Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)
#14
But to see this oscillating behaviour between I and -I you dont need to do numerical computations, as you already mentioned:

\( b^I=(I^I)^I=I^{-1}=-I \)
and the next step is to look at
\( b^{-I}=(I^I)^{-I}=I^1=I \)

Hence this oscillating behaviour for base \( b=I^I \).
And yes this is one non-converging case.
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RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by bo198214 - 05/24/2008, 08:55 AM

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