• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Generalized arithmetic operator MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 03/12/2014, 11:18 AM (This post was last modified: 03/13/2014, 08:40 AM by MphLee.) (03/11/2014, 11:20 PM)hixidom Wrote: I'm not familiar with the difference between left- and right-associative hyperoperators. I would say that the unary version is definitely righ-associative, as $[x][y][z]a=[x]([y]([z]a))$ but perhaps that is not what you are referring to. If I had to write a recursion relation along the lines of what has been posted so far, I would say it is: $a[x]b=(a[x]\frac{b}{2})[x-1](a[x]\frac{b}{2})$ which is neither left- nor right-associative, if I understand the meaning of that phrase. The ones that have been posted so far are $a [x] b=a[x-1](a [x] (b-1))$, and $a [x] b=(a [x] (b-1))[x-1]a$. which are very similar, with the only difference being that the [x-1] operation on the right side is flipped. But neither of these definitions result in the same [4] operator that I have. \begin{align*}a[4]2&=[3]a&=a^a\\a[4]4&=[3]^2a&=(a^a)^{(a^a)}\\a[4]8&=[3]^3a&=\left ((a^a)^{(a^a)}\right )^{\left ((a^a)^{(a^a)}\right )}\end{align*} etc. well , i think i got it. tell me if i'm right first lets notice that we have $a[4]2= {^{2}a}$ $a[4]4= {^{2}(^{2}a)}$ $a[4]8= {^{2}(^{2}(^{2}a))}$ so we have that if $t(a)=a^a={^2 a}$ (note that since $[3]=hyper 3$ we have that $[3]a=t(a)$) $a[4]2^k=t^{\circ k}(a)=[3]^{k}a$ and so $t(a[4]2^k)=a[4]((2^k) \times 2)=t^{\circ k+1}(a)=[3]^{k+1}a$ so the formulae that define all your hyperoperation family are Quote:1) $a[1]n:=a+n$ 2) $[s]a:=f_s(a)=a[s]a$ (that is A1) 3) $a[s+1]n=[s]^{\log_2(n)}a$ (that is A3) with this three difinitions (and if we already have defined the function iteration) we can get all the values of your hyperoperations. and if im right again we have too $a[s+1](2n)=(a[s+1]n)[s](a[s+1]n)$ I think i have already found this family somewhere, probably in the tetration FAQ pdf ----- I've jsut noticed that we can maybe write it as functional equation let $d(x):=2x$ Quote:1') $F_{1,a}(n):=a[1]n:=a+n$ 2') $f_s(a)=[s]a:=F_{s,a}(a)$ 3') $F_{s+1,a}(d(n))=f_s(F_{s+1,a}(n))$ the last can be rewritten in this way 3'') $F_{s+1,a}\circ d=f_s\circ F_{s+1,a}$ and we have this property $F_{s+1,a}\circ d^{\circ t}=f_s^{\circ t}\circ F_{s+1,a}$ and there is a name for this functional equation, I don't remember it but is a kind of abel functional equation i guess. I'm sure that someone here can help you more about this. Lee EDIT: BINGOOO found them. I knew it! bo198214 wrote about this family long ago (2008 ) BALANCED HYPEROPERATIONS http://math.eretrandre.org/tetrationforu...operations MathStackExchange account:MphLee « Next Oldest | Next Newest »

 Messages In This Thread Generalized arithmetic operator - by hixidom - 03/11/2014, 03:52 AM RE: Generalized arithmetic operator - by JmsNxn - 03/11/2014, 03:15 PM RE: Generalized arithmetic operator - by hixidom - 03/11/2014, 06:24 PM RE: Generalized arithmetic operator - by MphLee - 03/11/2014, 10:49 PM RE: Generalized arithmetic operator - by hixidom - 03/11/2014, 11:20 PM RE: Generalized arithmetic operator - by MphLee - 03/12/2014, 11:18 AM RE: Generalized arithmetic operator - by JmsNxn - 03/12/2014, 02:59 AM RE: Generalized arithmetic operator - by hixidom - 03/12/2014, 04:37 AM RE: Generalized arithmetic operator - by MphLee - 03/12/2014, 06:19 PM RE: Generalized arithmetic operator - by hixidom - 03/12/2014, 06:43 PM RE: Generalized arithmetic operator - by tommy1729 - 03/21/2014, 10:31 PM RE: Generalized arithmetic operator - by hixidom - 03/22/2014, 12:06 AM RE: Generalized arithmetic operator - by tommy1729 - 03/22/2014, 12:13 AM RE: Generalized arithmetic operator - by hixidom - 03/22/2014, 12:42 AM RE: Generalized arithmetic operator - by hixidom - 06/11/2014, 05:10 PM

 Possibly Related Threads... Thread Author Replies Views Last Post A fundamental flaw of an operator who's super operator is addition JmsNxn 4 6,450 06/23/2019, 08:19 PM Last Post: Chenjesu Tetration and Sign operator tetration101 0 302 05/15/2019, 07:55 PM Last Post: tetration101 Where is the proof of a generalized integral for integer heights? Chenjesu 2 733 03/03/2019, 08:55 AM Last Post: Chenjesu holomorphic binary operators over naturals; generalized hyper operators JmsNxn 15 15,629 08/22/2016, 12:19 AM Last Post: JmsNxn [rule 30] Is it possible to easily rewrite rule 30 in terms of modular arithmetic ? tommy1729 0 1,501 07/24/2014, 11:09 PM Last Post: tommy1729 Tetration and modular arithmetic. tommy1729 0 1,863 01/12/2014, 05:07 AM Last Post: tommy1729 Hyper operator space JmsNxn 0 1,819 08/12/2013, 10:17 PM Last Post: JmsNxn Generalized Bieberbach conjectures ? tommy1729 0 1,628 08/12/2013, 08:11 PM Last Post: tommy1729 Operator definition discrepancy? cacolijn 2 3,692 01/07/2013, 09:13 AM Last Post: cacolijn Generalized Wiener Ikehara for exp[1/2](n) instead of n ? tommy1729 0 1,645 12/17/2012, 05:01 PM Last Post: tommy1729

Users browsing this thread: 1 Guest(s)