{
  "index": "1950-B-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "2. Two obvious approximations to the length of the perimeter of the ellipse with semi-axes \\( a \\) and \\( b \\) are \\( \\pi(a+b) \\) and \\( 2 \\pi(a b)^{12} \\). Which one comes nearer the truth when the ratio \\( b / a \\) is very close to \\( 1 ? \\)",
  "solution": "Solution. Let the ellipse be taken in the parametric form \\( x=a \\cos t \\), \\( y=b \\sin t \\). Then the length \\( L \\) is given by\n\\[\nL=\\int_{0}^{2 \\pi} \\sqrt{\\left(\\frac{d x}{d t}\\right)^{2}+\\left(\\frac{d y}{d t}\\right)^{2}} d t=\\int_{0}^{2 \\pi} \\sqrt{a^{2} \\sin ^{2} t+b^{2} \\cos ^{2} t} d t\n\\]\na well-known elliptic integral. Since we are asked to consider \\( L \\) when \\( b / a \\) is nearly 1 we put \\( b=(1+\\lambda) a \\) and consider\n\\[\nL(\\lambda)=a \\int_{0}^{2 \\pi} \\sqrt{1+\\left(2 \\lambda+\\lambda^{2}\\right) \\cos ^{2} t} d t .\n\\]\n\nThis is evidently an analytic function of \\( \\lambda \\), and we calculate its power series to terms of degree 2\n\\[\n\\begin{aligned}\nL(\\lambda) & =a \\int_{0}^{2 \\pi}\\left(1+\\frac{1}{2}\\left(2 \\lambda+\\lambda^{2}\\right) \\cos ^{2} t-\\frac{1}{8}\\left(2 \\lambda+\\lambda^{2}\\right)^{2} \\cos ^{4} t+\\cdots\\right) d t \\\\\n& =2 \\pi a\\left[1+\\frac{1}{4}\\left(2 \\lambda+\\lambda^{2}\\right)-\\frac{3}{64}\\left(2 \\lambda+\\lambda^{2}\\right)^{2}+\\cdots\\right] \\\\\n& =2 \\pi a\\left[1+\\frac{1}{2} \\lambda+\\frac{1}{16} \\lambda^{2}+\\cdots\\right] .\n\\end{aligned}\n\\]\n\nThe first expression was obtained using the binomial expansion of \\( (1+z)^{1 / 2} \\). All the terms omitted are of degree at least three in \\( \\lambda \\).\n\nThe formal manipulation is justified because the series in question converges absolutely for small values of \\( \\lambda \\) (in fact if \\( |2 \\lambda|+\\lambda^{2}<1 \\) ). (We could also find the power series by differentiating under the integral sign.)\nThe proposed approximations to the perimeter are\n\\[\n\\pi(a+b)=2 \\pi a\\left(1+\\frac{1}{2} \\lambda\\right)\n\\]\nand\n\\[\n2 \\pi \\sqrt{a b}=2 \\pi a \\sqrt{1+\\lambda}=2 \\pi a\\left(1+\\frac{1}{2} \\lambda-\\frac{1}{8} \\lambda^{2} \\cdots\\right) .\n\\]\n\nSince the three functions have the same constant and first degree terms their differences are controlled by the second degree terms for small \\( \\lambda \\). We have\n\\[\nL(\\lambda)>2 \\pi a\\left(1+\\frac{1}{2} \\lambda\\right)>2 \\pi a\\left(1+\\frac{1}{2} \\lambda-\\frac{1}{8} \\lambda^{2}+\\cdots\\right)\n\\]\nfor small \\( \\lambda \\). Thus \\( \\pi(a+b) \\) is a better approximation to the length of an ellipse than \\( 2 \\pi \\sqrt{a b} \\); in fact it is roughly three times better since\n\\[\nL(\\lambda)-2 \\pi a\\left(1+\\frac{1}{2} \\lambda\\right) \\sim \\frac{1}{16} \\lambda^{2} \\text { and } L(\\lambda)-2 \\pi a \\sqrt{1+\\lambda} \\sim \\frac{3}{16} \\lambda^{2} .\n\\]\n\nContinuation. It is clear that \\( L \\) is a differentiable function of \\( \\lambda \\) for all values of \\( \\lambda>-1 \\). If we calculate the second derivative we find\n\\[\nL^{\\prime \\prime}(\\lambda)=a \\int_{0}^{2 \\pi} \\frac{\\sin ^{2} t \\cos ^{2} t}{\\left(1+\\left(2 \\lambda+\\lambda^{2}\\right) \\cos ^{2} t\\right)^{3 / 2}} d t,\n\\]\nwhich is evidently positive for all \\( \\lambda>-1 \\). Hence by Taylor's Theorem\n\\[\n\\begin{array}{c}\nL(\\lambda)=L(0)+\\lambda L^{\\prime}(0)+\\frac{1}{2} \\lambda^{2} L^{\\prime \\prime}(\\xi)> \\\\\nL(0)+\\lambda L^{\\prime}(0)=2 \\pi a\\left(1+\\frac{1}{2} \\lambda\\right)\n\\end{array}\n\\]\nfor all \\( \\lambda \\neq 0 \\). Thus the length of a (non-circular) ellipse always exceeds \\( \\pi(a \\) \\( +b) \\). By the arithmetic-geometric mean inequality we also have \\( \\pi(a+b) \\) \\( >2 \\pi \\sqrt{a b} \\), so \\( \\pi(a+b) \\) is always a better approximation to the perimeter of an ellipse than \\( 2 \\pi \\sqrt{a b} \\).\nThe inequality \\( L>\\pi(a+b) \\) can be demonstrated directly as follows: First note that\n\\[\n\\sqrt{\\mu A+(1-\\mu) B}+\\sqrt{(1-\\mu) A+\\mu B} \\geq \\sqrt{A}+\\sqrt{B}\n\\]\nwhenever \\( A \\) and \\( B \\) are positive and \\( 0 \\leq \\mu \\leq 1 \\), with strict inequality unless \\( A=B \\) or \\( \\mu=0 \\) or 1 . This is easy to prove by successive squaring and canceling.\n\nReplacing \\( t \\) by \\( t+\\pi / 2 \\) in the original integral for \\( L \\) we find that\n\\[\nL=\\int_{0}^{2 \\pi} \\sqrt{a^{2} \\cos ^{2} t+b^{2} \\sin ^{2} t} d t\n\\]\n\nThen\n\\[\n\\begin{array}{c}\n2 L=\\int_{0}^{2 \\pi}\\left(\\sqrt{a^{2}} \\cos ^{2} t \\overline{t b^{2}} \\sin ^{2} t+\\sqrt{ } a^{2} \\overline{\\sin ^{2} t+b^{2} \\cos ^{2} t}\\right) d t \\\\\n>\\int_{0}^{2 \\pi}(a+b) d t=2 \\pi(a+b)\n\\end{array}\n\\]\nusing the inequality just stated with \\( A=a^{2}, B=b^{2}, \\mu=\\cos ^{2} t \\). See M. S. Klamkin, \"Elementary Approximations to the Area of \\( N \\)-dimensional Ellipsoids,\" American Mathematical Monthly. vol. 78 (1971), pages 280-283.",
  "vars": [
    "x",
    "y",
    "t",
    "L",
    "z",
    "\\\\lambda",
    "\\\\mu",
    "A",
    "B",
    "N",
    "\\\\xi"
  ],
  "params": [
    "a",
    "b"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "xcoordinate",
        "y": "ycoordinate",
        "t": "tvariable",
        "L": "perimeter",
        "z": "zvariable",
