{
  "index": "1950-A-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "GEO"
  ],
  "difficulty": "",
  "question": "4. Answer either (i) or (ii).\n(i) In a right prism with triangular base, given the sum of the areas of three mutually adjacent faces (that is, of two lateral faces and one base), show that these faces are of equal area and perpendicular to each other when the volume attains its maximum.\n(page 292)\n(ii) Show that\n\\[\n\\frac{\\frac{x}{1}+\\frac{x^{3}}{1 \\cdot 3}+\\frac{x^{5}}{1 \\cdot 3 \\cdot 5}+\\frac{x^{7}}{1 \\cdot 3 \\cdot 5 \\cdot 7}+\\cdots}{1+\\frac{x^{2}}{2}+\\frac{x^{4}}{2 \\cdot 4}+\\frac{x^{6}}{2 \\cdot 4 \\cdot 6}+\\cdots}=\\int_{0}^{x} e^{-t^{2 / 2}} d t .\n\\]",
  "solution": "Solution. Let the base triangles have sides \\( a \\) and \\( b \\) with included angle \\( \\theta \\), and let the right prism have altitude \\( c \\). If \\( L \\) denotes the given sum of the three face areas, then \\( L=a c+b c+\\frac{1}{2} a b \\sin \\theta \\), and the volume is \\( V= \\) \\( \\frac{1}{2} a b c \\sin \\theta \\). Let \\( X=a c, Y=b c, Z=\\frac{1}{2} a b \\sin \\theta \\) be the areas of the three faces. Then \\( V^{2}=\\frac{1}{2} X Y Z \\sin \\theta \\) where \\( X, Y, Z \\) are three positive numbers whose sum is \\( L \\). Now the arithmetic mean-geometric mean inequality for positive numbers yields \\( (X Y Z)^{1 / 3} \\leq(X+Y+Z) / 3 \\), and hence\n\\[\nV^{2} \\leq \\frac{1}{2}\\left(\\frac{L}{3}\\right)^{3} \\sin \\theta, \\quad \\text { and } \\quad V \\leq \\frac{1}{\\sqrt{2}}\\left(\\frac{L}{3}\\right)^{3 / 2}\n\\]\nwith equality occurring only if \\( X=Y=Z=\\frac{1}{3} L \\) and \\( \\sin \\theta=1 \\). But if all these conditions are satisfied, the three faces all have equal area and are mutually perpendicular.\n\nIt remains to be shown that these conditions can be satisfied by a right prism. For this we need only take \\( a=b=2 c=\\sqrt{2 L / 3} \\) and \\( \\theta=\\pi / 2 \\).\n\nRemark. The Lagrange multiplier method leads quickly to the critical point, but the proof that it is a maximum seems to require an argument similar to the above.\n\nSolution. The series appearing in the numerator converges for all \\( \\boldsymbol{x} \\) by the ratio test. Let its sum be \\( f(x) \\). Since the denominator is immediately seen to be\n\\[\n\\sum_{n=0}^{\\infty} \\frac{1}{n!}\\left(\\frac{x^{2}}{2}\\right)^{n}=e^{x^{2} / 2},\n\\]\nthe problem is equivalent to proving that\n\\[\nf(x)=e^{x^{2} / 2} \\int_{0}^{x} e^{-t^{2} / 2} d t .\n\\]\n\nDifferentiating the series for \\( f \\) term by term, we find\n\\[\nf^{\\prime}(x)=1+x f(x)\n\\]\n\nThis first-order linear differential equation, together with the initial condition \\( f(0)=0 \\) determines the function \\( f \\), so we can establish (1) either by differentiating the right member of (1) and noting that it satisfies (2) and the initial condition, or by solving (2) by the usual method. (The corresponding homogeneous equation is \\( g^{\\prime}(x)-x g(x)=0 \\), with general solution \\( g(x)=c e^{x^{2} / 2} \\). Then \\( e^{-x^{2} / 2} \\) is an integrating factor for (2):\n\\[\n\\frac{d}{d x}\\left(f(x) e^{-x^{2} / 2}\\right)=e^{-x^{2} / 2}\n\\]\nand (1) follows immediately.)\nRemark. This is essentially Problem 4623, American Mathematical Monthly, vol. 63 (1956), page 260.",
  "vars": [
    "a",
    "b",
    "c",
    "L",
    "V",
    "X",
    "Y",
    "Z",
    "x",
    "t",
    "n",
    "g",
    "f",
    "\\\\theta"
  ],
  "params": [],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "a": "sidealpha",
        "b": "sidebeta",
        "c": "altitude",
        "L": "sumfaces",
        "V": "volumeprism",
        "X": "faceone",
        "Y": "facetwo",
        "Z": "facethree",
        "x": "inputreal",
        "t": "integvar",
        "n": "indexcount",
        "g": "genfunct",
        "f": "seriesfunc",
        "\\theta": "angletheta"
      },
