{
  "index": "1939-B-5",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG",
    "NT"
  ],
  "difficulty": "",
  "question": "12. Take either (i) or (ii).\n(i) Prove that\n\\[\n\\int_{1}^{a}[x] f^{\\prime}(x) d x=[a] f(a)-\\{f(1)+\\cdots+f([a])\\}\n\\]\nwhere \\( a \\) is greater than 1 and where [ \\( x \\) ] denotes the greatest of the integers not exceeding \\( x \\). Obtain a corresponding expression for\n\\[\n\\int_{1}^{a}\\left[x^{2}\\right] f^{\\prime}(x) d x\n\\]\n(ii) A particle moves on a straight line, the only force acting on it being a resistance proportional to the velocity. If it started with a velocity of \\( 1,000 \\mathrm{ft} \\). per sec. and had a velocity of 900 ft . per sec. when it had travelled \\( 1,200 \\mathrm{ft} \\)., calculate to the nearest hundredth of a second the time it took to travel this distance.",
  "solution": "Solution. We have\n\\[\n\\begin{aligned}\n\\int_{1}^{a}[x] f^{\\prime}(x) d x & =\\int_{1}^{2} 1 \\cdot f^{\\prime}(x) d x+\\int_{2}^{3} 2 \\cdot f^{\\prime}(x) d x+\\cdots+\\int_{[a \\mid}^{a}[a] \\cdot f^{\\prime}(x) d x \\\\\n& =f(2)-f(1)+2(f(3)-f(2))+\\cdots+[a](f(a)-f([a])) \\\\\n& =[a] f(a)-\\{f(1)+f(2)+\\cdots+f([a])\\} .\n\\end{aligned}\n\\]\n\nFor the second part, we have\n\\[\n\\begin{aligned}\n\\int_{1}^{a}\\left[x^{2}\\right] f^{\\prime}(x) d x= & \\int_{1}^{v 2} 1 \\cdot f^{\\prime}(x) d x+\\int_{v^{2}}^{v^{3}} 2 f^{\\prime}(x) d x+\\cdots+\\int_{v\\left|a^{2}\\right|}^{a}\\left[a^{2}\\right] f^{\\prime}(x) d x \\\\\n= & (f(\\sqrt{2})-f(1))+2(f(\\sqrt{3})-f(\\sqrt{2}))+\\cdots \\\\\n& \\left.+\\left[a^{2}\\right]\\left(f(a)-f\\left(\\sqrt{\\left[a^{2}\\right.}\\right]\\right)\\right) \\\\\n= & {\\left[a^{2}\\right] f(a)-\\left\\{f(1)+f(\\sqrt{2})+\\cdots+f\\left(\\sqrt{\\left[a^{2}\\right]}\\right)\\right\\} . }\n\\end{aligned}\n\\]\n\nRemark. These formulas result from integration by parts applied to Stieltjes integrals; for example:\n\\[\n\\begin{aligned}\n\\int_{1}^{a}[x] f^{\\prime}(x) d x & =\\int_{1 / 2}^{a}[x] f^{\\prime}(x) d x=\\left.[x] f(x)\\right|_{1 / 2} ^{a}-\\int_{1 / 2}^{a} f(x) d[x] \\\\\n& =[a] f(a)-(f(1)+f(2)+\\cdots+f([a])) .\n\\end{aligned}\n\\]\n\nSolution. The differential equation governing the motion is\n\\[\nm \\frac{d^{2} x}{d t^{2}}=-k \\frac{d x}{d t}\n\\]\nand the boundary conditions are\n\\[\n\\begin{array}{c}\nx=0, \\quad \\frac{d x}{d t}=1000, \\quad \\text { when } t=0 \\\\\nx=1200, \\quad \\frac{d x}{d t}=900, \\quad \\text { when } t=T\n\\end{array}\n\\]\nwhere \\( T \\) is the time required.\nLet \\( b=k / m \\). Then \\( d^{2} x / d t^{2}=-b d x / d t \\), which implies\n\\[\n\\frac{d x}{d t}=-b x+c\n\\]\n\nThe boundary conditions give\n\\[\n\\begin{aligned}\n1000 & =c \\\\\n900 & =-1200 b+c\n\\end{aligned}\n\\]\nwhence \\( b=\\frac{1}{12} \\). Using these values and (1), we have\n\\[\nT=\\int_{0}^{1200} \\frac{d t}{d x} d x=\\int_{0}^{1200} \\frac{d x}{1000-x / 12}=12 \\log \\frac{10}{9}\n\\]\n\nTo evaluate this it is convenient to write \\( T=-12 \\log \\frac{9}{10} \\) and use the series expansion\n\\[\n-\\log (1-x)=x+\\frac{1}{2} x^{2}+\\frac{1}{3} x^{3}+\\frac{1}{4} x^{4}+\\cdots\n\\]\n\nTaking \\( x=\\frac{1}{10} \\), we have\n\\[\n\\frac{1}{10}+\\frac{1}{200}+\\frac{1}{3000}<-\\log \\frac{9}{10}<\\frac{1}{10}+\\frac{1}{200}+\\frac{1}{3} \\sum_{n=3}^{\\infty}\\left|\\frac{1}{10}\\right|^{n} .