        "\\lambda": "lambdaparam",
        "\\mu": "muparameter",
        "A": "constalpha",
        "B": "constbeta",
        "N": "constgamma",
        "\\xi": "xiparameter",
        "a": "semimajor",
        "b": "semiminor"
      },
      "question": "2. Two obvious approximations to the length of the perimeter of the ellipse with semi-axes \\( semimajor \\) and \\( semiminor \\) are \\( \\pi(semimajor+semiminor) \\) and \\( 2 \\pi(semimajor semiminor)^{12} \\). Which one comes nearer the truth when the ratio \\( semiminor / semimajor \\) is very close to \\( 1 ? \\)",
      "solution": "Solution. Let the ellipse be taken in the parametric form \\( xcoordinate=semimajor \\cos tvariable \\), \\( ycoordinate=semiminor \\sin tvariable \\). Then the length \\( perimeter \\) is given by\n\\[\nperimeter=\\int_{0}^{2 \\pi} \\sqrt{\\left(\\frac{d xcoordinate}{d tvariable}\\right)^{2}+\\left(\\frac{d ycoordinate}{d tvariable}\\right)^{2}} d tvariable=\\int_{0}^{2 \\pi} \\sqrt{semimajor^{2} \\sin ^{2} tvariable+semiminor^{2} \\cos ^{2} tvariable} d tvariable\n\\]\na well-known elliptic integral. Since we are asked to consider \\( perimeter \\) when \\( semiminor / semimajor \\) is nearly 1 we put \\( semiminor=(1+lambdaparam) semimajor \\) and consider\n\\[\nperimeter(lambdaparam)=semimajor \\int_{0}^{2 \\pi} \\sqrt{1+\\left(2 lambdaparam+lambdaparam^{2}\\right) \\cos ^{2} tvariable} d tvariable .\n\\]\n\nThis is evidently an analytic function of \\( lambdaparam \\), and we calculate its power series to terms of degree 2\n\\[\n\\begin{aligned}\nperimeter(lambdaparam) & =semimajor \\int_{0}^{2 \\pi}\\left(1+\\frac{1}{2}\\left(2 lambdaparam+lambdaparam^{2}\\right) \\cos ^{2} tvariable-\\frac{1}{8}\\left(2 lambdaparam+lambdaparam^{2}\\right)^{2} \\cos ^{4} tvariable+\\cdots\\right) d tvariable \\\\\n& =2 \\pi semimajor\\left[1+\\frac{1}{4}\\left(2 lambdaparam+lambdaparam^{2}\\right)-\\frac{3}{64}\\left(2 lambdaparam+lambdaparam^{2}\\right)^{2}+\\cdots\\right] \\\\\n& =2 \\pi semimajor\\left[1+\\frac{1}{2} lambdaparam+\\frac{1}{16} lambdaparam^{2}+\\cdots\\right] .\n\\end{aligned}\n\\]\n\nThe first expression was obtained using the binomial expansion of \\( (1+zvariable)^{1 / 2} \\). All the terms omitted are of degree at least three in \\( lambdaparam \\).\n\nThe formal manipulation is justified because the series in question converges absolutely for small values of \\( lambdaparam \\) (in fact if \\( |2 lambdaparam|+lambdaparam^{2}<1 \\) ). (We could also find the power series by differentiating under the integral sign.)\nThe proposed approximations to the perimeter are\n\\[\n\\pi(semimajor+semiminor)=2 \\pi semimajor\\left(1+\\frac{1}{2} lambdaparam\\right)\n\\]\nand\n\\[\n2 \\pi \\sqrt{semimajor semiminor}=2 \\pi semimajor \\sqrt{1+lambdaparam}=2 \\pi semimajor\\left(1+\\frac{1}{2} lambdaparam-\\frac{1}{8} lambdaparam^{2} \\cdots\\right) .\n\\]\n\nSince the three functions have the same constant and first degree terms their differences are controlled by the second degree terms for small \\( lambdaparam \\). We have\n\\[\nperimeter(lambdaparam)>2 \\pi semimajor\\left(1+\\frac{1}{2} lambdaparam\\right)>2 \\pi semimajor\\left(1+\\frac{1}{2} lambdaparam-\\frac{1}{8} lambdaparam^{2}+\\cdots\\right)\n\\]\nfor small \\( lambdaparam \\). Thus \\( \\pi(semimajor+semiminor) \\) is a better approximation to the length of an ellipse than \\( 2 \\pi \\sqrt{semimajor semiminor} \\); in fact it is roughly three times better since\n\\[\nperimeter(lambdaparam)-2 \\pi semimajor\\left(1+\\frac{1}{2} lambdaparam\\right) \\sim \\frac{1}{16} lambdaparam^{2} \\text { and } perimeter(lambdaparam)-2 \\pi semimajor \\sqrt{1+lambdaparam} \\sim \\frac{3}{16} lambdaparam^{2} .\n\\]\n\nContinuation. It is clear that \\( perimeter \\) is a differentiable function of \\( lambdaparam \\) for all values of \\( lambdaparam>-1 \\). If we calculate the second derivative we find\n\\[\nperimeter^{\\prime \\prime}(lambdaparam)=semimajor \\int_{0}^{2 \\pi} \\frac{\\sin ^{2} tvariable \\cos ^{2} tvariable}{\\left(1+\\left(2 lambdaparam+lambdaparam^{2}\\right) \\cos ^{2} tvariable\\right)^{3 / 2}} d tvariable,\n\\]\nwhich is evidently positive for all \\( lambdaparam>-1 \\). Hence by Taylor's Theorem\n\\[\n\\begin{array}{c}\nperimeter(lambdaparam)=perimeter(0)+lambdaparam\\,perimeter^{\\prime}(0)+\\frac{1}{2} lambdaparam^{2} perimeter^{\\prime \\prime}(xiparameter)> \\\\\nperimeter(0)+lambdaparam\\,perimeter^{\\prime}(0)=2 \\pi semimajor\\left(1+\\frac{1}{2} lambdaparam\\right)\n\\end{array}\n\\]\nfor all \\( lambdaparam \\neq 0 \\). Thus the length of a (non-circular) ellipse always exceeds \\( \\pi(semimajor \\) \\( +semiminor) \\). By the arithmetic-geometric mean inequality we also have \\( \\pi(semimajor+semiminor) \\) \\( >2 \\pi \\sqrt{semimajor semiminor} \\), so \\( \\pi(semimajor+semiminor) \\) is always a better approximation to the perimeter of an ellipse than \\( 2 \\pi \\sqrt{semimajor semiminor} \\).\nThe inequality \\( perimeter>\\pi(semimajor+semiminor) \\) can be demonstrated directly as follows: First note that\n\\[\n\\sqrt{muparameter constalpha+(1-muparameter) constbeta}+\\sqrt{(1-muparameter) constalpha+muparameter constbeta} \\geq \\sqrt{constalpha}+\\sqrt{constbeta}\n\\]\nwhenever \\( constalpha \\) and \\( constbeta \\) are positive and \\( 0 \\leq muparameter \\leq 1 \\), with strict inequality unless \\( constalpha=constbeta \\) or \\( muparameter=0 \\) or 1 . This is easy to prove by successive squaring and canceling.