Login

***bo198214*** · 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*** · 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 $g^{\circ t}$ :
$<br /> {g^{\circ t}}_1 = \mbox{lna}^{t}\\<br /> {g^{\circ t}}_2 = \left(\frac{1}{2} \, \frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})}}{{(\mbox{lna}^{2} - \mbox{lna})}}\right) \mbox{lna}\\<br /> {g^{\circ t}}_3 = \left(\frac{1}{2} \, \frac{{(\frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})} \mbox{lna}^{t}}{{(\mbox{lna}^{2} - \mbox{lna})}} - \frac{{({(\mbox{lna})}^{t}^{2} - \mbox{lna}^{t})} \mbox{lna}}{{(\mbox{lna}^{2} - \mbox{lna})}})}}{{(\mbox{lna}^{3} - \mbox{lna})}}\right) \mbox{lna}^{2} + \left(\frac{1}{6} \, \frac{{({(\mbox{lna})}^{t}^{3} - \mbox{lna}^{t})}}{{(\mbox{lna}^{3} - \mbox{lna})}}\right) \mbox{lna}<br />$

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*** · (This post was last modified: 11/08/2009, 05:55 PM by bo198214.)

And now, finally, the picture of regular tetration!

Filename: tet.png Size: 23.15 KB 11/08/2009, 05:32 PM

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 $0<x<e^{1/e}$ .

I computed the graphs with the powerseries development with 20 summands and 500 bits precision.

The same picture with x and y equally scaled:

Filename: tet2.png Size: 14.35 KB 11/08/2009, 05:55 PM

mike3 · 11/08/2009, 08:25 PM

And $g^{\circ t}_n$ are the same g-coefficients as what are in the paper?

mike3 · 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*** · (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 · 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... $Smile$

***bo198214*** · 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).

Filename: tet_conv_radius.png Size: 13.21 KB 11/09/2009, 02:32 PM

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 · 11/09/2009, 09:07 PM

So what would this indicate? You said "some doubts are legitimate".

***bo198214*** · 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.

Login
Username:
Password:	Lost Password?
	Remember me

Possibly Related Threads...
Thread		Author	Replies	Views	Last Post
	Complex Tetration, to base exp(1/e)	Ember Edison	7	2,585	08/14/2019, 09:15 AM Last Post: sheldonison
	Can we get the holomorphic super-root and super-logarithm function?	Ember Edison	10	4,136	06/10/2019, 04:29 AM Last Post: Ember Edison
	Is bounded tetration is analytic in the base argument?	JmsNxn	0	1,460	01/02/2017, 06:38 AM Last Post: JmsNxn
	holomorphic binary operators over naturals; generalized hyper operators	JmsNxn	15	17,389	08/22/2016, 12:19 AM Last Post: JmsNxn
	tetration base sqrt(e)	tommy1729	2	3,533	02/14/2015, 12:36 AM Last Post: tommy1729
	Explicit formula for the tetration to base [tex]e^{1/e}[/tex]?	mike3	1	3,154	02/13/2015, 02:26 PM Last Post: Gottfried
	tetration base > exp(2/5)	tommy1729	2	3,317	02/11/2015, 12:29 AM Last Post: tommy1729
	regular tetration base sqrt(2) : an interesting(?) constant 2.76432104	Gottfried	7	9,751	06/25/2013, 01:37 PM Last Post: sheldonison
	tetration base conversion, and sexp/slog limit equations	sheldonison	44	58,314	02/27/2013, 07:05 PM Last Post: sheldonison
	simple base conversion formula for tetration	JmsNxn	0	3,411	09/22/2011, 07:41 PM Last Post: JmsNxn