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 Ackermann function and hyper operations andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 08/23/2009, 01:37 AM I wanted to say something about the first part, "1 Introduction". In this part of the paper, you equate the original Ackermann function with a[n]b, which (stictly speaking) is not true. Robert Munafo has discussed this on his website (http://www.mrob.com/pub/math/largenum.html), and I have also verified this for myself by reading the original paper Ackermann wrote. Ackermann's tetration is a function $f(n, m) = \phi(n, m, 3) = {}^{(m+1)}n$ which is an offset from tetration. Because it is an offset of tetration, $\phi(n, m, 4)$ is not pentation at all. It wasn't until Reuben Goodstein that the "offset" disappeared. I think we should take this into account when discussing the original Ackermann function, because its evolution is way more complicated than any introduction can really summarize. bo198214 Administrator Posts: 1,595 Threads: 101 Joined: Aug 2007 08/23/2009, 09:45 AM (This post was last modified: 08/23/2009, 12:34 PM by bo198214.) @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. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 08/24/2009, 01:09 AM Also, about a month ago, I redesigned the Hyperoperation page, to try and explain these differences. bo198214 Administrator Posts: 1,595 Threads: 101 Joined: Aug 2007 04/18/2011, 05:08 PM (This post was last modified: 04/18/2011, 05:09 PM by bo198214.) (08/23/2009, 09:45 AM)bo198214 Wrote: 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. Oh now I found out where this odd initial conditions comes from! I assert that Ackermann originally wanted to define left-braced hyperoperations! Then this initial condition $\alpha(a,n)=a$ for $n\ge 2$ makes sense! Left-braced hyperoperations $\psi$ would similarly be defined by: $\psi(a,b,0)=a+b$ $\psi(a,b,n+1)=(x\mapsto \psi(x,a,n))^{[b]}(\alpha(a,n))$ here again we have $\psi(a,b,1)=ab$ and $\psi(a,b,2)=a^b$. But the forth operation is not $a^{a^{b-1}}$ as one would obtain with the initial condition $\alpha(a,2)=1$, but it is $\psi(a,b,3)=a^{a^b}$ due to the initial value $\alpha(a,2)=a$! So this initial condition makes left-braced hyperoperations look simpler, while it makes right-braced hyperoperations looking odd. I think he started with the left-braced hyperoperations and then switched to the faster growing right-braced hyperoperations, perhaps it was more suitable for his proof of non-primitive recursiveness of $n\mapsto \phi(n,n,n)$ « Next Oldest | Next Newest »

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