      "question": "4. Answer either (i) or (ii).\n(i) In a right prism with triangular base, given the sum of the areas of three mutually adjacent faces (that is, of two lateral faces and one base), show that these faces are of equal area and perpendicular to each other when the volume attains its maximum.\n(page 292)\n(ii) Show that\n\\[\n\\frac{\\frac{inputreal}{1}+\\frac{inputreal^{3}}{1 \\cdot 3}+\\frac{inputreal^{5}}{1 \\cdot 3 \\cdot 5}+\\frac{inputreal^{7}}{1 \\cdot 3 \\cdot 5 \\cdot 7}+\\cdots}{1+\\frac{inputreal^{2}}{2}+\\frac{inputreal^{4}}{2 \\cdot 4}+\\frac{inputreal^{6}}{2 \\cdot 4 \\cdot 6}+\\cdots}=\\int_{0}^{inputreal} e^{-\\integvar^{2 / 2}} d \\integvar .\n\\]",
      "solution": "Solution. Let the base triangles have sides \\( sidealpha \\) and \\( sidebeta \\) with included angle \\( angletheta \\), and let the right prism have altitude \\( altitude \\). If \\( sumfaces \\) denotes the given sum of the three face areas, then \\( sumfaces=sidealpha altitude+sidebeta altitude+\\frac{1}{2} sidealpha sidebeta \\sin angletheta \\), and the volume is \\( volumeprism=\\frac{1}{2} sidealpha sidebeta altitude \\sin angletheta \\). Let \\( faceone=sidealpha altitude,\\; facetwo=sidebeta altitude,\\; facethree=\\frac{1}{2} sidealpha sidebeta \\sin angletheta \\) be the areas of the three faces. Then \\( volumeprism^{2}=\\frac{1}{2} faceone facetwo facethree \\sin angletheta \\) where \\( faceone, facetwo, facethree \\) are three positive numbers whose sum is \\( sumfaces \\). Now the arithmetic mean–geometric mean inequality for positive numbers yields \\( (faceone facetwo facethree)^{1 / 3}\\leq(faceone+facetwo+facethree)/3 \\), and hence\n\\[\nvolumeprism^{2}\\leq\\frac{1}{2}\\left(\\frac{sumfaces}{3}\\right)^{3}\\sin angletheta,\\quad \\text { and }\\quad volumeprism\\leq\\frac{1}{\\sqrt{2}}\\left(\\frac{sumfaces}{3}\\right)^{3 / 2}\n\\]\nwith equality occurring only if \\( faceone=facetwo=facethree=\\frac{1}{3} sumfaces \\) and \\( \\sin angletheta=1 \\). But if all these conditions are satisfied, the three faces all have equal area and are mutually perpendicular.\n\nIt remains to be shown that these conditions can be satisfied by a right prism. For this we need only take \\( sidealpha=sidebeta=2 altitude=\\sqrt{2 sumfaces / 3} \\) and \\( angletheta=\\pi / 2 \\).\n\nRemark. The Lagrange multiplier method leads quickly to the critical point, but the proof that it is a maximum seems to require an argument similar to the above.\n\nSolution. The series appearing in the numerator converges for all \\( \\mathbf{inputreal} \\) by the ratio test. Let its sum be \\( seriesfunc(inputreal) \\). Since the denominator is immediately seen to be\n\\[\n\\sum_{indexcount=0}^{\\infty} \\frac{1}{indexcount!}\\left(\\frac{inputreal^{2}}{2}\\right)^{indexcount}=e^{inputreal^{2} / 2},\n\\]\nthe problem is equivalent to proving that\n\\[\nseriesfunc(inputreal)=e^{inputreal^{2} / 2} \\int_{0}^{inputreal} e^{-\\integvar^{2} / 2} d\\integvar .\n\\]\n\nDifferentiating the series for \\( seriesfunc \\) term by term, we find\n\\[\nseriesfunc^{\\prime}(inputreal)=1+inputreal\\,seriesfunc(inputreal)\n\\]\n\nThis first-order linear differential equation, together with the initial condition \\( seriesfunc(0)=0 \\), determines the function \\( seriesfunc \\), so we can establish (1) either by differentiating the right member of (1) and noting that it satisfies (2) and the initial condition, or by solving (2) by the usual method. (The corresponding homogeneous equation is \\( genfunct^{\\prime}(inputreal)-inputreal\\,genfunct(inputreal)=0 \\), with general solution \\( genfunct(inputreal)=c e^{inputreal^{2} / 2} \\). Then \\( e^{-inputreal^{2} / 2} \\) is an integrating factor for (2):\n\\[\n\\frac{d}{d inputreal}\\left(seriesfunc(inputreal) e^{-inputreal^{2} / 2}\\right)=e^{-inputreal^{2} / 2}\n\\]\nand (1) follows immediately.)\nRemark. This is essentially Problem 4623, American Mathematical Monthly, vol. 63 (1956), page 260."
