{
  "index": "1962-B-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "6. Let\n\\[\nf(x)=\\sum_{k=0}^{n} a_{k} \\sin k x+b_{k} \\cos k x\n\\]\nwhere \\( a_{k} \\) and \\( b_{k} \\) are constants. Show that, if \\( |f(x)| \\leq 1 \\) for \\( 0 \\leq x \\leq 2 \\pi \\) and \\( \\left|f\\left(x_{i}\\right)\\right|=1 \\) for \\( 0 \\leq x_{1}<x_{2}<\\cdots<x_{2 n}<2 \\pi \\), then \\( f(x)=\\cos (n x+a) \\) for some constant \\( a \\).",
  "solution": "Solution. By writing \\( \\cos k x=\\frac{1}{2}\\left(e^{i k x}+e^{-i k x}\\right), \\sin k x=(1 / 2 i)\\left(e^{i k x}-e^{-i k x}\\right) \\) and \\( z=e^{i x} \\), any trigonometric polynomial \\( f(x) \\) of degree less than or equal to \\( n \\) (i.e., a function of the given form) can be written in the alternative form\n\\[\nf(x)=z^{-n} P(z)\n\\]\nwhere \\( P(z) \\) is an ordinary polynomial (with complex coefficients) of degree less than or equal to \\( 2 n \\).\n\nEach zero \\( b \\) of the trigonometric polynomial corresponds to a zero \\( e^{i b} \\) of \\( P(z) \\) on the unit circle in the complex plane. Differentiating, and noting that \\( d z / d x=i z \\), we find\n\\[\nf^{\\prime}(x)=i z^{-n}\\left(z P^{\\prime}(z)-n P(z)\\right) .\n\\]\n\nHence if \\( b \\) is a double root of \\( f\\left(\\right. \\) i.e., \\( f(b)=f^{\\prime}(b)=0 \\) ), then \\( P \\) has a double zero at \\( e^{\\text {ib }} \\). [And similarly for triple zeros, etc., but we shall not need that result.] Hence a non-zero trigonometric polynomial of degree \\( n \\) can have no more than \\( 2 n \\) zeros in one period counting multiplicities (because, if so, the ordinary polynomial \\( \\boldsymbol{P} \\) would have more zeros than its degree). Also if it has \\( 2 n \\) zeros, then it is determined by them, up to a constant multiple.\n\nNow consider the given trigonometric polynomial \\( f \\). We must assume that \\( f \\)\nhas real coefficients since for complex coefficients the result fails; e.g., for \\( f^{\\prime}(x)=\\cos x+i \\sin x \\). It also seems clear that the intent of the problem is that \\( f \\) is not a constant. Note that either \\( f(x) \\equiv 1 \\) or \\( f(x) \\equiv-1 \\) would satisfy the condition of the problem but not the conclusion if \\( n>0 \\).\n\nWe see that\n\\[\n1-f(x)^{2} \\text { and } f^{\\prime}(x)^{2}\n\\]\nare both non-negative trigonometric polynomials of degree \\( 2 n \\) with double zeros at \\( 2 n \\) distinct points \\( x_{1}, x_{2}, \\ldots, x_{2 n} \\). Since we are assuming that \\( 1- \\) [.\\( \\left.f^{\\prime}(x)\\right]^{2} \\) is not identically zero, there is a constant \\( m \\geq 0 \\) such that\n\\[\nf^{\\prime}(x)^{2}=m^{2}\\left\\lceil 1-f(x)^{2}\\right\\rceil .\n\\]\n\nThe possibility \\( m=0 \\) leads to the conclusion that \\( f \\) is a constant, so we assume \\( m \\neq 0 \\).\n\nThis differential equation has solutions of the form\n\\[\nf(x)=\\cos (m x+a)\n\\]\npieced together with segments of the form \\( f(x)= \\pm 1 \\). [The differential equation is singular for \\( f= \\pm 1 \\), and splitting of solutions may indeed occur along these lines; e.g.,\n\\[\n\\begin{aligned}\nf^{\\prime}(x) & =1 \\quad \\text { for } x \\leq 0 \\\\\n& =\\cos m x \\quad \\text { for } 0<x \\leq \\frac{\\pi}{m} \\\\\n& =-1 \\quad \\text { for } x \\geq \\frac{\\pi}{m}\n\\end{aligned}\n\\]\nis a solution.|\nWe are interested only in solutions that are trigonometric polynomials of degree \\( \\leq n \\). and since these are analytic functions of period \\( 2 \\pi \\) there can be no piecing and hence they must have the form (2) with \\( m \\) an integer. When \\( f \\) is of the form (2), \\( |f(x)|=1 \\) for exactly \\( 2 m \\) values of \\( x \\). Hence \\( m=n \\) and we have established the desired conclusion.",