\n\nReplacing \\( tvariable \\) by \\( tvariable+\\pi / 2 \\) in the original integral for \\( perimeter \\) we find that\n\\[\nperimeter=\\int_{0}^{2 \\pi} \\sqrt{semimajor^{2} \\cos ^{2} tvariable+semiminor^{2} \\sin ^{2} tvariable} d tvariable\n\\]\n\nThen\n\\[\n\\begin{array}{c}\n2 perimeter=\\int_{0}^{2 \\pi}\\left(\\sqrt{semimajor^{2}} \\cos ^{2} tvariable \\overline{tvariable semiminor^{2}} \\sin ^{2} tvariable+\\sqrt{ } semimajor^{2} \\overline{\\sin ^{2} tvariable+semiminor^{2} \\cos ^{2} tvariable}\\right) d tvariable \\\\\n>\\int_{0}^{2 \\pi}(semimajor+semiminor) d tvariable=2 \\pi(semimajor+semiminor)\n\\end{array}\n\\]\nusing the inequality just stated with \\( constalpha=semimajor^{2}, constbeta=semiminor^{2}, muparameter=\\cos ^{2} tvariable \\). See M. S. Klamkin, \"Elementary Approximations to the Area of \\( constgamma \\)-dimensional Ellipsoids,\" American Mathematical Monthly. vol. 78 (1971), pages 280-283."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "pinecones",
        "y": "lighthouse",
        "t": "raindrops",
        "L": "blueberry",
        "z": "windstorm",
        "\\lambda": "butterfly",
        "\\mu": "driftwood",
        "A": "waterfall",
        "B": "honeycomb",
        "N": "playground",
        "\\xi": "sandalwood",
        "a": "parsleyyy",
        "b": "sagebrush"
      },
      "question": "2. Two obvious approximations to the length of the perimeter of the ellipse with semi-axes \\( parsleyyy \\) and \\( sagebrush \\) are \\( \\pi(parsleyyy+sagebrush) \\) and \\( 2 \\pi(parsleyyy sagebrush)^{12} \\). Which one comes nearer the truth when the ratio \\( sagebrush / parsleyyy \\) is very close to 1?",
      "solution": "Solution. Let the ellipse be taken in the parametric form \\( pinecones=parsleyyy \\cos raindrops \\), \\( lighthouse=sagebrush \\sin raindrops \\). Then the length \\( blueberry \\) is given by\n\\[\nblueberry=\\int_{0}^{2 \\pi} \\sqrt{\\left(\\frac{d pinecones}{d raindrops}\\right)^{2}+\\left(\\frac{d lighthouse}{d raindrops}\\right)^{2}} d raindrops=\\int_{0}^{2 \\pi} \\sqrt{parsleyyy^{2} \\sin ^{2} raindrops+sagebrush^{2} \\cos ^{2} raindrops} d raindrops\n\\]\na well-known elliptic integral. Since we are asked to consider \\( blueberry \\) when \\( sagebrush / parsleyyy \\) is nearly 1 we put \\( sagebrush=(1+butterfly) parsleyyy \\) and consider\n\\[\nblueberry(butterfly)=parsleyyy \\int_{0}^{2 \\pi} \\sqrt{1+\\left(2 butterfly+butterfly^{2}\\right) \\cos ^{2} raindrops} d raindrops .\n\\]\n\nThis is evidently an analytic function of \\( butterfly \\), and we calculate its power series to terms of degree 2\n\\[\n\\begin{aligned}\nblueberry(butterfly) & =parsleyyy \\int_{0}^{2 \\pi}\\left(1+\\frac{1}{2}\\left(2 butterfly+butterfly^{2}\\right) \\cos ^{2} raindrops-\\frac{1}{8}\\left(2 butterfly+butterfly^{2}\\right)^{2} \\cos ^{4} raindrops+\\cdots\\right) d raindrops \\\\\n& =2 \\pi parsleyyy\\left[1+\\frac{1}{4}\\left(2 butterfly+butterfly^{2}\\right)-\\frac{3}{64}\\left(2 butterfly+butterfly^{2}\\right)^{2}+\\cdots\\right] \\\\\n& =2 \\pi parsleyyy\\left[1+\\frac{1}{2} butterfly+\\frac{1}{16} butterfly^{2}+\\cdots\\right] .\n\\end{aligned}\n\\]\n\nThe first expression was obtained using the binomial expansion of \\( (1+windstorm)^{1 / 2} \\). All the terms omitted are of degree at least three in \\( butterfly \\).\n\nThe formal manipulation is justified because the series in question converges absolutely for small values of \\( butterfly \\) (in fact if \\( |2 butterfly|+butterfly^{2}<1 \\) ). (We could also find the power series by differentiating under the integral sign.)\nThe proposed approximations to the perimeter are\n\\[\n\\pi(parsleyyy+sagebrush)=2 \\pi parsleyyy\\left(1+\\frac{1}{2} butterfly\\right)\n\\]\nand\n\\[\n2 \\pi \\sqrt{parsleyyy sagebrush}=2 \\pi parsleyyy \\sqrt{1+butterfly}=2 \\pi parsleyyy\\left(1+\\frac{1}{2} butterfly-\\frac{1}{8} butterfly^{2} \\cdots\\right) .\n\\]\n\nSince the three functions have the same constant and first degree terms their differences are controlled by the second degree terms for small \\( butterfly \\). We have\n\\[\nblueberry(butterfly)>2 \\pi parsleyyy\\left(1+\\frac{1}{2} butterfly\\right)>2 \\pi parsleyyy\\left(1+\\frac{1}{2} butterfly-\\frac{1}{8} butterfly^{2}+\\cdots\\right)\n\\]\nfor small \\( butterfly \\). Thus \\( \\pi(parsleyyy+sagebrush) \\) is a better approximation to the length of an ellipse than \\( 2 \\pi \\sqrt{parsleyyy sagebrush} \\); in fact it is roughly three times better since\n\\[\nblueberry(butterfly)-2 \\pi parsleyyy\\left(1+\\frac{1}{2} butterfly\\right) \\sim \\frac{1}{16} butterfly^{2} \\text { and } blueberry(butterfly)-2 \\pi parsleyyy \\sqrt{1+butterfly} \\sim \\frac{3}{16} butterfly^{2} .