generalizing the problem of fractional analytic Ackermann functions JmsNxn Ultimate Fellow Posts: 1,214 Threads: 126 Joined: Dec 2010 11/13/2011, 11:33 PM (This post was last modified: 11/14/2011, 02:36 AM by JmsNxn.) Well there's an obvious connection between the Ackermann function and superfunctions, but now I've found a way to generalize this to what I am labeling "meta-superfunctions". These have to do with iteration but in a very different complicated light. consider the definition: $f^{\diamond n}(g^{\diamond n}(x)) = g^{\diamond n}(f^{\diamond n}(x)) = x$ $f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1)$ or so that $f^{\diamond n + 1}(x)$ is the super function of $f^{\diamond n}(x)$ and $g^{\diamond n + 1}(x)$ is the abel function of $f^{\diamond n}(x)$ this gives the relation: $f^{\diamond n + 1}(x) = (f^{\diamond n} \circ f^{\diamond n} \circ ...\text{x amount of times}... \circ f^{\diamond n}) ( C ) = (f^{\diamond n})^{\circ x} ( C )$ for some arbitrary constant C, that gives no singularities and such. we'd center the equation at zero, so that: $f^{\diamond 0}(x) = f(x)$ $g^{\diamond 0}(x) = f^{\circ -1}(x)$ the question I ask is, how do we find fractional values of n? Or put more linguistically, what function is in between a function and its super function? The connection to the Ackermann function is simple: $\vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b$ $L_a^{\sigma}(\vartheta(a,b,\sigma)) = b$ so that we get $\vartheta(a,L_a^{\sigma + 1}(b) + 1, \sigma + 1) = \vartheta(a,b,\sigma)$ which is a simple exercise in super functions now to split confusion we'll rewrite this in "meta superfunction" notation, fixing a to some arbitrary constant $f^{\diamond \sigma}(b) = \vartheta(a,b,\sigma)$ $g^{\diamond \sigma}(b) = L_a^{\sigma}(b)$ we get the simple result from out previous formula above $f^{\diamond \sigma}(b) = f^{\diamond \sigma + 1}(g^{\diamond \sigma + 1}(b) + 1)$ and this defines the Ackermann function in terms of "meta superfunctions" Now, the difficulty with defining fractional values of these "meta superfunctions" is that we do not have a recurrence relation for fractional values... The difference between that and regular superfunctions is that we do, namely $f^{\circ \frac{1}{2}}(f^{\circ \frac{1}{2}}(x)) = f(x)$ there is no such property or relation for "meta superfunctions" the only property we know it must satisfy is: $f^{\diamond \frac{1}{2}}(b) = f^{\diamond \frac{3}{2}}(g^{\diamond \frac{3}{2}}(b) + 1)$ but this is no good because it doesn't give an expression in terms of an even integer superfunction. So my question is, are meta-superfunctions by nature discrete? and inconclusively continuous? Since there is no external property they need satisfy other than maybe $R(x) = f^{\diamond x}(b)$ with R(x) being analytic across x. So, I believe, before the Ackermann function can be continued analytically we must define a recurrence relation across meta superfunctions that applies to fractional values. My rather meagre attempt at this recurrence relation comes from defining a new operator. I call it "meta composition". It's rather vague and does not have a concrete definition, but it's the best I can come up with. $(f^{\diamond m} \diamond f^{\diamond n})(x) = f^{\diamond m + n}(x)$ this would mean that the identity of meta composition is the function itself $(f \diamond f^{\diamond n})(x) = f^{\diamond n}(x)$ which implies idempotency and also that $\diamond$ is defined relatively based on each function. Actually, $\diamond$ is very much like the composition operator $\circ$ except it is reformulated with $f$ as the identity function rather than the normal identity function $f(x) = x$ Actually thinking about it, I'm sure I've read somewhere of composition operators redefined to suit different identity functions. If this was the case this could perhaps be an approach to a solution for analytic fractional operators! So I'm wondering, is this just absurd? It's really the best I can come up with... any other possible "meta superfunction" recurrence relations would be welcomed with a warm heart! « Next Oldest | Next Newest »

 Messages In This Thread generalizing the problem of fractional analytic Ackermann functions - by JmsNxn - 11/13/2011, 11:33 PM RE: generalizing the problem of fractional analytic Ackermann functions - by JmsNxn - 11/14/2011, 01:18 AM RE: generalizing the problem of fractional analytic Ackermann functions - by tommy1729 - 11/14/2011, 08:38 PM RE: generalizing the problem of fractional analytic Ackermann functions - by JmsNxn - 11/16/2011, 12:11 AM RE: generalizing the problem of fractional analytic Ackermann functions - by JmsNxn - 11/16/2011, 04:09 PM RE: generalizing the problem of fractional analytic Ackermann functions - by tommy1729 - 11/16/2011, 06:48 PM RE: generalizing the problem of fractional analytic Ackermann functions - by tommy1729 - 11/16/2011, 07:48 PM RE: generalizing the problem of fractional analytic Ackermann functions - by JmsNxn - 11/16/2011, 10:57 PM RE: generalizing the problem of fractional analytic Ackermann functions - by tommy1729 - 11/17/2011, 07:52 PM RE: generalizing the problem of fractional analytic Ackermann functions - by JmsNxn - 11/18/2011, 12:41 AM RE: generalizing the problem of fractional analytic Ackermann functions - by Gottfried - 11/18/2011, 06:13 AM RE: generalizing the problem of fractional analytic Ackermann functions - by JmsNxn - 11/19/2011, 07:50 PM RE: generalizing the problem of fractional analytic Ackermann functions - by JmsNxn - 11/20/2011, 09:26 PM RE: generalizing the problem of fractional analytic Ackermann functions - by Gottfried - 11/21/2011, 10:33 AM RE: generalizing the problem of fractional analytic Ackermann functions - by tommy1729 - 11/21/2011, 08:08 PM RE: generalizing the problem of fractional analytic Ackermann functions - by JmsNxn - 11/22/2011, 02:13 AM objection to linear model for superfunctionoperator. - by tommy1729 - 11/23/2011, 06:00 PM RE: generalizing the problem of fractional analytic Ackermann functions - by JmsNxn - 11/24/2011, 01:18 AM

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