• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Superroots and a generalization for the Lambert-W tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 12/02/2015, 01:22 PM (12/02/2015, 03:49 AM)Gottfried Wrote: (12/01/2015, 11:58 PM)tommy1729 Wrote: The thing is solving (x_m ^ x_m)^[m] = y is only close to solving X_n^^[n] = y ( n = m in value )True. But having this way a (non-trivial) vector of different exponents (or better: bases) which comes out to be a meaningful "nested exponentiation" I'm curious, whether one can do something with it, for instance weighting, averaging, or multisecting that sequence of exponents/bases when re-combining them to a "nested exponential". We have not yet many examples of "nested exponentiations" with a meaningful outcome. For instance, the construction of the Schroeder-function is based on (ideally) infinite iteration of the base-function to get a linearization. If we iterate the h2()-function infinitely, the curve of the consecutive values in an x/y-diagram (where x is the iteration number) approach a horizontal line; don't know whether using that linearization shall prove useful for something similar. (When Euler found his version of the gamma-function, that was in one version putting together sequences of integer numbers weighting and repeating in a meaningful way; there is some infinite product-representation for his gamma-function I think I recall correctly... ) (see also the updates in my previous (introducing) posting) This reminds me of one of my posts on the OEIS many years ago , also under the pseudo tommy1729 ( i have other pseudo too ). Go to Oeis and enter tommy1729. Or use this link: https://oeis.org/search?q=Tommy1729&lang...&go=Search In particular https://oeis.org/A102575 In abuse notation this becomes (1 + 1/n)^^ And it was a special case of my investigations in the Tommy-Zeta functions between 2001 and 2009 given by (1+1^(-s))^(1+2^(-s)) ... This is somewhat similar and thus might intrest you. Regards Tommy1729 andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/09/2015, 06:34 AM I think super-roots are important. Iterated exponentials ($w = \exp_x^{y}(z)$) are a function of three variables (trivariate? function), and so they have 3 inverse functions: negatively iterated exponentials (solving for z), trivariate super-logarithms (solving for y), and trivariate super-roots (solving for x). Trivariate super-logarithms can be expressed with bivariate super-logarithms, and so are not fundamental operations, but trivariate super-roots have no known expression in terms of bivariate super-roots, and so are, so far, a fundamental operation so far as I know. My recent research into super-roots have convinced me that we know more about them than we think we know. We can calculate the derivatives of them to a rational number in some cases, and to any precision in other cases. Using a combination of power series and Lagrange inverse series, we can calculate many many things about them, but we still don't have a closed form for these apparently useful functions. I think that given enough time, effort, and insight, we can find at least a recurrence equation that expresses how to find super-root (n + 1) given complete knowledge of super-root (n). I'm going to go out on a limb and make a notation for these trivariate super-roots: $\sqrt[y]{w}^{(z)}_{\mathrm{s}} = x$ iff. $w = \exp_x^y(z)$ One of the advantages of trivariate super-roots is that they have more algebraic identities regarding them: $\sqrt[y]{w}^{(z)}_{\mathrm{s}} = \sqrt[y]{x^w}^{(x^z)}_{\mathrm{s}} = \sqrt[(y-1)]{w}^{(x^z)}_{\mathrm{s}} = x$ and there is one about the third super-root: $\sqrt[3]{w}^{(z)}_{\mathrm{s}} = \left(\sqrt[2]{w^z}^{\left(\frac{x^z}{z}\right)}_{\mathrm{s}}\right)^{1/z} = x$ and there is one about the second super-root: $\sqrt[2]{w}^{(z)}_{\mathrm{s}} = \left(\sqrt[2]{w^z}^{(1)}_{\mathrm{s}}\right)^{1/z} = x$ where $\sqrt[2]{w}^{(1)}_{\mathrm{s}}$ just means the bivariate super-root $\sqrt[2]{w}_{\mathrm{s}}$. If we could find a general way of expressing trivariate super-roots in terms of bivariate super-roots, then I think we would know much more about tetration than we do today. Perhaps along the way we will discover something new that will shed some light on super-logarithms, too, perhaps. Regards, Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/30/2015, 09:49 AM I believe I may have found a closed form for the power series of the third tetrate function as well. I'm not sure if these are known, but I just used the elementary properties of binomials and Stirling numbers to derive these: $ \begin{equation} {}^{3}x = \sum_{k=0}^{\infty} \log(x)^k \sum_{j=0}^{k} \sum_{i=0}^{k - j - 1} \frac{(k - j - i)^j j^i}{(k - j - i)!j!i!} \end{equation}$ $ \begin{equation} {}^{3}x = \sum_{k=0}^{\infty} (x - 1)^k \sum_{j=0}^{k} \sum_{J=0}^{j} \sum_{i=0}^{k} \sum_{I=0}^{i} {\left[{i \atop I}\right]} {\left[{j \atop J}\right]} {\left({J \atop {k - j - i}}\right)} \frac{J^I}{j!i!} \end{equation}$ The first one (logarithmic power series) reminds me of something in one of Galidakis' papers about tetration, but I don't remember which paper. The second one is derived from the fact that the generating function of the signed Stirling numbers the first kind is $(1 + x)^z$. « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Nixon-Banach-Lambert-Raes tetration is analytic , simple and “ closed form “ !! tommy1729 11 4,397 02/04/2021, 03:47 AM Last Post: JmsNxn Superroots (formal powerseries) Gottfried 10 22,399 04/05/2011, 03:22 AM Last Post: Stan Infinite towers & solutions to Lambert W-function brangelito 1 6,246 06/16/2010, 02:50 PM Last Post: bo198214 Lambert W function and the Super Square Root James Knight 3 12,896 10/29/2009, 06:30 AM Last Post: andydude the extent of generalization Matt D 11 21,145 10/15/2007, 04:52 PM Last Post: Matt D

Users browsing this thread: 1 Guest(s)