Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)
#9
Hi Henryk

But why would a^(a^(a^.......a^I) lead to the same limit as h(a), namely w such that a=w^(1/w) ? The way it goes step by step is totally different because iteration starts as complex number.... So we can map the trajectories, how they converge, whereas in purely real number case we can not.

why:

\( a^{(a^{(a^{...(a^I)}}}= h(a) \)???

I do not understand this, and is this even true?

On otherhand, it may be a useful tool to see the differences between various real bases a as trajectories will be different.

Can You please tell me how can I plot these trajectories for e.g. integer iterations? I am ready to learn another piece of software, perhaps PARI or even SAGE- this map is so simple that programming should not be too difficult.

Ivars
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RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Ivars - 05/21/2008, 10:29 AM

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