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  "body": "\n## Statement\n\nLet $f \\colon [0, 1] \\to [0, 1]$ be a [[continuous|0X3WGX95FE]] unimodal map.\nSo there is a turning point $c \\in (0, 1)$ such that $f$ is strictly increasing on $[0, c]$ and strictly decreasing on $[c, 1]$.\nAssume that $f^n(c) \\neq c$ for all $n \\ge 1$.\nFor $x \\in [0, 1]$ let $\\varepsilon(x) = 1$ if $x \u003c c$, and $\\varepsilon(x) = -1$ if $x \u003e c$.\nThe *kneading determinant* of $f$ is the [[formal power series|033WGN00BF]]\n$$D(t) = 1 + \\sum_{n=1}^{\\infty} \\varepsilon(f(c)) \\, \\varepsilon(f^2(c)) \\cdots \\varepsilon(f^n(c)) \\, t^n.$$\nLet $\\ell(f^n)$ be the number of maximal intervals on which $f^n$ is monotone (the lap number of $f^n$).\nThen:\n\n1. The limit $s = \\lim_{n \\to \\infty} \\ell(f^n)^{1/n}$ exists, $1 \\le s \\le 2$, and the [[topological entropy|BDZYBXBFDE]] of $f$ is $h(f) = \\log s$.\n2. The power series $D(t)$ converges for $\\abs{t} \u003c 1$ and has no zero in the disc $\\abs{t} \u003c 1/s$.\n3. If $s \u003e 1$, then $D(1/s) = 0$, so $1/s$ is the smallest positive zero of $D$.\n4. If $s \u003e 1$, then $f$ is [[semiconjugate|AEXJNRBASV]] to a piecewise linear unimodal map $g$ whose slopes are $s$ and $-s$. That is, there is a continuous nondecreasing surjection $h \\colon [0, 1] \\to [0, 1]$ with $h \\circ f = g \\circ h$.\n\n## Remarks\n\nThe coefficients of $D(t)$ are $\\pm 1$, which gives convergence for $\\abs{t} \u003c 1$.\nThe sign sequence $\\varepsilon(f(c)), \\varepsilon(f^2(c)), \\dots$, the kneading sequence, records on which side of the turning point the orbit of the critical value falls.\nMilnor and Thurston show that it determines the combinatorics of all orbits: the lap numbers, the numbers of periodic points, and the entropy can all be read off from it.\n\nExample: the full [[tent map|D0A22MEB1F]] $T(x) = 1 - \\abs{1 - 2x}$ with $c = 1/2$.\nHere $T(c) = 1 \u003e c$ and $T^n(c) = 0 \u003c c$ for $n \\ge 2$, so the signs are $-1, 1, 1, \\dots$ and\n$$D(t) = 1 - t - t^2 - t^3 - \\cdots = \\frac{1 - 2t}{1 - t}.$$\nIts only zero is $t = 1/2$, and indeed $\\ell(T^n) = 2^n$, so $s = 2$ and $h(T) = \\log 2$.\n\n## Explanations\n\n### L1 human\n\nA unimodal map folds an interval once, like kneading dough.\nEverything about its orbits is encoded in one list of signs: where the image of the fold point lands, left or right of the fold, after each step.\nMilnor and Thurston turned this list into a power series and showed that its first zero gives the entropy, the exponential rate at which the map creates complexity.\n\n### L2 human\n\nThe theorem reduces the topological dynamics of unimodal maps, such as the logistic family, to symbolic data.\nPart 4 says that, up to collapsing some intervals, every unimodal map with positive entropy is a tent-like map of constant slope, so the entropy is visible as a slope.\nKneading theory was a main tool in the proof that topological entropy of the logistic family $x \\mapsto r x (1 - x)$ is a nondecreasing function of $r$.\nIt also explains the order in which periodic orbits appear in such families, which refines [[Sharkovskii's theorem|YBCZ9VY7SX]].\n",
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