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between addition and multiplication
#23
If the operators over {q}, 0<=q<=1, are defined with an identity given by S(q), then they can define tetration if we allow:
q:log(x) = exp^[-q](x)

(1) x {q} y = -q:log(q:log(x) + q:log(y))
(2) x {1+q} y = -q:log(q:log(x) * y)

you yourself discussed these operators, I was pleasantly surprised to see someone come to the same formula as me.

Therefore, when I was talking about solving for tetration I was talking about defining x {q} y numerically as this modified Gauss mean and then conversely also allowing the two laws of logarithmic semi-operators (1). This would imply evaluations of rational values for tetration since semi-operators depend on tetration given (1) and (2).

Sadly however, this pseudo Gauss mean yields no identity, or no universal value S(q) for all x E C such that:
x {q} S(q) = x

this reduces the logarithmic laws (1) & (2) as null since q:log(S(q)) = 0 is an essential identity, and there is no value S(q)
(03/16/2011, 12:18 AM)tommy1729 Wrote: perhaps x {y} z is the mean ONLY for 0 < y < 1

and x {y} z for y > 1 is simply the (y-floor(y)) th iteration of x {floor(y)} z.

by lloyd rational exponentiation, or {y} for 1<y<=2 is defined by another limiting sequence.
by me, it would be defined as the superfunction of {q} and would only be solvable for natural values at this point.
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Messages In This Thread
between addition and multiplication - by lloyd - 03/10/2011, 09:10 PM
RE: between addition and multiplication - by JmsNxn - 03/16/2011, 12:20 AM

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