{
  "index": "1967-B-3",
  "type": "ANA",
  "tag": [
    "ANA"
  ],
  "difficulty": "",
  "question": "B-3. If \\( f \\) and \\( g \\) are continuous and periodic functions with period 1 on the real line, then \\( \\lim _{n \\rightarrow \\infty} \\int_{0}^{1} f(x) g(n x) d x=\\left(\\int_{0}^{1} f(x) d x\\right)\\left(\\int_{0}^{1} g(x) d x\\right) \\).",
  "solution": "B-3 Since \\( f \\) is uniformly continuous, for any \\( \\epsilon>0 \\) there is an integer \\( n \\) such that \\( |x-y|<1 / n \\) implies \\( |f(x)-f(y)|<\\epsilon \\).\n\\[\n\\begin{aligned}\n\\int_{0}^{1} f(x) g(n x) d x= & \\sum_{m=0}^{n-1} \\int_{m / n}^{m+1 / n} f(x) g(n x) d x=\\sum_{m=0}^{n-1} \\int_{m / n}^{m+1 / n} f(m / n) g(n x) d x \\\\\n& +\\sum_{m=0}^{n-1} \\int_{m / n}^{m+1 / n}(f(x)-f(m / n)) g(n x) d x\n\\end{aligned}\n\\]\n\nThe first term equals \\( \\sum_{m=0}^{n-1}(1 / n) f(m / n) \\int_{0}^{1} g(t) d t \\) and becomes \\( \\left(\\int_{0}^{1} f(x) d x\\right)\\left(\\int_{0}^{1} g(x) d x\\right) \\) as \\( n \\rightarrow \\infty \\). Furthermore,\n\\[\n\\begin{aligned}\n\\left|\\int_{m / n}^{m+1 / n}\\{f(x)-f(m / n)\\} g(n x) d x\\right| & <\\int_{m / n}^{m+1 / n}|f(x)-f(m / n)| \\cdot|g(n x)| d x \\\\\n& <\\int_{m / n}^{m+1 / n} \\epsilon|g(n x)| d x<\\epsilon / n \\int_{0}^{1}|g(t)| d t\n\\end{aligned}\n\\]\n\nThus the absolute value of the second term is less than or equal to \\( \\epsilon \\int_{0}^{1}|g(t)| d t \\) which becomes 0 as \\( \\epsilon \\rightarrow 0 \\).",
  "vars": [
    "x",
    "y",
    "t"
  ],
  "params": [
    "f",
    "g",
    "n",
    "m"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvarx",
        "y": "realvary",
        "t": "timevar",
        "f": "functionf",
        "g": "functiong",
        "n": "indexnum",
        "m": "indexsec"
      },
      "question": "B-3. If \\( functionf \\) and \\( functiong \\) are continuous and periodic functions with period 1 on the real line, then \\( \\lim _{indexnum \\rightarrow \\infty} \\int_{0}^{1} functionf(realvarx) functiong(indexnum\\, realvarx) d realvarx=\\left(\\int_{0}^{1} functionf(realvarx) d realvarx\\right)\\left(\\int_{0}^{1} functiong(realvarx) d realvarx\\right) \\).",
      "solution": "B-3 Since \\( functionf \\) is uniformly continuous, for any \\( \\epsilon>0 \\) there is an integer \\( indexnum \\) such that \\( |realvarx-realvary|<1 / indexnum \\) implies \\( |functionf(realvarx)-functionf(realvary)|<\\epsilon \\).\n\\[\n\\begin{aligned}\n\\int_{0}^{1} functionf(realvarx) functiong(indexnum\\, realvarx) d realvarx=& \\sum_{indexsec=0}^{indexnum-1} \\int_{indexsec / indexnum}^{indexsec+1 / indexnum} functionf(realvarx) functiong(indexnum\\, realvarx) d realvarx\\\\\n=&\\sum_{indexsec=0}^{indexnum-1} \\int_{indexsec / indexnum}^{indexsec+1 / indexnum} functionf(indexsec / indexnum) functiong(indexnum\\, realvarx) d realvarx\\\\\n&+\\sum_{indexsec=0}^{indexnum-1} \\int_{indexsec / indexnum}^{indexsec+1 / indexnum}(functionf(realvarx)-functionf(indexsec / indexnum)) functiong(indexnum\\, realvarx) d realvarx\n\\end{aligned}\n\\]\n\nThe first term equals \\( \\sum_{indexsec=0}^{indexnum-1}(1 / indexnum) functionf(indexsec / indexnum) \\int_{0}^{1} functiong(timevar) d timevar \\) and becomes \\( \\left(\\int_{0}^{1} functionf(realvarx) d realvarx\\right)\\left(\\int_{0}^{1} functiong(realvarx) d realvarx\\right) \\) as \\( indexnum \\rightarrow \\infty \\). Furthermore,\n\\[\n\\begin{aligned}\n\\left|\\int_{indexsec / indexnum}^{indexsec+1 / indexnum}\\{functionf(realvarx)-functionf(indexsec / indexnum)\\} functiong(indexnum\\, realvarx) d realvarx\\right|&<\\int_{indexsec / indexnum}^{indexsec+1 / indexnum}|functionf(realvarx)-functionf(indexsec / indexnum)| \\cdot|functiong(indexnum\\, realvarx)| d realvarx\\\\\n&<\\int_{indexsec / indexnum}^{indexsec+1 / indexnum} \\epsilon|functiong(indexnum\\, realvarx)| d realvarx<\\epsilon / indexnum \\int_{0}^{1}|functiong(timevar)| d timevar\n\\end{aligned}\n\\]\n\nThus the absolute value of the second term is less than or equal to \\( \\epsilon \\int_{0}^{1}|functiong(timevar)| d timevar \\) which becomes 0 as \\( \\epsilon \\rightarrow 0 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "blueberry",
        "y": "sailcloth",
        "t": "marigold",
        "f": "pendulum",
        "g": "labyrinth",
        "n": "waterfall",
        "m": "pinecone"
      },
      "question": "B-3. If \\( pendulum \\) and \\( labyrinth \\) are continuous and periodic functions with period 1 on the real line, then \\( \\lim _{waterfall \\rightarrow \\infty} \\int_{0}^{1} pendulum(blueberry) labyrinth(waterfall blueberry) d blueberry=\\left(\\int_{0}^{1} pendulum(blueberry) d blueberry\\right)\\left(\\int_{0}^{1} labyrinth(blueberry) d blueberry\\right) \\).",
      "solution": "B-3 Since \\( pendulum \\) is uniformly continuous, for any \\( \\epsilon>0 \\) there is an integer \\( waterfall \\) such that \\( |blueberry-sailcloth|<1 / waterfall \\) implies \\( |pendulum(blueberry)-pendulum(sailcloth)|<\\epsilon \\).\n\\[\n\\begin{aligned}\n\\int_{0}^{1} pendulum(blueberry) labyrinth(waterfall blueberry) d blueberry= & \\sum_{pinecone=0}^{waterfall-1} \\int_{pinecone / waterfall}^{pinecone+1 / waterfall} pendulum(blueberry) labyrinth(waterfall blueberry) d blueberry=\\sum_{pinecone=0}^{waterfall-1} \\int_{pinecone / waterfall}^{pinecone+1 / waterfall} pendulum(pinecone / waterfall) labyrinth(waterfall blueberry) d blueberry \\\\\n& +\\sum_{pinecone=0}^{waterfall-1} \\int_{pinecone / waterfall}^{pinecone+1 / waterfall}(pendulum(blueberry)-pendulum(pinecone / waterfall)) labyrinth(waterfall blueberry) d blueberry\n\\end{aligned}\n\\]\n\nThe first term equals \\( \\sum_{pinecone=0}^{waterfall-1}(1 / waterfall) pendulum(pinecone / waterfall) \\int_{0}^{1} labyrinth(marigold) d marigold \\) and becomes \\( \\left(\\int_{0}^{1} pendulum(blueberry) d blueberry\\right)\\left(\\int_{0}^{1} labyrinth(blueberry) d blueberry\\right) \\) as \\( waterfall \\rightarrow \\infty \\). Furthermore,\n\\[\n\\begin{aligned}\n\\left|\\int_{pinecone / waterfall}^{pinecone+1 / waterfall}\\{pendulum(blueberry)-pendulum(pinecone / waterfall)\\} labyrinth(waterfall blueberry) d blueberry\\right| & <\\int_{pinecone / waterfall}^{pinecone+1 / waterfall}|pendulum(blueberry)-pendulum(pinecone / waterfall)| \\cdot|labyrinth(waterfall blueberry)| d blueberry \\\\\n& <\\int_{pinecone / waterfall}^{pinecone+1 / waterfall} \\epsilon|labyrinth(waterfall blueberry)| d blueberry<\\epsilon / waterfall \\int_{0}^{1}|labyrinth(marigold)| d marigold\n\\end{aligned}\n\\]\n\nThus the absolute value of the second term is less than or equal to \\( \\epsilon \\int_{0}^{1}|labyrinth(marigold)| d marigold \\) which becomes 0 as \\( \\epsilon \\rightarrow 0 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "steadfast",
        "y": "unswerving",
        "t": "timeless",
        "f": "permanent",
        "g": "unchanging",
        "n": "finiteval",
        "m": "continuum"
      },
      "question": "B-3. If \\( permanent \\) and \\( unchanging \\) are continuous and periodic functions with period 1 on the real line, then \\( \\lim _{finiteval \\rightarrow \\infty} \\int_{0}^{1} permanent(steadfast) unchanging(finiteval steadfast) d steadfast=\\left(\\int_{0}^{1} permanent(steadfast) d steadfast\\right)\\left(\\int_{0}^{1} unchanging(steadfast) d steadfast\\right) \\).",
      "solution": "B-3 Since \\( permanent \\) is uniformly continuous, for any \\( \\epsilon>0 \\) there is an integer \\( finiteval \\) such that \\( |steadfast-unswerving|<1 / finiteval \\) implies \\( |permanent(steadfast)-permanent(unswerving)|<\\epsilon \\).\n\\[\n\\begin{aligned}\n\\int_{0}^{1} permanent(steadfast) unchanging(finiteval steadfast) d steadfast= & \\sum_{continuum=0}^{finiteval-1} \\int_{continuum / finiteval}^{continuum+1 / finiteval} permanent(steadfast) unchanging(finiteval steadfast) d steadfast=\\sum_{continuum=0}^{finiteval-1} \\int_{continuum / finiteval}^{continuum+1 / finiteval} permanent(continuum / finiteval) unchanging(finiteval steadfast) d steadfast \\\\\n& +\\sum_{continuum=0}^{finiteval-1} \\int_{continuum / finiteval}^{continuum+1 / finiteval}(permanent(steadfast)-permanent(continuum / finiteval)) unchanging(finiteval steadfast) d steadfast\n\\end{aligned}\n\\]\n\nThe first term equals \\( \\sum_{continuum=0}^{finiteval-1}(1 / finiteval) permanent(continuum / finiteval) \\int_{0}^{1} unchanging(timeless) d timeless \\) and becomes \\( \\left(\\int_{0}^{1} permanent(steadfast) d steadfast\\right)\\left(\\int_{0}^{1} unchanging(steadfast) d steadfast\\right) \\) as \\( finiteval \\rightarrow \\infty \\). Furthermore,\n\\[\n\\begin{aligned}\n\\left|\\int_{continuum / finiteval}^{continuum+1 / finiteval}\\{permanent(steadfast)-permanent(continuum / finiteval)\\} unchanging(finiteval steadfast) d steadfast\\right| & <\\int_{continuum / finiteval}^{continuum+1 / finiteval}|permanent(steadfast)-permanent(continuum / finiteval)| \\cdot|unchanging(finiteval steadfast)| d steadfast \\\\\n& <\\int_{continuum / finiteval}^{continuum+1 / finiteval} \\epsilon|unchanging(finiteval steadfast)| d steadfast<\\epsilon / finiteval \\int_{0}^{1}|unchanging(timeless)| d timeless\n\\end{aligned}\n\\]\n\nThus the absolute value of the second term is less than or equal to \\( \\epsilon \\int_{0}^{1}|unchanging(timeless)| d timeless \\) which becomes 0 as \\( \\epsilon \\rightarrow 0 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "t": "mndkslre",
        "f": "qlpkfntz",
        "g": "xcrtyhls",
        "n": "zmbqswle",
        "m": "kpdrtbah"
      },
      "question": "B-3. If \\( qlpkfntz \\) and \\( xcrtyhls \\) are continuous and periodic functions with period 1 on the real line, then \\( \\lim _{zmbqswle \\rightarrow \\infty} \\int_{0}^{1} qlpkfntz(qzxwvtnp) xcrtyhls(zmbqswle qzxwvtnp) d qzxwvtnp=\\left(\\int_{0}^{1} qlpkfntz(qzxwvtnp) d qzxwvtnp\\right)\\left(\\int_{0}^{1} xcrtyhls(qzxwvtnp) d qzxwvtnp\\right) \\).",
      "solution": "B-3 Since \\( qlpkfntz \\) is uniformly continuous, for any \\( \\epsilon>0 \\) there is an integer \\( zmbqswle \\) such that \\( |qzxwvtnp-hjgrksla|<1 / zmbqswle \\) implies \\( |qlpkfntz(qzxwvtnp)-qlpkfntz(hjgrksla)|<\\epsilon \\).\n\\[\n\\begin{aligned}\n\\int_{0}^{1} qlpkfntz(qzxwvtnp) xcrtyhls(zmbqswle qzxwvtnp) d qzxwvtnp= & \\sum_{kpdrtbah=0}^{zmbqswle-1} \\int_{kpdrtbah / zmbqswle}^{kpdrtbah+1 / zmbqswle} qlpkfntz(qzxwvtnp) xcrtyhls(zmbqswle qzxwvtnp) d qzxwvtnp=\\sum_{kpdrtbah=0}^{zmbqswle-1} \\int_{kpdrtbah / zmbqswle}^{kpdrtbah+1 / zmbqswle} qlpkfntz(kpdrtbah / zmbqswle) xcrtyhls(zmbqswle qzxwvtnp) d qzxwvtnp \\\\\n& +\\sum_{kpdrtbah=0}^{zmbqswle-1} \\int_{kpdrtbah / zmbqswle}^{kpdrtbah+1 / zmbqswle}(qlpkfntz(qzxwvtnp)-qlpkfntz(kpdrtbah / zmbqswle)) xcrtyhls(zmbqswle qzxwvtnp) d qzxwvtnp\n\\end{aligned}\n\\]\n\nThe first term equals \\( \\sum_{kpdrtbah=0}^{zmbqswle-1}(1 / zmbqswle) qlpkfntz(kpdrtbah / zmbqswle) \\int_{0}^{1} xcrtyhls(mndkslre) d mndkslre \\) and becomes \\( \\left(\\int_{0}^{1} qlpkfntz(qzxwvtnp) d qzxwvtnp\\right)\\left(\\int_{0}^{1} xcrtyhls(qzxwvtnp) d qzxwvtnp\\right) \\) as \\( zmbqswle \\rightarrow \\infty \\). Furthermore,\n\\[\n\\begin{aligned}\n\\left|\\int_{kpdrtbah / zmbqswle}^{kpdrtbah+1 / zmbqswle}\\{qlpkfntz(qzxwvtnp)-qlpkfntz(kpdrtbah / zmbqswle)\\} xcrtyhls(zmbqswle qzxwvtnp) d qzxwvtnp\\right| & <\\int_{kpdrtbah / zmbqswle}^{kpdrtbah+1 / zmbqswle}|qlpkfntz(qzxwvtnp)-qlpkfntz(kpdrtbah / zmbqswle)| \\cdot|xcrtyhls(zmbqswle qzxwvtnp)| d qzxwvtnp \\\\\n& <\\int_{kpdrtbah / zmbqswle}^{kpdrtbah+1 / zmbqswle} \\epsilon|xcrtyhls(zmbqswle qzxwvtnp)| d qzxwvtnp<\\epsilon / zmbqswle \\int_{0}^{1}|xcrtyhls(mndkslre)| d mndkslre\n\\end{aligned}\n\\]\n\nThus the absolute value of the second term is less than or equal to \\( \\epsilon \\int_{0}^{1}|xcrtyhls(mndkslre)| d mndkslre \\) which becomes 0 as \\( \\epsilon \\rightarrow 0 \\)."
