Generalized arithmetic operator
#8
(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. Cool

I knew it!
bo198214 wrote about this family long ago (2008 )

BALANCED HYPEROPERATIONS

http://math.eretrandre.org/tetrationforu...operations

Mother Law \(\sigma^+\circ 0=\sigma \circ \sigma^+ \)

\({\rm Grp}_{\rm pt} ({\rm RK}J,G)\cong \mathbb N{\rm Set}_{\rm pt} (J, \Sigma^G)\)
Reply


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 hixidom - 03/22/2014, 12:06 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 6 23,763 06/16/2022, 10:33 PM
Last Post: MphLee
  holomorphic binary operators over naturals; generalized hyper operators JmsNxn 15 54,339 08/22/2016, 12:19 AM
Last Post: JmsNxn
  Hyper operator space JmsNxn 0 6,651 08/12/2013, 10:17 PM
Last Post: JmsNxn
  Operator definition discrepancy? cacolijn 2 12,622 01/07/2013, 09:13 AM
Last Post: cacolijn
  the infinite operator, is there any research into this? JmsNxn 2 14,912 07/15/2011, 02:23 AM
Last Post: JmsNxn
  Generalized recursive operators Whiteknox 39 114,844 04/04/2011, 11:52 PM
Last Post: Stan



Users browsing this thread: 2 Guest(s)