# Tetration Forum

Full Version: base holomorphic tetration
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(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.
(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;
And now, finally, the picture of regular tetration!

[attachment=624]

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:
[attachment=625]
And $g^{\circ t}_n$ are the same g-coefficients as what are in the paper?
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.
(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.
(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...
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).

[attachment=628]

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.

[attachment=628]
So what would this indicate? You said "some doubts are legitimate".
(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.
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