Mizugadro, pentation, Book

[MphLee]

..I can't understand half of the contents..

Ask questions. I made also account at http://dmitriikouznets.livejournal.com
for questions. But if with formulas, here seems to be easier.
You may use also http://math.eretrandre.org/hyperops_wiki...Kouznetsov

Thank you very much for the kindness, so I'm happy to accept your invite and ask you something!
I wonder if you were able to explain to me some of your last results and achievements about tetration and the state of the research in laymen's terms... if is possible.
Given my poor knowledge of dynamics and basic analysis is pretty painful for me to go through your papers without "losing my mind" :

(1) for example how do you achieve uniqueness in easy words? I mean, what are the properties that the solution must satisfy and that allow us to derive/(imply) the uniqueness?
(2) what is/are your algorithm/s for tetration?

For the second question I did my best but I was never able to go through the overwhelming amount of formulas and find something..

Looking at Mizugadro's page of Tetration it says that three different algorithms are used for bases belonging to three different domains [A]\( b\in ]1;e^{1/e}[ \), [B]\( b=e^{1/e} \) and [C]\( b\in ]e^{1/e},\infty[ \) (let's skip the complex cases)

[A] For this case Mizugadro says that the related tetration function \( {\rm tet}_b \)( aka \( 1 \)-based supefunction) is constructed with regular iteration at smallest of the real fixed points \( L \) of the function \( \log_b \), even if I don't know much about how the regular iteration method works, it says that the function should be of the form

\( \displaystyle F(z) = L+\sum_{n=1}^{N} a_n {e^{kzn}} + o({e^{kxN}}) \)

...but I'm lost before finding the values for \( k \) and for the coefficients \( a_n \) in the case of tetration... references of Mizugadro sends me to "D.Kouznetsov, H.Trappmann. Portrait of the four regular super-exponentials to base sqrt(2). Mathematics of Computation, 2010, v.79, p.1727-1756. " but the link is broken and  D.Kouznetsov. (2009). Solutions of \( F(z+1)=\exp(F(z)) \) in the complex \( z \)-plane. is really too complex for me...

[B] For the case \( b=e^{1/e} \) is says that a modified formula is needed

\( \displaystyle
F(z)=\mathrm e\cdot\left(1-\frac{2}{z}\left(
1+\sum_{m=1}^{M} \frac{P_{m}\big(-\ln(\pm z) \big)}{(3z)^m}
+\mathcal{O}\!\left(\frac{|\ln(z)|^{m+1}}{z^{m+1}}\right)
\right) \right) \)
But since I can't understand how to evaluate the coefficents \( c_{m,n} \) of the polynomials \( P_m \)... I searched it in the reference [2] Trappmann-Kouznetsov-Computation of the Two Regular Super-Exponentials to base exp(1/e). Mathematics of computation, 2012 February 8.
In this paper you (and Trappmann) say that we have to apply the regular iteration to the function \( h(x)=e^x -1 \) and since it is conjugated to \( f(x)=(e^{1/e})^x \) by the function \( \tau(x)=e(x+1) \) iterating \( h \) gives us the iterates of \( f \)(hence tetration too) because \( (\tau h \tau^{-1})^n =\tau h^n \tau^{-1} \).
Later the text shows some known old methods (Lévy's, Newton limit formula, Fatou/Walker's ) and then (pag 9) the new expansion

\( \displaystyle
F(z)=\mathrm e\cdot\left(1-\frac{2}{z}\left(
1+\sum_{m=1}^{M} \frac{P_{m}\big(-\ln(\pm z) \big)}{(3z)^m}
+\mathcal{O}\!\left(\frac{|\ln(z)|^{m+1}}{z^{m+1}}\right)
\right) \right) \)

So if I'm not lost we have that for \( \eta=e^{1/e} \) and \( F(X_1)=1 \) the following is true

\( {\rm tet}_{\eta}(z)=F(z+X_1) \)

But again, what are the coefficents \( c_{m,n} \)?

[C] Ok, here with Cauchi integral, "iterated Cauchi algorithm" I'm totally lost.
What I've got is (reading Natural tet at Mizugadro) that superfunctions are identified up to a \( 1 \)-periodic function \( \theta(x) \) but inside the equivalence class of supefunctions (up to the equivalence relation given by \( f(x)=g(x+\theta(x)) \) we can achieve uniqueness looking for the asymptotic approach to the fixed point (of the transfer function?) so, for example, also \( {\rm tet}(z+\theta(x)) \) is a superfunction of exp but it is not... the right one?

If you could explain to me some of these points I would be infinitely grateful to you.

About Pentation: l have to notice that long ago, in 2006, Rubtsov and Romerio [1] were able to compute a first approximation of the fixed point \( L_{e,4,0}=-1.850354529... \). Their first approximation where denoted by \( \sigma=-1.84140566... \) and \( pent_e (-\infty)=sln(\sigma)={{}^\sigma}e=\sigma... \) (where pent is computed using their approximation). ..
[1] Rubtsov, Romerio - Notes on Hyper-Operations, Progress Report -NKS forum III, Final review 3, 2006.
http://math.eretrandre.org/tetrationforu...hp?aid=222

Thank you for the link, I add it to the article
http://mizugadro.mydns.jp/t/index.php/Superfunction
Do you understand, how do they calculate the tetration?
I think, my algorithms are more efficient, because Rubtsov and Romerio do not present any complex map of tetration, nor pentation.

Well, actually they didn't. Those values for sigma (the fixedpoint of natural tetration) are obtained with the fixed-point iteration method using approximations of natural superlogarithm (inverse of natural tetration) (linear and cubic).
The linear approximation uses \( x+1 \) (\( -1<x<0 \)) while the Cubic approximation is defined as follow (See [i]"Andrew Robbins - Solving for the Analytic Piecewise Extension of Tetration and the Super-logarithm[i]")

\( SE_3(x)=-{2\over 13}x^3+{3\over 13}x^2+{12\over 13}x-1 \) if \( -1<x<0 \)
\( SE_3(x)=\exp(SE_3(x-1)) \) if \( 0<x \)

So using the the linear approximation we get

\( \sigma_1\simeq -1.84140566043696... \)
using the cubic one

\( \sigma_3\simeq -1.8497049770580847... \)

so if we denote your approximation with \( \sigma_K \)

we have that \( \sigma_3-\sigma_K \simeq -0.00064931.. .. \)

close enough!