\n\\]\n\nThe lower bound exceeds \\( .1+.005+.0003=.1053 \\), and the upper bound is \\( .1+.005+1 / 2700<.1054 \\). Therefore,\n\\[\n1.2636<-12 \\log \\frac{9}{10}<1.2648\n\\]\nso, to the nearest hundreth of a second, \\( T \\simeq 1.26 \\mathrm{sec} \\).",
  "vars": [
    "x",
    "t",
    "n"
  ],
  "params": [
    "a",
    "f",
    "m",
    "k",
    "b",
    "c",
    "T"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "position",
        "t": "timevar",
        "n": "indexvar",
        "a": "upperlimit",
        "f": "function",
        "m": "massconst",
        "k": "resistc",
        "b": "decayrate",
        "c": "intconst",
        "T": "totaltime"
      },
      "question": "12. Take either (i) or (ii).\n(i) Prove that\n\\[\n\\int_{1}^{upperlimit}[position] function^{\\prime}(position) d position=[upperlimit] function(upperlimit)-\\{function(1)+\\cdots+function([upperlimit])\\}\n\\]\nwhere \\( upperlimit \\) is greater than 1 and where [ \\( position \\) ] denotes the greatest of the integers not exceeding \\( position \\). Obtain a corresponding expression for\n\\[\n\\int_{1}^{upperlimit}\\left[position^{2}\\right] function^{\\prime}(position) d position\n\\]\n(ii) A particle moves on a straight line, the only force acting on it being a resistance proportional to the velocity. If it started with a velocity of \\( 1,000 \\mathrm{ft} \\). per sec. and had a velocity of 900 ft . per sec. when it had travelled \\( 1,200 \\mathrm{ft} \\)., calculate to the nearest hundredth of a second the time it took to travel this distance.",
      "solution": "Solution. We have\n\\[\n\\begin{aligned}\n\\int_{1}^{upperlimit}[position] function^{\\prime}(position) d position & =\\int_{1}^{2} 1 \\cdot function^{\\prime}(position) d position+\\int_{2}^{3} 2 \\cdot function^{\\prime}(position) d position+\\cdots+\\int_{[upperlimit \\mid}^{upperlimit}[upperlimit] \\cdot function^{\\prime}(position) d position \\\\\n& =function(2)-function(1)+2(function(3)-function(2))+\\cdots+[upperlimit](function(upperlimit)-function([upperlimit])) \\\\\n& =[upperlimit] function(upperlimit)-\\{function(1)+function(2)+\\cdots+function([upperlimit])\\} .\n\\end{aligned}\n\\]\n\nFor the second part, we have\n\\[\n\\begin{aligned}\n\\int_{1}^{upperlimit}\\left[position^{2}\\right] function^{\\prime}(position) d position= & \\int_{1}^{\\sqrt{2}} 1 \\cdot function^{\\prime}(position) d position+\\int_{\\sqrt{2}}^{\\sqrt{3}} 2 function^{\\prime}(position) d position+\\cdots+\\int_{\\sqrt{[upperlimit^{2}]}}^{upperlimit}\\left[upperlimit^{2}\\right] function^{\\prime}(position) d position \\\\\n= & (function(\\sqrt{2})-function(1))+2(function(\\sqrt{3})-function(\\sqrt{2}))+\\cdots \\\\\n& \\left.