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 Superroots and a generalization for the Lambert-W Gottfried Ultimate Fellow     Posts: 767 Threads: 119 Joined: Aug 2007 11/09/2015, 01:17 PM (This post was last modified: 11/09/2015, 01:37 PM by Gottfried.) I experienced a bit with the problem of finding x in equations like and and where y is given. This was mainly motivated by frequent questions in MSE and/or MO for solutions where or What I mainly found was the requirement for a generalization of the Lambert-W (but a nicely straightforward one!) and some insight into the occuring power series. Although having now an accessible entry-point into the general problem, I did not yet find explicite, simple closed form expressions for the occuring coefficients except when (but those are already well known...), so it's an open field for pattern-detection and research on radii of convergence. It is too much to write it here in this limited box, so I made a pdf-file. I upload it as attachment but put it also on my webspace, see http://go.helms-net.de/math/tetdocs/Wexz...erroot.pdf Collaboration is appreciated... Gottfried Attached Files Wexzal_Superroot.pdf (Size: 156.93 KB / Downloads: 323) Gottfried Helms, Kassel nuninho1980 Fellow   Posts: 95 Threads: 6 Joined: Apr 2009 11/09/2015, 11:27 PM I'm interested. Can you solve x^^(1/2) = 3? but... using "new" formula tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 11/10/2015, 09:38 AM Conjecture 3.1 fails because the left hand side has radius going to 0. There are 2 ways to show that. On the other hand conjecture 3.1 does have meaning considering the first n derivatives of the LHS = RHS. It is a mystery how you intend to solve the cases x^^(3/2) = v though. Regards Tommy1729 Gottfried Ultimate Fellow     Posts: 767 Threads: 119 Joined: Aug 2007 11/10/2015, 12:04 PM (This post was last modified: 11/10/2015, 12:06 PM by Gottfried.) (11/10/2015, 09:38 AM)tommy1729 Wrote: Conjecture 3.1 fails because the left hand side has radius going to 0.Hmm, I've not yet settled everything about this in my mind. I've of course seen, that with increasing n the convergence-radius of the function decreases. However, as usually, if a function can be analytically continued (beyond its radius of convergence) for instance by Euler-summation, I assume, that the result is still meaningful. And we have here the possibility for Euler-summation, so I think there is a true analytic continuation. However, I don't know yet whether this can be correctly inserted in my conjecture-formula for the limit-case. Quote:It is a mystery how you intend to solve the cases x^^(3/2) = v though. As I understand this, this is using fractional iteration heights. As I described my exercises, I'm concerned with the unknown bases, and am using integer heights so far, not fractional heights ( superroot, not superlog) Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow     Posts: 767 Threads: 119 Joined: Aug 2007 11/10/2015, 12:06 PM (This post was last modified: 11/10/2015, 03:59 PM by Gottfried.) (11/09/2015, 11:27 PM)nuninho1980 Wrote: I'm interested. Can you solve x^^(1/2) = 3? but... using "new" formula Not with this elaboration. I'm on superroots of powertowers of integer heights so far. I've given no thought so far for analysis with fractional heights, except for some lazy tries to find an interpolation-formula for the rows in table 3.1 , but without easy success .... Mind you to step in for this? Gottfried [update] perhaps this is of interest: see MSE http://math.stackexchange.com/questions/...550#133550 Gottfried Helms, Kassel tommy1729 Ultimate Fellow     Posts: 1,372 Threads: 336 Joined: Feb 2009 11/10/2015, 11:19 PM Alot has already been said about the superroots. We know that lim n-> oo for x_n^^n = y for y > exp(1/e) [eta] gives x_n = eta. Also all results about slog and sexp relate. PROOF SKETCHES --- First i point out that when you have a nonzero radius , the eulersum = analytic continuation whenever and wherever both converge. BUT analytic continuation is USUALLY NOT the correct solution. For instance x^x^x^... = x^[oo] = y Has solution x = y^(1/y) IFF Dom y , range x are in the sheltron region. Clearly x=y^(1/y) is the analytic continuation , but thus false. Tommy's lemma : for all n -> W•n(0) = 0. Conjecture 3.1. Conjecture 3.1 conjectures lim n -> oo W•n (v) = -v exp(-v). Clearly the RHS has radius oo. But the algebra dictates that the radius can be at most : X^(1/x) = exp(ln(x) / x). Now v = ln(x) so x = exp(v). Therefore exp( ln(x) / x) = exp ( v exp(-v) ). Hence v exp(-v) = W•n(v) = - v exp(-v) => contradiction .. Unless for v satisfying V exp(-v) = v exp(-v) The solution set is v ={0,oo}. So the radius of lim W•n = 0. V= 0 implies x = 1. X = 1 is within the sheltron region so v = 0 is valid. Qed A second proof Let n = oo The fractal argument : W•n(v) = a <=> v = (b exp ( b exp ( ...