{
  "index": "1947-B-2",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "8. Let \\( f(x) \\) be a differentiable function defined in the closed interval \\( (0,1) \\) and such that\n\\[\n\\left|f^{\\prime}(x)\\right| \\leq M, \\quad 0<x<1\n\\]\n\nProve that\n\\[\n\\left|\\int_{0}^{1} f(x) d x-\\frac{1}{n} \\sum_{k=1}^{n} f\\left(\\frac{k}{n}\\right)\\right| \\leq \\frac{M}{n} .\n\\]",
  "solution": "Solution. Let\n\\[\nE_{k}=\\int_{(k-1) / n}^{k / n} f(x) d x-\\frac{1}{n} f\\left(\\frac{k}{n}\\right)\n\\]\nfor \\( k=1,2, \\ldots, n \\). Since \\( f \\) is differentiable, it is continuous, and therefore by the mean value theorem for integrals there exists a number \\( \\eta_{k} \\) such that\n\\[\n\\frac{k-1}{n}<\\eta_{k}<\\frac{k}{n}\n\\]\nand\n\\[\n\\int_{(k-1) / n}^{k / n} f(x) d x=\\frac{1}{n} f\\left(\\eta_{k}\\right) .\n\\]\n\nBy the mean value theorem for derivatives there exists a number \\( \\xi_{k} \\) such that \\( \\eta_{k}<\\xi_{k}<(k / n) \\) and\n\\[\nf\\left(\\eta_{k}\\right)-f\\left(\\frac{k}{n}\\right)=\\left(\\eta_{k}-\\frac{k}{n}\\right) f^{\\prime}\\left(\\xi_{k}\\right) .\n\\]\n\nThen\n\\[\n\\begin{aligned}\n\\left|E_{k}\\right| & =\\frac{1}{n}\\left|f\\left(\\eta_{k}\\right)-f\\left(\\frac{k}{n}\\right)\\right| \\\\\n& =\\frac{1}{n}\\left|\\eta_{k}-\\frac{k}{n}\\right| \\cdot\\left|f^{\\prime}\\left(\\xi_{k}\\right)\\right| \\leq \\frac{1}{n^{2}} M .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left|\\int_{0}^{1} f(x) d x-\\frac{1}{n} \\sum_{k=1}^{n} f\\left(\\frac{k}{n}\\right)\\right|=\\left|\\sum_{k=1}^{n} E_{k}\\right| \\leq \\frac{M}{n} .\n\\]\n\nRemark. The estimate can be improved by a factor of two. Let\n\\[\nF(x)=\\int_{0}^{x} f(t) d t .\n\\]\n\nExpanding by Taylor's theorem about \\( k / n \\), we obtain\n\\[\nF\\left(\\frac{k-1}{n}\\right)=F\\left(\\frac{k}{n}\\right)+\\left(-\\frac{1}{n}\\right) F^{\\prime}\\left(\\frac{k}{n}\\right)+\\frac{1}{2}\\left(-\\frac{1}{n}\\right)^{2} F^{\\prime \\prime}\\left(\\theta_{k}\\right),\n\\]\nwhere\n\\[\n\\frac{k-1}{n}<\\theta_{k}<\\frac{k}{n} .\n\\]\n\nSince \\( F^{\\prime}=f \\), this becomes\n\\[\nE_{k}=-\\frac{1}{2 n^{2}} f^{\\prime}\\left(\\theta_{k}\\right)\n\\]\nand we have gained a factor of two.",
  "vars": [
    "x",
    "t",
    "k",
    "f",
    "F",
    "E_k",
    "\\\\eta_k",
    "\\\\xi_k",
    "\\\\theta_k"
  ],
  "params": [
    "M",
    "n"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variablex",
        "t": "variablet",
        "k": "indexk",
        "f": "functionf",
        "F": "capitalf",
        "E_k": "errorterm",
        "\\eta_k": "etapoint",
        "\\xi_k": "xipoint",
        "\\theta_k": "thetapoint",
        "M": "slopebound",
        "n": "partitionnum"
      },
      "question": "8. Let \\( functionf(variablex) \\) be a differentiable function defined in the closed interval \\( (0,1) \\) and such that\n\\[\n\\left|functionf^{\\prime}(variablex)\\right| \\leq slopebound, \\quad 0<variablex<1\n\\]\n\nProve that\n\\[\n\\left|\\int_{0}^{1} functionf(variablex) d variablex-\\frac{1}{partitionnum} \\sum_{indexk=1}^{partitionnum} functionf\\left(\\frac{indexk}{partitionnum}\\right)\\right| \\leq \\frac{slopebound}{partitionnum} .\n\\]",
