• 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. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/27/2009, 12:52 AM I've heard that the series obtained for $f^t(z)$ for $f(z) = u^z - 1$ with respect to z at z = 0 do not converge but diverge and diverge very strongly. I suspect this is because at fractional z (iteration number), the fixed point z = 0 is also a branch point at such iteration numbers ("heights", though I prefer to reserve that term for only referring to the second parameter of tetration), and so f is not analytic there. I'm not sure if or how this would carry over to/affect the Schroeder function. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/27/2009, 06:21 AM (This post was last modified: 09/27/2009, 06:48 AM by bo198214.) Quote:I'm not sure if or how this would carry over to/affect the Schroeder function. You mean the Abel function? Schroeder function occurs only for fixed points p with |f'(p)|=0,1. (09/27/2009, 12:52 AM)mike3 Wrote: I've heard that the series obtained for $f^t(z)$ for $f(z) = u^z - 1$ with respect to z at z = 0 do not converge but diverge and diverge very strongly. This is true (provably true, there are only few exception of f'(0)=1 that converge). It was not completely clear to me whether this would affect the convergence of the powerseries in the Abel function. But it seems that it is also divergent, perhaps I can prove that in a next post. But this does not destroy us: 1. Often divergent series reach more quickly usable values than convergent series if you truncate in the right moment. 2. With help of the above formula there is a (dog-slowly but converging) limit formula. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 09/27/2009, 10:24 PM (09/27/2009, 06:21 AM)bo198214 Wrote: You mean the Abel function? Schroeder function occurs only for fixed points p with |f'(p)|=0,1. Yeah, I meant Abel function. (09/27/2009, 06:21 AM)bo198214 Wrote: 1. Often divergent series reach more quickly usable values than convergent series if you truncate in the right moment. Like an asymptotic series? (09/27/2009, 06:21 AM)bo198214 Wrote: 2. With help of the above formula there is a (dog-slowly but converging) limit formula. What's that? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 09/28/2009, 10:32 AM (09/27/2009, 10:24 PM)mike3 Wrote: Like an asymptotic series? yes, yes. The series you obtain is asymptotic at the fixed point and has 0 convergence radius. Quote: (09/27/2009, 06:21 AM)bo198214 Wrote: 2. With help of the above formula there is a (dog-slowly but converging) limit formula. What's that? Perhaps I was too fast here generalzing from the case of multiplier 1. But I give this case for reference: In the same way as above you can obtain the regular Abel function for $h(z)=e^{z}-1$: $\alpha(z)=\underbrace{\frac{1}{3}\log(z) - \frac{2}{z}}_{F(z)} + p(z)$, where $p(z)$ is analytic in the left sector where $h^{\circ -n}(z)\to 0$ and has the asymptotic expansion at the fixed point 0. Now we know that $\alpha(h^{\circ -n}(z))=\alpha(z)-n$ and hence $\alpha(z)=n+\alpha(h^{\circ -n}(z))=n+F(h^{\circ -n}(z)) + p(h^{\circ -n}(z))$. The rightmost summand converges to 0. So we can write: $\alpha(z)=\lim_{n\to\infty} n+F(h^{\circ -n}(z))$. But in your case we have two alternating sectors, I guess it needs some more thinking. Gottfried Ultimate Fellow Posts: 765 Threads: 119 Joined: Aug 2007 01/11/2010, 12:51 PM (This post was last modified: 01/11/2010, 01:22 PM by Gottfried.) (09/26/2009, 09:25 PM)bo198214 Wrote: In response to the thread Solving tetration for base 0 < b < e^-e. I will feature here what one could call regular superlogarithm. (...) And then we consider the first line of the matrix power $C^t$. These are the coefficients of $f^{\circ t}$: $0\\ \left(-1\right)^{t}\\ -\frac{1}{4} \, \left(-1\right)^{t} + \frac{1}{4}\\ \frac{1}{12} \, t \left(-1\right)^{{(t - 1)}} + \frac{1}{8} \, \left(-1\right)^{t} - \frac{1}{8}\\ -\frac{1}{16} \, t \left(-1\right)^{{(t - 1)}} - \frac{7}{96} \, \left(-1\right)^{t} - \frac{1}{24} \, t + \frac{7}{96}\\ \frac{1}{96} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} + \frac{7}{160} \, t \left(-1\right)^{{(t - 1)}} + \frac{3}{64} \, \left(-1\right)^{t} + \frac{1}{24} \, t - \frac{3}{64}\\ -\frac{5}{384} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} + \frac{1}{144} \, {(t - 1)} t - \frac{1}{32} \, t \left(-1\right)^{{(t - 1)}} - \frac{41}{1280} \, \left(-1\right)^{t} - \frac{181}{5760} \, t + \frac{41}{1280}\\ \frac{5}{3456} \, {(t - 2)} {(t - 1)} t \left(-1\right)^{{(t - 3)}} + \frac{1}{90} \, {(t - 1)} t \left(-1\right)^{{(t - 2)}} - \frac{1}{96} \, {(t - 1)} t + \frac{365}{16128} \, t \left(-1\right)^{{(t - 1)}} + \frac{263}{11520} \, \left(-1\right)^{t} + \frac{263}{11520} \, t - \frac{263}{11520}\\ \dots$ Each coefficient is a mixture of polynomials in $t$ containing $(-1)^t$. It has 0 