**8. Beyond Logistic**

Multiplication
cannot be defined from logistic operators, because those always have definite
values on zero and infinity, but zero times infinity equals the indefinite
ratio: 0 * 1/0 = 0/0.

Including
negative gives us access to the indefinite, for infinity minus infinity is
indefinite; 1/0 – 1/0 = 0/0.
Negative defines subtraction and “protraction”:

x –
y =
x + -y ; x {-} y = x
{+} -y

These
plus additions define multiplication,
squaring and inverse. For instance:

((u-v) {-} (u+v))
= ( u

^{2}– v^{2 }) / (2v)
((u{-}v)
- (u{+}v)) = (
2u

^{2 }{-} 2v^{2 }) / v
Therefore:

((a+b-c) {-} (a+b+c))
= (
a

^{2 }+ 2ab + b^{2}– c^{2 }) / (2c)
((a-b-c) {-} (a-b+c))
= (
a

^{2 }- 2ab + b^{2}– c^{2 }) / (2c)
((a{+}b{-}c) - (a{+}b{+}c)) = (
2a

^{2 }{+} ab {+} 2b^{2}{-} 2c^{2 }) / c
((a{-}b{-}c)
- (a{-}b{+}c)) = (
2a

^{2 }{-} ab {+} 2b^{2}{-} 2c^{2 }) / c
Therefore:

2ab/c = ((a+b-c){-}(a+b+c)) + ((a-b+c){-}(a-b-c))

ab/2c = ((a{+}b{-}c)-(a{+}b{+}c))
{+} ((a{-}b{+}c)-(a{-}b{-}c))

ab = ((a+b-2){-}(a+b+2)) +
((a-b+2){-}(a-b-2))

ab =
((a{+}b{-}½)-(a{+}b{+}½))
{+} ((a{-}b{+}½)-(a{-}b{-}½))

a/c = ((a+½-c){-}(a+½+c)) + ((a-½+c){-}(a-½-c))

a/c = ((a{+}2{-}c)-(a{+}2{+}c))
{+} ((a{-}2{+}c)-(a{-}2{-}c))

a

^{2}= 2((a-1){-}(a+1)) + 1 = ((a{-}2)-(a{+}2)) {+} 4
a

^{2}= ½((a{-}1)-(a{+}1)) {+} 1 = ((a-½){-} (a+½)) + ¼
There
are other formulas. For instance:

- a

^{2}/x = ((x-a){+}a)- a
= ((x+a){-}a)+ a

= ((x{-}a)+a){-}a

= ((x{+}a)-a){+} a

1/x = (((-x)-1){+}1)-1

= (((-x)+1){-}1)+1

= (((-x){-}1)+1){-}1

= (((-x){+}1)-1){+}
1

I
speculate that any formula defining x*y, x/y or x

^{2}from the additions must have at least one minus, at least one constant, and be at least three levels deep. Also, a formula defining x*y/z can lack constants, but it must have a minus, and be at least four levels deep.
Including
negative gets us multiplication, but then we lose logic, for –x is a quantity
that sums with x to zero:

x
+ -x = 0

In
logistic this would mean that the expression

X
and “minus X”

equals True for

*any*X!
We also lose stability:
iteration can be chaotic. Recall the iteration:

(x

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

_{N+1}, y_{N+1}) = ( 2(x_{N}{+}y_{N}) , (x_{N}+y_{N})/2 )
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
recall the 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}) )
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 recursion converges rapidly
to ( cuberoot(abc), cuberoot(abc), cuberoot(abc) ).

If a,
b and c do not all have the same sign, then the 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 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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