      "solution": "Solution. Let\n\\[\nerrorterm=\\int_{(indexk-1) / partitionnum}^{indexk / partitionnum} functionf(variablex) d variablex-\\frac{1}{partitionnum} functionf\\left(\\frac{indexk}{partitionnum}\\right)\n\\]\nfor \\( indexk=1,2, \\ldots, partitionnum \\). Since \\( functionf \\) is differentiable, it is continuous, and therefore by the mean value theorem for integrals there exists a number \\( etapoint \\) such that\n\\[\n\\frac{indexk-1}{partitionnum}<etapoint<\\frac{indexk}{partitionnum}\n\\]\nand\n\\[\n\\int_{(indexk-1) / partitionnum}^{indexk / partitionnum} functionf(variablex) d variablex=\\frac{1}{partitionnum} functionf\\left(etapoint\\right) .\n\\]\n\nBy the mean value theorem for derivatives there exists a number \\( xipoint \\) such that \\( etapoint<xipoint<(indexk / partitionnum) \\) and\n\\[\nfunctionf\\left(etapoint\\right)-functionf\\left(\\frac{indexk}{partitionnum}\\right)=\\left(etapoint-\\frac{indexk}{partitionnum}\\right) functionf^{\\prime}\\left(xipoint\\right) .\n\\]\n\nThen\n\\[\n\\begin{aligned}\n\\left|errorterm\\right| & =\\frac{1}{partitionnum}\\left|functionf\\left(etapoint\\right)-functionf\\left(\\frac{indexk}{partitionnum}\\right)\\right| \\\\\n& =\\frac{1}{partitionnum}\\left|etapoint-\\frac{indexk}{partitionnum}\\right| \\cdot\\left|functionf^{\\prime}\\left(xipoint\\right)\\right| \\leq \\frac{1}{partitionnum^{2}} slopebound .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left|\\int_{0}^{1} functionf(variablex) d variablex-\\frac{1}{partitionnum} \\sum_{indexk=1}^{partitionnum} functionf\\left(\\frac{indexk}{partitionnum}\\right)\\right|=\\left|\\sum_{indexk=1}^{partitionnum} errorterm\\right| \\leq \\frac{slopebound}{partitionnum} .\n\\]\n\nRemark. The estimate can be improved by a factor of two. Let\n\\[\ncapitalf(variablex)=\\int_{0}^{variablex} functionf(variablet) d variablet .\n\\]\n\nExpanding by Taylor's theorem about \\( indexk / partitionnum \\), we obtain\n\\[\ncapitalf\\left(\\frac{indexk-1}{partitionnum}\\right)=capitalf\\left(\\frac{indexk}{partitionnum}\\right)+\\left(-\\frac{1}{partitionnum}\\right) capitalf^{\\prime}\\left(\\frac{indexk}{partitionnum}\\right)+\\frac{1}{2}\\left(-\\frac{1}{partitionnum}\\right)^{2} capitalf^{\\prime \\prime}\\left(thetapoint\\right),\n\\]\nwhere\n\\[\n\\frac{indexk-1}{partitionnum}<thetapoint<\\frac{indexk}{partitionnum} .\n\\]\n\nSince \\( capitalf^{\\prime}=functionf \\), this becomes\n\\[\nerrorterm=-\\frac{1}{2 partitionnum^{2}} functionf^{\\prime}\\left(thetapoint\\right)\n\\]\nand we have gained a factor of two.\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandstone",
        "t": "peppermint",
        "k": "marzipans",
        "f": "blueberry",
        "F": "butternut",
        "E_k": "turnipseed",
        "\\eta_k": "saffronbud",
        "\\xi_k": "cinnamonbark",
        "\\theta_k": "cardamompod",
        "M": "moonlight",
        "n": "gingerroot"
      },
      "question": "8. Let \\( blueberry(sandstone) \\) be a differentiable function defined in the closed interval \\( (0,1) \\) and such that\n\\[\n\\left|blueberry^{\\prime}(sandstone)\\right| \\leq moonlight, \\quad 0<sandstone<1\n\\]\n\nProve that\n\\[\n\\left|\\int_{0}^{1} blueberry(sandstone) d sandstone-\\frac{1}{gingerroot} \\sum_{marzipans=1}^{gingerroot} blueberry\\left(\\frac{marzipans}{gingerroot}\\right)\\right| \\leq \\frac{moonlight}{gingerroot} .\n\\]",
      "solution": "Solution. Let\n\\[\nturnipseed=\\int_{(marzipans-1) / gingerroot}^{marzipans / gingerroot} blueberry(sandstone) d sandstone-\\frac{1}{gingerroot} blueberry\\left(\\frac{marzipans}{gingerroot}\\right)\n\\]\nfor \\( marzipans=1,2, \\ldots, gingerroot \\). Since \\( blueberry \\) is differentiable, it is continuous, and therefore by the mean value theorem for integrals there exists a number \\( saffronbud \\) such that\n\\[\n\\frac{marzipans-1}{gingerroot}<saffronbud<\\frac{marzipans}{gingerroot}\n\\]\nand\n\\[\n\\int_{(marzipans-1) / gingerroot}^{marzipans / gingerroot} blueberry(sandstone) d sandstone=\\frac{1}{gingerroot} blueberry\\left(saffronbud\\right) .\n\\]\n\nBy the mean value theorem for derivatives there exists a number \\( cinnamonbark \\) such that \\( saffronbud<cinnamonbark<(marzipans / gingerroot) \\) and\n\\[\nblueberry\\left(saffronbud\\right)-blueberry\\left(\\frac{marzipans}{gingerroot}\\right)=\\left(saffronbud-\\frac{marzipans}{gingerroot}\\right) blueberry^{\\prime}\\left(cinnamonbark\\right) .