Friday, May 14, 2021

Logistic Fractal

 

Consider this iterative system:

                    A = 1/A + B

                    B = 1/B + C

                    C = 1/C + 1/A

 

If we use the logistic interpretation:  0  =  True,  Infinity = False,  + = 'and',  1/x = 'not', then this translates into this paradox system:

A   =   B and not A

B   =   C and not B

C   =   not A and not C

 

I have experimented with this system, and find that whatever initial values I input for ABC, it quickly converges to a period-2 oscillation. But not the _same_ oscillation for different initial values!

 

My question is: what does the set of period-2 wobbles look like in 3d space? Can you do this: Program a computer to take a large number of random points, iterate all to a wobble, plot all the resulting point pairs, and display it, as seen along the three axes, as gifs. Or perhaps some stereograms?

 

Iterated twice, the system is:

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

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

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

 

    where b  =  1/B, and {+} is the reciprocal addition operator 1/((1/x)+(1/y)). This translates by logistic into this:

A  =  (A or b) and b and C

b  =   (b and C) or C or A

C  =  (C or A) and (A or b)

 

This converges to fixedpoints, but again not the same fixedpoint for different initial values. And this too I would like to see some gifs of. Can you do this?

 

The reciprocal dual of the systems noted above are:

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

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

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

 

          A: “If I’m right, then B is right.”

          B: “If I’m right, then C is right.”

          C: “If I’m right, then A is wrong.”

 

          This too converges to a period 2 wobble, but different ones for different initial conditions. This too I'd like to see a gif of.

 

 

          Its second iterate is:

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

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

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

         - where b = 1/B. It translates to:

 

A  =  (A and b) or b or C

b  =  (b  or C) and C and A

C  =  (C and A) or (A and b)

 

 

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