01/14/2015, 01:21 AM
01/14/2015, 04:08 AM
01/15/2015, 01:12 PM
Why isnt pentation defined as
 = sexp^{[z]}(x_0) =)
lim ) )
Do you really believe lim) =)
lim ) )
where 
is a simple elementary function ?
Afterall approximating slog^[n] with an elementary function seems wrong/divergent ?
Lim)
is usually very different from lim )
even if 
is small or getting smaller with growing 
.
Also its not defined as lim)
.
I assume its (your def of pentation in the paper) meant as an acceleration of lim.
If not that would appeal weird and dubious to me.
Is that acceleration really a big improvement ?
Still reading and thinking , I dont have much time ...
regards
tommy1729
Do you really believe lim
Afterall approximating slog^[n] with an elementary function seems wrong/divergent ?
Lim
Also its not defined as lim
I assume its (your def of pentation in the paper) meant as an acceleration of lim
If not that would appeal weird and dubious to me.
Is that acceleration really a big improvement ?
Still reading and thinking , I dont have much time ...
regards
tommy1729
01/15/2015, 03:09 PM
First, thank, you, Tommy, for your interest in that article.
I try to answer your questions, although I do not understand some notations.
and your slog is equivalent of my ate.
The natural pentation pen in my book and in my article is defined through the following way:
Some integer
is chosen.
)
=L_{e,4,0} + \sum_{m=0}^{M-1} a_m \varepsilon^m)
Where
is zeroth fixed point of the 4th Ackermann to base e, id est, real solution 
of equation =x)
.
Increment
and coefficients 
are determined by substitution of the asymptotic supertetration
=f(z)+O(\varepsilon^M))
into the transfer equation
=\mathrm{tet}(F(z)))
I define
=\mathrm{tet}^n(f(z-n)))
=\lim_{n\rightarrow \infty} F_n)
=F(x_1+z))
where
is real solution of equation =1)
My claim is, that resulting pen does not depend on the
chosen.
I define it in my way, because it seems to work.
=\mathrm{pen}(1+\mathrm{ape}(z)))
where
Is it that you mean?
After to construct the approximation, the researcher should substitute it into the equation, calculate the residual and plot the map of the agreement. Then we'll see, how many decimal digits may it keep for any given argument. If the numerical test passes, we may ask Henryk to write the long and complicated proof for some "Aequationes Mathematicae".
I doubt if these equations are useful for the evaluation of tetration.
Usually, such a limits are very slow to converge; I have not enough patience "to press a key, to have a tea".
There is no reason to compare acceleration of a car to the acceleration of the unmovable rock. The only, we may say, that the car moves, while the rock does not. However, one can refer to the theory of relativity, quantum mechanics, etc., but this does not look serious.
I repeat my old statement:
Until now, nobody could calculate and plot complex maps of tetration (nor pentation) with any algorithm, faster or simpler than those I had presented.
As soon, as you or anybody else present any algorithm, that does the same, we can plot the maps of the agreement and compare the algorithms.
Tommy, if you do not have much time, then, do not collect many questions in a bunch. Ask as soon as you have formulated one first question.
For me, it will be also easier to answer. Now, the preview does not fit one screen..
Best regard, Dima.
I try to answer your questions, although I do not understand some notations.
(01/15/2015, 01:12 PM)tommy1729 Wrote: [ -> ]Why isnt pentation defined asI think, your sexp is equivalent of my tet,
and your slog is equivalent of my ate.
The natural pentation pen in my book and in my article is defined through the following way:
Some integer
Where
Increment
are determined by substitution of the asymptotic supertetration
into the transfer equation
I define
where
My claim is, that resulting pen does not depend on the
I define it in my way, because it seems to work.
Quote:Do you really believe limNo. I do not understand the notation. I would say,
where
Is it that you mean?
Quote:Afterall approximating slog^[n] with an elementary function seems wrong/divergent ?It depends on the skills of the colleague who makes the approximation.
After to construct the approximation, the researcher should substitute it into the equation, calculate the residual and plot the map of the agreement. Then we'll see, how many decimal digits may it keep for any given argument. If the numerical test passes, we may ask Henryk to write the long and complicated proof for some "Aequationes Mathematicae".
Quote:LimWhat is L' ?
