{
  "index": "1958-2-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "5. The lengths of successive segments of a broken line are represented by the successive terms of the harmonic progression \\( 1,1 / 2,1 / 3, \\ldots, 1 / n, \\ldots \\). Each segment makes with the preceding segment a given angle \\( \\theta \\). What is the distance and what is the direction of the limiting point (if there is one) from the initial point of the first segment?",
  "solution": "First Solution. We may identify the plane with the complex number plane in such a way that the first segment extends from 0 to 1 . Then the next segment extends from 1 to \\( 1+\\frac{1}{2} e^{i \\theta} \\), since this segment represents the complex number \\( \\frac{1}{2} e^{i \\theta} \\).\n\nThe \\( n \\)th segment of the path represents the complex number ( \\( 1 / n \\) ) \\( e^{i(n-1) \\theta} \\), and when added to the previous ( \\( n-1 \\) ) segments, the sum ends at the point \\( \\sum_{p=1}^{n}(1 / p) e^{i(p-1) \\theta} \\). Thus the question really concerns the convergence and evaluation of\n\\[\n\\sum_{p=1}^{\\infty} \\frac{1}{p} e^{i(p-1) \\theta} .\n\\]\n\nThis becomes the harmonic series if \\( \\theta=0 \\) (or a multiple of \\( 2 \\pi \\) ) and does not converge in this case. We shall show that the series (1) converges in all other cases.\n\nA theorem of Abel asserts: If \\( b_{1}, b_{2}, b_{3}, \\ldots \\) is a real sequence that decreases monotonically to zero, and if the partial sums of the series \\( \\Sigma_{p=1}^{\\infty} a_{p} \\) are bounded, then \\( \\sum_{p=1}^{\\infty} a_{p} b_{p} \\) converges. A proof of this theorem was given in the discussion of Problem A.M. 7 of the Eleventh Competition (p. 325). It is also proved in a number of advanced calculus texts, for example, Louis Brand, Advanced Calculus: An Introduction to Classical Analysis, Wiley, New York, 1955, pages 418-420; or T. M. Apostol, Mathematical Analysis, Addison-Wesley, Reading, Mass., 1957, page 365.\n\nFor the present case, take \\( b_{p}=1 / p \\) and \\( a_{p}=e^{i(p-1) \\theta} \\). Since the \\( a \\) 's form a geometric progression, and since we are assuming \\( e^{i \\theta} \\neq 1 \\), we have\n\\[\n\\left|\\sum_{p=1}^{n} a_{p}\\right|=\\left|\\frac{1-e^{i n \\theta}}{1-e^{i \\theta}}\\right| \\leq \\frac{2}{\\left|1-e^{i \\theta}\\right|} .\n\\]\n\nThus the partial sums are bounded, Abel's theorem applies, and the series (1) converges whenever \\( e^{i \\theta} \\neq 1 \\).\n\nTo evaluate (1) we can use another theorem of Abel. If \\( \\sum_{p=1}^{\\infty} c_{p} \\) converges, then\n\\[\n\\lim _{r \\rightarrow 1} \\sum_{p=1}^{\\infty} c_{p} r^{p}=\\sum_{p=1}^{\\infty} c_{p}\n\\]\nwhere the limit is taken for \\( r \\) increasing to 1 through real values. See Brand, pages 423-424, or Apostol, page 421.\n\nIn the present case,\n\\[\n\\Sigma \\frac{1}{p} e^{i(p-1) \\theta}=e^{-i \\theta} \\lim _{r-1-} \\Sigma \\frac{1}{p}\\left(r e^{i \\theta}\\right)^{p} .\n\\]\n\nPutting \\( z=r e^{i \\theta} \\), we recognize the series on the right as the Taylor's series expansion for the principal value of \\( -\\log (1-z) \\), valid for \\( |z|<1 \\).\n\nHence\n\\[\n\\Sigma \\frac{1}{p} e^{i(p-1) \\theta}=-e^{i \\theta} \\lim _{r \\rightarrow 1} \\log \\left(1-r e^{i \\theta}\\right)=-e^{i \\theta} \\log \\left(1-e^{i \\theta}\\right) .\n\\]\n\nNow\n\\[\n1-e^{i \\theta}=2 \\sin \\frac{1}{2} \\theta e^{(1 / 2) i(\\theta-\\pi)} \\quad \\text { for } 0<\\theta<2 \\pi\n\\]\nand here \\( 2 \\sin \\frac{1}{2} \\theta>0 \\) and \\( \\frac{1}{2}(\\theta-\\pi) \\) is the principal value of the argument (since \\( \\left.-\\pi<\\frac{1}{2}(\\theta-\\pi)<\\pi\\right) \\). Therefore\n\\[\n\\log \\left(1-e^{i \\theta}\\right)=\\log \\left(2 \\sin \\frac{1}{2} \\theta\\right)+\\frac{1}{2} i(\\theta-\\pi)\n\\]\nand\n\\[\n\\Sigma \\frac{1}{p} e^{i(p-1) \\theta}=-e^{-i \\theta}\\left\\{\\log \\left(2 \\sin \\frac{1}{2} \\theta\\right)+\\frac{1}{2} i(\\theta-\\pi)\\right\\} .\n\\]\n\nHence the limit point is at a distance\n\\[\n\\left|\\Sigma \\frac{1}{p} e^{i(p-1) \\theta}\\right|=\\sqrt{\\left(\\log \\left(2 \\sin \\frac{\\theta}{2}\\right)\\right)^{2}+\\frac{1}{4}(\\theta-\\pi)^{2}}\n\\]\nfrom the origin; its argument is\n\\[\n\\pi-\\theta+\\arg \\left[\\log \\left(2 \\sin \\frac{\\theta}{2}\\right)+\\frac{i}{2}(\\theta-\\pi)\\right] .\n\\]\n\nRemark. If \\( \\theta=\\pi \\), the series to be summed is the well-known alternating series\n\\[\n\\sum_{n=1}^{\\infty} \\frac{(-1)^{n-1}}{n}=\\log 2 .\n\\]\n\nIn this case, (2) above reduces to \\( \\log 2 \\) and (3) becomes 0 .\nSecond Solution. One can prove the relation\n(4)\n\\[\n\\sum_{\\rho=1}^{\\infty} \\frac{1}{p} e^{i p \\theta}=-\\log \\left(1-e^{i \\theta}\\right)=-\\log \\left(2 \\sin \\frac{1}{2} \\theta\\right)+\\frac{i}{2}(\\pi-\\theta)\n\\]\nusing the theory of Fourier series. The Fourier series of \\( \\frac{1}{2}(\\pi-\\theta) \\) on the interval \\( [0,2 \\pi] \\) is readily found to be\n\\[\n\\sum_{p=1}^{\\infty} \\frac{1}{p} \\sin p \\theta .\n\\]\n\nWith slightly more difficulty the Fourier series of \\( \\log \\left(2 \\sin \\frac{1}{2} \\theta\\right) \\) is found to be\n\\[\n-\\sum_{p=1}^{\\infty} \\frac{1}{p} \\cos p \\theta .\n\\]\n\nSince both functions on the right side of (4) have continuous derivatives on the open interval \\( (0,2 \\pi) \\), they are represented by their Fourier series on this interval, i.e., the series converge and\n\\[\n\\begin{array}{c}\n\\Sigma \\frac{1}{p} \\cos p \\theta=-\\log \\left(2 \\sin \\frac{1}{2} \\theta\\right), \\\\\n\\Sigma \\frac{1}{p} \\sin p \\theta=\\frac{1}{2}(\\pi-\\dot{\\theta})\n\\end{array}\n\\]\nfor \\( 0<\\theta<2 \\pi \\).\nCombining these relations and using \\( e^{i p \\theta}=\\cos p \\theta+i \\sin p \\theta \\) we obtain (4).",
  "vars": [
    "n",
    "p",
    "r",
    "z",
    "\\\\rho",
    "b_p",
    "a_p",
    "c_p"
  ],
  "params": [
    "\\\\theta"
  ],
  "sci_consts": [
    "e",
    "i"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "n": "segmentindex",
        "p": "smallindex",
        "r": "radialvar",
        "z": "complexvar",
        "\\rho": "rhofactor",
        "b_p": "coeffbseq",
        "a_p": "coeffaseq",
        "c_p": "coeffcseq",
        "\\theta": "fixedangl"
      },
      "question": "5. The lengths of successive segments of a broken line are represented by the successive terms of the harmonic progression \\( 1,1 / 2,1 / 3, \\ldots, 1 / segmentindex, \\ldots \\). Each segment makes with the preceding segment a given angle \\( fixedangl \\). What is the distance and what is the direction of the limiting point (if there is one) from the initial point of the first segment?",