+\\left[upperlimit^{2}\\right]\\left(function(upperlimit)-function\\left(\\sqrt{\\left[upperlimit^{2}\\right]}\\right)\\right) \\\\\n= & {\\left[upperlimit^{2}\\right] function(upperlimit)-\\left\\{function(1)+function(\\sqrt{2})+\\cdots+function\\left(\\sqrt{\\left[upperlimit^{2}\\right]}\\right)\\right\\} . }\n\\end{aligned}\n\\]\n\nRemark. These formulas result from integration by parts applied to Stieltjes integrals; for example:\n\\[\n\\begin{aligned}\n\\int_{1}^{upperlimit}[position] function^{\\prime}(position) d position & =\\int_{1 / 2}^{upperlimit}[position] function^{\\prime}(position) d position=\\left.[position] function(position)\\right|_{1 / 2}^{upperlimit}-\\int_{1 / 2}^{upperlimit} function(position) d[position] \\\\\n& =[upperlimit] function(upperlimit)-(function(1)+function(2)+\\cdots+function([upperlimit])) .\n\\end{aligned}\n\\]\n\nSolution. The differential equation governing the motion is\n\\[\nmassconst \\frac{d^{2} position}{d timevar^{2}}=-resistc \\frac{d position}{d timevar}\n\\]\nand the boundary conditions are\n\\[\n\\begin{array}{c}\nposition=0, \\quad \\frac{d position}{d timevar}=1000, \\quad \\text { when } timevar=0 \\\\\nposition=1200, \\quad \\frac{d position}{d timevar}=900, \\quad \\text { when } timevar=totaltime\n\\end{array}\n\\]\nwhere \\( totaltime \\) is the time required.\nLet \\( decayrate=resistc / massconst \\). Then \\( d^{2} position / d timevar^{2}=-decayrate d position / d timevar \\), which implies\n\\[\n\\frac{d position}{d timevar}=-decayrate position+intconst\n\\]\n\nThe boundary conditions give\n\\[\n\\begin{aligned}\n1000 & =intconst \\\\\n900 & =-1200 decayrate+intconst\n\\end{aligned}\n\\]\nwhence \\( decayrate=\\frac{1}{12} \\). Using these values and (1), we have\n\\[\ntotaltime=\\int_{0}^{1200} \\frac{d timevar}{d position} d position=\\int_{0}^{1200} \\frac{d position}{1000-position / 12}=12 \\log \\frac{10}{9}\n\\]\n\nTo evaluate this it is convenient to write \\( totaltime=-12 \\log \\frac{9}{10} \\) and use the series expansion\n\\[\n-\\log (1-x)=x+\\frac{1}{2} x^{2}+\\frac{1}{3} x^{3}+\\frac{1}{4} x^{4}+\\cdots\n\\]\n\nTaking \\( x=\\frac{1}{10} \\), we have\n\\[\n\\frac{1}{10}+\\frac{1}{200}+\\frac{1}{3000}<-\\log \\frac{9}{10}<\\frac{1}{10}+\\frac{1}{200}+\\frac{1}{3} \\sum_{indexvar=3}^{\\infty}\\left|\\frac{1}{10}\\right|^{indexvar} .\n\\]\n\nThe lower bound exceeds \\( .1+.005+.0003=.1053 \\), and the upper bound is \\( .1+.005+1 / 2700<.1054 \\). Therefore,\n\\[\n1.2636<-12 \\log \\frac{9}{10}<1.2648\n\\]\nso, to the nearest hundreth of a second, \\( totaltime \\simeq 1.26 \\mathrm{sec} \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "blueprint",
        "t": "gemstone",
        "n": "lighthouse",
        "a": "compassrose",
        "f": "waterfall",
        "m": "sandcastle",
        "k": "northwind",
        "b": "driftwood",
        "c": "raincloud",
        "T": "thunderbolt"
      },
