{
  "index": "1954-B-5",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "5. Let \\( f(x) \\) be a real-valued function, defined for \\( -1<x<1 \\), such that \\( f^{\\prime}(0) \\) exists. Let \\( a_{n}, b_{n} \\) be two sequences such that\n\\[\n-1<a_{n}<0<b_{n}<1, \\lim _{n \\rightarrow \\infty} a_{n}=0, \\lim _{n \\rightarrow \\infty} b_{n}=0\n\\]\n\nProve that\n\\[\n\\lim _{n \\rightarrow \\infty} \\frac{f\\left(b_{n}\\right)-f\\left(a_{n}\\right)}{b_{n}-a_{n}}=f^{\\prime}(0) . \\quad \\quad \\quad \\text { page 397) }\n\\]",
  "solution": "Solution. Let \\( \\epsilon>0 \\) be given. Then we can choose \\( \\delta, 0<\\delta<1 \\), so that for all \\( x \\) with \\( 0<|x|<\\delta \\),\n\\[\n\\left|\\frac{f(x)-f(0)}{x}-f^{\\prime}(0)\\right|<\\epsilon .\n\\]\n\nLet \\( k \\) be chosen so that, for all \\( n \\geq k \\),\n\\[\n\\left|a_{n}\\right|<\\delta, \\quad \\text { and } \\quad\\left|b_{n}\\right|<\\delta .\n\\]\n\nThen for \\( m \\geq k \\) we have\n\\[\n\\left|\\frac{f\\left(a_{m}\\right)-f(0)}{a_{m}}-f^{\\prime}(0)\\right|<\\epsilon, \\quad\\left|\\frac{f\\left(b_{m}\\right)-f(0)}{b_{m}}-f^{\\prime}(0)\\right|<\\epsilon .\n\\]\n\nBut\n\\[\n\\begin{aligned}\n\\mid f\\left(b_{m}\\right)-f\\left(a_{m}\\right) & -\\left(b_{m}-a_{m}\\right) f^{\\prime}(0) \\mid \\\\\n& =\\left|\\left(f\\left(b_{m}\\right)-f(0)-b_{m} f^{\\prime}(0)\\right)-\\left(f\\left(a_{m}\\right)-f(0)-a_{m} f^{\\prime}(0)\\right)\\right| \\\\\n& \\leq\\left|f\\left(b_{m}\\right)-f(0)-b_{m} f^{\\prime}(0)\\right|+\\left|f\\left(a_{m}\\right)-f(0)-a_{m} f^{\\prime}(0)\\right| \\\\\n& \\leq \\epsilon\\left|b_{m}\\right|+\\epsilon\\left|a_{m}\\right|=\\epsilon\\left(b_{m}-a_{m}\\right) .\n\\end{aligned}\n\\]\n\nThe penultimate step follows from (1) and the last from \\( a_{m}<0<b_{m} \\).\nTherefore\n\\[\n\\left|\\frac{f\\left(b_{m}\\right)-f\\left(a_{m}\\right)}{b_{m}-a_{m}}-f^{\\prime}(0)\\right|<\\epsilon\n\\]\nfor all \\( m>k \\). Since this is true for any \\( \\epsilon>0 \\), we have proved that\n\\[\n\\lim _{n \\rightarrow \\infty} \\frac{f\\left(b_{n}\\right)-f\\left(a_{n}\\right)}{b_{n}-a_{n}}=f^{\\prime}(0) .\n\\]",
  "vars": [
    "x",
    "n",
    "m",
    "k",
    "a_n",
    "b_n"
  ],
  "params": [
    "f",
    "\\\\delta",
    "\\\\epsilon"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "varitem",
        "n": "indexer",
        "m": "subindex",
        "k": "threshold",
        "a_n": "seqleft",
        "b_n": "seqright",
        "f": "function",
        "\\delta": "smallrange",
        "\\epsilon": "accuracy"
      },
      "question": "5. Let \\( function(varitem) \\) be a real-valued function, defined for \\( -1<varitem<1 \\), such that \\( function^{\\prime}(0) \\) exists. Let \\( seqleft_{indexer}, seqright_{indexer} \\) be two sequences such that\n\\[\n-1<seqleft_{indexer}<0<seqright_{indexer}<1, \\lim _{indexer \\rightarrow \\infty} seqleft_{indexer}=0, \\lim _{indexer \\rightarrow \\infty} seqright_{indexer}=0\n\\]\n\nProve that\n\\[\n\\lim _{indexer \\rightarrow \\infty} \\frac{function\\left(seqright_{indexer}\\right)-function\\left(seqleft_{indexer}\\right)}{seqright_{indexer}-seqleft_{indexer}}=function^{\\prime}(0) . \\quad \\quad \\quad \\text { page 397) }\n\\]",