      "solution": "First Solution. We may identify the plane with the complex number plane in such a way that the first segment extends from 0 to 1. Then the next segment extends from 1 to \\( 1+\\frac{1}{2} e^{i fixedangl} \\), since this segment represents the complex number \\( \\frac{1}{2} e^{i fixedangl} \\).\n\nThe \\( segmentindex \\)th segment of the path represents the complex number \\( ( 1 / segmentindex ) e^{i(segmentindex-1) fixedangl} \\), and when added to the previous \\( ( segmentindex-1 ) \\) segments, the sum ends at the point \\( \\sum_{smallindex=1}^{segmentindex}(1 / smallindex) e^{i(smallindex-1) fixedangl} \\). Thus the question really concerns the convergence and evaluation of\n\\[\n\\sum_{smallindex=1}^{\\infty} \\frac{1}{smallindex} e^{i(smallindex-1) fixedangl}.\n\\tag{1}\n\\]\n\nThis becomes the harmonic series if \\( fixedangl=0 \\) (or a multiple of \\( 2 \\pi \\) ) and does not converge in this case. We shall show that the series (1) converges in all other cases.\n\nA theorem of Abel asserts: If \\( b_{1}, b_{2}, b_{3}, \\ldots \\) is a real sequence that decreases monotonically to zero, and if the partial sums of the series \\( \\sum_{smallindex=1}^{\\infty} coeffaseq \\) are bounded, then \\( \\sum_{smallindex=1}^{\\infty} coeffaseq\\, coeffbseq \\) converges. A proof of this theorem was given in the discussion of Problem A.M. 7 of the Eleventh Competition (p. 325). It is also proved in a number of advanced calculus texts, for example, Louis Brand, Advanced Calculus: An Introduction to Classical Analysis, Wiley, New York, 1955, pages 418–420; or T. M. Apostol, Mathematical Analysis, Addison-Wesley, Reading, Mass., 1957, page 365.\n\nFor the present case, take \\( coeffbseq = 1 / smallindex \\) and \\( coeffaseq = e^{i(smallindex-1) fixedangl} \\). Since the \\( coeffaseq \\)'s form a geometric progression, and since we are assuming \\( e^{i fixedangl}\\ne 1 \\), we have\n\\[\n\\left|\\sum_{smallindex=1}^{segmentindex} coeffaseq\\right|\n      =\\left|\\frac{1-e^{i\\,segmentindex\\,fixedangl}}{1-e^{i\\,fixedangl}}\\right|\n      \\le \\frac{2}{\\left|1-e^{i\\,fixedangl}\\right|}.\n\\]\n\nThus the partial sums are bounded, Abel's theorem applies, and the series (1) converges whenever \\( e^{i fixedangl}\\ne 1 \\).\n\nTo evaluate (1) we can use another theorem of Abel. If \\( \\sum_{smallindex=1}^{\\infty} coeffcseq \\) converges, then\n\\[\n\\lim_{radialvar\\to 1^-}\\sum_{smallindex=1}^{\\infty} coeffcseq\\, radialvar^{smallindex}\n      =\\sum_{smallindex=1}^{\\infty} coeffcseq,\n\\]\nwhere the limit is taken for \\( radialvar \\) increasing to 1 through real values. See Brand, pages 423–424, or Apostol, page 421.\n\nIn the present case,\n\\[\n\\sum_{smallindex=1}^{\\infty} \\frac{1}{smallindex} e^{i(smallindex-1) fixedangl}\n     = e^{-i fixedangl} \\lim_{radialvar\\to 1^-}\n       \\sum_{smallindex=1}^{\\infty} \\frac{1}{smallindex}\\bigl(radialvar e^{i fixedangl}\\bigr)^{smallindex}.\n\\]\n\nPutting \\( complexvar = radialvar e^{i fixedangl} \\), we recognize the series on the right as the Taylor series expansion for the principal value of \\( -\\log(1-complexvar) \\), valid for \\( |complexvar|<1 \\).\n\nHence\n\\[\n\\sum_{smallindex=1}^{\\infty} \\frac{1}{smallindex} e^{i(smallindex-1) fixedangl}\n      = -e^{i fixedangl}\\lim_{radialvar\\to 1} \\log\\bigl(1-radialvar e^{i fixedangl}\\bigr)\n      = -e^{i fixedangl}\\log\\bigl(1-e^{i fixedangl}\\bigr).\n\\]\n\nNow\n\\[\n1-e^{i fixedangl}=2\\sin\\frac{fixedangl}{2}\\,e^{\\frac{i}{2}(fixedangl-\\pi)}\n      \\quad\\text{for }0<fixedangl<2\\pi,\n\\]\nand here \\( 2\\sin\\frac{fixedangl}{2}>0 \\) and \\( \\frac{1}{2}(fixedangl-\\pi) \\) is the principal value of the argument (since \\( -\\pi<\\frac{1}{2}(fixedangl-\\pi)<\\pi \\)). Therefore\n\\[\n\\log\\bigl(1-e^{i fixedangl}\\bigr)\n      =\\log\\!\\Bigl(2\\sin\\frac{fixedangl}{2}\\Bigr)+\\frac{i}{2}(fixedangl-\\pi),\n\\]\nand\n\\[\n\\sum_{smallindex=1}^{\\infty} \\frac{1}{smallindex} e^{i(smallindex-1) fixedangl}\n      = -e^{-i fixedangl}\\Bigl\\{\\log\\!\\Bigl(2\\sin\\frac{fixedangl}{2}\\Bigr)\n                           +\\frac{i}{2}(fixedangl-\\pi)\\Bigr\\}.\n\\]\n\nHence the limiting point is at a distance\n\\[\n\\Bigl|\\sum_{smallindex=1}^{\\infty} \\frac{1}{smallindex} e^{i(smallindex-1) fixedangl}\\Bigr|\n      = \\sqrt{\\Bigl(\\log\\!\\bigl(2\\sin\\tfrac{fixedangl}{2}\\bigr)\\Bigr)^{2}\n              +\\tfrac{1}{4}(fixedangl-\\pi)^{2}}\n\\]\nfrom the origin; its argument is\n\\[\n\\pi - fixedangl\n   +\\arg\\!\\Bigl[\\log\\!\\bigl(2\\sin\\tfrac{fixedangl}{2}\\bigr)+\\tfrac{i}{2}(fixedangl-\\pi)\\Bigr].\n\\]\n\nRemark. If \\( fixedangl=\\pi \\), the series to be summed is the well-known alternating series\n\\[\n\\sum_{segmentindex=1}^{\\infty} \\frac{(-1)^{segmentindex-1}}{segmentindex}=\\log 2 .\n\\]\nIn this case, (2) above reduces to \\( \\log 2 \\) and (3) becomes \\( 0 \\).\n\nSecond Solution. One can prove the relation\n\\[\n\\sum_{rhofactor=1}^{\\infty} \\frac{1}{smallindex} e^{i\\,smallindex\\,fixedangl}\n      = -\\log\\bigl(1-e^{i fixedangl}\\bigr)\n      = -\\log\\!\\Bigl(2\\sin\\tfrac{fixedangl}{2}\\Bigr)+\\frac{i}{2}(\\pi-fixedangl)\n\\tag{4}\n\\]\nusing the theory of Fourier series. The Fourier series of \\( \\tfrac{1}{2}(\\pi-fixedangl) \\) on the interval \\( [0,2\\pi] \\) is readily found to be\n\\[\n\\sum_{smallindex=1}^{\\infty} \\frac{1}{smallindex} \\sin smallindex\\,fixedangl .\n\\]\n\nWith slightly more difficulty the Fourier series of \\( \\log\\!\\Bigl(2\\sin\\tfrac{fixedangl}{2}\\Bigr) \\) is found to be\n\\[\n-\\sum_{smallindex=1}^{\\infty} \\frac{1}{smallindex} \\cos smallindex\\,fixedangl .\n\\]\n\nSince both functions on the right side of (4) have continuous derivatives on the open interval \\( (0,2\\pi) \\), they are represented by their Fourier series on this interval, i.e. the series converge and\n\\[\n\\begin{array}{c}\n\\sum_{smallindex=1}^{\\infty} \\frac{1}{smallindex} \\cos smallindex\\,fixedangl\n     =-\\log\\!\\Bigl(2\\sin\\tfrac{fixedangl}{2}\\Bigr),\\\\[6pt]\n\\sum_{smallindex=1}^{\\infty} \\frac{1}{smallindex} \\sin smallindex\\,fixedangl\n     =\\tfrac{1}{2}(\\pi-\\dot{fixedangl})\n\\end{array}\n\\]\nfor \\( 0<fixedangl<2\\pi \\).\nCombining these relations and using \\( e^{i\\,smallindex\\,fixedangl}=\\cos smallindex\\,fixedangl+i\\sin smallindex\\,fixedangl \\) we obtain (4)."
