**9. Period Two Logistic**

[Note to blog
readers: this section is 9 of 8 because I wrote it up right after blogging the
8 sections, inspired by them. For now, with this, I catch up to myself; but
open questions remain, such as logistic chaos.]

Consider this logistic system:

A = 1/A {+} B = B /
(AB+1)

B = 1/B {+} C = C / (BC+1)

C = 1/C {+} 1/A = 1 / (C+A)

Moe: “If I’m right, then Larry is right.”

Larry: “If I’m right, then Curly is right.”

B = 1/B {+} C = C / (BC+1)

C = 1/C {+} 1/A = 1 / (C+A)

Moe: “If I’m right, then Larry is right.”

Larry: “If I’m right, then Curly is right.”

Curly: “If I’m right, then Moe is wrong.”

If you iterate from (1,1,1), then you converge to a wobble between (0.5346378109, 0.4619568588, 0.9633840464) and

(0.3704606374, 0.6666824455, 0.6675470022)

but if you start from (2,2,2), then you converge to a wobble between

(0.786389671, 0.3140677107, 1.417025223) and

(0.2518627488, 0.9806118803, 0.4538409914)

and I conjecture that any initial conditions converge to a period 2 wobble. Of course conjecture is not proof. So do we get a pair of strange attractors in 3-space? Perhaps someone could do a computer graphic of this.

The second iterate of a period-2 function has period 1; a fixedpoint. So second-iterate the function:

A = 1/(1/A {+} B) {+} (1/B {+} C) = (A + 1/B) {+} 1/B {+} C

B = 1/(1/B {+} C) {+} (1/C {+} 1/A) = (B + 1/C) {+} 1/C {+} 1/A

C = 1/(1/C {+} 1/A) {+} 1/(1/A {+} B) = (C + A) {+} (A + 1/B)

To simplify, let b = 1/B, so:

If you iterate from (1,1,1), then you converge to a wobble between (0.5346378109, 0.4619568588, 0.9633840464) and

(0.3704606374, 0.6666824455, 0.6675470022)

but if you start from (2,2,2), then you converge to a wobble between

(0.786389671, 0.3140677107, 1.417025223) and

(0.2518627488, 0.9806118803, 0.4538409914)

and I conjecture that any initial conditions converge to a period 2 wobble. Of course conjecture is not proof. So do we get a pair of strange attractors in 3-space? Perhaps someone could do a computer graphic of this.

The second iterate of a period-2 function has period 1; a fixedpoint. So second-iterate the function:

A = 1/(1/A {+} B) {+} (1/B {+} C) = (A + 1/B) {+} 1/B {+} C

B = 1/(1/B {+} C) {+} (1/C {+} 1/A) = (B + 1/C) {+} 1/C {+} 1/A

C = 1/(1/C {+} 1/A) {+} 1/(1/A {+} B) = (C + A) {+} (A + 1/B)

To simplify, let b = 1/B, so:

b = 1 / ((B + 1/C) {+} 1/C {+}
1/A) = (1/B {+} C) + C + A

Then we get this system:

Then we get this system:

A
= (A + b) {+} (b {+} C)

b = (b {+} C) + (C + A)

C = (C + A) {+} (A + b)

b = (b {+} C) + (C + A)

C = (C + A) {+} (A + b)

Experiment on a hand calculator shows that this system does
converge to fixedpoints, from a wide range of initial values for A,b,C; but not
the same fixedpoint for different initial values. The range of all fixedpoints
might be a strange attractor. Would graphing it be feasible?

Note the sub-sums and sub-reduction (A+b), (C+A), (b{+}C). Call these D, E, F; then:

A = D {+} E

b = E + F

C = F {+} D

D = A + b

E = b {+} C

F = C + A

If you start with D = 1/A, E = 1/b, F = 1/C, then these relations continue with this system’s iteration. Strange to say, this system iterates to period two!

A = (A + b) {+} (b {+} C)

b = (b {+} C) + (C+A)

C = (C+A) {+} (A + b)

D = (D {+} E) + (E + F)

E = (E + F) {+} (F {+} D)

F = (F {+} D) + (D {+} E)

In the second iterate, AbC and DEF are separate and reciprocal systems, matching additions with reductions and vice versa; both iterating to fixedpoints in a strange attractor.

You could redo the entire argument above with + and {+} swapped, starting with the period-2 system:

A = 1/A + B

B = 1/B + C

C = 1/C + 1/A

Moe: “I’m wrong, and Larry is right.”

Larry: “I’m wrong, and Curly is right.”

Note the sub-sums and sub-reduction (A+b), (C+A), (b{+}C). Call these D, E, F; then:

A = D {+} E

b = E + F

C = F {+} D

D = A + b

E = b {+} C

F = C + A

If you start with D = 1/A, E = 1/b, F = 1/C, then these relations continue with this system’s iteration. Strange to say, this system iterates to period two!

*Its*second iterate isA = (A + b) {+} (b {+} C)

b = (b {+} C) + (C+A)

C = (C+A) {+} (A + b)

D = (D {+} E) + (E + F)

E = (E + F) {+} (F {+} D)

F = (F {+} D) + (D {+} E)

In the second iterate, AbC and DEF are separate and reciprocal systems, matching additions with reductions and vice versa; both iterating to fixedpoints in a strange attractor.

You could redo the entire argument above with + and {+} swapped, starting with the period-2 system:

A = 1/A + B

B = 1/B + C

C = 1/C + 1/A

Moe: “I’m wrong, and Larry is right.”

Larry: “I’m wrong, and Curly is right.”

Curly:
“I’m wrong, and Moe’s wrong.”

which reiterates to this fixedpoint system:

A = (A {+} b) + (b + C)

b = (b + C) {+} (C {+} A)

C = (C {+} A) + (A {+} b)

which expands to this period-2 system:

A = D + E

b = E {+} F

C = F + D

D = A {+} b

E = b + C

F = C {+} A

which is the previous such system, re-labeled. Much symmetry here.

I have been trying to make a chaotic logistic system, but have yet to get period higher than two. Maybe two is the highest period?

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