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Rational operators (a {t} b); a,b > e solved
#13
(06/06/2011, 07:47 PM)JmsNxn Wrote: Alright, testing the left hand right hand limit I get different values....

I'll refer to from now on.

So therefore:


and
This matches my results (exactly, actually). The 1st derivatives are close, but they don't exactly match, at the transition between the function that defines operators between addition...multiplication, as compared to the function that defines multiplication...exponentiation.
I don't know why it works as well as it does, for base=eta. For other bases, which will also give the same results for integers, the resulting graphs are pretty ugly.

(06/06/2011, 07:47 PM)JmsNxn 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.
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
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Messages In This Thread
RE: Rational operators (a {t} b); a,b > e solved - by sheldonison - 06/06/2011, 08:43 PM

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