# Tetration Forum

Full Version:  New zeration and matrix log ?
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Ive been thinking about zeration lately.

ab = max(a,b) + 1.

This has some nice properties , however you cannot invert it.

For example , 24 = 4+1 = 5.

But 5? = 2 does not seem to have a solution.
Also inverting seems troublesome since 24 = 34.

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 22 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
222... = 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

{2}^[x + C_1](C_2) = 2 + x.

So we try to find the function T = T(z) = {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^(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

(03/24/2015, 12:17 AM)tommy1729 Wrote: [ -> ]ab = 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)