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  New zeration and matrix log ? tommy1729 Ultimate Fellow     Posts: 1,419 Threads: 345 Joined: Feb 2009 03/24/2015, 12:17 AM (This post was last modified: 03/24/2015, 12:20 AM by tommy1729.) 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 marraco Fellow   Posts: 100 Threads: 12 Joined: Apr 2011 03/24/2015, 07:07 AM (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) « Next Oldest | Next Newest »

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