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Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)
#19
Ivars Wrote:
Gottfried Wrote:Here are some plots of that tri-furcation, see uncommented list below.
I find the bi-,tri- and multifurcation an interesting subject, as we ask: can we assign an individual value if the iteration oscillates/is furcated - since this is somehow related to the partial evaluation of non-convergent oscillating series, to which we assign a value anyway.

Yes, that is an interesting idea, that these seemingly convergent iterations are actually divergent but get the value in the same way like e.g. series 1-1+1-1..........= 1/2. What then, one would assign to the point which when iterated with on top oscillates between and ?

From complex geometric series , would we have to assign value to that Iteration by analogy with divergent (?) sum:



whose module is and argument , so value would be:



No, the sum is not the same as . First I have to generate such sum where only odd powers of I are present.

Ivars
Hmm, for 1+I-1-I+1+I-1-I+.... I get 1/(1-I) = 1/2 + 1/2I by the formula just invoking Pari/GP on that, and also 1/2 + 1/2I by conventional Euler-summation.

For the second, to clarify the series you must assign an indexing-scheme, for instance powers of x.
Did you mean:
0 + Ix +0x^2 - Ix^3 + 0 x^4 + I x^5 .... = Ix *(1/(1+x^2)) -> 1/2 I
or
I - Ix + Ix^2 - Ix^3 .... = I*(1/(1+x)) -> 1/2 I

The limits should evaluate this way then (good to see in in the view of matrix-operations on formal powerseries, then the index is always clear under any transformation... ) Fortunately, they come out to be the same, but such equality may not be concluded only by the non-indexed (perhaps compressed by eliminating intermediate zeros) notation.

Gottfried
Gottfried Helms, Kassel
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Messages In This Thread
RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Gottfried - 05/25/2008, 10:12 PM

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