\n\\]\n\nThen\n\\[\n\\begin{aligned}\n\\left|turnipseed\\right| & =\\frac{1}{gingerroot}\\left|blueberry\\left(saffronbud\\right)-blueberry\\left(\\frac{marzipans}{gingerroot}\\right)\\right| \\\\\n& =\\frac{1}{gingerroot}\\left|saffronbud-\\frac{marzipans}{gingerroot}\\right| \\cdot\\left|blueberry^{\\prime}\\left(cinnamonbark\\right)\\right| \\leq \\frac{1}{gingerroot^{2}} moonlight .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left|\\int_{0}^{1} blueberry(sandstone) d sandstone-\\frac{1}{gingerroot} \\sum_{marzipans=1}^{gingerroot} blueberry\\left(\\frac{marzipans}{gingerroot}\\right)\\right|=\\left|\\sum_{marzipans=1}^{gingerroot} turnipseed\\right| \\leq \\frac{moonlight}{gingerroot} .\n\\]\n\nRemark. The estimate can be improved by a factor of two. Let\n\\[\nbutternut(sandstone)=\\int_{0}^{sandstone} blueberry(peppermint) d peppermint .\n\\]\n\nExpanding by Taylor's theorem about \\( marzipans / gingerroot \\), we obtain\n\\[\nbutternut\\left(\\frac{marzipans-1}{gingerroot}\\right)=butternut\\left(\\frac{marzipans}{gingerroot}\\right)+\\left(-\\frac{1}{gingerroot}\\right) butternut^{\\prime}\\left(\\frac{marzipans}{gingerroot}\\right)+\\frac{1}{2}\\left(-\\frac{1}{gingerroot}\\right)^{2} butternut^{\\prime \\prime}\\left(cardamompod\\right),\n\\]\nwhere\n\\[\n\\frac{marzipans-1}{gingerroot}<cardamompod<\\frac{marzipans}{gingerroot} .\n\\]\n\nSince \\( butternut^{\\prime}=blueberry \\), this becomes\n\\[\nturnipseed=-\\frac{1}{2 gingerroot^{2}} blueberry^{\\prime}\\left(cardamompod\\right)\n\\]\nand we have gained a factor of two."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "t": "steadystate",
        "k": "continuous",
        "f": "staticnumber",
        "F": "discreteform",
        "E_k": "exactvalue",
        "\\eta_k": "boundarypoint",
        "\\xi_k": "outerlimit",
        "\\theta_k": "finalpoint",
        "M": "variability",
        "n": "continuum"
      },
      "question": "8. Let \\( staticnumber(fixedvalue) \\) be a differentiable function defined in the closed interval \\( (0,1) \\) and such that\n\\[\n\\left|staticnumber^{\\prime}(fixedvalue)\\right| \\leq variability, \\quad 0<fixedvalue<1\n\\]\n\nProve that\n\\[\n\\left|\\int_{0}^{1} staticnumber(fixedvalue) d fixedvalue-\\frac{1}{continuum} \\sum_{continuous=1}^{continuum} staticnumber\\left(\\frac{continuous}{continuum}\\right)\\right| \\leq \\frac{variability}{continuum} .\n\\]",
      "solution": "Solution. Let\n\\[\nexactvalue=\\int_{(continuous-1) / continuum}^{continuous / continuum} staticnumber(fixedvalue) d fixedvalue-\\frac{1}{continuum} staticnumber\\left(\\frac{continuous}{continuum}\\right)\n\\]\nfor \\( continuous=1,2, \\ldots, continuum \\). Since \\( staticnumber \\) is differentiable, it is continuous, and therefore by the mean value theorem for integrals there exists a number \\( boundarypoint \\) such that\n\\[\n\\frac{continuous-1}{continuum}<boundarypoint<\\frac{continuous}{continuum}\n\\]\nand\n\\[\n\\int_{(continuous-1) / continuum}^{continuous / continuum} staticnumber(fixedvalue) d fixedvalue=\\frac{1}{continuum} staticnumber\\left(boundarypoint\\right) .\n\\]\n\nBy the mean value theorem for derivatives there exists a number \\( outerlimit \\) such that \\( boundarypoint<outerlimit<(continuous / continuum) \\) and\n\\[\nstaticnumber\\left(boundarypoint\\right)-staticnumber\\left(\\frac{continuous}{continuum}\\right)=\\left(boundarypoint-\\frac{continuous}{continuum}\\right) staticnumber^{\\prime}\\left(outerlimit\\right) .