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! JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 11/14/2011, 01:18 AM (This post was last modified: 11/14/2011, 02:24 AM by JmsNxn.) doing a little bit of algebra, it seems "meta superfunctions" are expressible by nested iteration: $f^{\diamond n}(x) = (f^{\diamond n-1} \circ f^{\diamond n})(x-1)$ which is obvious but it gives the relation: $f^{\diamond 2}(2) = (f^{\diamond 1} \circ f^{\diamond 1}) ( C )$ which, when plugged in with the iteration formula $f^{\diamond n}(x) = (f^{\diamond n-1})^{\circ x} ( C )$ gives $f^{\diamond 2}(2) = f^{\circ f^{\diamond 1} ( C )} ( C ) = f^{\circ f^{\circ C}( C )} ( C )$ which means $f^{\diamond 2}(x) = f^{\circ f^{\diamond 2}(x-1)} ( C )$ which to me resembles something like "compositional"/"iterational" tetration. $f^{\diamond 2}(x) = f^{\circ f^{\circ ...\,\, ^{\circ f ( C )}}} ( C )$ where there are x amount of nested iterations. actually, the generalized formula to give nested iterations is: $f^{\diamond n}(x) = (f^{\diamond n-2})^{\circ f^{\diamond n}(x-1)}( C )$ I think I just need a bit more time to make sense of this "meta compositional" operator $\diamond$ such that it gives a solution $(f^{\diamond \frac{1}{2}} \diamond f^{\diamond \frac{1}{2}})(x) = f^{\circ x}( C )$ which gives a recurrence relation which allows us to solve half -operators. tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 11/14/2011, 08:38 PM (This post was last modified: 11/14/2011, 08:47 PM by tommy1729.) (11/13/2011, 11:33 PM)JmsNxn Wrote: 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: $f^{\diamond n}(g^{\diamond n}(x)) = g^{\diamond n}(f^{\diamond n}(x)) = x$ $f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1)$ NO I DONT CONSIDER THE DEFINITION : the very first equation seems wrong from first glance already ( and im eating chocolate ! ) take the last equation take $f^{\diamond -n}$ on both sides => $x = f^{\diamond 1}(g^{\diamond n+1}(x) + 1)$ ( name this equation 3) this gives the same result as if we had used $g^{\diamond n}$ on both sides. hence together with the first equation we conclude f and g are inverse under the diamond operator. also now consider equation 3 and keep in mind all 3 equations and f and g are inverses then : take $g^{\diamond 1}$ on both sides : => $g^{\diamond 1}(x) = (g^{\diamond n+1}(x) + 1)$ for all n !!!! this looks really bad. for small n we get ( n = 0 ) $g^{\diamond 1}(x) = (g^{\diamond 1}(x) + 1)$ which seems paradoxal , and actually , it is. for larger n we get e.g. $g^{\diamond 1}(x) = (g^{\diamond 2}(x) + 1)$ $g^{\diamond 1}(x) = (g^{\diamond 3}(x) + 1)$ $g^{\diamond 1}(x) = (g^{\diamond 4}(x) + 1)$ $g^{\diamond 1}(x) = (g^{\diamond 5}(x) + 1)$ $g^{\diamond 1}(x) = (g^{\diamond 7}(x) + 1)$ $g^{\diamond 1}(x) = (g^{\diamond 1729}(x) + 1)$ ... till n approaching infinity !!! simultaneous ! it seems f and g are not functions , the diamond is not a function nor operator nor composition and none of them are finite !?!? i prefer different definitions and equations. once again your first post contains mistakes. all the rest that follows is even more paradoxal. no offense , but i prefer you look at it first ... mathematics requires thinking. at least twice. regards tommy1729 JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 11/16/2011, 12:11 AM (This post was last modified: 11/16/2011, 12:22 AM by JmsNxn.) you made a very major mistake... uhh $f^{\diamond - n}(f^{\diamond n + 1}(x)) \neq f^{\diamond 1}(x)$ this is exactly the assumption I was describing doesn't work. $(f^{\diamond n} \circ f^{\diamond m})(x) \neq f^{\diamond m +n}(x)$ instead, we'd need to create a new operator $(f^{\diamond n} \diamond f^{\diamond m})(x) = f^{\diamond m + n}(x)$ the diamond operator is NOT composition. And the diamond superscript is an index of superfunctions. $f^{\diamond n}(x) \neq (f_1 \circ f_2 \circ f_3 ... f_n)(x) = f^{\circ n}(x)$ instead, it's defined by $f(x) = f^{\diamond 1}(g^{\diamond 1}(x) + 1)$ $f^{\diamond 1}(x) = f^{\diamond 2}(g^{\diamond 2}(x) + 1)$ continued arbitrarily where $g^{\diamond n}(f^{\diamond n}(x)) = x$ seriously. You could try reading first. my definitions are perfectly consistent. I'll give you the benefit of the doubt for maybe misreading $\diamond$ for $\circ$ then again maybe you should try actually reading instead of