{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe restrict $f$ to the line through $a$ in the direction $h$ and use the one-variable Taylor theorem.\n\n1. Let $U$, $k$, $f$ and $a$ be as in the statement. For $h \\in \\R^n$ with $a + h \\in U$, let $R_k(h) = f(a + h) - \\sum_{\\abs{\\alpha} \\le k} \\frac{D^\\alpha f(a)}{\\alpha!} h^\\alpha$. We must show the two claims about $R_k$.\n2. Fix $h \\in \\R^n$ and let $O = \\set{t \\in \\R : a + t h \\in U}$. This set is open, since it is the preimage of the open set $U$ under the continuous map $t \\mapsto a + t h$. On $O$ define $g(t) = f(a + t h)$.\n3. Let $u$ be a [[partial derivative|02RSXVM1ZN]] of $f$ of order $j \u003c k$. The first partial derivatives of $u$ are partial derivatives of $f$ of order $j + 1 \\le k$. By the definition of [[class C^k|0TCBZMEYGN]], they exist and are [[continuous|0S9EK5MQPE]] on $U$.\n4. So $u$ is [[differentiable|0KXSK8TEEB]] on $U$. Each point of $O$ lies in an open interval contained in $O$. On such an interval, the [[chain rule for a curve|0A8TJ0VHC7]] applies. So $t \\mapsto u(a + t h)$ is differentiable on $O$ with derivative $\\sum_{i=1}^n \\partial_i u(a + t h) \\, h_i$.\n5. Claim: for $0 \\le j \\le k$, the derivative $g^{(j)}$ exists on $O$ and $g^{(j)}(t) = \\sum_{i_1, \\dots, i_j} \\partial_{i_j} \\cdots \\partial_{i_1} f(a + t h) \\, h_{i_1} \\cdots h_{i_j}$, where each index runs from $1$ to $n$. For $j = 0$ this is the definition of $g$.\n6. If the claim holds for some $j \u003c k$, apply step 4 to each term $u = \\partial_{i_j} \\cdots \\partial_{i_1} f$. This gives the claim for $j + 1$. By induction the claim holds for all $j \\le k$. Each $g^{(j)}$ is continuous, since the partial derivatives are continuous.\n7. Let $j \\le k$ and let $(i_1, \\dots, i_j)$ be a sequence of indices. Any rearrangement of this sequence is reached by finitely many swaps of two adjacent entries.\n8. Swapping $i_l$ and $i_{l+1}$ does not change the partial derivative. Indeed, $w = \\partial_{i_{l-1}} \\cdots \\partial_{i_1} f$ is of class $C^2$, since $l + 1 \\le j \\le k$. By [[Schwarz's theorem|0NNQY8WCW6]], $\\partial_{i_{l+1}} \\partial_{i_l} w = \\partial_{i_l} \\partial_{i_{l+1}} w$, and applying the remaining partial derivatives keeps this equality.\n9. For the sequence $(i_1, \\dots, i_j)$ let $\\alpha_l$ be the number of $r$ with $i_r = l$. Then $\\abs{\\alpha} = j$. By steps 7 and 8, $\\partial_{i_j} \\cdots \\partial_{i_1} f = D^\\alpha f$, in the [[multi-index notation|0PJXBR6QA3]] of the statement. Also $h_{i_1} \\cdots h_{i_j} = h^\\alpha$.\n10. For a fixed $\\alpha$ with $\\abs{\\alpha} = j$, the number of sequences that give $\\alpha$ is the [[multinomial coefficient|00K0RF9BFY]] $j! / \\alpha!$.\n11. Steps 5, 9 and 10 give, for $0 \\le j \\le k$ and $t \\in O$:\n$$g^{(j)}(t) = \\sum_{\\abs{\\alpha} = j} \\frac{j!}{\\alpha!} D^\\alpha f(a + t h) \\, h^\\alpha.$$\n12. In particular $\\sum_{j=0}^{k} g^{(j)}(0) / j! = \\sum_{\\abs{\\alpha} \\le k} \\frac{D^\\alpha f(a)}{\\alpha!} h^\\alpha$. So $R_k(h) = g(1) - \\sum_{j=0}^{k} g^{(j)}(0)/j!$ whenever $1 \\in O$.\n13. We now prove the limit. Since $U$ is open, choose $r \u003e 0$ with $\\set{x : \\norm{x - a} \u003c r} \\subseteq U$. Let $0 \u003c \\norm{h} \u003c r$. Then $a + t h \\in U$ for $t \\in [0, 1]$, so $[0, 1] \\subseteq O$.\n14. By step 6, $g^{(k)}$ exists and is continuous on $[0, 1]$. Apply [[Taylor's theorem with the integral remainder|0WJGR81HBD]] to $g$ on $[0, 1]$, with order $k - 1 \\ge 0$:\n$$g(1) = \\sum_{j=0}^{k-1} \\frac{g^{(j)}(0)}{j!} + \\int_0^1 \\frac{(1 - t)^{k-1}}{(k-1)!} \\, g^{(k)}(t) \\, dt.$$\n15. We have $\\int_0^1 \\frac{(1 - t)^{k-1}}{(k-1)!} \\, dt = \\frac{1}{k!}$. So, by [[linearity of the integral|080RWVVSB6]], steps 12 and 14 give\n$$R_k(h) = \\int_0^1 \\frac{(1 - t)^{k-1}}{(k-1)!} \\big( g^{(k)}(t) - g^{(k)}(0) \\big) \\, dt.$$\n16. Each $\\abs{h_l} \\le \\norm{h}$, so $\\abs{h^\\alpha} \\le \\norm{h}^k$ when $\\abs{\\alpha} = k$. Let $N = \\sum_{\\abs{\\alpha} = k} k!/\\alpha!