      "solution": "Solution. Let \\( accuracy>0 \\) be given. Then we can choose \\( smallrange, 0<smallrange<1 \\), so that for all \\( varitem \\) with \\( 0<|varitem|<smallrange \\),\n\\[\n\\left|\\frac{function(varitem)-function(0)}{varitem}-function^{\\prime}(0)\\right|<accuracy .\n\\]\n\nLet \\( threshold \\) be chosen so that, for all \\( indexer \\geq threshold \\),\n\\[\n\\left|seqleft_{indexer}\\right|<smallrange, \\quad \\text { and } \\quad\\left|seqright_{indexer}\\right|<smallrange .\n\\]\n\nThen for \\( subindex \\geq threshold \\) we have\n\\[\n\\left|\\frac{function\\left(seqleft_{subindex}\\right)-function(0)}{seqleft_{subindex}}-function^{\\prime}(0)\\right|<accuracy, \\quad\\left|\\frac{function\\left(seqright_{subindex}\\right)-function(0)}{seqright_{subindex}}-function^{\\prime}(0)\\right|<accuracy .\n\\]\n\nBut\n\\[\n\\begin{aligned}\n\\mid function\\left(seqright_{subindex}\\right)-function\\left(seqleft_{subindex}\\right) & -\\left(seqright_{subindex}-seqleft_{subindex}\\right) function^{\\prime}(0) \\mid \\\\\n& =\\left|\\left(function\\left(seqright_{subindex}\\right)-function(0)-seqright_{subindex} function^{\\prime}(0)\\right)-\\left(function\\left(seqleft_{subindex}\\right)-function(0)-seqleft_{subindex} function^{\\prime}(0)\\right)\\right| \\\\\n& \\leq\\left|function\\left(seqright_{subindex}\\right)-function(0)-seqright_{subindex} function^{\\prime}(0)\\right|+\\left|function\\left(seqleft_{subindex}\\right)-function(0)-seqleft_{subindex} function^{\\prime}(0)\\right| \\\\\n& \\leq accuracy\\left|seqright_{subindex}\\right|+accuracy\\left|seqleft_{subindex}\\right|=accuracy\\left(seqright_{subindex}-seqleft_{subindex}\\right) .\n\\end{aligned}\n\\]\n\nThe penultimate step follows from (1) and the last from \\( seqleft_{subindex}<0<seqright_{subindex} \\).\nTherefore\n\\[\n\\left|\\frac{function\\left(seqright_{subindex}\\right)-function\\left(seqleft_{subindex}\\right)}{seqright_{subindex}-seqleft_{subindex}}-function^{\\prime}(0)\\right|<accuracy\n\\]\nfor all \\( subindex>threshold \\). Since this is true for any \\( accuracy>0 \\), we have proved that\n\\[\n\\lim _{indexer \\rightarrow \\infty} \\frac{function\\left(seqright_{indexer}\\right)-function\\left(seqleft_{indexer}\\right)}{seqright_{indexer}-seqleft_{indexer}}=function^{\\prime}(0) .\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "lampposts",
        "n": "raincloud",
        "m": "sunflower",
        "k": "dandelion",
        "a_n": "breadcrumb",
        "b_n": "marigolds",
        "f": "hedgehog",
        "\\\\delta": "rainwater",
        "\\\\epsilon": "blueberry"
      },
      "question": "5. Let \\( hedgehog(lampposts) \\) be a real-valued function, defined for \\( -1<lampposts<1 \\), such that \\( hedgehog^{\\prime}(0) \\) exists. Let \\( breadcrumb_{raincloud}, marigolds_{raincloud} \\) be two sequences such that\n\\[\n-1<breadcrumb_{raincloud}<0<marigolds_{raincloud}<1, \\lim _{raincloud \\rightarrow \\infty} breadcrumb_{raincloud}=0, \\lim _{raincloud \\rightarrow \\infty} marigolds_{raincloud}=0\n\\]\n\nProve that\n\\[\n\\lim _{raincloud \\rightarrow \\infty} \\frac{hedgehog\\left(marigolds_{raincloud}\\right)-hedgehog\\left(breadcrumb_{raincloud}\\right)}{marigolds_{raincloud}-breadcrumb_{raincloud}}=hedgehog^{\\prime}(0) . \\quad \\quad \\quad \\text { page 397) }\n\\]",
