Base 'Enigma' iterative exponential, tetrational and pentational
#5
(07/02/2011, 02:00 AM)Cherrina_Pixie Wrote: Besides, I'm not totally sure about how to generate an analytic superfunction from a parabolic fixed point... From what I have read there is no way to construct an analytic non-integer iterate of a function that is analytic at a parabolic fixed point, is that correct?

Only the fixpoint is not analytic (though it is infinitely times differentiable).
For more details see my explanation on mathoverflow.
So generally we have 2 or more superfunctions (not even counting the constant superfunction, with value of the fixpoint).

I recently proved a rather simple formula to calculate the non-integer iterates:
\( f^t(x) = \lim_{n\to\infty} f^{-n}(f^n(x) - ta f^n(x)^{p}) \)
if \( f(x)=x-ax^{p}+o(x^{p}),x\to 0 \) and \( a,x>0 \), \( p\ge 2 \).
For example if \( f \) can be devloped into a powerseries \( f(x)=x-ax^{p} + c_{p+1}x^{p+1}+\dots \).

If \( a<0 \) and/or \( x<0 \) it possibly does not converge anymore.
You can use the identity \( f^t(x)=(f^{-1})^{-t}(x) \) to obtain a convergent version.

If the fixpoint is not 0 but \( x_0 \) then one considers \( g(x)=f(x+x_0)-x_0 \), the iterates of \( f \) are then:
\( f^t(x)=g^t(x-x_0)+x_0 \).

Originally I developed this formula to also have a direct formula for the superfuntion \( x\mapsto f^x(x_0) \), while the in my mathoverflow-answer mentioned last (quickly converging) formula for the Abel function of exp(x)-1 is not directly invertible opposing for example Lévy's formula.
Unfortunately it seems my formula is also not very quickly converging.

You can improve the initial exactness, when changing the \( \tau(x)=x-tax^p \) in the formula \( f^{-n}(\tau(f^n(x))) \).
\( \tau \) is a truncation of the iteration \( f^t \) of the formal powerseries of \( f \).
Though the whole formal powerseries of \( f^t \) has zero convergence radius, you can use any truncation \( \tau \) in the above formula, and it will converge.
With these non-converging powerseries there is a trick: If you truncate in the right moment, then you may very close to the real value.
This moment is usually the point where the indexed convergence radius \( r_n = 1/\sqrt[n]{|c_n|} \) is largest.

There is a recursive formula to calculate the coefficients of that formal iteration.
I wrote a sage-package formal_powerseries.py, which can be found here, which can calculate these iterates of formal powerseries.
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Messages In This Thread
RE: Base 'Enigma' iterative exponential, tetrational and pentational - by bo198214 - 07/02/2011, 07:13 AM

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