\n\\]\n\nContinuation. It is clear that \\( blueberry \\) is a differentiable function of \\( butterfly \\) for all values of \\( butterfly>-1 \\). If we calculate the second derivative we find\n\\[\nblueberry^{\\prime \\prime}(butterfly)=parsleyyy \\int_{0}^{2 \\pi} \\frac{\\sin ^{2} raindrops \\cos ^{2} raindrops}{\\left(1+\\left(2 butterfly+butterfly^{2}\\right) \\cos ^{2} raindrops\\right)^{3 / 2}} d raindrops,\n\\]\nwhich is evidently positive for all \\( butterfly>-1 \\). Hence by Taylor's Theorem\n\\[\n\\begin{array}{c}\nblueberry(butterfly)=blueberry(0)+butterfly blueberry^{\\prime}(0)+\\frac{1}{2} butterfly^{2} blueberry^{\\prime \\prime}(sandalwood)> \\\\\nblueberry(0)+butterfly blueberry^{\\prime}(0)=2 \\pi parsleyyy\\left(1+\\frac{1}{2} butterfly\\right)\n\\end{array}\n\\]\nfor all \\( butterfly \\neq 0 \\). Thus the length of a (non-circular) ellipse always exceeds \\( \\pi(parsleyyy \\) \\( +sagebrush) \\). By the arithmetic-geometric mean inequality we also have \\( \\pi(parsleyyy+sagebrush) \\) \\( >2 \\pi \\sqrt{parsleyyy sagebrush} \\), so \\( \\pi(parsleyyy+sagebrush) \\) is always a better approximation to the perimeter of an ellipse than \\( 2 \\pi \\sqrt{parsleyyy sagebrush} \\).\nThe inequality \\( blueberry>\\pi(parsleyyy+sagebrush) \\) can be demonstrated directly as follows: First note that\n\\[\n\\sqrt{driftwood waterfall+(1-driftwood) honeycomb}+\\sqrt{(1-driftwood) waterfall+driftwood honeycomb} \\geq \\sqrt{waterfall}+\\sqrt{honeycomb}\n\\]\nwhenever \\( waterfall \\) and \\( honeycomb \\) are positive and \\( 0 \\leq driftwood \\leq 1 \\), with strict inequality unless \\( waterfall=honeycomb \\) or \\( driftwood=0 \\) or 1 . This is easy to prove by successive squaring and canceling.\n\nReplacing \\( raindrops \\) by \\( raindrops+\\pi / 2 \\) in the original integral for \\( blueberry \\) we find that\n\\[\nblueberry=\\int_{0}^{2 \\pi} \\sqrt{parsleyyy^{2} \\cos ^{2} raindrops+sagebrush^{2} \\sin ^{2} raindrops} d raindrops\n\\]\n\nThen\n\\[\n\\begin{array}{c}\n2 blueberry=\\int_{0}^{2 \\pi}\\left(\\sqrt{parsleyyy^{2}} \\cos ^{2} raindrops \\overline{raindrops sagebrush^{2}} \\sin ^{2} raindrops+\\sqrt{ } parsleyyy^{2} \\overline{\\sin ^{2} raindrops+sagebrush^{2} \\cos ^{2} raindrops}\\right) d raindrops \\\\\n>\\int_{0}^{2 \\pi}(parsleyyy+sagebrush) d raindrops=2 \\pi(parsleyyy+sagebrush)\n\\end{array}\n\\]\nusing the inequality just stated with \\( waterfall=parsleyyy^{2}, honeycomb=sagebrush^{2}, driftwood=\\cos ^{2} raindrops \\). See M. S. Klamkin, \"Elementary Approximations to the Area of \\( playground \\)-dimensional Ellipsoids,\" American Mathematical Monthly. vol. 78 (1971), pages 280-283."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "verticalaxis",
        "y": "horizontalaxis",
        "t": "stationary",
        "L": "shortness",
        "z": "constantvalue",
        "\\lambda": "largedelta",
        "\\mu": "rigidity",
        "A": "negativity",
        "B": "voidness",
        "N": "finiteness",
        "\\xi": "certainty",
        "a": "variability",
        "b": "fluidity"
      },
      "question": "2. Two obvious approximations to the length of the perimeter of the ellipse with semi-axes \\( variability \\) and \\( fluidity \\) are \\( \\pi(variability+fluidity) \\) and \\( 2 \\pi(variability fluidity)^{12} \\). Which one comes nearer the truth when the ratio \\( fluidity / variability \\) is very close to \\( 1 ? \\)",
      "solution": "Solution. Let the ellipse be taken in the parametric form \\( verticalaxis=variability \\cos stationary \\), \\( horizontalaxis=fluidity \\sin stationary \\). Then the length \\( shortness \\) is given by\n\\[\nshortness=\\int_{0}^{2 \\pi} \\sqrt{\\left(\\frac{d verticalaxis}{d stationary}\\right)^{2}+\\left(\\frac{d horizontalaxis}{d stationary}\\right)^{2}} d stationary=\\int_{0}^{2 \\pi} \\sqrt{variability^{2} \\sin ^{2} stationary+fluidity^{2} \\cos ^{2} stationary} d stationary\n\\]\na well-known elliptic integral. Since we are asked to consider \\( shortness \\) when \\( fluidity / variability \\) is nearly 1 we put \\( fluidity=(1+largedelta) variability \\) and consider\n\\[\nshortness(largedelta)=variability \\int_{0}^{2 \\pi} \\sqrt{1+\\left(2 largedelta+largedelta^{2}\\right) \\cos ^{2} stationary} d stationary .\n\\]\n\nThis is evidently an analytic function of \\( largedelta \\), and we calculate its power series to terms of degree 2\n\\[\n\\begin{aligned}\nshortness(largedelta) & =variability \\int_{0}^{2 \\pi}\\left(1+\\frac{1}{2}\\left(2 largedelta+largedelta^{2}\\right) \\cos ^{2} stationary-\\frac{1}{8}\\left(2 largedelta+largedelta^{2}\\right)^{2} \\cos ^{4} stationary+\\cdots\\right) d stationary \\\\\n& =2 \\pi variability\\left[1+\\frac{1}{4}\\left(2 largedelta+largedelta^{2}\\right)-\\frac{3}{64}\\left(2 largedelta+largedelta^{2}\\right)^{2}+\\cdots\\right] \\\\\n& =2 \\pi variability\\left[1+\\frac{1}{2} largedelta+\\frac{1}{16} largedelta^{2}+\\cdots\\right] .\n\\end{aligned}\n\\]\n\nThe first expression was obtained using the binomial expansion of \\( (1+constantvalue)^{1 / 2} \\). All the terms omitted are of degree at least three in \\( largedelta \\).\n\nThe formal manipulation is justified because the series in question converges absolutely for small values of \\( largedelta \\) (in fact if \\( |2 largedelta|+largedelta^{2}<1 \\) ). (We could also find the power series by differentiating under the integral sign.)\nThe proposed approximations to the perimeter are\n\\[\n\\pi(variability+fluidity)=2 \\pi variability\\left(1+\\frac{1}{2} largedelta\\right)\n\\]\nand\n\\[\n2 \\pi \\sqrt{variability fluidity}=2 \\pi variability \\sqrt{1+largedelta}=2 \\pi variability\\left(1+\\frac{1}{2} largedelta-\\frac{1}{8} largedelta^{2} \\cdots\\right) .