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between addition and multiplication
#21
seems my quote was not relevant.

i dont know what you mean by x {y} z then.

with respect , but i find it confusingly explained.

first it was a mean , now its a kind of tetration ?

it wasnt the limiting mean i quoted , so what is it ?

perhaps define

x {y} z = ... with real math symbols instead of pseudocode.

i cannot discuss things i do not understand.
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#22
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.

but that has troubles : the definition is piecewise and the iterations are not uniquely defined for fractional iterations.

and x {y} z is not analytic.
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#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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#24
Hey James, unfortunately my functions don't meet your requirement :-(

x {1.5} 2 != x {0.5} x, here are some random examples for x = 2,3,4,11,134

> ./sesqui 2 1.5 2
4.000000000
> ./sesqui 2 0.5 2
4.000000000
> ./sesqui 3 1.5 2
7.310586452
> ./sesqui 3 0.5 3
7.424041271
> ./sesqui 4 1.5 2
11.173642159
> ./sesqui 4 0.5 4
11.654329300
> ./sesqui 11 1.5 2
49.238006592
> ./sesqui 11 0.5 11
61.141593933
> ./sesqui 134 1.5 2
1893.553344727
> ./sesqui 134 0.5 134
5044.535156250

Though I admit I don't totally see why your identity is required, it seems to be by analogy with the gamma extension to the factorial. Could you explain it in simple terms? Thanks!
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#25
(03/16/2011, 02:35 AM)lloyd Wrote: Hey James, unfortunately my functions don't meet your requirement :-(

x {1.5} 2 != x {0.5} x, here are some random examples for x = 2,3,4,11,134

> ./sesqui 2 1.5 2
4.000000000
> ./sesqui 2 0.5 2
4.000000000
> ./sesqui 3 1.5 2
7.310586452
> ./sesqui 3 0.5 3
7.424041271
> ./sesqui 4 1.5 2
11.173642159
> ./sesqui 4 0.5 4
11.654329300
> ./sesqui 11 1.5 2
49.238006592
> ./sesqui 11 0.5 11
61.141593933
> ./sesqui 134 1.5 2
1893.553344727
> ./sesqui 134 0.5 134
5044.535156250

Though I admit I don't totally see why your identity is required, it seems to be by analogy with the gamma extension to the factorial. Could you explain it in simple terms? Thanks!

well the identity isn't required per se, it's just aesthetically appealing for an operator to have an identity. The only reason we would need an identity is if we wanted to use semi-operators to solve for tetration.

under the following expansion:

x {q} y = -q:log(q:log(x) + q:log(y))
x {q} S(q) = x = -q:log(q:log(x) + q:log(S(q))),
therefore q:log(S(q))= 0

therefore S(q) = -q:log(0) = sexp(q-1)

but since there is no identity for the sesqui operator all this is null.
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#26
Sorry, didn't mean the identity function as in 1. I meant the identity (as in equality) that you said was necessary for the system, that x {1.5} 2 had to equal x {0.5} x.
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#27
x {1.5} 2 = x {0.5} x

This is the general requirement that rational operators be recursive.

Consider,
x {0} y = x + y
x {1} y = x * y
x {2} y = x ^ y
x {3} y = x ^^ y

x * 2 = x + x
x {1} 2 = x {0} x

x ^ 2 = x * x
x {2} 2 = x {1} x

x ^^ 2 = x ^ x
x {3} 2 = x {2} x

etc etc...
It only be natural that this law holds for rational operators.
Generally, if {r} is any operator, than {r+1} is the superfunction of {r}.

the law stated mathematically is:

(x {r+1} (n-1)) {r} x = x {r+1} n
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#28
(03/16/2011, 05:47 PM)JmsNxn Wrote: x * 2 = x + x
x {1} 2 = x {0} x

x ^ 2 = x * x
x {2} 2 = x {1} x

x ^^ 2 = x ^ x
x {3} 2 = x {2} x

Thanks, good explanation.
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