Ackermann function and hyper operations
#2
@Andrew: Good! Thats a very attentive observation.


While verifying myself I found that the deviation (up to a difference of 1 in the rank) from our operator sequence comes from forming an unnecessary odd initial condition. I dont know why he does, perhaps it is more suitable for his proof of non-primitive recursiveness.

In his article
Ackermann, W. (1928 ). Zum Hilbertschen Aufbau der reellen Zahlen. Math. Ann., 99, 118–133.

Ackermann defines:
\( \varphi(a,b,0)=a+b \)
\( \varphi(a,b,n+1)=\varrho_c(\varphi(a,c,n),\alpha(a,n),b) \).

Where \( \varrho_c(f( c),a,n)=f^{[n]}(a) \) in our notation (\( c \) is a free variable here to determine the argument of the function to be applied).

So everything would be good if the initial value was \( \alpha(a,n)=1 \) for \( n\ge 1 \). Then would \( \varphi(a,b,n)=a [n+1] b \). But instead Ackermann defines:

\( \alpha(a,n) = \iota(n,1)\cdot \iota(n,0) \cdot a + \lambda(n,1) \) where \( \iota(a,b)=1-\delta_{a,b} \) and \( \lambda(a,b)=\delta_{a,b} \) in today notation with the Kronecker-\( \delta \).
This definition is hence equivalent to:
\( \alpha(a,n) = 0 \) for \( n=0 \)
\( \alpha(a,n) = 1 \) for \( n=1 \)
\( \alpha(a,n) = a \) for \( n\ge 2 \)
as he also mentiones in his paper.

So he introduces a third initial value \( \alpha(a,n)=a \) besides 0 and 1 which causes the deviation from our operator sequence:
\( \varphi(a,b,1)=(c\mapsto\varphi(a,c,0)^{[b]}(\alpha(a,0))=(c\mapsto a+c)^{[b]}(0)=a\cdot b \).
\( \varphi(a,b,2)=(c\mapsto \varphi(a,c,1))^{[b]}(\alpha(a,1))=(c\mapsto a\cdot c)^{[b]}(1)=a^{b} \)
\( \varphi(a,b,3)=(c\mapsto \varphi(a,c,2))^{[b]}(\alpha(a,2))=(c\mapsto a^c)^{[b]}(a)=a [4] (b+1) \)

PS: Munafo gives a very detailed description of the different versions of the Ackermann-function here. It is a very good reference to show to someone for explaining about different versions of the Ackermann-function. All glory to Andrew for digging out such references.


Messages In This Thread
The Ackermann function - by bo198214 - 08/23/2009, 09:45 AM
RE: The Ackermann function - by andydude - 08/24/2009, 01:09 AM
RE: The Ackermann function - by bo198214 - 04/18/2011, 05:08 PM

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