  "vars": [
    "x",
    "k",
    "n",
    "z",
    "b"
  ],
  "params": [
    "f",
    "a_k",
    "b_k",
    "a",
    "P",
    "m"
  ],
  "sci_consts": [
    "e",
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "anglevar",
        "k": "indexer",
        "n": "degree",
        "z": "complexz",
        "b": "rootval",
        "f": "trigpoly",
        "a_k": "coeffa",
        "b_k": "coeffb",
        "a": "phasec",
        "P": "polyfun",
        "m": "scalar"
      },
      "question": "6. Let\n\\[\ntrigpoly(anglevar)=\\sum_{indexer=0}^{degree} coeffa \\sin indexer anglevar+coeffb \\cos indexer anglevar\n\\]\nwhere \\( coeffa \\) and \\( coeffb \\) are constants. Show that, if \\( |trigpoly(anglevar)| \\leq 1 \\) for \\( 0 \\leq anglevar \\leq 2 \\pi \\) and \\( \\left|trigpoly\\left(anglevar_{i}\\right)\\right|=1 \\) for \\( 0 \\leq anglevar_{1}<anglevar_{2}<\\cdots<anglevar_{2 degree}<2 \\pi \\), then \\( trigpoly(anglevar)=\\cos (degree anglevar+phasec) \\) for some constant \\( phasec \\).",
      "solution": "Solution. By writing \\( \\cos indexer anglevar=\\frac{1}{2}\\left(e^{i indexer anglevar}+e^{-i indexer anglevar}\\right), \\sin indexer anglevar=(1 / 2 i)\\left(e^{i indexer anglevar}-e^{-i indexer anglevar}\\right) \\) and \\( complexz=e^{i anglevar} \\), any trigonometric polynomial \\( trigpoly(anglevar) \\) of degree less than or equal to \\( degree \\) (i.e., a function of the given form) can be written in the alternative form\n\\[\ntrigpoly(anglevar)=complexz^{-degree} polyfun(complexz)\n\\]\nwhere \\( polyfun(complexz) \\) is an ordinary polynomial (with complex coefficients) of degree less than or equal to \\( 2 degree \\).\n\nEach zero \\( rootval \\) of the trigonometric polynomial corresponds to a zero \\( e^{i rootval} \\) of \\( polyfun(complexz) \\) on the unit circle in the complex plane. Differentiating, and noting that \\( d complexz / d anglevar=i complexz \\), we find\n\\[\ntrigpoly^{\\prime}(anglevar)=i complexz^{-degree}\\left(complexz polyfun^{\\prime}(complexz)-degree polyfun(complexz)\\right) .\n\\]\n\nHence if \\( rootval \\) is a double root of \\( trigpoly\\left(\\right. \\) i.e., \\( trigpoly(rootval)=trigpoly^{\\prime}(rootval)=0 \\) ), then \\( polyfun \\) has a double zero at \\( e^{\\text {i rootval }} \\). [And similarly for triple zeros, etc., but we shall not need that result.] Hence a non-zero trigonometric polynomial of degree \\( degree \\) can have no more than \\( 2 degree \\) zeros in one period counting multiplicities (because, if so, the ordinary polynomial \\( \\boldsymbol{polyfun} \\) would have more zeros than its degree). Also if it has \\( 2 degree \\) zeros, then it is determined by them, up to a constant multiple.\n\nNow consider the given trigonometric polynomial \\( trigpoly \\). We must assume that \\( trigpoly \\)\nhas real coefficients since for complex coefficients the result fails; e.g., for \\( trigpoly^{\\prime}(anglevar)=\\cos anglevar+i \\sin anglevar \\). It also seems clear that the intent of the problem is that \\( trigpoly \\) is not a constant. Note that either \\( trigpoly(anglevar) \\equiv 1 \\) or \\( trigpoly(anglevar) \\equiv-1 \\) would satisfy the condition of the problem but not the conclusion if \\( degree>0 \\).\n\nWe see that\n\\[\n1-trigpoly(anglevar)^{2} \\text { and } trigpoly^{\\prime}(anglevar)^{2}\n\\]\nare both non-negative trigonometric polynomials of degree \\( 2 degree \\) with double zeros at \\( 2 degree \\) distinct points \\( anglevar_{1}, anglevar_{2}, \\ldots, anglevar_{2 degree} \\). Since we are assuming that \\( 1- \\) [.\\( \\left.trigpoly^{\\prime}(anglevar)\\right]^{2} \\) is not identically zero, there is a constant \\( scalar \\geq 0 \\) such that\n\\[\ntrigpoly^{\\prime}(anglevar)^{2}=scalar^{2}\\left\\lceil 1-trigpoly(anglevar)^{2}\\right\\rceil .