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "strawberry",
        "b": "pineapple",
        "c": "watermelon",
        "L": "buttercup",
        "V": "honeycomb",
        "X": "marigold",
        "Y": "lilyflower",
        "Z": "daffodil",
        "x": "cantaloupe",
        "t": "elderberry",
        "n": "nightshade",
        "g": "dandelion",
        "f": "hibiscus",
        "\\\\theta": "lavender"
      },
      "question": "4. Answer either (i) or (ii).\n(i) In a right prism with triangular base, given the sum of the areas of three mutually adjacent faces (that is, of two lateral faces and one base), show that these faces are of equal area and perpendicular to each other when the volume attains its maximum.\n(page 292)\n(ii) Show that\n\\[\n\\frac{\\frac{cantaloupe}{1}+\\frac{cantaloupe^{3}}{1 \\cdot 3}+\\frac{cantaloupe^{5}}{1 \\cdot 3 \\cdot 5}+\\frac{cantaloupe^{7}}{1 \\cdot 3 \\cdot 5 \\cdot 7}+\\cdots}{1+\\frac{cantaloupe^{2}}{2}+\\frac{cantaloupe^{4}}{2 \\cdot 4}+\\frac{cantaloupe^{6}}{2 \\cdot 4 \\cdot 6}+\\cdots}=\\int_{0}^{cantaloupe} e^{-\\elderberry^{2 / 2}} d \\elderberry .\n\\]",
      "solution": "Solution. Let the base triangles have sides \\( strawberry \\) and \\( pineapple \\) with included angle \\( lavender \\), and let the right prism have altitude \\( watermelon \\). If \\( buttercup \\) denotes the given sum of the three face areas, then \\( buttercup=strawberry watermelon+pineapple watermelon+\\frac{1}{2} strawberry pineapple \\sin lavender \\), and the volume is \\( honeycomb= \\) \\( \\frac{1}{2} strawberry pineapple watermelon \\sin lavender \\). Let \\( marigold=strawberry watermelon, lilyflower=pineapple watermelon, daffodil=\\frac{1}{2} strawberry pineapple \\sin lavender \\) be the areas of the three faces. Then \\( honeycomb^{2}=\\frac{1}{2} marigold lilyflower daffodil \\sin lavender \\) where \\( marigold, lilyflower, daffodil \\) are three positive numbers whose sum is \\( buttercup \\). Now the arithmetic mean-geometric mean inequality for positive numbers yields \\( (marigold lilyflower daffodil)^{1 / 3} \\leq(marigold+lilyflower+daffodil) / 3 \\), and hence\n\\[\nhoneycomb^{2} \\leq \\frac{1}{2}\\left(\\frac{buttercup}{3}\\right)^{3} \\sin lavender, \\quad \\text { and } \\quad honeycomb \\leq \\frac{1}{\\sqrt{2}}\\left(\\frac{buttercup}{3}\\right)^{3 / 2}\n\\]\nwith equality occurring only if \\( marigold=lilyflower=daffodil=\\frac{1}{3} buttercup \\) and \\( \\sin lavender=1 \\). But if all these conditions are satisfied, the three faces all have equal area and are mutually perpendicular.\n\nIt remains to be shown that these conditions can be satisfied by a right prism. For this we need only take \\( strawberry=pineapple=2 watermelon=\\sqrt{2 buttercup / 3} \\) and \\( lavender=\\pi / 2 \\).\n\nRemark. The Lagrange multiplier method leads quickly to the critical point, but the proof that it is a maximum seems to require an argument similar to the above.\n\nSolution. The series appearing in the numerator converges for all \\( \\boldsymbol{cantaloupe} \\) by the ratio test. Let its sum be \\( hibiscus(cantaloupe) \\). Since the denominator is immediately seen to be\n\\[\n\\sum_{nightshade=0}^{\\infty} \\frac{1}{nightshade!}\\left(\\frac{cantaloupe^{2}}{2}\\right)^{nightshade}=e^{cantaloupe^{2} / 2},\n\\]\nthe problem is equivalent to proving that\n\\[\nhibiscus(cantaloupe)=e^{cantaloupe^{2} / 2} \\int_{0}^{cantaloupe} e^{-\\elderberry^{2} / 2} d \\elderberry .\n\\]\n\nDifferentiating the series for \\( hibiscus \\) term by term, we find\n\\[\nhibiscus^{\\prime}(cantaloupe)=1+cantaloupe\\, hibiscus(cantaloupe)\n\\]\n\nThis first-order linear differential equation, together with the initial condition \\( hibiscus(0)=0 \\) determines the function \\( hibiscus \\), so we can establish (1) either by differentiating the right member of (1) and noting that it satisfies (2) and the initial condition, or by solving (2) by the usual method. (The corresponding homogeneous equation is \\( dandelion^{\\prime}(cantaloupe)-cantaloupe\\, dandelion(cantaloupe)=0 \\), with general solution \\( dandelion(cantaloupe)=c e^{cantaloupe^{2} / 2} \\). Then \\( e^{-cantaloupe^{2} / 2} \\) is an integrating factor for (2):\n\\[\n\\frac{d}{d cantaloupe}\\left(hibiscus(cantaloupe) e^{-cantaloupe^{2} / 2}\\right)=e^{-cantaloupe^{2} / 2}\n\\]\nand (1) follows immediately.)\nRemark. This is essentially Problem 4623, American Mathematical Monthly, vol. 63 (1956), page 260."
