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 Number theory and hyper operators JmsNxn Ultimate Fellow Posts: 902 Threads: 111 Joined: Dec 2010 08/30/2012, 02:49 AM I'm wondering if anyone knows of any research of tetration and pentation and that whole gang in number theoretic areas. In the sense that we would only deal with natural x,y,n in the equation x [n] y. I'm interested more specifically in results to do with primes. It's interesting because the same prime factors from x are the only ones which occur for n > 2 and y > 0. I'm thinking however more characteristic patterns reoccur. Any help would be greatly appreciated! tommy1729 Ultimate Fellow Posts: 1,645 Threads: 369 Joined: Feb 2009 08/30/2012, 01:45 PM (This post was last modified: 08/30/2012, 01:49 PM by tommy1729.) i have asked that same question before in the past. im unaware of nontrivial work in that area. maybe its a silly question but what about convergeance questions f(p) = log(2)*sexp(slog(2)-4/3)/2 + log(3)*sexp(slog(3)-4/3)/3 + ... + log(p)*sexp(slog(p)-4/3)/p lim p -> oo f(p) = Constant ? or something better ... regards tommy1729 JmsNxn Ultimate Fellow Posts: 902 Threads: 111 Joined: Dec 2010 08/30/2012, 05:24 PM Well. The reason I ask is because I was structuring my semi operators around the distribution of the set: $\mathbb{I}_{y} = \{ s_0 | s_0 \in \mathbb{R}\,\,;\,\,x\,\,\bigtriangleup_{s_0}\,\,y\,\,\in \mathbb{N}}$ claim that there are operators unique to x and y which allow us to perform operations on elements of $\mathbb{I}_y$ instead of operations on $\mathbb{N}$. We then say that $x\,\,\bigtriangleup_s\,\,y$ is an isomorphism from $\mathbb{I}_y \to \mathbb{N}$ Then I found out I only needed to prove the recursive identity for primitive elements of $\mathbb{I}_y$; (i.e elements that return primes in N); and then do the rest by induction and breaking up the real argument into a product of primitive elements. However; this all and all sounded plausible but I hit some huge wall. Which is proving the recursive identity for primitive elements; mostly. I have a new technique now. It may or may not work. But having more information about how $x\,\,\bigtriangleup_n\,\,y$ behaves for naturals would really help. MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 05/25/2013, 10:15 PM Long ago I found this, I think is related, even if I don't know if can be usefull for your work. Patrick St-Amant - Number Theories He tries to build new number theories using a generalized concept of prime number based on the choice of the operations. Is quite interesting. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 05/27/2013, 01:18 PM (This post was last modified: 05/27/2013, 03:36 PM by MphLee.) (08/30/2012, 05:24 PM)JmsNxn Wrote: Well. The reason I ask is because I was structuring my semi operators around the distribution of the set: $\mathbb{I}_{y} = \{ s_0 | s_0 \in \mathbb{R}\,\,;\,\,x\,\,\bigtriangleup_{s_0}\,\,y\,\,\in \mathbb{N}}$ claim that there are operators unique to x and y which allow us to perform operations on elements of $\mathbb{I}_y$ instead of operations on $\mathbb{N}$. We then say that $n\,\,\bigtriangleup_{s_n}\,\,y$ is an isomorphism from $\mathbb{I}_y \to \mathbb{N}$Sorry if my question is stupid (I'm a newbie in math) but I'm a bit confused. From your definition of $\mathbb{I}_{y}$ (that is a set of the real ranks $s_0$ that satysfie $\,\,x\,\,\bigtriangleup_{s_0}\,\,y\,\,\in \mathbb{N}$ for a fixed $y$ ) so we have that $\mathbb{I}_{y}\subset \mathbb R$. In other words we can define a function $I_y:\mathbb {N} \rightarrow R$ whit the property $I_y(n)=s_n$ such that $n\,\,\bigtriangleup_{s_n}\,\,y\in \mathbb{N}$, at this point we have that the set of the $s_0, s_1, s_3, ...