• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 extension of the Ackermann function to operators less than addition JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 11/06/2011, 04:56 PM (This post was last modified: 11/06/2011, 08:10 PM by JmsNxn.) furthermore, if we define the identity function as: $a\,\,\bigtriangleup_\sigma\,\,S(\sigma) = a$ since S(1) = 0, and S(2) = 1. If we want S(x) to be analytic and continuous (which we do), by the mean value theorem there is an operator who's identity is 0.5. Let's call this q. $1 < q < 2; q \in \R$ $a\,\,\bigtriangleup_q\,\,0.5 = a$ doing some simple manipulations and we get $a\,\,\bigtriangleup_{q-1}\,\,a = a\,\,\bigtriangleup_q\,\,1.5$ this is by the first axiom. Given that. We can let a = 2, and get the contradictory result: $2\,\,\bigtriangleup_{q-1}\,\,2 = 2\,\,\bigtriangleup_{q}\,\,1.5$ if we want: $2\,\,\bigtriangleup_\sigma\,\,2 = 4$ we arrive at a contradiction or a result that is definitely not desired. Either $2\,\,\bigtriangleup_{\sigma}\,\,2 \neq 4$ or $2\,\,\bigtriangleup_{q}\,\, 1.5 = 2\,\,\bigtriangleup_{q}\,\,2 = 4$ this would imply our function at q has the following behaviour (if we simply reapply the first axiom recusively): $r \ge 1; r \in \mathbb{Z}$ $2\,\,\bigtriangleup_{q}\,\,r + 0.5 = 2\,\,\bigtriangleup_{q}\,\, (r + 1)$ this result can be extended to any value of sigma: $2\,\,\bigtriangleup_{\sigma}\,\,S(\sigma) = 2$ $2\,\,\bigtriangleup_{\sigma - 1}\,\,2 = 2\,\,\bigtriangleup_{\sigma}\,\,1 + S(\sigma)\,\, = 4 = 2\,\,\bigtriangleup_{\sigma}\,\,2$ giving: $2\,\,\bigtriangleup_{\sigma}\,\,r + S(\sigma) = 2\,\,\bigtriangleup_{\sigma}\,\, r + 1$ therefore we have three options 1. the identity function is not analytic 2. $2\,\,\bigtriangleup_\sigma\,\,2 \neq 4$ 3. or in general $f(x) = 2\,\,\bigtriangleup_{\sigma}\,\,x$ is not a smooth monotonic increasing function unless sigma is an integer. If we let 3 then we also get the result that operators extended less than addition are impossible. In this we imply $2\,\,\bigtriangleup_0\,\,2 = 4$ but we already know $2\,\,\bigtriangleup_0\,\,2 = 2 + 1 = 3$ The first option, that the identity function be not analytic isn't desirable at all. Furthermore, it's more than it just not be analytic, but it implies the only results possible are 0 and 1. And that it only takes 0 at addition and everywhere else we have 1. In conclusion, if we want to have the Ackermann function extended to the complex domain everywhere and $f(x) = 2\,\,\bigtriangleup_\sigma\,\,x$ be monotonically increasing everywhere we're going to have to lose the aesthetic property: $2\,\,\bigtriangleup_\sigma\,\,2 = 4$ « Next Oldest | Next Newest »

 Messages In This Thread extension of the Ackermann function to operators less than addition - by JmsNxn - 11/06/2011, 01:31 AM RE: extension of the Ackermann function to operators less than addition - by JmsNxn - 11/06/2011, 04:56 PM RE: extension of the Ackermann function to operators less than addition - by JmsNxn - 11/06/2011, 08:06 PM

 Possibly Related Threads... Thread Author Replies Views Last Post New mathematical object - hyperanalytic function arybnikov 4 1,056 01/02/2020, 01:38 AM Last Post: arybnikov Is there a function space for tetration? Chenjesu 0 667 06/23/2019, 08:24 PM Last Post: Chenjesu A fundamental flaw of an operator who's super operator is addition JmsNxn 4 7,526 06/23/2019, 08:19 PM Last Post: Chenjesu Degamma function Xorter 0 1,123 10/22/2018, 11:29 AM Last Post: Xorter 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 1,721 01/17/2017, 05:10 AM Last Post: JmsNxn Possible continuous extension of tetration to the reals Dasedes 0 1,499 10/10/2016, 04:57 AM Last Post: Dasedes Rational operators (a {t} b); a,b > e solved JmsNxn 30 43,074 09/02/2016, 02:11 AM Last Post: tommy1729 Andrew Robbins' Tetration Extension bo198214 32 46,486 08/22/2016, 04:19 PM Last Post: Gottfried holomorphic binary operators over naturals; generalized hyper operators JmsNxn 15 17,905 08/22/2016, 12:19 AM Last Post: JmsNxn

Users browsing this thread: 1 Guest(s)