10/13/2008, 07:15 PM
I was just thinking about the following for an arbitrary super exponential \( \text{sexp} \):
We surely have for natural numbers m and n that
\( \text{sexp}(n+m)\ge \text{sexp}(n) ^ {\text{sexp}(m)} \)
So why not demand this rule also for the super exponential extended to the reals?
For a super logarithm the rule would be:
\( \text{slog}(x^y) \le \text{slog}(x) + \text{slog}(y) \)
Note that this rule is not applicable to the left-bracketed super exponentials.
Because from the rule it follows already that:
\( \text{sexp}(n)\ge \exp^{\circ n}(1) \) which is not valid for left bracketed super exponentials because they grow more slowly.
I didnt verify the rule yet for our known tetration extensions. Do you think it will be valid?
However I dont think that this condition suffice as a uniqueness criterion. But at least it would reduce the set of valid candidates.
We surely have for natural numbers m and n that
\( \text{sexp}(n+m)\ge \text{sexp}(n) ^ {\text{sexp}(m)} \)
So why not demand this rule also for the super exponential extended to the reals?
For a super logarithm the rule would be:
\( \text{slog}(x^y) \le \text{slog}(x) + \text{slog}(y) \)
Note that this rule is not applicable to the left-bracketed super exponentials.
Because from the rule it follows already that:
\( \text{sexp}(n)\ge \exp^{\circ n}(1) \) which is not valid for left bracketed super exponentials because they grow more slowly.
I didnt verify the rule yet for our known tetration extensions. Do you think it will be valid?
However I dont think that this condition suffice as a uniqueness criterion. But at least it would reduce the set of valid candidates.