(*)) = (b exp( b exp ( ... (a)). So v = ( b exp(*))^oo. A fractal within the sheltron. V = fixpoint [b exp(*)] ==> solve b exp(A) = A. ==> A = - W(-b). => - W(-b) = v. --> b = v exp(-v) = W•n(v). Similar too previous proof ; v must be 0 ==> radius = 0. Qed. Too explain the fractal argument Notice (b exp(a))^ for fixed b and variabele a is NOT equal too (a exp(a))^ .. Even if we set a = b ! For a = b , The difference is in the second case a exp(a) exp( a exp(a) ) Whereas in the first case a exp(a exp(a) ). That is why I use a and b and then set them equal. That looks confusing but is correct. --- So we get 2 proofs with W•n(v) = v exp(-v). However that is only valid within the sheltron. V = ln(x) -> 0 = ln(1). Min( | ln(1) - ln(e^(-1/e)) | , | ln(1) - ln(e^(1/e)) | ) = 1/e. So W•n(v) = v exp(-v) is valid within " radius " 1/e. And analytic continuation does not help. Hope that helps. Regards Tommy1729 andydude Long Time Fellow    Posts: 509 Threads: 44 Joined: Aug 2007 11/13/2015, 05:58 PM I'm not sure if this should be in a different thread, but I just found a website with programming puzzles, and one of the puzzles is super-roots: http://www.checkio.org/mission/super-root/ this "mission" (programming puzzle) has a very large number of "solutions" (Python implementations) of super-roots. After you login and solve the mission, it will show you other people's solutions along with your own. I solved it using the infinite tetrate (x^x^x^...) which is probably not the fastest method, but I think it's more beautiful than using Newton's method with an unknown number of iterations until you find the right number. Most of the solutions are based on Newton's method or a similar bisection method, but I was considering going thru them to see if there's any new methods we didn't know about... Gottfried Ultimate Fellow     Posts: 767 Threads: 119 Joined: Aug 2007 11/13/2015, 07:05 PM (This post was last modified: 11/13/2015, 07:06 PM by Gottfried.) (11/13/2015, 05:58 PM)andydude Wrote: I'm not sure if this should be in a different thread, but I just found a website with programming puzzles, and one of the puzzles is super-roots: Hi Andy, nice to read from you... Yes the page looks interesting, I'll try to get logged in another day (I'm in bed because of some bacteria or whatever and am just lurking around here a bit). I had already other posts with superroots, for instance that one about the power series of (1+x)^(1+x)^...^(1+x). At the moment I was involved in that question of the supperroots -1 = x^x^x and had much fun already (see http://math.stackexchange.com/questions/...38#1415538 ) but again - cannot yet complete the discussion. Cordially - Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow     Posts: 767 Threads: 119 Joined: Aug 2007 11/16/2015, 01:08 AM An actual discussion/application is on MSE, see the question of Vladimir Reshetnikov where I tried to find an explanation /ansatz for a proof in my answer http://math.stackexchange.com/a/1530136/1714 Perhaps someone can help to make more progress... Gottfried Gottfried Helms, Kassel andydude Long Time Fellow    Posts: 509 Threads: 44 Joined: Aug 2007 11/21/2015, 05:05 AM I agree that it is a long-researched problem; trying to find a closed form for super-roots, or anything for that matter. Using a combination of known facts from the Tetration Ref I collected, I was able to find a simpler expression for the logarithmic power series expansion of than I remember from before. I think the ideal solution would be to find a recurrence equation similar to the one we know for n-th tetrates. I've attached a short discussion of the things we know that might help in finding a closed form. Attached Files 2015-11_SuperRoot3.pdf (Size: 123.09 KB / Downloads: 266) « Next Oldest | Next Newest »

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