{
  "index": "2000-B-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $f(x)$ be a continuous function such that $f(2x^2-1)=2xf(x)$ for\nall $x$.  Show that $f(x)=0$ for $-1\\leq x\\leq 1$.",
  "solution": "For $t$ real and not a multiple of $\\pi$, write $g(t) =\n\\frac{f(\\cos t)}{\\sin t}$.\nThen $g(t+\\pi) = g(t)$; furthermore, the given equation implies that\n\\[\ng(2t) = \\frac{f(2\\cos^2 t - 1)}{\\sin (2t)} =\n\\frac{2(\\cos t) f(\\cos t)}{\\sin(2t)} = g(t).\n\\]\nIn particular, for any integer $n$ and $k$, we have\n\\[\ng(1+n\\pi/2^k) = g(2^k + n\\pi) = g(2^k) = g(1).\n\\]\nSince $f$ is continuous, $g$ is continuous where it is defined;\nbut the set $\\{1+n\\pi/2^k | n,k\\in{\\mathbb{Z}}\\}$ is dense\nin the reals, and so $g$ must be constant on its domain.\nSince $g(-t) = -g(t)$ for all $t$, we must have $g(t) = 0$\nwhen $t$ is not a multiple of $\\pi$.\nHence $f(x) = 0$ for $x \\in (-1,1)$.  Finally,\nsetting $x=0$ and $x=1$ in the given equation yields\n$f(-1) = f(1) = 0$.",
  "vars": [
    "x",
    "t",
    "n",
    "k"
  ],
  "params": [
    "f",
    "g"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvar",
        "t": "anglevar",
        "n": "intindex",
        "k": "powindex",
        "f": "contifun",
        "g": "ratiofun"
      },
      "question": "Let $contifun(realvar)$ be a continuous function such that $contifun(2realvar^2-1)=2realvar\\, contifun(realvar)$ for all $realvar$.  Show that $contifun(realvar)=0$ for $-1\\leq realvar\\leq 1$.",
      "solution": "For $anglevar$ real and not a multiple of $\\pi$, write $ratiofun(anglevar) =\n\\frac{contifun(\\cos anglevar)}{\\sin anglevar}$.\nThen $ratiofun(anglevar+\\pi) = ratiofun(anglevar)$; furthermore, the given equation implies that\n\\[\nratiofun(2anglevar) = \\frac{contifun(2\\cos^2 anglevar - 1)}{\\sin (2anglevar)} =\n\\frac{2(\\cos anglevar) \\, contifun(\\cos anglevar)}{\\sin(2anglevar)} = ratiofun(anglevar).\n\\]\nIn particular, for any integer $intindex$ and $powindex$, we have\n\\[\nratiofun(1+intindex\\pi/2^{powindex}) = ratiofun(2^{powindex} + intindex\\pi) = ratiofun(2^{powindex}) = ratiofun(1).\n\\]\nSince $contifun$ is continuous, $ratiofun$ is continuous where it is defined;\nbut the set $\\{1+intindex\\pi/2^{powindex} \\mid intindex,powindex\\in{\\mathbb{Z}}\\}$ is dense\nin the reals, and so $ratiofun$ must be constant on its domain.\nSince $ratiofun(-anglevar) = -ratiofun(anglevar)$ for all $anglevar$, we must have $ratiofun(anglevar) = 0$\nwhen $anglevar$ is not a multiple of $\\pi$.\nHence $contifun(realvar) = 0$ for $realvar \\in (-1,1)$.  Finally,\nsetting $realvar=0$ and $realvar=1$ in the given equation yields\n$contifun(-1) = contifun(1) = 0$."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sequoiah",
        "t": "labyrinth",
        "n": "driftwood",
        "k": "parchment",
        "f": "sunflower",
        "g": "moonstone"
      },
