Tuesday, January 14, 2014

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)
                    (xN+1, yN+1)  =        ( 2(xN{+}yN) , (xN+yN)/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, yN  = ab/xN
          Therefore xN+1  =  (xN + ab/xN)/2 =  (xN2 + ab)/(2xN)
          This is the Newton’s-method recursion 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)
          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.

          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 )
            (xN+1, yN+1, zN+1)   =  ( (xN+yN+zN)/3, M(xN,yN,zN), 3(xN{+}yN{+}zN) )

          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:
                    xNyNzN = abc

          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)
                    (xN+1, yN+1)          =        ( (xN+yN)/2  ,  root(xN*yN) )

          Or in other terms:
               (x0, y0)               =     (a, b)
               (xN+1, yN+1)      =     ( (xN+yN){+}(xN+yN)  ,  M(xN, yN, yN+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 )
          (xN+1, yN+1, zN+1) =  ( (xN+yN+zN)/3, cuberoot(xNyNzN), 3(xN{+}yN{+}zN) )

          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

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