    },
    "descriptive_long_confusing": {
      "map": {
        "n": "waterfall",
        "p": "gemstone",
        "r": "labyrinth",
        "z": "parchment",
        "\\rho": "nebulaone",
        "b_p": "horizonstone",
        "a_p": "dragonfly",
        "c_p": "moonlitsea",
        "\\theta": "tempestsky"
      },
      "question": "5. The lengths of successive segments of a broken line are represented by the successive terms of the harmonic progression \\( 1,1 / 2,1 / 3, \\ldots, 1 / waterfall, \\ldots \\). Each segment makes with the preceding segment a given angle \\( tempestsky \\). What is the distance and what is the direction of the limiting point (if there is one) from the initial point of the first segment?",
      "solution": "First Solution. We may identify the plane with the complex number plane in such a way that the first segment extends from 0 to 1 . Then the next segment extends from 1 to \\( 1+\\frac{1}{2} e^{i tempestsky} \\), since this segment represents the complex number \\( \\frac{1}{2} e^{i tempestsky} \\).\n\nThe \\( waterfall \\)th segment of the path represents the complex number ( \\( 1 / waterfall \\) ) \\( e^{i(waterfall-1) tempestsky} \\), and when added to the previous ( \\( waterfall-1 \\) ) segments, the sum ends at the point \\( \\sum_{gemstone=1}^{waterfall}(1 / gemstone) e^{i(gemstone-1) tempestsky} \\). Thus the question really concerns the convergence and evaluation of\n\\[\n\\sum_{gemstone=1}^{\\infty} \\frac{1}{gemstone} e^{i(gemstone-1) tempestsky} .\n\\]\n\nThis becomes the harmonic series if \\( tempestsky=0 \\) (or a multiple of \\( 2 \\pi \\) ) and does not converge in this case. We shall show that the series (1) converges in all other cases.\n\nA theorem of Abel asserts: If \\( b_{1}, b_{2}, b_{3}, \\ldots \\) is a real sequence that decreases monotonically to zero, and if the partial sums of the series \\( \\Sigma_{gemstone=1}^{\\infty} dragonfly \\) are bounded, then \\( \\sum_{gemstone=1}^{\\infty} dragonfly horizonstone \\) converges. A proof of this theorem was given in the discussion of Problem A.M. 7 of the Eleventh Competition (p. 325). It is also proved in a number of advanced calculus texts, for example, Louis Brand, Advanced Calculus: An Introduction to Classical Analysis, Wiley, New York, 1955, pages 418-420; or T. M. Apostol, Mathematical Analysis, Addison-Wesley, Reading, Mass., 1957, page 365.\n\nFor the present case, take \\( horizonstone=1 / gemstone \\) and \\( dragonfly=e^{i(gemstone-1) tempestsky} \\). Since the \\( a \\)'s form a geometric progression, and since we are assuming \\( e^{i tempestsky} \\neq 1 \\), we have\n\\[\n\\left|\\sum_{gemstone=1}^{waterfall} dragonfly\\right|=\\left|\\frac{1-e^{i waterfall tempestsky}}{1-e^{i tempestsky}}\\right| \\leq \\frac{2}{\\left|1-e^{i tempestsky}\\right|} .\n\\]\n\nThus the partial sums are bounded, Abel's theorem applies, and the series (1) converges whenever \\( e^{i tempestsky} \\neq 1 \\).\n\nTo evaluate (1) we can use another theorem of Abel. If \\( \\sum_{gemstone=1}^{\\infty} moonlitsea \\) converges, then\n\\[\n\\lim _{labyrinth \\rightarrow 1} \\sum_{gemstone=1}^{\\infty} moonlitsea labyrinth^{gemstone}=\\sum_{gemstone=1}^{\\infty} moonlitsea\n\\]\nwhere the limit is taken for \\( labyrinth \\) increasing to 1 through real values. See Brand, pages 423-424, or Apostol, page 421.\n\nIn the present case,\n\\[\n\\Sigma \\frac{1}{gemstone} e^{i(gemstone-1) tempestsky}=e^{-i tempestsky} \\lim _{labyrinth-1-} \\Sigma \\frac{1}{gemstone}\\left(labyrinth e^{i tempestsky}\\right)^{gemstone} .\n\\]\n\nPutting \\( parchment=labyrinth e^{i tempestsky} \\), we recognize the series on the right as the Taylor's series expansion for the principal value of \\( -\\log (1-parchment) \\), valid for \\( |parchment|<1 \\).\n\nHence\n\\[\n\\Sigma \\frac{1}{gemstone} e^{i(gemstone-1) tempestsky}=-e^{i tempestsky} \\lim _{labyrinth \\rightarrow 1} \\log \\left(1-labyrinth e^{i tempestsky}\\right)=-e^{i tempestsky} \\log \\left(1-e^{i tempestsky}\\right) .\n\\]\n\nNow\n\\[\n1-e^{i tempestsky}=2 \\sin \\frac{1}{2} tempestsky e^{(1 / 2) i(tempestsky-\\pi)} \\quad \\text { for } 0<tempestsky<2 \\pi\n\\]\nand here \\( 2 \\sin \\frac{1}{2} tempestsky>0 \\) and \\( \\frac{1}{2}(tempestsky-\\pi) \\) is the principal value of the argument (since \\( -\\pi<\\frac{1}{2}(tempestsky-\\pi)<\\pi\\) ). Therefore\n\\[\n\\log \\left(1-e^{i tempestsky}\\right)=\\log \\left(2 \\sin \\frac{1}{2} tempestsky\\right)+\\frac{1}{2} i(tempestsky-\\pi)\n\\]\nand\n\\[\n\\Sigma \\frac{1}{gemstone} e^{i(gemstone-1) tempestsky}=-e^{-i tempestsky}\\left\\{\\log \\left(2 \\sin \\frac{1}{2} tempestsky\\right)+\\frac{1}{2} i(tempestsky-\\pi)\\right\\} .\n\\]\n\nHence the limit point is at a distance\n\\[\n\\left|\\Sigma \\frac{1}{gemstone} e^{i(gemstone-1) tempestsky}\\right|=\\sqrt{\\left(\\log \\left(2 \\sin \\frac{tempestsky}{2}\\right)\\right)^{2}+\\frac{1}{4}(tempestsky-\\pi)^{2}}\n\\]\nfrom the origin; its argument is\n\\[\n\\pi-tempestsky+\\arg \\left[\\log \\left(2 \\sin \\frac{tempestsky}{2}\\right)+\\frac{i}{2}(tempestsky-\\pi)\\right] .\n\\]\n\nRemark. If \\( tempestsky=\\pi \\), the series to be summed is the well-known alternating series\n\\[\n\\sum_{waterfall=1}^{\\infty} \\frac{(-1)^{waterfall-1}}{waterfall}=\\log 2 .\n\\]\n\nIn this case, (2) above reduces to \\( \\log 2 \\) and (3) becomes 0 .\nSecond Solution. One can prove the relation\n(4)\n\\[\n\\sum_{nebulaone=1}^{\\infty} \\frac{1}{gemstone} e^{i gemstone tempestsky}=-\\log \\left(1-e^{i tempestsky}\\right)=-\\log \\left(2 \\sin \\frac{1}{2} tempestsky\\right)+\\frac{i}{2}(\\pi-tempestsky)\n\\]\nusing the theory of Fourier series. The Fourier series of \\( \\frac{1}{2}(\\pi-tempestsky) \\) on the interval \\( [0,2 \\pi] \\) is readily found to be\n\\[\n\\sum_{gemstone=1}^{\\infty} \\frac{1}{gemstone} \\sin gemstone tempestsky .