Proof Ackermann function cannot have an analytic identity function JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 11/11/2011, 02:26 AM (This post was last modified: 11/11/2011, 02:32 AM by JmsNxn.) Well I've been having suspicions for quite a while that the Ackermann function cannot be analytic. I was having trouble visualizing it and then one step to the answer came to me through the identity function. It's actually incredibly simple the formulas involved. It's a proof by contradiction. First, we start off by defining what we mean by the "Ackermann function", so as not to cause any confusion. $\vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b$ where: $a\,\,\bigtriangleup_{\sigma - 1} (a\,\,\bigtriangleup_\sigma\,\,b) = a\,\,\bigtriangleup_\sigma\,\,b+1$ and if $a\,\,\bigtriangleup_{\sigma}\,\,S(\sigma) = a$ or $S(\sigma)$ is the identity function we set: $a\,\,\bigtriangleup_{1}\,\,b = a + b$ so that $S(1) = 0$ and we make the final two requirements: $n \in \mathbb{Z}\,\,;n \ge 2$ $S(n) = 1$ so that we have $\bigtriangleup_2$ as multiplication and $\bigtriangleup_3$ as exponentiation so on and so forth. if $L_a^{\sigma}(a\,\,\bigtriangleup_\sigma\,\,b) = b$ or $L_a^{\sigma}(b)$ is the "logarithmic" inverse operator of order $\sigma$ $L_a^{\sigma}(a) = S(\sigma)$ must be analytic across $\sigma$ therefore $L_a^{\sigma}(b)$ is analytic across $\sigma$ at all values of b if and only if our identity function ($S(\sigma)$) is analytic. We will work to prove that $L_a^{\sigma}(b)$ cannot be analytic by $\sigma$ at all values of b. So we come to the weaker conclusion than the Ackermann function cannot be analytic across all values, to the "logarithmic" Ackermann function cannot be analytic across all values. or the series $L_a^{1}(b) = b - a$ $L_a^{2}(b) = \frac{b}{a}$ $L_a^{3}(b) = \log_a(b)$ $L_a^{4}(b) = \text{slog}_a(b)$ cannot be analytic if it holds the property: $L_a^{\sigma}(a\,\,\bigtriangleup_\sigma\,\,b) = b$ Firstly, we make the argument that $S(\sigma)$ cannot be periodic if we want it to be analytic. Or simply put if: $S(\sigma + 1) = S(\sigma)$ this is instantly invalid for $S(1) = 0$ and $S(2) = 1$ and an analytic periodic function must be periodic for all values, however, we do have it to be true for integer values of $\sigma$ greater than 1. Now we write, since $S(2) = S(3) = S(4) = ... = 1$ and since $S(\sigma) \neq 1$, $S(\sigma)$ must oscillate similarly to a periodic function only it is not periodic since if it was periodic and analytic $S(1) = 1$ therefore we designate it cannot be analytic and periodic. So let us write what I call the "pseudo period" as $S(\sigma + \ell(\sigma)) = S(\sigma)$ evidently $\ell(\sigma)$ is multivalued and not analytic and does not always return values. for example $\ell(0)$ does not necessarily exist, because it is fully possible that $S(\sigma) > 0$ ; $\sigma > 1$ however, it is evident that $\ell(\sigma)$ must exist for some values of $\sigma$ in order that $S(\sigma)$ be analytic. it is trivial that $n, k \in \mathbb{Z} ; n \ge 2 ; k \ge 0$ $\ell(n) = k$ Now we must derive a few lemmas consider $(1 + S(\sigma))\,\, \bigtriangleup_{\sigma - 1}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\,\bigtriangleup_{\sigma - 1}\,\,((1 + S(\sigma))\,\,\bigtriangleup_{\sigma}\,\,S(\sigma))$ by the first axiom it's easy to deduce: $\text{I:}\,\,(1 + S(\sigma))\,\, \bigtriangleup_{\sigma-1}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\, \bigtriangleup_{\sigma}\,\, (1 + S(\sigma))$ continuing this process again we get: $(1 + S(\sigma))\,\, \bigtriangleup_{\sigma-1}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\, \bigtriangleup_{\sigma}\,\, (1 + S(\sigma)) = (1 + S(\sigma))\,\, \bigtriangleup_{\sigma+1}\,\, (1 + S(\sigma+1))$ this is how we get the famed identity $2\,\,\bigtriangleup_n\,\,2 = 4$ Now we move on to the "logarithmic" inverse function and defining a new function I call Q. $L_a^{\sigma}(b) + Q^{\sigma}(t, a, b) = L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma + t}\,\,b)$ it's obvious that $Q^{\sigma}(-1, a, b) = 