\n\\]\n\nThe possibility \\( scalar=0 \\) leads to the conclusion that \\( trigpoly \\) is a constant, so we assume \\( scalar \\neq 0 \\).\n\nThis differential equation has solutions of the form\n\\[\ntrigpoly(anglevar)=\\cos (scalar anglevar+phasec)\n\\]\npieced together with segments of the form \\( trigpoly(anglevar)= \\pm 1 \\). [The differential equation is singular for \\( trigpoly= \\pm 1 \\), and splitting of solutions may indeed occur along these lines; e.g.,\n\\[\n\\begin{aligned}\ntrigpoly^{\\prime}(anglevar) & =1 \\quad \\text { for } anglevar \\leq 0 \\\\\n& =\\cos scalar anglevar \\quad \\text { for } 0<anglevar \\leq \\frac{\\pi}{scalar} \\\\\n& =-1 \\quad \\text { for } anglevar \\geq \\frac{\\pi}{scalar}\n\\end{aligned}\n\\]\nis a solution.\nWe are interested only in solutions that are trigonometric polynomials of degree \\( \\leq degree \\), and since these are analytic functions of period \\( 2 \\pi \\) there can be no piecing and hence they must have the form (2) with \\( scalar \\) an integer. When \\( trigpoly \\) is of the form (2), \\( |trigpoly(anglevar)|=1 \\) for exactly \\( 2 scalar \\) values of \\( anglevar \\). Hence \\( scalar=degree \\) and we have established the desired conclusion."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "blueberry",
        "k": "cardboard",
        "n": "varnishings",
        "z": "rattlesnake",
        "b": "toothpicks",
        "f": "lighthouse",
        "a_k": "cappuccino",
        "b_k": "marshmallow",
        "a": "thunderstorms",
        "P": "caterpillar",
        "m": "chandelier"
      },
      "question": "6. Let\n\\[\nlighthouse(blueberry)=\\sum_{cardboard=0}^{varnishings} cappuccino \\sin cardboard blueberry+marshmallow \\cos cardboard blueberry\n\\]\nwhere \\( cappuccino \\) and \\( marshmallow \\) are constants. Show that, if \\( |lighthouse(blueberry)| \\leq 1 \\) for \\( 0 \\leq blueberry \\leq 2 \\pi \\) and \\( \\left|lighthouse\\left(blueberry_{i}\\right)\\right|=1 \\) for \\( 0 \\leq blueberry_{1}<blueberry_{2}<\\cdots<blueberry_{2 varnishings}<2 \\pi \\), then \\( lighthouse(blueberry)=\\cos (varnishings blueberry+thunderstorms) \\) for some constant \\( thunderstorms \\).",
      "solution": "Solution. By writing \\( \\cos cardboard blueberry=\\frac{1}{2}\\left(e^{i cardboard blueberry}+e^{-i cardboard blueberry}\\right), \\sin cardboard blueberry=(1 / 2 i)\\left(e^{i cardboard blueberry}-e^{-i cardboard blueberry}\\right) \\) and \\( rattlesnake=e^{i blueberry} \\), any trigonometric polynomial \\( lighthouse(blueberry) \\) of degree less than or equal to \\( varnishings \\) (i.e., a function of the given form) can be written in the alternative form\n\\[\nlighthouse(blueberry)=rattlesnake^{-varnishings} caterpillar(rattlesnake)\n\\]\nwhere \\( caterpillar(rattlesnake) \\) is an ordinary polynomial (with complex coefficients) of degree less than or equal to \\( 2 varnishings \\).\n\nEach zero \\( toothpicks \\) of the trigonometric polynomial corresponds to a zero \\( e^{i toothpicks} \\) of \\( caterpillar(rattlesnake) \\) on the unit circle in the complex plane. Differentiating, and noting that \\( d rattlesnake / d blueberry=i rattlesnake \\), we find\n\\[\nlighthouse^{\\prime}(blueberry)=i rattlesnake^{-varnishings}\\left(rattlesnake caterpillar^{\\prime}(rattlesnake)-varnishings caterpillar(rattlesnake)\\right) .