{
  "index": "2020-A-6",
  "type": "ANA",
  "tag": [
    "ANA",
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "For a positive integer $N$, let $f_N$\\footnote{Corrected from $F_N$ in the source.} be the function defined by \n\\[\nf_N(x) = \\sum_{n=0}^N \\frac{N+1/2-n}{(N+1)(2n+1)} \\sin((2n+1)x).\n\\]\nDetermine the smallest constant $M$ such that $f_N(x) \\leq M$ for all $N$ and all real $x$.",
  "solution": "The smallest constant $M$ is $\\pi/4$.\n\nWe start from the expression\n\\begin{equation} \\label{2020A6eq1}\nf_N(x) = \\sum_{n=0}^N \\frac{1}{2} \\left( \\frac{2}{2n+1} - \\frac{1}{N+1} \\right) \\sin((2n+1)x).\n\\end{equation}\nNote that if $\\sin(x) > 0$, then\n\\begin{align*}\n\\sum_{n=0}^N \\sin((2n+1)x) &= \\frac{1}{2i} \\sum_{n=0}^N (e^{i(2n+1)x} - e^{-i(2n+1)x}) \\\\\n&= \\frac{1}{2i} \\left( \\frac{e^{i(2N+3)x} - e^{ix}}{e^{2ix} - 1}  -\n\\frac{e^{-i(2N+3)x} - e^{-ix}}{e^{-2ix} - 1} \\right) \\\\\n&=\\frac{1}{2i} \\left( \\frac{e^{i(2N+2)x} - 1}{e^{ix} - e^{-ix}}  -\n\\frac{e^{-i(2N+2)x} - 1}{e^{-ix} - e^{ix}} \\right) \\\\\n&=\\frac{1}{2i} \\frac{e^{i(2N+2)x}+ e^{-i(2N+2)x} - 2}{e^{ix} - e^{-ix}} \\\\\n&= \\frac{2 \\cos ((2N+2)x) - 2}{2i(2i \\sin(x))} \\\\\n&= \\frac{1 - \\cos ((2N+2)x)}{2\\sin(x)} \\geq 0.\n\\end{align*}\nWe use this to compare the expressions of $f_N(x)$ and $f_{N+1}(x)$ given by \\eqref{2020A6eq1}.\nFor $x \\in (0, \\pi)$ with $\\sin((2N+3)x) \\geq 0$, we may omit the summand $n=N+1$ from $f_{N+1}(x)$ to obtain\n\\begin{align*}\n& f_{N+1}(x) - f_N(x) \\\\\n&\\geq \\frac{1}{2}  \\left( \\frac{1}{N+1} - \\frac{1}{N+2} \\right) \\sum_{n=0}^N \\sin((2n+1)x) \\geq 0.\n\\end{align*}\nFor $x \\in (0, \\pi)$ with $\\sin((2N+3)x) \\leq 0$, we may insert the summand $n=N+1$ into $f_{N+1}(x)$ to obtain\n\\begin{align*}\n&f_{N+1}(x) - f_N(x) \\\\\n&\\geq \\frac{1}{2}  \\left( \\frac{1}{N+1} - \\frac{1}{N+2} \\right) \\sum_{n=0}^{N+1} \\sin((2n+1)x) \\geq 0.\n\\end{align*}\nIn either case, we deduce that for $x \\in (0, \\pi)$, the sequence $\\{f_N(x)\\}_N$ is nondecreasing.\n\nNow rewrite \\eqref{2020A6eq1} as \n\\begin{equation} \\label{2020A6eq2}\nf_N(x) = \\sum_{n=0}^N \\frac{ \\sin((2n+1)x) }{2n+1}- \\frac{1-\\cos((2N+2)x)}{4(N+1) \\sin(x)}\n\\end{equation}\nand note that the last term tends to 0 as $N \\to \\infty$.\nConsequently, $\\lim_{N \\to \\infty} f_N(x)$ equals the sum of the series\n\\[\n\\sum_{n=0}^\\infty \\frac{1}{2n+1} \\sin((2n+1)x),\n\\]\nwhich is the Fourier series for the ``square wave'' function defined on $(-\\pi, \\pi]$ by\n\\[\nx \\mapsto \\begin{cases} -\\frac{\\pi}{4} & x \\in (-\\pi, 0) \\\\\n\\frac{\\pi}{4} & x \\in (0, \\pi) \\\\\n0 & x = 0, \\pi\n\\end{cases}\n\\]\nand extended periodically. Since this function is continuous on $(0, \\pi)$, we deduce that the Fourier series converges to the value of the function; that is,\n\\[\n\\lim_{N \\to \\infty} f_N(x) = \\frac{\\pi}{4} \\qquad (x \\in (0, \\pi)).\n\\]\nThis is enough to deduce the desired result as follows. \nSince\n\\[\nf_N(x+2\\pi) = f_N(x), \\qquad f_N(-x) = -f_N(x),\n\\]\nit suffices to check the bound $f_N(x) \\leq \\pi$ for $x \\in (-\\pi, \\pi]$.\nFor $x = 0, \\pi$ we have $f_N(x) = 0$ for all $N$.\nFor $x \\in (-\\pi, 0)$, the previous arguments imply that\n\\[\n0 \\geq f_0(x) \\geq f_1(x) \\geq \\cdots\n\\]\nFor $x \\in (0, \\pi)$, the previous arguments imply that\n\\[\n0 \\leq f_0(x) \\leq f_1(x) \\leq \\cdots \\leq \\frac{\\pi}{4}\n\\]\nand the limit is equal to $\\pi/4$. We conclude that $f_N(x) \\leq M$ holds for $M = \\pi/4$ but not for any smaller $M$, as desired.\n\n\\noindent\n\\textbf{Remark.}\nIt is also possible to replace the use of the convergence of the Fourier series with a more direct argument; it is sufficient to do this for $x$ in a dense subset of $(0, \\pi)$, such as the rational multiples of $\\pi$.\n\nAnother alternative (described at \\url{https://how-did-i-get-here.com/2020-putnam-a6/})\nis to deduce from \\eqref{2020A6eq2} and a second geometric series computation (omitted here) that\n\\begin{align*}\nf'_N(x) &= \\sum_{n=0}^N \\cos((2n+1)x) - \\frac{d}{dx} \\left( \\frac{1-\\cos((2N+2)x)}{4(N+1) \\sin(x)} \\right) \\\\\n&=\\frac{\\sin((2N+2)x)}{2\\sin(x)} \\\\\n&\\qquad - \\frac{(2N+2)\\sin((2N+2)x) - \\cos(x) (1-\\cos((N+2)x)}{4(N+1)\\sin(x)^2} \\\\\n&= \\frac{\\cos(x) (1-\\cos((N+2)x)}{4(N+1)\\sin(x)^2},\n\\end{align*}\nwhich is nonnegative for $x \\in (0, \\pi/2]$ and nonpositive for $x \\in [\\pi/2, \\pi)$.\nThis implies that $f_N(x)$ always has a global maximum at $x = \\pi/2$, so it suffices to check the\nconvergence of the Fourier series for the square wave at that point. This reduces to the Madhava--Gregory--Newton series evaluation\n\\[\n1 - \\frac{1}{3} + \\frac{1}{5} - \\frac{1}{7} + \\cdots = \\arctan(1) = \\frac{\\pi}{4}.\n\\]",
