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Infinite tetration and superroot of infinitesimal
No, it is a real number. Like e is also a real number, even when it is the "base" of an exponential. I don't see why it would be 2-dimensional when it is the "base" of a tetrational expression. I don't have any prejudice idea about "edges", but the name of that object is the "base" of the tetrational expression.

In y = b ^ x, b is the base, x is the (super)exponent and y is the result of b-tetra-x.

For x-> +oo, y -> h, which we may call "the height of the infinite tower", for a kind of nice word-joke. Both you and me we know that we don't have any "tower" around.

By the way, for my old-fashoned mind, every well simplified expression, if it is correctly done, is equivalent, or there is something wrong.


Messages In This Thread
RE: Infinite tetration and superroot of infinitesimal - by GFR - 01/31/2008, 05:06 PM

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