09/30/2007, 11:10 PM
UVIR Wrote:\( (x*n)@t=x \)
consequently it is easily seen that \( @t=*(1/n) \)
As I already explained before it is not "consequently easily seen" but a consequence of the law \( (xa)b=x(ab) \) and similarly for exponentiation of the law \( (x^a)^b=x^{ab} \).
This law holds for natural \( a \) and \( b \) and if we demand it to hold for fractional \( a \) and and \( b \) too, then necessarily \( \frac{1}{n} x \) is the inverse of \( nx \) and \( x^{1/n} \) is the inverse of \( x^n \) by:
\( (x\frac{1}{n})n=x(\frac{1}{n}n)=x1=x \) and
\( (x^{1/n})^n=x^{(1/n)n}=x^1=x \)
but we have not \( {^a(^bx)}={^{ab}x} \) 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:
\( {^{1/n}}({^{n}x)\neq x \)
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:
\( {^2}({^3 x})\neq {^6x} \).
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 \( {^0x}=1 \)). You showed already that \( \lim_{n\to\infty} {^{1/n}x}=x^{1/x}\neq 1={^0x} \) for \( x>1 \) in your definition.
