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)

                    (x_N+1, y_N+1)      =        ( 2(x_N<+>y_N) , (x_N+y_N)/2 )

          The recursion preserves the product of the components;

          therefore, by recurrence, y_N  = ab/x_N

          and therefore x_N+1  =  (x_N + ab/x_N)/2 =  (x_N² + ab)/(2x_N)

          - which is the Newton’s-method formula for the square root.

          If ab>0 then let z_N  = x_N/root(ab) ; this implies:

                    z_N+1 =  (z_N² + 1)/(2z_N)

          Compare that to the hyperbolic-cotangent identity:

                    coth(2A) =  (coth(A)² + 1)/(2coth(A))

          By recurrence we derive, for some A₀:

                    z_N+1 = coth(A₀*2^N)

          So therefore

                    x_N+1 = root(ab) coth(A₀*2^N)

          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 z_N  = x_N/(root(-ab)) ; this implies:

                    z_N+1 =  (z_N² - 1)/(2z_N)

          Compare that to the cotangent identity:

                    cot(2A) =  (cot(A)² - 1)/(2cot(A))

          By recurrence we derive, for some A₀:

                    z_N+1 = cot(A₀*2^N)

          So therefore

                    x_N+1 = (root(-ab)) cot(A₀*2^N)

          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)

          Therefore:

                    AM * mediant * HM    =    xyz

          Now consider this recursion:

                    (x₀,y₀,z₀)                =   ( a, b, c )

                    (x_N+1, y_N+1, z_N+1)   =  
                      ( (x_N+y_N+z_N)/3, x_N@y_N@z_N, 3(x_N<+>y_N<+>z_N) )

          This preserves product, and therefore by recurrence:

                    x_Ny_Nz_N = abc

          Conjectures:

          The number of positive entries in the triple (x_N, y_N, z_N) 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?