02/13/2015, 01:21 PM

(02/13/2015, 02:25 AM)Kouznetsov Wrote:(02/13/2015, 12:45 AM)tommy1729 Wrote: Im fascinated by the (subset) Taylor(x,ln(x)) (or transseries if you want) used for the parabolic fixpoint.Thank you for your interest, tommy1729.

Its ring structure makes solving the zoom equation ( F(f(x)) = f(ax) ) natural.

( polynomials of Taylor(x,ln(x)) are of same type and also Taylor(ax,ln(ax)) is of same type ! )

This needs more attention imho. ..

Have you write some similar expansions? Do they work?

Quote:Are all parabolic fixpoint expansions (for iterating analytic functions) like this ??I think so. You may compare it to expansion of superfunction of transferfunction zex(z)=z exp(z)

It is described in chapter 12 of book "Суперфункции".

The Mathematica code to caluate the expansion is suggested in section 12.2.

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

Best regard, Dmitrii

The term ln(x)^7 x^3 does not occur in your system because 7>3.

You work with P_n(ln(x)) x^n where P_n is a polynomial of degree n.

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.

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Could you give a link to the chapter 12 you mentioned ?

Maybe its in an earlier post but im unsure.

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I would like to see a proof that all parabolic fixpoint expansions are like this.

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IM not sure if we use / do things the same way.

I consider solving the zoom equation F(f(x)) = f(ax) because

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.

Im not sure if its solutions is unique , and think not because the Taylor series is a subset.

And even without the Taylor series subset Im not sure.

I also notice that an integral or derivative can be of DIFFERENT TYPE.

Contrary to Taylor series.

But the main issue is convergeance.

If im correct the series will only work if x > 1 because of the log parts ?

Or is it x > 0 ?

If we solve the zoom equation in terms of Taylor we get a formal Taylor with radius zero right ?

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

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Does this relate to the mittag-leffler expansion ?

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Alot to do.

regards

tommy1729