Wednesday, January 15, 2014

On Logistic, 8 of 8; Beyond Logistic

8. Beyond Logistic

          Multiplication cannot be defined from logistic operators, because those always have definite values on zero and infinity, but zero times infinity equals the indefinite ratio: 0 * 1/0 = 0/0.
          Including negative gives us access to the indefinite, for infinity minus infinity is indefinite; 1/0 – 1/0  =  0/0.  Negative defines subtraction and “protraction”:
          x – y  =  x + -y       ;         x {-} y   =  x {+} -y
          These plus additions define multiplication, squaring and inverse. For instance:

          ((u-v) {-} (u+v))              =        (  u2 – v2   ) / (2v)
          ((u{-}v) - (u{+}v))                    =        ( 2u2  {-} 2v2 ) /  v

          ((a+b-c) {-} (a+b+c))       =        ( a2 + 2ab + b2 – c2 ) / (2c)
          ((a-b-c) {-} (a-b+c))        =        ( a2 - 2ab + b2 – c2 ) / (2c)
          ((a{+}b{-}c) - (a{+}b{+}c))      =        ( 2a2 {+} ab {+} 2b2 {-} 2c2 ) / c
          ((a{-}b{-}c) - (a{-}b{+}c))       =        ( 2a2 {-} ab {+} 2b2 {-} 2c2 ) / c

          2ab/c            =    ((a+b-c){-}(a+b+c)) + ((a-b+c){-}(a-b-c))
          ab/2c            =    ((a{+}b{-}c)-(a{+}b{+}c)) {+} ((a{-}b{+}c)-(a{-}b{-}c))
          ab                =    ((a+b-2){-}(a+b+2)) + ((a-b+2){-}(a-b-2))  
          ab      =    ((a{+}b{-}½)-(a{+}b{+}½)) {+} ((a{-}b{+}½)-(a{-}b{-}½))  
          a/c      =    ((a+½-c){-}(a+½+c)) + ((a-½+c){-}(a-½-c))
          a/c      =    ((a{+}2{-}c)-(a{+}2{+}c)) {+} ((a{-}2{+}c)-(a{-}2{-}c))
          a2       =    2((a-1){-}(a+1)) + 1   =   ((a{-}2)-(a{+}2)) {+} 4
          a2       =    ½((a{-}1)-(a{+}1)) {+} 1    =   ((a-½){-} (a+½)) + ¼

          There are other formulas. For instance:
          - a2/x            =        ((x-a){+}a)- a
                              =        ((x+a){-}a)+ a
                              =        ((x{-}a)+a){-}a
                              =        ((x{+}a)-a){+} a

            1/x             =        (((-x)-1){+}1)-1    
=        (((-x)+1){-}1)+1 
=        (((-x){-}1)+1){-}1 
=        (((-x){+}1)-1){+} 1
          I speculate that any formula defining x*y, x/y or x2 from the additions must have at least one minus, at least one constant, and be at least three levels deep. Also, a formula defining x*y/z can lack constants, but it must have a minus, and be at least four levels deep.
          Including negative gets us multiplication, but then we lose logic, for –x is a quantity that sums with x to zero:
                    x + -x           =        0
          In logistic this would mean that the expression       
                    X and “minus X”  
equals True for any X!

We also lose stability: iteration can be chaotic. Recall the iteration:
                    (x0, y0)                   =        (a, b)
                    (xN+1, yN+1)             =        ( 2(xN{+}yN) , (xN+yN)/2 )

          If ab<0 then let zN  = xN/(root(-ab)) ; this implies:
                    zN+1 =  (zN2 - 1)/(2zN)
          Compare that to the cotangent identity:
                    cot(2A) =  (cot(A)2 - 1)/(2cot(A))
          By recurrence we derive, for some A0:
                    zN+1 = cot(A0*2N)
          So therefore
                    xN+1 = (root(-ab)) cot(A0*2N)

          This undergoes angle-doubling chaos. It is sensitive to initial conditions, and it has orbits that go arbitrarily close to any given countable set of values. It’s as if, when ab<0, the x’s are ‘trying to converge’ to root(ab), an imaginary number, even though the x’s are real.

          Experiments on a hand calculator (TI-83) show that if a and b have small imaginary parts, then the recursion quickly converges to one of the complex roots. The real line is a chaotic boundary between two basins of attraction.

          Now recall the recursion:
                    (x0,y0,z0)                =   ( a, b, c )
                    (xN+1, yN+1, zN+1)   =  ( (xN+yN+zN)/3, M(xN,yN,zN), 3(xN{+}yN{+}zN) )
          This preserves product, and therefore by recurrence:
                    xNyNzN = abc

          The number of positive entries in the triple (xN, yN, zN) is constant. That is, all-positive remains all-positive, two-positive-one-negative remains two-positive-one-negative, one-positive-two-negative remains one-positive-two-negative, and all-negative remains all-negative.
          If a, b and c are all positive or all negative, then the recursion converges rapidly to ( cuberoot(abc),  cuberoot(abc),  cuberoot(abc) ).
          If a, b and c do not all have the same sign, then the recursion is chaotic.
          If a, b and c do not all have the same sign, and to each we add a small imaginary part, and iterate from there, then the recursion converges rapidly to a complex cube root of the new abc.

          These claims are supported by numerical experimentation on a hand calculator (TI-83). But support is not proof. How to prove this?

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