{
  "index": "2000-B-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $f(t)=\\sum_{j=1}^N a_j \\sin(2\\pi jt)$, where each $a_j$ is real\nand\n$a_N$ is not equal to 0.  Let $N_k$ denote the number of zeroes (including\nmultiplicities) of $\\frac{d^k f}{dt^k}$.\nProve that\n\\[N_0\\leq N_1\\leq N_2\\leq \\cdots \\mbox{ and } \\lim_{k\\to\\infty} N_k =\n2N.\\]\n[Editorial clarification: only zeroes in $[0, 1)$ should be counted.]",
  "solution": "Put $f_k(t) = \\frac{df^k}{dt^k}$.\nRecall Rolle's theorem: if $f(t)$ is differentiable, then between any\ntwo zeroes of $f(t)$ there exists a zero of $f'(t)$. This also applies\nwhen the zeroes are not all distinct: if $f$ has a zero of multiplicity\n$m$ at $t=x$, then $f'$ has a zero of multiplicity at least $m-1$ there.\n\nTherefore, if $0 \\leq a_0 \\leq a_1 \\leq \\cdots \\leq a_r < 1$ are the roots\nof $f_k$ in $[0,1)$, then $f_{k+1}$ has a root\nin each of the intervals $(a_0, a_1), (a_1, a_2), \\dots, (a_{r-1}, a_r)$,\nso long as we\nadopt the convention that the empty interval $(t,t)$ actually contains\nthe point $t$ itself. There is also a root in the ``wraparound'' interval\n$(a_r, a_0)$. Thus $N_{k+1} \\geq N_k$.\n\nNext, note that if we set $z = e^{2\\pi i t}$; then\n\\[\nf_{4k}(t) = \\frac{1}{2i} \\sum_{j=1}^N j^{4k} a_j (z^j - z^{-j})\n\\]\nis equal to $z^{-N}$ times a polynomial of degree $2N$. Hence as a\nfunction of $z$, it has at most $2N$ roots; therefore $f_k(t)$ has\nat most $2N$ roots in $[0,1]$. That is, $N_k \\leq 2N$ for all $N$.\n\nTo establish that $N_k \\to 2N$, we make precise the observation that\n\\[\nf_k(t) = \\sum_{j=1}^N j^{4k} a_j \\sin(2\\pi j t)\n\\]\nis dominated by the term with $j=N$. At the points\n$t = (2i+1)/(2N)$ for $i=0,1, \\dots, N-1$, we have\n$N^{4k} a_N \\sin (2\\pi N t) = \\pm N^{4k} a_N$. If $k$ is chosen large enough\nso that\n\\[\n|a_N| N^{4k} > |a_1| 1^{4k} + \\cdots + |a_{N-1}| (N-1)^{4k},\n\\]\nthen $f_k((2i+1)/2N)$ has the same sign as $a_N \\sin (2\\pi N at)$,\nwhich is to say, the sequence $f_k(1/2N), f_k(3/2N), \\dots$ alternates\nin sign. Thus\nbetween these points (again including the ``wraparound'' interval) we find\n$2N$ sign changes of $f_k$. Therefore $\\lim_{k \\to \\infty} N_k = 2N$.",
  "vars": [
    "f",
    "f_k",
    "t",
    "j",
    "k",
    "x",
    "r",
    "i",
    "z"
  ],
  "params": [
    "N",
    "a_j",
    "a_N",
    "N_k",
    "a_0",
    "a_1",
    "a_N-1",
    "a_r"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "f": "trigpolyfun",
        "f_k": "kthderivative",
        "t": "unittime",
        "j": "harmonicindex",
        "k": "derivorder",
        "x": "rootpoint",
        "r": "rootindex",
        "i": "halfperiodindex",
        "z": "expvariable",
        "N": "highestfreq",
        "a_j": "amplitudelist",
        "a_N": "topamplitude",
        "N_k": "zerocountseq",
        "a_0": "firstamplitude",
        "a_1": "secondamplitude",
        "a_N-1": "nexttopelement",
        "a_r": "generalamp"
      },
      "question": "Let $\\trigpolyfun(\\unittime)=\\sum_{\\harmonicindex=1}^{\\highestfreq} \\amplitudelist \\sin(2\\pi \\harmonicindex \\unittime)$, where each $\\amplitudelist$ is real and $\\topamplitude$ is not equal to 0.  Let $\\zerocountseq$ denote the number of zeroes (including multiplicities) of $\\frac{d^{\\derivorder} \\trigpolyfun}{d\\unittime^{\\derivorder}}$. Prove that\n\\[\\zerocountseq_0\\leq \\zerocountseq_1\\leq \\zerocountseq_2\\leq \\cdots \\mbox{ and } \\lim_{\\derivorder\\to\\infty} \\zerocountseq = 2\\highestfreq.\\]\n[Editorial clarification: only zeroes in $[0, 1)$ should be counted.]",