\n\\]\n\nWith slightly more difficulty the Fourier series of \\( \\log \\left(2 \\sin \\frac{1}{2} tempestsky\\right) \\) is found to be\n\\[\n-\\sum_{gemstone=1}^{\\infty} \\frac{1}{gemstone} \\cos gemstone tempestsky .\n\\]\n\nSince both functions on the right side of (4) have continuous derivatives on the open interval \\( (0,2 \\pi) \\), they are represented by their Fourier series on this interval, i.e., the series converge and\n\\[\n\\begin{array}{c}\n\\Sigma \\frac{1}{gemstone} \\cos gemstone tempestsky=-\\log \\left(2 \\sin \\frac{1}{2} tempestsky\\right), \\\\\n\\Sigma \\frac{1}{gemstone} \\sin gemstone tempestsky=\\frac{1}{2}(\\pi-\\dot{tempestsky})\n\\end{array}\n\\]\nfor \\( 0<tempestsky<2 \\pi \\).\nCombining these relations and using \\( e^{i gemstone tempestsky}=\\cos gemstone tempestsky+i \\sin gemstone tempestsky \\) we obtain (4)."
    },
    "descriptive_long_misleading": {
      "map": {
        "n": "continuous",
        "p": "staticvalue",
        "r": "innercore",
        "z": "realaxis",
        "\\\\rho": "epsilonvalue",
        "b_p": "upperfixed",
        "a_p": "lowerfixed",
        "c_p": "nullfactor",
        "\\\\theta": "straightlen"
      },
      "question": "5. The lengths of successive segments of a broken line are represented by the successive terms of the harmonic progression \\( 1,1 / 2,1 / 3, \\ldots, 1 / continuous, \\ldots \\). Each segment makes with the preceding segment a given angle \\( straightlen \\). What is the distance and what is the direction of the limiting point (if there is one) from the initial point of the first segment?",
      "solution": "First Solution. We may identify the plane with the complex number plane in such a way that the first segment extends from 0 to 1 . Then the next segment extends from 1 to \\( 1+\\frac{1}{2} e^{i straightlen} \\), since this segment represents the complex number \\( \\frac{1}{2} e^{i straightlen} \\).\n\nThe \\( continuous \\)th segment of the path represents the complex number ( \\( 1 / continuous \\) ) \\( e^{i(continuous-1) straightlen} \\), and when added to the previous ( \\( continuous-1 \\) ) segments, the sum ends at the point \\( \\sum_{staticvalue=1}^{continuous}(1 / staticvalue) e^{i(staticvalue-1) straightlen} \\). Thus the question really concerns the convergence and evaluation of\n\\[\n\\sum_{staticvalue=1}^{\\infty} \\frac{1}{staticvalue} e^{i(staticvalue-1) straightlen} .\n\\]\n\nThis becomes the harmonic series if \\( straightlen=0 \\) (or a multiple of \\( 2 \\pi \\) ) and does not converge in this case. We shall show that the series (1) converges in all other cases.\n\nA theorem of Abel asserts: If \\( upperfixed_{1}, upperfixed_{2}, upperfixed_{3}, \\ldots \\) is a real sequence that decreases monotonically to zero, and if the partial sums of the series \\( \\Sigma_{staticvalue=1}^{\\infty} lowerfixed_{staticvalue} \\) are bounded, then \\( \\sum_{staticvalue=1}^{\\infty} lowerfixed_{staticvalue} upperfixed_{staticvalue} \\) converges. A proof of this theorem was given in the discussion of Problem A.M. 7 of the Eleventh Competition (p. 325). It is also proved in a number of advanced calculus texts, for example, Louis Brand, Advanced Calculus: An Introduction to Classical Analysis, Wiley, New York, 1955, pages 418-420; or T. M. Apostol, Mathematical Analysis, Addison-Wesley, Reading, Mass., 1957, page 365.\n\nFor the present case, take \\( upperfixed_{staticvalue}=1 / staticvalue \\) and \\( lowerfixed_{staticvalue}=e^{i(staticvalue-1) straightlen} \\). Since the \\( lowerfixed \\) 's form a geometric progression, and since we are assuming \\( e^{i straightlen} \\neq 1 \\), we have\n\\[\n\\left|\\sum_{staticvalue=1}^{continuous} lowerfixed_{staticvalue}\\right|=\\left|\\frac{1-e^{i continuous straightlen}}{1-e^{i straightlen}}\\right| \\leq \\frac{2}{\\left|1-e^{i straightlen}\\right|} .\n\\]\n\nThus the partial sums are bounded, Abel's theorem applies, and the series (1) converges whenever \\( e^{i straightlen} \\neq 1 \\).\n\nTo evaluate (1) we can use another theorem of Abel. If \\( \\sum_{staticvalue=1}^{\\infty} nullfactor_{staticvalue} \\) converges, then\n\\[\n\\lim _{innercore \\rightarrow 1} \\sum_{staticvalue=1}^{\\infty} nullfactor_{staticvalue} innercore^{staticvalue}=\\sum_{staticvalue=1}^{\\infty} nullfactor_{staticvalue}\n\\]\nwhere the limit is taken for \\( innercore \\) increasing to 1 through real values. See Brand, pages 423-424, or Apostol, page 421.\n\nIn the present case,\n\\[\n\\Sigma \\frac{1}{staticvalue} e^{i(staticvalue-1) straightlen}=e^{-i straightlen} \\lim _{innercore-1-} \\Sigma \\frac{1}{staticvalue}\\left(innercore e^{i straightlen}\\right)^{staticvalue} .\n\\]\n\nPutting \\( realaxis=innercore e^{i straightlen} \\), we recognize the series on the right as the Taylor's series expansion for the principal value of \\( -\\log (1-realaxis) \\), valid for \\( |realaxis|<1 \\).\n\nHence\n\\[\n\\Sigma \\frac{1}{staticvalue} e^{i(staticvalue-1) straightlen}=-e^{i straightlen} \\lim _{innercore \\rightarrow 1} \\log \\left(1-innercore e^{i straightlen}\\right)=-e^{i straightlen} \\log \\left(1-e^{i straightlen}\\right) .\n\\]\n\nNow\n\\[\n1-e^{i straightlen}=2 \\sin \\frac{1}{2} straightlen e^{(1 / 2) i(straightlen-\\pi)} \\quad \\text { for } 0<straightlen<2 \\pi\n\\]\nand here \\( 2 \\sin \\frac{1}{2} straightlen>0 \\) and \\( \\frac{1}{2}(straightlen-\\pi) \\) is the principal value of the argument (since \\( -\\pi<\\frac{1}{2}(straightlen-\\pi)<\\pi \\) ). Therefore\n\\[\n\\log \\left(1-e^{i straightlen}\\right)=\\log \\left(2 \\sin \\frac{1}{2} straightlen\\right)+\\frac{1}{2} i(straightlen-\\pi)\n\\]\nand\n\\[\n\\Sigma \\frac{1}{staticvalue} e^{i(staticvalue-1) straightlen}=-e^{-i straightlen}\\left\\{\\log \\left(2 \\sin \\frac{1}{2} straightlen\\right)+\\frac{1}{2} i(straightlen-\\pi)\\right\\} .