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 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

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