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Infinite tetration and superroot of infinitesimal
#33
andydude Wrote:
jaydfox Wrote:infinitesimals would necessarily have density measured with 2^C
If this is true, then there cannot be an isomorphism between the standard field of complex numbers where i is imaginary unit (Cardinality of the reals), and the non-standard field of where z is an infinitesimal (Cardinality of the powerset of reals), as Ivars suggests. If they are of fundamentally different cardinalities, then an isomorphism between the two is impossible.

@Ivars: This would mean you have to make the distinction between the imaginary unit and an infinitesimal.

Andrew Robbins


I do... But that will confuse things further:

Imaginary unit is a universal notation for infinitesimals if we do not care about differences they have in different scales.

In each scale though, infinitesimals are different... So they have more properties than just imaginary unit, internal degrees of freedom which i hope will be revealed by tetration.

A scale: A scale in my interpretation is what you get when You differentiate/integrate function using infinitesimals instead of limits:

dy and dx is infinitely smaller than x and y. If You differentiate y(x) the resulting function y'(x') (x' to remember scale has changed) is in fact, in infinitely smaller scale than y(x). But again, if You can differentiate it further , there is dy'/dx' which are again infinitesimals in the new scale, so the resulting y''(x'') lays in 2 infinities smaller scale than y(x). In scale where y(x) was, or even y'(x'), y''(x'') is nothing. 0. But it is not the smallest nothing in the world, since there are scales infinity below that.

The same goes for integration, in opposite way. Differentiation usually includes information loss about absolute placement of function in space. Integration recovers the scale, but information about placement is arbitrary-chaotic in a sense- it kind of wanders all over place - hence infinite speed since as we provide information about initial conditions, integration places function immedeately where it should be.

Now, if we ignore the fact there are infinitely different scales- and in mathematics we do, we may operate with imaginary unit over all scales as if there were no difference. e takes care of that. But actually, in each scale infinitesimals are different. An infinitesimal if looked from the scale we are in y(x) , is imaginary, since it does not exist in that scale, as it is 0, but it does exist in next or other scales , where it is finite.

So maybe You are right- normal i is an averaged infinitesimal over all scales ( as long as it is defined (or complies with) e^ipi/2+-i2pik. )

But each of values of i as it rotates by infinitesimal angles from one scale to next changes in complexity and matches pi and 1/2 ( I am not sure about e) on that scale and angle, providing a finite angle.

So e.g i= (i-1)/(i+1) = h(e^pi/2) defines only one infinitesimal in one scale depending on e, pi, 1/2. We can not distinguish them as for todays math, all e, i,pi, 1/2 are the same, but they are different in each scale, just results in the same angle.

See how bad it gets....
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Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by Ivars - 01/04/2008, 09:48 PM

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