Uniqueness Daniel Long Time Fellow    Posts: 279 Threads: 94 Joined: Aug 2007 12/27/2008, 08:29 AM Hello, I’ve been trying to assess the progress made lately in extending tetration. I am happy with the state of my own work on the subject, but worry that it is a happiness born of ignorance, since no one has critiqued my work. I haven’t been in a position to write my own work up on tetration and the Ackermann function over the last few years, but that has now changed. With so many excellent minds working on the subject, it has been somewhat of a mystery to me as to why someone hasn’t already published a paper extending tetration to everyone’s satisfaction. That is why I am interested in the idea presented on this forum that the blockage in extending tetration comes down to establishing the uniqueness of the solution. Is this really the only issue to proving the extension of tetration? Daniel Daniel bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 12/27/2008, 12:03 PM (This post was last modified: 12/27/2008, 12:04 PM by bo198214.) Hi Daniel, of course the uniqueness is one essentially important issue about tetration. Currently we have collected here perhaps 5 different methods to compute an analytic tetration. If they are different, which is 'the correct', 'the right', 'the proper' or 'the best' tetration? This question (though there were not different computational approaches at his time) already appeared to Szekeres in his article "Fractional iteration of exponentially growing functions" 1961. Where he also mentions that if we have one analytic superexponential sexp, then the function $f(x)=\text{sexp}(x+\frac{1}{2\pi}sin(2\pi x))$ is another analytic superexponential, which is even strictly increasing if sexp was. For overview here are the different established methods to compute an analytic tetration:Regular iteration at the lower real fixed point (for bases $e^{1/e}$). Jay's method via a direct limit (for bases $>^{1/e}$, is it analytic?). To my knowledge the methods except regular iteration are not well-investigated yet. There are for example no convergence proofs yet (though practically they all converge) and it is not known which methods yield the same tetration. In opposition to the numerous computation methods, uniqueness criterions are rare. To my knowledge the only ones existing are the ones I tried to present here on the forum. And I currently prepared an article about uniqueness with Dmitrii. My vision is that all (or at least most) analytic tetration methods become united by showing that they are equal, by satisfying the same uniqueness criterion. Only if this is achieved one can properly speak about "the" tetration. Otherwise its just a bunch of personal taste tetrations. Finitist Junior Fellow  Posts: 5 Threads: 0 Joined: Sep 2008 01/01/2009, 07:28 PM Hi Daniel. I'm not an expert on this topic but I'm guessing that the reason it's harder to find a "unique" solution, if one exists at all, is that, unlike addition and multiplication, exponentiation is neither commutative nor associative; hence there isn't the same level of "regularity" in the operations preceding tetration as there is in those preceding exponentiation. bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 01/01/2009, 10:12 PM (This post was last modified: 01/01/2009, 10:16 PM by bo198214.) Finitist Wrote:I'm guessing that the reason it's harder to find a "unique" solution, if one exists at all, is that, unlike addition and multiplication, exponentiation is neither commutative nor associative; hence there isn't the same level of "regularity" in the operations preceding tetration as there is in those preceding exponentiation. I see it similarly. The usual uniqueness condition for the multiplication $f(x)=bx$ is $f(x+y)=f(x)+f(y)$ and for the exponentiation $f(x)=b^x$ is $f(x+y)=f(x)f(y)$. However we can not continue on that path for tetration/superexponential $f$ as surely $f(x+y)\neq f(x)^{f(y)}$ for most integer $x,y$ (except if we switch to arborescent numbers, which I investigated in my phd thesis). Daniel Long Time Fellow    Posts: 279 Threads: 94 Joined: Aug 2007 01/04/2009, 03:05 AM Thanks for the feedback. The concern I have about uniqueness is a consequence of the logarithm being infinitely valued. Tetration is just as much iterated logarithm as it is iterated exponentiation. The technique I have developed for extending tetration is based on fixed points, but there are an infinite number of fixed points. Say you have an extension for tetration. What is the value of $^{-1} a$? It is usually assumed to be 0, but it can actually be $2 \pi k$ where k is an integer; it has an infinite number of values. Then $^{-2} a$ also has an infinite number of values $\omega^2 = \omega$. It seems to me that this leads to $^{-\omega} a$ having as many values as the continuum has points. This results in $^n a$ consisting on an $\aleph _1$ family of solutions. It is my understanding that there are $\aleph _2$ number of