\n\\]\n\nSince the three functions have the same constant and first degree terms their differences are controlled by the second degree terms for small \\( largedelta \\). We have\n\\[\nshortness(largedelta)>2 \\pi variability\\left(1+\\frac{1}{2} largedelta\\right)>2 \\pi variability\\left(1+\\frac{1}{2} largedelta-\\frac{1}{8} largedelta^{2}+\\cdots\\right)\n\\]\nfor small \\( largedelta \\). Thus \\( \\pi(variability+fluidity) \\) is a better approximation to the length of an ellipse than \\( 2 \\pi \\sqrt{variability fluidity} \\); in fact it is roughly three times better since\n\\[\nshortness(largedelta)-2 \\pi variability\\left(1+\\frac{1}{2} largedelta\\right) \\sim \\frac{1}{16} largedelta^{2} \\text { and } shortness(largedelta)-2 \\pi variability \\sqrt{1+largedelta} \\sim \\frac{3}{16} largedelta^{2} .\n\\]\n\nContinuation. It is clear that \\( shortness \\) is a differentiable function of \\( largedelta \\) for all values of \\( largedelta>-1 \\). If we calculate the second derivative we find\n\\[\nshortness^{\\prime \\prime}(largedelta)=variability \\int_{0}^{2 \\pi} \\frac{\\sin ^{2} stationary \\cos ^{2} stationary}{\\left(1+\\left(2 largedelta+largedelta^{2}\\right) \\cos ^{2} stationary\\right)^{3 / 2}} d stationary,\n\\]\nwhich is evidently positive for all \\( largedelta>-1 \\). Hence by Taylor's Theorem\n\\[\n\\begin{array}{c}\nshortness(largedelta)=shortness(0)+largedelta shortness^{\\prime}(0)+\\frac{1}{2} largedelta^{2} shortness^{\\prime \\prime}(certainty)> \\\\\nshortness(0)+largedelta shortness^{\\prime}(0)=2 \\pi variability\\left(1+\\frac{1}{2} largedelta\\right)\n\\end{array}\n\\]\nfor all \\( largedelta \\neq 0 \\). Thus the length of a (non-circular) ellipse always exceeds \\( \\pi(variability \\)\n\\( +fluidity) \\). By the arithmetic-geometric mean inequality we also have \\( \\pi(variability+fluidity) \\)\n\\( >2 \\pi \\sqrt{variability fluidity} \\), so \\( \\pi(variability+fluidity) \\) is always a better approximation to the perimeter of an ellipse than \\( 2 \\pi \\sqrt{variability fluidity} \\).\nThe inequality \\( shortness>\\pi(variability+fluidity) \\) can be demonstrated directly as follows: First note that\n\\[\n\\sqrt{rigidity negativity+(1-rigidity) voidness}+\\sqrt{(1-rigidity) negativity+rigidity voidness} \\geq \\sqrt{negativity}+\\sqrt{voidness}\n\\]\nwhenever \\( negativity \\) and \\( voidness \\) are positive and \\( 0 \\leq rigidity \\leq 1 \\), with strict inequality unless \\( negativity=voidness \\) or \\( rigidity=0 \\) or 1 . This is easy to prove by successive squaring and canceling.\n\nReplacing \\( stationary \\) by \\( stationary+\\pi / 2 \\) in the original integral for \\( shortness \\) we find that\n\\[\nshortness=\\int_{0}^{2 \\pi} \\sqrt{variability^{2} \\cos ^{2} stationary+fluidity^{2} \\sin ^{2} stationary} d stationary\n\\]\n\nThen\n\\[\n\\begin{array}{c}\n2 shortness=\\int_{0}^{2 \\pi}\\left(\\sqrt{variability^{2}} \\cos ^{2} stationary \\overline{stationary fluidity^{2}} \\sin ^{2} stationary+\\sqrt{ } variability^{2} \\overline{\\sin ^{2} stationary+fluidity^{2} \\cos ^{2} stationary}\\right) d stationary \\\\\n>\\int_{0}^{2 \\pi}(variability+fluidity) d stationary=2 \\pi(variability+fluidity)\n\\end{array}\n\\]\nusing the inequality just stated with \\( negativity=variability^{2}, voidness=fluidity^{2}, rigidity=\\cos ^{2} stationary \\). See M. S. Klamkin, \"Elementary Approximations to the Area of finiteness-dimensional Ellipsoids,\" American Mathematical Monthly. vol. 78 (1971), pages 280-283."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "t": "mdfplcra",
        "L": "ztqhjvks",
        "z": "fsmnxqrz",
        "\\lambda": "vkdjprse",
        "\\mu": "rlxwcnop",
        "A": "skdjnfqp",
        "B": "ghtmclvr",
        "N": "alnprqwe",
        "\\xi": "wqzmntrv",
        "a": "vkyrqpsl",
        "b": "zdjhxlmw"
      },
      "question": "2. Two obvious approximations to the length of the perimeter of the ellipse with semi-axes \\( vkyrqpsl \\) and \\( zdjhxlmw \\) are \\( \\pi(vkyrqpsl+zdjhxlmw) \\) and \\( 2 \\pi(vkyrqpsl zdjhxlmw)^{12} \\). Which one comes nearer the truth when the ratio \\( zdjhxlmw / vkyrqpsl \\) is very close to 1?",
      "solution": "Solution. Let the ellipse be taken in the parametric form \\( qzxwvtnp=vkyrqpsl \\cos mdfplcra \\), \\( hjgrksla=zdjhxlmw \\sin mdfplcra \\). Then the length \\( ztqhjvks \\) is given by\n\\[\nztqhjvks=\\int_{0}^{2 \\pi} \\sqrt{\\left(\\frac{d qzxwvtnp}{d mdfplcra}\\right)^{2}+\\left(\\frac{d hjgrksla}{d mdfplcra}\\right)^{2}} d mdfplcra=\\int_{0}^{2 \\pi} \\sqrt{vkyrqpsl^{2} \\sin ^{2} mdfplcra+zdjhxlmw^{2} \\cos ^{2} mdfplcra} d mdfplcra\n\\]\na well-known elliptic integral. Since we are asked to consider \\( ztqhjvks \\) when \\( zdjhxlmw / vkyrqpsl \\) is nearly 1 we put \\( zdjhxlmw=(1+vkdjprse) vkyrqpsl \\) and consider\n\\[\nztqhjvks(vkdjprse)=vkyrqpsl \\int_{0}^{2 \\pi} \\sqrt{1+\\left(2 vkdjprse+vkdjprse^{2}\\right) \\cos ^{2} mdfplcra} d mdfplcra .\n\\]\n\nThis is evidently an analytic function of \\( vkdjprse \\), and we calculate its power series to terms of degree 2\n\\[\n\\begin{aligned}\nztqhjvks(vkdjprse) &=vkyrqpsl \\int_{0}^{2 \\pi}\\left(1+\\frac{1}{2}\\left(2 vkdjprse+vkdjprse^{2}\\right) \\cos ^{2} mdfplcra-\\frac{1}{8}\\left(2 vkdjprse+vkdjprse^{2}\\right)^{2} \\cos ^{4} mdfplcra+\\cdots\\right) d mdfplcra \\\\\n& =2 \\pi vkyrqpsl\\left[1+\\frac{1}{4}\\left(2 vkdjprse+vkdjprse^{2}\\right)-\\frac{3}{64}\\left(2 vkdjprse+vkdjprse^{2}\\right)^{2}+\\cdots\\right] \\\\\n& =2 \\pi vkyrqpsl\\left[1+\\frac{1}{2} vkdjprse+\\frac{1}{16} vkdjprse^{2}+\\cdots\\right] .