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A more consistent definition of tetration of tetration for rational exponents
#12
UVIR Wrote:
consequently it is easily seen that

As I already explained before it is not "consequently easily seen" but a consequence of the law and similarly for exponentiation of the law .
This law holds for natural and and if we demand it to hold for fractional and and too, then necessarily is the inverse of and is the inverse of by:

and



but we have not for natural a and b and hence can not generally demand it for fractional a and b.

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:

.

Quote:Besides, there's no teling whether tetration as defined using tetraroots is or is not continuous.

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.
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Messages In This Thread
RE: A more consistent definition of tetration of tetration for rational exponents - by bo198214 - 09/30/2007, 11:10 PM

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