• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 Transseries, nest-series, and other exotic series representations for tetration bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 12/01/2009, 09:08 AM (12/01/2009, 02:56 AM)Daniel Wrote: I agree there are alternate ways to iterate $e^x-1$, there are at least three ways I know of from my own research.Well often there are lots of different ways to do things, the question is rather whether the outcome is the same. And I think it is important to note that your approach also just produces regular iteration (in the sense of Szekeres 1958) and not some other exotic iteration. Quote: Schroeder summations are not an efficient to iterate $e^x-1$. It requires 2312 summations in order to evaluate the tenth term. But this is only the lesser problem. Usually one is not interested to have an iterate as a non-convergent powerseries but one want to have a computable real/complex function. E.g. can you draw the graph of the regular half iterate of e^x-1 (or that of e^(x/e) which is equivalent)? Currently Dmitrii and I working on a paper showing the complex plot of the Abel function, its inverse and some iterates of e^(x/e). Quote: What they do is show that there is a combinatorial structure underlying all iterated functions, Schroeder's Fourth Problem http://www.research.att.com/~njas/sequences/A000311 . Also Schroeder summations are produced using Faà di Bruno's formula which is an example of a Hopf algebra which is important in several different areas of quantum field theory including renormalization. It is my hope that this might shine some light on how to show that our formulations of iterated functions and tetration are actually convergent. I am always a bit skeptical about too philosophical approaches. I think in mathematics there should always be some result that is applicable and helpful in solving some problem. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 12/01/2009, 10:22 PM (12/01/2009, 09:08 AM)bo198214 Wrote: (12/01/2009, 02:56 AM)Daniel Wrote: I agree there are alternate ways to iterate $e^x-1$, there are at least three ways I know of from my own research.Well often there are lots of different ways to do things, the question is rather whether the outcome is the same. And I think it is important to note that your approach also just produces regular iteration (in the sense of Szekeres 195 and not some other exotic iteration. Quote: Schroeder summations are not an efficient to iterate $e^x-1$. It requires 2312 summations in order to evaluate the tenth term. But this is only the lesser problem. Usually one is not interested to have an iterate as a non-convergent powerseries but one want to have a computable real/complex function. E.g. can you draw the graph of the regular half iterate of e^x-1 (or that of e^(x/e) which is equivalent)? Currently Dmitrii and I working on a paper showing the complex plot of the Abel function, its inverse and some iterates of e^(x/e). Quote: What they do is show that there is a combinatorial structure underlying all iterated functions, Schroeder's Fourth Problem http://www.research.att.com/~njas/sequences/A000311 . Also Schroeder summations are produced using Faà di Bruno's formula which is an example of a Hopf algebra which is important in several different areas of quantum field theory including renormalization. It is my hope that this might shine some light on how to show that our formulations of iterated functions and tetration are actually convergent. I am always a bit skeptical about too philosophical approaches. I think in mathematics there should always be some result that is applicable and helpful in solving some problem. Sir , i believe you are a constructivist. regards tommy1729 mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 12/08/2009, 10:32 AM (This post was last modified: 12/08/2009, 10:38 AM by mike3.) That's right. The trick here is finding a representation of a given analytical function $f(z)$ as some series of terms that can be continuum-summed (either they should be continuum-summable via direct application of Faulhaber's formula, or they should be reducable to some alternative representation that is summable via Faulhaber's formula (or some chain of such reductions)) so that the continuum sum series also converges, and even better if that continuum sum and the integral can be represented in the same form (so as to enable an iterative map on the series from the continuum-sum tetration