      "solution": "Put $\\kthderivative(\\unittime)=\\frac{d^{\\derivorder}\\trigpolyfun}{d\\unittime^{\\derivorder}}$.  \nRecall Rolle's theorem: if $\\trigpolyfun(\\unittime)$ is differentiable, then between any two zeroes of $\\trigpolyfun(\\unittime)$ there exists a zero of $\\trigpolyfun'(\\unittime)$. This also applies when the zeroes are not all distinct: if $\\trigpolyfun$ has a zero of multiplicity $m$ at $\\unittime=\\rootpoint$, then $\\trigpolyfun'$ has a zero of multiplicity at least $m-1$ there.\n\nTherefore, if $0 \\le \\firstamplitude \\le \\secondamplitude \\le \\cdots \\le \\generalamp_{\\rootindex} < 1$ are the roots of $\\kthderivative$ in $[0,1)$, then $\\kthderivative_{\\derivorder+1}$ has a root in each of the intervals $(\\firstamplitude,\\secondamplitude),(\\secondamplitude,\\generalamp_2),\\dots,(\\generalamp_{\\rootindex-1},\\generalamp_{\\rootindex})$, so long as we adopt the convention that the empty interval $(\\unittime,\\unittime)$ actually contains the point $\\unittime$ itself. There is also a root in the “wrap-around’’ interval $(\\generalamp_{\\rootindex},\\firstamplitude)$. Thus $\\zerocountseq_{\\derivorder+1}\\ge \\zerocountseq_{\\derivorder}$.\n\nNext, note that if we set $\\expvariable=e^{2\\pi i\\unittime}$, then\n\\[\n\\kthderivative_{4\\derivorder}(\\unittime)=\\frac{1}{2i}\\sum_{\\harmonicindex=1}^{\\highestfreq}\\harmonicindex^{4\\derivorder}\\amplitudelist\\,(\\expvariable^{\\harmonicindex}-\\expvariable^{-\\harmonicindex})\n\\]\nis equal to $\\expvariable^{-\\highestfreq}$ times a polynomial of degree $2\\highestfreq$. Hence, as a function of $\\expvariable$, it has at most $2\\highestfreq$ roots; therefore $\\kthderivative(\\unittime)$ has at most $2\\highestfreq$ roots in $[0,1]$. That is, $\\zerocountseq\\le 2\\highestfreq$ for all $\\derivorder$.\n\nTo establish that $\\zerocountseq\\to 2\\highestfreq$, we make precise the observation that\n\\[\n\\kthderivative(\\unittime)=\\sum_{\\harmonicindex=1}^{\\highestfreq}\\harmonicindex^{4\\derivorder}\\amplitudelist\\sin(2\\pi \\harmonicindex \\unittime)\n\\]\nis dominated by the term with $\\harmonicindex=\\highestfreq$. At the points $\\unittime=(2\\halfperiodindex+1)/(2\\highestfreq)$ for $\\halfperiodindex=0,1,\\dots,\\highestfreq-1$, we have\n\\[\n\\highestfreq^{4\\derivorder}\\topamplitude\\sin(2\\pi \\highestfreq \\unittime)=\\pm\\highestfreq^{4\\derivorder}\\topamplitude.\n\\]\nIf $\\derivorder$ is chosen large enough so that\n\\[\n|\\topamplitude|\\,\\highestfreq^{4\\derivorder}>|\\amplitudelist|\\,1^{4\\derivorder}+\\cdots+|\\nexttopelement|\\,(\\highestfreq-1)^{4\\derivorder},\n\\]\nthen $\\kthderivative((2\\halfperiodindex+1)/2\\highestfreq)$ has the same sign as $\\topamplitude\\sin(2\\pi \\highestfreq \\unittime)$, which is to say, the sequence $\\kthderivative(1/2\\highestfreq),\\kthderivative(3/2\\highestfreq),\\dots$ alternates in sign. Thus between these points (again including the “wrap-around’’ interval) we find $2\\highestfreq$ sign changes of $\\kthderivative$. Therefore $\\lim_{\\derivorder\\to\\infty}\\zerocountseq=2\\highestfreq$. "
    },
    "descriptive_long_confusing": {
      "map": {
        "f": "sunflower",
        "f_k": "gemstone",
        "t": "waterfall",
        "j": "lighthouse",
        "k": "tornadoes",
        "x": "chocolate",
        "r": "sandstorm",
        "z": "butterfly",
        "N": "telescope",
        "a_j": "pinecone",
        "a_N": "raincloud",
        "N_k": "marshmallow",
        "a_0": "driftwood",
        "a_1": "starfruit",
        "a_N-1": "aftershock",
        "a_r": "goldsmith"
      },
      "question": "Let $sunflower(waterfall)=\\sum_{lighthouse=1}^{telescope} pinecone \\sin(2\\pi lighthouse\\,waterfall)$, where each $pinecone$ is real and $raincloud$ is not equal to 0.  Let $marshmallow$ denote the number of zeroes (including multiplicities) of $\\frac{d^{tornadoes} sunflower}{dwaterfall^{tornadoes}}$. Prove that\n\\[marshmallow_0\\leq marhsmallow_1\\leq marshmallow_2\\leq \\cdots \\mbox{ and } \\lim_{tornadoes\\to\\infty} marshmallow = 2telescope.\\]\n[Editorial clarification: only zeroes in $[0, 1)$ should be counted.]",