      "question": "12. Take either (i) or (ii).\n(i) Prove that\n\\[\n\\int_{1}^{compassrose}[blueprint] waterfall^{\\prime}(blueprint) d blueprint=[compassrose] waterfall(compassrose)-\\{waterfall(1)+\\cdots+waterfall([compassrose])\\}\n\\]\nwhere \\( compassrose \\) is greater than 1 and where [ \\( blueprint \\) ] denotes the greatest of the integers not exceeding \\( blueprint \\). Obtain a corresponding expression for\n\\[\n\\int_{1}^{compassrose}\\left[blueprint^{2}\\right] waterfall^{\\prime}(blueprint) d blueprint\n\\]\n(ii) A particle moves on a straight line, the only force acting on it being a resistance proportional to the velocity. If it started with a velocity of \\( 1,000 \\mathrm{ft} \\). per sec. and had a velocity of 900 ft . per sec. when it had travelled \\( 1,200 \\mathrm{ft} \\)., calculate to the nearest hundredth of a second the time it took to travel this distance.",
      "solution": "Solution. We have\n\\[\n\\begin{aligned}\n\\int_{1}^{compassrose}[blueprint] waterfall^{\\prime}(blueprint) d blueprint & =\\int_{1}^{2} 1 \\cdot waterfall^{\\prime}(blueprint) d blueprint+\\int_{2}^{3} 2 \\cdot waterfall^{\\prime}(blueprint) d blueprint+\\cdots+\\int_{[compassrose \\mid}^{compassrose}[compassrose] \\cdot waterfall^{\\prime}(blueprint) d blueprint \\\\\n& =waterfall(2)-waterfall(1)+2(waterfall(3)-waterfall(2))+\\cdots+[compassrose](waterfall(compassrose)-waterfall([compassrose])) \\\\\n& =[compassrose] waterfall(compassrose)-\\{waterfall(1)+waterfall(2)+\\cdots+waterfall([compassrose])\\} .\n\\end{aligned}\n\\]\n\nFor the second part, we have\n\\[\n\\begin{aligned}\n\\int_{1}^{compassrose}\\left[blueprint^{2}\\right] waterfall^{\\prime}(blueprint) d blueprint= & \\int_{1}^{v 2} 1 \\cdot waterfall^{\\prime}(blueprint) d blueprint+\\int_{v^{2}}^{v^{3}} 2 waterfall^{\\prime}(blueprint) d blueprint+\\cdots+\\int_{v\\left|compassrose^{2}\\right|}^{compassrose}\\left[compassrose^{2}\\right] waterfall^{\\prime}(blueprint) d blueprint \\\\\n= & (waterfall(\\sqrt{2})-waterfall(1))+2(waterfall(\\sqrt{3})-waterfall(\\sqrt{2}))+\\cdots \\\\\n& \\left.+\\left[compassrose^{2}\\right]\\left(waterfall(compassrose)-waterfall\\left(\\sqrt{\\left[compassrose^{2}\\right.}\\right)\\right) \\\\\n= & {\\left[compassrose^{2}\\right] waterfall(compassrose)-\\left\\{waterfall(1)+waterfall(\\sqrt{2})+\\cdots+waterfall\\left(\\sqrt{\\left[compassrose^{2}\\right]}\\right)\\right\\} . }\n\\end{aligned}\n\\]\n\nRemark. These formulas result from integration by parts applied to Stieltjes integrals; for example:\n\\[\n\\begin{aligned}\n\\int_{1}^{compassrose}[blueprint] waterfall^{\\prime}(blueprint) d blueprint & =\\int_{1 / 2}^{compassrose}[blueprint] waterfall^{\\prime}(blueprint) d blueprint=\\left.[blueprint] waterfall(blueprint)\\right|_{1 / 2} ^{compassrose}-\\int_{1 / 2}^{compassrose} waterfall(blueprint) d[blueprint] \\\\\n& =[compassrose] waterfall(compassrose)-(waterfall(1)+waterfall(2)+\\cdots+waterfall([compassrose])) .\n\\end{aligned}\n\\]\n\nSolution. The differential equation governing the motion is\n\\[\nsandcastle \\frac{d^{2} blueprint}{d gemstone^{2}}=-northwind \\frac{d blueprint}{d gemstone}\n\\]\nand the boundary conditions are\n\\[\n\\begin{array}{c}\nblueprint=0, \\quad \\frac{d blueprint}{d gemstone}=1000, \\quad \\text { when } gemstone=0 \\\\\nblueprint=1200, \\quad \\frac{d blueprint}{d gemstone}=900, \\quad \\text { when } gemstone=thunderbolt\n\\end{array}\n\\]\nwhere \\( thunderbolt \\) is the time required.