      "solution": "Solution. Let \\( blueberry>0 \\) be given. Then we can choose \\( rainwater, 0<rainwater<1 \\), so that for all \\( lampposts \\) with \\( 0<|lampposts|<rainwater \\),\n\\[\n\\left|\\frac{hedgehog(lampposts)-hedgehog(0)}{lampposts}-hedgehog^{\\prime}(0)\\right|<blueberry .\n\\]\n\nLet \\( dandelion \\) be chosen so that, for all \\( raincloud \\geq dandelion \\),\n\\[\n\\left|breadcrumb_{raincloud}\\right|<rainwater, \\quad \\text { and } \\quad\\left|marigolds_{raincloud}\\right|<rainwater .\n\\]\n\nThen for \\( sunflower \\geq dandelion \\) we have\n\\[\n\\left|\\frac{hedgehog\\left(breadcrumb_{sunflower}\\right)-hedgehog(0)}{breadcrumb_{sunflower}}-hedgehog^{\\prime}(0)\\right|<blueberry, \\quad\\left|\\frac{hedgehog\\left(marigolds_{sunflower}\\right)-hedgehog(0)}{marigolds_{sunflower}}-hedgehog^{\\prime}(0)\\right|<blueberry .\n\\]\n\nBut\n\\[\n\\begin{aligned}\n\\mid hedgehog\\left(marigolds_{sunflower}\\right)-hedgehog\\left(breadcrumb_{sunflower}\\right) & -\\left(marigolds_{sunflower}-breadcrumb_{sunflower}\\right) hedgehog^{\\prime}(0) \\mid \\\\\n& =\\left|\\left(hedgehog\\left(marigolds_{sunflower}\\right)-hedgehog(0)-marigolds_{sunflower} hedgehog^{\\prime}(0)\\right)-\\left(hedgehog\\left(breadcrumb_{sunflower}\\right)-hedgehog(0)-breadcrumb_{sunflower} hedgehog^{\\prime}(0)\\right)\\right| \\\\\n& \\leq\\left|hedgehog\\left(marigolds_{sunflower}\\right)-hedgehog(0)-marigolds_{sunflower} hedgehog^{\\prime}(0)\\right|+\\left|hedgehog\\left(breadcrumb_{sunflower}\\right)-hedgehog(0)-breadcrumb_{sunflower} hedgehog^{\\prime}(0)\\right| \\\\\n& \\leq blueberry\\left|marigolds_{sunflower}\\right|+blueberry\\left|breadcrumb_{sunflower}\\right|=blueberry\\left(marigolds_{sunflower}-breadcrumb_{sunflower}\\right) .\n\\end{aligned}\n\\]\n\nThe penultimate step follows from (1) and the last from \\( breadcrumb_{sunflower}<0<marigolds_{sunflower} \\).\nTherefore\n\\[\n\\left|\\frac{hedgehog\\left(marigolds_{sunflower}\\right)-hedgehog\\left(breadcrumb_{sunflower}\\right)}{marigolds_{sunflower}-breadcrumb_{sunflower}}-hedgehog^{\\prime}(0)\\right|<blueberry\n\\]\nfor all \\( sunflower>dandelion \\). Since this is true for any \\( blueberry>0 \\), we have proved that\n\\[\n\\lim _{raincloud \\rightarrow \\infty} \\frac{hedgehog\\left(marigolds_{raincloud}\\right)-hedgehog\\left(breadcrumb_{raincloud}\\right)}{marigolds_{raincloud}-breadcrumb_{raincloud}}=hedgehog^{\\prime}(0) .\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "n": "fixedcount",
        "m": "rigidmark",
        "k": "motionless",
        "a_n": "positivepeak",
        "b_n": "negativepit",
        "f": "malfunction",
        "\\delta": "largespread",
        "\\epsilon": "vastamount"
      },