    },
    "descriptive_long_misleading": {
      "map": {
        "a": "interiorangle",
        "b": "vertexangle",
        "c": "baselength",
        "L": "facedifference",
        "V": "surfacearea",
        "X": "perimeter",
        "Y": "circumference",
        "Z": "borderlength",
        "x": "constantvalue",
        "t": "stillpoint",
        "n": "continuum",
        "g": "constant",
        "f": "scalarvalue",
        "\\theta": "straightline"
      },
      "question": "4. Answer either (i) or (ii).\n(i) In a right prism with triangular base, given the sum of the areas of three mutually adjacent faces (that is, of two lateral faces and one base), show that these faces are of equal area and perpendicular to each other when the volume attains its maximum.\n(page 292)\n(ii) Show that\n\\[\n\\frac{\\frac{constantvalue}{1}+\\frac{constantvalue^{3}}{1 \\cdot 3}+\\frac{constantvalue^{5}}{1 \\cdot 3 \\cdot 5}+\\frac{constantvalue^{7}}{1 \\cdot 3 \\cdot 5 \\cdot 7}+\\cdots}{1+\\frac{constantvalue^{2}}{2}+\\frac{constantvalue^{4}}{2 \\cdot 4}+\\frac{constantvalue^{6}}{2 \\cdot 4 \\cdot 6}+\\cdots}=\\int_{0}^{constantvalue} e^{-stillpoint^{2 / 2}} d stillpoint .\n\\]",
      "solution": "Solution. Let the base triangles have sides \\( interiorangle \\) and \\( vertexangle \\) with included angle \\( straightline \\), and let the right prism have altitude \\( baselength \\). If \\( facedifference \\) denotes the given sum of the three face areas, then \\( facedifference=interiorangle baselength+vertexangle baselength+\\frac{1}{2} interiorangle vertexangle \\sin straightline \\), and the volume is \\( surfacearea= \\frac{1}{2} interiorangle vertexangle baselength \\sin straightline \\). Let \\( perimeter=interiorangle baselength, circumference=vertexangle baselength, borderlength=\\frac{1}{2} interiorangle vertexangle \\sin straightline \\) be the areas of the three faces. Then \\( surfacearea^{2}=\\frac{1}{2} perimeter circumference borderlength \\sin straightline \\) where \\( perimeter, circumference, borderlength \\) are three positive numbers whose sum is \\( facedifference \\). Now the arithmetic mean-geometric mean inequality for positive numbers yields \\( (perimeter circumference borderlength)^{1 / 3} \\leq(perimeter+circumference+borderlength) / 3 \\), and hence\n\\[\nsurfacearea^{2} \\leq \\frac{1}{2}\\left(\\frac{facedifference}{3}\\right)^{3} \\sin straightline, \\quad \\text { and } \\quad surfacearea \\leq \\frac{1}{\\sqrt{2}}\\left(\\frac{facedifference}{3}\\right)^{3 / 2}\n\\]\nwith equality occurring only if \\( perimeter=circumference=borderlength=\\frac{1}{3} facedifference \\) and \\( \\sin straightline=1 \\). But if all these conditions are satisfied, the three faces all have equal area and are mutually perpendicular.\n\nIt remains to be shown that these conditions can be satisfied by a right prism. For this we need only take \\( interiorangle=vertexangle=2 baselength=\\sqrt{2 facedifference / 3} \\) and \\( straightline=\\pi / 2 \\).\n\nRemark. The Lagrange multiplier method leads quickly to the critical point, but the proof that it is a maximum seems to require an argument similar to the above.\n\nSolution. The series appearing in the numerator converges for all \\( \\boldsymbol{constantvalue} \\) by the ratio test. Let its sum be \\( scalarvalue(constantvalue) \\). Since the denominator is immediately seen to be\n\\[\n\\sum_{continuum=0}^{\\infty} \\frac{1}{continuum!}\\left(\\frac{constantvalue^{2}}{2}\\right)^{continuum}=e^{constantvalue^{2} / 2},\n\\]\nthe problem is equivalent to proving that\n\\[\nscalarvalue(constantvalue)=e^{constantvalue^{2} / 2} \\int_{0}^{constantvalue} e^{-stillpoint^{2} / 2} d stillpoint .\n\\]\n\nDifferentiating the series for \\( scalarvalue \\) term by term, we find\n\\[\nscalarvalue^{\\prime}(constantvalue)=1+constantvalue scalarvalue(constantvalue)\n\\]\n\nThis first-order linear differential equation, together with the initial condition \\( scalarvalue(0)=0 \\) determines the function \\( scalarvalue \\), so we can establish (1) either by differentiating the right member of (1) and noting that it satisfies (2) and the initial condition, or by solving (2) by the usual method. (The corresponding homogeneous equation is \\( constant^{\\prime}(constantvalue)-constantvalue constant(constantvalue)=0 \\), with general solution \\( constant(constantvalue)=baselength e^{constantvalue^{2} / 2} \\). Then \\( e^{-constantvalue^{2} / 2} \\) is an integrating factor for (2):\n\\[\n\\frac{d}{d constantvalue}\\left(scalarvalue(constantvalue) e^{-constantvalue^{2} / 2}\\right)=e^{-constantvalue^{2} / 2}\n\\]\nand (1) follows immediately.)\nRemark. This is essentially Problem 4623, American Mathematical Monthly, vol. 63 (1956), page 260."
