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+--- Thread: Additional super exponential condition (/showthread.php?tid=204)
Additional super exponential condition - bo198214 - 10/13/2008
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.
RE: Additional super exponential condition - andydude - 10/14/2008
bo198214 Wrote:For a super logarithm the rule would be:
\( \text{slog}(x^y) \le \text{slog}(x) + \text{slog}(y) \)
This is certainly consistent. For example:
\( \text{slog}(e^y) \le \text{slog}(e) + \text{slog}(y) \)
\( \text{slog}(y) + 1 \le \text{slog}(e) + \text{slog}(y) \)
\( 1 \le \text{slog}(e) \)
\( 1 \le 1 \)
which is true.
Andrew Robbins
Extension by mean values - bo198214 - 10/21/2008
Ansus Wrote:By the way, I had an idea to extend hyper-operator based on the sequence of mean values:And how? I.e. what is \( \text{mean}_4(a,b) \)?
RE: Additional super exponential condition - bo198214 - 10/21/2008
Ansus Wrote:\(
\text{mean}_n(a,b)=\text{hroot}_{n+1}(\text{hyper}_n(a,b),2)
\)
Ya of course, but what is \( \text{hyper}_4(a,b) \)? You said you have an idea how to extend it to real \( b \) via those means.
RE: Additional super exponential condition - martin - 10/21/2008
I doubt this is an option to extend the mean value operations.
In a given set of data (say a1, a2, ...), ordering is irrelevant for calculating a mean value. But a1^a2(^a3...an) is different from a2^a1(^a3...). Well, at least most of the time.