\nLet \\( driftwood=northwind / sandcastle \\). Then \\( d^{2} blueprint / d gemstone^{2}=-driftwood d blueprint / d gemstone \\), which implies\n\\[\n\\frac{d blueprint}{d gemstone}=-driftwood blueprint+raincloud\n\\]\n\nThe boundary conditions give\n\\[\n\\begin{aligned}\n1000 & =raincloud \\\\\n900 & =-1200 driftwood+raincloud\n\\end{aligned}\n\\]\nwhence \\( driftwood=\\frac{1}{12} \\). Using these values and (1), we have\n\\[\nthunderbolt=\\int_{0}^{1200} \\frac{d gemstone}{d blueprint} d blueprint=\\int_{0}^{1200} \\frac{d blueprint}{1000-blueprint / 12}=12 \\log \\frac{10}{9}\n\\]\n\nTo evaluate this it is convenient to write \\( thunderbolt=-12 \\log \\frac{9}{10} \\) and use the series expansion\n\\[\n-\\log (1-blueprint)=blueprint+\\frac{1}{2} blueprint^{2}+\\frac{1}{3} blueprint^{3}+\\frac{1}{4} blueprint^{4}+\\cdots\n\\]\n\nTaking \\( blueprint=\\frac{1}{10} \\), we have\n\\[\n\\frac{1}{10}+\\frac{1}{200}+\\frac{1}{3000}<-\\log \\frac{9}{10}<\\frac{1}{10}+\\frac{1}{200}+\\frac{1}{3} \\sum_{lighthouse=3}^{\\infty}\\left|\\frac{1}{10}\\right|^{lighthouse} .\n\\]\n\nThe lower bound exceeds \\( .1+.005+.0003=.1053 \\), and the upper bound is \\( .1+.005+1 / 2700<.1054 \\). Therefore,\n\\[\n1.2636<-12 \\log \\frac{9}{10}<1.2648\n\\]\nso, to the nearest hundreth of a second, \\( thunderbolt \\simeq 1.26 \\mathrm{sec} \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "t": "timelessness",
        "n": "irrational",
        "a": "lowerbound",
        "f": "argument",
        "m": "weightless",
        "k": "assistance",
        "b": "acceleration",
        "c": "variable",
        "T": "distance"
      },
      "question": "12. Take either (i) or (ii).\n(i) Prove that\n\\[\n\\int_{1}^{lowerbound}[constantvalue] argument^{\\prime}(constantvalue) d constantvalue=[lowerbound] argument(lowerbound)-\\{argument(1)+\\cdots+argument([lowerbound])\\}\n\\]\nwhere \\( lowerbound \\) is greater than 1 and where [ \\( constantvalue \\) ] denotes the greatest of the integers not exceeding \\( constantvalue \\). Obtain a corresponding expression for\n\\[\n\\int_{1}^{lowerbound}\\left[constantvalue^{2}\\right] argument^{\\prime}(constantvalue) d constantvalue\n\\]\n(ii) A particle moves on a straight line, the only force acting on it being a resistance proportional to the velocity. If it started with a velocity of 1,000 ft. per sec. and had a velocity of 900 ft. per sec. when it had travelled 1,200 ft., calculate to the nearest hundredth of a second the time it took to travel this distance.",
      "solution": "Solution. We have\n\\[\n\\begin{aligned}\n\\int_{1}^{lowerbound}[constantvalue] argument^{\\prime}(constantvalue) d constantvalue & =\\int_{1}^{2} 1 \\cdot argument^{\\prime}(constantvalue) d constantvalue+\\int_{2}^{3} 2 \\cdot argument^{\\prime}(constantvalue) d constantvalue+\\cdots+\\int_{[lowerbound \\mid}^{lowerbound}[lowerbound] \\cdot argument^{\\prime}(constantvalue) d constantvalue \\\\\n& =argument(2)-argument(1)+2(argument(3)-argument(2))+\\cdots+[lowerbound](argument(lowerbound)-argument([lowerbound])) \\\\\n& =[lowerbound] argument(lowerbound)-\\{argument(1)+argument(2)+\\cdots+argument([lowerbound])\\} .