$ is your $\mathbb{I}_{y}$ The question is, how do you know that $I_y:\mathbb {N} \rightarrow R$ is injective? It can be maybe a surjection on the reals (not-injective, multivalued)? In fact from your definiton of $\mathbb{I}_{y}$ seems me that you except it to be a countable subset of $\mathbb{R}$ then you are asuming (is an hypothesis?) that $I_y:\mathbb {N} \rightarrow \mathbb{I}_{y}$ is a bijection (is this what you mean when you say that they are isomorphic? And which is the funtion you use for the isomorphism? maybe you use $I^{\circ -1}_y:\mathbb{I}_{y} \rightarrow \mathbb {N}$ ? ) Thanks in advance, and sorry for my bad english. 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: 902 Threads: 111 Joined: Dec 2010 05/27/2013, 11:33 PM (This post was last modified: 05/27/2013, 11:33 PM by JmsNxn.) I meant to say that $\forall x,y > 2\,\,\forall s \in \mathbb{R} \,\,s\ge 0$ $x\,\bigtriangleup_{s_1}\,y = x\,\bigtriangleup_{s_2}\,y\,\,\,\,\Leftrightarrow\,\,\,s_1 = s_2$ $\forall n \in \mathbb{N}\,\,n \ge x+y\,\,\,\,\exists s_0\,\,x\,\,\bigtriangleup_{s_0}\,\,y = n$ Now the distribution of the set $\mathbb{I}_{x,y}$ (which depends on x and y) is pivotal insofar as, if we extend hyper operators for all natural arguments and real operators they have to obey the recursion on these points and these points alone. The recursive Identity is written: $x\,\,\bigtriangleup_{s_0-1}\,\,(x\,\,\bigtriangleup_{s_0}\,\,y) = x\,\,\bigtriangleup_{s_0}\,\,(y+1)$ And we need to only worry when $x\,\,\bigtriangleup_{s_0}\,\,y$ is an integer, which is when $s_0 \in \mathbb{I}_{x,y}$ I was trying to think of ways to talk only about recursion on this discrete set rather than the whole plane. What I was thinking, which I am no longer is that there maybe binary operations on this set $\mathbb{I}_{x,y}$ which allows us to talk about the operators more freely. I haven't really had much luck in uncovering much of anything. By isomorphism I meant one to one and onto insofar as $x\,\bigtriangleup_s\,y$ obeys the rules I laid out at the beginning. MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 05/28/2013, 10:40 AM Ok Now is clear. Anyways the first two statemens are intuitively acceptable, the problem is the I can't think how to prove them for the hypoeroperator starting from the recursive identity, that is the fundamental axiom of the hyperoperations. You mean to use them as axioms and study the "generated" hyperoperators, or do you have any clues? Are them really acceptable from a philosophical point of view in your opinion? MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 05/29/2013, 09:24 PM (This post was last modified: 05/31/2013, 08:49 AM by MphLee.) Yea, you're a genius! Now I understand what you tried to do. But I don't know if it will be really fruitfull, but at least sound interesting. Tell me if I got it right: You was right about the bijection (only if you first statements hold). Anyways we have