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 Rational operators (a {t} b); a,b > e solved JmsNxn Ultimate Fellow Posts: 895 Threads: 110 Joined: Dec 2010 06/06/2011, 05:44 PM (This post was last modified: 06/06/2011, 05:55 PM by JmsNxn.) (06/06/2011, 11:59 AM)tommy1729 Wrote: what is the idea or intention behind working with base eta ? First I noticed that at base 2 humps would appear at around 0.5 and 1.5, and if we increased the base, the humps would get sharper and sharper and taller and taller. So I figured, there must be a base value where the humps disappear altogether. My first guess was eta, since everyone talks about how closely related it is to tetration, but when I made a graph of $f(t) = \eta \bigtriangleup_t 2$ there were still visible humps at 0.5 and 1.5. Disappointed, and intrigued by the cheta function, I tried using the upper superfunction of eta, and voila, the values converged miraculously. So it was more just a shot in the dark as opposed to a real mathematical deduction as to why we used base eta. (06/06/2011, 05:16 PM)nuninho1980 Wrote: (06/06/2011, 04:39 AM)sheldonison Wrote: I wonder what it means that fatb(3,-1,4)=5.429897..?no, it isn't correct, sorry. fatb(2,-1,2) = 2º2=4 -> 2+2=4 fatb(3,-1,3) = 3º3=5 -> 3+2=5 fatb(4,-1,3) = 4º3=5 - it's correct. because that's here down: aºb if a>b then a+1 if b>a then b+1 if a=b then a+2 or b+2 no, this isn't zeration, this a different operator designed to preserve the ring structure of operators [t-1] and [t]. (06/06/2011, 09:23 AM)bo198214 Wrote: Well not on the whole complex plane, but on the real axis, wouldnt that be nice? I anyway wonder whether thats possible at all. As $h_b$ is not even differentiable at 1, I wonder whether f is. Did you compare the derivations from left and right?Yes it would be nice to have it analytic on the real axis, I think it should be potentially analytic for at least (-oo, 1). I'm just not sure about the convergence radius. No I haven't compared the derivations, I'll do that though. It's hard for me to think up tests I can do with only highschool math under my belt (06/06/2011, 09:23 AM)bo198214 Wrote: This is also in sync with the convention for quasigroups, i.e. groups with non-associative operation but with left- and right-inverse. There the left- and right inverse are written as / and \. hmm, I really understand where you're coming from with using a standard notation instead of the triangles, it saves a lot of confusion, but I think that for the same reason the gamma function is written $\Gamma(n+1)$ instead of $n!$, logarithmic semi-operators should be written using their own notation--just to be clear this is only one extension of hyper operators. I know that simply stating this is the natural extension of hyper operators will step on a lot of people's toes. Also, I have something very interesting to report! My conjecture $\text{cheta}(t) = e\, \bigtriangleup_t \,e$ can be proved for $t \in (-\infty, 2], t \in Z$. $\text{cheta}(0) = 2e\\ \text{cheta}(-1) = \log_\eta(2e) = \log_\eta(2) + e$ and if anyone knows their deltation, like I just explained $a + \log_\eta(2) = a \,\,\bigtriangleup_{-1}\,\, a$ therefore: $\text{cheta}(-1) = e\,\, \bigtriangleup_{-1}\,\, e$ The proof can actually be made even simpler, consider, $\R(t),\R(p) \le 1; p,t \in C$: $log_\eta^{\circ -p}(a\,\, \bigtriangleup_t\,\, b) = log_\eta^{\circ -p}(a)\,\, \bigtriangleup_{t-p}\,\, \log_\eta^{\circ -p}(b)$ therefore since: $\text{cheta}(0) = e\,\, \bigtriangleup_0\,\, e$, $\log_\eta^{\circ -p}\text{cheta}(0) = \log_\eta^{\circ -p}(e\,\, \bigtriangleup_0\,\, e)$ $\text{cheta}(p) = \log_\eta^{\circ -p}(e)\,\, \bigtriangleup_p \,\,\log_\eta^{\circ -p}(e)$, and since there's a fixpoint at e, $\text{cheta}(p) = e\,\, \bigtriangleup_p\,\, e$ This proves a beautiful connection between logarithmic semi operators and the cheta function. And also gives me a new beautiful identity: $e^{\large{\frac{e\,\, \bigtriangleup_{t-1}\,\, e}{e}}} = e\, \bigtriangleup_t\, e$ « 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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