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Another new person! :)
#3
(07/18/2019, 04:05 PM)bo198214 Wrote: Hi Syzithryx, welcome on the forum. For me zeration is the increment. But yes I remember fiery discussions on this board with Gianfranco, there are surely other opinions about zeration. About the topic of fractional rank of hyperoperations I don't remember much said here. Do you have an approach to that?

My particular view on zeration is that a ∘ b = max(a,b)+1, the variant which Cesco Reale used - I just finished his paper and he came to many of the same conclusions as I have. His way of producing the inverse operation by extending the reals with "stigma-reals" is slightly different than what I would have done though, but it may be better! I was thinking that given a ∘ b = c < (b+1), the resulting a must then be greater than b but have a successor less than b's successor, so my version of "stigma numbers" would have had ςx>y for all real x,y - but Reale decided instead to have ςx>y iff x<y - his method however may actually be more "symmetrical" than mine; not sure.

Either way, though, I think that it makes more sense to define zeration as a truly binary operation relying on both its arguments, rather than unary or just the successor of the right argument. This definition actually also obeys the Mother Law, properly generalized - here are two equivalent versions, one basing the definition of an operation on the next lower operation, one on the next higher:

a [n] b = a [n-1] (a [n] (b - 1)) OR a [n] b = (a [n] (b - 1)) [n-1] a
a [n] b = a [n+1] ((b [n+1]\ a) + 1) OR a [n] b = b [n+1] ((a [n+1]\ b) + 1)

Notice that OR. This statement is true for all n≥1 - in particular it doesn't matter which of the two possibilities you choose for n=1 or 2 because addition and multiplication are commutative - it DOES matter which for 0 or 3+, but zeration can still be made commutative by having zerate(a,b) = max(a,b)+1.

Though note, there are other operations that also obey this law, such as min(a,b)+1, or even {floor(a+b) is odd: a+1; floor(a+b) is even: b+1}! So, the choice of exactly which pseudo-zeration is the "correct" one still depends on adding other identities it must obey, and I think given all the things Reale proved in his paper, it seems clear max(a,b)+1 is the most useful. (Note also btw that my new version of the Mother Law, which is more permissive, *still* doesn't accept Rubtsov and Romerio's variant!)

As for intermediate / fractional ops, I have thought of various approaches. My initial concept relied on using averages. For instance, with 1.5ation, the arithmetic average of (a+b) and (ab) obviously in some sense "favors" addition, because it's summing them before dividing by two; but the geometric average "favors" multiplication because it multiplies them before taking the square root; so properly we want to go right between those two averages.

So, define two functions: f(a,b) = (a+b)/2, and g(a,b) = ²√(ab). Then make an iterative sequence: c₀ = a+b, d₀ = ab; c_i = f(c_(i-1), d_(i-1)); d_i = g(c_(i-1), d_(i-1)). Then c_i and d_i should get closer together with each iteration and converge on a single value, then defined to be the 1.5ation of a and b; but unfortunately I don't know enough math to actually *prove* that or find any kind of closed expression for what that actually is.

Another way of doing 1.5ation, which probably wouldn't end up with the same values (but of course, who knows), is something I came up with yesterday looking at graphs of y=x+a versus y=xa - they are both lines, which meet at some point except when a=1; so, one could define a new line which bisects the angle they form, or which is exactly intermediate between them in the a=1 case, and that would be 1.5ation. I haven't really had a chance to analyze the properties of this variant yet though - and I really don't know how to extend this method to other pairs of ops.
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Messages In This Thread
Another new person! :) - by Syzithryx - 07/18/2019, 02:09 PM
RE: Another new person! :) - by bo198214 - 07/18/2019, 04:05 PM
RE: Another new person! :) - by Syzithryx - 07/18/2019, 05:13 PM



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