    },
    "kernel_variant": {
      "question": "Let $f$ and $g$ be continuous real–valued functions on $\\mathbb R$, each having period $2\\pi$.  Prove that\n\\[\n\\lim_{n\\to\\infty}\\int_{-\\pi}^{\\pi} f(x)\\,g(nx)\\,dx \n   = \\frac{1}{2\\pi}\\Bigl(\\int_{-\\pi}^{\\pi} f(x)\\,dx\\Bigr)\n     \\Bigl(\\int_{-\\pi}^{\\pi} g(x)\\,dx\\Bigr).\n\\]",
      "solution": "1.  Uniform continuity.\nBecause f is continuous on the compact set [−π,π] and 2π–periodic, it is uniformly continuous on ℝ.  Fix ε>0 and choose an integer N so large that\n     |x−y|<2π/N  ⇒  |f(x)−f(y)|<ε.\n\n2.  Partition of one period.\nSet Δ=2π/N and let\n     Iₘ=[−π+mΔ,\n           −π+(m+1)Δ]   (0≤m≤N−1).\nThen\n     ∫_{−π}^{π} f(x)g(Nx)\n       dx  =  ∑_{m=0}^{N−1} ∫_{Iₘ} f(x)g(Nx)\n       dx.\n\n3.  Freeze f on each subinterval.\nWrite the integral as a main term plus an error term:\n     ∑_{m=0}^{N−1} ∫_{Iₘ} f(aₘ)g(Nx)\n       dx\n   + ∑_{m=0}^{N−1} ∫_{Iₘ} [f(x)−f(aₘ)] g(Nx)\n       dx,\nwhere aₘ=−π+mΔ.  Call these M_N and E_N, respectively.\n\n4.  Evaluate the main term M_N.\nFor x∈Iₘ set t= N x; then dx=dt/N and t runs over an interval of length NΔ=2π, i.e.\none full period of g:\n     ∫_{Iₘ} g(Nx)\n       dx  =  (1/N) ∫_{−π}^{π} g(t)\n       dt.\nHence\n     M_N  =  (1/N)\n               (∫_{−π}^{π} g)\n              ∑_{m=0}^{N−1} f(aₘ)\n           =  (∫_{−π}^{π} g)/(2π)\n              ∑_{m=0}^{N−1} f(aₘ) Δ.\nThe sum ∑f(aₘ)Δ is a Riemann sum for ∫_{−π}^{π}f, so\n     lim_{N→∞} M_N\n       = (1/2π)(∫_{−π}^{π}f)(∫_{−π}^{π}g).\n\n5.  Bound the error term E_N.\nSet M:=supₓ |g(x)|<∞.  If x∈Iₘ then |x−aₘ|≤Δ, so |f(x)−f(aₘ)|<ε.  Therefore\n     |E_N|\n       ≤ ∑_{m=0}^{N−1} ∫_{Iₘ} ε|g(Nx)|\n       dx\n       ≤ ε M ∑_{m=0}^{N−1} ∫_{Iₘ} dx\n       = ε M (2π).\nSince ε>0 was arbitrary, E_N→0 as N→∞.\n\n6.  Conclusion.\nCombining the limits of M_N and E_N gives\n     lim_{n→∞} ∫_{−π}^{π} f(x)g(nx)\n       dx\n       = (1/2π)(∫_{−π}^{π}f(x)dx)(∫_{−π}^{π}g(x)dx),\nas required.",
      "_meta": {
        "core_steps": [
          "Invoke uniform continuity of f to control its variation on sufficiently small intervals.",
          "Partition one full period of integration into n equal subintervals and write the integral as a sum over them.",
          "Freeze f at a representative point on each subinterval, splitting the integral into a main term and an error term.",
          "Recognize the main term as a Riemann sum that converges to (∫f)(∫g).",
          "Bound the error term by ε via the uniform-continuity estimate, letting ε→0 to make it vanish."
        ],
        "mutable_slots": {
          "slot_period_length": {
            "description": "Length of the common period assumed for both functions.",
            "original": 1
          },
          "slot_integration_interval": {
            "description": "Specific interval of one full period over which the integral is taken.",
            "original": "[0,1]"
          },
          "slot_subinterval_length": {
            "description": "Width of each partition subinterval used to exploit uniform continuity.",
            "original": "1/n"
          },
          "slot_g_norm_constant": {
            "description": "Constant used to bound the error term (any finite bound on g would work).",
            "original": "∫_0^1 |g(t)| dt"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}