\n\\]\n\nThen\n\\[\n\\begin{aligned}\n\\left|exactvalue\\right| & =\\frac{1}{continuum}\\left|staticnumber\\left(boundarypoint\\right)-staticnumber\\left(\\frac{continuous}{continuum}\\right)\\right| \\\\\n& =\\frac{1}{continuum}\\left|boundarypoint-\\frac{continuous}{continuum}\\right| \\cdot\\left|staticnumber^{\\prime}\\left(outerlimit\\right)\\right| \\leq \\frac{1}{continuum^{2}} variability .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left|\\int_{0}^{1} staticnumber(fixedvalue) d fixedvalue-\\frac{1}{continuum} \\sum_{continuous=1}^{continuum} staticnumber\\left(\\frac{continuous}{continuum}\\right)\\right|=\\left|\\sum_{continuous=1}^{continuum} exactvalue\\right| \\leq \\frac{variability}{continuum} .\n\\]\n\nRemark. The estimate can be improved by a factor of two. Let\n\\[\ndiscreteform(fixedvalue)=\\int_{0}^{fixedvalue} staticnumber(steadystate) d steadystate .\n\\]\n\nExpanding by Taylor's theorem about \\( continuous / continuum \\), we obtain\n\\[\ndiscreteform\\left(\\frac{continuous-1}{continuum}\\right)=discreteform\\left(\\frac{continuous}{continuum}\\right)+\\left(-\\frac{1}{continuum}\\right) discreteform^{\\prime}\\left(\\frac{continuous}{continuum}\\right)+\\frac{1}{2}\\left(-\\frac{1}{continuum}\\right)^{2} discreteform^{\\prime \\prime}\\left(finalpoint\\right),\n\\]\nwhere\n\\[\n\\frac{continuous-1}{continuum}<finalpoint<\\frac{continuous}{continuum} .\n\\]\n\nSince \\( discreteform^{\\prime}=staticnumber \\), this becomes\n\\[\nexactvalue=-\\frac{1}{2 continuum^{2}} staticnumber^{\\prime}\\left(finalpoint\\right)\n\\]\nand we have gained a factor of two."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "t": "hjgrksla",
        "k": "mplqzxwe",
        "f": "nzxmbrte",
        "F": "cvpldakj",
        "E_k": "qlsrmnpo",
        "\\eta_k": "zpwrtghq",
        "\\xi_k": "rjklmhvg",
        "\\theta_k": "vnsldkju",
        "M": "pkrsngtw",
        "n": "qvnjxlsa"
      },
      "question": "Let \\( nzxmbrte(qzxwvtnp) \\) be a differentiable function defined in the closed interval \\( (0,1) \\) and such that\n\\[\n\\left|nzxmbrte^{\\prime}(qzxwvtnp)\\right| \\leq pkrsngtw, \\quad 0<qzxwvtnp<1\n\\]\n\nProve that\n\\[\n\\left|\\int_{0}^{1} nzxmbrte(qzxwvtnp) d qzxwvtnp-\\frac{1}{qvnjxlsa} \\sum_{mplqzxwe=1}^{qvnjxlsa} nzxmbrte\\left(\\frac{mplqzxwe}{qvnjxlsa}\\right)\\right| \\leq \\frac{pkrsngtw}{qvnjxlsa} .\n\\]",
      "solution": "Solution. Let\n\\[\nqlsrmnpo=\\int_{(mplqzxwe-1) / qvnjxlsa}^{mplqzxwe / qvnjxlsa} nzxmbrte(qzxwvtnp) d qzxwvtnp-\\frac{1}{qvnjxlsa} nzxmbrte\\left(\\frac{mplqzxwe}{qvnjxlsa}\\right)\n\\]\nfor \\( mplqzxwe=1,2, \\ldots, qvnjxlsa \\). Since \\( nzxmbrte \\) is differentiable, it is continuous, and therefore by the mean value theorem for integrals there exists a number \\( zpwrtghq \\) such that\n\\[\n\\frac{mplqzxwe-1}{qvnjxlsa}<zpwrtghq<\\frac{mplqzxwe}{qvnjxlsa}\n\\]\nand\n\\[\n\\int_{(mplqzxwe-1) / qvnjxlsa}^{mplqzxwe / qvnjxlsa} nzxmbrte(qzxwvtnp) d qzxwvtnp=\\frac{1}{qvnjxlsa} nzxmbrte\\left(zpwrtghq\\right) .\n\\]\n\nBy the mean value theorem for derivatives there exists a number \\( rjklmhvg \\) such that \\( zpwrtghq<rjklmhvg<(mplqzxwe / qvnjxlsa) \\) and\n\\[\nnzxmbrte\\left(zpwrtghq\\right)-nzxmbrte\\left(\\frac{mplqzxwe}{qvnjxlsa}\\right)=\\left(zpwrtghq-\\frac{mplqzxwe}{qvnjxlsa}\\right) nzxmbrte^{\\prime}\\left(rjklmhvg\\right) .\n\\]\n\nThen\n\\[\n\\begin{aligned}\n\\left|qlsrmnpo\\right| & =\\frac{1}{qvnjxlsa}\\left|nzxmbrte\\left(zpwrtghq\\right)-nzxmbrte\\left(\\frac{mplqzxwe}{qvnjxlsa}\\right)\\right| \\\\\n& =\\frac{1}{qvnjxlsa}\\left|zpwrtghq-\\frac{mplqzxwe}{qvnjxlsa}\\right| \\cdot\\left|nzxmbrte^{\\prime}\\left(rjklmhvg\\right)\\right| \\leq \\frac{1}{qvnjxlsa^{2}} pkrsngtw .