that the sets $\mathbb{I}_ {m,n}$ and $\mathbb{I}_ {n}$ are really differents but they have interesting links. Lets re-start with a new version of your definitons: let $m, n \in \mathbb{N}$ Ia) $\mathbb{I} _{m,n} = \{ s: m \bigtriangleup_{s} n \in\mathbb{N} \}$ for fixed $m, n$ IIa) $\mathbb{I} _{n} = \{ s: m \bigtriangleup_{s} n \in\mathbb{N} \}$ for fixed $n$ second step: Ib) $\Pi _{m,n} (s) = m \bigtriangleup_{s} n$ IIb) $I _{n} (m) = s_m \text{ only if } (m \bigtriangleup_{s _m} n) \in\mathbb{N}$ First: $\Pi _{m,n}$ maps $\mathbb {N}$ on a subset of the naturals that now we call $P _{m,n}$ and like with the primes there are naturals numbers cutted off. We want $\Pi _{m,n} : \mathbb{N} \rightarrow \mathbb{N}$ to be injective (that is your condition $\Pi _{m,n} (s_1)=\Pi _{m,n} (s_2) \Longleftrightarrow s_1=s_2$ ) but is not invertible because is restricted to $P _{m,n}$: in fact $\Pi _{m,n} : \mathbb{N} \rightarrow P _{m,n}$ is bijective. Your idea is to extend $\Pi _{m,n}(s)$ to make it invertible adding more elements in the domain, in other words extending it to $\mathbb{I} _{m,n}$ (but you want do it adding real numbers) Finally we have a bijection, in fact $\mathbb{I} _{m,n} \approx^{\Pi_{m,n}} \mathbb{N}$ holds! for example using $0$ for the addition we have: $P _{2,3}=\{ \Pi _{2,3}(0); \Pi _{2,3}(1); \Pi _{2,3}(2); \Pi _{2,3}(3); ... \} = \{ 5; 6; 8; 16; ...\}$ then we build $\mathbb{I} _{2,3}$ inverting the naturals using $\Pi ^{\circ -1}_{2,3}$ $ \mathbb{I} _{2,3}= \{\Pi ^{\circ -1}_{2,3}(0);\Pi ^{\circ -1}_{2,3}(1);\Pi ^{\circ -1}_{2,3}(2);\Pi ^{\circ -1}_{2,3}(3);\Pi ^{\circ -1}_{2,3}(4); \Pi ^{\circ -1}_{2,3}(5); \Pi ^{\circ -1}_{2,3}(6); \Pi ^{\circ -1}_{2,3}(7); \Pi ^{\circ -1}_{2,3}(\text{8}); \Pi ^{\circ -1}_{2,3}(9); \Pi ^{\circ -1}_{2,3}(10); \Pi ^{\circ -1}_{2,3}(11); \Pi ^{\circ -1}_{2,3}(12); \Pi ^{\circ -1}_{2,3}(13); \Pi ^{\circ -1}_{2,3}(14); \Pi ^{\circ -1}_{2,3}(15);\Pi ^{\circ -1}_{2,3}(16); ... \} $ $\Pi ^{\circ -1}_{2,3}$ maps the naturals only if the domain is $P_{2,3}$ then in our set we have numbers that are not naturals $\mathbb{I} _{2,3}= \{r_0; r_1; r_2; r_3; r_4; 0; 1; r_7; 2; r_9; r_{10} ;r_{11} ;r_{12} ;r_{13} ;r_{14} ;r_{16} ; 3; ... \}$ To be honest I don't know if the extension of the naturals to $\mathbb{I} _{m,n}$ is a subset of the reals (as you defined) but if we have that $\Pi _{m,n} (s)$ preserves the order we can find this result $m \bigtriangleup_{s_0} n \lt m \bigtriangleup_{s_1} n \Leftrightarrow s_0 \lt s_1$ in our example this lead us to this $1 \lt r_7 \lt 2$ so the rank is not a natural number. My question is: is it rational, irrational? Maybe it is trascendental, but this is beyond my limits. Second: About the relations with $\mathbb{I} _{n}$ now the differencese are clear. We have that $\forall m,n \in\mathbb {N}$ $P_{m,n} \subset \mathbb{N} \subset \mathbb{I} _{m,n} \subset \mathbb{I} _{n}$ because $I_n [ \{ k \} ]= \mathbb{I} _{k,n}$ in other words the function IIb) is multivalued, and the set $\mathbb{I} _{n}$ is the set of all possibles solution, in symbols: $I_n [ \mathbb{N} ]= \mathbb{I} _{n}$ that is equivalent to $\mathbb{I} _{n}= \bigcup _{i\in \mathbb{N} } \mathbb{I} _{i, n}$ MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ « Next Oldest | Next Newest »

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