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?
No comments:
Post a Comment