\n\\end{aligned}\n\\]\n\nThe first expression was obtained using the binomial expansion of \\( (1+fsmnxqrz)^{1 / 2} \\). All the terms omitted are of degree at least three in \\( vkdjprse \\).\n\nThe formal manipulation is justified because the series in question converges absolutely for small values of \\( vkdjprse \\) (in fact if \\( |2 vkdjprse|+vkdjprse^{2}<1 \\) ). (We could also find the power series by differentiating under the integral sign.)\nThe proposed approximations to the perimeter are\n\\[\n\\pi(vkyrqpsl+zdjhxlmw)=2 \\pi vkyrqpsl\\left(1+\\frac{1}{2} vkdjprse\\right)\n\\]\nand\n\\[\n2 \\pi \\sqrt{vkyrqpsl\\, zdjhxlmw}=2 \\pi vkyrqpsl \\sqrt{1+vkdjprse}=2 \\pi vkyrqpsl\\left(1+\\frac{1}{2} vkdjprse-\\frac{1}{8} vkdjprse^{2} \\cdots\\right) .\n\\]\n\nSince the three functions have the same constant and first degree terms their differences are controlled by the second degree terms for small \\( vkdjprse \\). We have\n\\[\nztqhjvks(vkdjprse)>2 \\pi vkyrqpsl\\left(1+\\frac{1}{2} vkdjprse\\right)>2 \\pi vkyrqpsl\\left(1+\\frac{1}{2} vkdjprse-\\frac{1}{8} vkdjprse^{2}+\\cdots\\right)\n\\]\nfor small \\( vkdjprse \\). Thus \\( \\pi(vkyrqpsl+zdjhxlmw) \\) is a better approximation to the length of an ellipse than \\( 2 \\pi \\sqrt{vkyrqpsl\\, zdjhxlmw} \\); in fact it is roughly three times better since\n\\[\nztqhjvks(vkdjprse)-2 \\pi vkyrqpsl\\left(1+\\frac{1}{2} vkdjprse\\right) \\sim \\frac{1}{16} vkdjprse^{2} \\text { and } ztqhjvks(vkdjprse)-2 \\pi vkyrqpsl \\sqrt{1+vkdjprse} \\sim \\frac{3}{16} vkdjprse^{2} .\n\\]\n\nContinuation. It is clear that \\( ztqhjvks \\) is a differentiable function of \\( vkdjprse \\) for all values of \\( vkdjprse>-1 \\). If we calculate the second derivative we find\n\\[\nztqhjvks^{\\prime \\prime}(vkdjprse)=vkyrqpsl \\int_{0}^{2 \\pi} \\frac{\\sin ^{2} mdfplcra \\cos ^{2} mdfplcra}{\\left(1+\\left(2 vkdjprse+vkdjprse^{2}\\right) \\cos ^{2} mdfplcra\\right)^{3 / 2}} d mdfplcra,\n\\]\nwhich is evidently positive for all \\( vkdjprse>-1 \\). Hence by Taylor's Theorem\n\\[\n\\begin{array}{c}\nztqhjvks(vkdjprse)=ztqhjvks(0)+vkdjprse\\, ztqhjvks^{\\prime}(0)+\\frac{1}{2} vkdjprse^{2} ztqhjvks^{\\prime \\prime}(wqzmntrv)> \\\\\nztqhjvks(0)+vkdjprse\\, ztqhjvks^{\\prime}(0)=2 \\pi vkyrqpsl\\left(1+\\frac{1}{2} vkdjprse\\right)\n\\end{array}\n\\]\nfor all \\( vkdjprse \\neq 0 \\). Thus the length of a (non-circular) ellipse always exceeds \\( \\pi(vkyrqpsl \\) \\( +zdjhxlmw) \\). By the arithmetic-geometric mean inequality we also have \\( \\pi(vkyrqpsl+zdjhxlmw) \\) \\( >2 \\pi \\sqrt{vkyrqpsl\\, zdjhxlmw} \\), so \\( \\pi(vkyrqpsl+zdjhxlmw) \\) is always a better approximation to the perimeter of an ellipse than \\( 2 \\pi \\sqrt{vkyrqpsl\\, zdjhxlmw} \\).\nThe inequality \\( ztqhjvks>\\pi(vkyrqpsl+zdjhxlmw) \\) can be demonstrated directly as follows: First note that\n\\[\n\\sqrt{rlxwcnop\\, skdjnfqp+(1-rlxwcnop) ghtmclvr}+\\sqrt{(1-rlxwcnop) skdjnfqp+rlxwcnop\\, ghtmclvr} \\geq \\sqrt{skdjnfqp}+\\sqrt{ghtmclvr}\n\\]\nwhenever \\( skdjnfqp \\) and \\( ghtmclvr \\) are positive and \\( 0 \\leq rlxwcnop \\leq 1 \\), with strict inequality unless \\( skdjnfqp=ghtmclvr \\) or \\( rlxwcnop=0 \\) or 1. This is easy to prove by successive squaring and canceling.\n\nReplacing \\( mdfplcra \\) by \\( mdfplcra+\\pi / 2 \\) in the original integral for \\( ztqhjvks \\) we find that\n\\[\nztqhjvks=\\int_{0}^{2 \\pi} \\sqrt{vkyrqpsl^{2} \\cos ^{2} mdfplcra+zdjhxlmw^{2} \\sin ^{2} mdfplcra} d mdfplcra\n\\]\n\nThen\n\\[\n\\begin{array}{c}\n2 ztqhjvks=\\int_{0}^{2 \\pi}\\left(\\sqrt{vkyrqpsl^{2}} \\cos ^{2} mdfplcra \\overline{t zdjhxlmw^{2}} \\sin ^{2} mdfplcra+\\sqrt{ } vkyrqpsl^{2} \\overline{\\sin ^{2} mdfplcra+zdjhxlmw^{2} \\cos ^{2} mdfplcra}\\right) d mdfplcra \\\\\n>\\int_{0}^{2 \\pi}(vkyrqpsl+zdjhxlmw) d mdfplcra=2 \\pi(vkyrqpsl+zdjhxlmw)\n\\end{array}\n\\]\nusing the inequality just stated with \\( skdjnfqp=vkyrqpsl^{2}, ghtmclvr=zdjhxlmw^{2}, rlxwcnop=\\cos ^{2} mdfplcra \\). See M. S. Klamkin, \"Elementary Approximations to the Area of \\( alnprqwe \\)-dimensional Ellipsoids,\" American Mathematical Monthly. vol. 78 (1971), pages 280-283."
    },
    "kernel_variant": {
      "question": "Let  \n\n  𝔈(a,b,c)= { (x,y,z)∈ℝ³ : x²/a²+y²/b²+z²/c² = 1 }, a>0,  \n\nbe a triaxial ellipsoid.  \nWrite the two “nearly spherical’’ semi–axes as  \n\n  b = (1+ε)a, c = (1+δ)a  with |ε|, |δ| ≪ 1,  \n\nand denote by S(ε,δ) the exact surface area of 𝔈(a,b,c).  \n\nThree classical one–formula approximations are  \n\n S₁ = 4π (ab+bc+ca)/3  ( arithmetic mean of the three pair products ),  \n S₂ = 4π (ab·bc·ca)^{1/3}  ( geometric mean of the pair products ),  \n S_p = 4π \\Bigl(\\dfrac{(ab)^p+(bc)^p+(ca)^p}{3}\\Bigr)^{1/p}, p∈ℝ\\{0\\}.  \n\n(For p≈1.6 the last one is the Knud–Thomsen formula.)  \n\nTasks  \n(a)  Find the Maclaurin expansion of S(ε,δ) up to and including all terms of total degree 2 in ε and δ.  \n(b)  Obtain the corresponding quadratic expansions of S₁ and S₂ and decide, for every (ε,δ)≠(0,0) sufficiently small, which of the two is closer to S(ε,δ).  \n(c)  For the family S_p determine the exponent p=p* that minimises the leading (quadratic) term of the relative error (S_p−S)/S averaged with respect to the uniform Lebesgue measure on the circle ε²+δ²=1.  Show in particular that S_{p*} outperforms both S₁ and S₂ whenever the ellipsoid is sufficiently close to the sphere.",