  "vars": [
    "x",
    "n"
  ],
  "params": [
    "N",
    "f_N"
  ],
  "sci_consts": [
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "realvar",
        "n": "indexvar",
        "N": "upperindex",
        "f_N": "finitefunc"
      },
      "question": "For a positive integer $\\upperindex$, let $\\finitefunc$\\footnote{Corrected from $F_N$ in the source.} be the function defined by \n\\[\n\\finitefunc(\\realvar) = \\sum_{\\indexvar=0}^{\\upperindex} \\frac{\\upperindex+1/2-\\indexvar}{(\\upperindex+1)(2\\indexvar+1)} \\sin((2\\indexvar+1)\\realvar).\n\\]\nDetermine the smallest constant $M$ such that $\\finitefunc(\\realvar) \\leq M$ for all $\\upperindex$ and all real $\\realvar$.",
      "solution": "The smallest constant $M$ is $\\pi/4$.\n\nWe start from the expression\n\\begin{equation} \\label{2020A6eq1}\n\\finitefunc(\\realvar) = \\sum_{\\indexvar=0}^{\\upperindex} \\frac{1}{2} \\left( \\frac{2}{2\\indexvar+1} - \\frac{1}{\\upperindex+1} \\right) \\sin((2\\indexvar+1)\\realvar).\n\\end{equation}\nNote that if $\\sin(\\realvar) > 0$, then\n\\begin{align*}\n\\sum_{\\indexvar=0}^{\\upperindex} \\sin((2\\indexvar+1)\\realvar) &= \\frac{1}{2i} \\sum_{\\indexvar=0}^{\\upperindex} (e^{i(2\\indexvar+1)\\realvar} - e^{-i(2\\indexvar+1)\\realvar}) \\\\\n&= \\frac{1}{2i} \\left( \\frac{e^{i(2\\upperindex+3)\\realvar} - e^{i\\realvar}}{e^{2i\\realvar} - 1}  -\n\\frac{e^{-i(2\\upperindex+3)\\realvar} - e^{-i\\realvar}}{e^{-2i\\realvar} - 1} \\right) \\\\\n&=\\frac{1}{2i} \\left( \\frac{e^{i(2\\upperindex+2)\\realvar} - 1}{e^{i\\realvar} - e^{-i\\realvar}}  -\n\\frac{e^{-i(2\\upperindex+2)\\realvar} - 1}{e^{-i\\realvar} - e^{i\\realvar}} \\right) \\\\\n&=\\frac{1}{2i} \\frac{e^{i(2\\upperindex+2)\\realvar}+ e^{-i(2\\upperindex+2)\\realvar} - 2}{e^{i\\realvar} - e^{-i\\realvar}} \\\\\n&= \\frac{2 \\cos ((2\\upperindex+2)\\realvar) - 2}{2i(2i \\sin(\\realvar))} \\\\\n&= \\frac{1 - \\cos ((2\\upperindex+2)\\realvar)}{2\\sin(\\realvar)} \\geq 0.\n\\end{align*}\nWe use this to compare the expressions of $\\finitefunc(\\realvar)$ and $\\finitefunc_{\\upperindex+1}(\\realvar)$ given by \\eqref{2020A6eq1}.\nFor $\\realvar \\in (0, \\pi)$ with $\\sin((2\\upperindex+3)\\realvar) \\geq 0$, we may omit the summand $\\indexvar=\\upperindex+1$ from $\\finitefunc_{\\upperindex+1}(\\realvar)$ to obtain\n\\begin{align*}\n& \\finitefunc_{\\upperindex+1}(\\realvar) - \\finitefunc(\\realvar) \\\\\n&\\geq \\frac{1}{2}  \\left( \\frac{1}{\\upperindex+1} - \\frac{1}{\\upperindex+2} \\right) \\sum_{\\indexvar=0}^{\\upperindex} \\sin((2\\indexvar+1)\\realvar) \\geq 0.\n\\end{align*}\nFor $\\realvar \\in (0, \\pi)$ with $\\sin((2\\upperindex+3)\\realvar) \\leq 0$, we may insert the summand $\\indexvar=\\upperindex+1$ into $\\finitefunc_{\\upperindex+1}(\\realvar)$ to obtain\n\\begin{align*}\n&\\finitefunc_{\\upperindex+1}(\\realvar) - \\finitefunc(\\realvar) \\\\\n&\\geq \\frac{1}{2}  \\left( \\frac{1}{\\upperindex+1} - \\frac{1}{\\upperindex+2} \\right) \\sum_{\\indexvar=0}^{\\upperindex+1} \\sin((2\\indexvar+1)\\realvar) \\geq 0.\n\\end{align*}\nIn either case, we deduce that for $\\realvar \\in (0, \\pi)$, the sequence $\\{\\finitefunc(\\realvar)\\}_{\\upperindex}$ is nondecreasing.\n\nNow rewrite \\eqref{2020A6eq1} as \n\\begin{equation} \\label{2020A6eq2}\n\\finitefunc(\\realvar) = \\sum_{\\indexvar=0}^{\\upperindex} \\frac{ \\sin((2\\indexvar+1)\\realvar) }{2\\indexvar+1}- \\frac{1-\\cos((2\\upperindex+2)\\realvar)}{4(\\upperindex+1) \\sin(\\realvar)}\n\\end{equation}\nand note that the last term tends to 0 as $\\upperindex \\to \\infty$.\nConsequently, $\\lim_{\\upperindex \\to \\infty} \\finitefunc(\\realvar)$ equals the sum of the series\n\\[\n\\sum_{\\indexvar=0}^{\\infty} \\frac{1}{2\\indexvar+1} \\sin((2\\indexvar+1)\\realvar),\n\\]\nwhich is the Fourier series for the ``square wave'' function defined on $(-\\pi, \\pi]$ by\n\\[\n\\realvar \\mapsto \\begin{cases} -\\frac{\\pi}{4} & \\realvar \\in (-\\pi, 0) \\\\\n\\frac{\\pi}{4} & \\realvar \\in (0, \\pi) \\\\\n0 & \\realvar = 0, \\pi\n\\end{cases}\n\\]\nand extended periodically. Since this function is continuous on $(0, \\pi)$, we deduce that the Fourier series converges to the value of the function; that is,\n\\[\n\\lim_{\\upperindex \\to \\infty} \\finitefunc(\\realvar) = \\frac{\\pi}{4} \\qquad (\\realvar \\in (0, \\pi)).