1$ which is a simple exercise in super operators and $Q^{\sigma}(t, a, b) = L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma + t}\,\,b) - L_a^{\sigma}(b)$ before continuing we need to acknowledge that $L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma+1}\,\,b) = a\,\,\bigtriangleup_{\sigma+1} (b-1)$ which is again a simple exercise. now we have to make some odd observations that may get confusing $L_{1+S(\sigma)}^{\sigma + \ell(\sigma)}((1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma) + 1}\,\,(1+S(\sigma + \ell(\sigma) + 1)) = (1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma) + 1}\,\,S(\sigma + \ell(\sigma) + 1) = 1 + S(\sigma)$ from the "logarithmic" inverse law $L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma+1}\,\,b) = a\,\,\bigtriangleup_{\sigma+1} (b-1)$ this implies: $L_{1+S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma)) + Q^{\sigma + \ell(\sigma)}(1, 1 + S(\sigma), 1 + S(\sigma + \ell(\sigma) + 1)) = 1 + S(\sigma)$ which in turn gives $Q^{\sigma + \ell(\sigma)}(1, 1 + S(\sigma), 1 + S(\sigma + \ell(\sigma) + 1)) = 1$ since $L_{1+S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma)) = S(\sigma + \ell(\sigma)) = S(\sigma)$ by the identity $L_a^{\sigma}(a) = S(\sigma)$ but we also know by Q's definition $Q^{\sigma}(t, a, b) = L_a^{\sigma}(a\,\,\bigtriangleup_{\sigma + t}\,\,b) - L_a^{\sigma}(b)$ so that $Q^{\sigma + \ell(\sigma)}(1, 1 + S(\sigma), 1 + S(\sigma + \ell(\sigma) + 1)) = 1 = L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}((1 + S(\sigma))\,\,\bigtriangleup_{1 + \sigma + \ell(\sigma)} (1 + S(\sigma + \ell(\sigma) + 1))) - L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))$ we know that from earlier $L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}((1 + S(\sigma))\,\,\bigtriangleup_{1 + \sigma + \ell(\sigma)} (1 + S(\sigma + \ell(\sigma) + 1))) = 1 + S(\sigma)$ so that we get $1 = 1 + S(\sigma) - L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))$ $S(\sigma) = L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))$ but we also know, by the definition of the "logarithmic" inverse: $a\,\,\bigtriangleup_\sigma\,\,L_a^{\sigma}(b) = b$ so therefore $(1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma)}\,\,S(\sigma) = (1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell(\sigma)}\,\,L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1))$ which since $S(\sigma) = S(\sigma + \ell(\sigma))$ and $(1 + S(\sigma))\,\,\bigtriangleup_{\sigma + \ell}\,\,L_{1 + S(\sigma)}^{\sigma + \ell(\sigma)}(1 + S(\sigma + \ell(\sigma) + 1)) = 1 + S(\sigma + \ell(\sigma) + 1)$ we get the beautiful result: $S(\sigma) = S(\sigma + \ell(\sigma) + 1)$ which is the core identity of our proof since $\ell(\sigma)$ is multivalued, we find that for some value $q > 0$ if $S(\sigma + q) = S(\sigma)$ then $S(\sigma + q + 1) = S(\sigma)$ which in turn applies $\ell(\sigma) = q + 1$ which iterated gives $k \in \mathbb{Z} ; k \ge 0$ $S(\sigma + q + k) = S(\sigma + q)$ or that $S(\sigma + q)$ is periodic. However we already noted that if $S(\sigma)$ is periodic then $S(\sigma)$ cannot be analytic since by definition of an analytic function that is periodic it must be periodic over every value in the domain and $S(1) \neq S(2)$ Therefore $L_a^{\sigma}(b)$ is not analytic by $\sigma$ for all values of b and $S(\sigma)$ is not analytic by $\sigma$ Q.E.D. This puts me one step closer to proving non analycity across $\sigma$ for some value of a and b in the Ackermann function $\vartheta(a,b,\sigma)$ Personally I think that non-analycity is intuitively implied since $\vartheta(a,b,\sigma)$ is not primitive recursive, but this is yet to be rigorously decided. To disprove universal analycity, it suffices to show that for some fixed $u, v \in C$, $\vartheta(u,v,\sigma)$ is not analytic across $\sigma$. I understand the notation in this proof maybe a bit confusing, but when working with three variables I think that's inevitable. any questions, comments? « Next Oldest | Next Newest »

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