    },
    "garbled_string": {
      "map": {
        "a": "qzxwvtnp",
        "b": "hjgrksla",
        "c": "frlqzmpd",
        "L": "pkmnxdwa",
        "V": "vdrgqsle",
        "X": "blmqzktp",
        "Y": "znxprhva",
        "Z": "cjrvmqse",
        "x": "tksznpqa",
        "t": "glqrzvnh",
        "n": "rmpqvszl",
        "g": "vkqrhlmn",
        "f": "wlzpkqre",
        "\\theta": "\\ndjvqfza"
      },
      "question": "4. Answer either (i) or (ii).\n(i) In a right prism with triangular base, given the sum of the areas of three mutually adjacent faces (that is, of two lateral faces and one base), show that these faces are of equal area and perpendicular to each other when the volume attains its maximum.\n(page 292)\n(ii) Show that\n\\[\n\\frac{\\frac{tksznpqa}{1}+\\frac{tksznpqa^{3}}{1 \\cdot 3}+\\frac{tksznpqa^{5}}{1 \\cdot 3 \\cdot 5}+\\frac{tksznpqa^{7}}{1 \\cdot 3 \\cdot 5 \\cdot 7}+\\cdots}{1+\\frac{tksznpqa^{2}}{2}+\\frac{tksznpqa^{4}}{2 \\cdot 4}+\\frac{tksznpqa^{6}}{2 \\cdot 4 \\cdot 6}+\\cdots}=\\int_{0}^{tksznpqa} e^{-glqrzvnh^{2 / 2}} d glqrzvnh .\n\\]",
      "solution": "Solution. Let the base triangles have sides \\( qzxwvtnp \\) and \\( hjgrksla \\) with included angle \\( \\ndjvqfza \\), and let the right prism have altitude \\( frlqzmpd \\). If \\( pkmnxdwa \\) denotes the given sum of the three face areas, then \\( pkmnxdwa=qzxwvtnp frlqzmpd+hjgrksla frlqzmpd+\\frac{1}{2} qzxwvtnp hjgrksla \\sin \\ndjvqfza \\), and the volume is \\( vdrgqsle= \\) \\( \\frac{1}{2} qzxwvtnp hjgrksla frlqzmpd \\sin \\ndjvqfza \\). Let \\( blmqzktp=qzxwvtnp frlqzmpd, znxprhva=hjgrksla frlqzmpd, cjrvmqse=\\frac{1}{2} qzxwvtnp hjgrksla \\sin \\ndjvqfza \\) be the areas of the three faces. Then \\( vdrgqsle^{2}=\\frac{1}{2} blmqzktp znxprhva cjrvmqse \\sin \\ndjvqfza \\) where \\( blmqzktp, znxprhva, cjrvmqse \\) are three positive numbers whose sum is \\( pkmnxdwa \\). Now the arithmetic mean-geometric mean inequality for positive numbers yields \\( (blmqzktp znxprhva cjrvmqse)^{1 / 3} \\leq(blmqzktp+znxprhva+cjrvmqse) / 3 \\), and hence\n\\[\nvdrgqsle^{2} \\leq \\frac{1}{2}\\left(\\frac{pkmnxdwa}{3}\\right)^{3} \\sin \\ndjvqfza, \\quad \\text { and } \\quad vdrgqsle \\leq \\frac{1}{\\sqrt{2}}\\left(\\frac{pkmnxdwa}{3}\\right)^{3 / 2}\n\\]\nwith equality occurring only if \\( blmqzktp=znxprhva=cjrvmqse=\\frac{1}{3} pkmnxdwa \\) and \\( \\sin \\ndjvqfza=1 \\). But if all these conditions are satisfied, the three faces all have equal area and are mutually perpendicular.\n\nIt remains to be shown that these conditions can be satisfied by a right prism. For this we need only take \\( qzxwvtnp=hjgrksla=2 frlqzmpd=\\sqrt{2 pkmnxdwa / 3} \\) and \\( \\ndjvqfza=\\pi / 2 \\).\n\nRemark. The Lagrange multiplier method leads quickly to the critical point, but the proof that it is a maximum seems to require an argument similar to the above.\n\nSolution. The series appearing in the numerator converges for all \\( \\boldsymbol{tksznpqa} \\) by the ratio test. Let its sum be \\( wlzpkqre(tksznpqa) \\). Since the denominator is immediately seen to be\n\\[\n\\sum_{rmpqvszl=0}^{\\infty} \\frac{1}{rmpqvszl!}\\left(\\frac{tksznpqa^{2}}{2}\\right)^{rmpqvszl}=e^{tksznpqa^{2} / 2},\n\\]\nthe problem is equivalent to proving that\n\\[\nwlzpkqre(tksznpqa)=e^{tksznpqa^{2} / 2} \\int_{0}^{tksznpqa} e^{-glqrzvnh^{2} / 2} d glqrzvnh .\n\\]\n\nDifferentiating the series for \\( wlzpkqre \\) term by term, we find\n\\[\nwlzpkqre^{\\prime}(tksznpqa)=1+tksznpqa \\, wlzpkqre(tksznpqa)\n\\]\n\nThis first-order linear differential equation, together with the initial condition \\( wlzpkqre(0)=0 \\) determines the function \\( wlzpkqre \\), so we can establish (1) either by differentiating the right member of (1) and noting that it satisfies (2) and the initial condition, or by solving (2) by the usual method. (The corresponding homogeneous equation is \\( vkqrhlmn^{\\prime}(tksznpqa)-tksznpqa \\, vkqrhlmn(tksznpqa)=0 \\), with general solution \\( vkqrhlmn(tksznpqa)=c e^{tksznpqa^{2} / 2} \\). Then \\( e^{-tksznpqa^{2} / 2} \\) is an integrating factor for (2):\n\\[\n\\frac{d}{d tksznpqa}\\left(wlzpkqre(tksznpqa) e^{-tksznpqa^{2} / 2}\\right)=e^{-tksznpqa^{2} / 2}\n\\]\nand (1) follows immediately.)\nRemark. This is essentially Problem 4623, American Mathematical Monthly, vol. 63 (1956), page 260."