\n\\]\nThis is enough to deduce the desired result as follows. \nSince\n\\[\n\\finitefunc(\\realvar+2\\pi) = \\finitefunc(\\realvar), \\qquad \\finitefunc(-\\realvar) = -\\finitefunc(\\realvar),\n\\]\nit suffices to check the bound $\\finitefunc(\\realvar) \\leq \\pi$ for $\\realvar \\in (-\\pi, \\pi]$.\nFor $\\realvar = 0, \\pi$ we have $\\finitefunc(\\realvar) = 0$ for all $\\upperindex$.\nFor $\\realvar \\in (-\\pi, 0)$, the previous arguments imply that\n\\[\n0 \\geq \\finitefunc_{0}(\\realvar) \\geq \\finitefunc_{1}(\\realvar) \\geq \\cdots\n\\]\nFor $\\realvar \\in (0, \\pi)$, the previous arguments imply that\n\\[\n0 \\leq \\finitefunc_{0}(\\realvar) \\leq \\finitefunc_{1}(\\realvar) \\leq \\cdots \\leq \\frac{\\pi}{4}\n\\]\nand the limit is equal to $\\pi/4$. We conclude that $\\finitefunc(\\realvar) \\leq M$ holds for $M = \\pi/4$ but not for any smaller $M$, as desired.\n\n\\noindent\n\\textbf{Remark.}\nIt is also possible to replace the use of the convergence of the Fourier series with a more direct argument; it is sufficient to do this for $\\realvar$ in a dense subset of $(0, \\pi)$, such as the rational multiples of $\\pi$.\n\nAnother alternative (described at \\url{https://how-did-i-get-here.com/2020-putnam-a6/})\nis to deduce from \\eqref{2020A6eq2} and a second geometric series computation (omitted here) that\n\\begin{align*}\n\\finitefunc'(\\realvar) &= \\sum_{\\indexvar=0}^{\\upperindex} \\cos((2\\indexvar+1)\\realvar) - \\frac{d}{d\\realvar} \\left( \\frac{1-\\cos((2\\upperindex+2)\\realvar)}{4(\\upperindex+1) \\sin(\\realvar)} \\right) \\\\\n&=\\frac{\\sin((2\\upperindex+2)\\realvar)}{2\\sin(\\realvar)} \\\\\n&\\qquad - \\frac{(2\\upperindex+2)\\sin((2\\upperindex+2)\\realvar) - \\cos(\\realvar) (1-\\cos((\\upperindex+2)\\realvar))}{4(\\upperindex+1)\\sin(\\realvar)^2} \\\\\n&= \\frac{\\cos(\\realvar) (1-\\cos((\\upperindex+2)\\realvar))}{4(\\upperindex+1)\\sin(\\realvar)^2},\n\\end{align*}\nwhich is nonnegative for $\\realvar \\in (0, \\pi/2]$ and nonpositive for $\\realvar \\in [\\pi/2, \\pi)$.\nThis implies that $\\finitefunc(\\realvar)$ always has a global maximum at $\\realvar = \\pi/2$, so it suffices to check the\nconvergence of the Fourier series for the square wave at that point. This reduces to the Madhava–Gregory–Newton series evaluation\n\\[\n1 - \\frac{1}{3} + \\frac{1}{5} - \\frac{1}{7} + \\cdots = \\arctan(1) = \\frac{\\pi}{4}.\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "quartzine",
        "n": "crescendo",
        "N": "wanderlust",
        "f_N": "cloudburst"
      },
      "question": "For a positive integer $wanderlust$, let $cloudburst$\\footnote{Corrected from $F_N$ in the source.} be the function defined by \n\\[\ncloudburst_{wanderlust}(quartzine) = \\sum_{crescendo=0}^{wanderlust} \\frac{wanderlust+1/2-crescendo}{(wanderlust+1)(2crescendo+1)} \\sin((2crescendo+1)quartzine).\n\\]\nDetermine the smallest constant $M$ such that $cloudburst_{wanderlust}(quartzine) \\leq M$ for all $wanderlust$ and all real $quartzine$.",
      "solution": "The smallest constant $M$ is $\\pi/4$.\n\nWe start from the expression\n\\begin{equation} \\label{2020A6eq1}\ncloudburst_{wanderlust}(quartzine) = \\sum_{crescendo=0}^{wanderlust} \\frac{1}{2} \\left( \\frac{2}{2crescendo+1} - \\frac{1}{wanderlust+1} \\right) \\sin((2crescendo+1)quartzine).\n\\end{equation}\nNote that if $\\sin(quartzine) > 0$, then\n\\begin{align*}\n\\sum_{crescendo=0}^{wanderlust} \\sin((2crescendo+1)quartzine) &= \\frac{1}{2i} \\sum_{crescendo=0}^{wanderlust} \\bigl(e^{i(2crescendo+1)quartzine} - e^{-i(2crescendo+1)quartzine}\\bigr) \\\\\n&= \\frac{1}{2i} \\left( \\frac{e^{i(2wanderlust+3)quartzine} - e^{i\\,quartzine}}{e^{2i\\,quartzine} - 1}  -\n\\frac{e^{-i(2wanderlust+3)quartzine} - e^{-i\\,quartzine}}{e^{-2i\\,quartzine} - 1} \\right) \\\\\n&=\\frac{1}{2i} \\left( \\frac{e^{i(2wanderlust+2)quartzine} - 1}{e^{i\\,quartzine} - e^{-i\\,quartzine}}  -\n\\frac{e^{-i(2wanderlust+2)quartzine} - 1}{e^{-i\\,quartzine} - e^{i\\,quartzine}} \\right) \\\\\n&=\\frac{1}{2i} \\frac{e^{i(2wanderlust+2)quartzine}+ e^{-i(2wanderlust+2)quartzine} - 2}{e^{i\\,quartzine} - e^{-i\\,quartzine}} \\\\\n&= \\frac{2 \\cos ((2wanderlust+2)quartzine) - 2}{2i(2i \\sin(quartzine))} \\\\\n&= \\frac{1 - \\cos ((2wanderlust+2)quartzine)}{2\\sin(quartzine)} \\ge 0.