Sorry, I have no clue what you linked me to? I don't think you understand what I am getting at. I see the similarity but I see nothing outright stating that these are operators inbetween addition multiplication and exponentiation.
And as I said, it's simply a hypotheses that S(q) = q, actually, it's less so than a hypothesis.
Consider: 0 <= q <= 1
m {q} S(q) = m
take the rational iterated log q times, and we get
q:log(m) + q:log(S(q)) = q:log(m)
therefore q:log(S(q)) = 0
And everywhere I've checked,
q:log(q) = 0
Absolutely S(g) = 1, if g >= 1
The proof for operators less than {2} is so incredibly simple:
m {1 + q} S(1+q) = m
Take the rational iterated log q times, and we get:
q:log(m) * S(1+q) = q:log(m)
Therefore S(1+q) = 1
It is a direct result of the axiom which states that logarithms are distributive over any operator less than or equal to {1}, and exclusive for operators greater than {1}. Exclusive in the sense that only the base is placed within the logarithm.
The function I stated was not an Ackerman clone, it's the Ackerman functioned defined for Real operator values over period [0, 2]. It's a generalization.
1 - 1/ln2 comes from a long and clunky proof, not really important enough to repeat it here.
To be honest, I never considered differentiating S(z). There is only a critical section [0,1] that is worth underlining, and its value is marked by
S(q) = b {3} (q -1)
And so therefore if tetration is not linear over this domain S(q) is dependent upon a logarithm base.
And we should never consider the quantity q > 1, because operators follow recursion and and adding 1 to q is like stepping one step up on the recursion ladder. q should always be the rational part of a number.
I thank you for reading this over and I didn't mean to step on your toes if you already had this idea. I've been trying to get at the heart of what rational operators are for two years now :/
And as I said, it's simply a hypotheses that S(q) = q, actually, it's less so than a hypothesis.
Consider: 0 <= q <= 1
m {q} S(q) = m
take the rational iterated log q times, and we get
q:log(m) + q:log(S(q)) = q:log(m)
therefore q:log(S(q)) = 0
And everywhere I've checked,
q:log(q) = 0
Absolutely S(g) = 1, if g >= 1
The proof for operators less than {2} is so incredibly simple:
m {1 + q} S(1+q) = m
Take the rational iterated log q times, and we get:
q:log(m) * S(1+q) = q:log(m)
Therefore S(1+q) = 1
It is a direct result of the axiom which states that logarithms are distributive over any operator less than or equal to {1}, and exclusive for operators greater than {1}. Exclusive in the sense that only the base is placed within the logarithm.
The function I stated was not an Ackerman clone, it's the Ackerman functioned defined for Real operator values over period [0, 2]. It's a generalization.
1 - 1/ln2 comes from a long and clunky proof, not really important enough to repeat it here.
To be honest, I never considered differentiating S(z). There is only a critical section [0,1] that is worth underlining, and its value is marked by
S(q) = b {3} (q -1)
And so therefore if tetration is not linear over this domain S(q) is dependent upon a logarithm base.
And we should never consider the quantity q > 1, because operators follow recursion and and adding 1 to q is like stepping one step up on the recursion ladder. q should always be the rational part of a number.
I thank you for reading this over and I didn't mean to step on your toes if you already had this idea. I've been trying to get at the heart of what rational operators are for two years now :/