      "question": "5. Let \\( malfunction(constantval) \\) be a real-valued function, defined for \\( -1<constantval<1 \\), such that \\( malfunction^{\\prime}(0) \\) exists. Let \\( positivepeak, negativepit \\) be two sequences such that\n\\[\n-1<positivepeak<0<negativepit<1, \\lim _{fixedcount \\rightarrow \\infty} positivepeak=0, \\lim _{fixedcount \\rightarrow \\infty} negativepit=0\n\\]\n\nProve that\n\\[\n\\lim _{fixedcount \\rightarrow \\infty} \\frac{malfunction\\left(negativepit\\right)-malfunction\\left(positivepeak\\right)}{negativepit-positivepeak}=malfunction^{\\prime}(0) . \\quad \\quad \\quad \\text { page 397) }\n\\]",
      "solution": "Solution. Let \\( vastamount>0 \\) be given. Then we can choose \\( largespread, 0<largespread<1 \\), so that for all \\( constantval \\) with \\( 0<|constantval|<largespread \\),\n\\[\n\\left|\\frac{malfunction(constantval)-malfunction(0)}{constantval}-malfunction^{\\prime}(0)\\right|<vastamount .\n\\]\n\nLet \\( motionless \\) be chosen so that, for all \\( fixedcount \\geq motionless \\),\n\\[\n\\left|positivepeak\\right|<largespread, \\quad \\text { and } \\quad\\left|negativepit\\right|<largespread .\n\\]\n\nThen for \\( rigidmark \\geq motionless \\) we have\n\\[\n\\left|\\frac{malfunction\\left(positivepeak\\right)-malfunction(0)}{positivepeak}-malfunction^{\\prime}(0)\\right|<vastamount, \\quad\\left|\\frac{malfunction\\left(negativepit\\right)-malfunction(0)}{negativepit}-malfunction^{\\prime}(0)\\right|<vastamount .\n\\]\n\nBut\n\\[\n\\begin{aligned}\n\\mid malfunction\\left(negativepit\\right)-malfunction\\left(positivepeak\\right) & -\\left(negativepit-positivepeak\\right) malfunction^{\\prime}(0) \\mid \\\\\n& =\\left|\\left(malfunction\\left(negativepit\\right)-malfunction(0)-negativepit malfunction^{\\prime}(0)\\right)-\\left(malfunction\\left(positivepeak\\right)-malfunction(0)-positivepeak malfunction^{\\prime}(0)\\right)\\right| \\\\\n& \\leq\\left|malfunction\\left(negativepit\\right)-malfunction(0)-negativepit malfunction^{\\prime}(0)\\right|+\\left|malfunction\\left(positivepeak\\right)-malfunction(0)-positivepeak malfunction^{\\prime}(0)\\right| \\\\\n& \\leq vastamount\\left|negativepit\\right|+vastamount\\left|positivepeak\\right|=vastamount\\left(negativepit-positivepeak\\right) .\n\\end{aligned}\n\\]\n\nThe penultimate step follows from (1) and the last from \\( positivepeak<0<negativepit \\).\nTherefore\n\\[\n\\left|\\frac{malfunction\\left(negativepit\\right)-malfunction\\left(positivepeak\\right)}{negativepit-positivepeak}-malfunction^{\\prime}(0)\\right|<vastamount\n\\]\nfor all \\( rigidmark>motionless \\). Since this is true for any \\( vastamount>0 \\), we have proved that\n\\[\n\\lim _{fixedcount \\rightarrow \\infty} \\frac{malfunction\\left(negativepit\\right)-malfunction\\left(positivepeak\\right)}{negativepit-positivepeak}=malfunction^{\\prime}(0) .\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "n": "hjgrksla",
        "m": "vdlqczno",
        "k": "tmbsjepr",
        "a_n": "fgdmrkyu",
        "b_n": "wqcslzha",
        "f": "lbntxeyo",
        "\\delta": "gvtmskui",
        "\\epsilon": "zpjrwhad"
      },