eating chocolate... and thinking instead of claiming other people arent... JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 11/16/2011, 04:09 PM (This post was last modified: 11/16/2011, 09:21 PM by JmsNxn.) if perhaps I read you wrong, and maybe you meant to say $(f^{\diamond -n} \diamond f^{\diamond n})(x)= f(x)$ $(f^{\diamond -n} \diamond f^{\diamond n+1}(g^{\diamond n + 1}(x) + 1))(x) = f^{\diamond 1}(g^{\diamond n + 1}(x) + 1)$ this would imply inconsistency. Which to me is no surprise, the diamond operator was just suggested in a makeshift attempt to come up with a recurrence relation for fractional values of superfunction indexes. It was by no means canon. It hardly makes the rest of my post "inconsistent" as if you read carefully I even asked [in reference to the diamond operator]: Quote:So I'm wondering, is this just absurd? It's really the best I can come up with... That being said, the concept of $g^{\diamond n}(f^{\diamond n}(x)) = x$ $f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1)$ is still perfectly consistent which was the main point of the post. And it is hardly wrong at first glance. therefore I still can't help but think by your post that you mistook $\diamond$ for $\circ$ tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 11/16/2011, 06:48 PM (11/16/2011, 04:09 PM)JmsNxn Wrote: if perhaps I read you wrong, and maybe you meant to say $(f^{\diamond -n} \diamond f^{\diamond n})(x)= f(x)$ $(f^{\diamond -n} \diamond f^{\diamond n+1}(g^{\diamond n + 1}(x) + 1))(x) = f^{\diamond 1}(g^{\diamond n + 1}(x) + 1)$ this would imply inconsistency. Which to me is no surprise, it was just suggested in a makeshift attempt to come up with a recurrence relation for fractional values of superfunction indexes. It was by no means canon. It hardly makes the rest of my post "inconsistent" as if you read carefully I even asked [in reference to the diamond operator]: Quote:So I'm wondering, is this just absurd? It's really the best I can come up with... That being said, the concept of $g^{\diamond n}(f^{\diamond n}(x)) = x$ $f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1)$ is still perfectly consistent which was the main point of the post. And it is hardly wrong at first glance. therefore I still can't help but think by your post that you mistook $\diamond$ for $\circ$ ?? first you say you would not be surprised if it was inconsistant. then you continue , the rest of the post remains correct. but everything is based on the beginning equations ? as for composition , i know your diamond isnt composition. but it does share many properties. and you yourself gave a relation between the 0 th diamond of g and the compositional inverse of f. i did not even mention composition. assuming an equation is welldefined , it must be assumed that when inverses are introduced , they are introduced without switching branches. hence if we do not loose information or do forbidden operations , we get a group structure. this group structure has the diamond operator. and you write - before your diamonds , suggesting it has inverses. all diamond operators , f and g give a result that belongs to the same structure ( so we can continu taking diamonds ) hence all conditions are met for representation by groups. if the diamond is your group operator and the inverse is also a group operator in that group ( what must be as explained above ) , then we can do left-inverses of the group operators or aka taking diamonds and anti-diamonds on the left side. since by your first equation we can reduce the group operations (diamonds) we get at the contradion i posted. ( note what i did is more general then working with functional compositions of continu functions ... also a counterargument with integrals will fail because of the " + constant " after integrals ) so im far from convinced i made a mistake , if i did plz let me know clearly and keep in mind , you are the one proposing something so in fact it is actually up to you to prove that it exists. i might apologize if im wrong. but i doubt it that i am if you insist on all the equation you wrote down and x being real. sorry for the emotional shit regards tommy1729 tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 11/16/2011, 07:48 PM >> 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! btw , there already exist fractional