• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 generalizing the problem of fractional analytic Ackermann functions JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 11/24/2011, 01:18 AM (This post was last modified: 11/24/2011, 01:44 AM by JmsNxn.) Oh I know this is by no means concrete yet, and your concerns are viable and duly noted. I probably should've made it explicit that there will no doubt be restrictions on $f$ and $t$ when we talk about $F^t \{ f \} (x)$; what those restrictions are though, that's still up for debate. For the consideration of the superfunction fix point at successorship I thought of a way of accommodating it, but it only really works in definition of the hyperoperation sequence. We define the superfunctions by the identity function: $S(\sigma)$, so that $S(1) = 0; S(2) = 1; S(3) = 1 ... S(n) = 1; n \in \mathbb{Z}/ \{ 0 \}$ and then for negative integers we define it relative to the base value, i.e: $S(-n) = a -1$ then we get the result: $a\,\,\bigtriangleup_\sigma\,\,S(\sigma) = a$ $a\,\,\bigtriangleup_{\sigma -1} \,\,(a\,\,\bigtriangleup_\sigma\,\,b) = a\,\,\bigtriangleup_\sigma\,\,(b+1)$ this defines $a\,\,\bigtriangleup_{-n}\,\,b = b+1$ $a\,\,\bigtriangleup_{0}\,\,b = b + 1$ and $a\,\,\bigtriangleup_{1} b = a + b$ and so on and so forth as the hyperoperation sequence continues. This is more easily expressed using the diamond notation: $f^{\diamond \sigma}(x) = ((f^{\diamond \sigma-1})^{\circ x})(f^{\diamond \sigma}(0))$ and now the definition of setting the identity function for $S(1) = 0$ gives us the result that $f^{\diamond \sigma}(b) = a\,\,\bigtriangleup_\sigma\,\,b$ $f^{\diamond 1}(0) = a$ where as for all negative integers and zero we get $f^{\diamond 0}(0) = 1 ; f^{\diamond -1}(0) = 1 ; f^{\diamond -2}(0) = 1 ... f^{\diamond -n}(0) = 1$ It's necessary that we'd make the requirement for all integers except 1 and 2 $k \in \mathbb{Z}/ \{ 1,2 \}; f^{\diamond k}(0) = 1$ where $f^{\diamond 1}(0) = a$ and $f^{\diamond 2}(0) = 0$ now apply that to $f^{\diamond n}(x) = ((f^{\diamond n-1})^{\circ x})(f^{\diamond n}(0))$ and we get successorship for negative integers and zero, and addition at one, multiplication at two etc. etc... It just requires a little bit of algebraic manipulation and specification as to what we really mean when we say $F^t \{ f \} (x)$ I think this answers your question earlier about generating the sequence of superfunctions. Though rather simply and not with much usefulness. also, $S(\sigma) = a\,\,\bigtriangleup_{\sigma + 1}\,\,(S(\sigma + 1) - 1)$; this is how I derive that all the positive integers greater than two produce one at zero $f^{\diamond k}(0) = 1$. And two produces zero at zero $f^{\diamond 2}(0) = 0$ Furthermore, I've been figuring that it's perfectly possible to have: $a\,\,\bigtriangleup_{-0.5}\,\,b \neq a\,\,\bigtriangleup_{-1.5}\,\,b$ even though, technically they both equal: $F^{-0.5} \{ f \} (b)$ where $f(b) = b + 1$ The methods by which we would deduce this to me seem purely based on an "analytic solution". Quite obviously the hyper operation sequence cannot start being periodic once it hits zero and the negative real numbers; and also, it cannot simply become a constant successorship for all negative reals; so we'll have to create a formula that satisfies: $a\,\,\bigtriangleup_{-n-\frac{1}{2}}\,\,(a\,\,\bigtriangleup_{-n + \frac{1}{2}}\,\,b) = a\,\,\bigtriangleup_{-n + \frac{1}{2}}\,\,b+1$ and even just looking at this equation we cannot have $\bigtriangleup_{-\frac{1}{2}}$ be successorship because then $\bigtriangleup_{\frac{1}{2}}$ would be a variation of addition and would undoubtedly not be analytic across sigma. Partially, I believe, where as you say "it isn't true"--which I agree with, I think it's more "abuse of notation". And