convergence radius. (...) Hi Henryk - I was curious how you derived that coefficients but am unable to apply your definitions using integral and Julia-function from the beginning of your post. Unfortunately I have written my own interpretation of the coefficients for the iterates of exp(x)-1 in a form, where (u-1),(u^2-1) etc occur in the denominators and thus cannot easily be evaluated for -for instance- u=-1 which is the case you're discussing here. I derived them using the symbolic eigensystem-decomposition and have documented them here http: go.helms-net.de/math/tetdocs/APT.htm On the other hand, there is also a method to derive them directly from the list of coefficients for increasing powers using a special scheme of weighted forward differences. This all seems finally to be connected if expressed in terms of q-analogues, and such a formulation seems then to allow to express the coefficients for any base-logarithm "u", including u=1, u=-1, and all complex unit-roots - using the matrix-method I had to choose between the cases u=1 (mat-log) |u|<>1 (diagonalization) and |u|=1,u=/=1 (impossible to express) so far. I see, how your polynomials can be used to compute the coefficients for any power, the results agree with the list of coefficients as they occur for the integer iterates of exp(-x)-1, so they must be correct, but seem to be compressed telling only the sums of the numerically indistinguishable powers of (-1) To make things waterproof for me I'd like to see your formula (or better the list of coefficients) in generalization , where the variable u is inserted for the parameter (-1) of your current example. So that for u=log(2) we can generate the coefficients for the t'th iterate of 2^x-1 ,for u=1 we can generate the coefficients of the t'th iterate of exp(x)-1 and for u=-1 we can generate your coefficients for the t'th iterate of exp(-x)-1. Does your computation-scheme allow to insert the general expression "u" (I am used to "u" for the log of the base for the dxp-function )? Gottfried P.s. My own new expression is very simple. define the bell-matrix Ut for exp(u*x) - 1 (keeping "u" as symbol), then generate a list of the coefficients for the k'th iterated function using the k'th integer power of Ut and extract its second columns into the k'th column of the list. Then define the triangular matrix of ones, scaled by inverse powers of u: Code:DR =   1    .      .      .      .   1  1/u      .      .      .   1  1/u  1/u^2      .      .   1  1/u  1/u^2  1/u^3      .   1  1/u  1/u^2  1/u^3  1/u^4 ...and also, to form the difference-operator, DRI = DR^-1 Then, using k as index for the column in list and also for the iteration-index for exp(u*x)-1 , DRI^k*list[,k] = column(a0,a1,...,ak, 0,0,0,0,...) reduces the coefficients to the sequence of zeros with some parameters a0,a1,...ak at the beginning and thus define a finite recursion for the computation of coefficients using the (powers of) DR with that parameters as initial values. Here, DRI consists of the q-binomials to base -1 . While I could reproduce some examples from your coefficients by a q&d-check with that procedure I did not yet set up a clean description, I'll come up with it later, perhaps wednesday or friday. Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 765 Threads: 119 Joined: Aug 2007 08/23/2010, 05:45 PM Just for the record. If we expand the numerators of the constant terms (1,1/4,-1/8,7/96,...) by powers of 2 and additionally by factorials then we get the sequence 1,1,2,7,36,246,2104,... This sequence is also known in OEIS A007889 and related to some "esoteric" functions/sequences... http://www.research.att.com/~njas/sequences/A007889 Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Regular iteration using matrix-Jordan-form Gottfried 7 9,313 09/29/2014, 11:39 PM Last Post: Gottfried regular tetration base sqrt(2) : an interesting(?) constant 2.76432104 Gottfried 7 10,083 06/25/2013, 01:37 PM Last Post: sheldonison regular iteration of sqrt(2)^x (was: eta as branchpoint of tetrational) JmsNxn 5 8,023 06/15/2011, 12:27 PM Last Post: Gottfried Regular "pentation"? mike3 12 21,969 04/04/2011, 03:16 AM Last Post: BenStandeven closed form for regular superfunction expressed as a periodic function sheldonison 31 34,137 09/09/2010, 10:18 PM Last Post: tommy1729 [Regular tetration] [Iteration series] norming fixpoint-dependencies Gottfried 11 14,525 08/31/2010, 11:55 PM Last Post: tommy1729 [Regular tetration] bases arbitrarily near eta Gottfried 0 2,964 08/22/2010, 09:01 AM Last Post: Gottfried "Natural boundary", regular tetration, and Abel matrix mike3 9 15,772 06/24/2010, 07:19 AM Last Post: Gottfried Regular slog for base sqrt(2) - Using z=2 jaydfox 13 19,452 03/10/2010, 12:47 PM Last Post: Gottfried regular slog bo198214 18 21,917 07/31/2009, 08:55 AM Last Post: bo198214

Users browsing this thread: 1 Guest(s)