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 holomorphic binary operators over naturals; generalized hyper operators tommy1729 Ultimate Fellow Posts: 1,505 Threads: 358 Joined: Feb 2009 08/06/2012, 03:32 PM that proof of convergeance seems valid. ( and used 1/n^2 as i expected ) congrats ! im not sure if we have differentiability though ... if we do im betting on uniqueness. i think we will get closer to an answer of uniqueness if we find a good solution. ( differentiable or not ) regards tommy1729 ps : off topic , but im sick JmsNxn Long Time Fellow Posts: 624 Threads: 102 Joined: Dec 2010 08/08/2012, 11:23 AM (This post was last modified: 08/11/2012, 12:46 AM by JmsNxn.) Let's start by restricting ourselves to the following set: $\mathbb{L}_{x,s} = \{\ell | \,\,x\,\,\bigtriangleup_s\,\,\ell\,\,\in \,\,\mathbb{N}\}$ We find that these numbers can be sequenced by $\ell_c$ such that $\ell_c < \ell_{c+1}$ Our required theorem is the following: $x\,\,\bigtriangleup_{s-1}\,\,(x\,\,\bigtriangleup_s\,\,\ell_c) = x \,\,\bigtriangleup_s\,\,(\ell_c + 1)$ I'll rephrase this algebraically as: $\circ: \mathbb{C} \times \mathbb{N} \to \mathbb{C}$ $\circ$ is non-commutative and non-associative $s \in \mathbb{C}$ $s \circ \ell_c \in \mathbb{N}$ $(s-1) \circ (s \circ \ell_c) = s \circ (\ell_c+1)$ I'm keeping this as a foot note. It may be beneficial to consider operators as such. It seems far less gargantuan and much more as an algebraic equation. Another formula I'm thinking I'll have to make use of is: $\forall m\forall x \forall y \,\,m,x,y \in \mathbb{N}\,\,;\,\,m \ge x+y\,\,x,y>1\,\,(x \vee y) > 2$ $\exists \sigma\,\,\sigma \in \mathbb{R}$ $x\,\,\bigtriangleup_\sigma\,\,y = m$ It's sort of like a quick fundamental theorem of arithmetic. I think it encodes more data then it's letting on. Although this seems a bit trivial now. But with the inverse operator function; the inverse of: $\Pi(s) = x\,\,\bigtriangleup_s\,\,y$ Or at least a discrete point set of when $\phi$ returns a natural number we may be able to have some conversations with number theory. I feel like if this isn't the right answer for semi operators; it's a big leap in the right direction--the analytic direction. I like to think of it aesthetically as every complex operator is a unique factorization of the natural operators. Hopefully this can be sustained with the recursive law and maybe an identity function. I'm hoping to sort of web together the recursion at all the points that are natural; showing from the mere fact of their existence (which requires the two requirements before) rather than computing them. I think it's pretty clear that computing this function to any degree of accuracy would need a quantum computer. Lol. I'm sorry to hear that you're sick. Hope you get well. YES!!!! I have a taylor series! I have recursion written out as a requirement using typical analytic expressions! It's all down to a recursive pattern in $\psi$ I'm very close to obtaining a solution. I have a feeling uniqueness will be the difficult task. From here on however. I don't want to post too much until I have it all organized. I'm going to start writing up a short fifteen page or so paper on these. I'll be sure to give everyone at this forum credit. I've been posting stuff about this for a year and you guys have helped me into the environment tremendously--even just by forcing myself to keep up with you guys ^_^. When I have a rough draft I'll post another thread. I may ask a few questions here and there as well. I'm very confident this is going to be fruitful in some way. Gottfried Ultimate Fellow Posts: 795 Threads: 121 Joined: Aug 2007 08/09/2012, 08:59 PM JmsNxn - that sounds as very nice news! I hope you can go on and find something valueable! Gottfried Gottfried Helms, Kassel JmsNxn Long Time Fellow Posts: 624 Threads: 102 Joined: Dec 2010 08/10/2012, 10:57 PM (This post was last modified: 08/11/2012, 09:38 PM by JmsNxn.) Right now I'm writing out some assumptions we have to put away. For example: $2 \,\,\bigtriangleup_s\,\,2 = 4 \,\,\Rightarrow\,\,2\,\,\bigtriangleup_{s+1}\,\,1 = 2$ This implies that $2\,\,\bigtriangleup_s\,\,1$ is not analytic because it is 2 for all s with real part greater than or equal to 1 and 3 when s is 0. Another one is that any continuous segment of operators is commutative or associative all the operators have to be. As well; operators in a continuous segment cannot have the same identity. The functional requirement is the following: $\vartheta_n(s+1) = \sum_{k=0}^{\infty} \vartheta_n(\mu_k) \vartheta_k(s)$ where we have: $\Pi(s) = x\,\,\bigtriangleup_s\,\,y$ $\Pi(\mu_k) = x\,\,\bigtriangleup_k\,\,\ell$ $\ell = x\,\,\bigtriangleup_{s+1}\,\,(y-1)\,\,\in\,\mathbb{N}$ And $\vartheta$ is as before. I can obtain $\Pi^{-1}(s)$ as a taylor series using lagrange inversion. So all of these functions are theoretically computable besides $\psi_n$ which is in $\vartheta_n$. So the requirement is restricted to it. I'm writing this all out trying to solve for the taylor series coefficients of $\psi$. $\vartheta_n$ is entire if $\psi_n$ is entire; so I hope it is. Thanks for the encouragement Gottfried. Like all math; it's slow progress. Little breakthroughs from time to time. Xorter Fellow Posts: 93 Threads: 30 Joined: Aug 2016 08/18/2016, 04:40 PM All it sounds so interesting. But somewhy I cannot evaluate non-trivial problems. E. g. 3[0.5]3 or something like this. Could show me more examples, please? Xorter Unizo JmsNxn Long Time Fellow Posts: 624 Threads: 102 Joined: Dec 2010 08/22/2016, 12:19 AM this is an old scrapped idea, it ended up falling apart upon close analysis. « Next Oldest | Next Newest »

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