• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Initial values for hyper operations bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/14/2008, 03:09 PM (This post was last modified: 03/14/2008, 03:19 PM by bo198214.) The problem can be equivalently (and even more canonically) posed with argument 0 a [0] 0 = 1, no right neutral element e, x[0]e=x => e=x-1 a [1] 0 = a, neutral element 0 a [2] 0 = 0, neutral element 1 a [3] 0 = 1, right neutral element 1, no left neutral element e[3]x=x => e=x^(1/x) a [4] 0 = 1 a [3L] 0 = a And now we also see the pattern. For each operation we choose as initial value the neutral element of the corresponding side of the sub/preceding operation, if there is any. Otherwise we choose a, the left operand, as initial value. (*) "the corresponding side" means the left side for the left-bracketed/lower hyperoperation and means the right side for the right-bracketed/upper/normal hyperoperation. To look at some deviant hyper operations, let us define (only for this post, hopefully we elaborate a more refined notation later): a[nr]0=right neutral elment of [n] a[nR]0=a a[nl]0=left neutral element of [n] a[nL]0=a If we now start with the initial operation a[0]x=x+1, we have the following hierarchy: a[1]x=a[0R]x=a+x a[2]x=a[1r]x=ax a[3]x=a[2r]x=a^x a[3L]x=a[3L]x=a^a^x and the following deviant operations: a[1R]0=a, a[1R]1=a[1]a=2a, ... a[1R]n=a(n+1) There is no left neutral element e, e[1R]x=x => e=x/(x+1). 0 is the right neutral element a[1Rr]0=0, a[1Rr]1=a[1R]0=a, a[1Rr]2=a[1R]a=a(a+1)=a^2+a, a[1Rr]3=a[1R](a^2+a)=a^3+a^2+a, ... a[1Rr]n=$\sum_{i=1}^n a^i$ There is no left neutral element e, e[1Rr]x=x. 1 is the right neutral element How can this be extended to the real n? Deviant tetration a[1Rrr]0=1, a[1Rrr]1=a[1Rr]1=a, a[1Rrr]2=a[1Rr]a=? a[2R]0=a a[2R]1=a a=a^2 a[2R]n=a^(n+1) There is no left neutral element 0 is the right neutral element Another deviant tetration: a[2Rr]0=0, a[2Rr]1=a[2R]0=a, a[2Rr]2=a[2R]a=a^(a+1), a[2Rr]3=a^(1+a^(a+1)), ... We saw already that the sub operations below [0] are all equal to [0], i.e. increments. And those are unfortunately the operation(s) that dont fit into this scheme. Because the initial value to go from a[-1]x=x+1 to a[0]x=x+1, is a[0]0=1, though neither 1 is a right neutral element of [-1] nor is 1=a. But this can be reformulated that there are no operations below [0] that obey our general rule (*), i.e. the hierarchy starts at 0. « Next Oldest | Next Newest »

 Messages In This Thread Initial values for hyper operations - by bo198214 - 03/14/2008, 10:44 AM RE: Initial values for hyper operations - by GFR - 03/19/2008, 11:03 AM

 Possibly Related Threads... Thread Author Replies Views Last Post Thoughts on hyper-operations of rational but non-integer orders? VSO 2 662 09/09/2019, 10:38 PM Last Post: tommy1729 Could there be an "arctic geometry" by raising the rank of all operations? Syzithryx 2 804 07/24/2019, 05:59 PM Last Post: Syzithryx What has been published on absolute values of tetration? 11Keith22 0 833 11/10/2018, 05:02 AM Last Post: 11Keith22 Hyper-volume by integration Xorter 0 1,446 04/08/2017, 01:52 PM Last Post: Xorter Hyper operators in computability theory JmsNxn 5 4,213 02/15/2017, 10:07 PM Last Post: MphLee Recursive formula generating bounded hyper-operators JmsNxn 0 1,562 01/17/2017, 05:10 AM Last Post: JmsNxn holomorphic binary operators over naturals; generalized hyper operators JmsNxn 15 16,931 08/22/2016, 12:19 AM Last Post: JmsNxn Intresting ternary operations ? tommy1729 0 1,661 06/11/2015, 08:18 AM Last Post: tommy1729 on constructing hyper operations for bases > eta JmsNxn 1 2,766 04/08/2015, 09:18 PM Last Post: marraco Bounded Analytic Hyper operators JmsNxn 25 20,613 04/01/2015, 06:09 PM Last Post: MphLee

Users browsing this thread: 1 Guest(s)