operators : fractional calculus. where we have e.g. fractional derivatives. JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 11/16/2011, 10:57 PM (This post was last modified: 11/16/2011, 11:04 PM by JmsNxn.) okay okay we seem to just have a big miscommunication. the definition I was putting forth was $f^{\diamond n}(g^{\diamond n}(x)) = g^{\diamond n}(f^{\diamond n}(x)) = x$ so that $f^{\diamond n}(x)$ is the inverse of $g^{\diamond n}(x)$ $f^{\diamond n}(x) = f^{\diamond n+1}(g^{\diamond n+1}(x) + 1)$ which defines a sequence of super functions therefore $f^{\diamond 1}(x)$ is the superfunction of $f(x)$ and $f^{\diamond 2}(x)$ is the superfunction of $f^{\diamond 1}(x)$ so on and so forth this is all perfectly consistent. Now I related that to the Ackermann function, which is self-evident, and sort of generalizes the question of operators inbetween addition and multiplication that are analytic, to functions inbetween a subfunction and a superfunction. namely, what is $f^{\diamond \frac{1}{2}}(x)$, which if made sense of puts us a lot closer to an analytic Ackermann function. I wrote about the "diamond operator" which would have this relation: $(f^{\diamond m} \diamond f^{\diamond n})(x) = f^{\diamond m + n}(x)$ just as a shot in the dark to try and see if we can come up with a formula that relates half-superfunctions $f^{\diamond \frac{1}{2}}(x)$ and superfunctions $f^{\diamond 1}(x)$ I posed this operator not as if it would be true, but just as a "does this work?" Clearly it doesn't work, which frankly, doesn't surprise me. The point of this thread was to try to come up with a formula that will relate half-superfunctions $f^{\diamond \frac{1}{2}}(x)$ and superfunctions $f^{\diamond 1}(x)$ in the same way that we have a relation between half-iterates $f^{\circ \frac{1}{2}}(x)$ and a function itself $f(x)$ Until we can agree upon a proper relation between half-superfunctions and superfunctions (even if that relation is the fact that there is no relation) then we ultimately cannot solve the Ackermann function for fractional values. I was merely pointing out that if we do not have a relation between half-superfunctions and superfunctions then it will be about 100 times harder and will give about a 100 times more solutions. And I am well aware about fractional calculus operators. By fractional operators I was referring to operators inbetween addition and multiplication and exponentiation and tetration etc etc that form a smooth analytic transition for all arguments (save maybe a few holes or what have you). In conclusion. I just wanted to ask the question how can we make the transition from function to superfunction to super-superfunction etc etc analytic? And what qualifications should a half-superfunction have in terms of the function and its superfunction itself? that's it. The fact that the diamond operator is inconsistent doesn't really matter. That was just thrown in there to try to come up with a solution. Frankly I'm a little embarrassed I didn't see the obvious fault with it, lol I just got the feeling that you were implying $f^{\diamond n}(x)$ is inconsistent seperate from the diamond operator; which is not true. It's very simple and nothing about it implies inconsistency, unless you think an infinite sequence of superfunctions is inconsistent. tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 11/17/2011, 07:52 PM (11/16/2011, 10:57 PM)JmsNxn Wrote: I just got the feeling that you were implying $f^{\diamond n}(x)$ is inconsistent seperate from the diamond operator; which is not true. It's very simple and nothing about it implies inconsistency, unless you think an infinite sequence of superfunctions is inconsistent. hmm ... how does a converging infinite sequence of uniquely defined superfunctions look like any way ? fractal ?? another question or remark .. if a superfunction look like ( carleman f(z) ) ^ x , does that imply that fractional superfunction operators look like ( carleman f(z) ) ^ ^ x ? if so , we need to extend the bases E [1,eta]U[sqrt(e),oo[ to matrix bases ? does this imply that the (carleman) matrices require |det| = E[1,eta]U[sqrt(e),oo[ ?? again - as usual - many questions , more questions than answers. regards tommy1729 JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 11/18/2011, 12:41 AM (This post was last modified: 11/18/2011, 