# Tetration Forum

Full Version: Additional super exponential condition
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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.
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
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 $\text{mean}_4(a,b)$?
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.
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.