On Logisitic, 7 of 8; Means by Logistic

7. Means by Logistic

         

          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)

          The arithmetic mean corresponds to these logistic expressions:

                    (x and y) or (x and y)

                    (x or x) and (y or y)

-        and the harmonic mean corresponds to these:

     (x or y) and (x or y)

     (x and x) or (y and y)

          Now recall this equation:

          x {+} y   =    (xy) / (x+y)

          Therefore: (x+y) (x{+}y)   =   xy;

                           ((x+y)/2)(2(x{+}y))   =   xy;

                            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 )

          This corresponds to the logistic system:

          Tweedledee: “I and my brother are true, or I and my brother are true.”

          Tweedledum: “I or my brother are true, and I or my brother are true.”

          The recursion preserves the product of the components, since AM*HM=xy.

          Therefore, by recurrence, y_N  = ab/x_N

          Therefore x_N+1  =  (x_N + ab/x_N)/2 =  (x_N² + ab)/(2x_N)

          This is the Newton’s-method recursion 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)

          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.

          That iteration depends upon initial conditions. Here's one that doesn't:

                    x = M(a,x,b)          

“Most of a, b and this is true.”

          It converges to root(a*b). When a = zero and b = infinity, then that root equals the indefinite ratio; fittingly, for M(0,x,∞) = x for any number x.

          Now for the cubic case. Consider these three means for three numbers; arithmetic mean, harmonic mean, and mediant:

                    AM = (x+y+z)/3  = (x+y+z){+}(x+y+z){+}(x+y+z)

HM = 3(x{+}y{+}z)  = (x{+}y{+}z)+(x{+}y{+}z)+(x{+}y{+}z)

                    M(x,y,z)       =        x*y*z  /  ((x+y+z)*(x{+}y{+}z))

          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, M(x_N,y_N,z_N), 3(x_N{+}y_N{+}z_N) )

          Alice: “I and Bob and Carol are true, or I and Bob and Carol are true, or I and Bob and Carol are true.”

          Bob: “Most of Alice, Carol and I are true.”

          Carol: “I or Alice or Bob is true, and I or Alice or Bob is true, and I or Alice or Bob is true.”

          This preserves product;   

AM * mediant * HM    =    xyz

          Therefore by recurrence:

                    x_Ny_Nz_N = abc

          Conjecture:

          If a, b and c are all positive, then the above recursion converges rapidly to (cuberoot(abc),  cuberoot(abc),  cuberoot(abc) ).

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

          We can combine arithmetic and geometric means:

                    (x₀, y₀)                   =        (a, b)

                    (x_N+1, y_N+1)          =        ( (x_N+y_N)/2  ,  root(x_N*y_N) )

          Or in other terms:

               (x₀, y₀)               =     (a, b)

               (x_N+1, y_N+1)      =     ( (x_N+y_N){+}(x_N+y_N)  ,  M(x_N, y_N, y_N+1) )

          This is self-reference within iteration; or iteration within iteration. The result is the “arithmetic-geometric mean”; by means of which we can compute elliptic integrals; therefore elliptic integrals are logistic.

          Exercise for the ambitious reader:

          What if we combined arithmetic, harmonic and geometric means?

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

          (x_N+1, y_N+1, z_N+1) =  ( (x_N+y_N+z_N)/3, cuberoot(x_Ny_Nz_N), 3(x_N{+}y_N{+}z_N) )

          Call the limit AGH: the “arithmetic-geometric-harmonic mean”. In general:

          K * AGH(x,y,z)      =        AGH(Kx, Ky, Kz)

          K / AGH(x,y,z)       =        AGH(K/x, K/y, K/z)

          AGH(a, ab, ab²)     =        ab