      "solution": "Put $gemstone(waterfall) = \\frac{d sunflower^{tornadoes}}{dwaterfall^{tornadoes}}$.\nRecall Rolle's theorem: if $sunflower(waterfall)$ is differentiable, then between any two zeroes of $sunflower(waterfall)$ there exists a zero of $sunflower'(waterfall)$. This also applies when the zeroes are not all distinct: if $sunflower$ has a zero of multiplicity $m$ at $waterfall=chocolate$, then $sunflower'$ has a zero of multiplicity at least $m-1$ there.\n\nTherefore, if $0 \\le driftwood \\le starfruit \\le \\cdots \\le goldsmith < 1$ are the roots of $gemstone$ in $[0,1)$, then $gemstone_{tornadoes+1}$ has a root in each of the intervals $(driftwood, starfruit), (starfruit, a_2), \\dots, (a_{sandstorm-1}, goldsmith)$, so long as we adopt the convention that the empty interval $(waterfall,waterfall)$ actually contains the point $waterfall$ itself. There is also a root in the ``wraparound'' interval $(goldsmith, driftwood)$. Thus $marshmallow_{tornadoes+1} \\ge marshmallow$.\n\nNext, note that if we set $butterfly = e^{2\\pi i\\,waterfall}$, then\n\\[\n gemston_{4tornadoes}(waterfall) = \\frac{1}{2i}\\sum_{lighthouse=1}^{telescope} lighthouse^{4tornadoes}\\, pinecone\\,(butterfly^{lighthouse}-butterfly^{-lighthouse})\n\\]\nis equal to $butterfly^{-telescope}$ times a polynomial of degree $2telescope$. Hence as a function of $butterfly$, it has at most $2telescope$ roots; therefore $gemstone(waterfall)$ has at most $2telescope$ roots in $[0,1]$. That is, $marshmallow \\le 2telescope$ for all $telescope$.\n\nTo establish that $marshmallow \\to 2telescope$, we make precise the observation that\n\\[\n gemston_{tornadoes}(waterfall)=\\sum_{lighthouse=1}^{telescope} lighthouse^{4tornadoes}\\, pinecone\\,\\sin(2\\pi lighthouse\\,waterfall)\n\\]\nis dominated by the term with $lighthouse=telescope$.  At the points $waterfall=(2i+1)/(2telescope)$ for $i=0,1,\\dots,telescope-1$, we have $telescope^{4tornadoes} raincloud \\sin(2\\pi telescope\\,waterfall)=\\pm telescope^{4tornadoes} raincloud$.  If $tornadoes$ is chosen large enough so that\n\\[\n |raincloud|\\,telescope^{4tornadoes} > |starfruit|\\,1^{4tornadoes}+\\cdots+|aftershock|\\,(telescope-1)^{4tornadoes},\n\\]\nthen $gemston_{tornadoes}((2i+1)/2telescope)$ has the same sign as $raincloud \\sin(2\\pi telescope\\,waterfall)$, which is to say, the sequence $gemston_{tornadoes}(1/2telescope), gemston_{tornadoes}(3/2telescope), \\dots$ alternates in sign. Thus between these points (again including the ``wraparound'' interval) we find $2telescope$ sign changes of $gemstone$. Therefore $\\lim_{tornadoes\\to\\infty}marshmallow = 2telescope$.}",
      "confidence": "0.10"
    },
    "descriptive_long_misleading": {
      "map": {
        "f": "nonmapping",
        "f_k": "nonderive",
        "t": "timeless",
        "j": "aggregate",
        "k": "staticindex",
        "x": "voidpoint",
        "r": "peakindex",
        "z": "realvalue",
        "N": "minimumbound",
        "a_j": "zeroeffect",
        "a_N": "nullterminal",
        "N_k": "rootlessnumber",
        "a_0": "farpointzero",
        "a_1": "farpointone",
        "a_N-1": "farpointprev",
        "a_r": "farpointr"
      },
      "question": "Let $nonmapping(timeless)=\\sum_{aggregate=1}^{minimumbound} zeroeffect \\sin(2\\pi aggregate timeless)$, where each $zeroeffect$ is real\nand\n$nullterminal$ is not equal to 0.  Let $rootlessnumber$ denote the number of zeroes (including\nmultiplicities) of $\\frac{d^{staticindex} nonmapping}{d timeless^{staticindex}}$.\nProve that\n\\[minimumbound_0\\leq minimumbound_1\\leq minimumbound_2\\leq \\cdots \\mbox{ and } \\lim_{staticindex\\to\\infty} rootlessnumber =\n2 minimumbound.\\]\n[Editorial clarification: only zeroes in $[0, 1)$ should be counted.]",
