Fibonacci Sampling and the Coarse Time Quotient

Abstract

Finite Fibonacci samples determine exactly the coarse time observations; a nonzero sampling kernel prevents an autonomous one-step update.

Theorem 1.1 (Coarse time equivalence and the one-step obstruction).

$$(\forall x,\forall j,u\left(j+2,x\right)=L\cdot u\left(j+1,x\right)-(-1{)}^g\cdot u\left(j,x\right))\wedge (\forall z,((O\left(z\right)=0)⟺((r\left(s,z\right)=0)\wedge (r\left(s+g,z\right)=0)))\wedge ((O\left(z\right)=0)⟺((\mathrm{snd}\left(\mathrm{iterate}\left(S,s,z\right)\right)=0)\wedge (\mathrm{fib}\left(g\right)\cdot \mathrm{fst}\left(\mathrm{iterate}\left(S,s,z\right)\right)=0))))\wedge (\forall x,\forall y,(O\left(x\right)=O\left(y\right))⟺(\forall j,u\left(j,x\right)=u\left(j,y\right)))\wedge (\forall Phi,(\forall x,O\left(S\left(x\right)\right)=\mathrm{Phi}\left(O\left(x\right)\right))⟹\forall z,O\left(z\right)=0⟹O\left(S\left(z\right)\right)=0)\wedge (\mathrm{Injective}\left(x\mapsto (r\left(s,x\right),r\left(s+1,x\right))\right))\wedge (\det \left(\mathrm{adjacentRows}\left(s\right)\right)=(-1{)}^{s+1})\wedge ((\exists z,(z\ne 0)\wedge (O\left(z\right)=0))⟹\neg (\exists Phi,\forall x,O\left(S\left(x\right)\right)=\mathrm{Phi}\left(O\left(x\right)\right)))\wedge ((M{3}^4=2\cdot I)\wedge (\forall j,\forall x,r3\left(4\cdot j,x\right)=2^j\cdot \mathrm{snd}\left(x\right))\wedge (\forall j,\forall x,(r3\left(8\cdot j,x\right)=\mathrm{snd}\left(x\right))\wedge (r3\left(8\cdot j+4,x\right)=2\cdot \mathrm{snd}\left(x\right)))\wedge (\forall j,r3\left(4\cdot j,\mathrm{pair}\left(0,0\right)\right)=r3\left(4\cdot j,\mathrm{pair}\left(1,0\right)\right))\wedge (r3\left(1,\mathrm{pair}\left(0,0\right)\right)=0)\wedge (r3\left(1,\mathrm{pair}\left(1,0\right)\right)=1)\wedge (0\ne 1)\wedge (\exists Psi,\forall x,O4\left(\mathrm{iterate}\left(S,4,x\right)\right)=\mathrm{Psi}\left(O4\left(x\right)\right))\wedge (\neg (\exists Phi,\forall x,O4\left(\mathrm{iterate}\left(S,1,x\right)\right)=\mathrm{Phi}\left(O4\left(x\right)\right)))\wedge (\neg ((\exists Psi,\forall x,O4\left(\mathrm{iterate}\left(S,4,x\right)\right)=\mathrm{Psi}\left(O4\left(x\right)\right))⟹(\exists Phi,\forall x,O4\left(\mathrm{iterate}\left(S,1,x\right)\right)=\mathrm{Phi}\left(O4\left(x\right)\right)))))$$

Proof. Machine-checked in Lean as D5/S3/Arith/FibonacciAtomic/SamplingQuotient.sampling_quotient (✓ std3). ∎

Source. Repository-derived.

Commentary.

Let n be a positive natural number, including n=1, and let m be at least two. The time family t : Fin(m) -> N is strictly increasing. Set s=t(0), g=gcd(t(i)-s : i is nonzero), and O(x)(i)=r(t(i),x). All states x=(a,b) belong to (Z/nZ)^2 and r(k,x)=F(k)a+F(k+1)b, where F(0)=0 and F(1)=1. The step S(a,b)=(b,a+b) has matrix M=[[0,1],[1,1]]. Every occurrence of a time observation refers to this same state and keeps its time label. In the displayed formula, adjacentRows(s) has rows (F(s),F(s+1)) and (F(s+1),F(s+2)); M3 and r3 denote M and r over Z/3Z, and I is the identity matrix.

Write u(j,x)=r(s+jg,x) and L=trace(M^g). For every natural j and every x, u(j+2,x)=Lu(j+1,x)-(-1)^gu(j,x). The kernel of O equals the joint kernel of r(s,-) and r(s+g,-). In the shifted coordinates (a’,b’)=S^s(z), it is exactly b’=0 and F(g)*a’=0. Consequently O(x)=O(y) if and only if u(j,x)=u(j,y) for all natural j. The first two coarse readings therefore determine the entire coarse sequence.

For every function Phi on the sample space, the identity O(S(x))=Phi(O(x)) for all states implies that O(z)=0 entails O(S(z))=0. The pair x -> (r(s,x),r(s+1,x)) is injective: its matrix has determinant (-1)^(s+1). Thus a state in a stable sampling kernel must be zero. If any nonzero state lies in the sampling kernel, there is no such Phi, even without a linearity assumption on Phi.

Over Z/3Z, M^4=2I and r(4j,x)=2^j*b for every natural j. In particular r(8j,x)=b and r(8j+4,x)=2b. The states (0,0) and (1,0) agree at every time 4j, but their time-one readings are respectively 0 and 1, which are distinct.

For this modulus define O4(x)=(r(0,x),r(4,x)). There exists a function Psi with O4(S^4(x))=Psi(O4(x)) for every x; multiplication of both sample coordinates by 2 gives one. There is no function Phi with O4(S(x))=Phi(O4(x)) for all x. The theorem explicitly negates the implication from the existence of the four-step update to the existence of the one-step update: in this example the coarse sample closes under the four-step clock but not under the one-step clock.

Cayley-Hamilton gives the coarse recurrence. Fibonacci divisibility places the two-reading kernel inside the sampling kernel. Their equal finite cardinalities, both gcd(n,F(g)), give equality. Applying that equality to differences yields the coarse observation equivalence. The invertible Fibonacci step gives adjacent-reading injectivity and the obstruction to a one-step update.

References

• Truth anchor: D5/S3/Arith/FibonacciAtomic/SamplingQuotient.sampling_quotient
• Dependency: D5/S1/Recurrence/FiniteSamplingSmithDefect
• Dependency: D5/S3/Arith/FibonacciAtomic/TimeSampling