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  "body": "\n## Statement\n\nLet $d \\ge 1$ and let $c_0, \\dots, c_{d-1}$ be complex numbers with $c_0 \\ne 0$.\nConsider the [[linear recurrence|AA7XRZ4VAG]]\n$$x_{n+d} = c_{d-1} x_{n+d-1} + \\cdots + c_1 x_{n+1} + c_0 x_n, \\qquad n \\ge 0,$$\nand its [[characteristic polynomial|0YA7X86K05]] $\\chi(z) = z^d - c_{d-1} z^{d-1} - \\cdots - c_0$.\nThe following are equivalent.\n\n1. Every solution $(x_n)$ satisfies $x_n \\to 0$ as $n \\to \\infty$.\n2. The zero solution is asymptotically [[stable|CHEQ235HW9]].\n3. The polynomial $\\chi$ is [[Schur stable|X11Q0G8PNA]], that is, every root $\\lambda$ of $\\chi$ satisfies $\\abs{\\lambda} \u003c 1$.\n\n## Remarks\n\nEvery solution is a finite sum of terms $p(n) \\lambda^n$, where $\\lambda$ is a root of $\\chi$ and $p$ is a polynomial of degree less than the multiplicity of $\\lambda$.\nIf all roots satisfy $\\abs{\\lambda} \u003c 1$, each such term tends to $0$, because $\\abs{\\lambda}^n$ decays faster than any polynomial grows.\nIf some root has $\\abs{\\lambda} \\ge 1$, then $x_n = \\lambda^n$ is a solution that does not tend to $0$.\nFor example, $x_{n+2} = x_{n+1} - \\tfrac{1}{2} x_n$ has characteristic polynomial $z^2 - z + \\tfrac{1}{2}$ with roots of modulus $1/\\sqrt{2}$, so all its solutions tend to $0$.\nIn contrast, $x_{n+2} = x_n$ has roots $\\pm 1$ on the unit circle, and its solution $x_n = 1$ stays bounded but does not tend to $0$.\nSuppose all roots satisfy $\\abs{\\lambda} \\le 1$, some root lies on the unit circle, and the roots on the circle are simple.\nThen the zero solution is stable but not asymptotically stable.\n\n## Explanations\n\n### L1 human\n\nA linear recurrence builds each term from a fixed combination of the previous ones.\nThis theorem says when every such sequence dies out, whatever the starting values.\nThe answer is read off from one polynomial: all its roots must lie strictly inside the unit circle.\nIt turns a question about infinitely many sequences into a question about finitely many numbers.\n\n### L2 human\n\nThe result is the discrete counterpart of the fact that $x' = Ax$ is asymptotically stable exactly when all eigenvalues of $A$ have negative real part.\nWriting the recurrence as $X_{n+1} = C X_n$ with the companion matrix $C$, the condition says that the spectral radius of $C$ is less than $1$.\nIn practice one does not compute the roots; tests such as the [[Jury stability criterion|VGV9FAPCHN]] decide Schur stability from the coefficients alone.\nThe same criterion governs the local stability of a fixed point of a nonlinear map through its linearization.\n",
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