      "question": "5. Let \\( lbntxeyo(qzxwvtnp) \\) be a real-valued function, defined for \\( -1<qzxwvtnp<1 \\), such that \\( lbntxeyo^{\\prime}(0) \\) exists. Let \\( fgdmrkyu, wqcslzha \\) be two sequences such that\n\\[\n-1<fgdmrkyu<0<wqcslzha<1, \\lim _{hjgrksla \\rightarrow \\infty} fgdmrkyu=0, \\lim _{hjgrksla \\rightarrow \\infty} wqcslzha=0\n\\]\n\nProve that\n\\[\n\\lim _{hjgrksla \\rightarrow \\infty} \\frac{lbntxeyo\\left(wqcslzha\\right)-lbntxeyo\\left(fgdmrkyu\\right)}{wqcslzha-fgdmrkyu}=lbntxeyo^{\\prime}(0) . \\quad \\quad \\quad \\text { page 397) }\n\\]",
      "solution": "Solution. Let \\( zpjrwhad>0 \\) be given. Then we can choose \\( gvtmskui, 0<gvtmskui<1 \\), so that for all \\( qzxwvtnp \\) with \\( 0<|qzxwvtnp|<gvtmskui \\),\n\\[\n\\left|\\frac{lbntxeyo(qzxwvtnp)-lbntxeyo(0)}{qzxwvtnp}-lbntxeyo^{\\prime}(0)\\right|<zpjrwhad .\n\\]\n\nLet \\( tmbsjepr \\) be chosen so that, for all \\( hjgrksla \\geq tmbsjepr \\),\n\\[\n\\left|fgdmrkyu\\right|<gvtmskui, \\quad \\text { and } \\quad\\left|wqcslzha\\right|<gvtmskui .\n\\]\n\nThen for \\( vdlqczno \\geq tmbsjepr \\) we have\n\\[\n\\left|\\frac{lbntxeyo\\left(fgdmrkyu\\right)-lbntxeyo(0)}{fgdmrkyu}-lbntxeyo^{\\prime}(0)\\right|<zpjrwhad, \\quad\\left|\\frac{lbntxeyo\\left(wqcslzha\\right)-lbntxeyo(0)}{wqcslzha}-lbntxeyo^{\\prime}(0)\\right|<zpjrwhad .\n\\]\n\nBut\n\\[\n\\begin{aligned}\n\\mid lbntxeyo\\left(wqcslzha\\right)-lbntxeyo\\left(fgdmrkyu\\right) & -\\left(wqcslzha-fgdmrkyu\\right) lbntxeyo^{\\prime}(0) \\mid \\\\\n& =\\left|\\left(lbntxeyo\\left(wqcslzha\\right)-lbntxeyo(0)-wqcslzha\\, lbntxeyo^{\\prime}(0)\\right)-\\left(lbntxeyo\\left(fgdmrkyu\\right)-lbntxeyo(0)-fgdmrkyu\\, lbntxeyo^{\\prime}(0)\\right)\\right| \\\\\n& \\leq\\left|lbntxeyo\\left(wqcslzha\\right)-lbntxeyo(0)-wqcslzha\\, lbntxeyo^{\\prime}(0)\\right|+\\left|lbntxeyo\\left(fgdmrkyu\\right)-lbntxeyo(0)-fgdmrkyu\\, lbntxeyo^{\\prime}(0)\\right| \\\\\n& \\leq zpjrwhad\\left|wqcslzha\\right|+zpjrwhad\\left|fgdmrkyu\\right|=zpjrwhad\\left(wqcslzha-fgdmrkyu\\right) .\n\\end{aligned}\n\\]\n\nThe penultimate step follows from (1) and the last from \\( fgdmrkyu<0<wqcslzha \\).\nTherefore\n\\[\n\\left|\\frac{lbntxeyo\\left(wqcslzha\\right)-lbntxeyo\\left(fgdmrkyu\\right)}{wqcslzha-fgdmrkyu}-lbntxeyo^{\\prime}(0)\\right|<zpjrwhad\n\\]\nfor all \\( vdlqczno>tmbsjepr \\). Since this is true for any \\( zpjrwhad>0 \\), we have proved that\n\\[\n\\lim _{hjgrksla \\rightarrow \\infty} \\frac{lbntxeyo\\left(wqcslzha\\right)-lbntxeyo\\left(fgdmrkyu\\right)}{wqcslzha-fgdmrkyu}=lbntxeyo^{\\prime}(0) .\n\\]"
    },
    "kernel_variant": {
      "question": "Let k ≥ 1 be an integer and let  \n\n  f : ℝᵏ ⟶ ℝ  \n\nbe defined on an open neighbourhood of the origin.  \nAssume that f is twice Fréchet–differentiable at 0; i.e. there exist  \n\n a linear map A : ℝᵏ ⟶ ℝ   and a symmetric bilinear form B : ℝᵏ × ℝᵏ ⟶ ℝ  \n\nsuch that  \n\n  lim_{h→0}  [ f(h) − f(0) − A(h) − ½ B(h,h) ] ⁄ ‖h‖² = 0. (★)\n\nFor two sequences {a_n}, {b_n} ⊂ ℝᵏ suppose  \n\n(i) a_n ≠ 0, b_n ≠ 0 for every n;  \n(ii) a_n → 0 and b_n → 0;  \n(iii) r_n := ‖b_n − a_n‖ ⟶ 0;  \n(iv) u_n := (b_n − a_n) ⁄ r_n ⟶ u  with ‖u‖ = 1;  \n(v) m_n := (a_n + b_n)/2 satisfies ‖m_n‖ = o(r_n)  (the midpoint approaches 0 much faster than the chord–length).