Comparing the Known Tetration Solutions

[bo198214]

No, the slog function is only defined for bases greater than 1. For bases between 0 and 1, the tetration function is not one-to-one, so its inverse is not a function, because there are multiple values. It's much like saying that y=x^2 does not have an inverse function. y=sqrt(x) only covers part of the domain of the original function.

Surely, by "arbitrary" I meant: also for \( b<\eta \).
I was referring to

and I suppose would you could call the elliptic? interval \( (e^{1/e}, \infty) \), which is what my definition of the super-logarithm is valid for

however in his pdf he defined it for bases \( >1 \).
(Quite ridiculous how misunderstandings reach its maximum in the communication between Jay and me ...)

Also, there's a question about whether you consider the slog function to only apply to the inverse of the function of iterated exponentials/logarithms from 1. For b=2, for example, the domain of slog is negative infinity to 2.

Dont understand this, \( t\mapsto \exp_2^{\circ t}(1.0) \) maps \( (-2,\infty) \) to \( (-\infty,\infty) \), so the slog is defined on \( (-\infty,\infty) \)?

However, you can perform iterated exponentials/logarithms from any real number as a starting point, so you could also include the graph for x>4 and the corridor between 2 and 4.

If you use the fixed point method however for base \( <\eta \) you have to specify which fixed point you use in case you start with \( x_0 \) between these fixed points.