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 Initial values for hyper operations bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/14/2008, 10:44 AM During the introduction of zeration we saw already that there are some inconsistencies with the initial values. The addition as the successor operation of the successor operation (I consider here a[0]x=x+1 according to the rule a[n+1](x+1)=a[n](a[n+1]x) ) differs in its initial value from all the higher operations: a [1] 1 = a + 1 a [2] 1 = a a [3] 1 = a a [4] 1 = a .... If we would choose a[1]1=a we would get a[1]x=a+(x-1) I see a quite similar phenomenon when considering a[3L]x, the lower super operation of exponentiation, defined by a[3L](x+1)=(a[3L]x)[3]a. If we chose the initial condition in the used to a[3L]1=a manner then we get a[3L]x=a^(a^(x-1)). Especially in comparision with the similarly wrong defined addition a[1]x=a+(x-1) I would prefer to be a[3L]x=a^a^x a[3L]1=a^a Then we had unified a[3L]1=a[3]a a[1]1=a[0]a So does anyone see a strictly describable pattern when to use a ** 1 = a and when to use a ** 1 = a * a, where * is the sub operation of **. 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. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 03/16/2008, 10:39 PM bo198214 Wrote: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 = aThank you Henryk, for your comments. I know that KAR is carefully following this, but he is extremely busy with his job, sharing his time between Belgorod (Russia) and Martin (Slovakia). In the meantime, I shall try to keep contacts within this thread, awaiting that the "father" of zeration will react personally. I agree that there are a lot of (infinite?) operations with a rank lower than (that of) addition. You proposed a variant of definition (I am still thinking of it) and Quickfur also defined one, by correctly adding that it was not zeration. The problem, as you also correctly said, is to define an operation (ONE hyperop) that would be the unique operation, exactly fitting in the hyperops hierarchy, at rank 0. Agreed. I am carefully studying all your comments and observations. As a matter of documenting the KAR version of zeration, until now .... heavily supported also by me (no problem, I can easily change of mind, if necessary), I should like to supply the following tables, always supposing to take the following definition for zeration: a[0]x = max[a,x] + 1 /; a >< x a[0]x = a + 2 = x + 2 /; a ==x. We should have: a[0]0 = a + 1, for a >< 0 a[1]0 = a a[2]0 = 0 a[3]0 = 1, for a >< 0 a[4]0 = 1, for a >< 0 .... 0[0]a = a + 1 for a >< 0 0[1]a = a 0[2]a = 0 0[3]a = 0, for a < 0 ; 0[3]a = 1, for a > 0; 0[4]a = 1, for a > 0 (?) .... a[0]1 = a + 1, for a >< 0 a[1]1 = a + 1 a[2]1 = a a[3]1 = a a[4]1 = a .... 1[0]a = a + 1 for a >< 0 1[1]a = a + 1 1[2]a = a 1[3]a = 1 1[4]a = 1, for a > 0 (!) Concerning the uniformity of behaviour of the hyperops at what we might call their "initial values", the overall situation is rather messy, exactly for the classical ranks: 1, 2, 3. Quote:......... And now we also see the pattern. But this can be reformulated that there are no operations below [0] that obey our general rule (*), i.e. the hierarchy starts at 0. So, no possibility to have: a[-1]a = a[0]2 ? Schade ...! Aber, warum nicht. It might even be justified by the Ackermann Function. Nevertheless, this is not the KAR's opinion, as far as I know. GFR bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/17/2008, 10:54 AM I would suggest to move the discussion about the "proper" zeration to the zeration thread. For this discussion it makes anyway not that much difference which zeration we choose: GFR Wrote:a[0]0 = a + 1, for a >< 0 a[1]0 = a a[2]0 = 0 a[3]0 = 1, for a >< 0 a[4]0 = 1, for a >< 0 because the condition $a\neq 0$ for [3] and [4] is anyway satisfied. If we consider exponentiation and tetration on the real numbers we must anyway impose that $a>0$ for [3] and $a>1$ for [4]. I called the values a[n]0 initial values because they, together with the rule a[n+1](x+1)=a[n](a[n+1]x), determine the operation (at least on the natural numbers for the right operand) like an initial condition would do for a differential/difference equation. Thatswhy I also dont consider the values 1[n]x, 0[n]x here. And the values a[n]1 follow then from a[n]0 by a[n]1=a[n-1](a[n]0). Quote:So, no possibility to have: a[-1]a = a[0]2 ? Schade ...! Aber, warum nicht. It was already shown in the zeration thread that if we obey the law a[n+1](b+1)=a[n](a[n+1]b) then it must a[-k]b=b+1, $k\ge 0$. No room for a[-1]a=a[0]2=3 or a[0]a=a+2. If we now backwards search for the initial condition that makes a[-k]b=b+1, given that a[-k-1]b=b+1, then we see that it is: a[-k]0=1. Quite in accordance with a[n]0=1 for $n\ge 3$. The difference why [-k] is an increment but [1] is addition, though both are the super operation of the increment, lies only in the initial condition: a[-k]0=1 vs. a[1]0=a. The only thing is that a[-k]0=1 does not fit in my rules for chosing the initial value. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 03/19/2008, 11:03 AM OK! I see, Thank you for your explanations. « Next Oldest | Next Newest »

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