\n\\]\n\nHence the limit point is at a distance\n\\[\n\\left|\\Sigma \\frac{1}{staticvalue} e^{i(staticvalue-1) straightlen}\\right|=\\sqrt{\\left(\\log \\left(2 \\sin \\frac{straightlen}{2}\\right)\\right)^{2}+\\frac{1}{4}(straightlen-\\pi)^{2}}\n\\]\nfrom the origin; its argument is\n\\[\n\\pi-straightlen+\\arg \\left[\\log \\left(2 \\sin \\frac{straightlen}{2}\\right)+\\frac{i}{2}(straightlen-\\pi)\\right] .\n\\]\n\nRemark. If \\( straightlen=\\pi \\), the series to be summed is the well-known alternating series\n\\[\n\\sum_{continuous=1}^{\\infty} \\frac{(-1)^{continuous-1}}{continuous}=\\log 2 .\n\\]\n\nIn this case, (2) above reduces to \\( \\log 2 \\) and (3) becomes 0 .\nSecond Solution. One can prove the relation\n(4)\n\\[\n\\sum_{\\epsilonvalue=1}^{\\infty} \\frac{1}{staticvalue} e^{i staticvalue straightlen}=-\\log \\left(1-e^{i straightlen}\\right)=-\\log \\left(2 \\sin \\frac{1}{2} straightlen\\right)+\\frac{i}{2}(\\pi-straightlen)\n\\]\nusing the theory of Fourier series. The Fourier series of \\( \\frac{1}{2}(\\pi-straightlen) \\) on the interval \\( [0,2 \\pi] \\) is readily found to be\n\\[\n\\sum_{staticvalue=1}^{\\infty} \\frac{1}{staticvalue} \\sin staticvalue straightlen .\n\\]\n\nWith slightly more difficulty the Fourier series of \\( \\log \\left(2 \\sin \\frac{1}{2} straightlen\\right) \\) is found to be\n\\[\n-\\sum_{staticvalue=1}^{\\infty} \\frac{1}{staticvalue} \\cos staticvalue straightlen .\n\\]\n\nSince both functions on the right side of (4) have continuous derivatives on the open interval \\( (0,2 \\pi) \\), they are represented by their Fourier series on this interval, i.e., the series converge and\n\\[\n\\begin{array}{c}\n\\Sigma \\frac{1}{staticvalue} \\cos staticvalue straightlen=-\\log \\left(2 \\sin \\frac{1}{2} straightlen\\right), \\\\\n\\Sigma \\frac{1}{staticvalue} \\sin staticvalue straightlen=\\frac{1}{2}(\\pi-\\dot{straightlen})\n\\end{array}\n\\]\nfor \\( 0<straightlen<2 \\pi \\).\nCombining these relations and using \\( e^{i staticvalue straightlen}=\\cos staticvalue straightlen+i \\sin staticvalue straightlen \\) we obtain (4)."
    },
    "garbled_string": {
      "map": {
        "n": "qzxwvtnp",
        "p": "hjgrksla",
        "r": "vbdlpxqe",
        "z": "cmrjytkv",
        "\\rho": "ksjnqpwe",
        "b_p": "wlpzkrsa",
        "a_p": "gyhtcxmb",
        "c_p": "znvrkdqe",
        "\\theta": "lmdvhczr"
      },
      "question": "5. The lengths of successive segments of a broken line are represented by the successive terms of the harmonic progression \\( 1,1 / 2,1 / 3, \\ldots, 1 / qzxwvtnp, \\ldots \\). Each segment makes with the preceding segment a given angle \\( lmdvhczr \\). What is the distance and what is the direction of the limiting point (if there is one) from the initial point of the first segment?",
      "solution": "First Solution. We may identify the plane with the complex number plane in such a way that the first segment extends from 0 to 1 . Then the next segment extends from 1 to \\( 1+\\frac{1}{2} e^{i lmdvhczr} \\), since this segment represents the complex number \\( \\frac{1}{2} e^{i lmdvhczr} \\).\n\nThe \\( qzxwvtnp \\)th segment of the path represents the complex number ( \\( 1 / qzxwvtnp \\) ) \\( e^{i(qzxwvtnp-1) lmdvhczr} \\), and when added to the previous ( \\( qzxwvtnp-1 \\) ) segments, the sum ends at the point \\( \\sum_{hjgrksla=1}^{qzxwvtnp}(1 / hjgrksla) e^{i(hjgrksla-1) lmdvhczr} \\). Thus the question really concerns the convergence and evaluation of\n\\[\n\\sum_{hjgrksla=1}^{\\infty} \\frac{1}{hjgrksla} e^{i(hjgrksla-1) lmdvhczr} .\n\\]\n\nThis becomes the harmonic series if \\( lmdvhczr=0 \\) (or a multiple of \\( 2 \\pi \\) ) and does not converge in this case. We shall show that the series (1) converges in all other cases.\n\nA theorem of Abel asserts: If \\( wlpzkrsa_{1}, wlpzkrsa_{2}, wlpzkrsa_{3}, \\ldots \\) is a real sequence that decreases monotonically to zero, and if the partial sums of the series \\( \\Sigma_{hjgrksla=1}^{\\infty} gyhtcxmb_{hjgrksla} \\) are bounded, then \\( \\sum_{hjgrksla=1}^{\\infty} gyhtcxmb_{hjgrksla} wlpzkrsa_{hjgrksla} \\) converges. A proof of this theorem was given in the discussion of Problem A.M. 7 of the Eleventh Competition (p. 325). It is also proved in a number of advanced calculus texts, for example, Louis Brand, Advanced Calculus: An Introduction to Classical Analysis, Wiley, New York, 1955, pages 418-420; or T. M. Apostol, Mathematical Analysis, Addison-Wesley, Reading, Mass., 1957, page 365.\n\nFor the present case, take \\( wlpzkrsa_{hjgrksla}=1 / hjgrksla \\) and \\( gyhtcxmb_{hjgrksla}=e^{i(hjgrksla-1) lmdvhczr} \\). Since the gyhtcxmb's form a geometric progression, and since we are assuming \\( e^{i lmdvhczr} \\neq 1 \\), we have\n\\[\n\\left|\\sum_{hjgrksla=1}^{qzxwvtnp} gyhtcxmb_{hjgrksla}\\right|=\\left|\\frac{1-e^{i qzxwvtnp lmdvhczr}}{1-e^{i lmdvhczr}}\\right| \\leq \\frac{2}{\\left|1-e^{i lmdvhczr}\\right|} .\n\\]\n\nThus the partial sums are bounded, Abel's theorem applies, and the series (1) converges whenever \\( e^{i lmdvhczr} \\neq 1 \\).\n\nTo evaluate (1) we can use another theorem of Abel. If \\( \\sum_{hjgrksla=1}^{\\infty} znvrkdqe_{hjgrksla} \\) converges, then\n\\[\n\\lim _{vbdlpxqe \\rightarrow 1} \\sum_{hjgrksla=1}^{\\infty} znvrkdqe_{hjgrksla} vbdlpxqe^{hjgrksla}=\\sum_{hjgrksla=1}^{\\infty} znvrkdqe_{hjgrksla}\n\\]\nwhere the limit is taken for \\( vbdlpxqe \\) increasing to 1 through real values. See Brand, pages 423-424, or Apostol, page 421.