• 1 Vote(s) - 5 Average
• 1
• 2
• 3
• 4
• 5
 regular tetration at b=e^(-e) bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/26/2009, 09:25 PM (This post was last modified: 09/26/2009, 09:35 PM by bo198214.) In response to the thread Solving tetration for base 0 < b < e^-e. I will feature here what one could call regular superlogarithm. The idea is the following: One can obtain the regular Abel function $\alpha$ of $f$ by the Julia function $j$: (*) $\alpha(z)=\int \frac{1}{j(z)}$ The Julia function (or also called iterative logarithm) can be obtained by (**) $j(z)=\left.\frac{\partial f^{\circ t}}{\partial t}\right|_{t=0}$ where we mean $f^{\circ t}$ to be the *regular* iteration of $f$. It is similar to the logarithm: $\left.\frac{\partial z^t}{\partial t}\right|_{t=0}=\ln(z)$ And we have a way to obtain the regular iteration $f^{\circ t}$ by matrix powers. Before we begin we reduce the problem of iterating $f(z)=e^{-ez}$ to iterating $h(z)=e^{-z}-1$. This is a linear conjugation with $\tau(z)=ez-1$ and $\tau^{\circ -1}(z)=\frac{z+1}{e}$, i.e. $h=\tau\circ f\circ \tau^{\circ -1}$. It moves the fixed point from $1/e$ to $0$. $h$ has the powerseries coefficients: Code:0, -1, 1/2, -1/6, 1/24, -1/120, 1/720, -1/5040, 1/40320, -1/362880, ... Then we take the Carleman matrix of this series (I truncate it to 10 here): $C=\left(\begin{array}{rrrrrrrrrr} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & \frac{1}{2} & -\frac{1}{6} & \frac{1}{24} & -\frac{1}{120} & \frac{1}{720} & -\frac{1}{5040} & \frac{1}{40320} & -\frac{1}{362880} \\ 0 & 0 & 1 & -1 & \frac{7}{12} & -\frac{1}{4} & \frac{31}{360} & -\frac{1}{40} & \frac{127}{20160} & -\frac{17}{12096} \\ 0 & 0 & 0 & -1 & \frac{3}{2} & -\frac{5}{4} & \frac{3}{4} & -\frac{43}{120} & \frac{23}{160} & -\frac{605}{12096} \\ 0 & 0 & 0 & 0 & 1 & -2 & \frac{13}{6} & -\frac{5}{3} & \frac{81}{80} & -\frac{37}{72} \\ 0 & 0 & 0 & 0 & 0 & -1 & \frac{5}{2} & -\frac{10}{3} & \frac{25}{8} & -\frac{331}{144} \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & -3 & \frac{19}{4} & -\frac{21}{4} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & \frac{7}{2} & -\frac{77}{12} \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & -4 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \end{array}\right)$ And then we consider the first line of the matrix power $C^t$. These are the coefficients of $f^{\circ t}$: $0\\ \left(-1\right)^{t}\\ -\frac{1}{4} \, \left(-1\right)^{t} + \frac{1}{4}\\ \frac{1}{12} \, t \left(-1\right)^{{(t - 1)}} + \frac{1}{8} \, \left(-1\right)^{t} - \frac{1}{8}\\ -\frac{1}{16} \, t \left(-1\right)^{{(t - 1)}} - \frac{7}{96} \, \left(-1\right)^{t} - \frac{1}{24} \, t + \frac{7}{96}\\ \frac{1}{96} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} + \frac{7}{160} \, t \left(-1\right)^{{(t - 1)}} + \frac{3}{64} \, \left(-1\right)^{t} + \frac{1}{24} \, t - \frac{3}{64}\\ -\frac{5}{384} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} + \frac{1}{144} \, {(t - 1)} t - \frac{1}{32} \, t \left(-1\right)^{{(t - 1)}} - \frac{41}{1280} \, \left(-1\right)^{t} - \frac{181}{5760} \, t + \frac{41}{1280}\\ \frac{5}{3456} \, {(t - 2)} {(t - 1)} t \left(-1\right)^{{(t - 3)}} + \frac{1}{90} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} - \frac{1}{96} \, {(t - 1)} t + \frac{365}{16128} \, t \left(-1\right)^{{(t - 1)}} + \frac{263}{11520} \, \left(-1\right)^{t} + \frac{263}{11520} \, t - \frac{263}{11520}\\ -\frac{35}{13824} \, {(t - 2)} {(t - 1)} t \left(-1\right)^{{(t - 3)}} - \frac{1}{864} \, {(t - 2)} {(t - 1)} t - \frac{11}{1280} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} + \frac{43}{4608} \, {(t - 1)} t - \frac{769}{46080} \, t \left(-1\right)^{{(t - 1)}} - \frac{901}{53760} \, \left(-1\right)^{t} - \frac{1807}{107520} \, t + \frac{901}{53760}\\ \frac{35}{165888} \, {(t - 3)} {(t - 2)} {(t - 1)} t \left(-1\right)^{{(t - 4)}} + \frac{259}{103680} \, {(t - 2)} {(t - 1)} t \left(-1\right)^{{(t - 3)}} + \frac{1}{432} \, {(t - 2)} {(t - 1)} t + \frac{37691}{5806080} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} - \frac{1}{144} \, {(t - 1)} t + \frac{5197}{414720} \, t \left(-1\right)^{{(t - 1)}} + \dots$ Each coefficient is a mixture of polynomials in $t$ containing $(-1)^t$. It has 0 convergence radius. We apply (**) and get the coefficients of the Julia function $j$: Code:0, I*pi, -1/4*I*pi, 1/8*I*pi - 1/12, -7/96*I*pi + 1/48, 3/64*I*pi - 1/80, -41/1280*I*pi + 17/2880, 263/11520*I*pi - 41/12096, -901/53760*I*pi + 229/120960, 3245/258048*I*pi - 227/207360 Now we apply (*): When we take the reciprocal of the julia function we get a formal Laurent series starting at -1. The coefficient at -1 is $-\frac{I}{\pi}$, the other coefficients are: $-\left(\frac{1}{4} I\right) \, \frac{1}{\pi}\\ \left(\frac{1}{16} I\right) \, \frac{1}{\pi} - \frac{1}{12} \, \frac{1}{\pi^{2}}\\ -\left(\frac{5}{192} I\right) \, \frac{1}{\pi} - \frac{1}{48} \, \frac{1}{\pi^{2}}\\ \left(\frac{11}{768} I\right) \, \frac{1}{\pi} + \frac{1}{320} \, \frac{1}{\pi^{2}} + \left(\frac{1}{144} I\right) \, \frac{1}{\pi^{3}}\\ -\left(\frac{137}{15360} I\right) \, \frac{1}{\pi} - \frac{13}{3840} \, \frac{1}{\pi^{2}} + \left(\frac{1}{576} I\right) \, \frac{1}{\pi^{3}}\\ \left(\frac{367}{61440} I\right) \, \frac{1}{\pi} + \frac{1697}{967680} \, \frac{1}{\pi^{2}} - \left(\frac{1}{11520} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{1728} \, \frac{1}{\pi^{4}}\\ -\left(\frac{21557}{5160960} I\right) \, \frac{1}{\pi} - \frac{1669}{1290240} \, \frac{1}{\pi^{2}} + \left(\frac{53}{138240} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{6912} \, \frac{1}{\pi^{4}}\\ \left(\frac{62171}{20643840} I\right) \, \frac{1}{\pi} + \frac{42083}{46448640} \, \frac{1}{\pi^{2}} - \left(\frac{10943}{58060800} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{138240} \, \frac{1}{\pi^{4}} - \left(\frac{1}{20736} I\right) \, \frac{1}{\pi^{5}}\\ \left(\frac{52187}{7077888} I\right) \, \frac{1}{\pi} - \frac{401003}{309657600} \, \frac{1}{\pi^{2}} + \left(\frac{12203}{77414400} I\right) \, \frac{1}{\pi^{3}} + \frac{67}{1658880} \, \frac{1}{\pi^{4}} - \left(\frac{1}{82944} I\right) \, \frac{1}{\pi^{5}}\\ -\left(\frac{4720619}{4954521600} I\right) \, \frac{1}{\pi} + \frac{95027}{176947200} \, \frac{1}{\pi^{2}} - \left(\frac{309629}{2786918400} I\right) \, \frac{1}{\pi^{3}} - \frac{487}{25804800} \, \frac{1}{\pi^{4}} - \left(\frac{1}{552960} I\right) \, \frac{1}{\pi^{5}} - \frac{1}{248832} \, \frac{1}{\pi^{6}}$ When integrating the coefficient of $z^{-1}$ becomes the coefficient of the logarithm. The other coefficients of the Abel function are obtained by formal integration. The coefficients of the Abel function $\alpha$ of $h$ are: $ \log(z):\quad -I\,\frac{1}{\pi}\\ z:\quad-\left(\frac{1}{4} I\right) \, \frac{1}{\pi}\\ z^{2}:\quad\left(\frac{1}{32} I\right) \, \frac{1}{\pi} - \frac{1}{24} \, \frac{1}{\pi^{2}}\\ z^{3}:\quad-\left(\frac{5}{576} I\right) \, \frac{1}{\pi} - \frac{1}{144} \, \frac{1}{\pi^{2}}\\ z^{4}:\quad\left(\frac{11}{3072} I\right) \, \frac{1}{\pi} + \frac{1}{1280} \, \frac{1}{\pi^{2}} + \left(\frac{1}{576} I\right) \, \frac{1}{\pi^{3}}\\ z^{5}:\quad-\left(\frac{137}{76800} I\right) \, \frac{1}{\pi} - \frac{13}{19200} \, \frac{1}{\pi^{2}} + \left(\frac{1}{2880} I\right) \, \frac{1}{\pi^{3}}\\ z^{6}:\quad\left(\frac{367}{368640} I\right) \, \frac{1}{\pi} + \frac{1697}{5806080} \, \frac{1}{\pi^{2}} - \left(\frac{1}{69120} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{10368} \, \frac{1}{\pi^{4}}\\ z^{7}:\quad-\left(\frac{21557}{36126720} I\right) \, \frac{1}{\pi} - \frac{1669}{9031680} \, \frac{1}{\pi^{2}} + \left(\frac{53}{967680} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{48384} \, \frac{1}{\pi^{4}}\\ z^{8}:\quad\left(\frac{62171}{165150720} I\right) \, \frac{1}{\pi} + \frac{42083}{371589120} \, \frac{1}{\pi^{2}} - \left(\frac{10943}{464486400} I\right) \, \frac{1}{\pi^{3}} + \frac{1}{1105920} \, \frac{1}{\pi^{4}} - \left(\frac{1}{165888} I\right) \, \frac{1}{\pi^{5}}\\ z^{9}:\quad\left(\frac{52187}{63700992} I\right) \, \frac{1}{\pi} - \frac{401003}{2786918400} \, \frac{1}{\pi^{2}} + \left(\frac{12203}{696729600} I\right) \, \frac{1}{\pi^{3}} + \frac{67}{14929920} \, \frac{1}{\pi^{4}} - \left(\frac{1}{746496} I\right) \, \frac{1}{\pi^{5}}$ The superlogarithm is then just $\operatorname{slog}(z)=\alpha(ez-1)$. Notes: 1. The provided values are just for comparison if you want to reproduce this approach. I worked with Sage and my own powerseries libary. I can provide more coefficients as numerics if you specify some format. I didnt test my computations against obvious errors nor plotted the corresponding graph, nor do I know whether the series converges (thoug it looks quite so). 2. The powers $(-1)^t$ in the first row of the matrix power $C^t$ are of course ambigous. I choose the standard branch $k=0$ of the logarithm in $(-1)^t=\exp((i\cdot \pi + 2\pi i k) z)$. 3. In the formula for $\alpha$ you can of course choose arbitrary cuts and branches for the logarithm. The corresponding cut is then the cut for the Abel function and the cut after application of $\frac{z+1}{e}$ is the cut of the superlogarithm in the point $\frac{1}{e}$. 4. You can add arbitrary constants to the superlogarithm and obtain again a superlogarithm. « Next Oldest | Next Newest »

 Messages In This Thread regular tetration at b=e^(-e) - by bo198214 - 09/26/2009, 09:25 PM RE: regular tetration at b=e^(-e) - by mike3 - 09/27/2009, 12:52 AM RE: regular tetration at b=e^(-e) - by bo198214 - 09/27/2009, 06:21 AM RE: regular tetration at b=e^(-e) - by mike3 - 09/27/2009, 10:24 PM RE: regular tetration at b=e^(-e) - by Gottfried - 01/11/2010, 12:51 PM RE: regular tetration at b=e^(-e) - by Gottfried - 08/23/2010, 05:45 PM

 Possibly Related Threads... Thread Author Replies Views Last Post Regular iteration using matrix-Jordan-form Gottfried 7 9,320 09/29/2014, 11:39 PM Last Post: Gottfried regular tetration base sqrt(2) : an interesting(?) constant 2.76432104 Gottfried 7 10,105 06/25/2013, 01:37 PM Last Post: sheldonison regular iteration of sqrt(2)^x (was: eta as branchpoint of tetrational) JmsNxn 5 8,037 06/15/2011, 12:27 PM Last Post: Gottfried Regular "pentation"? mike3 12 22,031 04/04/2011, 03:16 AM Last Post: BenStandeven closed form for regular superfunction expressed as a periodic function sheldonison 31 34,204 09/09/2010, 10:18 PM Last Post: tommy1729 [Regular tetration] [Iteration series] norming fixpoint-dependencies Gottfried 11 14,554 08/31/2010, 11:55 PM Last Post: tommy1729 [Regular tetration] bases arbitrarily near eta Gottfried 0 2,972 08/22/2010, 09:01 AM Last Post: Gottfried "Natural boundary", regular tetration, and Abel matrix mike3 9 15,807 06/24/2010, 07:19 AM Last Post: Gottfried Regular slog for base sqrt(2) - Using z=2 jaydfox 13 19,530 03/10/2010, 12:47 PM Last Post: Gottfried regular slog bo198214 18 21,949 07/31/2009, 08:55 AM Last Post: bo198214

Users browsing this thread: 1 Guest(s)