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On the existence of rational operators
#6
(12/19/2010, 08:23 PM)sheldonison Wrote:
(12/14/2010, 01:41 AM)JmsNxn Wrote: And now if the critical strip of tetration is defined as:
-1 <= f <= 0

b {3} f = f + 1

S(q) = q

and therefore:
m {q} q = m
Further notes:
Consider the function
A(x) = m {x} n

Which is a generalization of the Ackerman function, extending it to domain real.
...
I'm a little slow at catching on to the jist of your post, but the Ackermann function is A(m=4,n)=2^^(n+3) - 3, or roughly base(2) tetration for m=4. So I assume you're trying to define an extension to the Ackermann function for real numbers, where A(x) = m {x} n, where "m" is the base, and x is a rational operator.
So,
A(x=2)= m (2) n = m*n.
A(x=3)= m (3) n = m^n.
A(x=4)= m (4) n = m^^n

Is this the basic idea, where we are extending it to allow for for real values of "x" as well? Then Henryk's request is to see a graph of
f(q) = 2 {q} 3.
so f(2)=2*3=6, f(3)=2^3=8, f(4)=2^^3=16 ..... Sounds interesting!

I don't think the linear approximation for the critical strip for [-1..0] for tetration is a good idea. There are many approaches to extending tetration to real numbers, that are analytic on the complex plane, and they all seem to agree with each other.
- Sheldon

Yes, everything you posted was completely right, except I like to begin our operation sequence with {0} as addition. My argument is because there are no well defined operators below {0} (besides successorship which is invalid with rational numbers). However, I do have a few "suggestions" on how to extend to negative and complex operators.

To be honest, I had only chosen the linear approximation model because it was on wikipedia, and I had the desperate urge to evaluate operators. It also makes the algebra simple. But now that I see the wavy lines I am a bit against it. But, if tetration is not linear over domain [0, 1] rational operators depend on a logarithm base for their identity--and it no longer has universality.

The essential axioms are as follows:
0<= q <= 1
q:log(m {1+q} n) = q:log(m) {1} n = q:log(m) * n
q:log(m {q} n) = q:log(m) {0} q:log(n) = q:log(m) + q:log(n)

They follow recursion, and therefore {q} can be thought of as multiplication, and {1+q} can be thought of as exponentiation. And therefore multiplication is to addition as exponentiation is to multiplication.
Therefore, rational tetration, which occurs over domain (2, 3] can be defined for natural numbers as recursive {1+q}.
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RE: On the existence of rational operators - by JmsNxn - 12/20/2010, 02:16 AM

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