• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Rational operators (a {t} b); a,b > e solved sheldonison Long Time Fellow Posts: 641 Threads: 22 Joined: Oct 2008 06/06/2011, 08:43 PM (This post was last modified: 06/06/2011, 09:38 PM by sheldonison.) (06/06/2011, 07:47 PM)JmsNxn Wrote: Alright, testing the left hand right hand limit I get different values.... I'll refer to $\vartheta(a, b, \sigma) = a\,\,\bigtriangle_\sigma\,\,b$ $\{a,b, \sigma| \R(a), \R(b) > e\,;\,a,b,\sigma \in C\}$ from now on. So therefore: $\lim_{h\to 0^+} \frac{\large \vartheta(3, 4, 1 + h) - \vartheta(3, 4, 1)}{h} = 10.7633\\ \lim_{h\to 0^-} \frac{\large \vartheta(3, 4, 1 + h) - \vartheta(3, 4, 1)}{h} = 9.9206$ and $\lim_{h\to 0^+} \frac{\large \vartheta(3, 3, 1 + h) - \vartheta(3, 3, 1)}{h} = 5.4105\\ \lim_{h\to 0^-} \frac{\large \vartheta(3, 3, 1 + h) - \vartheta(3, 3, 1)}{h} = 5.3674$This matches my results (exactly, actually). The 1st derivatives are close, but they don't exactly match, at the transition between the function that defines operators between addition...multiplication, as compared to the function that defines multiplication...exponentiation. I don't know why it works as well as it does, for base=eta. For other bases, which will also give the same results for integers, the resulting graphs are pretty ugly. (06/06/2011, 07:47 PM)JmsNxn Wrote: $\lim_{h\to 0}\, \vartheta(e+h, e+h, 1 + q)\, =\, \text{cheta}(1+q)$If you could get a definition about a complex circle around h=1, at a,b=e, that might be a big start. This would be 1+q, 1-q, 1+qi, 1-qi, also matching both of your initial definitions (which I haven't checked). If that were the case, you already have analytic functions defied for 0<=h<=1, and analytic functions defined for 1<=h<=2. Then, for one case, a=b=e, you might have a function defined for 0<=h<=2. Then the key is to morph this function, perhaps starting with the case a=b, as a=b becomes less than e, and greater than e, in such a way that it remains analytic. Of course, there is the small issue that the inverse superfunctions of eta have singularities at z=e, and the issue of the upper/lower superfunctions of eta, so there are many many challenges on this path. By the way, I agree with Henryk, that exponentiation should be rational operator three, and multiplication, rational operator 2, and addition rational operator 1. - Sheldon « Next Oldest | Next Newest »

 Messages In This Thread Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 02:45 AM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 04:39 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 05:34 AM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 06:02 AM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 07:03 AM RE: Rational operators (a {t} b); a,b > e solved - by nuninho1980 - 06/06/2011, 05:16 PM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/06/2011, 06:53 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 08:47 AM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/06/2011, 09:23 AM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 06/06/2011, 11:59 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 05:44 PM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/06/2011, 09:28 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/06/2011, 07:47 PM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 08:43 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/07/2011, 02:45 AM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/07/2011, 06:59 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/08/2011, 04:54 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/08/2011, 07:31 PM RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/08/2011, 08:32 PM RE: Rational operators (a {t} b); a,b > e solved - by bo198214 - 06/08/2011, 09:14 PM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 09/02/2016, 01:50 AM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/08/2011, 11:47 PM RE: Rational operators (a {t} b); a,b > e solved - by Gottfried - 06/11/2011, 02:33 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/12/2011, 07:55 PM RE: Rational operators (a {t} b); a,b > e solved - by Xorter - 08/21/2016, 06:56 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 08/22/2016, 12:36 AM RE: Rational operators (a {t} b); a,b > e solved - by Xorter - 08/24/2016, 07:24 PM RE: Rational operators (a {t} b); a,b > e solved - by Xorter - 08/29/2016, 02:06 PM RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 09/01/2016, 06:47 PM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 09/02/2016, 02:04 AM RE: Rational operators (a {t} b); a,b > e solved - by tommy1729 - 09/02/2016, 02:11 AM

 Possibly Related Threads... Thread Author Replies Views Last Post Thoughts on hyper-operations of rational but non-integer orders? VSO 2 950 09/09/2019, 10:38 PM Last Post: tommy1729 Hyper operators in computability theory JmsNxn 5 4,751 02/15/2017, 10:07 PM Last Post: MphLee Recursive formula generating bounded hyper-operators JmsNxn 0 1,722 01/17/2017, 05:10 AM Last Post: JmsNxn holomorphic binary operators over naturals; generalized hyper operators JmsNxn 15 17,924 08/22/2016, 12:19 AM Last Post: JmsNxn The bounded analytic semiHyper-operators JmsNxn 2 3,982 05/27/2016, 04:03 AM Last Post: JmsNxn Bounded Analytic Hyper operators JmsNxn 25 22,549 04/01/2015, 06:09 PM Last Post: MphLee Incredible reduction for Hyper operators JmsNxn 0 2,464 02/13/2014, 06:20 PM Last Post: JmsNxn interpolating the hyper operators JmsNxn 3 5,554 06/07/2013, 09:03 PM Last Post: JmsNxn Number theory and hyper operators JmsNxn 7 8,438 05/29/2013, 09:24 PM Last Post: MphLee Number theoretic formula for hyper operators (-oo, 2] at prime numbers JmsNxn 2 4,372 07/17/2012, 02:12 AM Last Post: JmsNxn

Users browsing this thread: 1 Guest(s)