      "solution": "Put $nonderive(timeless) = \\frac{d nonmapping^{staticindex}}{d timeless^{staticindex}}$.\nRecall Rolle's theorem: if $nonmapping(timeless)$ is differentiable, then between any\ntwo zeroes of $nonmapping(timeless)$ there exists a zero of $nonmapping'(timeless)$. This also applies\nwhen the zeroes are not all distinct: if nonmapping has a zero of multiplicity\n$m$ at $timeless=voidpoint$, then $nonmapping'$ has a zero of multiplicity at least $m-1$ there.\n\nTherefore, if $0 \\leq farpointzero \\leq farpointone \\leq \\cdots \\leq farpointr < 1$ are the roots\nof nonderive in $[0,1)$, then $nonmapping_{staticindex+1}$ has a root\nin each of the intervals $(farpointzero, farpointone), (farpointone, a_2), \\dots, (a_{peakindex-1}, farpointr)$,\nso long as we\nadopt the convention that the empty interval $(timeless,timeless)$ actually contains\nthe point $timeless$ itself. There is also a root in the ``wraparound'' interval\n$(farpointr, farpointzero)$. Thus $minimumbound_{staticindex+1} \\geq rootlessnumber$.\n\nNext, note that if we set $realvalue = e^{2\\pi i timeless}$; then\n\\[\nnonmapping_{4staticindex}(timeless) = \\frac{1}{2i} \\sum_{aggregate=1}^{minimumbound} aggregate^{4staticindex} zeroeffect (realvalue^{aggregate} - realvalue^{-aggregate})\n\\]\nis equal to $realvalue^{-minimumbound}$ times a polynomial of degree $2 minimumbound$. Hence as a\nfunction of $realvalue$, it has at most $2 minimumbound$ roots; therefore nonderive(timeless) has\nat most $2 minimumbound$ roots in $[0,1]$. That is, $rootlessnumber \\leq 2 minimumbound$ for all $minimumbound$.\n\nTo establish that $rootlessnumber \\to 2 minimumbound$, we make precise the observation that\n\\[\nnonderive(timeless) = \\sum_{aggregate=1}^{minimumbound} aggregate^{4staticindex} zeroeffect \\sin(2\\pi aggregate timeless)\n\\]\nis dominated by the term with $aggregate=minimumbound$. At the points\n$timeless = (2i+1)/(2 minimumbound)$ for $i=0,1, \\dots, minimumbound-1$, we have\n$minimumbound^{4staticindex} nullterminal \\sin (2\\pi minimumbound timeless) = \\pm minimumbound^{4staticindex} nullterminal$. If $staticindex$ is chosen large enough\nso that\n\\[\n|nullterminal| minimumbound^{4staticindex} > |farpointone| 1^{4staticindex} + \\cdots + |farpointprev| (minimumbound-1)^{4staticindex},\n\\]\nthen nonderive((2i+1)/2 minimumbound) has the same sign as $nullterminal \\sin (2\\pi minimumbound timeless)$,\nwhich is to say, the sequence nonderive(1/2 minimumbound), nonderive(3/2 minimumbound), \\dots\\ alternates\nin sign. Thus\nbetween these points (again including the ``wraparound'' interval) we find\n$2 minimumbound$ sign changes of nonderive. Therefore $\\lim_{staticindex \\to \\infty} rootlessnumber = 2 minimumbound$.}",
      "confidence": "0.16"
    },
    "garbled_string": {
      "map": {
        "f": "qzxwvtnp",
        "f_k": "hjgrksla",
        "t": "bnmloiuy",
        "j": "plkmijnb",
        "k": "asdfghjk",
        "x": "zxcvbnml",
        "r": "ghjklpoi",
        "z": "cvbnhgfd",
        "N": "qwerasdf",
        "a_j": "poiulkjh",
        "a_N": "lkjhgfds",
        "N_k": "mnbvcxzq",
        "a_0": "zpoiuytr",
        "a_1": "asdfqwer",
        "a_N-1": "hgfdrewq",
        "a_r": "tyuioplk"
      },
      "question": "Let $qzxwvtnp(bnmloiuy)=\\sum_{plkmijnb=1}^{qwerasdf} poiulkjh \\sin(2\\pi plkmijnb bnmloiuy)$, where each $poiulkjh$ is real\nand\n$lkjhgfds$ is not equal to 0.  Let $mnbvcxzq$ denote the number of zeroes (including\nmultiplicities) of $\\frac{d^{asdfghjk} qzxwvtnp}{d bnmloiuy^{asdfghjk}}$.\nProve that\n\\[qwerasdf_0\\leq qwerasdf_1\\leq qwerasdf_2\\leq \\cdots \\mbox{ and } \\lim_{asdfghjk\\to\\infty} mnbvcxzq =\n2qwerasdf.\\]\n[Editorial clarification: only zeroes in $[0, 1)$ should be counted.]",
