• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 base holomorphic tetration bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 11/07/2009, 09:55 AM (11/07/2009, 08:21 AM)mike3 Wrote: So then ${g^m}_n$ is nth coefficient of mth power of g (truncated).yes, while truncated is equal to untruncated. Quote:I thought I also saw somethign ike ${f_n}^m$. Note the positions of the super/subscripts are different.. would that mean the same thing or would that mean to raise the nth coefficient of f to the power m? yes, the latter. Like one would read it, first take the index then take the power. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 11/07/2009, 04:47 PM (11/06/2009, 11:29 PM)bo198214 Wrote: $g(x)=\ln(a) (e^x -1 )$, $\tau(x)=\frac{x}{\ln(b)}+a$, $\tau^{-1}(x)=\ln(b)(x - a)$, $g = \tau^{-1}\circ f\circ \tau$. $g(x)=\ln(a)x + \frac{\ln(a)}{2} x^2 + \frac{\ln(a)}{6} x^3 + \dots$ This gives the following unsimplified coefficients of $g^{\circ t}$: $ {g^{\circ t}}_1 = \mbox{lna}^{t}\\ {g^{\circ t}}_2 = \left(\frac{1}{2} \, \frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})}}{{(\mbox{lna}^{2} - \mbox{lna})}}\right) \mbox{lna}\\ {g^{\circ t}}_3 = \left(\frac{1}{2} \, \frac{{(\frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})} \mbox{lna}^{t}}{{(\mbox{lna}^{2} - \mbox{lna})}} - \frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})} \mbox{lna}}{{(\mbox{lna}^{2} - \mbox{lna})}})}}{{(\mbox{lna}^{3} - \mbox{lna})}}\right) \mbox{lna}^{2} + \left(\frac{1}{6} \, \frac{{({(\mbox{lna})}^{t}^{3} - \mbox{lna}^{t})}}{{(\mbox{lna}^{3} - \mbox{lna})}}\right) \mbox{lna}$ One can see that the coefficients are polynomials in $\ln(a)^t$ with rational coefficients in $\ln(a)$. One needs to investigate whether $f^{\circ t}(1)=\tau\circ g^{\circ t}\circ\tau^{-1}(1)=a+\frac{1}{\ln(b)}\sum_{n=1}^\infty {g^{\circ t}}_n (\ln(b)(1 - a))^n$ is analytic in $b=e^{1/e}$ with $a=\exp(-W(-\ln(b)))$. I just wanted to unify the variables: with $a = \ln(a)/\ln(b)$ we can write: $b[4]t = f^{\circ t}(1)=\tau\circ g^{\circ t}\circ\tau^{-1}(1)=\frac{1}{\ln(b)}\left(\ln(a)+\sum_{n=1}^\infty {g^{\circ t}}_n (\ln(b) - \ln(a))^n\right)$, $\ln(b)\in (0,1/e)$, $\ln(a) = - W(-\ln(b))\in (0,1)$ or shorter, setting $x=\ln(b)$ and $y=\ln(a)$ $e^x [4] t = \frac{1}{x}\left(y+\sum_{n=1}^\infty {g^{\circ t}}_n (x - y)^n\right)$, $x\in (0,1/e)$, $y=-W(-x)\in (0,1)$. The thing is now that Lambert $W$ has a singularity at $-1/e$, i.e. if $x=\ln(b)$ approaches $1/e$. The question is whether this singularity gets compensated somehow by the infinite sum. I want to further simplify the formula: with $x = y e^{-y} = y/e^y$ $e^{ye^{-y}} [4] t = e^y\left(1+\sum_{n=1}^\infty {g^{\circ t}}_n (e^{-y} - 1)^n y^{n-1}\right)$, $y\in (0,1)$ where ${g^{\circ t}}_n$ are polynomials in $y^t$ with coefficients that are rational functions in $y$. I hope i dint put errors somewhere; bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 11/08/2009, 05:39 PM (This post was last modified: 11/08/2009, 05:55 PM by bo198214.) And now, finally, the picture of regular tetration!     The red lines are ${^{0.5} x}$, ${^{1.5} x}$, ${^{2.5} x}$, ${^{3.5} x}$. The blue lines are $x, x^x, x^{x^x}, x^{x^{x^x}}$ And the green line is the limit $\lim_{n\to\infty} ({^n x})$. In the range $0. I computed the graphs with the powerseries development with 20 summands and 500 bits precision. The same picture with x and y equally scaled:     mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/08/2009, 08:25 PM And $g^{\circ t}_n$ are the same g-coefficients as what are in the paper? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/08/2009, 08:27 PM You sure that's actually the regular tetration $^{y} x$ against x or against the fixed point? Because the graph's x-coordinate looks to go way past $e^{1/e}$ if that scale is right. