{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe expand the product by induction on the number of factors and compare coefficients.\n\n1. Let $R$ be a [[commutative ring|0GYGHKX834]] and let $r_1, \\dots, r_n \\in R$. For $0 \\le m \\le n$ and $0 \\le k \\le m$, let $e^{(m)}_k = e_k(r_1, \\dots, r_m)$ be the [[elementary symmetric polynomial|0ZMY8441GW]] $e_k$ evaluated at $r_1, \\dots, r_m$. Put $e^{(m)}_0 = 1$, and $e^{(m)}_k = 0$ for $k \u003c 0$ or $k \u003e m$.\n2. So $e^{(m)}_k$ is the sum, over the $k$-element subsets $S$ of $\\set{1, \\dots, m}$, of $\\prod_{i \\in S} r_i$.\n3. Claim: $P_m = (x - r_1) \\cdots (x - r_m)$ equals $\\sum_{k=0}^{m} (-1)^k e^{(m)}_k x^{m-k}$ in the [[polynomial ring|0Q9EZS90GY]] $R[x]$. For $m = 0$ both sides are $1$.\n4. The $k$-element subsets of $\\set{1, \\dots, m+1}$ either omit $m+1$, or are $T \\cup \\set{m+1}$ with $T$ a $(k-1)$-element subset of $\\set{1, \\dots, m}$. So $e^{(m+1)}_k = e^{(m)}_k + r_{m+1} e^{(m)}_{k-1}$ for $0 \\le k \\le m+1$.\n5. Assume the claim for $m$. Then $P_{m+1} = (x - r_{m+1}) P_m = \\sum_k (-1)^k e^{(m)}_k x^{m+1-k} - \\sum_k (-1)^k r_{m+1} e^{(m)}_k x^{m-k}$.\n6. In the second sum, put $j = k + 1$. The coefficient of $x^{m+1-j}$ in $P_{m+1}$ is then $(-1)^j e^{(m)}_j + (-1)^{j} r_{m+1} e^{(m)}_{j-1}$.\n7. By step 4 this is $(-1)^j e^{(m+1)}_j$, for $0 \\le j \\le m+1$. This proves the claim for $m + 1$.\n8. Now let $f = a_n x^n + \\dots + a_0 = a_n (x - r_1) \\cdots (x - r_n)$. By the claim with $m = n$, $f = \\sum_{k=0}^{n} a_n (-1)^k e_k(r_1, \\dots, r_n) x^{n-k}$.\n9. Two polynomials are equal exactly when their coefficients are equal. Comparing the coefficients of $x^{n-k}$ gives $a_{n-k} = (-1)^k a_n e_k(r_1, \\dots, r_n)$.\n10. Multiplying by $(-1)^k$ gives $a_n \\, e_k(r_1, \\dots, r_n) = (-1)^k a_{n-k}$ for $1 \\le k \\le n$.\n\n## Idea\n\nMultiplying out $(x - r_1) \\cdots (x - r_n)$ means choosing $x$ or $-r_i$ from each factor. Choosing $-r_i$ from $k$ factors gives $(-1)^k$ times a product of $k$ roots in front of $x^{n-k}$. Summing over all choices gives $e_k$, and comparing with the coefficients of $f$ gives the formulas.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "12D10"
    ],
    "name": "Proof of Vieta's formulas",
    "proves": "0M8EWQJ1YV",
    "sources": [
      {
        "locator": "Basic formulas",
        "url": "https://en.wikipedia.org/wiki/Vieta%27s_formulas"
      }
    ],
    "tag": "FW67GK92SR",
    "uses": [
      "0GYGHKX834",
      "0ZMY8441GW",
      "0Q9EZS90GY"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/FW67GK92SR/",
  "links": [
    {
      "field": "proves",
      "target": "0M8EWQJ1YV"
    },
    {
      "field": "uses",
      "target": "0GYGHKX834"
    },
    {
      "field": "uses",
      "target": "0ZMY8441GW"
    },
    {
      "field": "uses",
      "target": "0Q9EZS90GY"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/FW67GK92SR.md",
  "msc": [
    "12D10"
  ],
  "name": "Proof of Vieta's formulas",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/FW6/FW67GK92SR.md",
  "status": null,
  "tag": "FW67GK92SR",
  "uri": "m2://FW67GK92SR",
  "url": "https://math2.tamnd.com/wiki/proof-of-vietas-formulas/",
  "writing": {
    "tag": "FW67GK92SR",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}