      "question": "Let $sunflower(sequoiah)$ be a continuous function such that $sunflower(2sequoiah^2-1)=2sequoiah sunflower(sequoiah)$ for\nall $sequoiah$.  Show that $sunflower(sequoiah)=0$ for $-1\\leq sequoiah\\leq 1$.",
      "solution": "For $labyrinth$ real and not a multiple of $\\pi$, write $moonstone(labyrinth) =\n\\frac{sunflower(\\cos labyrinth)}{\\sin labyrinth}$. Then $moonstone(labyrinth+\\pi) = moonstone(labyrinth)$; furthermore, the given equation implies that\n\\[\nmoonstone(2labyrinth) = \\frac{sunflower(2\\cos^2 labyrinth - 1)}{\\sin (2labyrinth)} =\n\\frac{2(\\cos labyrinth) sunflower(\\cos labyrinth)}{\\sin(2labyrinth)} = moonstone(labyrinth).\n\\]\nIn particular, for any integer $driftwood$ and $parchment$, we have\n\\[\nmoonstone(1+driftwood\\pi/2^{parchment}) = moonstone(2^{parchment} + driftwood\\pi) = moonstone(2^{parchment}) = moonstone(1).\n\\]\nSince $sunflower$ is continuous, $moonstone$ is continuous where it is defined;\nbut the set $\\{1+driftwood\\pi/2^{parchment} | driftwood,parchment\\in{\\mathbb{Z}}\\}$ is dense\nin the reals, and so $moonstone$ must be constant on its domain.\nSince $moonstone(-labyrinth) = -moonstone(labyrinth)$ for all $labyrinth$, we must have $moonstone(labyrinth) = 0$\nwhen $labyrinth$ is not a multiple of $\\pi$.\nHence $sunflower(sequoiah) = 0$ for $sequoiah \\in (-1,1)$.  Finally,\nsetting $sequoiah=0$ and $sequoiah=1$ in the given equation yields\n$sunflower(-1) = sunflower(1) = 0$.} }\n"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "t": "steadyvalue",
        "n": "continuousindex",
        "k": "fractionalindex",
        "f": "constantfunc",
        "g": "motionlessfun"
      },
      "question": "Let $constantfunc(fixedvalue)$ be a continuous function such that $constantfunc(2fixedvalue^2-1)=2fixedvalue\\,constantfunc(fixedvalue)$ for all $fixedvalue$.  Show that $constantfunc(fixedvalue)=0$ for $-1\\leq fixedvalue\\leq 1$.",
      "solution": "For $steadyvalue$ real and not a multiple of $\\pi$, write $motionlessfun(steadyvalue) = \\frac{constantfunc(\\cos steadyvalue)}{\\sin steadyvalue}$.\\par\nThen $motionlessfun(steadyvalue+\\pi) = motionlessfun(steadyvalue)$; furthermore, the given equation implies that\n\\[\nmotionlessfun(2steadyvalue) = \\frac{constantfunc(2\\cos^2 steadyvalue - 1)}{\\sin (2steadyvalue)} =\n\\frac{2(\\cos steadyvalue)\\,constantfunc(\\cos steadyvalue)}{\\sin(2steadyvalue)} = motionlessfun(steadyvalue).\n\\]\nIn particular, for any integer $continuousindex$ and $fractionalindex$, we have\n\\[\nmotionlessfun(1+continuousindex\\pi/2^{fractionalindex}) = motionlessfun(2^{fractionalindex} + continuousindex\\pi) = motionlessfun(2^{fractionalindex}) = motionlessfun(1).\n\\]\nSince $constantfunc$ is continuous, $motionlessfun$ is continuous where it is defined; but the set $\\{1+continuousindex\\pi/2^{fractionalindex}\\mid continuousindex,fractionalindex\\in{\\mathbb{Z}}\\}$ is dense in the reals, and so $motionlessfun$ must be constant on its domain.\\par\nSince $motionlessfun(-steadyvalue) = -motionlessfun(steadyvalue)$ for all $steadyvalue$, we must have $motionlessfun(steadyvalue) = 0$ when $steadyvalue$ is not a multiple of $\\pi$.\\par\nHence $constantfunc(fixedvalue) = 0$ for $fixedvalue \\in (-1,1)$.  Finally, setting $fixedvalue=0$ and $fixedvalue=1$ in the given equation yields $constantfunc(-1) = constantfunc(1) = 0$.  Thus $constantfunc(fixedvalue)=0$ for $-1 \\le fixedvalue \\le 1$.",
