Friday, November 29, 2013

Roots and Chaos by Means

Roots and Chaos by Means

          First define ‘harmonic addition’ <+> this way:
          x <+> y   = 1 / ( 1/x + 1/y )   =    (xy) / (x+y)

          Therefore (x+y)(x<+>y)   =   xy.

          Now consider the arithmetic and harmonic means:

          AM    =    (x+y)/2         =     (x+y) <+> (x+y)    =   (x<+>x)+(y<+>y)
          HM    =    2(x <+> y)   =     (x<+>y) + (x<+>y)  =    (x+x)<+>(y+y)

          Recall that (x+y)(x <+> y) = xy:
          this implies that AM*HM = xy.

          Now consider this iteration:
                    (x0, y0)               =        (a, b)
                    (xN+1, yN+1)      =        ( 2(xN<+>yN) , (xN+yN)/2 )

          The recursion preserves the product of the components;
          therefore, by recurrence, yN  = ab/xN
          and therefore xN+1  =  (xN + ab/xN)/2 =  (xN2 + ab)/(2xN)
          - which is the Newton’s-method formula for the square root.

          If ab>0 then let zN  = xN/root(ab) ; this implies:
                    zN+1 =  (zN2 + 1)/(2zN)
          Compare that to the hyperbolic-cotangent identity:
                    coth(2A) =  (coth(A)2 + 1)/(2coth(A))
          By recurrence we derive, for some A0:
                    zN+1 = coth(A0*2N)
          So therefore
                    xN+1 = root(ab) coth(A0*2N)
          This converges to root(ab) if A0 > 0, which happens if a>0;  and it converges to -root(ab) if A0 < 0, which happens if a<0.

          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 for the cubic case.

          Consider the arithmetic and harmonic means for three numbers:
                    AM = (x+y+z)/3    ;     
                    HM = 3(x<+>y<+>z)  =  3 / ( 1/x + 1/y + 1/z )
          Define the mediant as:
                    X @ Y @ Z  =    X*Y*Z  /  ((X+Y+Z)*(X<+>Y<+>Z))
                                        =     (XY+YZ+ZX) / (X+Y+Z)

                    AM * mediant * HM    =    xyz

          Now consider this recursion:
                    (x0,y0,z0)                =   ( a, b, c )
                    (xN+1, yN+1, zN+1)   =  
                      ( (xN+yN+zN)/3, 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 above recursion converges rapidly to ( cuberoot(abc),  cuberoot(abc),  cuberoot(abc) ).

          If a, b and c do not all have the same sign, then the above 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 above  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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