    },
    "kernel_variant": {
      "question": "Let an integer d ≥ 2 be fixed and set  \n    c_d = 2 Γ((d+1)/2)/√π.      (1)\n\nFor x∈ℝ introduce the two power series  \n\n  A_d(x) =  Σ_{n=0}^{∞}  (d/2)_n (−1)^n\n               ─────────────── (x²/d)^n,   (2)\n                   (n!)²\n\n  B_d(x) =  c_d Σ_{n=0}^{∞} (d/2)_n (−1)^n\n                ──────────────────────── x^{d+2n}. (3)\n                 (n!)² d^{\\,n}(d+2n)\n\nHere (a)_n = a(a+1)…(a+n−1) denotes the rising factorial.\n\n(a) Prove that the series (2)–(3) converge absolutely for every x∈ℝ and hence define entire functions.\n\n(b) Show that the derivative identity  \n    B′_d(x) = c_d x^{d−1} A_d(x)           (4)  \nholds for all real x.\n\n(c) Deduce from (4) that  \n    B_d(x)=c_d∫_0^x t^{d−1}A_d(t)\\,dt       (5)  \nand that the quotient  \n    F_d(x)=B_d(x)/A_d(x)              (6)  \nsatisfies the first-order linear ODE  \n    F′_d(x)+F_d(x)\\,A′_d(x)/A_d(x)=c_d x^{d−1}.  (7)\n\n(d) Specialise your results to d = 2.  Prove that  \n    A₂(x)=e^{−x²/2},  B₂(x)=∫_0^x t e^{−t²/2}\\,dt = 1−e^{−x²/2}, (8)  \nhence  \n    F₂(x)=B₂(x)/A₂(x)=e^{x²/2}−1.          (9)\n\nExplain why B₂(x)=1−e^{−x²/2} is the cumulative distribution function of the Maxwell–Boltzmann speed, i.e. of the χ-distribution with two degrees of freedom.\n\n(Full derivations of (4)–(9) are required; quoting special-function identities without proof earns no credit.)\n\n\n--------------------------------------------------------------------",
      "solution": "Step 1. Absolute convergence and analyticity of A_d and B_d.  \nUsing Γ(n+d/2)=Γ(d/2)·(d/2)_n and Stirling’s formula,\n\n  (d/2)_n = Γ(n+d/2)/Γ(d/2) ∼ n^{d/2}\\,n!     (n→∞).\n\nHence the n-th coefficient of (2) has size\n\n  |(d/2)_n| |x|^{2n}/d^n /(n!)² = O(n^{d/2}|x|^{2n}/(d^n n!)),\n\nwhich tends to 0 because n! grows faster than c^n for any fixed c.  \nBy the Cauchy–Hadamard formula the radius of convergence is ∞, so (2) defines an entire function; adding the extra factor 1/(d+2n) in (3) only improves convergence, and B_d is entire as well.  Uniform convergence on every compact subset of ℂ follows from the Weierstrass M–test and will justify all forthcoming term-wise manipulations.\n\nStep 2. A confluent hypergeometric representation for A_d.  \nInsert the harmless factor n!/n! = (1)_n/(1)_n into (2):\n\n  A_d(x)=Σ_{n=0}^{∞} (d/2)_n(−x²/d)^n\n        ──────────────────────────── = Σ_{n=0}^{∞}\n            (1)_n·n!\n\n        (d/2)_n      (−x²/d)^n\n      = Σ_{n=0}^{∞} ──────────────── ,      (10)\n         (1)_n n!\n\nwhose right–hand side is precisely the series of Kummer’s confluent\nhypergeometric function  \n\n  A_d(x)= {}_1F_1(d/2; 1; −x²/d).        (11)\n\nFor generic integer d this function has no simpler closed form in\nelementary functions.\n\nWhen d=2 one has (d/2)=1.  Because for every z\n\n  {}_1F_1(1;1;z)=e^{z}\n\n(a consequence of term-by-term comparison with the series of e^{z}),\n(11) yields\n\n  A₂(x)=e^{−x²/2},             (12)\n\nthe first identity in (8).\n\nStep 3. Term-wise differentiation and the identity (4).  \nThe series (3) is uniformly convergent on compact subsets, hence\n\n  B′_d(x)=c_d Σ_{n=0}^{∞}(d/2)_n(−1)^n\n           ────────────── x^{d+2n−1}.\n            (n!)² d^{\\,n}\n\nFactor x^{d−1} and recognise A_d(x):\n\n  B′_d(x)=c_d x^{d−1} Σ_{n=0}^{∞}(d/2)_n(−x²/d)^n /(n!)²\n        = c_d x^{d−1} A_d(x),      (13)\n\nestablishing (4).\n\nStep 4. Integral representation of B_d.  \nSince B_d(0)=0, integration of (4) from 0 to x gives (5):\n\n  B_d(x)=c_d∫_0^x t^{d−1}A_d(t)\\,dt.     (14)\n\nStep 5. The quotient F_d and its ODE.  \nPut F_d=B_d/A_d.  Differentiating,\n\n  B′_d = A_d F′_d + A′_d F_d.        (15)\n\nInsert (4) and divide by A_d to obtain (7):\n\n  F′_d(x)+F_d(x) A′_d(x)/A_d(x)=c_d x^{d−1}.  (16)\n\nBecause A_d is entire and non-vanishing at x=0 (A_d(0)=1), (16) is a\nlinear ODE with continuous coefficients; the integrating-factor method\nreproduces (5) and therefore B_d/A_d as its unique solution that obeys\nF_d(0)=0.\n\nStep 6. The special case d = 2.  \nFirst, c_2=2 Γ(3/2)/√π = 1.  \nWith (12) the integral formula (14) becomes\n\n  B₂(x)=∫_0^x t e^{−t²/2}\\,dt\n     =[−e^{−t²/2}]_0^x\n     =1−e^{−x²/2},         (17)\n\nthe second identity in (8).  Dividing by A₂(x) gives (9):\n\n  F₂(x)= (1−e^{−x²/2})/e^{−x²/2}=e^{x²/2}−1.  (18)\n\nStep 7. Probabilistic interpretation.  \nFor x≥0 the χ-distribution with two degrees of freedom has density\nf(x)=x e^{−x²/2}.  Hence its cumulative distribution function is\n\n  P(X≤x)=∫_0^x t e^{−t²/2}\\,dt =1−e^{−x²/2}=B₂(x). (19)\n\nThus B₂—not F₂—is the Maxwell–Boltzmann speed CDF, as required.\n\nAll series manipulations are justified by uniform convergence on\ncompact sets; the solution is therefore rigorous.