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Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I)
#22
Ivars Wrote:while I try to understand increasing height series...

I think, that is not too difficult. Just take a base, where the trajectory converges (for instance b=sqrt(2)) and alternate sum the points of the trajectory. (In pari/gp there is the sumalt()-function).
The problem is then only, if b>e^(1/e). Then the trajectory begins to strongly diverge - but for a certain range of b you may still be able to apply one common summation-method, for instance the Shanks-method.

But while even this method allows bases a little bit greater, the diagonalization method seems to be able to sum that to arbitrary bases...
:cool:

Best is, you try with convergent example - just add the points of your trajectory in the list-matrix with alternating signs (even Cesaro-sum should be sufficient for this)
Quote:I was similarly(?) to You wandering that perhaps we need infinite product power series to evaluate tetration ( because of its speed)?

Yes, sometimes I was speculating about this. But I doubt it is of any use... Infinite product-series can be evaluated by the additive series of their logarithms. In the context of tetration this means, if we get an improvement at all (by this reformulation) then it is only occuring for height 1. For greater heights we are still lost. So everytime this idea pops up in my head, I just dismiss it...

Gottfried
Gottfried Helms, Kassel
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RE: Fractals from calculations of 2^I, 2^(2^I), 2^(2^(2^I).. a^(a^(...a^I) - by Gottfried - 05/26/2008, 09:01 AM

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