      "solution": "Put $hjgrksla(bnmloiuy) = \\frac{d qzxwvtnp^{asdfghjk}}{d bnmloiuy^{asdfghjk}}$.\n\nRecall Rolle's theorem: if $qzxwvtnp(bnmloiuy)$ is differentiable, then between any\ntwo zeroes of $qzxwvtnp(bnmloiuy)$ there exists a zero of $qzxwvtnp'(bnmloiuy)$. This also applies\nwhen the zeroes are not all distinct: if $qzxwvtnp$ has a zero of multiplicity\n$m$ at $bnmloiuy=zxcvbnml$, then $qzxwvtnp'$ has a zero of multiplicity at least $m-1$ there.\n\nTherefore, if $0 \\leq zpoiuytr \\leq asdfqwer \\leq \\cdots \\leq tyuioplk < 1$ are the roots\nof $hjgrksla$ in $[0,1)$, then $qzxwvtnp_{asdfghjk+1}$ has a root\nin each of the intervals $(zpoiuytr, asdfqwer), (asdfqwer, a_{ghjklpoi-1}), \\dots, (a_{ghjklpoi-1}, tyuioplk)$,\nso long as we\nadopt the convention that the empty interval $(bnmloiuy,bnmloiuy)$ actually contains\nthe point $bnmloiuy$ itself. There is also a root in the ``wraparound'' interval\n$(tyuioplk, zpoiuytr)$. Thus $qwerasdf_{asdfghjk+1} \\geq mnbvcxzq$.\n\nNext, note that if we set $cvbnhgfd = e^{2\\pi i bnmloiuy}$; then\n\\[\nqzxwvtnp_{4asdfghjk}(bnmloiuy) = \\frac{1}{2i} \\sum_{plkmijnb=1}^{qwerasdf} plkmijnb^{4asdfghjk} poiulkjh (cvbnhgfd^{plkmijnb} - cvbnhgfd^{-plkmijnb})\n\\]\nis equal to $cvbnhgfd^{-qwerasdf}$ times a polynomial of degree $2qwerasdf$. Hence as a\nfunction of $cvbnhgfd$, it has at most $2qwerasdf$ roots; therefore $hjgrksla(bnmloiuy)$ has\nat most $2qwerasdf$ roots in $[0,1]$. That is, $mnbvcxzq \\leq 2qwerasdf$ for all $qwerasdf$.\n\nTo establish that $mnbvcxzq \\to 2qwerasdf$, we make precise the observation that\n\\[\nhjgrksla(bnmloiuy) = \\sum_{plkmijnb=1}^{qwerasdf} plkmijnb^{4asdfghjk} poiulkjh \\sin(2\\pi plkmijnb bnmloiuy)\n\\]\nis dominated by the term with $plkmijnb=qwerasdf$. At the points\n$bnmloiuy = (2i+1)/(2qwerasdf)$ for $i=0,1, \\dots, qwerasdf-1$, we have\n$qwerasdf^{4asdfghjk} lkjhgfds \\sin (2\\pi qwerasdf bnmloiuy) = \\pm qwerasdf^{4asdfghjk} lkjhgfds$. If $asdfghjk$ is chosen large enough\nso that\n\\[\n|lkjhgfds| qwerasdf^{4asdfghjk} > |asdfqwer| 1^{4asdfghjk} + \\cdots + |a_{qwerasdf-1}| (qwerasdf-1)^{4asdfghjk},\n\\]\nthen $hjgrksla((2i+1)/2qwerasdf)$ has the same sign as $lkjhgfds \\sin (2\\pi qwerasdf bnmloiuy)$,\nwhich is to say, the sequence $hjgrksla(1/2qwerasdf), hjgrksla(3/2qwerasdf), \\dots$ alternates\nin sign. Thus\nbetween these points (again including the ``wraparound'' interval) we find\n$2qwerasdf$ sign changes of $hjgrksla$. Therefore $\\lim_{asdfghjk \\to \\infty} mnbvcxzq = 2qwerasdf$."
    },
    "kernel_variant": {
      "question": "Let      \n  h(t)=∑_{j=1}^{N}\\bigl(a_{j}\\cos 2πjt+b_{j}\\sin 2πjt\\bigr),  a_{j},b_{j}\\inℝ,\\;N≥2,\\;a_{N}^{2}+b_{N}^{2}\\neq0.    \n\nAll functions are regarded as 1-periodic; zeros are counted with their multiplicities and the point  t=−½ is included while  t=½ is not.  \n\nFor k=0,1,2,… write  \n\n  h_{k}(t):=d^{k}h/dt^{k},  and put  \n\n  M_{k}:=# {t∈[−½,½): h_{k}(t)=0}.  \n\nList the (repeated) zeros of h_{k} in increasing order  \n\n  −½≤t_{k,1}≤t_{k,2}≤⋯≤t_{k,M_{k}}<½.\n\n1.  (Monotonicity) Show that the sequence (M_{k})_{k≥0} is non-decreasing.\n\n2.  (Asymptotic number of zeros) Prove that lim_{k→∞}M_{k}=2N.\n\n3.  (Quantitative asymptotics – exponential localisation)  \n  Put A_{N}:=a_{N}−ib_{N} and, for k≥0, set  \n\n   σ_{k}:=−arg(i^{k}A_{N}) /(2πN) ∈(−1/(2N),1/(2N)].  \n\n  Show that there exist constants C>0 and 0<ρ<1 (depending only on h) such that for every sufficiently large k one has\n\n  (a) M_{k}=2N and all zeros of h_{k} are simple;  \n\n  (b) after a suitable labelling of the zeros,  \n    |t_{k,m} − σ_{k} − (2m−1)/(4N)| ≤ Cρ^{k} (m=1,…,2N).\n\n(In words: the 2N zeros of the k-th derivative lie exponentially close to the rigid grid σ_{k}+(2m−1)/(4N); the whole grid is translated by σ_{k}, which depends only on k (mod 4) and on arg A_{N}.  For instance, for h(t)=cos 2πNt one gets σ_{2ℓ}=0, σ_{2ℓ+1}=1/(4N).)",
