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  "body": "\n## Statement\n\nLet $A$ be the [[shift operator|MBK6AE6734]] on functions $f:\\mathbb{Z}\\to\\mathbb{C}$, so that $(Af)(n)=f(n+1)$.\nLet $r$ be a nonzero number, and let $f$ satisfy the operator equation $(A-r)f=0$, that is, $f(n+1)=r\\,f(n)$ for all integers $n$.\nPut $c=f(0)$.\nThen $f(n)=c\\,r^n$ for every integer $n$.\n\n## Remarks\n\nThe values of $f$ at nonnegative $n$ follow from the rule $f(n+1)=rf(n)$ by induction upward from $0$.\nThe values at negative $n$ follow by induction downward from $0$, using $f(n)=r^{-1}f(n+1)$. This step is where $r\\neq 0$ is needed.\nIf $r=0$, the claim can fail: a function with $f(0)=c$ and $f(n)=0$ for $n\\ge 1$ is a solution, but its values at negative $n$ are not fixed by $c$.\n",
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    "name": "A solution of (A-r)f=0 is a geometric sequence fixed by its value at 0",
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        "locator": "source/s_recurrence_rigorous.ptx:62",
        "url": "https://github.com/mitchkeller/applied-combinatorics/blob/ebd95d6bd59e23c633c65cf560bcb93efa912aae/source/s_recurrence_rigorous.ptx#L62"
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  "markdown": "https://math2.tamnd.com/node/GGSHNC4CGZ.md",
  "msc": [
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  "name": "A solution of (A-r)f=0 is a geometric sequence fixed by its value at 0",
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  "role": "lemma",
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  "tag": "GGSHNC4CGZ",
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  "url": "https://math2.tamnd.com/wiki/a-solution-of-a-r-f-0-is-a-geometric-sequence-fixed-by-its-value-at-0/",
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