\n\\]\n\nHence if \\( toothpicks \\) is a double root of \\( lighthouse\\left(\\right. \\) i.e., \\( lighthouse(toothpicks)=lighthouse^{\\prime}(toothpicks)=0 \\) ), then \\( caterpillar \\) has a double zero at \\( e^{\\text {itoothpicks }} \\). [And similarly for triple zeros, etc., but we shall not need that result.] Hence a non-zero trigonometric polynomial of degree \\( varnishings \\) can have no more than \\( 2 varnishings \\) zeros in one period counting multiplicities (because, if so, the ordinary polynomial \\( \\boldsymbol{caterpillar} \\) would have more zeros than its degree). Also if it has \\( 2 varnishings \\) zeros, then it is determined by them, up to a constant multiple.\n\nNow consider the given trigonometric polynomial \\( lighthouse \\). We must assume that \\( lighthouse \\)\nhas real coefficients since for complex coefficients the result fails; e.g., for \\( lighthouse^{\\prime}(blueberry)=\\cos blueberry+i \\sin blueberry \\). It also seems clear that the intent of the problem is that \\( lighthouse \\) is not a constant. Note that either \\( lighthouse(blueberry) \\equiv 1 \\) or \\( lighthouse(blueberry) \\equiv-1 \\) would satisfy the condition of the problem but not the conclusion if \\( varnishings>0 \\).\n\nWe see that\n\\[\n1-lighthouse(blueberry)^{2} \\text { and } lighthouse^{\\prime}(blueberry)^{2}\n\\]\nare both non-negative trigonometric polynomials of degree \\( 2 varnishings \\) with double zeros at \\( 2 varnishings \\) distinct points \\( blueberry_{1}, blueberry_{2}, \\ldots, blueberry_{2 varnishings} \\). Since we are assuming that \\( 1- \\) [.\\( \\left.lighthouse^{\\prime}(blueberry)\\right]^{2} \\) is not identically zero, there is a constant \\( chandelier \\geq 0 \\) such that\n\\[\nlighthouse^{\\prime}(blueberry)^{2}=chandelier^{2}\\left\\lceil 1-lighthouse(blueberry)^{2}\\right\\rceil .\n\\]\n\nThe possibility \\( chandelier=0 \\) leads to the conclusion that \\( lighthouse \\) is a constant, so we assume \\( chandelier \\neq 0 \\).\n\nThis differential equation has solutions of the form\n\\[\nlighthouse(blueberry)=\\cos (chandelier blueberry+thunderstorms)\n\\]\npieced together with segments of the form \\( lighthouse(blueberry)= \\pm 1 \\). [The differential equation is singular for \\( lighthouse= \\pm 1 \\), and splitting of solutions may indeed occur along these lines; e.g.,\n\\[\n\\begin{aligned}\nlighthouse^{\\prime}(blueberry) & =1 \\quad \\text { for } blueberry \\leq 0 \\\\\n& =\\cos chandelier blueberry \\quad \\text { for } 0<blueberry \\leq \\frac{\\pi}{chandelier} \\\\\n& =-1 \\quad \\text { for } blueberry \\geq \\frac{\\pi}{chandelier}\n\\end{aligned}\n\\]\nis a solution.| We are interested only in solutions that are trigonometric polynomials of degree \\( \\leq varnishings \\). and since these are analytic functions of period \\( 2 \\pi \\) there can be no piecing and hence they must have the form (2) with \\( chandelier \\) an integer. When \\( lighthouse \\) is of the form (2), \\( |lighthouse(blueberry)|=1 \\) for exactly \\( 2 chandelier \\) values of \\( blueberry \\). Hence \\( chandelier=varnishings \\) and we have established the desired conclusion."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "fixedvalue",
        "k": "continuum",
        "n": "leastvalue",
        "z": "realnumber",
        "b": "peakvalue",
        "f": "constant",
        "a_k": "consequence",
        "b_k": "invariant",
        "a": "steadystate",
        "P": "transcendental",
        "m": "noninteger"
      },
      "question": "6. Let\n\\[\nconstant(fixedvalue)=\\sum_{continuum=0}^{leastvalue} consequence \\sin continuum fixedvalue+invariant \\cos continuum fixedvalue\n\\]\nwhere \\( consequence \\) and \\( invariant \\) are constants. Show that, if \\( |constant(fixedvalue)| \\leq 1 \\) for \\( 0 \\leq fixedvalue \\leq 2 \\pi \\) and \\( \\left|constant\\left(fixedvalue_{i}\\right)\\right|=1 \\) for \\( 0 \\leq fixedvalue_{1}<fixedvalue_{2}<\\cdots<fixedvalue_{2 leastvalue}<2 \\pi \\), then \\( constant(fixedvalue)=\\cos (leastvalue fixedvalue+steadystate) \\) for some constant \\( steadystate \\).",
