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  "body": "\n## Statement\n\nLet $R$ be a commutative ring and let $d \\ge 1$ be an integer.\nLet $E$ be a [[linear recurrence with constant coefficients|081THG13JZ]] of order $d$ over $R$, with coefficients $c_0, \\dots, c_{d-1} \\in R$.\nLet $u$ and $v$ be [[solutions|0ZV2YKKZ58]] of $E$, that is, sequences $\\N \\to R$ with $u_{n+d} = \\sum_{i=0}^{d-1} c_i u_{n+i}$ and $v_{n+d} = \\sum_{i=0}^{d-1} c_i v_{n+i}$ for all $n \\in \\N$.\nIf $u_i = v_i$ for $i = 0, 1, \\dots, d-1$, then $u_n = v_n$ for all $n \\in \\N$.\n\n## Remarks\n\nThe proof is a strong induction on $n$.\nFor $n \u003c d$ the claim is the hypothesis.\nFor $n \\ge d$, the recurrence writes $u_n$ and $v_n$ by the same formula in the $d$ previous terms, which agree by induction.\nExistence also holds: for each choice of $u_0, \\dots, u_{d-1}$ in $R$, the recurrence defines the later terms one by one.\nSo the map $u \\mapsto (u_0, \\dots, u_{d-1})$ is a bijection from the solutions onto $R^d$; see the [[space of solutions|099GXMHV40]].\nExample: let $\\varphi$ and $\\psi$ be the roots of $X^2 - X - 1$, and let $v_n = (\\varphi^n - \\psi^n)/\\sqrt{5}$.\nThen $v$ solves $u_{n+2} = u_{n+1} + u_n$ over $\\R$, and $v_0 = 0$, $v_1 = 1$.\nThe [[Fibonacci sequence|0GA9K57A39]] solves the same recurrence with the same two initial values, so it equals $v$; this proves Binet's formula.\nFewer initial values are not enough: $(0, 1, 1, 2, 3, \\dots)$ and $(0, 2, 2, 4, 6, \\dots)$ both solve $u_{n+2} = u_{n+1} + u_n$ and agree in the first term only.\n\n## Explanations\n\n### L1 human\n\nA linear recurrence of order $d$ tells you how to compute each term from the $d$ terms before it.\nSo once the first $d$ terms are known, there is no freedom left: every later term is forced.\nTwo sequences that obey the same rule and start the same way are the same sequence.\n\n### L2 human\n\nThis fact is the basic tool for proving closed formulas for recurrent sequences.\nTo check a formula, one checks that it satisfies the recurrence and that it has the right first $d$ terms.\nTogether with existence, it shows that the solutions form a free module of rank $d$ with the initial values as coordinates.\nIn Mathlib it is `LinearRecurrence.eq_mk_of_is_sol_of_eq_init`.\n",
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  "name": "Uniqueness of solutions given initial values",
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