Tetration.org What Lies Beyond Exponentiation?

# Ackermann Function

The Ackermann function consist of the of addition, multiplication, exponentiation, tetration, pentation, hexation[1], ... and so on; forming an infinite series of arithmetic operators, with those beyond addition recursively defined from their predecessor. Thanks to Andrew Robbins for correcting my usage of arrow notation and for pointing out that Mathworld has relevent entries for Arrow Notation, Chained Arrow Notation, and Power Tower,

 Arithmetic Standard Ackermann Knuth Conway Addition a+b ack(a,b,0) Multiplication a*b ack(a,b,1) Exponentiation ab ack(a,b,2) a↑b a→b→1 Tetration ba ack(a,b,3) a↑↑b a→b→2 Pentation ba ack(a,b,4) a↑↑↑b a→b→3 Hexation ack(a,b,5) a↑↑↑↑b a→b→4 ... ... ... Circulation ack(a,b,∞) a↑∞b a→b→∞

Let f(x) a→x→k; then
f(1) =
a→1→k = a
f
2(1) = f(a) = a→a→k = a→2→k+1
f3(1) = f(a→a→k) = a→(a→a→k)→k = a→3→k+1

This is the mechanism by which each arithmetic operator is defined recursively from its predecessor. The notation is odd but it points out an important fact, that the Ackermann function is founded on dynamics. The work over the last decade in defining continuous iteration takes the well known expression fn(x) and allows n to be extended from the integers to the real numbers. The immediate ramification is that a→b→k now exists for real values of a and b instead of them being constrained to the whole numbers.

Each arithmetic operator of the Ackermann function eventually grows faster than any precedeing operator but tetration displays a curious reflection of this principle. Consider the tetrational parabolic fixed point at ee-1≈ 1.444 where (ee-1)e = ee e-1 = e.
Even though 1→∞→1 is the largest value that a real number can converge under expondentiation, 1.444→∞→2 converges.The infinite tower of faster growing operators turns into a tower of ever slower growing operators, they beome slack operators. The arithmetic operator k+1 is built on operator k, it uses operator k's trasportation system so to speak. But the operator k+1 can't go where operator k  is unable to go. If 1.444→∞→2 is finite then 1.444→∞→k is finite where k ≥ 2 and therefore 1.444→∞→∞ is finite. Consider that 1→∞→∞ = 1. What is the largest real number x such that x→∞→∞ is finite?

The following table is a first attempt at extending the Ackermann function to the real numbers. Just as the iteration of a→z→c normally follows a logarithmic spiral into a→∞→c, I hypothesize that as the iteration of a→∞→x typically follows a logarithmic spiral into a→∞→∞ and that a→∞→∞=a where 1<=a<=1.444.

The Mathematica notebook used to compute the preceding numbers.

### Ackermann Expansions

2→2→∞ = 4

 2→3→∞ = 2→(2→2→∞)→∞ = 2→4→∞ = 2→(2→(2→2→∞)→∞)→∞ = 2→(2→4→∞)→∞ = ∞

 3→2→∞ = 3→3→∞ = 3→(3→3→∞)→∞ = 3→(3→(3→3→∞)→∞)→∞ = ∞

## References

[1]Reuben Louis Goodstein
Transfinite ordinals in recursive number theory
Journal of Symbolic Logic 12 (1947)
[Jeff Miller, Samuel S. Kutler, Dave L. Renfro]

All is Arithmetic A responce to a question posed by Stephen Wolfram regarding continuously iterated functions and the Ackermann function.

Hyper-operations Posting on A New Kind of Science Forum discussing the arithmetic operators of the Ackermann function and their inverse operators with link to an extended article on the subject.