      "solution": "Solution. By writing \\( \\cos continuum fixedvalue=\\frac{1}{2}\\left(e^{i continuum fixedvalue}+e^{-i continuum fixedvalue}\\right), \\sin continuum fixedvalue=(1 / 2 i)\\left(e^{i continuum fixedvalue}-e^{-i continuum fixedvalue}\\right) \\) and \\( realnumber=e^{i fixedvalue} \\), any trigonometric polynomial \\( constant(fixedvalue) \\) of degree less than or equal to \\( leastvalue \\) (i.e., a function of the given form) can be written in the alternative form\n\\[\nconstant(fixedvalue)=realnumber^{-leastvalue} transcendental(realnumber)\n\\]\nwhere \\( transcendental(realnumber) \\) is an ordinary polynomial (with complex coefficients) of degree less than or equal to \\( 2 leastvalue \\).\n\nEach zero \\( peakvalue \\) of the trigonometric polynomial corresponds to a zero \\( e^{i peakvalue} \\) of \\( transcendental(realnumber) \\) on the unit circle in the complex plane. Differentiating, and noting that \\( d realnumber / d fixedvalue=i realnumber \\), we find\n\\[\nconstant^{\\prime}(fixedvalue)=i realnumber^{-leastvalue}\\left(realnumber transcendental^{\\prime}(realnumber)-leastvalue transcendental(realnumber)\\right) .\n\\]\n\nHence if \\( peakvalue \\) is a double root of \\( constant\\left(\\right. \\) i.e., \\( constant(peakvalue)=constant^{\\prime}(peakvalue)=0 \\) ), then \\( transcendental \\) has a double zero at \\( e^{\\text {i peakvalue }} \\). [And similarly for triple zeros, etc., but we shall not need that result.] Hence a non-zero trigonometric polynomial of degree \\( leastvalue \\) can have no more than \\( 2 leastvalue \\) zeros in one period counting multiplicities (because, if so, the ordinary polynomial \\( \\boldsymbol{transcendental} \\) would have more zeros than its degree). Also if it has \\( 2 leastvalue \\) zeros, then it is determined by them, up to a constant multiple.\n\nNow consider the given trigonometric polynomial \\( constant \\). We must assume that \\( constant \\)\nhas real coefficients since for complex coefficients the result fails; e.g., for \\( constant^{\\prime}(fixedvalue)=\\cos fixedvalue+i \\sin fixedvalue \\). It also seems clear that the intent of the problem is that \\( constant \\) is not a constant. Note that either \\( constant(fixedvalue) \\equiv 1 \\) or \\( constant(fixedvalue) \\equiv-1 \\) would satisfy the condition of the problem but not the conclusion if \\( leastvalue>0 \\).\n\nWe see that\n\\[\n1-constant(fixedvalue)^{2} \\text { and } constant^{\\prime}(fixedvalue)^{2}\n\\]\nare both non-negative trigonometric polynomials of degree \\( 2 leastvalue \\) with double zeros at \\( 2 leastvalue \\) distinct points \\( fixedvalue_{1}, fixedvalue_{2}, \\ldots, fixedvalue_{2 leastvalue} \\). Since we are assuming that \\( 1- \\) [.\\( \\left.constant^{\\prime}(fixedvalue)\\right]^{2} \\) is not identically zero, there is a constant \\( noninteger \\geq 0 \\) such that\n\\[\nconstant^{\\prime}(fixedvalue)^{2}=noninteger^{2}\\left\\lceil 1-constant(fixedvalue)^{2}\\right\\rceil .\n\\]\n\nThe possibility \\( noninteger=0 \\) leads to the conclusion that \\( constant \\) is a constant, so we assume \\( noninteger \\neq 0 \\).\n\nThis differential equation has solutions of the form\n\\[\nconstant(fixedvalue)=\\cos (noninteger fixedvalue+steadystate)\n\\]\npieced together with segments of the form \\( constant(fixedvalue)= \\pm 1 \\). [The differential equation is singular for \\( constant= \\pm 1 \\), and splitting of solutions may indeed occur along these lines; e.g.,\n\\[\n\\begin{aligned}\nconstant^{\\prime}(fixedvalue) & =1 \\quad \\text { for } fixedvalue \\leq 0 \\\\\n& =\\cos noninteger fixedvalue \\quad \\text { for } 0<fixedvalue \\leq \\frac{\\pi}{noninteger} \\\\\n& =-1 \\quad \\text { for } fixedvalue \\geq \\frac{\\pi}{noninteger}\n\\end{aligned}\n\\]\nis a solution.