Talk:Hyper operator

message: This article is very muddy to me. It needs some love. -- imbaczek
:It may be helpful if you can define "love" mathematically.  OmegaMan
The definition previous given was incomplete. It defines hyper(a,n,b)=hyper(a,n-1,hyper(a,n,b-1)) without ever defining a base case to which hyper(a,n,b-1) can reduce. This C++ code demostrates why the definition without the stipulation that hyper(a,n,0)=1 for n>1 is invalid. The deletion of the lines containing "if(b==0) return 1;" results in infinite recursion.
 template<class T>
 T Hyper(T a, T n, T b)
 {
      switch(n)
      {
      case 1:
           return a+b; // intentionally grouped with also infrequent n=2 branch
      case 2:
           return a*b;
      case 3:
           return IntegerPow(a,b); // intentionally grouped with also common n>3 branch
      default:
           if(b==0)
                return 1;
           return Hyper( a , n-1 , Hyper(a,n,b-1) );
      }
 }
130.39.153.46 04:17, 25 Sep 2004 (UTC)

message: We should investigate funtions f(x) such that f(hyper(a,b,n))=hyper(f(a),b,n-1).  Such funtions would allow definition of hyper with real numbers by f-1(hyper(f(a),b,n-1)).--SurrealWarrior
That sounds good, and I've recently been working on the same thing. So far I've defined it over all integers n. In the article on Knuth arrow notation, it is mentioned that a^{(n)}b = a^{(n-1)}a^{(n)}(b-1). From that I've found for all even negative n, it is simply b+1, and all odd negative n give simply b.That gave me, so far:
  \forall -n\in\mathbb{N};
  a ^ {(n)} b=
  \left\{
    \begin{matrix}
      b+1, & \mbox{if }-n\in\mathbb{E} \\
      b,   & \mbox{if }-n\in\mathbb{O} \\
    \end{matrix}
  \right.
Where O and E are the sets of all positive odd and even numbers, respectively. This aside, I've made no progress yet to generalizing over all reals, or complex n\in\mathbb{C}
He Who Is 01:00, 23 April 2006 (UTC)
thread_title:  Cool Funtions 

message: I'm not sure if someone said this already, but what about a hypothetical hyper 0? I think I have a defintion, for any a, the hyper 0 operation of a would be a# = a + 1. This fits in with the idea that addition is hyper one, it's chained counting, and that would be a "count" function.
Also, what about hyper negative numbers? They could be the inverse functions, hyper negative one would be subtraction, hyper -2 division, then roots, super logs, etc.
Sorry if I'm doing something wrong here, the whole wiki thing is a bit new to me.  —Preceding unsigned comment added by 68.198.122.160 (talk) 01:04, 12 December 2007 (UTC)
replies: text: There are many problems with what you are talking about. One is that there are multiple definitions of hyper0 depending on which axioms you hold to carry over. If you think that the property \text{hy}_n(2, 2) = 4 is more important than anything else, then you get \text{hy}_0(y, y) = y + 2. If you think that the inverse relation \text{hy}_{n-1}(x, y) = \text{hy}_n(x, \text{hylog}_n(x, y) + 1) is more important than anything else, then you will get \text{hy}_0(x, y) = y + 1, and this will also give the same answer for all negative hyper-operators, so \text{hy}_{(-1)}(x, y) = y + 1 and \text{hy}_{(-2)}(x, y) = y + 1as well, and so on. The problem with this is that the iteration of these operators give addition prematurely, and so it is debatable whether these can be considered hyper-operators as well. So with these included we actually have 3 different definitions for negative hyper-operators: (1) All the successor function, (2) the hyper-roots, and (3) the hyper-logarithms as you suggest (although you mention roots with super-logs which are not of the same type), since all hyper-operators addition and above have two inverses. Either way is arbitrary, and debatable. AJRobbins (talk) 19:30, 12 December 2007 (UTC)
thread_title: Hyper 0 and other hyper integers