\n\\end{aligned}\n\\]\n\nHence\n\\[\n\\left|\\int_{0}^{1} nzxmbrte(qzxwvtnp) d qzxwvtnp-\\frac{1}{qvnjxlsa} \\sum_{mplqzxwe=1}^{qvnjxlsa} nzxmbrte\\left(\\frac{mplqzxwe}{qvnjxlsa}\\right)\\right|=\\left|\\sum_{mplqzxwe=1}^{qvnjxlsa} qlsrmnpo\\right| \\leq \\frac{pkrsngtw}{qvnjxlsa} .\n\\]\n\nRemark. The estimate can be improved by a factor of two. Let\n\\[\ncvpldakj(qzxwvtnp)=\\int_{0}^{qzxwvtnp} nzxmbrte(hjgrksla) d hjgrksla .\n\\]\n\nExpanding by Taylor's theorem about \\( mplqzxwe / qvnjxlsa \\), we obtain\n\\[\ncvpldakj\\left(\\frac{mplqzxwe-1}{qvnjxlsa}\\right)=cvpldakj\\left(\\frac{mplqzxwe}{qvnjxlsa}\\right)+\\left(-\\frac{1}{qvnjxlsa}\\right) cvpldakj^{\\prime}\\left(\\frac{mplqzxwe}{qvnjxlsa}\\right)+\\frac{1}{2}\\left(-\\frac{1}{qvnjxlsa}\\right)^{2} cvpldakj^{\\prime \\prime}\\left(vnsldkju\\right),\n\\]\nwhere\n\\[\n\\frac{mplqzxwe-1}{qvnjxlsa}<vnsldkju<\\frac{mplqzxwe}{qvnjxlsa} .\n\\]\n\nSince \\( cvpldakj^{\\prime}=nzxmbrte \\), this becomes\n\\[\nqlsrmnpo=-\\frac{1}{2 qvnjxlsa^{2}} nzxmbrte^{\\prime}\\left(vnsldkju\\right)\n\\]\nand we have gained a factor of two."
    },
    "kernel_variant": {
      "question": "Let d be a fixed integer with d ≥ 2 and put h := 1/n for n ∈ ℕ.  \nIntroduce the uniform lattice  \n\n  G_n := { k h : k = (k₁,…,k_d), k_j = 0,1,…,n−1 } ⊂ [0,1]^d,\n\nand for every multi–index k the closed cube  \n\n  Q_k := ∏_{j=1}^{d} [k_j h,(k_j+1)h].\n\n(1)  (First–order rule)  \n Let f : [0,1]^d → ℝ be C¹ and assume  \n\n  |∂_j f(x)| ≤ M_j  (x ∈ [0,1]^d, j = 1,…,d).                                           \n\n Define the left–endpoint Riemann sum  \n\n  S_n(f) := h^d ∑_{k ∈ {0,…,n−1}^d} f(k h).                                                 (♠)\n\n Prove the sharp error estimate  \n\n  | ∫_[0,1]^d f(x) dx − S_n(f) | ≤ h·½·∑_{j=1}^{d} M_j.                                     (⋆)\n\n(2)  (First–order–corrected rule)  \n Suppose now f is C² and that for all 1 ≤ i,j ≤ d  \n\n  |∂²_{ij} f(x)| ≤ K_{ij}  (x ∈ [0,1]^d).\n\n Define  \n\n  T_n(f) := S_n(f) + h^{d+1} ½ ∑_{j=1}^{d}∑_{k} (∂_j f)(k h).                             (♣)\n\n Show that  \n\n  | ∫_[0,1]^d f − T_n(f) |  \n   ≤ h² · [ (1/6) ∑_{j=1}^{d} K_{jj}  +  (1/4) ∑_{1≤ i<j ≤d} K_{ij} ].                 (★★)\n\n(3)  Optimality of the constants  \n (a)  For each ε > 0 there exists a C¹ function whose error in (⋆) is at least  \n   (1−ε)·h·½ ∑_{j} M_j for every n.  \n (b)  For each ε > 0 there is a C² function for which the error in (★★) exceeds  \n\n   (1−ε)·h²·[ (1/6) ∑_{j} K_{jj} + (1/4) ∑_{i<j} K_{ij} ] for all n.\n\nHence none of the coefficients ½, 1/6 (diagonal) and 1/4 (off–diagonal) can be improved uniformly in n and d.",
      "solution": "Throughout we put h := 1/n and denote by ∫ the integral over [0,1]^d.  \nFor a cube Q_k we freely switch to the rescaled variable \n\n x = k h + h y, y = (y₁,…,y_d) ∈ [0,1]^d, dx = h^d dy.                                     (0)\n\n────────────────────────────────────────────────\n1. Proof of (⋆).\n\nFor every cube Q_k set  \n\n E_k := ∫_{Q_k} f(x) dx − h^d f(k h).                                                       (1)\n\nFix k and write x as in (0).  \nFor each y ∈ [0,1]^d consider the line γ(t) = k h + t h y  (0 ≤ t ≤ 1).  \nBy the one–dimensional mean–value theorem\n\n f(k h + h y) − f(k h) = ∫_0^1 h y · ∇f(γ(t)) dt.                                          (2)\n\nWith the bound |∂_j f| ≤ M_j we get  \n\n |f(k h + h y) − f(k h)| ≤ h ∑_{j=1}^{d} M_j y_j.                                         (3)\n\nInsert (3) into (1) and use (0):\n\n |E_k| ≤ h^d · h ∑_{j=1}^{d} M_j ∫_{[0,1]^d} y_j dy  \n    = h^{d+1} ½ ∑_{j=1}^{d} M_j, because ∫_0^1 y_j dy_j = ½.                       (4)\n\nSumming (4) over the n^d cubes yields  \n\n |∫ f − S_n(f)| ≤ n^d h^{d+1} ½ ∑_{j} M_j = h ½ ∑_{j} M_j,                     (5)\n\nwhich is exactly the estimate (⋆).  \nSharpness is obtained from the linear functions f_j(x) = M_j x_j as before.\n\n────────────────────────────────────────────────\n2. Proof of (★★).\n\nFor every k let  \n\n F_k(y) := f(k h + h y) − f(k h) − (h/2) ∑_{j=1}^{d} ∂_j f(k h).                          (6)\n\nBecause of (0) the error over Q_k is  \n\n E_k := ∫_{Q_k} f(x) dx − h^d f(k h) − h^{d+1} ½ ∑_{j} ∂_j f(k h)  \n      = h^d ∫_{[0,1]^d} F_k(y) dy.                                                        (7)\n\nStep 1.  Second-order expansion along a line.  \nFor fixed y set g(t) = f(k h + t h y) (0 ≤ t ≤ 1).  \nTwice applying the one–dimensional mean–value theorem gives\n\n g(1) = g(0) + h ∑_{j} y_j ∂_j f(k h)  \n           + ½ h² ∑_{i,j} y_i y_j ∂²_{ij} f(k h + θ_{y} h y),                             (8)\n\nfor some θ_{y} ∈ (0,1) depending on y.  \nSubtracting f(k h) + (h/2)∑_j ∂_j f(k h) and comparing with (6) we obtain\n\n F_k(y) = h ∑_{j} (y_j − ½) ∂_j f(k h)  \n        + ½ h² ∑_{i,j} y_i y_j ∂²_{ij} f(k h + θ_{y} h y).                               (9)\n\nStep 2.  Averaging over the unit cube.  \nBecause ∫_0^1 (y_j − ½) dy_j = 0, the first (gradient) term in (9) vanishes after\nintegration in y.  Hence\n\n E_k = h^{d+2}/2 ∑_{i,j} ∂²_{ij} f(k h + θ_{y} h y) ∫_{[0,1]^d} y_i y_j dy.              (10)\n\nCompute the elementary integrals  \n\n ∫_{[0,1]^d} y_i² dy = 1/3,  ∫_{[0,1]^d} y_i y_j dy = 1/4 (i ≠ j).                    (11)\n\nUsing the uniform bounds |∂²_{ij} f| ≤ K_{ij} in (10)–(11) yields\n\n |E_k| ≤ h^{d+2} [ (1/6) ∑_{j=1}^{d} K_{jj} + (1/4) ∑_{1≤ i<j ≤d} K_{ij} ].          (12)\n\nStep 3.  Summation over all cubes.  \nSince there are n^d cubes and n^d h^{d+2} = h², we finally get\n\n |∫ f − T_n(f)| ≤ h² [ (1/6) ∑_{j} K_{jj} + (1/4) ∑_{i<j} K_{ij} ],                  (13)\n\nwhich is the desired inequality (★★).\n\n────────────────────────────────────────────────\n3. Optimality of the constants.\n\n(a) Diagonal term.  \nFor a fixed r define f(x) = ½ K_{rr} x_r².  Then ∂²_{rr}f ≡ K_{rr}, all other\nsecond derivatives vanish, and a direct calculation with n = 1 gives\n\n |∫ f − T_1(f)| = K_{rr}/6 = h² K_{rr}/6,                                                (14)\n\nso 1/6 cannot be reduced.\n\n(b) Off–diagonal term.  \nFor p < q set f(x) = K_{pq} x_p x_q.  Here ∂²_{pq}f ≡ K_{pq}; with n = 1 we find\n\n |∫ f − T_1(f)| = K_{pq}/4 = h² K_{pq}/4,                                                (15)\n\nproving that 1/4 is best possible.  \nSmooth bump functions agreeing with these polynomials on the interior of [0,1]^d\ntransfer the conclusion to every n, completing the proof.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.411940",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension.  The problem moves from a single interval to a d-cube, \n  introducing multi–indices, facet geometry, and mixed partial derivatives.  \n• Additional constraints.  Separate bounds for every first and second partial \n  derivative must be tracked through the argument.  \n• Two distinct quadrature rules are analysed, the second demanding a full \n  second–order Taylor expansion with multidimensional remainder-estimates.  \n• Optimal constants are proved, requiring construction of extremal functions \n  and ε-perturbation reasoning, not just an a-priori bound.  \n• The solution combines:  \n  – Multidimensional FTC/Taylor with integral remainder,  \n  – Fubini-type decompositions over a cubical mesh,  \n  – Careful bookkeeping of combinatorial factors arising from mixed terms,  \n  – Sharpness/optimality arguments (a mini extremal problem).  \n\nThese layers of technique and theory go well beyond the one-dimensional mean-value argument in the original and the current kernel variant, making the enhanced variant substantially more challenging."
