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Infinite tetration and superroot of infinitesimal
#85
Well when I read the defintion of hypersurface in Wikipedia:

In algebraic geometry, a hypersurface in projective space of dimension n is an algebraic set that is purely of dimension n − 1. It is then defined by a single equation F = 0, a homogeneous polynomial in the homogeneous coordinates. (It may have singularities, so not in fact be a submanifold in the strict sense.)

Letting out the subtleties, the hypersurface of hypervolume -I must be dI= - I lnI = -pi/2
And, for hypervolume I -> dI = IlnI = pi/2.

Because there is nothing else given, the only way to reduce infinite dimensionality of hypervolume I to (infinity -1 ) is by taking a differential of it.
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RE: Infinite tetration and superroot of infinitesimal - by Ivars - 01/31/2008, 11:13 AM

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