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 Rational operators (a {t} b); a,b > e solved JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 06/06/2011, 02:45 AM (This post was last modified: 06/06/2011, 04:42 AM by JmsNxn.) Well alas, logarithmic semi operators have paid off and have given a beautiful smooth curve over domain $(-\infty, 2]$. This solution for rational operators is given by $t \in (-\infty, 2]$ $\{a, b \in \R| a, b > e\}$: $ f(t) = a\, \{t\}\, b = \left\{ \begin{array}{c l} \exp_\eta^{\alpha t}(\exp_\eta^{\alpha-t}(a) + \exp_\eta^{\alpha -t}(b)) & t \in (-\infty,1]\\ \exp_\eta^{\alpha t-1}(b * \exp_\eta^{\alpha 1-t}(a)) & t \in [1,2]\\ \end{array} \right.$ Which extends the ackerman function to domain real (given the restrictions provided). the upper superfunction of $\exp_\eta(x)$ is used (i.e: the cheta function). $\exp_\eta^{\alpha t}(a) = \text{cheta}(\text{cheta}^{-1}(a) + t)$ Logarithmic semi-operators contain infinite rings and infinite abelian groups. In so far as {t} and {t-1} always form a ring and {t-1} is always an abelian group (therefore any operator greater than {1} is not commutative and is not abelian). There is an identity function S(t), however its values occur below e and are therefore still unknown for operators less than {1} (except at negative integers where it is a variant of infinity (therefore difficult to play with) and at 0 where it is 0). Greater than {1} operators have identity 1. The difficulty is, if we use the lower superfunction of $\exp_\eta(x)$ to define values less than e we get a hump in the middle of our transformation from $a\,\{t\}\,b$. Therefore we have difficulty in defining an inverse for rational exponentiation. however, we still have a piecewise formula: $g(t) = a\, \}t \{ \,$$b = \left\{ \begin{array}{c l} \exp_\eta^{\alpha t}(\exp_\eta^{\alpha-t}(a) - \exp_\eta^{\alpha -t}(b)) & t \in (-\infty,1]\\ \exp_\eta^{\alpha t-1}(\frac{1}{b} \exp_\eta^{\alpha 1-t}(a)) & t \in [1,2]\\ \end{array} \right.$ therefore rational roots, the inverse of rational exponentiation is defined so long as $a > e$ and $\frac{1}{b} \exp_\eta^{\alpha 1-t}(a) > e$. rational division and rational subtraction is possible if $a,b > e$ and $\exp_\eta^{\alpha-t}(a) - \exp_\eta^{\alpha -t}(b) > e$. Here are some graphs, I'm sorry about their poor quality but I'm rather new to pari-gp so I don't know how to draw graphs using it. I'm stuck using python right now. Nonetheless here are the graphs. the window for these ones is xmin = -1, xmax = 2, ymin = 0, ymax = 100 If there's any transformation someone would like to see specifically, please just ask me. I wanted to do the transformation of $f(x) = x\, \{t\}\, 3$ as we slowly raise t, but the graph doesn't look too good since x > e. Some numerical values: $4\,\{1.5\} 3 = 4\,\{0.5\}\,4\,\{0.5\}\, 4= 21.01351835\\ 3\, \{1.25\}\, 2 = 3\, \{0.25\}\, 3 = 6.46495791\\ 5\, \{1.89\}\, 3 = 5\, \{0.89\}\,5\,\{0.89\}\, 5 = 81.307046337\\$ (I know I'm not supposed to be able to calculate the second one, but that's the power of recursion) I'm very excited by this, I wonder if anyone has any questions comments? for more on rational operators in general, see the identities they follow on this thread http://math.eretrandre.org/tetrationforu...hp?tid=546 thanks, James PS: thanks go to Sheldon for the taylor series approximations of cheta and its inverse which allowed for the calculations. « 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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