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 On the existence of rational operators JmsNxn Long Time Fellow    Posts: 291 Threads: 67 Joined: Dec 2010 12/20/2010, 02:16 AM (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}. « Next Oldest | Next Newest »

 Messages In This Thread On the existence of rational operators - by JmsNxn - 12/14/2010, 01:41 AM RE: On the existence of rational operators - by bo198214 - 12/19/2010, 05:23 AM RE: On the existence of rational operators - by tommy1729 - 12/19/2010, 05:21 PM RE: On the existence of rational operators - by JmsNxn - 12/19/2010, 06:47 PM RE: On the existence of rational operators - by sheldonison - 12/19/2010, 08:23 PM RE: On the existence of rational operators - by JmsNxn - 12/20/2010, 02:16 AM RE: On the existence of rational operators - by sheldonison - 12/20/2010, 04:53 AM RE: On the existence of rational operators - by JmsNxn - 12/20/2010, 07:01 PM RE: On the existence of rational operators - by sheldonison - 12/20/2010, 08:28 PM RE: On the existence of rational operators - by JmsNxn - 12/20/2010, 09:38 PM RE: On the existence of rational operators - by bo198214 - 12/20/2010, 09:17 AM RE: On the existence of rational operators - by JmsNxn - 03/11/2011, 07:23 PM RE: On the existence of rational operators - by JmsNxn - 03/19/2011, 05:36 PM

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