10/01/2007, 05:56 PM
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:
\( {^{1/n}}({^{n}x)\neq x \)
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:
\( {^2}({^3 x})\neq {^6x} \).
That's right. There is a BIG difference however: The law \( {^m}({^n x})\neq {^{mn}x} \) fails because of the fundamental LAW of tetration for naturals and there's no remedy for the failure. \( {^{1/n}}({^n x})\neq x \) 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 \( {^0x}=1 \)). You showed already that \( \lim_{n\to\infty} {^{1/n}x}=x^{1/x}\neq 1={^0x} \) for \( x>1 \) in your definition.
*IF* we assume that tetration is defined as \( {^0x}=1 \). If we don't assume that (in particular if we assume that \( {^0x}=x^{1/x} \)), 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]).
