On Logistic, 6 of 8: Logistic Fixedpoints

          6. Logistic Fixedpoints

          Let F denote the generic finite positive quantity. The triple 0,F,∞ are then logistic ‘seen at low-resolution’,  for all finite positives are confused. In this low-rez view, here are tables for addition, reduction, Einstein addition and reciprocal:

+     0  F  ∞     {+}  0  F  ∞     ~   0  F  ∞   1/x

0     0  F  ∞      0   0  0  0     0   0  F  ∞    ∞  

F     F  F  ∞      F   0  F  F     F   F  F  F    F  

∞     ∞  ∞  ∞      ∞   0  F  ∞     ∞   ∞  F  0    0

These are isomorphic to ‘and’, ‘or’, ‘iff’ and ‘not’, in three-valued Kleenean logic, if we identify 0 with True, Finite with Intermediate, and ∞ with False. So logistic, at low resolution, looks Kleenean.

Kleenean logic has a fixedpoint property:

The following self-referential system, if Kleenean, has a solution:

x₁ =   F₁(x₁, x₂, … x_n,…)

x₂ =   F₂(x₁, x₂, … x_n,…)

…

x_n =   F_n(x₁, x₂, … x_n,…)

…

          We can find such a fixedpoint by iteration. Start by setting all the x’s to the intermediate truth value I; evaluate the functions; substitute these values back into the x’s; repeat. The iteration will converge to a fixedpoint in finitely many steps if the system is finite. If the system is infinite, then it converges after sufficiently large-ordinal transfinite iteration.

          Self-referential Kleenean systems have solutions; logistic is Kleenean at low resolution; therefore logistic self-refers at low-rez. Does it self-refer at high-rez?

          Define a logistic function as “positive” if it is defined from variables, constants, addition and reduction only; no reciprocals. Then you can prove by induction that positive logistic functions preserve linear order; and this implies that any self-referential system of positive logistic functions has a fixedpoint. Just iterate from (0,0,0,…); each component must increase with every step; this converges to a fixedpoint in the closed interval [0, ∞]. Therefore positive logistic systems of any dimension can self-refer. This implies, among other things, infinite sums and reductions.

          Full logistic systems, which include reciprocal as well as the additions, do not preserve linear order; nonetheless they can still self-refer. The interval [0,∞] is compact and convex; so are finite powers of it; so by Brouwer’s theorem, any finite-dimensional logistic system can self-refer. If the system is infinite-dimensional, then a fixedpoint exists by Tychonoff’s theorem.

          Here are some logistic systems:

          x        =        1/x

          “This sentence is false.”

          It has a solution: x = 1. 

          x        =        x+1

          “This sentence and the Liar are true.”

          Solution: x = ∞. Infinite, or false.

          x        =        x{+}1

          “This sentence or the Liar are true.”

          Solution: x = 0.  Void, or true.

          x        =        1  +  1/x

          “The Liar is true and this sentence is false.”

          Solution: x = the golden mean, about 1.6108;  therefore falser than the Liar.

          x        =        1  {+}  1/x

          “The Liar is true or this sentence is false.”

          “This sentence implies the Liar.”

          Solution: x = 1/phi, about 0.6108;  therefore truer than the Liar.

          x        =        4(x {+} 1/x)

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

          “This sentence is true or false, and this sentence is true or false, and this sentence is true or false, and this sentence is true or false.”

          Solutions: x = zero, or root 3 (about 1.732).

                     

          x        =        (x + 1/x)/4

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

“This sentence is true and false, or this sentence is true and false, or this sentence is true and false, or this sentence is true and false.”

          Solutions: x = infinity, or root 1/3 (about 0.577).

x        =        (x + 2/x) / 2 

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

          “This sentence is true, and it is false, and it is false,

                                        or

          this sentence is true, and it is false, and it is false.”

          Solutions: x = infinity, or root 2 (about 1.414).

          x        =        2(x {+} 1/(2x))      

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

          “This sentence is true, or it is false, or it is false,

                                        and

          this sentence is true, or it is false, or it is false.”

          Solutions: x = zero, or root ½ (about 0.7071).

          x        =        1/x + (x~x)

=        1/x + (x{+}1/x) + (x{+}1/x)

          “This sentence is false, and it is equivalent to itself.”

          “This sentence is false, and it is true or false, and it is true or false.”

          Solution: same as the solution to x⁴-2x²-1=0:

                    x = sqrt( 1 + sqrt(2) )

About 1.553773974

          x        =        1 + (x~x)     

=        1 + (x{+}1/x) + (x{+}1/x)

          “The Liar is true, and this sentence is equivalent to itself.”

          “The Liar is true, and this sentence is true or false, and it is true or false.”

          Solution: same as the solution to x³-x²-x-1=0:

                    x = ( cuberoot(19+sqrt(297)) + cuberoot(19-sqrt(297)) + 1 ) / 3

About 1.839286755

x        =        1/x  {+} (x + 1/x)

“This sentence is false, or it is true and false.”

“If this sentence is true, then it is true and false.”

Solution: same as the solution to x⁴+x²-1=0:

                    x = sqrt(1/phi),   where phi is the golden mean

x is about 0.7861513778

Here is a two-component fixedpoint:

          A  =  1/A {+} 1/B
                    B  =  1/B {+}  A
          Tweedledee: “If I’m not mistaken, then Tweedledum is mistaken.”
          Tweedledum: “If I’m not mistaken, then Tweedledee is not mistaken.”
                    Iteration solution:  (0.7653668647... , 0.5411961001...)
          This matches the algebraic solution:
                    A = sqrt( 2 - sqrt(2) )
                    B = sqrt( 1 - 1 / sqrt(2) )