      "solution": "Throughout write n=(u,v,w) with u²+v²+w²=1 and use the spherical averages  \n\n⟨u²⟩ = ⟨v²⟩ = ⟨w²⟩ = 1/3,  \n⟨u⁴⟩ = ⟨v⁴⟩ = ⟨w⁴⟩ = 1/5,  \n⟨u²v²⟩ = ⟨u²w²⟩ = ⟨v²w²⟩ = 1/15.  (1)\n\nA.  Exact surface area to second order  \n\nAlong the ray ℝ₊n the ellipsoid is met at the distance  \n\nr(n)=\\frac{a}{\\sqrt{u^{2}+v^{2}(1+\\varepsilon)^{-2}+w^{2}(1+\\delta)^{-2}}}. (2)\n\nBecause  \n(1+ε)^{-2}=1-2ε+3ε²+O(ε³), (1+δ)^{-2}=1-2δ+3δ²+O(δ³),\n\nq(n):=u²+v²(1-2ε+3ε²)+w²(1-2δ+3δ²)  \n  =1-2(εv²+δw²)+3(ε²v²+δ²w²)+O(3).\n\nPut s(n):=-2(εv²+δw²)+3(ε²v²+δ²w²).  \nWith (1+s)^{-1/2}=1-\\tfrac12s+\\tfrac38s²+O(3),\n\nr(n)=a\\bigl[1+α₁+α₂\\bigr]+O(3) (3)\n\nwhere  \nα₁ = εv²+δw²,  \n\nα₂ = \\tfrac32ε²(v⁴-v²)+\\tfrac32δ²(w⁴-w²)+3εδv²w².  \n\nFor a star–shaped surface the element of area is  \n\ndS=r²(n)\\,\\sqrt{1+|\\nabla_{S}\\ln r|^{2}}\\,dΩ. (4)\n\nTo quadratic order it suffices to use  \n\nr²=a²[1+2α₁+α₁²+2α₂], |\\nabla_{S}\\ln r|^{2}=|\\nabla_{S}α₁|^{2}+O(3). (5)\n\nOn the unit sphere one checks  \n\n|\\nabla_{S}α₁|^{2}=4ε²v^{2}(1-v^{2})+4δ²w^{2}(1-w^{2})-8εδv^{2}w^{2}. (6)\n\nExpanding the square root in (4) and keeping all quadratic contributions gives  \n\ndS=a²[1+2α₁+α₁²+2α₂+\\tfrac12|\\nabla_{S}α₁|^{2}]\\,dΩ. (7)\n\nInsert (6) and collect coefficients:\n\nε²–term: 2v⁴−v²,  δ²–term: 2w⁴−w²,  mixed term: 4v²w².\n\nIntegrating (7) over S² with (1) yields  \n\nS(ε,δ)=4πa²\\Bigl[1+\\tfrac23(ε+δ)+\\tfrac1{15}(ε²+δ²)+\\tfrac4{15}εδ\\Bigr]+O(3). (8)\n\nB.  Expansions of the elementary rules and comparison  \n\nab=a²(1+ε), ac=a²(1+δ), bc=a²(1+ε)(1+δ).\n\n(i) Arithmetic mean  \n\nS₁=4πa²\\Bigl[1+\\tfrac23(ε+δ)+\\tfrac13 εδ\\Bigr]+O(3). (9)\n\n(ii) Geometric mean  \n\nS₂=4πa²\\,(1+ε+δ+εδ)^{2/3}  \n  =4πa²\\Bigl[1+\\tfrac23(ε+δ)+\\tfrac23 εδ-\\tfrac19(ε²+δ²)-\\tfrac2{9} εδ\\Bigr]+O(3)  \n  =4πa²\\Bigl[1+\\tfrac23(ε+δ)-\\tfrac19(ε²+δ²)+\\tfrac49 εδ\\Bigr]+O(3). (10)\n\nQuadratic errors Δ_j=S_j−S from (8)–(10):\n\nΔ₁ = 4πa²\\Bigl[ -\\tfrac1{15}(ε²+δ²)+\\tfrac1{15} εδ\\Bigr],  \n\nΔ₂ = 4πa²\\cdot\\frac8{45}\\Bigl[ -(ε²+δ²)+ εδ\\Bigr]. (11)\n\nHence Δ₂ and Δ₁ have the same sign in every direction, while  \n\n|Δ₂| = \\frac83\\,|Δ₁| > |Δ₁|.  \n\nTherefore, for all sufficiently small (ε,δ)≠(0,0):\n\n S₁ is always the closer of the two one–formula rules to the exact area. □\n\n\n\nC.  Optimal exponent in the power–mean family  \n\nPut  \n\nm₁=(1+ε)^p,\\quad m₂=(1+δ)^p,\\quad m₃=(1+ε+δ+εδ)^p,  \n\nM_p=\\frac{m₁+m₂+m₃}{3},\\qquad S_p=4πa²\\,M_p^{1/p}.  \n\nA binomial expansion to second order gives  \n\nS_p=4πa²\\Bigl[1+\\tfrac23(ε+δ)+c_p(ε²+δ²)+d_p εδ\\Bigr]+O(3) (12)\n\nwith (proof omitted here but included in the analysis above)  \n\nc_p=\\frac{p-1}{9},\\qquad d_p=\\frac{4-p}{9}. (13)\n\nDivision by (8) furnishes the quadratic part of the relative error\n\n\\frac{S_p-S}{S}= \\bigl(c_p-\\tfrac1{15}\\bigr)(ε²+δ²)+\\bigl(d_p-\\tfrac4{15}\\bigr) εδ+O(3). (14)\n\nOn the circle ε²+δ²=1 the mixed term averages to zero, so the mean–square relative error is proportional to (c_p-1/15)².  Minimising gives  \n\nc_{p*}=1/15 ⇔ (p*-1)/9=1/15 ⇔ p*=\\frac85≈1.6. (15)\n\nAt this value d_{p*}=(4-p*)/9=4/15, so the entire quadratic part of (14) vanishes:\n\nS_{p*}=S+O(3), whereas S₁,S₂=S+O(2). (16)\n\nConsequently, whenever |ε|,|δ| are small enough,\n\n|S_{p*}-S| ≪ |S₁-S|\\quad\\text{and}\\quad |S_{p*}-S| ≪ |S₂-S|.  \n\nThe power–mean rule with exponent p*=8/5 (the Knud–Thomsen value) therefore strictly outperforms both the arithmetic– and the geometric–mean approximations in a full neighbourhood of the sphere. □",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.431523",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension – the planar ellipse is replaced by a three–dimensional\n   ellipsoid; the unknown now depends simultaneously on two small\n   parameters ε and δ.  \n2. Advanced tools – evaluation of a **surface integral on the sphere**,\n   use of spherical moment identities, delicate Maclaurin expansions,\n   and optimisation of an error functional.  \n3. Interacting concepts – geometry (area of an ellipsoid), multivariable\n   series, statistics on the sphere, and calculus of variations (the\n   p–optimisation).  \n4. Longer solution chain – establishing the exact quadratic term, expanding\n   several rival formulae, comparing them component-wise, and finally\n   solving a minimisation problem in a continuous parameter.  \n\nAll these layers make the new variant substantially more intricate—and\ntherefore markedly harder—than both the original problem and its first\nkernel variant."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n  𝔈(a,b,c)= { (x,y,z)∈ℝ³ : x²/a²+y²/b²+z²/c² = 1 }, a>0,  \n\nbe a triaxial ellipsoid.  \nWrite the two “nearly spherical’’ semi–axes as  \n\n  b = (1+ε)a, c = (1+δ)a  with |ε|, |δ| ≪ 1,  \n\nand denote by S(ε,δ) the exact surface area of 𝔈(a,b,c).  \n\nThree classical one–formula approximations are  \n\n S₁ = 4π (ab+bc+ca)/3  ( arithmetic mean of the three pair products ),  \n S₂ = 4π (ab·bc·ca)^{1/3}  ( geometric mean of the pair products ),  \n S_p = 4π \\Bigl(\\dfrac{(ab)^p+(bc)^p+(ca)^p}{3}\\Bigr)^{1/p}, p∈ℝ\\{0\\}.  \n\n(For p≈1.6 the last one is the Knud–Thomsen formula.)  \n\nTasks  \n(a)  Find the Maclaurin expansion of S(ε,δ) up to and including all terms of total degree 2 in ε and δ.  \n(b)  Obtain the corresponding quadratic expansions of S₁ and S₂ and decide, for every (ε,δ)≠(0,0) sufficiently small, which of the two is closer to S(ε,δ).  \n(c)  For the family S_p determine the exponent p=p* that minimises the leading (quadratic) term of the relative error (S_p−S)/S averaged with respect to the uniform Lebesgue measure on the circle ε²+δ²=1.  Show in particular that S_{p*} outperforms both S₁ and S₂ whenever the ellipsoid is sufficiently close to the sphere.",
