Additional super exponential condition - Printable Version +- Tetration Forum ( https://math.eretrandre.org/tetrationforum)+-- Forum: Tetration and Related Topics ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)+--- Forum: Mathematical and General Discussion ( https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)+--- 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 : We surely have for natural numbers m and n that So why not demand this rule also for the super exponential extended to the reals? For a super logarithm the rule would be: Note that this rule is not applicable to the left-bracketed super exponentials. Because from the rule it follows already that: 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: This is certainly consistent. For example: 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 ? RE: Additional super exponential condition - bo198214 - 10/21/2008
Ansus Wrote: Ya of course, but what is ? You said you have an idea how to extend it to real 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. |