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Infinite tetration and superroot of infinitesimal
Ivars Wrote:When looking at x^y as hypercube in dimension y with edge x, I have made implicit assumption that the ANGLE in which each edge of such hypercube creates with another is 90 degrees in any dimension y, just by analogy, or Pi/2, as in general definitions of hypercube.

However, this angle may be another,may be negative, may be imaginary.

This leads to interpretation of real etc. extensions of hyperoperation number z in x[z]y as angle between such edges x of a hypercube in dimension y either directly, or via some exponential relation like log (pi/2)^n/log (pi/2) = n for integer z, and z*log pi/2/log (pi/2) =z for all other. , or n= log (e^n*I*pi/2)/log I , z= log (e^z*I*pi/2)/log I.


In expression :

x[z]y

x- edge of hypervolume in y dimensions
y- dimensions of hypervolume with edge x
z- is related to varying angle between edges x of hypervolume in y dimensions, perhaps via logarithm or some trigonometric function. A meaningful definition should recover usual angles pi/2 in case of hypercube in integer dimensions.

The geometric interpretation of hyperoperation in general and its real, complex extensions is related to projective geometry in imaginary form extended to non-integer, negative and imaginary dimensions.

To give a more concrete test example:

I[I]I is a hypervolume with edge I in I dimensions with angle between imaginary edges (e^(I*I*pi/2) = e^(-pi/2)
I[-I]I is a hypervolume with edge I in I dimensions with angle between imaginary edges e^pi/2. In I dimensional space, such angle might have a meaning.

Ivars

For example in I dimensional space, angles (?) e^(-pi/2) and e^(pi/2) are both solutions of equation:

z^(2*I)+1=0 which is equivalent to :

(z^I+I)*(z^I-I)=0 and has solutions:

z1=e^pi/2
z2=e^-pi/2

It is easy to see that these solutions are such that z1*z2=1 or
z1= 1/z2. This is involution, and z1 , z2 are 2 real fixed points of involution on imaginary line=dimension I.

All REAL circles in projective plane have in common circular points I, -I. Obviously (??) , ALL IMAGINARY circles in projective plane have in common 2 real points (in angle space? or e^( angle) space) ?) e^pi/2, e^-pi/2.

Infinite tetration transforms one of these real common points of Imaginary circles (e^pi/2) into both common points of real circles (+I, -I) since h( e^pi/2) =

a) h( I^(1/I) = I
b) h((1/I)^I) =-I

But it does not transform the other common point of imaginary circles, e^-pi/2, into anything so interesting, as h( e^-pi/2) = 0,47541..


Ivars
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Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by Ivars - 10/16/2008, 10:39 AM

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