## Wednesday, January 15, 2014

### On Logistic, 8 of 8; Beyond Logistic

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))              =        (  u2 – v2   ) / (2v)
((u{-}v) - (u{+}v))                    =        ( 2u2  {-} 2v2 ) /  v

Therefore:
((a+b-c) {-} (a+b+c))       =        ( a2 + 2ab + b2 – c2 ) / (2c)
((a-b-c) {-} (a-b+c))        =        ( a2 - 2ab + b2 – c2 ) / (2c)
((a{+}b{-}c) - (a{+}b{+}c))      =        ( 2a2 {+} ab {+} 2b2 {-} 2c2 ) / c
((a{-}b{-}c) - (a{-}b{+}c))       =        ( 2a2 {-} ab {+} 2b2 {-} 2c2 ) / 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))
a2       =    2((a-1){-}(a+1)) + 1   =   ((a{-}2)-(a{+}2)) {+} 4
a2       =    ½((a{-}1)-(a{+}1)) {+} 1    =   ((a-½){-} (a+½)) + ¼

There are other formulas. For instance:
- a2/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 x2 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:
(x0, y0)                   =        (a, b)
(xN+1, yN+1)             =        ( 2(xN{+}yN) , (xN+yN)/2 )

If ab<0 then let zN  = xN/(root(-ab)) ; this implies:
zN+1 =  (zN2 - 1)/(2zN)
Compare that to the cotangent identity:
cot(2A) =  (cot(A)2 - 1)/(2cot(A))
By recurrence we derive, for some A0:
zN+1 = cot(A0*2N)
So therefore
xN+1 = (root(-ab)) cot(A0*2N)

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:
(x0,y0,z0)                =   ( a, b, c )
(xN+1, yN+1, zN+1)   =  ( (xN+yN+zN)/3, M(xN,yN,zN), 3(xN{+}yN{+}zN) )
This preserves product, and therefore by recurrence:
xNyNzN = abc

Conjectures:
The number of positive entries in the triple (xN, yN, zN) 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?