Also its not defined as lim
I doubt if these equations are useful for the evaluation of tetration.
Usually, such a limits are very slow to converge; I have not enough patience "to press a key, to have a tea".
Quote:I assume its (your def of pentation in the paper) meant as an acceleration of limI do not see any equivalent of this formula in my article. Nor in my book..
If not that would appeal weird and dubious to me.
Quote: Is that acceleration really a big improvement ?Hmm... As soon, as anybody reproduces the complex maps from my Book or from my articles with any alternative algorithms, it will be possible to compare the efficiency.
There is no reason to compare acceleration of a car to the acceleration of the unmovable rock. The only, we may say, that the car moves, while the rock does not. However, one can refer to the theory of relativity, quantum mechanics, etc., but this does not look serious.
I repeat my old statement:
Until now, nobody could calculate and plot complex maps of tetration (nor pentation) with any algorithm, faster or simpler than those I had presented.
As soon, as you or anybody else present any algorithm, that does the same, we can plot the maps of the agreement and compare the algorithms.
Tommy, if you do not have much time, then, do not collect many questions in a bunch. Ask as soon as you have formulated one first question.
For me, it will be also easier to answer. Now, the preview does not fit one screen..
Best regard, Dima.
01/16/2015, 12:31 AM
02/09/2015, 08:19 PM
(01/11/2015, 05:01 PM)Kouznetsov Wrote: [ -> ]Quote:..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]
[A] For this case Mizugadro says that the related tetration function
...but I'm lost before finding the values for
[B] For the case
But since I can't understand how to evaluate the coefficents
In this paper you (and Trappmann) say that we have to apply the regular iteration to the function
Later the text shows some known old methods (Lévy's, Newton limit formula, Fatou/Walker's ) and then (pag 9) the new expansion
So if I'm not lost we have that for
But again, what are the coefficents
[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
If you could explain to me some of these points I would be infinitely grateful to you.
Quote:Quote: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 pointThank you for the link, I add it to the article. Their first approximation where denoted by
and
(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
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
So using the the linear approximation we get
using the cubic one
so if we denote your approximation with
we have that
close enough!
02/09/2015, 10:35 PM
02/10/2015, 12:07 AM
Thank you, MphLee, for your interest.
I postulate the specific behaviour at infinity.
These postulates allow the efficient evaluation.
What happens, it you substitute the representation above into the transfer equation
=\exp_b(F(z)))
?
Please, express the left hand side and the right hand side
as series with respect to the small parameter
 )
and collect terms with the same power of
.
You may compare the result with general formulas (6.3) and (6.4)
of my book
"Суперфункции"
for the specific transfer function
and indicate, which ULR does not work.
I shall try to fix it.
Then we'll consider the other your questions.
Sincerely. Dmitrii.
(02/09/2015, 08:19 PM)MphLee Wrote: [ -> ].. I wonder if you were able to explain to me some of your last results..Yes, I shall try.
Quote:(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?I postulate the holomorphism in so wide range as I can.
I postulate the specific behaviour at infinity.
These postulates allow the efficient evaluation.
Quote:(2) what is/are your algorithm/s for tetration?Yes. Let us consider these three cases one by one.
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]
and [C]
(let's skip the complex cases)
Quote:[A] For this case Mizugadro says that the related tetration functionLet us find you.( aka
-based supefunction) is constructed with regular iteration at smallest of the real fixed points
of the function
, even if I don't know much about how the regular iteration method works, it says that the function should be of the form
...but I'm lost before finding the values for
What happens, it you substitute the representation above into the transfer equation
?
Please, express the left hand side and the right hand side
as series with respect to the small parameter
and collect terms with the same power of
You may compare the result with general formulas (6.3) and (6.4)
of my book
"Суперфункции"
for the specific transfer function
Quote:and for the coefficientsPlease, type the source that indicates the wrong linkin 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 indicate, which ULR does not work.
I shall try to fix it.
Then we'll consider the other your questions.
Sincerely. Dmitrii.
02/10/2015, 12:18 AM
02/10/2015, 12:12 PM
Ok, let's try it.
