02/14/2015, 01:22 AM

(02/13/2015, 01:21 PM)tommy1729 Wrote:Hence, the corresponding coefficient is zero. Please, look:(02/13/2015, 02:25 AM)Kouznetsov Wrote:The term ln(x)^7 x^3 does not occur in your system because 7>3.Quote:.. what about terms ln(x)^7 x^3 then ?You may extract the coefficient from the C++ implementation at

http://mizugadro.mydns.jp/t/index.php/E1etf.cin

My expansion use ln(x)^m x^n for positive m and negative n.

Quote:You work with P_n(ln(x)) x^n where P_n is a polynomial of degree n.I doubt, if the expansion with powers of ln(x) and x should be called "taylor series".

hence my comment about ln(x)^7 x^3.

and also the term " subset " of the taylor series (x,ln(x)).

This gives me some question mark feelings.

The Taylor series correspond to expansion of holomorphic function in a point, that is inside the range of holomorphism. Neither logarithm, nor negative powers of the argument are allowed there.

Perhaps, this observation resolves your "question mark feelings".

Quote: Could you give a link to the chapter 12 you mentioned ?Now the book is in "one piece". At the beginning, I had suggested the link

Maybe its in an earlier post but im unsure.

http://mizugadro.mydns.jp/t/index.php/Суперфункции

does it open from your country?

The book can be ordered at

https://www.morebooks.de/store/ru/book/С...59-56202-0

and/or loaded at

http://www.ils.uec.ac.jp/~dima/BOOK/202.pdf

http://mizugadro.mydns.jp/BOOK/202.pdf

Chapter 12 begins at page 147 of the printed version.

Numeration of pages begins with -3, this is standard for that Editorial.

The book is a little bit in Russian, but the formulas are in English, and the codes are in C++ and in Mathematica; hope you understand the last three languages mentioned.

Quote:I would like to see a proof that all parabolic fixpoint expansions are like this.Some approaches to this can be found in the book

5+ methods for real analytic tetration

Henryk Trappmann, Dmitrii Kouznetsov June 28, 2010.

https://bbuseruploads.s3.amazonaws.com/b...2Z1HF1TZ82

Also, some hints are in the article

H.Trappmann, D.Kouznetsov. Computation of the Two Regular Super-Exponentials to base exp(1/e). Mathematics of Computation. Math. Comp., v.81 (2012), p. 2207-2227. ISSN 1088-6842(e) ISSN 0025-5718(p)

http://www.ams.org/journals/mcom/0000-00...2590-7.pdf

http://mizugadro.mydns.jp/PAPERS/2012e1eMcom2590.pdf offprint

http://mizugadro.mydns.jp/PAPERS/2011e1e.pdf (preprint)

https://bitbucket.org/bo198214/e1e/raw/6...3/main.pdf

I do not like that "proof" by Henryk, described there, and I do not include it into the "Суперфункции".

I suspect, if you want some accurate

"proof that all parabolic fixpoint expansions are like this",

then you have to write it by yourself.

I think, before to deal with the "general proof",

you should consider one or two special cases.

Quote:IM not sure if we use / do things the same way.We do not do things in the same way.

We do things in different ways.

I consider argument z and treat 1/z as small parameter.

Using this approach,

I suggest the simple algorithms to evaluate superfunctions;

I implement them in C++,

I plot complex maps of the function and maps of the agreement,

I supply them with generators.

You consider argument x as small parameter,

You suggest the complicated algorithms,

You do not implement them in C++ (nor in Fortran),

You do not plot the complex maps of superfunctions,

You do not plot the maps of the agreement at the substitution of your approximations into the transfer equation.

Quote:I consider solving the zoom equation F(f(x)) = f(ax) becauseConsider to plot the map of the solution and the map of the agreement.

polynomial/Taylor of P_n(ln(x)) x^n = P*_n(ln(x)) x^n and

P_n(ln(ax)) ax^n = P_n(ln(x)+ln(a)) a^n x^n = P°_n(ln(x)) x^n.

Therefore we get a ring structure that is completely solvable.

I mean expansion of superfunction at infinity, for large |x|.

This expansion provides the efficient algorithm for the evaluation of superfunction.

You seem to deal with the Schroeder rquation and the Schroeder functions.

In some cases, the superfunction can be expressed through the Schroeder function and vice versa. But it is not a case of tetration to base exp(1/e).

Quote:.. But the main issue is convergeance.No convergence. The series are asymptotic.

With my expansion, using the transfer equation, for any argument you like, you can get so many decimal digits as you like. In this sense, the solution is exact. We can use it to plot the complex maps.

With your expansion, I think, you cannot do the same.

Quote:If im correct the series will only work if x > 1 because of the log parts ?I do not know about your series. I doubt, if your series "work", in the sense of efficient algorithm to evaluate the superfunction.

Or is it x > 0 ?

As for my series, for the evaluation, you should shift the argument to deal with large values of |x|.

Note, that in my expansion, the argument appear in NEGATIVE powers.

Quote:If we solve the zoom equation in terms of Taylor we get a formal Taylor with radius zero right ?Not. I think, always it does NOT.

[quote]

About radius zero, I think so.

About Taylor, I think, "not".

[quote]

But does this new series expansion give a nonzero radius then ? ALWAYS ?

I think, all the "primary" series in this case have radius of convergence zero.

These are typical perturbation series; there is empiric observation that all non–trivial perturbation series diverge.

(There is common confusion among physicists: many colleagues believe, that, at small value of the expansion parameter, the perturbation series converge; usually, it costs a lot of efforts in a hard discussion, to convince them, that it is not so.)

Quote:Does this relate to the mittag-leffler expansion ?Do you mean http://en.wikipedia.org/wiki/Mittag-Leffler's_theorem

?

It may give you some hints, but I do not think you can use it "as is" for tetration, because the tetration is not a meromorphic function.

Best regard, Dmitrii.