$. Then step 11 gives, for $t \\in [0, 1]$,\n$$\\abs{g^{(k)}(t) - g^{(k)}(0)} \\le N \\norm{h}^k \\max_{\\abs{\\alpha} = k} \\abs{D^\\alpha f(a + t h) - D^\\alpha f(a)}.$$\n17. Let $\\varepsilon \u003e 0$. There are finitely many $\\alpha$ with $\\abs{\\alpha} = k$, and each $D^\\alpha f$ is continuous at $a$. So there is $\\rho$ with $0 \u003c \\rho \\le r$ and $\\abs{D^\\alpha f(x) - D^\\alpha f(a)} \u003c \\varepsilon$ for all such $\\alpha$ and all $x$ with $\\norm{x - a} \u003c \\rho$.\n18. Let $0 \u003c \\norm{h} \u003c \\rho$. Then $\\norm{(a + t h) - a} \u003c \\rho$ for $t \\in [0, 1]$. By steps 15 to 17 and [[monotonicity of the integral|0MBWQKM8QR]], $\\abs{R_k(h)} \\le \\int_0^1 \\frac{(1 - t)^{k-1}}{(k-1)!} N \\varepsilon \\norm{h}^k \\, dt = \\frac{N \\varepsilon}{k!} \\norm{h}^k$.\n19. Since $\\varepsilon \u003e 0$ was arbitrary and $N / k!$ does not depend on $\\varepsilon$, this shows $R_k(h) / \\norm{h}^k \\to 0$ as $h \\to 0$.\n20. Now let $f$ be of class $C^{k+1}$, and let the segment from $a$ to $a + h$ lie in $U$. Then $[0, 1] \\subseteq O$.\n21. Steps 3 to 11 hold with $k + 1$ in place of $k$. So $g^{(k+1)}$ exists and is continuous on $[0, 1]$, and it is given by step 11 with $j = k + 1$.\n22. Taylor's theorem with the integral remainder, now with order $k$, gives\n$$g(1) = \\sum_{j=0}^{k} \\frac{g^{(j)}(0)}{j!} + \\int_0^1 \\frac{(1 - t)^{k}}{k!} \\, g^{(k+1)}(t) \\, dt.$$\n23. By step 12, the integral is $R_k(h)$. Insert step 11 with $j = k + 1$ and use linearity of the integral. Since $(k+1)!/(k! \\, \\beta!) = (k+1)/\\beta!$, this gives\n$$R_k(h) = \\sum_{\\abs{\\beta} = k + 1} \\frac{k + 1}{\\beta!} \\, h^\\beta \\int_0^1 (1 - t)^k \\, D^\\beta f(a + t h) \\, dt.$$\n\n## Idea\n\nAlong the line $t \\mapsto a + t h$, the function $f$ becomes a function $g$ of one variable. The chain rule writes the $j$-th derivative of $g$ with the partial derivatives of $f$ of order $j$. Schwarz's theorem groups equal terms, and the multinomial coefficients turn $1/j!$ into $1/\\alpha!$. The one-variable Taylor theorem then gives both the expansion and the remainder.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "26B12"
    ],
    "name": "Proof of Taylor's theorem in several variables by restriction to a line",
    "proves": "0J8T4K4YTE",
    "sources": [
      {
        "locator": "Derivation for the remainder of multivariate Taylor polynomials",
        "url": "https://en.wikipedia.org/wiki/Taylor%27s_theorem"
      }
    ],
    "tag": "Q58E952DVF",
    "uses": [
      "0TCBZMEYGN",
      "02RSXVM1ZN",
      "0PJXBR6QA3",
      "0KXSK8TEEB",
      "0A8TJ0VHC7",
      "0NNQY8WCW6",
      "00K0RF9BFY",
      "0WJGR81HBD",
      "0S9EK5MQPE",
      "0MBWQKM8QR",
      "080RWVVSB6"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/Q58E952DVF/",
  "links": [
    {
      "field": "proves",
      "target": "0J8T4K4YTE"
    },
    {
      "field": "uses",
      "target": "0TCBZMEYGN"
    },
    {
      "field": "uses",
      "target": "02RSXVM1ZN"
    },
    {
      "field": "uses",
      "target": "0PJXBR6QA3"
    },
    {
      "field": "uses",
      "target": "0KXSK8TEEB"
    },
    {
      "field": "uses",
      "target": "0A8TJ0VHC7"
    },
    {
      "field": "uses",
      "target": "0NNQY8WCW6"
    },
    {
      "field": "uses",
      "target": "00K0RF9BFY"
    },
    {
      "field": "uses",
      "target": "0WJGR81HBD"
    },
    {
      "field": "uses",
      "target": "0S9EK5MQPE"
    },
    {
      "field": "uses",
      "target": "0MBWQKM8QR"
    },
    {
      "field": "uses",
      "target": "080RWVVSB6"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/Q58E952DVF.md",
  "msc": [
    "26B12"
  ],
  "name": "Proof of Taylor's theorem in several variables by restriction to a line",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/Q58/Q58E952DVF.md",
  "status": null,
  "tag": "Q58E952DVF",
  "uri": "m2://Q58E952DVF",
  "url": "https://math2.tamnd.com/wiki/proof-of-taylors-theorem-in-several-variables-by-restriction-to-a-line/",
  "writing": {
    "tag": "Q58E952DVF",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}