formula). That's why I mentioned the "transseries", which gives one possible approach. Transseries actually forms a somewhat complicated theory and there's a lot of stuff in that paper, but it includes more sophisticated and exotic forms of series than just simple power series, such as nested sums of power series, exp-series, polynomial-series (incl. Newton and Mittag-Leffler series in a star), and more. So far, I've found 2 types of transseries representation such that if the series converges, the continuum sum does as well (proof given earlier here): exp-series: $f(z) = \sum_{n=0}^{\infty} a_n e^{nz}$ $\sum_{n=0}^{z-1} f(n) = a_0 z + \sum_{n=1}^{\infty} \frac{a_n}{e^n - 1} \left(e^{nz} - 1\right) = -\left(\sum_{n=1}^{\infty} \frac{a_n}{e^n - 1}\right) + a_0 z + \sum_{n=1}^{\infty} \frac{a_n}{e^n - 1} e^{nz}$ Newton series or umbral series(*): $f(z) = \sum_{n=0}^{\infty} a_n (z)_n$ $\sum_{n=0}^{z-1} f(n) = \sum_{n=0}^{\infty} \frac{a_n}{n+1} (z)_{n+1}$. where $(z)_n$ is the falling factorial. If $f(z)$ can be given by a Newton series on some part of the plane, then $a_n = \frac{\Delta^n[f](0)}{n!}$ (where $\Delta$ is the forward difference operator.) (*) I call this umbral series because this series is akin to a Taylor series with finite differences, see e.g. "umbral calculus". However neither are very useful for the extension of tetration. The exp-series always have a periodicity of $2 \pi i$, thus they cannot represent aperiodic functions or functions of different period. Also, the exp-series does not continuum-sum to another exp-series, as one should be able to see from the above. The Newton series can only represent functions with that grow at most exponential-type -- tetration grows too fast. Tetration at bases $1 < b < e^{1/e}$ does, however expand to a Newton series, and we don't even need to run the sum formula to find this out. I think this is equivalent to the regular iteration. The sums for $e^{nz}$ and $(x)_n$ can be recovered from Faulhaber's formula. Another possible series is the Mittag-Leffler star expansion, from here: http://eom.springer.de/s/s087230.htm (special case for expansion at 0) $f(z) = \sum_{n=0}^{\infty} \sum_{\nu=0}^{k_n} c_\nu^{(n)} a_\nu z^\nu$ $\sum_{n=0}^{z-1} f(n) = \sum_{n=0}^{\infty} \sum_{\nu=0}^{k_n} c_\nu^{(n)} \frac{a_\nu}{\nu+1} \left(B_{\nu+1}(z) - B_{\nu+1}(0)\right)$ (appl. Faulhaber's formula, so $B$ are the Bernoulli polynomials) where $a_\nu = \frac{f^{(\nu)}(0)}{\nu!}$, which are the Taylor coefficients of $f(z)$. This series converges in the entire Mittag-Leffler star of the function. Whether the continuum sum converges will likely depend on the characteristics of the magic numbers $c_\nu^{(n)}$, which I have not been able to find. The site mentions they can be "evaluated once and for all". Two refs were given. I was emailed a snapshot of the proof from the book by Markushevich, but it did not detail how to actually obtain the coefficients. The second reference by Borel was in French, which I don't know, so that was not very useful. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 12/09/2009, 06:25 AM I like the sound of "magic numbers" they sound so ... magical. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 12/14/2009, 01:33 PM (This post was last modified: 12/14/2009, 01:36 PM by bo198214.) (12/08/2009, 10:32 AM)mike3 Wrote: So far, I've found 2 types of transseries representation such that if the series converges, the continuum sum does as well (proof given earlier here): I was a bit puzzled why $e^{e^x}$ converges via transseries while $e^{x^2}$ does not converge. Actually $e^{\frac{x^2}{2}}$ must also converge as we have $\frac{x^2}{2}, and indeed it also converges! But it seems that this already the limit, i.e. every series $e^{c x^2}$ for $c>\frac{1}{2}$ does not converge. I guess that this is the same for $e^{\frac{x^n}{n!}$, $n>1$, while Faulhaber always works for $n=1$, i.e. $e^{c x}$. Do you have any idea how to sum $e^{x^2}$? kobi_78 Junior Fellow Posts: 11 Threads: 3 Joined: Dec 2009 12/14/2009, 02:11 PM (This post was last modified: 12/14/2009, 02:12 PM by kobi_78.) (12/14/2009, 01:33 PM)bo198214 Wrote: (12/08/2009, 10:32 AM)mike3 Wrote: So far, I've found 2 types of transseries representation such that if the series converges, the continuum sum does as well (proof given earlier here): I was a bit puzzled why $e^{e^x}$ converges via transseries while $e^{x^2}$ does not converge. Actually $e^{\frac{x^2}{2}}$ must also converge as we have $\frac{x^2}{2}, and indeed it also converges! But it seems that this already the limit, i.e. every series $e^{c x^2}$ for $c>\frac{1}{2}$ does not converge. I guess that this is the same for $e^{\frac{x^n}{n!