+- Tetration Forum (https://tetrationforum.org)
+-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1)
+--- Forum: Hyperoperations and Related Studies (https://tetrationforum.org/forumdisplay.php?fid=11)
+--- Thread: [2015] New zeration and matrix log ? (/showthread.php?tid=969)
[2015] New zeration and matrix log ? - tommy1729 - 03/24/2015
Ive been thinking about zeration lately.
a[0]b = max(a,b) + 1.
This has some nice properties , however you cannot invert it.
For example , 2[0]4 = 4+1 = 5.
But 5[0]? = 2 does not seem to have a solution.
Also inverting seems troublesome since 2[0]4 = 3[0]4.
I was able to find additional arguments/properties for max(a,b)+1.
But its still the same function.
Then there is ln(exp(a) + exp(b)).
However this does not belong to the family q^x , q x , q + x , ... where x is variable and q is fixed(base).
but rather to the family x^ln(y) = y^ln(x) , x y , x+ y , ... where both x,y are variables and everything is commutative.
( and then there is offcourse the meaningfull but boring opinion that zeration is ALSO addition ; the inverse super of x+1 is x+1 => argument )
What else could exist ?
I got inspired by myself when I was considering equations like
f^[A(x)] (B(x)) = C(x) in my early teen years.
To keep a long story short here is the logic :
Base 2 is "holy" here.
2^^2 = 4
2^2 = 4
2*2 = 4
2+2 = 4
However 2[0]2 is not necessarily 4.
This turned out to be a wasted effort to zeration.
so zeration is not x+1 and not x+2.
So we need a new way to look at things without going to the max(a,b)+1 and ln(exp(a)+exp(b)) solutions.
And that logic is this :
...
2^2^2^... = 2^^x
2*2*2*... = 2^x
2+2+2+... = 2 x
2[0]2[0]2[0]... = 2 + x
I use {} for function names. C_1 ,C_2 , ... are constants.
The trend is {[q]2}^[x + C_1](C_2) = 2[q+1]x + f(q)
where f(q) = 0 for integer q.
SO for zeration we get
{[0]2}^[x + C_1](C_2) = 2 + x.
So we try to find the function T = T(z) = {[0]2}(z).
T^[x + C_1](C_2) = 2 + x. [equation 1]
or
C_2 = T^[ - x - C_1] (2 + x) [equation 2]
However solving equation 2 seems like a mistake , solving equation 1 seems like the correct way ;
From equation 1 we get
C_3 = T^[1](C_2) = {2 + x}^[1/(x + C_1)]
Now let CARL_2 be the carleman matrix for 2 + x , and
Carl(") be the carleman matrix of ".
Then we get the matrix equation
Carl(C_3) = CARL_2 ^ [1/(x + C_1)]
Let EXP be the matrix exponential and LOG be the matrix ln of CARL_2.
Carl(C_3) = EXP( LOG / (x + C_1) ) or = EXP ( 1/(x + C_1) * LOG ).
If this equation holds in SOME WAY then we have solution to zeration.
But there may be issues with the matrix ideas.
Or others ?
I wonder what you guys think.
Gottfried and myself have investigated the matrix logarithm and similar problems ... as did others.
The matrix log is " semi-classical " as I like to call it.
It is classical as the inverse of EXP but if A^B = exp(ln(A)*ln(B)) or if A^B = exp(ln(B)*ln(A)) ... what is the log of a nilpotent ... connections to tetration and other controversial research ... makes it non-classical.
This might lead to a new zeration ?
Or maybe a variation of this idea will ?
regards
tommy1729
RE: [2015] New zeration and matrix log ? - marraco - 03/24/2015
(03/24/2015, 12:17 AM)tommy1729 Wrote: a[0]b = max(a,b) + 1.
Then there is ln(exp(a) + exp(b)).
Maybe there is more than one zeration the way that there is more than one tetration (right bracket and left bracket)