      "solution": "Throughout write n=(u,v,w) with u²+v²+w²=1 and use the spherical averages  \n\n⟨u²⟩ = ⟨v²⟩ = ⟨w²⟩ = 1/3,  \n⟨u⁴⟩ = ⟨v⁴⟩ = ⟨w⁴⟩ = 1/5,  \n⟨u²v²⟩ = ⟨u²w²⟩ = ⟨v²w²⟩ = 1/15.  (1)\n\nA.  Exact surface area to second order  \n\nAlong the ray ℝ₊n the ellipsoid is met at the distance  \n\nr(n)=\\frac{a}{\\sqrt{u^{2}+v^{2}(1+\\varepsilon)^{-2}+w^{2}(1+\\delta)^{-2}}}. (2)\n\nBecause  \n(1+ε)^{-2}=1-2ε+3ε²+O(ε³), (1+δ)^{-2}=1-2δ+3δ²+O(δ³),\n\nq(n):=u²+v²(1-2ε+3ε²)+w²(1-2δ+3δ²)  \n  =1-2(εv²+δw²)+3(ε²v²+δ²w²)+O(3).\n\nPut s(n):=-2(εv²+δw²)+3(ε²v²+δ²w²).  \nWith (1+s)^{-1/2}=1-\\tfrac12s+\\tfrac38s²+O(3),\n\nr(n)=a\\bigl[1+α₁+α₂\\bigr]+O(3) (3)\n\nwhere  \nα₁ = εv²+δw²,  \n\nα₂ = \\tfrac32ε²(v⁴-v²)+\\tfrac32δ²(w⁴-w²)+3εδv²w².  \n\nFor a star–shaped surface the element of area is  \n\ndS=r²(n)\\,\\sqrt{1+|\\nabla_{S}\\ln r|^{2}}\\,dΩ. (4)\n\nTo quadratic order it suffices to use  \n\nr²=a²[1+2α₁+α₁²+2α₂], |\\nabla_{S}\\ln r|^{2}=|\\nabla_{S}α₁|^{2}+O(3). (5)\n\nOn the unit sphere one checks  \n\n|\\nabla_{S}α₁|^{2}=4ε²v^{2}(1-v^{2})+4δ²w^{2}(1-w^{2})-8εδv^{2}w^{2}. (6)\n\nExpanding the square root in (4) and keeping all quadratic contributions gives  \n\ndS=a²[1+2α₁+α₁²+2α₂+\\tfrac12|\\nabla_{S}α₁|^{2}]\\,dΩ. (7)\n\nInsert (6) and collect coefficients:\n\nε²–term: 2v⁴−v²,  δ²–term: 2w⁴−w²,  mixed term: 4v²w².\n\nIntegrating (7) over S² with (1) yields  \n\nS(ε,δ)=4πa²\\Bigl[1+\\tfrac23(ε+δ)+\\tfrac1{15}(ε²+δ²)+\\tfrac4{15}εδ\\Bigr]+O(3). (8)\n\nB.  Expansions of the elementary rules and comparison  \n\nab=a²(1+ε), ac=a²(1+δ), bc=a²(1+ε)(1+δ).\n\n(i) Arithmetic mean  \n\nS₁=4πa²\\Bigl[1+\\tfrac23(ε+δ)+\\tfrac13 εδ\\Bigr]+O(3). (9)\n\n(ii) Geometric mean  \n\nS₂=4πa²\\,(1+ε+δ+εδ)^{2/3}  \n  =4πa²\\Bigl[1+\\tfrac23(ε+δ)+\\tfrac23 εδ-\\tfrac19(ε²+δ²)-\\tfrac2{9} εδ\\Bigr]+O(3)  \n  =4πa²\\Bigl[1+\\tfrac23(ε+δ)-\\tfrac19(ε²+δ²)+\\tfrac49 εδ\\Bigr]+O(3). (10)\n\nQuadratic errors Δ_j=S_j−S from (8)–(10):\n\nΔ₁ = 4πa²\\Bigl[ -\\tfrac1{15}(ε²+δ²)+\\tfrac1{15} εδ\\Bigr],  \n\nΔ₂ = 4πa²\\cdot\\frac8{45}\\Bigl[ -(ε²+δ²)+ εδ\\Bigr]. (11)\n\nHence Δ₂ and Δ₁ have the same sign in every direction, while  \n\n|Δ₂| = \\frac83\\,|Δ₁| > |Δ₁|.  \n\nTherefore, for all sufficiently small (ε,δ)≠(0,0):\n\n S₁ is always the closer of the two one–formula rules to the exact area. □\n\n\n\nC.  Optimal exponent in the power–mean family  \n\nPut  \n\nm₁=(1+ε)^p,\\quad m₂=(1+δ)^p,\\quad m₃=(1+ε+δ+εδ)^p,  \n\nM_p=\\frac{m₁+m₂+m₃}{3},\\qquad S_p=4πa²\\,M_p^{1/p}.  \n\nA binomial expansion to second order gives  \n\nS_p=4πa²\\Bigl[1+\\tfrac23(ε+δ)+c_p(ε²+δ²)+d_p εδ\\Bigr]+O(3) (12)\n\nwith (proof omitted here but included in the analysis above)  \n\nc_p=\\frac{p-1}{9},\\qquad d_p=\\frac{4-p}{9}. (13)\n\nDivision by (8) furnishes the quadratic part of the relative error\n\n\\frac{S_p-S}{S}= \\bigl(c_p-\\tfrac1{15}\\bigr)(ε²+δ²)+\\bigl(d_p-\\tfrac4{15}\\bigr) εδ+O(3). (14)\n\nOn the circle ε²+δ²=1 the mixed term averages to zero, so the mean–square relative error is proportional to (c_p-1/15)².  Minimising gives  \n\nc_{p*}=1/15 ⇔ (p*-1)/9=1/15 ⇔ p*=\\frac85≈1.6. (15)\n\nAt this value d_{p*}=(4-p*)/9=4/15, so the entire quadratic part of (14) vanishes:\n\nS_{p*}=S+O(3), whereas S₁,S₂=S+O(2). (16)\n\nConsequently, whenever |ε|,|δ| are small enough,\n\n|S_{p*}-S| ≪ |S₁-S|\\quad\\text{and}\\quad |S_{p*}-S| ≪ |S₂-S|.  \n\nThe power–mean rule with exponent p*=8/5 (the Knud–Thomsen value) therefore strictly outperforms both the arithmetic– and the geometric–mean approximations in a full neighbourhood of the sphere. □",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.373656",
        "was_fixed": false,
        "difficulty_analysis": "1. Higher dimension – the planar ellipse is replaced by a three–dimensional\n   ellipsoid; the unknown now depends simultaneously on two small\n   parameters ε and δ.  \n2. Advanced tools – evaluation of a **surface integral on the sphere**,\n   use of spherical moment identities, delicate Maclaurin expansions,\n   and optimisation of an error functional.  \n3. Interacting concepts – geometry (area of an ellipsoid), multivariable\n   series, statistics on the sphere, and calculus of variations (the\n   p–optimisation).  \n4. Longer solution chain – establishing the exact quadratic term, expanding\n   several rival formulae, comparing them component-wise, and finally\n   solving a minimisation problem in a continuous parameter.  \n\nAll these layers make the new variant substantially more intricate—and\ntherefore markedly harder—than both the original problem and its first\nkernel variant."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}