\n\\end{align*}\nWe use this to compare the expressions of $cloudburst_{wanderlust}(quartzine)$ and $cloudburst_{wanderlust+1}(quartzine)$ given by \\eqref{2020A6eq1}.\nFor $quartzine \\in (0, \\pi)$ with $\\sin((2wanderlust+3)quartzine) \\ge 0$, we may omit the summand $crescendo=wanderlust+1$ from $cloudburst_{wanderlust+1}(quartzine)$ to obtain\n\\begin{align*}\n& cloudburst_{wanderlust+1}(quartzine) - cloudburst_{wanderlust}(quartzine) \\\\\n&\\ge \\frac{1}{2}  \\left( \\frac{1}{wanderlust+1} - \\frac{1}{wanderlust+2} \\right) \\sum_{crescendo=0}^{wanderlust} \\sin((2crescendo+1)quartzine) \\ge 0.\n\\end{align*}\nFor $quartzine \\in (0, \\pi)$ with $\\sin((2wanderlust+3)quartzine) \\le 0$, we may insert the summand $crescendo=wanderlust+1$ into $cloudburst_{wanderlust+1}(quartzine)$ to obtain\n\\begin{align*}\n&cloudburst_{wanderlust+1}(quartzine) - cloudburst_{wanderlust}(quartzine) \\\\\n&\\ge \\frac{1}{2}  \\left( \\frac{1}{wanderlust+1} - \\frac{1}{wanderlust+2} \\right) \\sum_{crescendo=0}^{wanderlust+1} \\sin((2crescendo+1)quartzine) \\ge 0.\n\\end{align*}\nIn either case, we deduce that for $quartzine \\in (0, \\pi)$, the sequence $\\{cloudburst_{wanderlust}(quartzine)\\}_{wanderlust}$ is nondecreasing.\n\nNow rewrite \\eqref{2020A6eq1} as \n\\begin{equation} \\label{2020A6eq2}\ncloudburst_{wanderlust}(quartzine) = \\sum_{crescendo=0}^{wanderlust} \\frac{ \\sin((2crescendo+1)quartzine) }{2crescendo+1}- \\frac{1-\\cos((2wanderlust+2)quartzine)}{4(wanderlust+1) \\sin(quartzine)}\n\\end{equation}\nand note that the last term tends to $0$ as $wanderlust \\to \\infty$.\nConsequently,\n\\[\n\\lim_{wanderlust \\to \\infty} cloudburst_{wanderlust}(quartzine)= \\sum_{crescendo=0}^\\infty \\frac{1}{2crescendo+1} \\sin((2crescendo+1)quartzine),\n\\]\nwhich is the Fourier series for the ``square wave'' function defined on $(-\\pi, \\pi]$ by\n\\[\nquartzine \\mapsto \\begin{cases} -\\dfrac{\\pi}{4} & quartzine \\in (-\\pi, 0),\\\\[6pt]\n\\dfrac{\\pi}{4} & quartzine \\in (0, \\pi),\\\\[6pt]\n0 & quartzine = 0,\\pi.\n\\end{cases}\n\\]\nSince this function is continuous on $(0, \\pi)$, the Fourier series converges to its value; that is,\n\\[\n\\lim_{wanderlust \\to \\infty} cloudburst_{wanderlust}(quartzine)=\\frac{\\pi}{4}\\qquad (quartzine \\in (0, \\pi)).\n\\]\n\nThis is enough to deduce the desired result as follows.\nSince\n\\[\ncloudburst_{wanderlust}(quartzine+2\\pi)=cloudburst_{wanderlust}(quartzine),\\qquad\ncloudburst_{wanderlust}(-quartzine)=-cloudburst_{wanderlust}(quartzine),\n\\]\nit suffices to check the bound $cloudburst_{wanderlust}(quartzine)\\le \\pi$ for $quartzine\\in(-\\pi,\\pi]$.\nFor $quartzine=0,\\pi$ we have $cloudburst_{wanderlust}(quartzine)=0$ for all $wanderlust$.\nFor $quartzine\\in(-\\pi,0)$, the previous arguments imply that\n\\[\n0 \\ge cloudburst_{0}(quartzine)\\ge cloudburst_{1}(quartzine)\\ge\\cdots\n\\]\nFor $quartzine\\in(0,\\pi)$, they imply that\n\\[\n0 \\le cloudburst_{0}(quartzine)\\le cloudburst_{1}(quartzine)\\le\\cdots\\le\\frac{\\pi}{4},\n\\]\nand the limit is $\\pi/4$. We conclude that $cloudburst_{wanderlust}(quartzine)\\le M$ holds for $M=\\pi/4$ but not for any smaller $M$, as desired.\n\n\\noindent\n\\textbf{Remark.}\nIt is also possible to replace the use of the convergence of the Fourier series with a more direct argument; it is sufficient to do this for $quartzine$ in a dense subset of $(0, \\pi)$, such as the rational multiples of $\\pi$.\n\nAnother alternative (described at \\url{https://how-did-i-get-here.com/2020-putnam-a6/})\nis to deduce from \\eqref{2020A6eq2} and a second geometric series computation (omitted here) that\n\\begin{align*}\ncloudburst'_{wanderlust}(quartzine) &= \\sum_{crescendo=0}^{wanderlust} \\cos((2crescendo+1)quartzine) - \\frac{d}{dquartzine} \\left( \\frac{1-\\cos((2wanderlust+2)quartzine)}{4(wanderlust+1) \\sin(quartzine)} \\right) \\\\\n&=\\frac{\\sin((2wanderlust+2)quartzine)}{2\\sin(quartzine)} \\\\\n&\\qquad - \\frac{(2wanderlust+2)\\sin((2wanderlust+2)quartzine) - \\cos(quartzine) \\bigl(1-\\cos((wanderlust+2)quartzine)\\bigr)}{4(wanderlust+1)\\sin(quartzine)^2} \\\\\n&= \\frac{\\cos(quartzine)\\bigl(1-\\cos((wanderlust+2)quartzine)\\bigr)}{4(wanderlust+1)\\sin(quartzine)^2},\n\\end{align*}\nwhich is nonnegative for $quartzine \\in (0, \\pi/2]$ and nonpositive for $quartzine \\in [\\pi/2, \\pi)$.\nThis implies that $cloudburst_{wanderlust}(quartzine)$ always has a global maximum at $quartzine=\\pi/2$, so it suffices to check the convergence of the Fourier series for the square wave at that point. This reduces to the Madhava--Gregory--Newton series evaluation\n\\[\n1 - \\frac{1}{3} + \\frac{1}{5} - \\frac{1}{7} + \\cdots = \\arctan(1) = \\frac{\\pi}{4}.\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constant",