that the actual means by which a superfunction is created is far more complicated then a simple transformation, as my notation implies. BUT, and this is an important but, the notation very much adds to the visualization and interpretation of how such a superfunction sequence would be generated. And until we have more rigorous ties to how such a sequence would be generated we're left with primitive formations. If you find errors, and full on problems with the notation, we can address them and carve out how to work around them, or if necessary, scrap the notation and start all over again. Because, quite frankly, I don't think there is much research out there on "half-superfunctions". Partly because superfunctions themselves are still very much unexplored. And by all means, I see your argument, if superfunctions are created through fixpoints and can only be created from the previous "sub function" (that's what I think people have been calling the inverse superfunction), how is it possible to iterate that process from only a single function with, presumably, only one fixpoint? It seems totally unfeasible! And the question of periodic superfunctions. Well that would be interesting, and I think I wrote out a few notes on functions that satisfy identities of that sort: $g(x) = f(f^{-1}(x) + 1)$ $f(x) = g(g^{-1}(x) + 1)$ this would give $g(x) = g(g^{-1}(f^{-1}(x) + 1) + 1) = g(g^{-1}(g(g^{-1}(x) - 1) + 1) + 1)$ I'm not sure if I managed to prove any lemmas about it, it may have gotten too convoluted, and I generally lose these notes; there more just done to keep my brain active. I think you can extend that to iteration as: $g^{\circ t}(x) = f(f^{-1}(x) + t) = g(g^{-1}(g(g^{-1}(x) - 1) + t) + 1)$ I'm sure you can deduce tons of equations like that for integer periodic superfunctions. I'm totally clueless as to what you would do for fractional or complex periods, let alone quasi periods! And the question of $\lim_{n \to \infty} F^{n} \{ f \} (x)$; I can't even begin to interpret the meaning of that question! And someone in another thread said the Ackermann function will not be continued analytically in our lifetime. I can agree with that, considering that the Abel function (which is essentially the superfunction) wasn't investigated till the early 1800s, and only now are we really creating powerful methods for actually determining abel functions of special functions. I'm just hoping to clear some of the path and prove a few lemmas about some of the requirements of half-superfunctions It's almost all speculative. « 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

 Possibly Related Threads... Thread Author Replies Views Last Post Using a family of asymptotic tetration functions... JmsNxn 15 4,644 08/06/2021, 01:47 AM Last Post: JmsNxn On my old fractional calculus approach to hyper-operations JmsNxn 14 3,826 07/07/2021, 07:35 AM Last Post: JmsNxn Brute force tetration A_k(s) is analytic ! tommy1729 9 2,870 03/22/2021, 11:39 PM Last Post: JmsNxn [exercise] fractional iteration of f(z)= 2*sinh (log(z)) ? Gottfried 4 1,977 03/14/2021, 05:32 PM Last Post: tommy1729 Nixon-Banach-Lambert-Raes tetration is analytic , simple and “ closed form “ !! tommy1729 11 4,406 02/04/2021, 03:47 AM Last Post: JmsNxn Math overflow question on fractional exponential iterations sheldonison 4 9,516 04/01/2018, 03:09 AM Last Post: JmsNxn [repost] A nowhere analytic infinite sum for tetration. tommy1729 0 3,104 03/20/2018, 12:16 AM Last Post: tommy1729 Analytic matrices and the base units Xorter 2 5,525 07/19/2017, 10:34 AM Last Post: Xorter The AB functions ! tommy1729 0 3,221 04/04/2017, 11:00 PM Last Post: tommy1729 THE problem with dynamics tommy1729 1 4,787 04/04/2017, 10:52 PM Last Post: tommy1729

Users browsing this thread: 1 Guest(s)