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Rational operators (a {t} b); a,b > e solved
#15
(06/06/2011, 08:43 PM)sheldonison Wrote: If you could get a definition about a complex circle around h=1, at a,b=e, that might be a big start. This would be 1+q, 1-q, 1+qi, 1-qi, also matching both of your initial definitions (which I haven't checked). If that were the case, you already have analytic functions defied for 0<=h<=1, and analytic functions defined for 1<=h<=2. Then, for one case, a=b=e, you might have a function defined for 0<=h<=2. Then the key is to morph this function, perhaps starting with the case a=b, as a=b becomes less than e, and greater than e, in such a way that it remains analytic. Of course, there is the small issue that the inverse superfunctions of eta have singularities at z=e, and the issue of the upper/lower superfunctions of eta, so there are many many challenges on this path.

I don't understand what you mean about the first part, of making a definition about a complex circle, however I do understand what you mean by morphing the cheta function, and creating a generalization to .

(06/06/2011, 08:43 PM)sheldonison Wrote: By the way, I agree with Henryk, that exponentiation should be rational operator three, and multiplication, rational operator 2, and addition rational operator 1.
- Sheldon
Hmm, this is just like the Gamma function predicament. Well, hear my arguments for centering addition at 0. Firstly I wanted to center the Identity function so that S(0) = 0, and S(1) = 1.
Secondly, the function designed is naturally centered with addition as zero. Shifting it over by one is an offset.




Thirdly, the negative integers of operators were specifically designed to be negative because they aren't proper recursive. i.e:
, having 0 as the first non-proper recursive operator seems off. I feel it should be -1 (or technically -0.0000000001). Also, again you can see the series is centered with addition as 0.

I don't really have any other arguments, it just seems simpler to center it at 0. It's just unfortunate that someone decided to call it "tetration" instead of "tertiation".


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RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/07/2011, 02:45 AM

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