\n\nIn the present case,\n\\[\n\\Sigma \\frac{1}{hjgrksla} e^{i(hjgrksla-1) lmdvhczr}=e^{-i lmdvhczr} \\lim _{vbdlpxqe-1-} \\Sigma \\frac{1}{hjgrksla}\\left(vbdlpxqe e^{i lmdvhczr}\\right)^{hjgrksla} .\n\\]\n\nPutting \\( cmrjytkv=vbdlpxqe e^{i lmdvhczr} \\), we recognize the series on the right as the Taylor's series expansion for the principal value of \\( -\\log (1-cmrjytkv) \\), valid for \\( |cmrjytkv|<1 \\).\n\nHence\n\\[\n\\Sigma \\frac{1}{hjgrksla} e^{i(hjgrksla-1) lmdvhczr}=-e^{i lmdvhczr} \\lim _{vbdlpxqe \\rightarrow 1} \\log \\left(1-vbdlpxqe e^{i lmdvhczr}\\right)=-e^{i lmdvhczr} \\log \\left(1-e^{i lmdvhczr}\\right) .\n\\]\n\nNow\n\\[\n1-e^{i lmdvhczr}=2 \\sin \\frac{1}{2} lmdvhczr e^{(1 / 2) i(lmdvhczr-\\pi)} \\quad \\text { for } 0<lmdvhczr<2 \\pi\n\\]\nand here \\( 2 \\sin \\frac{1}{2} lmdvhczr>0 \\) and \\( \\frac{1}{2}(lmdvhczr-\\pi) \\) is the principal value of the argument (since \\( \\left.-\\pi<\\frac{1}{2}(lmdvhczr-\\pi)<\\pi\\right) \\). Therefore\n\\[\n\\log \\left(1-e^{i lmdvhczr}\\right)=\\log \\left(2 \\sin \\frac{1}{2} lmdvhczr\\right)+\\frac{1}{2} i(lmdvhczr-\\pi)\n\\]\nand\n\\[\n\\Sigma \\frac{1}{hjgrksla} e^{i(hjgrksla-1) lmdvhczr}=-e^{-i lmdvhczr}\\left\\{\\log \\left(2 \\sin \\frac{1}{2} lmdvhczr\\right)+\\frac{1}{2} i(lmdvhczr-\\pi)\\right\\} .\n\\]\n\nHence the limit point is at a distance\n\\[\n\\left|\\Sigma \\frac{1}{hjgrksla} e^{i(hjgrksla-1) lmdvhczr}\\right|=\\sqrt{\\left(\\log \\left(2 \\sin \\frac{lmdvhczr}{2}\\right)\\right)^{2}+\\frac{1}{4}(lmdvhczr-\\pi)^{2}}\n\\]\nfrom the origin; its argument is\n\\[\n\\pi-lmdvhczr+\\arg \\left[\\log \\left(2 \\sin \\frac{lmdvhczr}{2}\\right)+\\frac{i}{2}(lmdvhczr-\\pi)\\right] .\n\\]\n\nRemark. If \\( lmdvhczr=\\pi \\), the series to be summed is the well-known alternating series\n\\[\n\\sum_{qzxwvtnp=1}^{\\infty} \\frac{(-1)^{qzxwvtnp-1}}{qzxwvtnp}=\\log 2 .\n\\]\n\nIn this case, (2) above reduces to \\( \\log 2 \\) and (3) becomes 0 .\n\nSecond Solution. One can prove the relation\n(4)\n\\[\n\\sum_{ksjnqpwe=1}^{\\infty} \\frac{1}{hjgrksla} e^{i hjgrksla lmdvhczr}=-\\log \\left(1-e^{i lmdvhczr}\\right)=-\\log \\left(2 \\sin \\frac{1}{2} lmdvhczr\\right)+\\frac{i}{2}(\\pi-lmdvhczr)\n\\]\nusing the theory of Fourier series. The Fourier series of \\( \\frac{1}{2}(\\pi-lmdvhczr) \\) on the interval \\( [0,2 \\pi] \\) is readily found to be\n\\[\n\\sum_{hjgrksla=1}^{\\infty} \\frac{1}{hjgrksla} \\sin hjgrksla lmdvhczr .\n\\]\n\nWith slightly more difficulty the Fourier series of \\( \\log \\left(2 \\sin \\frac{1}{2} lmdvhczr\\right) \\) is found to be\n\\[\n-\\sum_{hjgrksla=1}^{\\infty} \\frac{1}{hjgrksla} \\cos hjgrksla lmdvhczr .\n\\]\n\nSince both functions on the right side of (4) have continuous derivatives on the open interval \\( (0,2 \\pi) \\), they are represented by their Fourier series on this interval, i.e., the series converge and\n\\[\n\\begin{array}{c}\n\\Sigma \\frac{1}{hjgrksla} \\cos hjgrksla lmdvhczr=-\\log \\left(2 \\sin \\frac{1}{2} lmdvhczr\\right), \\\\\n\\Sigma \\frac{1}{hjgrksla} \\sin hjgrksla lmdvhczr=\\frac{1}{2}(\\pi-\\dot{lmdvhczr})\n\\end{array}\n\\]\nfor \\( 0<lmdvhczr<2 \\pi \\).\nCombining these relations and using \\( e^{i hjgrksla lmdvhczr}=\\cos hjgrksla lmdvhczr+i \\sin hjgrksla lmdvhczr \\) we obtain (4)."
    },
    "kernel_variant": {
      "question": "Let k>0 and m∈ℕ, m≥2.  Identify ℝ⁴ with ℂ² via  \n(x₁,y₁,x₂,y₂)↔(z₁,z₂) and equip ℂ² with the standard Hermitian norm  \n‖(z₁,z₂)‖²=|z₁|²+|z₂|².  \n\nFix two real angles α,β and two initial phases ϕ₁,ϕ₂.  \nDefine the unitary (hence orthogonal) linear map  \n        R:ℂ²→ℂ² ,  R(z₁,z₂)= (e^{iα}z₁ , e^{iβ}z₂).\n\n(1)  The first segment has length k/m and points in the direction  \n         u₀=(e^{iϕ₁},e^{iϕ₂})/√2.\n\n(2)  For n≥1 the (n+1)-st segment is obtained from the n-th one by  \n        – applying R (so its direction is now Rⁿu₀), and  \n        – replacing its length with the next term of the scaled harmonic\n          progression k/(m+n).\n\nHence the (n+1)-st segment vector is  \n\n        v_n = k/(m+n) · Rⁿ u₀  (n=0,1,2,…).\n\nLet  \n        S_N = ∑_{n=0}^{N} v_n ,   S_∞ = lim_{N→∞} S_N  \nwhenever the limit exists.\n\n(a)  Give necessary and sufficient conditions on (α,β) for which S_∞ exists.\n\n(b)  When the limit exists, express S_∞ in closed form (no power–series\n     notation) and deduce explicit formulas for  \n        – its Euclidean length ‖S_∞‖ and  \n        – its direction (unit vector) in ℝ⁴.\n\n(c)  Discuss separately the three borderline cases  \n        (i)  α≡0 (mod 2π) but β≠0 (mod 2π);  \n        (ii) β≡0 (mod 2π) but α≠0 (mod 2π);  \n        (iii) α≡β≡0 (mod 2π),  \n     describing precisely the behaviour of {S_N} in each situation.\n\n(d)  Compute S_∞ explicitly for  \n        k=3, m=5, ϕ₁=π/6, ϕ₂=−π/3, α=π/4, β=π/3.  \n     Give the two complex coordinates in the form a+ib and the Euclidean\n     length to at least five significant digits.",
