**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:

(x

_{0}, y_{0}) = (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_{N}^{2}+ 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_{N}^{2}+ 1)/(2z_{N})
Compare that to the
hyperbolic-cotangent 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})
So 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.
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 cotangent
identity:

cot(2A) = (cot(A)

^{2}- 1)/(2cot(A))
By recurrence we derive, for some A

_{0}:
z

_{N+1}= cot(A_{0}*2^{N})
So therefore

x

_{N+1}= (root(-ab)) cot(A_{0}*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 )

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:

(x

_{0},y_{0},z_{0}) = ( a, b, c )
(x

( (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

_{N}y_{N}z_{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?

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