| We are interested only in solutions that are trigonometric polynomials of degree \\( \\leq leastvalue \\). and since these are analytic functions of period \\( 2 \\pi \\) there can be no piecing and hence they must have the form (2) with \\( noninteger \\) an integer. When \\( constant \\) is of the form (2), \\( |constant(fixedvalue)|=1 \\) for exactly \\( 2 noninteger \\) values of \\( fixedvalue \\). Hence \\( noninteger=leastvalue \\) and we have established the desired conclusion."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "k": "hjgrksla",
        "n": "vdflqeru",
        "z": "wmnlkyop",
        "b": "cmrjtfsa",
        "f": "lksnehtv",
        "a_k": "qprncyxa",
        "b_k": "zhtmeure",
        "a": "xzgdlwqh",
        "P": "nvaeokcy",
        "m": "udksifgh"
      },
      "question": "6. Let\n\\[\nlksnehtv(qzxwvtnp)=\\sum_{hjgrksla=0}^{vdflqeru} qprncyxa \\sin hjgrksla qzxwvtnp+zhtmeure \\cos hjgrksla qzxwvtnp\n\\]\nwhere \\( qprncyxa \\) and \\( zhtmeure \\) are constants. Show that, if \\( |lksnehtv(qzxwvtnp)| \\leq 1 \\) for \\( 0 \\leq qzxwvtnp \\leq 2 \\pi \\) and \\( \\left|lksnehtv\\left(qzxwvtnp_{i}\\right)\\right|=1 \\) for \\( 0 \\leq qzxwvtnp_{1}<qzxwvtnp_{2}<\\cdots<qzxwvtnp_{2 vdflqeru}<2 \\pi \\), then \\( lksnehtv(qzxwvtnp)=\\cos (vdflqeru qzxwvtnp+xzgdlwqh) \\) for some constant \\( xzgdlwqh \\).",
      "solution": "Solution. By writing \\( \\cos hjgrksla qzxwvtnp=\\frac{1}{2}\\left(e^{i hjgrksla qzxwvtnp}+e^{-i hjgrksla qzxwvtnp}\\right), \\sin hjgrksla qzxwvtnp=(1 / 2 i)\\left(e^{i hjgrksla qzxwvtnp}-e^{-i hjgrksla qzxwvtnp}\\right) \\) and \\( wmnlkyop=e^{i qzxwvtnp} \\), any trigonometric polynomial \\( lksnehtv(qzxwvtnp) \\) of degree less than or equal to \\( vdflqeru \\) (i.e., a function of the given form) can be written in the alternative form\n\\[\nlksnehtv(qzxwvtnp)=wmnlkyop^{-vdflqeru} nvaeokcy(wmnlkyop)\n\\]\nwhere \\( nvaeokcy(wmnlkyop) \\) is an ordinary polynomial (with complex coefficients) of degree less than or equal to \\( 2 vdflqeru \\).\n\nEach zero \\( cmrjtfsa \\) of the trigonometric polynomial corresponds to a zero \\( e^{i cmrjtfsa} \\) of \\( nvaeokcy(wmnlkyop) \\) on the unit circle in the complex plane. Differentiating, and noting that \\( d wmnlkyop / d qzxwvtnp=i wmnlkyop \\), we find\n\\[\nlksnehtv^{\\prime}(qzxwvtnp)=i wmnlkyop^{-vdflqeru}\\left(wmnlkyop nvaeokcy^{\\prime}(wmnlkyop)-vdflqeru nvaeokcy(wmnlkyop)\\right) .\n\\]\n\nHence if \\( cmrjtfsa \\) is a double root of \\( lksnehtv\\left(\\right. \\) i.e., \\( lksnehtv(cmrjtfsa)=lksnehtv^{\\prime}(cmrjtfsa)=0 \\) ), then \\( nvaeokcy \\) has a double zero at \\( e^{\\text {icmrjtfsa }} \\). [And similarly for triple zeros, etc., but we shall not need that result.] Hence a non-zero trigonometric polynomial of degree \\( vdflqeru \\) can have no more than \\( 2 vdflqeru \\) zeros in one period counting multiplicities (because, if so, the ordinary polynomial \\( \\boldsymbol{nvaeokcy} \\) would have more zeros than its degree). Also if it has \\( 2 vdflqeru \\) zeros, then it is determined by them, up to a constant multiple.\n\nNow consider the given trigonometric polynomial \\( lksnehtv \\). We must assume that \\( lksnehtv \\)\nhas real coefficients since for complex coefficients the result fails; e.g., for \\( lksnehtv^{\\prime}(qzxwvtnp)=\\cos qzxwvtnp+i \\sin qzxwvtnp \\). It also seems clear that the intent of the problem is that \\( lksnehtv \\) is not a constant. Note that either \\( lksnehtv(qzxwvtnp) \\equiv 1 \\) or \\( lksnehtv(qzxwvtnp) \\equiv-1 \\) would satisfy the condition of the problem but not the conclusion if \\( vdflqeru>0 \\).