        "n": "continuous",
        "N": "infinite",
        "f_N": "nonfunction"
      },
      "question": "For a positive integer $infinite$, let $nonfunction$ be the function defined by \n\\[\nnonfunction(constant) = \\sum_{continuous=0}^{infinite} \\frac{infinite+1/2-continuous}{(infinite+1)(2continuous+1)} \\sin((2continuous+1)constant).\n\\]\nDetermine the smallest constant $M$ such that $nonfunction(constant) \\leq M$ for all $infinite$ and all real $constant$.",
      "solution": "The smallest constant $M$ is $\\pi/4$.\n\nWe start from the expression\n\\begin{equation} \\label{2020A6eq1}\nnonfunction(constant) = \\sum_{continuous=0}^{infinite} \\frac{1}{2} \\left( \\frac{2}{2continuous+1} - \\frac{1}{infinite+1} \\right) \\sin((2continuous+1)constant).\n\\end{equation}\nNote that if $\\sin(constant) > 0$, then\n\\begin{align*}\n\\sum_{continuous=0}^{infinite} \\sin((2continuous+1)constant) &= \\frac{1}{2i} \\sum_{continuous=0}^{infinite} (e^{i(2continuous+1)constant} - e^{-i(2continuous+1)constant}) \\\\\n&= \\frac{1}{2i} \\left( \\frac{e^{i(2infinite+3)constant} - e^{i constant}}{e^{2i constant} - 1}  -\n\\frac{e^{-i(2infinite+3)constant} - e^{-i constant}}{e^{-2i constant} - 1} \\right) \\\\\n&=\\frac{1}{2i} \\left( \\frac{e^{i(2infinite+2)constant} - 1}{e^{i constant} - e^{-i constant}}  -\n\\frac{e^{-i(2infinite+2)constant} - 1}{e^{-i constant} - e^{i constant}} \\right) \\\\\n&=\\frac{1}{2i} \\frac{e^{i(2infinite+2)constant}+ e^{-i(2infinite+2)constant} - 2}{e^{i constant} - e^{-i constant}} \\\\\n&= \\frac{2 \\cos ((2infinite+2)constant) - 2}{2i(2i \\sin(constant))} \\\\\n&= \\frac{1 - \\cos ((2infinite+2)constant)}{2\\sin(constant)} \\geq 0.\n\\end{align*}\nWe use this to compare the expressions of $nonfunction(constant)$ and $f_{infinite+1}(constant)$ given by \\eqref{2020A6eq1}.\nFor $constant \\in (0, \\pi)$ with $\\sin((2infinite+3)constant) \\geq 0$, we may omit the summand $continuous=infinite+1$ from $f_{infinite+1}(constant)$ to obtain\n\\begin{align*}\n& f_{infinite+1}(constant) - nonfunction(constant) \\\\\n&\\geq \\frac{1}{2}  \\left( \\frac{1}{infinite+1} - \\frac{1}{infinite+2} \\right) \\sum_{continuous=0}^{infinite} \\sin((2continuous+1)constant) \\geq 0.\n\\end{align*}\nFor $constant \\in (0, \\pi)$ with $\\sin((2infinite+3)constant) \\leq 0$, we may insert the summand $continuous=infinite+1$ into $f_{infinite+1}(constant)$ to obtain\n\\begin{align*}\n&f_{infinite+1}(constant) - nonfunction(constant) \\\\\n&\\geq \\frac{1}{2}  \\left( \\frac{1}{infinite+1} - \\frac{1}{infinite+2} \\right) \\sum_{continuous=0}^{infinite+1} \\sin((2continuous+1)constant) \\geq 0.\n\\end{align*}\nIn either case, we deduce that for $constant \\in (0, \\pi)$, the sequence $\\{nonfunction(constant)\\}_{infinite}$ is nondecreasing.\n\nNow rewrite \\eqref{2020A6eq1} as \n\\begin{equation} \\label{2020A6eq2}\nnonfunction(constant) = \\sum_{continuous=0}^{infinite} \\frac{ \\sin((2continuous+1)constant) }{2continuous+1}- \\frac{1-\\cos((2infinite+2)constant)}{4(infinite+1) \\sin(constant)}\n\\end{equation}\nand note that the last term tends to 0 as $infinite \\to \\infty$.\nConsequently, $\\lim_{infinite \\to \\infty} nonfunction(constant)$ equals the sum of the series\n\\[\n\\sum_{continuous=0}^\\infty \\frac{1}{2continuous+1} \\sin((2continuous+1)constant),\n\\]\nwhich is the Fourier series for the ``square wave'' function defined on $(-\\pi, \\pi]$ by\n\\[\nconstant \\mapsto \\begin{cases} -\\frac{\\pi}{4} & constant \\in (-\\pi, 0) \\\\\n\\frac{\\pi}{4} & constant \\in (0, \\pi) \\\\\n0 & constant = 0, \\pi\n\\end{cases}\n\\]\nand extended periodically. Since this function is continuous on $(0, \\pi)$, we deduce that the Fourier series converges to the value of the function; that is,\n\\[\n\\lim_{infinite \\to \\infty} nonfunction(constant) = \\frac{\\pi}{4} \\qquad (constant \\in (0, \\pi)).