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A more consistent definition of tetration of tetration for rational exponents
bo198214 Wrote:
Quote:Now, the very subtle problem which I guess nobody sees (for some strange reason) is that if the operator for tetrating to (1/n) *is NOT* the same operator as that of the tetraroot of order n, then we have an operator discrepancy at a very low level in the hierarchy of operators:

Yes, well we have to live with it. But except that we have to deal with more different operations on the tetra level, I see no problems arising from the inequality. We also have to live with for example:


That's right. There is a BIG difference however: The law fails because of the fundamental LAW of tetration for naturals and there's no remedy for the failure. doesn't HAVE to fail. It fails because of the way *WE* have defined tetration for the (1/n)-th iterate. That's an additional *introduced* discrepancy, which does NOT depend on the fundamental law of tetration for naturals, and which HAS a remedy. Do you see the difference?

Quote:There is a telling. It is not continuous at (the exponent) 0 (if we assume that tetration is defined on natural numbered exponents in the usual way, particularly ). You showed already that for in your definition.

*IF* we assume that tetration is defined as . If we don't assume that (in particular if we assume that ), nothing has been shown. In particular, you haven't shown anything about continuity of the proposed way to do tetration (inside and outside the interval [0,1]).

Messages In This Thread
RE: A more consistent definition of tetration of tetration for rational exponents - by UVIR - 10/01/2007, 05:56 PM

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