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Infinite tetration and superroot of infinitesimal
#46
andydude Wrote:I think trying to understand that statement intuitively is doomed, as recursive interpretation may lead to madness...

Andrew Robbins

May be not if You try to understand at least just one: h(e^(pi/2))=-+i.

This means e^(-i*(pi/2)) = e^h(e^(pi/2))*(pi/2).

So infinitely working beginning with end ( from right) in infinite number of dimensions on e^(pi/2) is equal to -imaginary unit.

If we look at e^(-i*pi/2) as turn 90 degrees clockwise in complex plane, this turn involves going through infinite tetration in other dimensions. What could that be?

So log(log(log(...... ) is tetration? if we take e^pi/2=i^-i, and log with base i, we get (from right) (only main value) :

log i (.................log i ((i^-i)) = log i( ....( log i ( -i) = log i (....(log i (-i))= log i ( ... log i (logi ( -1) = log i ( ...log i ( log i^2) = log i (......log i (2) = log i(...........log i(log i (i) + log i(i)) = log i(..............logi(log i i^2 = log i (......log i (2)=log i(...........log i(log i (i) + log i(i)) = log i(..............logi(log i i^2 = log i (......log i (2)=............

So,if we ever get to the end, we might have either log i(i^2) = 2 or log i (2) = 2ln2/i*pi.

Is this the same as tetration of (e^pi/2)? How do we calculate it in this case?
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Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by Ivars - 01/08/2008, 03:44 PM

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