\n\nShow that  \n\n  lim_{n→∞} 4 · [ f(b_n) − 2 f(m_n) + f(a_n) ] ⁄ r_n² = B(u,u). (1)\n\nIn words: after the appropriate normalisation by the factor 4, the discrete second difference of f along any shrinking chord whose midpoint collapses sufficiently fast to the base point reproduces the quadratic form determined by the Hessian at 0.\n\n——————————————————————————————————————————",
      "solution": "Step 0.  Basic notation  \nPut  \n\n h_n := b_n − a_n (so that r_n = ‖h_n‖ and u_n = h_n / r_n),   \n m_n := (a_n + b_n)/2.\n\nHence a_n = m_n − ½ h_n and b_n = m_n + ½ h_n.\n\nStep 1.  Second–order expansion about the single point 0  \nBecause of (★) we have, for every vector x near 0,  \n\n f(x) = f(0) + A(x) + ½ B(x,x) + o(‖x‖²). (2)\n\nApply (2) to x = a_n, m_n, b_n:\n\n f(a_n) = f(0) + A(a_n) + ½ B(a_n,a_n) + o(‖a_n‖²),  \n f(m_n) = f(0) + A(m_n) + ½ B(m_n,m_n) + o(‖m_n‖²), (3)  \n f(b_n) = f(0) + A(b_n) + ½ B(b_n,b_n) + o(‖b_n‖²).\n\nThe three little-o terms are with respect to the single limit h↘0; they depend only on the arguments a_n, m_n, b_n and hence are legitimate.\n\nStep 2.  Form the discrete second difference  \nSet  \n\n Δ_n := f(b_n) − 2 f(m_n) + f(a_n).\n\nInsert (3):\n\n Δ_n = [A(b_n) − 2A(m_n) + A(a_n)]  \n    + ½ [ B(b_n,b_n) − 2B(m_n,m_n) + B(a_n,a_n) ]  \n    + o(‖a_n‖²) + o(‖m_n‖²) + o(‖b_n‖²).   (4)\n\n(a)  Cancellation of the linear terms.  \nBecause A is linear and m_n = (a_n + b_n)/2,\n\n A(b_n) − 2A(m_n) + A(a_n)  \n = A(b_n) + A(a_n) − A(a_n + b_n) = 0.   (5)\n\n(b)  Evaluation of the quadratic bracket.  \nWrite b_n = m_n + ½ h_n and a_n = m_n − ½ h_n.  Using bilinearity and symmetry of B,\n\n B(b_n,b_n) = B(m_n,m_n) + B(m_n,h_n) + ¼ B(h_n,h_n),  \n B(a_n,a_n) = B(m_n,m_n) − B(m_n,h_n) + ¼ B(h_n,h_n).\n\nAdding and subtracting 2 B(m_n,m_n) we get\n\n B(b_n,b_n) − 2B(m_n,m_n) + B(a_n,a_n) = ½ B(h_n,h_n).   (6)\n\nCombining (4), (5) and (6),\n\n Δ_n = ¼ B(h_n,h_n) + ρ_n,                          (7)\n\nwhere ρ_n := o(‖a_n‖²) + o(‖m_n‖²) + o(‖b_n‖²).\n\nStep 3.  Size of the error term ρ_n  \nBecause ‖a_n‖ ≤ ‖m_n‖ + ½‖h_n‖ = o(r_n) + O(r_n) = O(r_n) and likewise ‖b_n‖ = O(r_n), each of ‖a_n‖², ‖b_n‖², ‖m_n‖² is O(r_n²).  Consequently\n\n ρ_n = o(r_n²).                                  (8)\n\nStep 4.  Normalisation  \n\n 4 Δ_n / r_n² = 4·[¼ B(h_n,h_n)] / r_n² + 4ρ_n / r_n²  \n              = B(h_n/‖h_n‖, h_n/‖h_n‖) + o(1)  \n              = B(u_n,u_n) + o(1).                (9)\n\nStep 5.  Passage to the limit  \nBecause u_n → u by assumption, continuity of the quadratic form yields\n\n lim_{n→∞} B(u_n,u_n) = B(u,u).                  (10)\n\nFrom (9) and (10) we finally obtain\n\n lim_{n→∞} 4 · [ f(b_n) − 2 f(m_n) + f(a_n) ] ⁄ r_n²  \n         = B(u,u),                                 as required. ∎\n\n\n\n——————————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.472106",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher order: The original and kernel variants concern first derivatives; the enhanced problem requires understanding and manipulating second Fréchet derivatives (Hessians).