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 Additional super exponential condition bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 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. andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 10/14/2008, 08:11 PM 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 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 10/21/2008, 07:53 AM 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 $\text{mean}_4(a,b)$? bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 10/21/2008, 09:16 AM 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. martin Junior Fellow Posts: 14 Threads: 1 Joined: Jul 2008 10/21/2008, 03:40 PM 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. « Next Oldest | Next Newest »

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