A more consistent definition of tetration of tetration for rational exponents

[bo198214]

I have examined with interest the "details" shown in the UVIR's posting. Unfortunately, I cannot agree with putting:
b # n = y => b = y # (1/n).

There has already been at least one peer reviewed paper on this definition of tetraroots, so whether you "agree" or not is irrelevant.

The tetra root is defined as the inverse of the x#n but it has nothing to do with x#(1/n).

Moreover, I think that these two "tetration" formulas don't have any point in common. Their sets are disjoint. There is no similarity with the "exponentiation" case, where the power and the root functions are described in the same domain and we can find a root function representable as a power. I know but, unfortunately, ... this is the real problem.

I absolutely agree with Gianfranco.
The reason why \( x^{1/n} \) and \( \sqrt[n]{x} \) coincide is that \( (x^a)^b=x^{ab} \). If we demand this law to be valid also for rational exponents then \( (x^{1/n})^n = x^{n/n}=x^1=x \) and thatswhy \( x^{1/n} \) must be the inverse of \( x^n \). Unfortunately \( {}^a({}^b x)={}^{ab}x \) is no more valid.

The tetraroot is the exact reverse operation of tetration, since

Nobody denies that the tetraroot is the inverse of \( {}^nx \) (by definition) however to define \( {}^{1/n}x \) as being the tetraroot is quite arbitrary and additionally does not coincide with our other methods. As far as I have seen the 3 definitions of \( {}^y x \) (via Daniel's continuous iteration at the first fixed point, via Gottfried's matrixoperator method and via Andrew's natural slog) coincide for \( 1<x<e^{1/e} \). As an example I graphed \( {}^{1/2}x \) (via the method of continuous iteration at the first fixed point) (red) in comparison with the inverse of \( {}^{2}x \) (blue):

   
Curve 1 is \( {}^{1/2}x \)
Curve 2 is the identity function.
Curve 3 is \( {}^2x \) and
Curve 4 is the inverse of \( {}^2x \) (2nd superroot)

\(
lim_{n\rightarrow \infty}{^{1/n}}e=y\Leftrightarrow\\
e=lim_{n\rightarrow \infty}{^n} y\Leftrightarrow\\
y=e^{1/e}
\)

Sorry Ioannis, but this rather proves that it is the wrong definition. As \( {}^xe \) should be a function continuous in \( x \) it must
\( \lim_{n\to\infty}{}^{1/n}e={}^0e=1\neq e^{1/e} \)