\n\n2. Higher dimensions: The setting is ℝᵏ (arbitrary dimension) instead of the real line.\n\n3. Additional sequences & geometry: Two sequences produce a shrinking chord whose midpoint also approaches the base point subject to a subtle “midpoint =o(chord)” condition, forcing control of non-collinearity and necessitating a careful geometric decomposition a_n = m_n − ½ h_n, b_n = m_n + ½ h_n.\n\n4. Sophisticated tools: The solution demands\n   • Multivariable Taylor expansions with Peano-type remainders,  \n   • Uniform control of remainders under simultaneous limits,  \n   • Operator–norm continuity of derivatives, and  \n   • Coordination of several converging objects (h_n, m_n, u_n).\n\n5. More steps & deeper insight: One must (i) reformulate the chord in terms of midpoint and direction, (ii) expand f at a moving base point m_n (not fixed at 0), (iii) show cancellation of first-order terms, (iv) normalise correctly, and (v) pass limits through bilinear forms—each a non-trivial step absent from the original exercise.\n\nAltogether these features make the enhanced variant substantially harder than both the original and the current kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let k ≥ 1 be an integer and let  \n\n  f : ℝᵏ ⟶ ℝ  \n\nbe defined on an open neighbourhood of the origin.  \nAssume that f is twice Fréchet–differentiable at 0; i.e. there exist  \n\n a linear map A : ℝᵏ ⟶ ℝ   and a symmetric bilinear form B : ℝᵏ × ℝᵏ ⟶ ℝ  \n\nsuch that  \n\n  lim_{h→0}  [ f(h) − f(0) − A(h) − ½ B(h,h) ] ⁄ ‖h‖² = 0. (★)\n\nFor two sequences {a_n}, {b_n} ⊂ ℝᵏ suppose  \n\n(i) a_n ≠ 0, b_n ≠ 0 for every n;  \n(ii) a_n → 0 and b_n → 0;  \n(iii) r_n := ‖b_n − a_n‖ ⟶ 0;  \n(iv) u_n := (b_n − a_n) ⁄ r_n ⟶ u  with ‖u‖ = 1;  \n(v) m_n := (a_n + b_n)/2 satisfies ‖m_n‖ = o(r_n)  (the midpoint approaches 0 much faster than the chord–length).\n\nShow that  \n\n  lim_{n→∞} 4 · [ f(b_n) − 2 f(m_n) + f(a_n) ] ⁄ r_n² = B(u,u). (1)\n\nIn words: after the appropriate normalisation by the factor 4, the discrete second difference of f along any shrinking chord whose midpoint collapses sufficiently fast to the base point reproduces the quadratic form determined by the Hessian at 0.\n\n——————————————————————————————————————————",
      "solution": "Step 0.  Basic notation  \nPut  \n\n h_n := b_n − a_n (so that r_n = ‖h_n‖ and u_n = h_n / r_n),   \n m_n := (a_n + b_n)/2.\n\nHence a_n = m_n − ½ h_n and b_n = m_n + ½ h_n.\n\nStep 1.  Second–order expansion about the single point 0  \nBecause of (★) we have, for every vector x near 0,  \n\n f(x) = f(0) + A(x) + ½ B(x,x) + o(‖x‖²). (2)\n\nApply (2) to x = a_n, m_n, b_n:\n\n f(a_n) = f(0) + A(a_n) + ½ B(a_n,a_n) + o(‖a_n‖²),  \n f(m_n) = f(0) + A(m_n) + ½ B(m_n,m_n) + o(‖m_n‖²), (3)  \n f(b_n) = f(0) + A(b_n) + ½ B(b_n,b_n) + o(‖b_n‖²).\n\nThe three little-o terms are with respect to the single limit h↘0; they depend only on the arguments a_n, m_n, b_n and hence are legitimate.\n\nStep 2.  Form the discrete second difference  \nSet  \n\n Δ_n := f(b_n) − 2 f(m_n) + f(a_n).