{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe expand $f(x)$ around a moving point $t$ and apply the mean value theorem in $t$.\n\n1. Let $\\alpha \u003c \\beta$ be the endpoints of $J$, so $\\set{\\alpha, \\beta} = \\set{a, x}$. The hypothesis means that $f, f', \\dots, f^{(k)}$ exist on $J$, with [[one-sided derivatives|0HXEX0DPRA]] at $\\alpha$ and $\\beta$.\n2. For $t \\in J$ define $F(t) = \\sum_{j=0}^{k} \\dfrac{f^{(j)}(t)}{j!} (x - t)^j$.\n3. For $j \u003c k$, $f^{(j)}$ is differentiable on $J$, so it is continuous on $J$. Inside $J$ this is [[differentiability implies continuity|07ZMX4ANJS]].\n4. At an endpoint, write $f^{(j)}(s) - f^{(j)}(\\alpha) = (s - \\alpha) \\cdot \\dfrac{f^{(j)}(s) - f^{(j)}(\\alpha)}{s - \\alpha}$. The quotient has a finite limit, so the product tends to $0$. The endpoint $\\beta$ is similar.\n5. Also $f^{(k)}$ is continuous on $J$ by hypothesis. So $F$ is continuous on $J$, as a sum of products of continuous functions.\n6. By the [[power rule|0JZGW75C5Z]] and the [[chain rule|0RXX2PT1MQ]], $\\dfrac{d}{dt} \\dfrac{(x - t)^j}{j!} = -\\dfrac{(x - t)^{j-1}}{(j-1)!}$ for $j \\ge 1$.\n7. For $0 \\le j \\le k$ and $t \\in (\\alpha, \\beta)$ let $A_j(t) = \\dfrac{f^{(j+1)}(t)}{j!} (x - t)^j$. This uses the hypothesis that [[$f^{(k+1)}$|084NGSTWM7]] exists on $(\\alpha, \\beta)$.\n8. By the [[product rule|0V9A43H7WV]] and step 6, the $j$-th term of $F$ has derivative $A_j - A_{j-1}$ for $j \\ge 1$. The term with $j = 0$ is $f$, with derivative $A_0$.\n9. Adding, the sum telescopes: $F'(t) = A_0 + \\sum_{j=1}^{k} (A_j - A_{j-1}) = A_k = \\dfrac{f^{(k+1)}(t)}{k!} (x - t)^k$ on $(\\alpha, \\beta)$.\n10. By the [[mean value theorem|0BARQY52N5]], there is $\\xi \\in (\\alpha, \\beta)$ with $F(\\beta) - F(\\alpha) = F'(\\xi)(\\beta - \\alpha)$.\n11. Exchanging $\\alpha$ and $\\beta$ multiplies both sides by $-1$. So $F(x) - F(a) = F'(\\xi)(x - a)$, and $\\xi$ lies strictly between $a$ and $x$.\n12. At $t = x$, every term of $F$ with $j \\ge 1$ vanishes, so $F(x) = f(x)$. At $t = a$, $F(a) = P_k(x)$, the value at $x$ of the [[Taylor polynomial|0HFSFVMWQB]] of $f$ at $a$.\n13. Substituting steps 9 and 12 into step 11 gives $f(x) = P_k(x) + \\dfrac{f^{(k+1)}(\\xi)}{k!} (x - \\xi)^k (x - a)$.\n\n## Idea\n\nThe function $F(t)$ is the Taylor polynomial of degree $k$ at $t$, evaluated at $x$. It moves from $P_k(x)$ at $t = a$ to $f(x)$ at $t = x$, and its derivative telescopes to a single term. The plain mean value theorem applied to $F$ then gives the Cauchy form directly.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "41A80"
    ],
    "name": "Proof of Taylor's theorem with the Cauchy remainder by the mean value theorem",
    "proves": "005Q84BMAD",
    "sources": [
      {
        "locator": "Derivation for the mean value forms of the remainder",
        "url": "https://en.wikipedia.org/wiki/Taylor%27s_theorem"
      }
    ],
    "tag": "C9NEEBEG0K",
    "uses": [
      "0HFSFVMWQB",
      "0BARQY52N5",
      "084NGSTWM7",
      "0HXEX0DPRA",
      "07ZMX4ANJS",
      "0V9A43H7WV",
      "0JZGW75C5Z",
      "0RXX2PT1MQ"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/C9NEEBEG0K/",
  "links": [
    {
      "field": "proves",
      "target": "005Q84BMAD"
    },
    {
      "field": "uses",
      "target": "0HFSFVMWQB"
    },
    {
      "field": "uses",
      "target": "0BARQY52N5"
    },
    {
      "field": "uses",
      "target": "084NGSTWM7"
    },
    {
      "field": "uses",
      "target": "0HXEX0DPRA"
    },
    {
      "field": "uses",
      "target": "07ZMX4ANJS"
    },
    {
      "field": "uses",
      "target": "0V9A43H7WV"
    },
    {
      "field": "uses",
      "target": "0JZGW75C5Z"
    },
    {
      "field": "uses",
      "target": "0RXX2PT1MQ"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/C9NEEBEG0K.md",
  "msc": [
    "41A80"
  ],
  "name": "Proof of Taylor's theorem with the Cauchy remainder by the mean value theorem",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/C9N/C9NEEBEG0K.md",
  "status": null,
  "tag": "C9NEEBEG0K",
  "uri": "m2://C9NEEBEG0K",
  "url": "https://math2.tamnd.com/wiki/proof-of-taylors-theorem-with-the-cauchy-remainder-by-the-mean-value-theorem/",
  "writing": {
    "tag": "C9NEEBEG0K",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}