\n\\]\nThis is enough to deduce the desired result as follows. \nSince\n\\[\nnonfunction(constant+2\\pi) = nonfunction(constant), \\qquad nonfunction(-constant) = -nonfunction(constant),\n\\]\nit suffices to check the bound $nonfunction(constant) \\leq \\pi$ for $constant \\in (-\\pi, \\pi]$.\nFor $constant = 0, \\pi$ we have $nonfunction(constant) = 0$ for all $infinite$.\nFor $constant \\in (-\\pi, 0)$, the previous arguments imply that\n\\[\n0 \\geq f_0(constant) \\geq f_1(constant) \\geq \\cdots\n\\]\nFor $constant \\in (0, \\pi)$, the previous arguments imply that\n\\[\n0 \\leq f_0(constant) \\leq f_1(constant) \\leq \\cdots \\leq \\frac{\\pi}{4}\n\\]\nand the limit is equal to $\\pi/4$. We conclude that $nonfunction(constant) \\leq M$ holds for $M = \\pi/4$ but not for any smaller $M$, as desired.\n\n\\noindent\n\\textbf{Remark.}\nIt is also possible to replace the use of the convergence of the Fourier series with a more direct argument; it is sufficient to do this for $constant$ in a dense subset of $(0, \\pi)$, such as the rational multiples of $\\pi$.\n\nAnother alternative (described at \\url{https://how-did-i-get-here.com/2020-putnam-a6/})\nis to deduce from \\eqref{2020A6eq2} and a second geometric series computation (omitted here) that\n\\begin{align*}\nf'_{infinite}(constant) &= \\sum_{continuous=0}^{infinite} \\cos((2continuous+1)constant) - \\frac{d}{dconstant} \\left( \\frac{1-\\cos((2infinite+2)constant)}{4(infinite+1) \\sin(constant)} \\right) \\\\\n&=\\frac{\\sin((2infinite+2)constant)}{2\\sin(constant)} \\\\\n&\\qquad - \\frac{(2infinite+2)\\sin((2infinite+2)constant) - \\cos(constant) (1-\\cos((infinite+2)constant)}{4(infinite+1)\\sin(constant)^2} \\\\\n&= \\frac{\\cos(constant) (1-\\cos((infinite+2)constant)}{4(infinite+1)\\sin(constant)^2},\n\\end{align*}\nwhich is nonnegative for $constant \\in (0, \\pi/2]$ and nonpositive for $constant \\in [\\pi/2, \\pi)$.\nThis implies that $nonfunction(constant)$ always has a global maximum at $constant = \\pi/2$, so it suffices to check the\nconvergence of the Fourier series for the square wave at that point. This reduces to the Madhava--Gregory--Newton series evaluation\n\\[\n1 - \\frac{1}{3} + \\frac{1}{5} - \\frac{1}{7} + \\cdots = \\arctan(1) = \\frac{\\pi}{4}.\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "grmbltxq",
        "n": "kjdspvla",
        "N": "zclhrmdo",
        "f_N": "qzxwvtnp"
      },
      "question": "Problem:\n<<<\nFor a positive integer $zclhrmdo$, let $qzxwvtnp_{zclhrmdo}$\\footnote{Corrected from $F_N$ in the source.} be the function defined by \n\\[\nqzxwvtnp_{zclhrmdo}(grmbltxq) = \\sum_{kjdspvla=0}^{zclhrmdo} \\frac{zclhrmdo+1/2-kjdspvla}{(zclhrmdo+1)(2kjdspvla+1)} \\sin((2kjdspvla+1)grmbltxq).\n\\]\nDetermine the smallest constant $M$ such that $qzxwvtnp_{zclhrmdo}(grmbltxq) \\leq M$ for all $zclhrmdo$ and all real $grmbltxq$.\n>>>\n",
      "solution": "Solution:\n<<<\nThe smallest constant $M$ is $\\pi/4$.\n\nWe start from the expression\n\\begin{equation} \\label{2020A6eq1}\nqzxwvtnp_{zclhrmdo}(grmbltxq) = \\sum_{kjdspvla=0}^{zclhrmdo} \\frac{1}{2} \\left( \\frac{2}{2kjdspvla+1} - \\frac{1}{zclhrmdo+1} \\right) \\sin((2kjdspvla+1)grmbltxq).\n\\end{equation}\nNote that if $\\sin(grmbltxq) > 0$, then\n\\begin{align*}\n\\sum_{kjdspvla=0}^{zclhrmdo} \\sin((2kjdspvla+1)grmbltxq) &= \\frac{1}{2i} \\sum_{kjdspvla=0}^{zclhrmdo} (e^{i(2kjdspvla+1)grmbltxq} - e^{-i(2kjdspvla+1)grmbltxq}) \\\\\n&= \\frac{1}{2i} \\left( \\frac{e^{i(2zclhrmdo+3)grmbltxq} - e^{igrmbltxq}}{e^{2igrmbltxq} - 1}  -\n\\frac{e^{-i(2zclhrmdo+3)grmbltxq} - e^{-igrmbltxq}}{e^{-2igrmbltxq} - 1} \\right) \\\\\n&=\\frac{1}{2i} \\left( \\frac{e^{i(2zclhrmdo+2)grmbltxq} - 1}{e^{igrmbltxq} - e^{-igrmbltxq}}  -\n\\frac{e^{-i(2zclhrmdo+2)grmbltxq} - 1}{e^{-igrmbltxq} - e^{igrmbltxq}} \\right) \\\\\n&=\\frac{1}{2i} \\frac{e^{i(2zclhrmdo+2)grmbltxq}+ e^{-i(2zclhrmdo+2)grmbltxq} - 2}{e^{igrmbltxq} - e^{-igrmbltxq}} \\\\\n&= \\frac{2 \\cos ((2zclhrmdo+2)grmbltxq) - 2}{2i(2i \\sin(grmbltxq))} \\\\\n&= \\frac{1 - \\cos ((2zclhrmdo+2)grmbltxq)}{2\\sin(grmbltxq)} \\geq 0.\n\\end{align*}\nWe use this to compare the expressions of $qzxwvtnp_{zclhrmdo}(grmbltxq)$ and $qzxwvtnp_{zclhrmdo+1}(grmbltxq)$ given by \\eqref{2020A6eq1}.