}$, $n>1$, while Faulhaber always works for $n=1$, i.e. $e^{c x}$. Do you have any idea how to sum $e^{x^2}$? Hi, I've been reading on this forum for a while, and just now I've signed up. I have been exploring the sum operator, what you guys call "continuum sum" for a couple of years as a hobby. I think I have an idea how to sum $e^{x^2}$. Recall that $e^{x^2} = \lim_{n \to \infty} \left( 1 + \frac{x^2}{n} \right)^n$ Now calculate the polynomial sum of $\left( 1 + \frac{x^2}{n} \right)^n$ (using Faulhaber's formula). This seems to convergent to a nice function. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 12/14/2009, 02:19 PM (12/14/2009, 02:11 PM)kobi_78 Wrote: I have been exploring the sum operator, what you guys call "continuum sum" for a couple of years as a hobby. I think I have an idea how to sum $e^{x^2}$. Recall that $e^{x^2} = \lim_{n \to \infty} \left( 1 + \frac{x^2}{n} \right)^n$ Now calculate the polynomial sum of $\left( 1 + \frac{x^2}{n} \right)^n$ (using Faulhaber's formula). This seems to convergent to a nice function. Wow, what a debut! Hey Kobi, welcome to the forum! Did you also find out something about the uniqueness of the obtained functions? Are there perhaps examples where you group the terms in a different way and get different results? bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 12/14/2009, 02:45 PM (This post was last modified: 12/14/2009, 02:48 PM by bo198214.) (12/14/2009, 02:11 PM)kobi_78 Wrote: Now calculate the polynomial sum of $\left( 1 + \frac{x^2}{n} \right)^n$ (using Faulhaber's formula). This seems to convergent to a nice function. But wait, first it *seems* to converge, however later it takes off. E.g. if I compute the coefficient of x, I get rounded to two significant digits (scroll down): Code:n= 60 : 1.2 n= 61 : 1.2 n= 62 : 1.2 n= 63 : 1.1 n= 64 : 1.2 n= 65 : 1.1 n= 66 : 1.2 n= 67 : 0.96 n= 68 : 1.7 n= 69 : -0.14 n= 70 : 4.5 n= 71 : -7.8 n= 72 : 25. n= 73 : -64. n= 74 : 180. n= 75 : -500. n= 76 : 1400. n= 77 : -4100. n= 78 : 12000. n= 79 : -35000. This is not a rounding error as I compute with fractions and do the rounding only for better visibility. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 12/14/2009, 03:16 PM (This post was last modified: 12/14/2009, 03:21 PM by bo198214.) Mike, I have another bad news, I found out that not even the Faulhaber sum of $e^{\frac{x^2}{2}}$ is convergent, contrary to what I said before because it looks convergent up to n=100 terms; so your transseries sum $e^{e^x}$ can not be convergent. The first coefficient of the Faulhaber sum for $n$ terms: Code:n= 100 : 1.1 n= 101 : 1.0 n= 102 : 1.0 n= 103 : 1.2 n= 104 : 1.2 n= 105 : 0.68 n= 106 : 0.68 n= 107 : 2.2 n= 108 : 2.2 n= 109 : -1.8 n= 110 : -1.8 n= 111 : 9.2 n= 112 : 9.2 n= 113 : -22. n= 114 : -22. n= 115 : 67. n= 116 : 67. n= 117 : -190. n= 118 : -190. n= 119 : 570. kobi_78 Junior Fellow Posts: 11 Threads: 3 Joined: Dec 2009 12/14/2009, 07:17 PM Hi, I think I know how to solve it. I'll think about it and update you guys later. Kob « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Taylor series of i[x] Xorter 12 10,114 02/20/2018, 09:55 PM Last Post: Xorter An explicit series for the tetration of a complex height Vladimir Reshetnikov 13 10,369 01/14/2017, 09:09 PM Last Post: Vladimir Reshetnikov Complaining about MSE ; attitude against tetration and iteration series ! tommy1729 0 1,402 12/26/2016, 03:01 AM Last Post: tommy1729 2 fixpoints , 1 period --> method of iteration series tommy1729 0 1,242 12/21/2016, 01:27 PM Last Post: tommy1729 Taylor series of cheta Xorter 13 10,788 08/28/2016, 08:52 PM Last Post: sheldonison Tetration series for integer exponent. Can you find the pattern? marraco 20 14,898 02/21/2016, 03:27 PM Last Post: marraco [AIS] (alternating) Iteration series: Half-iterate using the AIS? Gottfried 33 34,675 03/27/2015, 11:28 PM Last Post: tommy1729 [2014] Representations by 2sinh^[0.5] tommy1729 1 2,126 11/16/2014, 07:40 PM Last Post: tommy1729 [integral] How to integrate a fourier series ? tommy1729 1 2,278 05/04/2014, 03:19 PM Last Post: tommy1729 Wonderful new form of infinite series; easy solve tetration JmsNxn 1 4,089 09/06/2012, 02:01 AM Last Post: JmsNxn

Users browsing this thread: 1 Guest(s)