\n\nInsert (3):\n\n Δ_n = [A(b_n) − 2A(m_n) + A(a_n)]  \n    + ½ [ B(b_n,b_n) − 2B(m_n,m_n) + B(a_n,a_n) ]  \n    + o(‖a_n‖²) + o(‖m_n‖²) + o(‖b_n‖²).   (4)\n\n(a)  Cancellation of the linear terms.  \nBecause A is linear and m_n = (a_n + b_n)/2,\n\n A(b_n) − 2A(m_n) + A(a_n)  \n = A(b_n) + A(a_n) − A(a_n + b_n) = 0.   (5)\n\n(b)  Evaluation of the quadratic bracket.  \nWrite b_n = m_n + ½ h_n and a_n = m_n − ½ h_n.  Using bilinearity and symmetry of B,\n\n B(b_n,b_n) = B(m_n,m_n) + B(m_n,h_n) + ¼ B(h_n,h_n),  \n B(a_n,a_n) = B(m_n,m_n) − B(m_n,h_n) + ¼ B(h_n,h_n).\n\nAdding and subtracting 2 B(m_n,m_n) we get\n\n B(b_n,b_n) − 2B(m_n,m_n) + B(a_n,a_n) = ½ B(h_n,h_n).   (6)\n\nCombining (4), (5) and (6),\n\n Δ_n = ¼ B(h_n,h_n) + ρ_n,                          (7)\n\nwhere ρ_n := o(‖a_n‖²) + o(‖m_n‖²) + o(‖b_n‖²).\n\nStep 3.  Size of the error term ρ_n  \nBecause ‖a_n‖ ≤ ‖m_n‖ + ½‖h_n‖ = o(r_n) + O(r_n) = O(r_n) and likewise ‖b_n‖ = O(r_n), each of ‖a_n‖², ‖b_n‖², ‖m_n‖² is O(r_n²).  Consequently\n\n ρ_n = o(r_n²).                                  (8)\n\nStep 4.  Normalisation  \n\n 4 Δ_n / r_n² = 4·[¼ B(h_n,h_n)] / r_n² + 4ρ_n / r_n²  \n              = B(h_n/‖h_n‖, h_n/‖h_n‖) + o(1)  \n              = B(u_n,u_n) + o(1).                (9)\n\nStep 5.  Passage to the limit  \nBecause u_n → u by assumption, continuity of the quadratic form yields\n\n lim_{n→∞} B(u_n,u_n) = B(u,u).                  (10)\n\nFrom (9) and (10) we finally obtain\n\n lim_{n→∞} 4 · [ f(b_n) − 2 f(m_n) + f(a_n) ] ⁄ r_n²  \n         = B(u,u),                                 as required. ∎\n\n\n\n——————————————————————————————————————————",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.396797",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher order: The original and kernel variants concern first derivatives; the enhanced problem requires understanding and manipulating second Fréchet derivatives (Hessians).\n\n2. Higher dimensions: The setting is ℝᵏ (arbitrary dimension) instead of the real line.\n\n3. Additional sequences & geometry: Two sequences produce a shrinking chord whose midpoint also approaches the base point subject to a subtle “midpoint =o(chord)” condition, forcing control of non-collinearity and necessitating a careful geometric decomposition a_n = m_n − ½ h_n, b_n = m_n + ½ h_n.\n\n4. Sophisticated tools: The solution demands\n   • Multivariable Taylor expansions with Peano-type remainders,  \n   • Uniform control of remainders under simultaneous limits,  \n   • Operator–norm continuity of derivatives, and  \n   • Coordination of several converging objects (h_n, m_n, u_n).\n\n5. More steps & deeper insight: One must (i) reformulate the chord in terms of midpoint and direction, (ii) expand f at a moving base point m_n (not fixed at 0), (iii) show cancellation of first-order terms, (iv) normalise correctly, and (v) pass limits through bilinear forms—each a non-trivial step absent from the original exercise.\n\nAltogether these features make the enhanced variant substantially harder than both the original and the current kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}