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Proof Ackermann function cannot have an analytic identity function - JmsNxn - 11/11/2011
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. where: and if or is the identity function we set: so that and we make the final two requirements: so that we have as multiplication and as exponentiation so on and so forth. if or is the "logarithmic" inverse operator of order must be analytic across therefore is analytic across at all values of b if and only if our identity function () is analytic. We will work to prove that cannot be analytic by 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 cannot be analytic if it holds the property: Firstly, we make the argument that cannot be periodic if we want it to be analytic. Or simply put if: this is instantly invalid for and and an analytic periodic function must be periodic for all values, however, we do have it to be true for integer values of greater than 1. Now we write, since and since , must oscillate similarly to a periodic function only it is not periodic since if it was periodic and analytic therefore we designate it cannot be analytic and periodic. So let us write what I call the "pseudo period" as evidently is multivalued and not analytic and does not always return values. for example does not necessarily exist, because it is fully possible that ; however, it is evident that must exist for some values of in order that be analytic. it is trivial that Now we must derive a few lemmas consider by the first axiom it's easy to deduce: continuing this process again we get: this is how we get the famed identity Now we move on to the "logarithmic" inverse function and defining a new function I call Q. it's obvious that which is a simple exercise in super operators and before continuing we need to acknowledge that which is again a simple exercise. now we have to make some odd observations that may get confusing from the "logarithmic" inverse law this implies: which in turn gives since by the identity but we also know by Q's definition so that we know that from earlier so that we get but we also know, by the definition of the "logarithmic" inverse: so therefore which since and we get the beautiful result: which is the core identity of our proof since is multivalued, we find that for some value if then which in turn applies which iterated gives or that is periodic. However we already noted that if is periodic then cannot be analytic since by definition of an analytic function that is periodic it must be periodic over every value in the domain and Therefore is not analytic by for all values of b and is not analytic by Q.E.D. This puts me one step closer to proving non analycity across for some value of a and b in the Ackermann function Personally I think that non-analycity is intuitively implied since is not primitive recursive, but this is yet to be rigorously decided. To disprove universal analycity, it suffices to show that for some fixed , is not analytic across . 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? |