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 Rational operators (a {t} b); a,b > e solved JmsNxn Long Time Fellow Posts: 739 Threads: 104 Joined: Dec 2010 06/07/2011, 02:45 AM (This post was last modified: 06/07/2011, 04:59 PM by JmsNxn.) (06/06/2011, 08:43 PM)sheldonison Wrote: 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. I don't understand what you mean about the first part, of making a definition about a complex circle, however I do understand what you mean by morphing the cheta function, and creating a generalization to $a\,\,\bigtriangle_\sigma\,\,a$. (06/06/2011, 08:43 PM)sheldonison Wrote: By the way, I agree with Henryk, that exponentiation should be rational operator three, and multiplication, rational operator 2, and addition rational operator 1. - SheldonHmm, this is just like the Gamma function predicament. Well, hear my arguments for centering addition at 0. Firstly I wanted to center the Identity function so that S(0) = 0, and S(1) = 1. Secondly, the function designed is naturally centered with addition as zero. Shifting it over by one is an offset. $\vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b = \exp_\eta^{\circ \sigma}(\exp_\eta^{\circ -\sigma}(a) + h_b(\sigma))$ $h_b(\sigma)=\left{\begin{array}{c l} \exp_\eta^{\circ -\sigma}(b) & \sigma \le 1\\ \exp_\eta^{\circ -1}(b) & \sigma \in [1,2] \end{array}\right.$ Thirdly, the negative integers of operators were specifically designed to be negative because they aren't proper recursive. i.e: $p<=0$ $a\,\,\bigtriangleup_{p+1}\,\,\log_\eta^{\circ -p}(n) = a_1\,\,\bigtriangleup_{p}\,\,a_2\,\,\bigtriangleup_{p}\,\,a_3...\bigtriangleup_{p}\,\,a_n$, having 0 as the first non-proper recursive operator seems off. I feel it should be -1 (or technically -0.0000000001). Also, again you can see the series is centered with addition as 0. I don't really have any other arguments, it just seems simpler to center it at 0. It's just unfortunate that someone decided to call it "tetration" instead of "tertiation". « 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

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