\nFor $grmbltxq \\in (0, \\pi)$ with $\\sin((2zclhrmdo+3)grmbltxq) \\geq 0$, we may omit the summand $kjdspvla=zclhrmdo+1$ from $qzxwvtnp_{zclhrmdo+1}(grmbltxq)$ to obtain\n\\begin{align*}\n& qzxwvtnp_{zclhrmdo+1}(grmbltxq) - qzxwvtnp_{zclhrmdo}(grmbltxq) \\\\\n&\\geq \\frac{1}{2}  \\left( \\frac{1}{zclhrmdo+1} - \\frac{1}{zclhrmdo+2} \\right) \\sum_{kjdspvla=0}^{zclhrmdo} \\sin((2kjdspvla+1)grmbltxq) \\geq 0.\n\\end{align*}\nFor $grmbltxq \\in (0, \\pi)$ with $\\sin((2zclhrmdo+3)grmbltxq) \\leq 0$, we may insert the summand $kjdspvla=zclhrmdo+1$ into $qzxwvtnp_{zclhrmdo+1}(grmbltxq)$ to obtain\n\\begin{align*}\n&qzxwvtnp_{zclhrmdo+1}(grmbltxq) - qzxwvtnp_{zclhrmdo}(grmbltxq) \\\\\n&\\geq \\frac{1}{2}  \\left( \\frac{1}{zclhrmdo+1} - \\frac{1}{zclhrmdo+2} \\right) \\sum_{kjdspvla=0}^{zclhrmdo+1} \\sin((2kjdspvla+1)grmbltxq) \\geq 0.\n\\end{align*}\nIn either case, we deduce that for $grmbltxq \\in (0, \\pi)$, the sequence $\\{qzxwvtnp_{zclhrmdo}(grmbltxq)\\}_{zclhrmdo}$ is nondecreasing.\n\nNow rewrite \\eqref{2020A6eq1} as \n\\begin{equation} \\label{2020A6eq2}\nqzxwvtnp_{zclhrmdo}(grmbltxq) = \\sum_{kjdspvla=0}^{zclhrmdo} \\frac{ \\sin((2kjdspvla+1)grmbltxq) }{2kjdspvla+1}- \\frac{1-\\cos((2zclhrmdo+2)grmbltxq)}{4(zclhrmdo+1) \\sin(grmbltxq)}\n\\end{equation}\nand note that the last term tends to 0 as $zclhrmdo \\to \\infty$.\nConsequently, $\\lim_{zclhrmdo \\to \\infty} qzxwvtnp_{zclhrmdo}(grmbltxq)$ equals the sum of the series\n\\[\n\\sum_{kjdspvla=0}^\\infty \\frac{1}{2kjdspvla+1} \\sin((2kjdspvla+1)grmbltxq),\n\\]\nwhich is the Fourier series for the ``square wave'' function defined on $(-\\pi, \\pi]$ by\n\\[\ngr mbltxq \\mapsto \\begin{cases} -\\frac{\\pi}{4} & grmbltxq \\in (-\\pi, 0) \\\\\n\\frac{\\pi}{4} & grmbltxq \\in (0, \\pi) \\\\\n0 & grmbltxq = 0, \\pi\n\\end{cases}\n\\]\nand extended periodically. Since this function is continuous on $(0, \\pi)$, we deduce that the Fourier series converges to the value of the function; that is,\n\\[\n\\lim_{zclhrmdo \\to \\infty} qzxwvtnp_{zclhrmdo}(grmbltxq) = \\frac{\\pi}{4} \\qquad (grmbltxq \\in (0, \\pi)).\n\\]\nThis is enough to deduce the desired result as follows. \nSince\n\\[\nqzxwvtnp_{zclhrmdo}(grmbltxq+2\\pi) = qzxwvtnp_{zclhrmdo}(grmbltxq), \\qquad qzxwvtnp_{zclhrmdo}(-grmbltxq) = -qzxwvtnp_{zclhrmdo}(grmbltxq),\n\\]\nit suffices to check the bound $qzxwvtnp_{zclhrmdo}(grmbltxq) \\leq \\pi$ for $grmbltxq \\in (-\\pi, \\pi]$.\nFor $grmbltxq = 0, \\pi$ we have $qzxwvtnp_{zclhrmdo}(grmbltxq) = 0$ for all $zclhrmdo$.\nFor $grmbltxq \\in (-\\pi, 0)$, the previous arguments imply that\n\\[\n0 \\geq qzxwvtnp_{0}(grmbltxq) \\geq qzxwvtnp_{1}(grmbltxq) \\geq \\cdots\n\\]\nFor $grmbltxq \\in (0, \\pi)$, the previous arguments imply that\n\\[\n0 \\leq qzxwvtnp_{0}(grmbltxq) \\leq qzxwvtnp_{1}(grmbltxq) \\leq \\cdots \\leq \\frac{\\pi}{4}\n\\]\nand the limit is equal to $\\pi/4$. We conclude that $qzxwvtnp_{zclhrmdo}(grmbltxq) \\leq M$ holds for $M = \\pi/4$ but not for any smaller $M$, as desired.\n\n\\noindent\n\\textbf{Remark.}\nIt is also possible to replace the use of the convergence of the Fourier series with a more direct argument; it is sufficient to do this for $grmbltxq$ in a dense subset of $(0, \\pi)$, such as the rational multiples of $\\pi$.\n\nAnother alternative (described at \\url{https://how-did-i-get-here.com/2020-putnam-a6/})\nis to deduce from \\eqref{2020A6eq2} and a second geometric series computation (omitted here) that\n\\begin{align*}\nqzxwvtnp'_{zclhrmdo}(grmbltxq) &= \\sum_{kjdspvla=0}^{zclhrmdo} \\cos((2kjdspvla+1)grmbltxq) - \\frac{d}{dgrmbltxq} \\left( \\frac{1-\\cos((2zclhrmdo+2)grmbltxq)}{4(zclhrmdo+1) \\sin(grmbltxq)} \\right) \\\\\n&=\\frac{\\sin((2zclhrmdo+2)grmbltxq)}{2\\sin(grmbltxq)} \\\\\n&\\qquad - \\frac{(2zclhrmdo+2)\\sin((2zclhrmdo+2)grmbltxq) - \\cos(grmbltxq) (1-\\cos((zclhrmdo+2)grmbltxq))}{4(zclhrmdo+1)\\sin(grmbltxq)^2} \\\\\n&= \\frac{\\cos(grmbltxq) (1-\\cos((zclhrmdo+2)grmbltxq))}{4(zclhrmdo+1)\\sin(grmbltxq)^2},\n\\end{align*}\nwhich is nonnegative for $grmbltxq \\in (0, \\pi/2]$ and nonpositive for $grmbltxq \\in [\\pi/2, \\pi)$.\nThis implies that $qzxwvtnp_{zclhrmdo}(grmbltxq)$ always has a global maximum at $grmbltxq = \\pi/2$, so it suffices to check the\nconvergence of the Fourier series for the square wave at that point. This reduces to the Madhava--Gregory--Newton series evaluation\n\\[\n1 - \\frac{1}{3} + \\frac{1}{5} - \\frac{1}{7} + \\cdots = \\arctan(1) = \\frac{\\pi}{4}.\n\\]\n>>>\n"