LEFT SIDE Substituting it in)
it gives
 = L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N))
 = L+\sum_{n=1}^{N} a_n {e^{k(z+1)n}} + o({e^{k(z+1)N}}))
=L+\sum_{n=1}^{N}(a_n{e^{kn}})\varepsilon^n+ o(\varepsilon^N{e^{kN}}))
Let's pretend for a second that I know how the little-o notation works (even if I don't even know what it is), now we have a new series that instead of the coefficents has the coefficents 
that looks really similar to the one we started with
=L+\sum_{n=1}^{N}b_n\varepsilon^n+ o(\varepsilon^N{e^{kN}}))
LEFT
---
RIGHT SIDE Now I substitute it in))
) = \exp_b(L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N)))
Im not really good with series but maybe I could use the definition of exponentiation (for T=exp_b) and obtain this
} =e^{\ln(b)\cdot F(z)}= \exp_b(L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N)))
^i\over i!}F(z)^i= \sum_{n=0}^{\infty}{\ln(b)^i\over i!}\left (L+\sum_{n=1}^{N} a_n {\varepsilon^n} + o(\varepsilon^N) \right)^i)
At this point with my poor knowledge of the subject I can't continue, so looking at your book it says that series has the property that (6.4)
) = L+T^{[1]}(L)\sum_{n=1}^{\infty}a_n {\varepsilon^n} +{T^{[2]}(L)\over 2}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^2+...)
The summation in (6.4) seems an infinite series because I saw the dots “...”, so if I understand well the formulas, assuming that the sequence 1,2,6... is the factorial and with transfer function (with )
) the final form should be something like that
) =L+\sum_{i=1}^{\infty}{T^{[i]}(L)\over i!}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^i)
RIGHT
---
So the question is how can be the left side the same as the right side?
=L+\sum_{n=1}^{N}b_n\varepsilon^n+ o(\varepsilon^N{e^{kN}})=L+\sum_{i=1}^{\infty}{T^{[i]}(L) \over i!}\left(\sum_{n=1}^{\infty}a_n {\varepsilon^n}\right )^i=T(F(z)))
So what the book is probably trying to say is that is possible to manipulate algebraically the RIGHT side in order to obtain a series of the form
)=L+\sum_{n=1}^{\infty}\tau_n\varepsilon^n)
with the coefficients
defined in the formulas (6.5), (6.6) and (6.7) of your book
)
a_2+T^{[2]}(L)/2)
a_3+T^{[2]}(L)a_2+T^{[3]}(L)/6)
and probably
a_n+T^{[2]}(L)a_{n-1}+...+T^{[n-1]}(L)a_2+T^{[n]}(L)/n!)
I don't know the rules that makes you able to turn the series on the RIGHT in a series with coefficients but if is possible and we assume that )=F(z+1))
then we should have that
=L+b_1\varepsilon+b_2\varepsilon^2+b_3\varepsilon^3+...=L+ {{\tau}}_1 \varepsilon + {\tau}_2 \varepsilon^2 + {\tau}_3 \varepsilon^3+...=T(F(z)))
and thus
. This makes us able to find
=\exp'_b(L)=\ln(b)b^L)
so
b^L\over a_1}))
Is this correct? And since we have and 
the other are given by your formulas (6.8 ), (6.9) and (6.10)...
Thank you alot for the effort and your help. I'm really curious about this but I'm also sorry if is so hard, I'm a donkey in analysis and im not good with powerseries as well.
LEFT SIDE Substituting it in
Let's pretend for a second that I know how the little-o notation works (even if I don't even know what it is), now we have a new series that instead of the coefficents
---
RIGHT SIDE Now I substitute it in
Im not really good with series but maybe I could use the definition of exponentiation (for T=exp_b) and obtain this
At this point with my poor knowledge of the subject I can't continue, so looking at your book it says that series has the property that (6.4)
The summation in (6.4) seems an infinite series because I saw the dots “...”, so if I understand well the formulas, assuming that the sequence 1,2,6... is the factorial and with transfer function
---
So the question is how can be the left side the same as the right side?
So what the book is probably trying to say is that is possible to manipulate algebraically the RIGHT side in order to obtain a series of the form
with the coefficients
and probably
I don't know the rules that makes you able to turn the series on the RIGHT in a series with coefficients
and thus
so
Is this correct? And since we have
Thank you alot for the effort and your help. I'm really curious about this but I'm also sorry if is so hard, I'm a donkey in analysis and im not good with powerseries as well.