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.430498",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension & additional parameter.  \n   • The original identity involved one real variable and a fixed\n      Gaussian $e^{-t^{2}/3}$.  \n   • Here the exponent, the power of $t$ in the integrand, and all\n      series coefficients depend on a new discrete parameter $d$,\n      representing the ambient dimension in a Maxwell–Boltzmann\n      model.  The proof must hold simultaneously for every\n      integer $d\\ge2$.\n\n2.  Sophisticated special-function machinery.  \n   • The argument uses Pochhammer symbols, double factorials, the\n      Gamma function, and the constant $\\Gamma(\\frac{d+1}{2})$ from\n      high-dimensional geometry.  \n   • Identifying $A_{d}$ with a Gauss hypergeometric $\\,{}_{1}F_{0}\\,$\n      and exploiting its closed form $(1-x^{2}/d)^{-d/2}$\n      introduces non-elementary functions that never appear in the\n      original problem.\n\n3.  Non-trivial differential-equation analysis.  \n   • Deriving (7) requires careful manipulation of\n      $\\,(2n+1)!!\\,$ and of the ratio of successive Pochhammer\n      symbols.  \n   • The quotient $F_{d}=B_{d}/A_{d}$ obeys a variable-coefficient\n      linear ODE (9); solving it needs an integrating factor that\n      itself depends on $d$.\n\n4.  Interplay of combinatorics and analysis.  \n   • One must juggle three different families of coefficients\n      (Pochhammer, ordinary factorials, and double factorials) and\n      show that their combined recurrence produces the simple\n      right-hand side in (7).  \n   • Asymptotic estimates for the general terms are necessary to\n      justify the term-wise differentiation that underpins the whole\n      argument.\n\n5.  Conceptual depth.  \n   • While the original kernel identity can be settled with a\n      first-order ODE and the elementary exponential series, the\n      present variant demands familiarity with higher-dimensional\n      Gaussian integrals, the Maxwell–Boltzmann distribution,\n      hypergeometric series, and Gamma-function identities.  \n   • Hence the solution is significantly longer, uses richer\n      mathematical structures, and combines combinatorial,\n      analytic, and probabilistic insights—well beyond simple\n      pattern matching."
      }
    },
    "original_kernel_variant": {
      "question": "Let an integer d ≥ 2 be fixed and set  \n    c_d = 2 Γ((d+1)/2)/√π.      (1)\n\nFor x∈ℝ introduce the two power series  \n\n  A_d(x) =  Σ_{n=0}^{∞}  (d/2)_n (−1)^n\n               ─────────────── (x²/d)^n,   (2)\n                   (n!)²\n\n  B_d(x) =  c_d Σ_{n=0}^{∞} (d/2)_n (−1)^n\n                ──────────────────────── x^{d+2n}. (3)\n                 (n!)² d^{\\,n}(d+2n)\n\nHere (a)_n = a(a+1)…(a+n−1) denotes the rising factorial.\n\n(a) Prove that the series (2)–(3) converge absolutely for every x∈ℝ and hence define entire functions.\n\n(b) Show that the derivative identity  \n    B′_d(x) = c_d x^{d−1} A_d(x)           (4)  \nholds for all real x.\n\n(c) Deduce from (4) that  \n    B_d(x)=c_d∫_0^x t^{d−1}A_d(t)\\,dt       (5)  \nand that the quotient  \n    F_d(x)=B_d(x)/A_d(x)              (6)  \nsatisfies the first-order linear ODE  \n    F′_d(x)+F_d(x)\\,A′_d(x)/A_d(x)=c_d x^{d−1}.  (7)\n\n(d) Specialise your results to d = 2.  Prove that  \n    A₂(x)=e^{−x²/2},  B₂(x)=∫_0^x t e^{−t²/2}\\,dt = 1−e^{−x²/2}, (8)  \nhence  \n    F₂(x)=B₂(x)/A₂(x)=e^{x²/2}−1.          (9)\n\nExplain why B₂(x)=1−e^{−x²/2} is the cumulative distribution function of the Maxwell–Boltzmann speed, i.e. of the χ-distribution with two degrees of freedom.\n\n(Full derivations of (4)–(9) are required; quoting special-function identities without proof earns no credit.)\n\n\n--------------------------------------------------------------------",
      "solution": "Step 1. Absolute convergence and analyticity of A_d and B_d.  \nUsing Γ(n+d/2)=Γ(d/2)·(d/2)_n and Stirling’s formula,\n\n  (d/2)_n = Γ(n+d/2)/Γ(d/2) ∼ n^{d/2}\\,n!     (n→∞).\n\nHence the n-th coefficient of (2) has size\n\n  |(d/2)_n| |x|^{2n}/d^n /(n!)² = O(n^{d/2}|x|^{2n}/(d^n n!)),\n\nwhich tends to 0 because n! grows faster than c^n for any fixed c.  \nBy the Cauchy–Hadamard formula the radius of convergence is ∞, so (2) defines an entire function; adding the extra factor 1/(d+2n) in (3) only improves convergence, and B_d is entire as well.  Uniform convergence on every compact subset of ℂ follows from the Weierstrass M–test and will justify all forthcoming term-wise manipulations.\n\nStep 2. A confluent hypergeometric representation for A_d.  \nInsert the harmless factor n!/n! = (1)_n/(1)_n into (2):\n\n  A_d(x)=Σ_{n=0}^{∞} (d/2)_n(−x²/d)^n\n        ──────────────────────────── = Σ_{n=0}^{∞}\n            (1)_n·n!