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Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)
#15
Then it seems, that for each real point in the region of convergence of h(a) we can check regions of z where convergence still holds.
That should give a deeper view, perhaps split the region into subregions with similar behaviours?

As You said, does not converge if z>4, while I have checked that does not converge at .

So how would it be possible to map such values of z = f(a) where h(a) with z on top diverges and where converges?

I can imagine that requires some 3D picture, with a on vertical axis, and z regions on complex plane drawn at every a. But I do not know how to make such calculations for all a and all z simultaneously and plot it.

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 - 05/24/2008, 04:09 PM

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