## 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

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)
(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.