    },
    "kernel_variant": {
      "question": "For every non–negative integer N define the 2π–periodic, odd trigonometric polynomial\n\ng_N(x)=\\sum_{n=0}^{N}\\frac{N+\\dfrac32-n}{(N+2)(2n+1)}\\,\\sin\\bigl((2n+1)x\\bigr),\\qquad x\\in\\mathbb R.\n\nDetermine the smallest real number M for which the inequality\n\ng_N(x)\\le M\\qquad(\\forall N\\ge 0,\\;\\forall x\\in\\mathbb R)\n\nholds simultaneously for every choice of N and x.",
      "solution": "Answer.\nM = \\dfrac{\\pi}{4}.\n\nThroughout we put\n\nS_m(x):=\\sum_{k=0}^{m}\\frac{\\sin((2k+1)x)}{2k+1}\\qquad(m\\ge 0),\n\nso that S_m is the (2m+1)-st Dirichlet partial sum of the Fourier series of the 2π-periodic square-wave\n\nF(x):=\\tfrac{\\pi}{4}\\,\\operatorname{sgn}(\\sin x)\\quad(-\\pi<x<\\pi).\n\n1.  Abel–summation identity.\n\nWrite\nc_n:=\\dfrac{N+\\tfrac32-n}{N+2}\\;(0\\le n\\le N),\\qquad d_n(x):=\\dfrac{\\sin((2n+1)x)}{2n+1}.\nSince c_n-c_{n+1}=\\dfrac1{N+2}, Abel summation gives\n\n\\begin{aligned}\ng_N(x)\n &=\\sum_{n=0}^{N}c_n d_n(x)\n   =c_{N}S_{N}(x)+\\sum_{n=0}^{N-1}(c_{n}-c_{n+1})S_{n}(x)\\\\[4pt]\n &=\\frac{\\tfrac32}{N+2}S_{N}(x)+\\frac1{N+2}\\sum_{n=0}^{N-1}S_{n}(x)\\\\[4pt]\n &=\\frac{N+1}{N+2}\\,\\sigma_{N}(x)+\\frac{1}{2(N+2)}S_{N}(x),\\tag{1}\n\\end{aligned}\nwhere\n\n\\sigma_{N}(x):=\\frac1{N+1}\\sum_{n=0}^{N}S_{n}(x)\\qquad(N\\ge0)\n\nis the N-th Cesàro (Fejér) mean of the sine series of F.\n\n2.  A uniform bound for the Dirichlet sums.\n\nLemma 2.1   For every integer m≥0 and every real number x,\n|S_m(x)|\\le \\tfrac{\\pi}{2}.\n\n(The standard proof uses the identity\nS_m(x)=\\displaystyle\\int_{0}^{x}\\frac{\\sin\\bigl(2(m+1)t\\bigr)}{2\\sin t}\\,dt.)\n\n3.  A uniform bound for the Cesàro means σ_N.\n\nThe Fejér kernel\nK_{N}(t):=\\frac{1}{N+1}\\Bigl(\\frac{\\sin\\frac{N+1}{2}t}{\\sin\\frac{t}{2}}\\Bigr)^2 \\ge 0\nsatisfies (1/2π)∫_{-π}^{π}K_{N}(t)dt = 1.  A standard computation shows\n\nσ_N(x)=\\frac1{2π}\\int_{-π}^{π} F(x-t)K_{N}(t)\\,dt.\\tag{2}\n\nBecause K_{N}\\ge0 and |F|\\le\\pi/4 everywhere, (2) implies\n\n|σ_N(x)|\\le\\frac{\\pi}{4}\\qquad(\\forall N\\ge0,\\;\\forall x).\\tag{3}\n\n4.  The global upper bound for g_N.\n\nInsert the bounds (3) and Lemma 2.1 into identity (1):\n\n\\begin{aligned}\n|g_N(x)|&\\le \\frac{N+1}{N+2}\\,|σ_N(x)|+\\frac{1}{2(N+2)}|S_{N}(x)|\\\\[4pt]\n&\\le \\frac{N+1}{N+2}\\cdot\\frac{\\pi}{4}+\\frac{1}{2(N+2)}\\cdot\\frac{\\pi}{2}\\\\[4pt]\n&= \\frac{\\pi}{4}\\Bigl(\\frac{N+1}{N+2}+\\frac{1}{N+2}\\Bigr)=\\frac{\\pi}{4}.\\tag{4}\n\\end{aligned}\n\nSince g_N is odd, inequality (4) yields in particular\n\ng_N(x)\\le \\frac{\\pi}{4}\\qquad(\\forall N,\\,\\forall x).\\tag{5}\n\n5.  Sharpness of π/4.\n\nTake x = π/2.  Because \\sin((2n+1)π/2)=(-1)^n,\n\n\\begin{aligned}\ng_N(\\tfrac{π}{2})&=\\sum_{n=0}^{N}\\frac{(-1)^n}{2n+1}-\\frac{1-(-1)^{N+1}}{4(N+2)}.\\tag{6}\n\\end{aligned}\n\nWhen N = 2m+1 is odd, the second term in (6) vanishes, and letting m→∞ gives\n\n\\lim_{m\\to\\infty} g_{2m+1}(\\tfrac{π}{2}) = \\sum_{n=0}^{\\infty}\\frac{(-1)^n}{2n+1} = \\arctan(1)=\\frac{\\pi}{4}.\\tag{7}\n\nThus values of g_N(x) come arbitrarily close to π/4 from below, so no smaller constant can work in (5).\n\nCombining (5) with (7) we conclude that the smallest admissible constant is\n\nM = \\boxed{\\dfrac{\\pi}{4}}.",
      "_meta": {
        "core_steps": [
          "Algebraically split each coefficient so that (N+½−n)/((N+1)(2n+1)) = ½(2/(2n+1) − 1/(N+1))",
          "Compute the finite sum Σ_{n=0}^N sin((2n+1)x) via a geometric-series trick and use it to prove f_{N+1}(x) − f_N(x) ≥ 0 on (0,π)",
          "Conclude that, for each fixed x, {f_N(x)}_N is monotone; hence its supremum equals its limit",
          "Rewrite f_N(x)=Σ_{n=0}^N sin((2n+1)x)/(2n+1) − vanishing error, so the limit is the odd-harmonic Fourier series of the square wave",
          "Evaluate that series (Gregory–Leibniz/Madhava) to get π/4 and extend the bound to all x by periodicity and oddness"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "The constant added to N in the denominator (N+1) can be replaced by any fixed positive real c",
            "original": "1"
          },
          "slot2": {
            "description": "The constant added to N in the numerator must track slot1 as (N+c−½) so that the algebraic split in step 1 still works",
            "original": "1/2"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}