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  "body": "\n## Definition\n\nLet $D \\subseteq \\R^m$, let $\\norm{\\cdot}$ be a [[norm|0V2WP9B1MF]] on $\\R^m$, and let $f \\colon D \\to D$ be a map.\nLet $x^* \\in D$ be an [[equilibrium|D62PEKHRV5]] of $x_{n+1} = f(x_n)$, so $f(x^*) = x^*$.\nWrite $f^n$ for the $n$-th iterate of $f$.\n\n1. The equilibrium $x^*$ is *stable* if for every $\\varepsilon \u003e 0$ there is $\\delta \u003e 0$ such that $\\norm{x_0 - x^*} \u003c \\delta$ and $x_0 \\in D$ imply $\\norm{f^n(x_0) - x^*} \u003c \\varepsilon$ for all $n \\ge 0$.\n2. It is *attracting* if there is $\\eta \u003e 0$ such that $\\norm{x_0 - x^*} \u003c \\eta$ and $x_0 \\in D$ imply $f^n(x_0) \\to x^*$ as $n \\to \\infty$.\n3. It is *asymptotically stable* if it is stable and attracting.\n4. It is *unstable* if it is not stable.\n\n## Remarks\n\nExample: for $f(x) = x/2$ on $\\R$, the equilibrium $0$ is asymptotically stable: take $\\delta = \\varepsilon$, since $\\abs{f^n(x_0)} = \\abs{x_0}/2^n \\le \\abs{x_0}$, and $f^n(x_0) \\to 0$ for every $x_0$.\nNon-example: for $f(x) = 2x$, the equilibrium $0$ is not stable, since $\\abs{f^n(x_0)} = 2^n \\abs{x_0}$ exceeds any $\\varepsilon$ for $x_0 \\ne 0$.\nFor $f(x) = -x$, the equilibrium $0$ is stable but not attracting, as solutions jump between $x_0$ and $-x_0$.\nSince all norms on $\\R^m$ are equivalent, the notions do not depend on the choice of norm.\nFor a higher-order equation, the same definitions apply to the equivalent first-order system on $\\R^{mk}$.\nThey are the discrete analogues of the definitions of [[Lyapunov stability|092S6VBECQ]] and [[asymptotic stability|05Y1JAM3Y5]] for flows.\n",
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