On the existence of rational operators

[tommy1729]

for starters i would like to comment that i find this whole thread suspiciously like my own recent threads

http://math.eretrandre.org/tetrationforu...hp?tid=543

http://math.eretrandre.org/tetrationforu...hp?tid=520

and i dont see what the new benefit or new intention of this similar idea is.

also , i dont see why we need a new ackermann clone.

furthermore im not totally sure everything is correct.

also , i dont think we should use totally new terminology just like that.

0 <= q <= 1
q:log(m {1} n) = q:log(m) {1-q} q:log(n)
q:log(m {2} n) = q:log(m) {2-q} n

These operators would have the following property; if S(q) returns the identity of any operator:
m {q} S(q) = m
q:log(m) + q:log(S(q)) = q:log(m)
therefore
q:log(S(q)) = 0
which is true regardless of logarithm base.
m {1+q} S(1+q) = m
q:log(m) * S(1+q) = q:log(m)
therefore:
S(1+q) = 1
and which in general becomes all operators greater than or equal to one have identity one.



Rational operators are not distributive over addition, however, as multiplication is to exponentiation and as exponentiation is to multiplication {q} is to {1+q} as {1+q} is to {q}
therefore:
(m {q} n) {1 + q} l = (m {1+q} l) {q} (n {1+q} l)

(m {1+q} n) {q} m = m {1+q} (n+1)


m {1+q} 0 = S(q)
since
q:log(m) * 0 = q:log(S(q)) = 0

and therefore:
r:log(m) = b {3} (slog(m) - r)

Where slog(x) is the inverse function of tetration.

now, since:
-r:log(r:log(x)) = x
m {q} n = -q:log(q:log(m) + q:log(n))
m {q} n = b {3} (slog( (b {3} (slog(m) - q)) + (b {3} (slog(n) - q))) + q)

m {1+q} n = -q:log(q:log(m) * n)
m {1+q} n = b {3} (slog((b {3} (slog(m)-q)) * n) + q)


Now, further observing the identity function:
since:
q:log(S(q)) = 0

b {3} (slog(S(q)) -q) = 0

slog(S(q)) - q = -1

slog(S(q)) = q - 1

S(q) = b {3} (q-1)

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.

Interesting results are:
A(x) = 2 {x} 2 peaks at A(1 - 1/ln(2)) = 4.248828844

The fact that it's squiggly bewilders me and leaves me in awe.
I believe it has something to do with the extension of tetration I use...

im not sure that S(q) = q and S(1+q) = 1 .. in fact , isnt that a contradiction already combining those two ?

( we want S(z) to be differentiable not ? )

you started to define your identities for 0 =< q =< 1 , what makes me doubt if the proclaimed things for q > 1 must hold.

where does 1 - 1/ln(2) come from btw ??

yes , it has to do with the extension of tetration you used , your slog is defined piecewise and linear, hence no nice properties are satisfied and no new slog is discussed.

regards

tommy1729