\n\nWe see that\n\\[\n1-lksnehtv(qzxwvtnp)^{2} \\text { and } lksnehtv^{\\prime}(qzxwvtnp)^{2}\n\\]\nare both non-negative trigonometric polynomials of degree \\( 2 vdflqeru \\) with double zeros at \\( 2 vdflqeru \\) distinct points \\( qzxwvtnp_{1}, qzxwvtnp_{2}, \\ldots, qzxwvtnp_{2 vdflqeru} \\). Since we are assuming that \\( 1- \\) [.\\( \\left.lksnehtv^{\\prime}(qzxwvtnp)\\right]^{2} \\) is not identically zero, there is a constant \\( udksifgh \\geq 0 \\) such that\n\\[\nlksnehtv^{\\prime}(qzxwvtnp)^{2}=udksifgh^{2}\\left\\lceil 1-lksnehtv(qzxwvtnp)^{2}\\right\\rceil .\n\\]\n\nThe possibility \\( udksifgh=0 \\) leads to the conclusion that \\( lksnehtv \\) is a constant, so we assume \\( udksifgh \\neq 0 \\).\n\nThis differential equation has solutions of the form\n\\[\nlksnehtv(qzxwvtnp)=\\cos (udksifgh qzxwvtnp+xzgdlwqh)\n\\]\npieced together with segments of the form \\( lksnehtv(qzxwvtnp)= \\pm 1 \\). [The differential equation is singular for \\( lksnehtv= \\pm 1 \\), and splitting of solutions may indeed occur along these lines; e.g.,\n\\[\n\\begin{aligned}\nlksnehtv^{\\prime}(qzxwvtnp) & =1 \\quad \\text { for } qzxwvtnp \\leq 0 \\\\\n& =\\cos udksifgh qzxwvtnp \\quad \\text { for } 0<qzxwvtnp \\leq \\frac{\\pi}{udksifgh} \\\\\n& =-1 \\quad \\text { for } qzxwvtnp \\geq \\frac{\\pi}{udksifgh}\n\\end{aligned}\n\\]\nis a solution.|\nWe are interested only in solutions that are trigonometric polynomials of degree \\( \\leq vdflqeru \\). and since these are analytic functions of period \\( 2 \\pi \\) there can be no piecing and hence they must have the form (2) with \\( udksifgh \\) an integer. When \\( lksnehtv \\) is of the form (2), \\( |lksnehtv(qzxwvtnp)|=1 \\) for exactly \\( 2 udksifgh \\) values of \\( qzxwvtnp \\). Hence \\( udksifgh=vdflqeru \\) and we have established the desired conclusion."
    },
    "kernel_variant": {
      "question": "Let n be a positive integer and define\ng(x)=\\displaystyle\\sum_{k=-n}^{n} c_k e^{ikx}, \\qquad c_{-k}=\\overline{c_k}\\;(0\\le k\\le n),\nso that g is a real trigonometric polynomial of exact degree n (i.e. c_{\\pm n}\\ne0).\nAssume that on the interval I=[-\\pi ,\\pi]\n|g(x)|\\le 2 \\qquad(x\\in I),\nand that for 2n pairwise different points\n-\\pi<y_1<y_2<\\dots <y_{2n}<\\pi \\quad\\text{one has}\\quad |g(y_j)|=2\\; (j=1,\\dots ,2n).\nProve that there is a real constant \\beta such that\n\\boxed{\\;g(x)=2\\cos (nx+\\beta)\\;}\\qquad(\\text{for every real }x).",
      "solution": "Throughout we write z=e^{ix}\\;(x\\in\\mathbb R).  Indices always run over their indicated ranges.\n\nSTEP 0 –  Non-constancy.\nBecause the coefficients c_{\\pm n} are non–zero, g has degree n\\,(\\ge 1) and is therefore not constant.  We shall use this fact repeatedly.\n\nSTEP 1 –  From a trigonometric to an algebraic polynomial.\nSince e^{ikx}=z^{k},\n g(x)=\\sum_{k=-n}^{n}c_k z^{k}=z^{-n}P(z), \\qquad P(z):=\\sum_{j=0}^{2n} c_{j-n}z^{j}.\nBecause c_{-k}=\\overline{c_k}, the coefficients of P satisfy\n \\overline{c_{j-n}}=c_{n-j},\nwhence the {\nself-reciprocal} identity\n \\boxed{\\;z^{2n}\\,\\overline{P(1/\\bar z)}=P(z)\\;} \\qquad(z\\in\\mathbb C\\setminus\\{0\\}).\nConsequently every zero of P that is not on the unit circle occurs in a reciprocal pair (\\xi,1/\\bar\\xi).  Zeros on |z|=1 occur either singly or with their complex conjugates.\n\nSTEP 2 –  How many zeros can a non-constant real trigonometric polynomial have?\nFor a zero b of g, the point z=e^{ib} is a zero of P.  Differentiating,\n g'(x)=iz^{-n}\\bigl(zP'(z)-nP(z)\\bigr),\nso multiplicities are preserved: if b is a root of multiplicity m for g, e^{ib} is a root of multiplicity m for P.  