curves, but holomorphic functions are much more restricted. I wonder if it isn’t true that any holomorphic function agreeing with the values of $^k a$, where k is a natural number, comes arbitrarily close to one of the infinite family of tetration solutions. Daniel Daniel bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 01/04/2009, 10:47 AM (This post was last modified: 01/04/2009, 10:48 AM by bo198214.) Daniel Wrote:The technique I have developed for extending tetration is based on fixed points, So what is your technique? As far as I remember you used always regular iteration at a real fixed point. And regular iteration at a complex fixed point leads to comlex values for real arguments. Which I would exclude from feasible solutions for tetrationals. Quote:Say you have an extension for tetration. What is the value of $^{-1} a$? It is usually assumed to be 0, but it can actually be $2 \pi k$ where k is an integer; it has an infinite number of values. Yes, but you know if you consider the global behaviour of holomorphic functions, they can be many valued, actually infinitely many valued. Best example is the logarithm. It has one powerseries development at 1, but you can continue it along any path not crossing 0 and you get infinitely many values for one point, depending on how often the path to the point revolves around 0. Same with a tetrational. It has singularities at each negative integer number below or equal -2. So if you start with a powerseries development at 0 you can continue to any non-singular point along every path not hitting a singular point. So you have again infintely many values at every non-singular point depending how the path to that point revolves around the singularities. Quote:Then $^{-2} a$ also has an infinite number of values $\omega^2 = \omega$. It seems to me that this leads to $^{-\omega} a$ having as many values as the continuum has points. I am reluctant to discuss this before there is not a proper definition and meaning of what is meant by $^{-\omega} a$. Either we work with complex numbers or we work with cardinalities, or we work with ordinalities. But I never heard about a number system that unites complex numbers with cardinalities or ordinalities. The only thing about infinity that is allowed with the complex numbers is to add an infinity which makes the complex plane into a complex sphere. But this infinity is not a cardinality and not an ordinality. Quote: This results in $^n a$ consisting on an $\aleph _1$ family of solutions. Yes and no. Seeing this in the context of global holomorphic functions: For any point you have a path to this point and this path winds finitely many times around each singularity. For (countably) infinitely many singularities we get $\aleph_0^{\aleph_0}=\aleph_1$ many possibilities for the path to wind around the singularities. And hence we can have possibly so many values for each point. But dont forget that it is one global holomorphic function (many-valued). The whole function can be given by one power-series development for example at 0. And probably there is only one out of the many at 0, that is real analytic. So what you are speaking about can be summarized as many-valuedness of one holomorphic function. Imho it makes no sense to diversify this into different holomorphic solutions on cutted domains. (If you see the logarithm as a global function you dont need the cut at the negative real axis. But if you put a cut somewhere than it looks whether you have infinitely many logarithms.) Quote: but there are an infinite number of fixed points. And this is a completely different uniqueness concern. This is about whether there are indeed different global functions. I.e. two real analytic powerseries developments at 0 which have somewhere different coefficients. Quote: It is my understanding that there are $\aleph _2$ number of curves, but holomorphic functions are much more restricted. Yes, I would agree; You get all real anlytic global functions (defined at 0) by going through all real powerseries developments at 0. And this is $|\mathbb{R}|^{\aleph_0}=\aleph_1$ Quote: I wonder if it isn’t true that any holomorphic function agreeing with the values of $^k a$, where k is a natural number, comes arbitrarily close to one of the infinite family of tetration solutions. This question perhaps becomes obsolete after distinguishing many-valuedness of one holomorphic function from different holomorphic tetrationals. (Note also that for a tetrational I would have the stronger condition $f(x+1)=a^{f(x)}$ for all real/complex $x$ instead of just for integers $x=k$). Daniel Long Time Fellow    Posts: 279 Threads: 94 Joined: Aug 2007 01/05/2009, 03:06 AM Once again, thanks for your reply Henryk. bo198214 Wrote:Daniel Wrote:The technique I have developed for extending tetration is based on fixed points,So what is your technique? As far as I remember you used always regular iteration at a real fixed point. And regular iteration at a complex fixed point leads to comlex values for real arguments. Which I would exclude from feasible solutions for tetrationals.Yes, you are correct in how my technique works. I do disagree with your a priori exclusion of my approach as not being a feasible solution. Rather, I consider the constraining of complex values for real arguments as a potential axiom that is worthy of exploration for extending tetration. Hopefully, different approaches to extending tetration that use the same axioms will produce consistent results. bo198214 Wrote:Daniel Wrote:Say you have an extension for tetration. What is the value of $^{-1} a$? It is usually assumed to be 0, but it can actually be $2 \pi k$ where k is an integer; it has an infinite number of values.