want to extend the validity of the recursion step formula to them: as you noted the sub-function of the successor is the successor itself. $S \circ S \circ S^{\circ -1}=S \circ (S \circ S^{\circ -1})=S \circ Id=S$ And, for every function that commutes with the successor (like the addition), its subfucntion coincides with the successor: $f \circ S \circ f^{\circ -1}=(f \circ S )\circ f^{\circ -1}=(S\circ f )\circ f^{\circ -1}=S\circ( f \circ f^{\circ -1})=S \circ Id=S$ Some choices, thus, "forces", in some sense, Zeration to be trivial (the successor function). But we could aswell change our concept of Hyperoperation and consider the Commutative hyperoperations (like the Bennet's family). In that case there is a binary operations on the real below the addition. About this I have already said something in your thread about the distributive property: Quote:As you know these are a generalization of the Bennet Hyperoperations (Commutative Hyperoperations). I usually use this notation for them because i find it very confortable $a \odot_r^k b:={\exp}_k^{\circ r}(log_k^{\circ r}(a)+log_k^{\circ r}(b))$ Bennet Hyperoperations are a special case of these (with the natural base $a \odot_r^e b$) $\odot_0^{K} =+$ $\odot_1^{K} =\cdot$ and $a \odot_{-1}^{+\infty} b$ is the max operation while $a \odot_{-1}^{0} b$ is the min operation(this limit process is related with the litinov-maslov dequantization of the semifield of non-negative real numbers in to the Tropical semifield $\mathbb{T}_{max}$) from: http://math.eretrandre.org/tetrationforu...hp?tid=520 MathStackExchange account:MphLee « Next Oldest | Next Newest »

 Messages In This Thread Zeration = inconsistant ? - by tommy1729 - 10/01/2014, 08:40 AM RE: Zeration = inconsistant ? - by MphLee - 10/01/2014, 11:27 AM RE: Zeration = inconsistant ? - by GFR - 10/02/2014, 02:44 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/02/2014, 09:27 PM RE: Zeration = inconsistant ? - by MphLee - 10/02/2014, 10:02 PM RE: Zeration = inconsistant ? - by GFR - 10/04/2014, 09:58 AM RE: Zeration = inconsistant ? - by tommy1729 - 10/02/2014, 10:58 PM RE: Zeration = inconsistant ? - by GFR - 10/03/2014, 11:29 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 12:11 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/02/2014, 11:11 PM RE: Zeration = inconsistant ? - by GFR - 10/03/2014, 11:39 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 12:12 PM RE: Zeration = inconsistant ? - by MphLee - 10/03/2014, 09:20 AM RE: Zeration = inconsistant ? - by tommy1729 - 10/03/2014, 09:32 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/03/2014, 09:41 PM RE: Zeration = inconsistant ? - by GFR - 10/04/2014, 10:19 AM RE: Zeration = inconsistant ? - by MphLee - 10/04/2014, 11:24 AM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 12:16 PM RE: Zeration = inconsistant ? - by MphLee - 10/04/2014, 12:58 PM RE: Zeration = inconsistant ? - by tommy1729 - 10/04/2014, 10:20 PM RE: Zeration = inconsistant ? - by MphLee - 10/05/2014, 03:36 PM

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