The logistic map at r = 4 has Lyapunov exponent ln 2

Adam Zettel · September 21, 2026 · written with an agent

λ = n \to \infty lim \frac{1}{n} k < n \sum ln ∣ r (1 - 2 x_{k}) ∣

Iterate $x \to r x (1 - x)$ . At $r = 3.2$ it settles on two values you can write down. At $r = 4$ it never settles, nearby starts separate by a factor of two per step on average, and that factor is exact: the Lyapunov exponent is $ln 2$ .

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At $r = 4$ , substitute $x = sin^{2} θ$ and the map becomes $θ \to 2 θ$ : angle doubling, which is where $ln 2$ comes from. The two-cycle for $3 < r < 1 + 6$ solves $f (f (x)) = x$ with the fixed points divided out: $x_{\pm} = (r + 1 \pm (r - 3) (r + 1)) /2 r$ .

Printed

reached_a: 0.513045

reached_b: 0.799455

chaos: 0.693145

order: -0.916291

Attractor of x -> r x (1 - x), by r

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Tests 3 of 3 pass

at r = 3.2 the orbit lands on the closed-form two-cycle (passes)
there the Lyapunov exponent is negative: nearby starts converge (passes)
at r = 4 it is ln 2 (passes)