02:24 AM by JmsNxn.) (11/17/2011, 07:52 PM)tommy1729 Wrote: (11/16/2011, 10:57 PM)JmsNxn Wrote: I just got the feeling that you were implying $f^{\diamond n}(x)$ is inconsistent seperate from the diamond operator; which is not true. It's very simple and nothing about it implies inconsistency, unless you think an infinite sequence of superfunctions is inconsistent. hmm ... how does a converging infinite sequence of uniquely defined superfunctions look like any way ? fractal ?? another question or remark .. if a superfunction look like ( carleman f(z) ) ^ x , does that imply that fractional superfunction operators look like ( carleman f(z) ) ^ ^ x ? if so , we need to extend the bases E [1,eta]U[sqrt(e),oo[ to matrix bases ? does this imply that the (carleman) matrices require |det| = E[1,eta]U[sqrt(e),oo[ ?? again - as usual - many questions , more questions than answers. regards tommy1729 That's a good way to approach the question. I'm not too familiar with carleman matrices but I think it goes something like this Beware: the following maybe extremely inconsistent... Enter at own discretion $M[(f \circ g) (x)] = M[f(x)]M[g(x)]$ and by the recurrence relation $f^{\diamond n+1}(x) = (f^{\diamond n})^{\circ x} ( C_n )$ for any appropriate constant $f^{\diamond n+1}(0) = C_n$ that does not produce singularities or other problems. therefore $M[f^{\diamond 1}(x)] = M[f^{\circ x} ( C_0 )] = M[f( C_0 )]^x$ which is the traditional superfunction equation. continuing this we would get $M[f^{\diamond n + 1} ( x ) ] = M[(f^{\diamond n})^{\circ x}(C_n)] = M[f^{\diamond n} ( C_n )]^x$ $M[f^{\diamond n+1}(x)] = M[(f^{\diamond n-1})^{\circ C_{n}} ( C_{n-1} ) ]^x = (M[f^{\diamond n-1} ( C_{n-1} ) ]^{ C_n })^x$ which would give: $M[f^{\diamond n + 1}(x)] = ((...(M[f(C_0)]^{C_1})^{C_2})...)^{C_n})^x$ This isn't good though, this would imply the matrix method doesn't work for the Ackermann function since $C_k = 1$ Arghhhh if only I knew more about matrices. I hope you can help me out. Hopefully your guess is right, that it somehow turns into tetration across matrices. At least that way we get something non-contradictory. but there is hope, if $M[f^{\diamond n + 1}(x)] = ((...(M[f(C_0)]^{C_1})^{C_2})...)^{C_n})^x$ is valid then we can have a solution for the modified Ackermann function; namely $a\,\,\bigtriangleup_\sigma^K\,\,b = a\,\,\bigtriangleup_{\sigma - 1}^K\,\,(a\,\bigtriangleup_\sigma^K\,\,b-1)$ where $a\,\,\bigtriangleup_{1}^K\,\,b = a + b$ and $a\,\,\bigtriangleup_{n}^K\,\,K = a$ for $n \in \mathbb{Z};\,\,n > 1$ for multiplication $a\,\,\bigtriangleup_{2}^K\,\,b = a \cdot (b - K + 1)$ and for exponentiation $a\,\,\bigtriangleup_{3}^K\,\,b + K = a^{b+1} + (1-K) \cdot \sum_{c = 1}^{b} a^c$ $a\,\,\bigtriangleup_{3}^K\,\,b = a^{b-K+1} + (1-K) \cdot \sum_{c = 1}^{b-K} a^c$ tetration would be ridiculously complex. with this we have the relation $f^{\diamond \sigma + 1}(b) = a\,\,\bigtriangleup_{\sigma + 1}^K\,\,b = ((...(M[f(a)])^{f^{\diamond 1}(0)})^{f^{\diamond 2}(0)})...)^{f^{\diamond \sigma}(0)})^b$ with $f(x) = x + 1$ and $f(a) = f^{\diamond 1}(1)$ Hopefully you or somebody can correct me if I'm wrong. I think that I'm wrong; this seems too contradictory and fishy to be right... The fact that it fails for the usual Ackermann function makes me angry D:< If this is right though, we may have a tool for solving half-superfunctions given certain parameters that $f^{\diamond n}(0) \neq 1$ which is the equivalent of $f^{\diamond n}(x) \neq (f^{\diamond n - 1})^{\circ x}(1)$ In a sense you were right about the tetration intuition. Except that it's like a left-handed tetration! which forms an independent operator over matrices since matrix multiplication is not commutative! EDIT: The formula seems to change from the original one $M[f^{\diamond n + 1}(x)] = ((...(M[f(C_0)]^{C_1})^{C_2})...)^{C_n})^x = ((...(M[f^{\diamond 1}(1)]^{f^{\diamond 2}(0)})^{f^{\diamond 3}(0)})...)^{f^{\diamond n+1}(0)})^x$   to $((...(M[f^{\diamond 1}(1)]^{f^{\diamond 2}(0)})^{f^{\diamond 3}(0)})...)^{f^{\diamond n}(0)})^x$ not sure why that is, must've made a minor mistake somewhere... Too tired right now to figure out where. I think I've slashed through enough mathematical jungle for one day. Maybe you can clarify my errors... « Next Oldest | Next Newest »

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