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:
(x_{0}, y_{0}) = (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_{N}^{2} + ab)/(2x_{N})
This
is the Newton’smethod recursion for the square root.
If
ab>0 then let z_{N} = x_{N}/root(ab)
; this implies:
z_{N+1}
= (z_{N}^{2} + 1)/(2z_{N})
Compare
that to the hyperboliccotangent identity:
coth(2A)
= (coth(A)^{2} + 1)/(2coth(A))
By
recurrence we derive, for some A_{0}:
z_{N+1}
= coth(A_{0}*2^{N})
Therefore x_{N+1} = root(ab) coth(A_{0}*2^{N})
This
converges to root(ab) if A_{0} > 0, which happens if a>0; and it converges to root(ab) if A_{0}
< 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:
(x_{0},y_{0},z_{0})
= ( 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_{N}y_{N}z_{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 (TI83).
But support is not proof. How to prove this?
We
can combine arithmetic and geometric means:
(x_{0}, y_{0}) = (a, b)
(x_{N+1}, y_{N+1}) =
( (x_{N}+y_{N})/2 , root(x_{N}*y_{N})
)
Or in
other terms:
(x_{0}, y_{0}) = (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 selfreference within iteration; or iteration within iteration. The result
is the “arithmeticgeometric 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?
(x_{0},y_{0},z_{0})
= ( a, b, c )
(x_{N+1},
y_{N+1}, z_{N+1})
= ( (x_{N}+y_{N}+z_{N})/3,
cuberoot(x_{N}y_{N}z_{N}), 3(x_{N}{+}y_{N}{+}z_{N})
)
Call
the limit AGH: the “arithmeticgeometricharmonic 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, ab^{2}) = ab
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