      "solution": "Throughout we put  \n\n  A_{j}:=a_{j}-ib_{j}, ω_{j}:=2πj, \n  h(t)=Re ∑_{j=1}^{N}A_{j}e^{iω_{j}t}. (★)\n\n\n\nStep 0.  k-th derivative and dominant harmonic.  \nDifferentiating (★) gives  \n\n  h_{k}(t)=Re ∑_{j=1}^{N}A_{j}(iω_{j})^{k}e^{iω_{j}t}  \n    =ω_{N}^{k}\\,Re\\bigl(B_{k}e^{iω_{N}t}+R_{k}(t)\\bigr), (1)\n\nwhere  \n\n  B_{k}:=i^{k}A_{N}, \n  R_{k}(t):=∑_{j=1}^{N-1}(j/N)^{k}i^{k}A_{j}e^{iω_{j}t}. (2)\n\nBecause j/N≤1−1/N<1, there is a constant C_{0}>0 and a ratio  \n\n  ρ:=1−1/N∈(0,1)  \n\nwith  \n\n  |R_{k}(t)| ≤ C_{0}ρ^{k} (∀t,∀k). (3)\n\n\n\nStep 1.  Monotonicity of (M_{k}).  \nThe circle version of Rolle’s theorem yields M_{k+1}≥M_{k}; hence (M_{k}) is non-decreasing.\n\n\n\nStep 2.  Uniform upper bound  M_{k}≤2N  (repaired proof).  \nFix k.  Write  \n\n  h_{k}(t)=½∑_{j=-N}^{N}C_{j}e^{iω_{j}t},  C_{-j}=\\overline{C_{j}}, C_{N}=2B_{k}.  \n\nMultiply by e^{-iω_{N}t} and set z=e^{i2πt}.  We obtain  \n\n  e^{-iω_{N}t}h_{k}(t)=P_{k}(z):=½∑_{j=-N}^{N}C_{j}z^{j-N}\n        =∑_{m=0}^{2N}c_{m,k}z^{m},  deg P_{k}≤2N.  \n\nThus e^{-iω_{N}t}h_{k}(t) is the restriction to the unit circle |z|=1 of the ordinary complex polynomial P_{k}.  \n\nIf t₀ is a zero of h_{k} of multiplicity m, then all derivatives ∂^{r}_{t}h_{k}(t₀) (0≤r≤m−1) vanish; because dt/dz=(1/2π i)z^{-1}, this is equivalent to P_{k} and its first m−1 complex derivatives vanishing at z₀=e^{i2πt₀}. Hence z₀ is a root of P_{k} of multiplicity at least m.  Since deg P_{k}≤2N, the sum of all such multiplicities cannot exceed 2N, and therefore  \n\n  M_{k}≤2N  (∀k). (4)\n\n\n\nStep 3.  Phase shift σ_{k} and an alternating sign pattern.  \nWrite B_{k}=|A_{N}|e^{iφ_{k}} and set  \n\n  σ_{k}:=−φ_{k}/ω_{N}. (5)\n\nBecause φ_{k}+ω_{N}σ_{k}=0,  \n\n  cos(ω_{N}(t−σ_{k})) = cos(ω_{N}t+φ_{k}). (6)\n\nInserting t=σ_{k}+m/(2N) into (1) and using (3) yields  \n\n  h_{k}(σ_{k}+m/(2N)) = ω_{N}^{k}|A_{N}|\\,[(-1)^{m}+O(ρ^{k})]. (7)\n\nChoose k≥k₀ large enough that C_{0}ρ^{k}≤|A_{N}|/2; then  \n\n  sgn h_{k}(σ_{k}+m/(2N)) = (-1)^{m}. (8)\n\n\n\nStep 4.  Lower bound  M_{k}≥2N  for k≥k₀.  \nPut  \n\n  I_{k,m}:=[σ_{k}+(m-1)/(2N),\\,σ_{k}+m/(2N)] (mod 1), m=1,…,2N.\n\nBy (8) the endpoints of I_{k,m} have opposite signs, so every I_{k,m} contains at least one zero; hence  \n\n  M_{k} ≥ 2N  (k≥k₀). (9)\n\n\n\nStep 5.  Exclusion of extra zeros and simplicity.  \nFix m∈{1,…,2N} and write points of I_{k,m} as  \n\n  t=σ_{k}+\\frac{2m-1}{4N}+\\frac{u}{2πN}, |u|≤π/2. (10)\n\nUsing (1)–(3) we obtain  \n\n  h_{k}(t)=ω_{N}^{k}|A_{N}|\\,[(-1)^{m}\\sin u+E_{k}(u)], |E_{k}(u)|≤C_{0}ρ^{k}. (11)\n\nChoose k≥k₁≥k₀ so that C_{0}ρ^{k₁}≤|A_{N}|ρ^{k₁/2}/2 and let  \n\n  δ_{k}:=ρ^{k/2},  (0<δ_{k}<π/4 for k≥k₁). (12)\n\nSplit I_{k,m} into an inner core |u|≤δ_{k} and an outer ring δ_{k}<|u|≤π/2.\n\nOuter ring:  |sin u|≥(2/π)δ_{k}; with (11)–(12) this gives |h_{k}(t)|≥const·δ_{k}>0, so no zeros occur there.\n\nInner core:  sin u=u+O(u^{3}); inserting into (11) shows that h_{k}(t) has exactly one zero, situated at |u|≪ρ^{k}.  Differentiating (11) proves that h_{k}' does not vanish there, hence the zero is simple.  Thus each I_{k,m} contains precisely one simple zero, establishing (a).\n\n\n\nStep 6.  Exponential localisation.  \nFrom |u|≪ρ^{k} and (10) we deduce  \n\n  |t_{k,m} − σ_{k} − (2m−1)/(4N)| ≤ Cρ^{k}, (13)\n\ni.e. assertion (b).\n\n\n\nStep 7.  Summary.  \n• Step 1 gives monotonicity of (M_{k}).  \n• Steps 2 and 4, combined, yield 2N ≤ M_{k} ≤ 2N for k≥k₀, hence lim_{k→∞}M_{k}=2N.  \n• Steps 5–6 provide simplicity and exponential localisation of the 2N limiting zeros. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.770139",