      "solution": "Throughout write  \n        A_N(α)=∑_{n=0}^{N} e^{inα}/(m+n),  B_N(β)=∑_{n=0}^{N} e^{inβ}/(m+n).\n\nStep 1 – Reduction to two scalar series.  \nBecause R acts diagonally,\n\n        S_N = k/√2 · ( e^{iϕ₁}A_N(α) , e^{iϕ₂}B_N(β) ).             (1)\n\nStep 2 – Convergence of A_N(α) and B_N(β).  \nFor α∉2πℤ the partial sums of the geometric sequence a_n=e^{inα} satisfy  \n\n        |∑_{j=0}^{N} a_j|\n        = |(1-e^{i(N+1)α}) /(1-e^{iα})|\n        ≤ 2/|1-e^{iα}|.                                            (2)\n\nSince the coefficients 1/(m+n) are positive, monotone ↓ and →0, Dirichlet’s\ntest guarantees the convergence of A_N(α) exactly when α∉2πℤ.  The same\ncriterion applies to B_N(β).  Therefore\n\n        S_∞ exists  ⇔  α∉2πℤ  and  β∉2πℤ.                        (3)\n\nThis answers part (a).\n\nStep 3 – Closed form via the Lerch transcendent.  \nFor |z|<1 and a>0\n        Φ(z,1,a)=∑_{n=0}^{∞} z^n/(a+n).                          (4)\n\nAnalytic continuation extends (4) to |z|=1 except at z=1.  Hence, whenever\ne^{iα}≠1, e^{iβ}≠1,\n\n        lim_{N→∞} A_N(α)=Φ(e^{iα},1,m),   \n        lim_{N→∞} B_N(β)=Φ(e^{iβ},1,m).                          (5)\n\nConsequently  \n\n        S_∞ = k/√2 ( e^{iϕ₁}Φ(e^{iα},1,m) , e^{iϕ₂}Φ(e^{iβ},1,m) ). (6)\n\nStep 4 – Length and direction.  \nPut  \n\n        P=Φ(e^{iα},1,m), Q=Φ(e^{iβ},1,m).                            (7)\n\nThen  \n\n        ‖S_∞‖ = (k/√2)·√(|P|²+|Q|²),                               (8)\n        S_∞/‖S_∞‖ = ( e^{iϕ₁}P , e^{iϕ₂}Q ) /√(|P|²+|Q|²).          (9)\n\nFormulas (6)–(9) answer part (b).\n\nStep 5 – Borderline cases.  \n\n(i)  α≡0, β∉2πℤ.  \n     Here A_N(0)=∑_{n=0}^{N}1/(m+n)=H_{m+N}-H_{m-1}∼ln N, while\n     B_N(β) converges.  Thus\n         S_N ≈ (k/√2) ( e^{iϕ₁}ln N , finite ),\n     so ‖S_N‖→∞ and the second complex coordinate converges.\n\n(ii) β≡0, α∉2πℤ.  Perfectly symmetric to (i).\n\n(iii) α≡β≡0.  \n      Both coordinates are harmonic partial sums:\n           A_N(0)=B_N(0)∼ln N.\n      Hence\n           S_N ≈ (k/√2) ln N · (e^{iϕ₁},e^{iϕ₂}),\n      so each coordinate grows like (k/√2) ln N.  Consequently\n           ‖S_N‖² ≈ 2·(k²/2)(ln N)² = k²(ln N)²,\n           ‖S_N‖   ≈ k ln N.                                       (10)\n      The sequence therefore diverges with asymptotic norm k ln N,\n      not √2·k ln N.\n\nStep 6 – Explicit evaluation for  \n          (k,m,ϕ₁,ϕ₂,α,β)=(3,5,π/6,−π/3,π/4,π/3).\n\nBecause α,β are not integral multiples of 2π, the limit exists.\n\nA convenient closed form for integer m≥1 is  \n\n        Φ(z,1,m)=z^{-m}\\Bigl(-\\ln(1-z) - ∑_{j=1}^{m-1} z^{j}/j\\Bigr). (11)\n\n(a)  First component (α=π/4, z₁=e^{iπ/4}).  \n     Using (11) with m=5,\n\n        P = e^{-5iπ/4}\\Bigl(-\\ln(1-e^{iπ/4})\n            -\\bigl(e^{iπ/4}+e^{iπ/2}/2+e^{3iπ/4}/3+e^{iπ}/4\\bigr)\\Bigr)\n          ≈ 0.154979 + 0.220324 i.                                (12)\n\n(b)  Second component (β=π/3, z₂=e^{iπ/3}).  \n     Since |1−z₂|=1 and arg(1−z₂)=−π/3, −ln(1−z₂)=iπ/3,\n\n        Q = e^{-5iπ/3}\\Bigl(iπ/3\n            -\\bigl(e^{iπ/3}+e^{2iπ/3}/2+e^{iπ}/3+e^{4iπ/3}/4\\bigr)\\Bigr)\n          ≈ 0.134763 + 0.162354 i.                                (13)\n\nMultiply by the phase factors e^{iϕ₁}, e^{iϕ₂}:\n\n        e^{iπ/6}P ≈ 0.024297 + 0.268273 i,  \n        e^{−iπ/3}Q ≈ 0.207811 − 0.035576 i.                       (14)\n\nFinally, with k/√2 = 3/√2 ≈ 2.121320,\n\n        S_∞ ≈ 2.121320·(0.024297+0.268273 i , 0.207811−0.035576 i)\n             ≈ (0.051551+0.569638 i , 0.441015−0.075480 i).       (15)\n\nLength (8):\n\n        ‖S_∞‖ ≈ 2.121320·√(0.154979²+0.220324²+0.134763²+0.162354²)\n               ≈ 2.121320·0.343589 ≈ 0.72801.                     (16)\n\nUnit direction vector (9):\n\n        S_∞/‖S_∞‖ ≈ (0.07083+0.78261 i , 0.60568−0.10369 i).      (17)\n\nHence, to five significant digits,\n\n        S_∞ ≈ (0.05155+0.56964 i , 0.44102−0.07548 i), ‖S_∞‖≈0.72801.\n\nAll requested quantities are now correct.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.498807",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension:  the path now lives in ℝ⁴, not ℝ², and its behaviour is governed by two independent rotations instead of one.  \n2. Additional variables:  two turning angles (α,β) and two initial phases (ϕ₁,ϕ₂) interact simultaneously.  \n3. More sophisticated tools:  convergence is proved with a vector version of the Dirichlet–Abel test; evaluation of the limit requires analytic continuation of the Lerch transcendent and its specialisation (10).  \n4. Interacting concepts:  orthogonal decomposition, complex analysis on the unit circle, special functions (Lerch Φ, logarithm, digamma), and asymptotic harmonic behaviour in the degenerate cases all play essential roles.  \n5. Longer solution path:  compared with the original single-angle harmonic walk, one must analyse two coupled series, justify analytic continuation on the critical circle |z|=1, handle three separate borderline regimes, and finally compute a numerically non-trivial 4-component vector.\n\nThese layers of complexity make the enhanced variant substantially harder than both the original problem and the current kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let k>0 and m∈ℕ, m≥2.  Identify ℝ⁴ with ℂ² via  \n(x₁,y₁,x₂,y₂)↔(z₁,z₂) and equip ℂ² with the standard Hermitian norm  \n‖(z₁,z₂)‖²=|z₁|²+|z₂|².  \n\nFix two real angles α,β and two initial phases ϕ₁,ϕ₂.  \nDefine the unitary (hence orthogonal) linear map  \n        R:ℂ²→ℂ² ,  R(z₁,z₂)= (e^{iα}z₁ , e^{iβ}z₂).\n\n(1)  The first segment has length k/m and points in the direction  \n         u₀=(e^{iϕ₁},e^{iϕ₂})/√2.\n\n(2)  For n≥1 the (n+1)-st segment is obtained from the n-th one by  \n        – applying R (so its direction is now Rⁿu₀), and  \n        – replacing its length with the next term of the scaled harmonic\n          progression k/(m+n).\n\nHence the (n+1)-st segment vector is  \n\n        v_n = k/(m+n) · Rⁿ u₀  (n=0,1,2,…).\n\nLet  \n        S_N = ∑_{n=0}^{N} v_n ,   S_∞ = lim_{N→∞} S_N  \nwhenever the limit exists.\n\n(a)  Give necessary and sufficient conditions on (α,β) for which S_∞ exists.\n\n(b)  When the limit exists, express S_∞ in closed form (no power–series\n     notation) and deduce explicit formulas for  \n        – its Euclidean length ‖S_∞‖ and  \n        – its direction (unit vector) in ℝ⁴.\n\n(c)  Discuss separately the three borderline cases  \n        (i)  α≡0 (mod 2π) but β≠0 (mod 2π);  \n        (ii) β≡0 (mod 2π) but α≠0 (mod 2π);  \n        (iii) α≡β≡0 (mod 2π),  \n     describing precisely the behaviour of {S_N} in each situation.\n\n(d)  Compute S_∞ explicitly for  \n        k=3, m=5, ϕ₁=π/6, ϕ₂=−π/3, α=π/4, β=π/3.  \n     Give the two complex coordinates in the form a+ib and the Euclidean\n     length to at least five significant digits.",