      "}": "",
      "note": ""
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "t": "hjgrksla",
        "n": "vplezthm",
        "k": "rfsdchua",
        "f": "xprbgtle",
        "g": "clzwakdn"
      },
      "question": "Let $xprbgtle(qzxwvtnp)$ be a continuous function such that $xprbgtle(2qzxwvtnp^2-1)=2 qzxwvtnp xprbgtle(qzxwvtnp)$ for\nall $qzxwvtnp$.  Show that $xprbgtle(qzxwvtnp)=0$ for $-1\\leq qzxwvtnp\\leq 1$.",
      "solution": "For $hjgrksla$ real and not a multiple of $\\pi$, write $clzwakdn(hjgrksla) =\n\\frac{xprbgtle(\\cos hjgrksla)}{\\sin hjgrksla}$.\nThen $clzwakdn(hjgrksla+\\pi) = clzwakdn(hjgrksla)$; furthermore, the given equation implies that\n\\[\nclzwakdn(2 hjgrksla) = \\frac{xprbgtle(2\\cos^2 hjgrksla - 1)}{\\sin (2 hjgrksla)} =\n\\frac{2(\\cos hjgrksla) xprbgtle(\\cos hjgrksla)}{\\sin(2 hjgrksla)} = clzwakdn(hjgrksla).\n\\]\nIn particular, for any integer $vplezthm$ and $rfsdchua$, we have\n\\[\nclzwakdn(1+vplezthm\\pi/2^{rfsdchua}) = clzwakdn(2^{rfsdchua} + vplezthm\\pi) = clzwakdn(2^{rfsdchua}) = clzwakdn(1).\n\\]\nSince $xprbgtle$ is continuous, $clzwakdn$ is continuous where it is defined;\nbut the set $\\{1+vplezthm\\pi/2^{rfsdchua} | vplezthm,rfsdchua\\in{\\mathbb{Z}}\\}$ is dense\nin the reals, and so $clzwakdn$ must be constant on its domain.\nSince $clzwakdn(-hjgrksla) = -clzwakdn(hjgrksla)$ for all $hjgrksla$, we must have $clzwakdn(hjgrksla) = 0$\nwhen $hjgrksla$ is not a multiple of $\\pi$.\nHence $xprbgtle(qzxwvtnp) = 0$ for $qzxwvtnp \\in (-1,1)$.  Finally,\nsetting $qzxwvtnp=0$ and $qzxwvtnp=1$ in the given equation yields\n$xprbgtle(-1) = xprbgtle(1) = 0$.",
      "extra": ""
    },
    "kernel_variant": {
      "question": "Let f:[−1,1]→ℝ be a continuous function that satisfies\n\n              f( cos 2θ ) = 2 cos θ · f( cos θ )      for every real θ.\n\nProve that f(x) = 0 for all x ∈ [−1,1].",
      "solution": "We show that the only continuous function f:[−1,1]→ℝ that fulfils\n              f( cos 2θ ) = 2 cos θ · f( cos θ ) (∀θ∈ℝ)\nis the zero–function.\n\nStep 1 – Reformulation and oddness of f.\nPut x = cos θ.  Because cos θ takes every value of [−1,1] when θ runs through ℝ, the equation may also be written\n              f( 2x² − 1 ) = 2x · f(x) (∀x∈[−1,1]).          (1)\nReplacing x by −x in (1) gives\n              f( 2x² − 1 ) = −2x · f(−x).\nComparing this with (1) yields 2x·f(x) = −2x·f(−x) for x≠0, hence f(−x)=−f(x).  Continuity at 0 then forces f(0)=0, so f is odd on [−1,1].\n\nStep 2 – Introducing g.\nFor an angle t with sin t ≠ 0 define\n              g(t) := f( cos t ) / sin t .                     (2)\nBecause f is continuous, g is continuous on its domain\n              D := ℝ \\ { kπ | k∈ℤ }.\n\nStep 3 – Two elementary identities for g.\n(a) π-periodicity.  Using cos(t+π)=−cos t, sin(t+π)=−sin t and the oddness of f,\n              g(t+π) = f(−cos t)/(−sin t)=g(t).\n(b) Doubling.  With (2) and sin2t = 2 sin t cos t,\n              g(2t)= f(cos2t)/sin2t = 2cos t·f(cos t)/(2 sin t cos t)=g(t).\n\nStep 4 – Fixing one angle and propagating its value.\nChoose once and for all an angle\n              t₀ = 1 (radian).\nBecause 1/π is irrational, none of the numbers 2^m t₀ + nπ (m,n∈ℤ) is a multiple of π; hence g is defined at all those points.  Put\n              C := g(t₀).\n\nClaim.  For every pair of integers m≥0 and n we have\n              g( t₀ + nπ/2^m ) = C.                               (3)\n\nProof of the claim.  Fix m,n and set\n              u := t₀ + nπ/2^m.\nFor j = 0,1,…,m define u_j := 2^j u.  Then u_0 = u and u_m = 2^m u = 2^m t₀ + nπ.  None of the u_j is a multiple of π, so g is defined at each u_j.  By the doubling property,\n              g(u_{j+1}) = g(2u_j) = g(u_j) (j=0,…,m−1).\nConsequently g(u_0)=g(u_m).  Using the π-periodicity at the last equality,\n              g(u) = g(2^m t₀ + nπ) = g(2^m t₀).\nApplying the doubling property m times to t₀ gives g(2^m t₀)=g(t₀)=C, whence (3).\n∎\n\nStep 5 – Density and constancy of g.\nThe set\n              S := { t₀ + nπ/2^m | m≥0, n∈ℤ }\nis dense in ℝ (it is a translate of the dyadic rationals multiplied by π).  Formula (3) shows that g is constant and equal to C on the dense subset S of its domain D.  Because g is continuous on D, it follows that\n              g(t) ≡ C for every t∈D.                            (4)\n\nStep 6 – The constant C is zero.\nFor t∈D we have, by (2),\n              g(−t) = f(cos(−t))/sin(−t) = f(cos t)/(−sin t) = −g(t).\nCombining this with (4) yields C = −C, hence C = 0.  Therefore\n              g(t)=0 (∀t∈D).\n\nStep 7 – Vanishing of f on (−1,1).\nIf x∈(−1,1) we can write x = cos t with t∈D, so by (2)\n              f(x) = f(cos t) = g(t)·sin t = 0.\nThus f(x)=0 for every x in (−1,1).\n\nStep 8 – The endpoints ±1.\nUsing (1) with x = 1 gives f(1) = 2·1·f(1), hence f(1)=0.  Using x = −1 gives f(−1) = −2·1·f(−1), so f(−1)=0 as well.\n\nConclusion.  f(x)=0 for all x ∈ [−1,1], as was to be shown.",
      "_meta": {
        "core_steps": [
          "Substitute x = cos t and set g(t) = f(cos t)/sin t",
          "Use cos 2t = 2cos²t−1 and sin 2t = 2sin t cos t to get g(t+π)=g(t) and g(2t)=g(t)",
          "Iterate those two relations to show g is equal on a dense set; continuity ⇒ g is constant",
          "Since g(−t) = −g(t), the only constant is 0",
          "Back-substitute to obtain f(x)=0 on (−1,1) and check endpoints with the original equation"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The fixed starting point t₀ at which g is first evaluated; it only needs to avoid multiples of π",
            "original": "1"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}