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Infinite tetration and superroot of infinitesimal
I was thinking how infinitesimal Volume of any dimension ( including 3) may reach 0.

Given the dimensionality of space, at each dimension x there exists a smallest indivisible which can not be divided further. Further division can happen only by reducing its dimensionality, i.e. x.

Since we are used to work in 3D space, the smallest spatial indivisible could be found as a thing when dimensionality of space is x=3 which still has 3D shape and area but 0 volume in 3D- it would have volume in space with less than 3D , but not in 3D. As such, since its volume in 3D is 0, it is not further divisible spatially, but it can be divided further only by reducing space dimensionality.

The reasoning why this should be possible is based on fact that if You reduce small volume of space in 3D to infinitesimal, its volume ( since it is a cube of edge, roughly) reduces to 0 infinitely faster than its area, which in turn reduces to zero infinitely faster than edge. So at each dimensionality, there has to be an smallest indivisible just on border between previous and next smaller dimensionality.


The thought of the order in which infinitesimal volume disappears - first dimension, then edge, leads to a notion of continuos space dimensionalities (which in language of AN number of variables would mean non-integer number of variables, which should not be impossible, since there are non-itnteger number of terms in sums, non-integer iteration of functions, non-integer Euler characteristics (or measures) of polytopes etc.) where integer ones are just very important special cases. If there are continuos space dimensionalities y, then there can be also imaginary
and negative.

We may assume in the beginning these continuos dimensionalities to be real number, but very soon ( e.g. via derivatives or tetration) some special cases will start to pop out, perhaps in the end leading to some prime polytype structures in space being more important than others.

This leads to natural (?) interpretation of exponentiation as hypercube of
edge x and dimension y= x^y.

Particularly interesting seems case where y<1 in which case it seems to me space
will become disconnected, or ? I have not read anything about this yet.

Which in turn allows to consider case x^x but here both x have
DIFFERENT geometrical meaning- one is edge, another SPACE dimension.

The speed of growth or decline of hypervolume x^x depends DIFFERENTLY on each x, so the derivative of the function makes them distinquishable. For some
reason, if the edge of such hypercube is e, (so e^x) all its
derivatives are e^x.

If we take partial derivatives of x^y, then they are:

y*x^y-1 - this is derivative by length of edge x, dimensionality y=constant

and

x^y*ln(y) - this is derivative by dimension y, edge x= constant.

If we now replace y with x, these derivatives are equal with oppossite signs at x= 1/e. The value of hypervolume is (1/e)^(1/e)=0,69... At x=e,at hypervolume e^e=15,.. the derivative by edge length gets smaller than derivative by dimension, so if edge is smaller than e AND dimension is smaller than e, then volume disapppear first by reducing edge in dimension x, than dimension x itself. If edge is bigger than e AND dimension is bigger than e, then increase of hypervolume happens first by increase in DIMENSION x , and then by increase of EDGE to x in THAT dimension x.

Understanding case x^y and exponentiation in general in dimensional
terms should allow to interpret tetration x^x^x^ ........
geometrically, and what infinite tetration of e^pi/2 = +-i might mean
geometrically and physically.

Of interest are also other geometrical figures in X dimensional space,
like sphere, and its volume/area values as function of radius.

Here are separated derivatives in picture:

   


Ivars
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Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by Ivars - 07/04/2008, 07:02 PM

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