\n\n        (d/2)_n      (−x²/d)^n\n      = Σ_{n=0}^{∞} ──────────────── ,      (10)\n         (1)_n n!\n\nwhose right–hand side is precisely the series of Kummer’s confluent\nhypergeometric function  \n\n  A_d(x)= {}_1F_1(d/2; 1; −x²/d).        (11)\n\nFor generic integer d this function has no simpler closed form in\nelementary functions.\n\nWhen d=2 one has (d/2)=1.  Because for every z\n\n  {}_1F_1(1;1;z)=e^{z}\n\n(a consequence of term-by-term comparison with the series of e^{z}),\n(11) yields\n\n  A₂(x)=e^{−x²/2},             (12)\n\nthe first identity in (8).\n\nStep 3. Term-wise differentiation and the identity (4).  \nThe series (3) is uniformly convergent on compact subsets, hence\n\n  B′_d(x)=c_d Σ_{n=0}^{∞}(d/2)_n(−1)^n\n           ────────────── x^{d+2n−1}.\n            (n!)² d^{\\,n}\n\nFactor x^{d−1} and recognise A_d(x):\n\n  B′_d(x)=c_d x^{d−1} Σ_{n=0}^{∞}(d/2)_n(−x²/d)^n /(n!)²\n        = c_d x^{d−1} A_d(x),      (13)\n\nestablishing (4).\n\nStep 4. Integral representation of B_d.  \nSince B_d(0)=0, integration of (4) from 0 to x gives (5):\n\n  B_d(x)=c_d∫_0^x t^{d−1}A_d(t)\\,dt.     (14)\n\nStep 5. The quotient F_d and its ODE.  \nPut F_d=B_d/A_d.  Differentiating,\n\n  B′_d = A_d F′_d + A′_d F_d.        (15)\n\nInsert (4) and divide by A_d to obtain (7):\n\n  F′_d(x)+F_d(x) A′_d(x)/A_d(x)=c_d x^{d−1}.  (16)\n\nBecause A_d is entire and non-vanishing at x=0 (A_d(0)=1), (16) is a\nlinear ODE with continuous coefficients; the integrating-factor method\nreproduces (5) and therefore B_d/A_d as its unique solution that obeys\nF_d(0)=0.\n\nStep 6. The special case d = 2.  \nFirst, c_2=2 Γ(3/2)/√π = 1.  \nWith (12) the integral formula (14) becomes\n\n  B₂(x)=∫_0^x t e^{−t²/2}\\,dt\n     =[−e^{−t²/2}]_0^x\n     =1−e^{−x²/2},         (17)\n\nthe second identity in (8).  Dividing by A₂(x) gives (9):\n\n  F₂(x)= (1−e^{−x²/2})/e^{−x²/2}=e^{x²/2}−1.  (18)\n\nStep 7. Probabilistic interpretation.  \nFor x≥0 the χ-distribution with two degrees of freedom has density\nf(x)=x e^{−x²/2}.  Hence its cumulative distribution function is\n\n  P(X≤x)=∫_0^x t e^{−t²/2}\\,dt =1−e^{−x²/2}=B₂(x). (19)\n\nThus B₂—not F₂—is the Maxwell–Boltzmann speed CDF, as required.\n\nAll series manipulations are justified by uniform convergence on\ncompact sets; the solution is therefore rigorous.\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.372877",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher dimension & additional parameter.  \n   • The original identity involved one real variable and a fixed\n      Gaussian $e^{-t^{2}/3}$.  \n   • Here the exponent, the power of $t$ in the integrand, and all\n      series coefficients depend on a new discrete parameter $d$,\n      representing the ambient dimension in a Maxwell–Boltzmann\n      model.  The proof must hold simultaneously for every\n      integer $d\\ge2$.\n\n2.  Sophisticated special-function machinery.  \n   • The argument uses Pochhammer symbols, double factorials, the\n      Gamma function, and the constant $\\Gamma(\\frac{d+1}{2})$ from\n      high-dimensional geometry.  \n   • Identifying $A_{d}$ with a Gauss hypergeometric $\\,{}_{1}F_{0}\\,$\n      and exploiting its closed form $(1-x^{2}/d)^{-d/2}$\n      introduces non-elementary functions that never appear in the\n      original problem.\n\n3.  Non-trivial differential-equation analysis.  \n   • Deriving (7) requires careful manipulation of\n      $\\,(2n+1)!!\\,$ and of the ratio of successive Pochhammer\n      symbols.  \n   • The quotient $F_{d}=B_{d}/A_{d}$ obeys a variable-coefficient\n      linear ODE (9); solving it needs an integrating factor that\n      itself depends on $d$.\n\n4.  Interplay of combinatorics and analysis.  \n   • One must juggle three different families of coefficients\n      (Pochhammer, ordinary factorials, and double factorials) and\n      show that their combined recurrence produces the simple\n      right-hand side in (7).  \n   • Asymptotic estimates for the general terms are necessary to\n      justify the term-wise differentiation that underpins the whole\n      argument.\n\n5.  Conceptual depth.  \n   • While the original kernel identity can be settled with a\n      first-order ODE and the elementary exponential series, the\n      present variant demands familiarity with higher-dimensional\n      Gaussian integrals, the Maxwell–Boltzmann distribution,\n      hypergeometric series, and Gamma-function identities.  \n   • Hence the solution is significantly longer, uses richer\n      mathematical structures, and combines combinatorial,\n      analytic, and probabilistic insights—well beyond simple\n      pattern matching."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}