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Infinite tetration of the imaginary unit
#4
I am more wondering about inverse operation- how do we get e^pi/2 out of -i - that is self root of -i - but self root of - i - what does it does? If tetration of e^(pi/2) creates hypervolumes from base e^(pi/2), than self root starts from that hypervolume -i and ?

When we do infinite tetration of e^(pi/2), we can do it step by step, like a limit when number of steps-> infinity- which is of course a speed -up operation ( limit taking) but than we should be able to come back also somehow step by step- but there is none- just selfroot of - i =-i^(1/-i) =e^(pi/2) in one jump.

That was the reason I suggested a step dI, hypersurface dI=lnI/I=-pi/2 etc- because differentiation is also a limit operation , so that by infinite differentation of -I we would reach e^(pi/2). But that seems a little out of usual approach. Are there any other?

The reason I thought pentation must lead to e^-pi was, on one hand, the idea that MAYBE pentation of e^(pi/2) would lead to further updimensioning of hypervolume - I, which might be somewhat special case ( getting -i from e^(pi/2) seems special to me.)

So I thought, Ok, pentation of (e^pi/2) would lead to hypervolumer -i^2 =I^2 = -1= e^-I*pi=e^I*pi.
Then sextation would be -I^3= I=e^(-3pi/2) and heptation -I^4 = 1 = e^(-2pi*I). But may be it is not so simple.

I think tetration may be special in this sequence as it kind of connects the "normal" operations with hyperoperations. It works with both, in the middle.

If so , there has to be inverse superslow operations of similarly important place. e.g 1/2 -ation? The first slower operation between addition and zeration?
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Messages In This Thread
Infinite tetration of the imaginary unit - by GFR - 02/10/2008, 12:09 AM
RE: Infinite tetration of the imaginary unit - by Ivars - 02/10/2008, 01:28 PM

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