
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,
Systems of Notation for Arithmetic Operators
Arithmetic 
Standard 
Ackermann

Knuth 
Conway 
Addition 
a+b 
ack(a,b,0)



Multiplication 
a*b 
ack(a,b,1) 


Exponentiation 
a^{b} 
ack(a,b,2) 
a↑b 
a→b→1 
Tetration 
^{b}a 
ack(a,b,3) 
a↑↑b 
a→b→2 
Pentation 
_{b}a 
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
f^{3}(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 f^{n}(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 e^{e1}≈ 1.444 where (e^{e1})^{e} = e^{e e1} = 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]
Wikipedia: Ackermann function
All is Arithmetic A responce to a question posed by Stephen Wolfram regarding continuously iterated functions and the Ackermann function.
Hyperoperations 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.
