{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe remove the last element from its cycle and count each case by a bijection.\n\n1. Let $\\sigma$ be a [[permutation|0BQMWCNS8S]] of a finite set $X$. For $x \\in X$, the cycle of $\\sigma$ through $x$ is the set $\\set{\\sigma^t(x) : t \\ge 0}$.\n2. By the [[decomposition into disjoint cycles|0YWDAZPRPQ]], these sets are the moved sets of the cycles of $\\sigma$, together with the singletons of fixed points. They partition $X$, and their number is the number of cycles of $\\sigma$ in the definition of the [[Stirling numbers of the first kind|0Z22VQH17V]].\n3. If $x$ lies in a cycle with $L$ elements, then $x, \\sigma(x), \\dots, \\sigma^{L-1}(x)$ are its elements, all distinct, and $\\sigma^L(x) = x$.\n4. Let $n$ and $k$ be natural numbers. Let $X = \\set{1, \\dots, n + 1}$, let $z = n + 1$, and let $Y = \\set{1, \\dots, n}$. We count the permutations of $X$ with $k + 1$ cycles.\n5. Case $\\sigma(z) = z$. Then $\\set{z}$ is a cycle, and $\\sigma$ restricts to a permutation of $Y$. Its cycles are the other $k$ cycles of $\\sigma$.\n6. Conversely, a permutation of $Y$ with $k$ cycles, extended by $z \\mapsto z$, gives such a $\\sigma$. These maps are inverse to each other, so this case has $c(n, k)$ permutations.\n7. Case $\\sigma(z) \\ne z$. Let $j = \\sigma^{-1}(z)$, so $j \\in Y$. Define $\\tau : Y \\to Y$ by $\\tau(j) = \\sigma(z)$ and $\\tau(x) = \\sigma(x)$ for $x \\in Y \\setminus \\set{j}$.\n8. The map $\\tau$ is a bijection of $Y$. Indeed, $\\sigma$ maps $Y \\setminus \\set{j}$ onto $X \\setminus \\set{z, \\sigma(z)}$, and $\\tau(j) = \\sigma(z)$ supplies the missing value.\n9. Let $C$ be the cycle of $\\sigma$ through $z$, with $L$ elements. Write $a_t = \\sigma^t(z)$ for $1 \\le t \\le L - 1$. Since $\\sigma(z) \\ne z$, $L \\ge 2$, and $a_{L-1} = j$ because $\\sigma(a_{L-1}) = z$.\n10. For $1 \\le t \u003c L - 1$, $a_t \\ne j$, so $\\tau(a_t) = a_{t+1}$. Also $\\tau(a_{L-1}) = \\tau(j) = a_1$.\n11. So $C \\setminus \\set{z} = \\set{a_1, \\dots, a_{L-1}}$ is a cycle of $\\tau$, with $L - 1 \\ge 1$ elements.\n12. Any other cycle $C'$ of $\\sigma$ avoids $z$ and $j$, since $j \\in C$. On $C'$ the maps $\\tau$ and $\\sigma$ agree, so $C'$ is also a cycle of $\\tau$.\n13. The sets of steps 11 and 12 partition $Y$, so they are all the cycles of $\\tau$. Hence $\\tau$ has as many cycles as $\\sigma$, namely $k + 1$.\n14. Map $\\sigma$ to the pair $(\\tau, j)$, where $\\tau$ is a permutation of $Y$ with $k + 1$ cycles and $j \\in Y$.\n15. Conversely, let $\\tau$ be a permutation of $Y$ with $k + 1$ cycles, and let $j \\in Y$. Define $\\sigma$ on $X$ by $\\sigma(j) = z$, $\\sigma(z) = \\tau(j)$, and $\\sigma(x) = \\tau(x)$ for the other $x$.\n16. This $\\sigma$ is a permutation of $X$, because $\\tau$ maps $Y \\setminus \\set{j}$ onto $Y \\setminus \\set{\\tau(j)}$. Also $\\sigma(z) = \\tau(j) \\ne z$ and $\\sigma^{-1}(z) = j$. Applying step 7 to it gives back $\\tau$, because $\\tau(j) = \\sigma(z)$. So by step 13, $\\sigma$ has $k + 1$ cycles.\n17. Starting from $\\sigma$, step 15 applied to $(\\tau, j)$ gives back $\\sigma$, since $\\sigma(j) = z$ and $\\sigma(z) = \\tau(j)$. So the maps of steps 14 and 15 are inverse to each other.\n18. By the [[bijection principle|0GV2573FP4]] and the [[rule of sum|0EE2V3CHFK]] over the $n$ choices of $j$, this case has $n\\, c(n, k + 1)$ permutations.\n19. The two cases are disjoint and cover all permutations of $X$ with $k + 1$ cycles. By the rule of sum,\n$$c(n + 1, k + 1) = n\\, c(n, k + 1) + c(n, k).$$\n\n## Idea\n\nDelete $z$ from the permutation: if $z$ is a fixed point, one cycle disappears. Otherwise, link the element before $z$ directly to the element after $z$, which keeps the number of cycles. This element before $z$ can be any of the $n$ other elements, which gives the factor $n$.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "11B73"
    ],
    "name": "Proof of the recurrence for Stirling numbers of the first kind",
    "proves": "0C5BFX94J9",
    "sources": [
      {
        "locator": "Recurrence relation",
        "url": "https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind"
      }
    ],
    "tag": "TFWBM9VE3V",
    "uses": [
      "0Z22VQH17V",
      "0BQMWCNS8S",
      "0YWDAZPRPQ",
      "0EE2V3CHFK",
      "0GV2573FP4"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/TFWBM9VE3V/",
  "links": [
    {
      "field": "proves",
      "target": "0C5BFX94J9"
    },
    {
      "field": "uses",
      "target": "0Z22VQH17V"
    },
    {
      "field": "uses",
      "target": "0BQMWCNS8S"
    },
    {
      "field": "uses",
      "target": "0YWDAZPRPQ"
    },
    {
      "field": "uses",
      "target": "0EE2V3CHFK"
    },
    {
      "field": "uses",
      "target": "0GV2573FP4"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/TFWBM9VE3V.md",
  "msc": [
    "11B73"
  ],
  "name": "Proof of the recurrence for Stirling numbers of the first kind",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/TFW/TFWBM9VE3V.md",
  "status": null,
  "tag": "TFWBM9VE3V",
  "uri": "m2://TFWBM9VE3V",
  "url": "https://math2.tamnd.com/wiki/proof-of-the-recurrence-for-stirling-numbers-of-the-first-kind/",
  "writing": {
    "tag": "TFWBM9VE3V",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}