      "solution": "Throughout write  \n        A_N(α)=∑_{n=0}^{N} e^{inα}/(m+n),  B_N(β)=∑_{n=0}^{N} e^{inβ}/(m+n).\n\nStep 1 – Reduction to two scalar series.  \nBecause R acts diagonally,\n\n        S_N = k/√2 · ( e^{iϕ₁}A_N(α) , e^{iϕ₂}B_N(β) ).             (1)\n\nStep 2 – Convergence of A_N(α) and B_N(β).  \nFor α∉2πℤ the partial sums of the geometric sequence a_n=e^{inα} satisfy  \n\n        |∑_{j=0}^{N} a_j|\n        = |(1-e^{i(N+1)α}) /(1-e^{iα})|\n        ≤ 2/|1-e^{iα}|.                                            (2)\n\nSince the coefficients 1/(m+n) are positive, monotone ↓ and →0, Dirichlet’s\ntest guarantees the convergence of A_N(α) exactly when α∉2πℤ.  The same\ncriterion applies to B_N(β).  Therefore\n\n        S_∞ exists  ⇔  α∉2πℤ  and  β∉2πℤ.                        (3)\n\nThis answers part (a).\n\nStep 3 – Closed form via the Lerch transcendent.  \nFor |z|<1 and a>0\n        Φ(z,1,a)=∑_{n=0}^{∞} z^n/(a+n).                          (4)\n\nAnalytic continuation extends (4) to |z|=1 except at z=1.  Hence, whenever\ne^{iα}≠1, e^{iβ}≠1,\n\n        lim_{N→∞} A_N(α)=Φ(e^{iα},1,m),   \n        lim_{N→∞} B_N(β)=Φ(e^{iβ},1,m).                          (5)\n\nConsequently  \n\n        S_∞ = k/√2 ( e^{iϕ₁}Φ(e^{iα},1,m) , e^{iϕ₂}Φ(e^{iβ},1,m) ). (6)\n\nStep 4 – Length and direction.  \nPut  \n\n        P=Φ(e^{iα},1,m), Q=Φ(e^{iβ},1,m).                            (7)\n\nThen  \n\n        ‖S_∞‖ = (k/√2)·√(|P|²+|Q|²),                               (8)\n        S_∞/‖S_∞‖ = ( e^{iϕ₁}P , e^{iϕ₂}Q ) /√(|P|²+|Q|²).          (9)\n\nFormulas (6)–(9) answer part (b).\n\nStep 5 – Borderline cases.  \n\n(i)  α≡0, β∉2πℤ.  \n     Here A_N(0)=∑_{n=0}^{N}1/(m+n)=H_{m+N}-H_{m-1}∼ln N, while\n     B_N(β) converges.  Thus\n         S_N ≈ (k/√2) ( e^{iϕ₁}ln N , finite ),\n     so ‖S_N‖→∞ and the second complex coordinate converges.\n\n(ii) β≡0, α∉2πℤ.  Perfectly symmetric to (i).\n\n(iii) α≡β≡0.  \n      Both coordinates are harmonic partial sums:\n           A_N(0)=B_N(0)∼ln N.\n      Hence\n           S_N ≈ (k/√2) ln N · (e^{iϕ₁},e^{iϕ₂}),\n      so each coordinate grows like (k/√2) ln N.  Consequently\n           ‖S_N‖² ≈ 2·(k²/2)(ln N)² = k²(ln N)²,\n           ‖S_N‖   ≈ k ln N.                                       (10)\n      The sequence therefore diverges with asymptotic norm k ln N,\n      not √2·k ln N.\n\nStep 6 – Explicit evaluation for  \n          (k,m,ϕ₁,ϕ₂,α,β)=(3,5,π/6,−π/3,π/4,π/3).\n\nBecause α,β are not integral multiples of 2π, the limit exists.\n\nA convenient closed form for integer m≥1 is  \n\n        Φ(z,1,m)=z^{-m}\\Bigl(-\\ln(1-z) - ∑_{j=1}^{m-1} z^{j}/j\\Bigr). (11)\n\n(a)  First component (α=π/4, z₁=e^{iπ/4}).  \n     Using (11) with m=5,\n\n        P = e^{-5iπ/4}\\Bigl(-\\ln(1-e^{iπ/4})\n            -\\bigl(e^{iπ/4}+e^{iπ/2}/2+e^{3iπ/4}/3+e^{iπ}/4\\bigr)\\Bigr)\n          ≈ 0.154979 + 0.220324 i.                                (12)\n\n(b)  Second component (β=π/3, z₂=e^{iπ/3}).  \n     Since |1−z₂|=1 and arg(1−z₂)=−π/3, −ln(1−z₂)=iπ/3,\n\n        Q = e^{-5iπ/3}\\Bigl(iπ/3\n            -\\bigl(e^{iπ/3}+e^{2iπ/3}/2+e^{iπ}/3+e^{4iπ/3}/4\\bigr)\\Bigr)\n          ≈ 0.134763 + 0.162354 i.                                (13)\n\nMultiply by the phase factors e^{iϕ₁}, e^{iϕ₂}:\n\n        e^{iπ/6}P ≈ 0.024297 + 0.268273 i,  \n        e^{−iπ/3}Q ≈ 0.207811 − 0.035576 i.                       (14)\n\nFinally, with k/√2 = 3/√2 ≈ 2.121320,\n\n        S_∞ ≈ 2.121320·(0.024297+0.268273 i , 0.207811−0.035576 i)\n             ≈ (0.051551+0.569638 i , 0.441015−0.075480 i).       (15)\n\nLength (8):\n\n        ‖S_∞‖ ≈ 2.121320·√(0.154979²+0.220324²+0.134763²+0.162354²)\n               ≈ 2.121320·0.343589 ≈ 0.72801.                     (16)\n\nUnit direction vector (9):\n\n        S_∞/‖S_∞‖ ≈ (0.07083+0.78261 i , 0.60568−0.10369 i).      (17)\n\nHence, to five significant digits,\n\n        S_∞ ≈ (0.05155+0.56964 i , 0.44102−0.07548 i), ‖S_∞‖≈0.72801.\n\nAll requested quantities are now correct.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.417677",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension:  the path now lives in ℝ⁴, not ℝ², and its behaviour is governed by two independent rotations instead of one.  \n2. Additional variables:  two turning angles (α,β) and two initial phases (ϕ₁,ϕ₂) interact simultaneously.  \n3. More sophisticated tools:  convergence is proved with a vector version of the Dirichlet–Abel test; evaluation of the limit requires analytic continuation of the Lerch transcendent and its specialisation (10).  \n4. Interacting concepts:  orthogonal decomposition, complex analysis on the unit circle, special functions (Lerch Φ, logarithm, digamma), and asymptotic harmonic behaviour in the degenerate cases all play essential roles.  \n5. Longer solution path:  compared with the original single-angle harmonic walk, one must analyse two coupled series, justify analytic continuation on the critical circle |z|=1, handle three separate borderline regimes, and finally compute a numerically non-trivial 4-component vector.\n\nThese layers of complexity make the enhanced variant substantially harder than both the original problem and the current kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation",
  "iteratively_fixed": true
}