      }
    },
    "original_kernel_variant": {
      "question": "Let d be a fixed integer with d ≥ 2 and put h := 1/n for n ∈ ℕ.  \nIntroduce the uniform lattice  \n\n  G_n := { k h : k = (k₁,…,k_d), k_j = 0,1,…,n−1 } ⊂ [0,1]^d,\n\nand for every multi–index k the closed cube  \n\n  Q_k := ∏_{j=1}^{d} [k_j h,(k_j+1)h].\n\n(1)  (First–order rule)  \n Let f : [0,1]^d → ℝ be C¹ and assume  \n\n  |∂_j f(x)| ≤ M_j  (x ∈ [0,1]^d, j = 1,…,d).                                           \n\n Define the left–endpoint Riemann sum  \n\n  S_n(f) := h^d ∑_{k ∈ {0,…,n−1}^d} f(k h).                                                 (♠)\n\n Prove the sharp error estimate  \n\n  | ∫_[0,1]^d f(x) dx − S_n(f) | ≤ h·½·∑_{j=1}^{d} M_j.                                     (⋆)\n\n(2)  (First–order–corrected rule)  \n Suppose now f is C² and that for all 1 ≤ i,j ≤ d  \n\n  |∂²_{ij} f(x)| ≤ K_{ij}  (x ∈ [0,1]^d).\n\n Define  \n\n  T_n(f) := S_n(f) + h^{d+1} ½ ∑_{j=1}^{d}∑_{k} (∂_j f)(k h).                             (♣)\n\n Show that  \n\n  | ∫_[0,1]^d f − T_n(f) |  \n   ≤ h² · [ (1/6) ∑_{j=1}^{d} K_{jj}  +  (1/4) ∑_{1≤ i<j ≤d} K_{ij} ].                 (★★)\n\n(3)  Optimality of the constants  \n (a)  For each ε > 0 there exists a C¹ function whose error in (⋆) is at least  \n   (1−ε)·h·½ ∑_{j} M_j for every n.  \n (b)  For each ε > 0 there is a C² function for which the error in (★★) exceeds  \n\n   (1−ε)·h²·[ (1/6) ∑_{j} K_{jj} + (1/4) ∑_{i<j} K_{ij} ] for all n.\n\nHence none of the coefficients ½, 1/6 (diagonal) and 1/4 (off–diagonal) can be improved uniformly in n and d.",
      "solution": "Throughout we put h := 1/n and denote by ∫ the integral over [0,1]^d.  \nFor a cube Q_k we freely switch to the rescaled variable \n\n x = k h + h y, y = (y₁,…,y_d) ∈ [0,1]^d, dx = h^d dy.                                     (0)\n\n────────────────────────────────────────────────\n1. Proof of (⋆).\n\nFor every cube Q_k set  \n\n E_k := ∫_{Q_k} f(x) dx − h^d f(k h).                                                       (1)\n\nFix k and write x as in (0).  \nFor each y ∈ [0,1]^d consider the line γ(t) = k h + t h y  (0 ≤ t ≤ 1).  \nBy the one–dimensional mean–value theorem\n\n f(k h + h y) − f(k h) = ∫_0^1 h y · ∇f(γ(t)) dt.                                          (2)\n\nWith the bound |∂_j f| ≤ M_j we get  \n\n |f(k h + h y) − f(k h)| ≤ h ∑_{j=1}^{d} M_j y_j.                                         (3)\n\nInsert (3) into (1) and use (0):\n\n |E_k| ≤ h^d · h ∑_{j=1}^{d} M_j ∫_{[0,1]^d} y_j dy  \n    = h^{d+1} ½ ∑_{j=1}^{d} M_j, because ∫_0^1 y_j dy_j = ½.                       (4)\n\nSumming (4) over the n^d cubes yields  \n\n |∫ f − S_n(f)| ≤ n^d h^{d+1} ½ ∑_{j} M_j = h ½ ∑_{j} M_j,                     (5)\n\nwhich is exactly the estimate (⋆).  \nSharpness is obtained from the linear functions f_j(x) = M_j x_j as before.\n\n────────────────────────────────────────────────\n2. Proof of (★★).\n\nFor every k let  \n\n F_k(y) := f(k h + h y) − f(k h) − (h/2) ∑_{j=1}^{d} ∂_j f(k h).                          (6)\n\nBecause of (0) the error over Q_k is  \n\n E_k := ∫_{Q_k} f(x) dx − h^d f(k h) − h^{d+1} ½ ∑_{j} ∂_j f(k h)  \n      = h^d ∫_{[0,1]^d} F_k(y) dy.                                                        (7)\n\nStep 1.  Second-order expansion along a line.  \nFor fixed y set g(t) = f(k h + t h y) (0 ≤ t ≤ 1).  \nTwice applying the one–dimensional mean–value theorem gives\n\n g(1) = g(0) + h ∑_{j} y_j ∂_j f(k h)  \n           + ½ h² ∑_{i,j} y_i y_j ∂²_{ij} f(k h + θ_{y} h y),                             (8)\n\nfor some θ_{y} ∈ (0,1) depending on y.  \nSubtracting f(k h) + (h/2)∑_j ∂_j f(k h) and comparing with (6) we obtain\n\n F_k(y) = h ∑_{j} (y_j − ½) ∂_j f(k h)  \n        + ½ h² ∑_{i,j} y_i y_j ∂²_{ij} f(k h + θ_{y} h y).                               (9)\n\nStep 2.  Averaging over the unit cube.  \nBecause ∫_0^1 (y_j − ½) dy_j = 0, the first (gradient) term in (9) vanishes after\nintegration in y.  Hence\n\n E_k = h^{d+2}/2 ∑_{i,j} ∂²_{ij} f(k h + θ_{y} h y) ∫_{[0,1]^d} y_i y_j dy.              (10)\n\nCompute the elementary integrals  \n\n ∫_{[0,1]^d} y_i² dy = 1/3,  ∫_{[0,1]^d} y_i y_j dy = 1/4 (i ≠ j).                    (11)\n\nUsing the uniform bounds |∂²_{ij} f| ≤ K_{ij} in (10)–(11) yields\n\n |E_k| ≤ h^{d+2} [ (1/6) ∑_{j=1}^{d} K_{jj} + (1/4) ∑_{1≤ i<j ≤d} K_{ij} ].          (12)\n\nStep 3.  Summation over all cubes.  \nSince there are n^d cubes and n^d h^{d+2} = h², we finally get\n\n |∫ f − T_n(f)| ≤ h² [ (1/6) ∑_{j} K_{jj} + (1/4) ∑_{i<j} K_{ij} ],                  (13)\n\nwhich is the desired inequality (★★).\n\n────────────────────────────────────────────────\n3. Optimality of the constants.\n\n(a) Diagonal term.  \nFor a fixed r define f(x) = ½ K_{rr} x_r².  Then ∂²_{rr}f ≡ K_{rr}, all other\nsecond derivatives vanish, and a direct calculation with n = 1 gives\n\n |∫ f − T_1(f)| = K_{rr}/6 = h² K_{rr}/6,                                                (14)\n\nso 1/6 cannot be reduced.\n\n(b) Off–diagonal term.  \nFor p < q set f(x) = K_{pq} x_p x_q.  Here ∂²_{pq}f ≡ K_{pq}; with n = 1 we find\n\n |∫ f − T_1(f)| = K_{pq}/4 = h² K_{pq}/4,                                                (15)\n\nproving that 1/4 is best possible.  \nSmooth bump functions agreeing with these polynomials on the interior of [0,1]^d\ntransfer the conclusion to every n, completing the proof.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.354801",
        "was_fixed": false,
        "difficulty_analysis": "• Higher dimension.  The problem moves from a single interval to a d-cube, \n  introducing multi–indices, facet geometry, and mixed partial derivatives.  \n• Additional constraints.  Separate bounds for every first and second partial \n  derivative must be tracked through the argument.  \n• Two distinct quadrature rules are analysed, the second demanding a full \n  second–order Taylor expansion with multidimensional remainder-estimates.  \n• Optimal constants are proved, requiring construction of extremal functions \n  and ε-perturbation reasoning, not just an a-priori bound.  \n• The solution combines:  \n  – Multidimensional FTC/Taylor with integral remainder,  \n  – Fubini-type decompositions over a cubical mesh,  \n  – Careful bookkeeping of combinatorial factors arising from mixed terms,  \n  – Sharpness/optimality arguments (a mini extremal problem).  \n\nThese layers of technique and theory go well beyond the one-dimensional mean-value argument in the original and the current kernel variant, making the enhanced variant substantially more challenging."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}