Hence a non-constant real trigonometric polynomial of degree n has at most 2n zeros (counted with multiplicities) in any interval of length 2\\pi.\n\nSTEP 3 –  Two auxiliary non-negative trigonometric polynomials.\nDefine\n T_1(x):=4-g(x)^2, \\qquad  T_2(x):=g'(x)^2.\nBoth are real and satisfy T_i(x)\\ge 0.  At every point y_j we have |g(y_j)|=2, so g attains a local extremum there; consequently g'(y_j)=0.  Thus\n T_1(y_j)=T_2(y_j)=0, \\; T_1'(y_j)=T_2'(y_j)=0 \\;(j=1,\\dots ,2n),\nso each y_j is a double zero of both T_1 and T_2.  Because the degree of g is n, the trigonometric degree of T_1 and T_2 is at most 2n; therefore on every 2\\pi-interval each T_i has at most 4n zeros but actually has \\emph{at least} 4n (the doubles at the y_j), hence exactly 4n.\n\nSTEP 4 –  Algebraic description of T_1 and T_2.\nWrite the Fourier series T_i(x)=\\sum_{k=-2n}^{2n}d_k^{(i)}z^{k}.  Multiplying by z^{2n} gives\n z^{2n}T_i(x)=Q_i(z), \\qquad Q_i(z):=\\sum_{k=0}^{4n}d_{k-2n}^{(i)}z^{k}.\nEquivalently,  T_i(x)=z^{-2n}Q_i(z).  For z=\\zeta_j:=e^{iy_j}\\,(|\\zeta_j|=1) we have Q_i(\\zeta_j)=0, and \\zeta_j is at least a double zero.  Because \\deg Q_i\\le 4n and Q_i possesses 4n zeros counted with multiplicity, we must have\n Q_i(z)=\\kappa_i\\prod_{j=1}^{2n}(z-\\zeta_j)^2\\quad(\\kappa_i\\ne0,\\;i=1,2).\nThus Q_2 and Q_1 are proportional:\n Q_2(z)=\\lambda Q_1(z)\\quad(z\\in\\mathbb C)\nfor some constant \\lambda\\ge0.  Since g is not constant, T_2\\not\\equiv0, hence \\lambda>0.  Returning to x we obtain\n \\boxed{\\;g'(x)^2=\\lambda\\bigl(4-g(x)^2\\bigr)\\;} \\tag{1}\nfor every real x.\n\nSTEP 5 –  Solving the differential equation.\nWrite \\lambda=m^2 with m>0.  Equation (1) is separable:\n \\frac{g'(x)}{\\sqrt{4-g(x)^2}}=\\pm m.\nSelecting the plus sign (the minus sign merely changes the phase) and integrating, we get\n \\arcsin\\!\\frac{g(x)}{2}=mx+\\beta\\quad(\\beta\\in\\mathbb R),\nhence\n \\boxed{\\;g(x)=2\\cos(mx+\\beta)\\;} \\tag{2}\nfor all real x.  The analyticity of g guarantees that (2) holds globally.\n\nSTEP 6 –  Identifying m.\nOver any interval of length 2\\pi the function (2) attains the level |g|=2 precisely 2m times (the interior points where mx+\\beta\\equiv k\\pi).  On the other hand g'(y_j)=0 for j=1,\\dots ,2n, so g' has at least 2n zeros in such an interval.  But g' is a trigonometric polynomial of degree n and therefore has at most 2n zeros (Step 2).  Hence g' has \\emph{exactly} those 2n zeros, forcing 2m=2n and therefore m=n.\n\nSTEP 7 –  Conclusion.\nThere exists \\beta\\in\\mathbb R such that\n g(x)=2\\cos(nx+\\beta)\\qquad(\\forall x\\in\\mathbb R),\nas desired.",
      "_meta": {
        "core_steps": [
          "Rewrite f(x) as z^{-n} P(z) (z=e^{ix});  deg P ≤ 2n",
          "Degree bound ⇒ trigonometric poly has ≤ 2n zeros per period",
          "At extremal points  |f|=1 ⇒ 1−f² and (f′)² are ≥0 trigon. polys of deg 2n with the same double zeros, so  (f′)² = m²(1−f²)",
          "Solve ODE  (f′)² = m²(1−f²)  → analytic periodic solutions are  f(x)=cos(mx+a) (m∈ℤ, m≠0 for non-constant f)",
          "Count extremal points: |f| reaches its bound 2m times ⇒ m=n, hence f(x)=cos(nx+a)"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Numerical height of the bound on |f(x)|; any positive constant works after rescaling",
            "original": "1"
          },
          "slot2": {
            "description": "Specific 2π-length interval chosen for one period (e.g. [0,2π]); any contiguous length-2π interval would do",
            "original": "[0, 2π]"
          },
          "slot3": {
            "description": "Real-trigonometric basis (sin, cos) used to present f; any equivalent exponential basis e^{ikx} is acceptable",
            "original": "∑_{k=0}^{n} a_k sin kx + b_k cos kx"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}