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 11/08/2009, 08:44 PM (This post was last modified: 11/08/2009, 08:49 PM by bo198214.) (11/08/2009, 08:25 PM)mike3 Wrote: And $g^{\circ t}_n$ are the same g-coefficients as what are in the paper? yes. (11/08/2009, 08:27 PM)mike3 Wrote: You sure that's actually the regular tetration $^{y} x$ against x or against the fixed point? Because the graph's x-coordinate looks to go way past $e^{1/e}$ if that scale is right. No, thats this damn sage scale. As I wrote the x-axis starts at 1 (so does the y-axis). 1.22 is roughly the middle of $1.44 \approx e^{1/e}$. But sage just doesnt get it managed that at least two numbers are shown at every axis, sometimes there is not even one number at the scale. mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/08/2009, 09:51 PM (11/08/2009, 08:44 PM)bo198214 Wrote: No, thats this damn sage scale. As I wrote the x-axis starts at 1 (so does the y-axis). 1.22 is roughly the middle of $1.44 \approx e^{1/e}$. But sage just doesnt get it managed that at least two numbers are shown at every axis, sometimes there is not even one number at the scale. Ah. I thought it started at 0... but I suppose that'd be wrong, as the regular iteration only goes down to $b = e^{-e} > 0$ and is complex-valued for $e^{-e} \le b < 1$. Oops, my bad... bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 11/09/2009, 02:41 PM Actually some doubts are legitimate, as the convergence radius for bases near $e^{1/e}$ is too small than being able to compute the value at 1. This is due to the fact that the non-integer iterates have a singularity at the upper fixed point. Thatswhy the convergence radius around the lower fixed point can be at most the distance to the upper fixed point. In the following picture I show this distance from the lower to the upper fixed point (red) - which is the convergence radius - and compares it with the distance of the lower fixed point to 1 - which is the needed convergence radius (in dependency of b at the x-axis).     That means that for b right from the intersection of the both curves, the point 1 is not inside the convergence radius of the tetra-power (which is developed at the lower fixed point). BUT, it seems that the divergent summation above that value is till precise enough.     mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/09/2009, 09:07 PM So what would this indicate? You said "some doubts are legitimate". bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 11/10/2009, 12:09 AM (11/09/2009, 09:07 PM)mike3 Wrote: So what would this indicate? You said "some doubts are legitimate". I just mean the powerseries convergence at the fixed point if it is near e. « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Complex Tetration, to base exp(1/e) Ember Edison 7 4,703 08/14/2019, 09:15 AM Last Post: sheldonison Can we get the holomorphic super-root and super-logarithm function? Ember Edison 10 7,276 06/10/2019, 04:29 AM Last Post: Ember Edison Is bounded tetration is analytic in the base argument? JmsNxn 0 1,906 01/02/2017, 06:38 AM Last Post: JmsNxn holomorphic binary operators over naturals; generalized hyper operators JmsNxn 15 20,417 08/22/2016, 12:19 AM Last Post: JmsNxn tetration base sqrt(e) tommy1729 2 4,251 02/14/2015, 12:36 AM Last Post: tommy1729 Explicit formula for the tetration to base $$e^{1/e}$$? mike3 1 3,748 02/13/2015, 02:26 PM Last Post: Gottfried tetration base > exp(2/5) tommy1729 2 4,042 02/11/2015, 12:29 AM Last Post: tommy1729 regular tetration base sqrt(2) : an interesting(?) constant 2.76432104 Gottfried 7 11,609 06/25/2013, 01:37 PM Last Post: sheldonison tetration base conversion, and sexp/slog limit equations sheldonison 44 68,135 02/27/2013, 07:05 PM Last Post: sheldonison simple base conversion formula for tetration JmsNxn 0 3,758 09/22/2011, 07:41 PM Last Post: JmsNxn

Users browsing this thread: 1 Guest(s)