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On the existence of rational operators
#5
(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
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Messages In This Thread
RE: On the existence of rational operators - by sheldonison - 12/19/2010, 08:23 PM

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