… Same with a tetrational. It has singularities at each negative integer number below or equal -2. So if you start with a powerseries development at 0 you can continue to any non-singular point along every path not hitting a singular point. So you have again infintely many values at every non-singular point depending how the path to that point revolves around the singularities. The $log_a(log_a(1))$ used to define a tetrational for –2 is infinitely multi-valued. Yes, if you restrict complex values from real arguments, then you just obtain singularities. This is a weakness of my technique in that it doesn’t provide a way to compute tetrationals that contain singularities. My guess is that these singularities have lead people to extend tetration using superlogarithms instead of superexponentiation. bo198214 Wrote:Daniel Wrote:I wonder if it isn’t true that any holomorphic function agreeing with the values of $^k a$, where k is a natural number, comes arbitrarily close to one of the infinite family of tetration solutions. … (Note also that for a tetrational I would have the stronger condition $f(x+1)=a^{f(x)}$ for all real/complex $x$ instead of just for integers $x=k$).Yes, I agree. I show this by demonstrating $f^a(f^b(z))-f^{a+b}(z)=O(n)$ in Mathematica. Daniel Daniel bo198214 Administrator Posts: 1,624 Threads: 103 Joined: Aug 2007 01/05/2009, 05:04 PM Daniel Wrote:Yes, you are correct in how my technique works. I do disagree with your a priori exclusion of my approach as not being a feasible solution. Yes of course not a priori. If we had no methods for real analytic tetrationals we would would of course fall back to other methods that are still good, that is they have complex values on the real axis, or are not analytic on the real axis. But we have even 5 methods for a real analytic tetrational, i.e. a function that hopefully satisfies: * real on the real axis (>2) (on one branch) * strictly increasing on the real axis * holomorphic on the whole complex plane with exception of $z\le -2$. * satisifying $f(z+1)=\exp_b(f(z))$ and $f(0)=1$ for complex all $z$ (though one has to be careful with branching, so this maybe true only when considering $f$ as a global holomorphic function). Quote:The $log_a(log_a(1))$ used to define a tetrational for –2 is infinitely multi-valued. Yes, if you restrict complex values from real arguments, then you just obtain singularities. Well this possible non-singularity at -2 is an interesting phenomenon when considering tetration as a global function. It means that only on one (or some) branch(es) of the global function there is a singularity at -2. If the path to the point -2 winds differently around the other singularities, then there may be a continuation to -2 (yielding no singularity there). Thats really an intriguing topological manifold where the singularities change depending on which branch we reside. Where on the other hand the branches depend on the singularities, as they come to existence by different winding around the (other) singularities. Quote:My guess is that these singularities have lead people to extend tetration using superlogarithms instead of superexponentiation. Not that I would know of. The superlogarithm to base $b$ corresponds to the Abel function of $b^x$. Its just in the theory of regular fractional iterates that the Abel function plays a bigger role and is easier usable and constructable than its inverse, the superexponentiation in this case. bo198214 Wrote:Yes, I agree. I show this by demonstrating $f^a(f^b(z))-f^{a+b}(z)=O(n)$ in Mathematica.Ya but that is already done without Mathematica. Which is necessary as it is no proof for arbitrary $n$, but you can only verify it for some given big $n$. However there is a thorough mathematical base of regular iteration, this is for example presented in the book: Jean Écalle, Théorie des invariants holomorphes, Publ. Math. Orsay No. 67–7409, 1974 and also to some extent in the book: Kuczma, Iterative functional equations, 1990 And it is the definition of fractional iterates that they satisfy $f^1=f$ $f^{s+t}=f^{s}\circ f^{t}$ and regular fractional iterates are shown to be fractional iterates. « Next Oldest | Next Newest »

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