        "was_fixed": false,
        "difficulty_analysis": "•  Extra conclusions (3)(a)(b) demand localisation and quantitative\n   estimates—not just counting zeros.\n•  The proof now mixes three advanced tools absent from the original\n   kernel: complex-variable reformulation (Step 2), uniform Rouché\n   estimates on families of disks (Steps 3, 5), and quantitatively\n   controlled implicit-function arguments to obtain exponential\n   convergence (Step 6).\n•  Handling both sine and cosine terms (complex amplitudes A_j) and\n   allowing N≥2 removes the even/odd symmetry exploited in simpler\n   versions, adding technical work in sign arguments and localisation.\n•  Establishing simplicity of the zeros requires bounds on derivatives,\n   not needed in the original.\n•  The final exponential–rate statement is strictly stronger than mere\n   convergence of the count; it forces a precise asymptotic geometry\n   of the nodal set.\n\nThese layers of additional structure and analysis make the enhanced\nproblem substantially more intricate than both the original problem and\nthe previous kernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let      \n  h(t)=∑_{j=1}^{N}\\bigl(a_{j}\\cos 2πjt+b_{j}\\sin 2πjt\\bigr),  a_{j},b_{j}\\inℝ,\\;N≥2,\\;a_{N}^{2}+b_{N}^{2}\\neq0.    \n\nAll functions are regarded as 1-periodic; zeros are counted with their multiplicities and the point  t=−½ is included while  t=½ is not.  \n\nFor k=0,1,2,… write  \n\n  h_{k}(t):=d^{k}h/dt^{k},  and put  \n\n  M_{k}:=# {t∈[−½,½): h_{k}(t)=0}.  \n\nList the (repeated) zeros of h_{k} in increasing order  \n\n  −½≤t_{k,1}≤t_{k,2}≤⋯≤t_{k,M_{k}}<½.\n\n1.  (Monotonicity) Show that the sequence (M_{k})_{k≥0} is non-decreasing.\n\n2.  (Asymptotic number of zeros) Prove that lim_{k→∞}M_{k}=2N.\n\n3.  (Quantitative asymptotics – exponential localisation)  \n  Put A_{N}:=a_{N}−ib_{N} and, for k≥0, set  \n\n   σ_{k}:=−arg(i^{k}A_{N}) /(2πN) ∈(−1/(2N),1/(2N)].  \n\n  Show that there exist constants C>0 and 0<ρ<1 (depending only on h) such that for every sufficiently large k one has\n\n  (a) M_{k}=2N and all zeros of h_{k} are simple;  \n\n  (b) after a suitable labelling of the zeros,  \n    |t_{k,m} − σ_{k} − (2m−1)/(4N)| ≤ Cρ^{k} (m=1,…,2N).\n\n(In words: the 2N zeros of the k-th derivative lie exponentially close to the rigid grid σ_{k}+(2m−1)/(4N); the whole grid is translated by σ_{k}, which depends only on k (mod 4) and on arg A_{N}.  For instance, for h(t)=cos 2πNt one gets σ_{2ℓ}=0, σ_{2ℓ+1}=1/(4N).)",
      "solution": "Throughout we put  \n\n  A_{j}:=a_{j}-ib_{j}, ω_{j}:=2πj, \n  h(t)=Re ∑_{j=1}^{N}A_{j}e^{iω_{j}t}. (★)\n\n\n\nStep 0.  k-th derivative and dominant harmonic.  \nDifferentiating (★) gives  \n\n  h_{k}(t)=Re ∑_{j=1}^{N}A_{j}(iω_{j})^{k}e^{iω_{j}t}  \n    =ω_{N}^{k}\\,Re\\bigl(B_{k}e^{iω_{N}t}+R_{k}(t)\\bigr), (1)\n\nwhere  \n\n  B_{k}:=i^{k}A_{N}, \n  R_{k}(t):=∑_{j=1}^{N-1}(j/N)^{k}i^{k}A_{j}e^{iω_{j}t}. (2)\n\nBecause j/N≤1−1/N<1, there is a constant C_{0}>0 and a ratio  \n\n  ρ:=1−1/N∈(0,1)  \n\nwith  \n\n  |R_{k}(t)| ≤ C_{0}ρ^{k} (∀t,∀k). (3)\n\n\n\nStep 1.  Monotonicity of (M_{k}).  \nThe circle version of Rolle’s theorem yields M_{k+1}≥M_{k}; hence (M_{k}) is non-decreasing.\n\n\n\nStep 2.  Uniform upper bound  M_{k}≤2N  (repaired proof).  \nFix k.  Write  \n\n  h_{k}(t)=½∑_{j=-N}^{N}C_{j}e^{iω_{j}t},  C_{-j}=\\overline{C_{j}}, C_{N}=2B_{k}.  \n\nMultiply by e^{-iω_{N}t} and set z=e^{i2πt}.  We obtain  \n\n  e^{-iω_{N}t}h_{k}(t)=P_{k}(z):=½∑_{j=-N}^{N}C_{j}z^{j-N}\n        =∑_{m=0}^{2N}c_{m,k}z^{m},  deg P_{k}≤2N.  \n\nThus e^{-iω_{N}t}h_{k}(t) is the restriction to the unit circle |z|=1 of the ordinary complex polynomial P_{k}.  \n\nIf t₀ is a zero of h_{k} of multiplicity m, then all derivatives ∂^{r}_{t}h_{k}(t₀) (0≤r≤m−1) vanish; because dt/dz=(1/2π i)z^{-1}, this is equivalent to P_{k} and its first m−1 complex derivatives vanishing at z₀=e^{i2πt₀}. Hence z₀ is a root of P_{k} of multiplicity at least m.  Since deg P_{k}≤2N, the sum of all such multiplicities cannot exceed 2N, and therefore  \n\n  M_{k}≤2N  (∀k). (4)\n\n\n\nStep 3.  Phase shift σ_{k} and an alternating sign pattern.  \nWrite B_{k}=|A_{N}|e^{iφ_{k}} and set  \n\n  σ_{k}:=−φ_{k}/ω_{N}. (5)\n\nBecause φ_{k}+ω_{N}σ_{k}=0,  \n\n  cos(ω_{N}(t−σ_{k})) = cos(ω_{N}t+φ_{k}). (6)\n\nInserting t=σ_{k}+m/(2N) into (1) and using (3) yields  \n\n  h_{k}(σ_{k}+m/(2N)) = ω_{N}^{k}|A_{N}|\\,[(-1)^{m}+O(ρ^{k})]. (7)\n\nChoose k≥k₀ large enough that C_{0}ρ^{k}≤|A_{N}|/2; then  \n\n  sgn h_{k}(σ_{k}+m/(2N)) = (-1)^{m}. (8)\n\n\n\nStep 4.  Lower bound  M_{k}≥2N  for k≥k₀.  \nPut  \n\n  I_{k,m}:=[σ_{k}+(m-1)/(2N),\\,σ_{k}+m/(2N)] (mod 1), m=1,…,2N.\n\nBy (8) the endpoints of I_{k,m} have opposite signs, so every I_{k,m} contains at least one zero; hence  \n\n  M_{k} ≥ 2N  (k≥k₀). (9)\n\n\n\nStep 5.  Exclusion of extra zeros and simplicity.  \nFix m∈{1,…,2N} and write points of I_{k,m} as  \n\n  t=σ_{k}+\\frac{2m-1}{4N}+\\frac{u}{2πN}, |u|≤π/2. (10)\n\nUsing (1)–(3) we obtain  \n\n  h_{k}(t)=ω_{N}^{k}|A_{N}|\\,[(-1)^{m}\\sin u+E_{k}(u)], |E_{k}(u)|≤C_{0}ρ^{k}. (11)\n\nChoose k≥k₁≥k₀ so that C_{0}ρ^{k₁}≤|A_{N}|ρ^{k₁/2}/2 and let  \n\n  δ_{k}:=ρ^{k/2},  (0<δ_{k}<π/4 for k≥k₁). (12)\n\nSplit I_{k,m} into an inner core |u|≤δ_{k} and an outer ring δ_{k}<|u|≤π/2.\n\nOuter ring:  |sin u|≥(2/π)δ_{k}; with (11)–(12) this gives |h_{k}(t)|≥const·δ_{k}>0, so no zeros occur there.\n\nInner core:  sin u=u+O(u^{3}); inserting into (11) shows that h_{k}(t) has exactly one zero, situated at |u|≪ρ^{k}.  Differentiating (11) proves that h_{k}' does not vanish there, hence the zero is simple.  Thus each I_{k,m} contains precisely one simple zero, establishing (a).\n\n\n\nStep 6.  Exponential localisation.  \nFrom |u|≪ρ^{k} and (10) we deduce  \n\n  |t_{k,m} − σ_{k} − (2m−1)/(4N)| ≤ Cρ^{k}, (13)\n\ni.e. assertion (b).\n\n\n\nStep 7.  Summary.  \n• Step 1 gives monotonicity of (M_{k}).  \n• Steps 2 and 4, combined, yield 2N ≤ M_{k} ≤ 2N for k≥k₀, hence lim_{k→∞}M_{k}=2N.  \n• Steps 5–6 provide simplicity and exponential localisation of the 2N limiting zeros. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.589972",
        "was_fixed": false,
        "difficulty_analysis": "•  Extra conclusions (3)(a)(b) demand localisation and quantitative\n   estimates—not just counting zeros.\n•  The proof now mixes three advanced tools absent from the original\n   kernel: complex-variable reformulation (Step 2), uniform Rouché\n   estimates on families of disks (Steps 3, 5), and quantitatively\n   controlled implicit-function arguments to obtain exponential\n   convergence (Step 6).\n•  Handling both sine and cosine terms (complex amplitudes A_j) and\n   allowing N≥2 removes the even/odd symmetry exploited in simpler\n   versions, adding technical work in sign arguments and localisation.\n•  Establishing simplicity of the zeros requires bounds on derivatives,\n   not needed in the original.\n•  The final exponential–rate statement is strictly stronger than mere\n   convergence of the count; it forces a precise asymptotic geometry\n   of the nodal set.\n\nThese layers of additional structure and analysis make the enhanced\nproblem substantially more intricate than both the original problem and\nthe previous kernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}