[AIS] (alternating) Iteration series: Half-iterate using the AIS?
#5
(12/06/2012, 12:10 AM)Gottfried Wrote: I'm still looking at the whereabouts of series of iterates, like \( s(x)=x-b^x+b^{b^x} -b^{b^{b^x} } + ... - ... \) and consider the potential of such series for the definition of half-iterates, independently of the selection of fixpoints and the fractional "regular tetration" and similar goodies. The best approach so far to make this compatible with criteria like summability (we cannot have convergence in the following definitions) seems to be the following ansatz.

We take a base, say b=1.3, and use the decremented exponentiation \( f:x \to b^x-1 \), such that we have two real fixpoints. Then we can select any abscissa x_0 between that fixpoints and iterate infinitely to positive heights and to negative heights as well, getting \( x_1,x_2,... \) and \( x_{-1},x_{-2},... \). The alternating series of all that iterates is then Cesaro- or Euler-summable to some finite value \( S(x) \).

Obviously this is periodic with \( S(f^{[2+h]}(x))=S(f^{[h]}(x)) \) and also sinusoidal. With that base b=1.3 I find for instance at \( x_0 \sim 0.427734366938 \) that \( S(x_0)=0 \).

After this it is surely natural to assume, that beginning at the first iterate we have also that \( S(x_1)=0 \), but it seems also natural to assume, that then the maxima or the minima of the sinusoidal curve of all \( S(x_0)...S(x_1) \) are at the half-iterates between them.

Well, this is the crucial assumtion for my discussion here, which must prove to be sensical. Now if I let Pari/GP search for the first extremum in that curve I get the abcissa \( x_{min(serial)} \sim 0.2273401 546757 \) . If I compute the halfiterate using the regular tetration via the squareroot of the formal powerseries/the Schröder-function mechanism, I get \( x_{0.5(regular)} \sim 0.2273401 704241 \) which is very close, but only to some leading digits. The values of the infinite series beginning at these values differ only by 1e-16 and smaller, so maybe the non-match is an artifact (which I do not believe).

Do you have any opinion about this or even any idea how to proceed to make it an interesing item (say we find some method where \( x_{0.5(your_method)} \sim x_{min(serial)} \)) or simply - that it would be better to put this all aside for a good reason?

(I can provide the Pari/GP-routines if this would be convenient)

Gottfried

I'm interested in your method, but I can't seem to reproduce the results. I tried using a series of positive and negative iterates like

\( S(x)\ \stackrel{\mathrm{pseudo}}{=}\ \cdots - f^{-3}(x) + f^{-2}(x) - f^{-1}(x) + x - f(x) + f^2(x) - f^3(x) + \cdots \)

(where the "pseudo-equality" implies that we are considering equality via divergent summation as opposed to a "true" equality of a convergent series)

and summing via Cesaro summation by averaging the partial sums

\( P_1(x) = x \)
\( P_2(x) = -f^{-1}(x) + x - f(x) \)
\( P_3(x) = f^{-2}(x) - f^{-1}(x) + x - f(x) + f^2(x) \)
...

all where \( f(x) = b^x - 1 \) and \( f^{-1}(x) = \log_b(x + 1) \) for \( b = 1.3 \), but it seems to approach 0 or something close to it. Now, this is not strictly wrong since \( S(x) = 0 \) would satisfy \( S(f^{h+2}(x)) = S(f^h(x)) \), but it's not what we want. I've tried other ways of arranging the partial sums (like Cesaro-summing the positive-"degree" terms and the negative-"degree" terms separately), but it doesn't seem to help. Do I need to use Euler summation?


Messages In This Thread
RE: Half-iterate using the infinite iteration-series? - by mike3 - 12/10/2012, 03:33 AM

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