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Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)
#20
Refering to the header of this thread and your obvious love to iteration and trajectories, I challenge you to compute the alternating sum of all the intermediate steps of the trajectory
I, 2^I, 2^(2^I), 2^(2^(2^I)...

AS(2,I) = I - 2^I + 2^2^I - 2^2^2^I + ... -... = ??

My proposal is

AS(2,I) = -0.440033096027+0.928380628227*I

What do you think?

Gottfried

[update 4'2014] The value was computed using "regular tetration" which includes a fixpoint-shift of the exponential-series (which also leads to triangular matrices). When I wrote this posting I was not aware of the deviance of the evaluation when repelling or attracting fixpoints were chosen. A short explanation of this and a better/more meaningful estimate for this sum one can find in the mail from yesterday a couple of post below this one. 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, 10:51 PM

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