\n\\end{aligned}\n\\]\n\nFor the second part, we have\n\\[\n\\begin{aligned}\n\\int_{1}^{lowerbound}\\left[constantvalue^{2}\\right] argument^{\\prime}(constantvalue) d constantvalue= & \\int_{1}^{v 2} 1 \\cdot argument^{\\prime}(constantvalue) d constantvalue+\\int_{v^{2}}^{v^{3}} 2 argument^{\\prime}(constantvalue) d constantvalue+\\cdots+\\int_{v\\left|lowerbound^{2}\\right|}^{lowerbound}\\left[lowerbound^{2}\\right] argument^{\\prime}(constantvalue) d constantvalue \\\\\n= & (argument(\\sqrt{2})-argument(1))+2(argument(\\sqrt{3})-argument(\\sqrt{2}))+\\cdots \\\\\n& \\left.+\\left[lowerbound^{2}\\right]\\left(argument(lowerbound)-argument\\left(\\sqrt{\\left[lowerbound^{2}\\right.}\\right)\\right)\\right) \\\\\n= & {\\left[lowerbound^{2}\\right] argument(lowerbound)-\\left\\{argument(1)+argument(\\sqrt{2})+\\cdots+argument\\left(\\sqrt{\\left[lowerbound^{2}\\right]}\\right)\\right\\} . }\n\\end{aligned}\n\\]\n\nRemark. These formulas result from integration by parts applied to Stieltjes integrals; for example:\n\\[\n\\begin{aligned}\n\\int_{1}^{lowerbound}[constantvalue] argument^{\\prime}(constantvalue) d constantvalue & =\\int_{1 / 2}^{lowerbound}[constantvalue] argument^{\\prime}(constantvalue) d constantvalue=\\left.[constantvalue] argument(constantvalue)\\right|_{1 / 2} ^{lowerbound}-\\int_{1 / 2}^{lowerbound} argument(constantvalue) d[constantvalue] \\\\\n& =[lowerbound] argument(lowerbound)-(argument(1)+argument(2)+\\cdots+argument([lowerbound])) .\n\\end{aligned}\n\\]\n\nSolution. The differential equation governing the motion is\n\\[\nweightless \\frac{d^{2} constantvalue}{d timelessness^{2}}=-assistance \\frac{d constantvalue}{d timelessness}\n\\]\nand the boundary conditions are\n\\[\n\\begin{array}{c}\nconstantvalue=0, \\quad \\frac{d constantvalue}{d timelessness}=1000, \\quad \\text { when } timelessness=0 \\\\\nconstantvalue=1200, \\quad \\frac{d constantvalue}{d timelessness}=900, \\quad \\text { when } timelessness=distance\n\\end{array}\n\\]\nwhere \\( distance \\) is the time required.\nLet \\( acceleration=assistance / weightless \\). Then \\( d^{2} constantvalue / d timelessness^{2}=-acceleration d constantvalue / d timelessness \\), which implies\n\\[\n\\frac{d constantvalue}{d timelessness}=-acceleration constantvalue+variable\n\\]\n\nThe boundary conditions give\n\\[\n\\begin{aligned}\n1000 & =variable \\\\\n900 & =-1200 acceleration+variable\n\\end{aligned}\n\\]\nwhence \\( acceleration=\\frac{1}{12} \\). Using these values and (1), we have\n\\[\ndistance=\\int_{0}^{1200} \\frac{d timelessness}{d constantvalue} d constantvalue=\\int_{0}^{1200} \\frac{d constantvalue}{1000-constantvalue / 12}=12 \\log \\frac{10}{9}\n\\]\n\nTo evaluate this it is convenient to write \\( distance=-12 \\log \\frac{9}{10} \\) and use the series expansion\n\\[\n-\\log (1-x)=x+\\frac{1}{2} x^{2}+\\frac{1}{3} x^{3}+\\frac{1}{4} x^{4}+\\cdots\n\\]\n\nTaking \\( x=\\frac{1}{10} \\), we have\n\\[\n\\frac{1}{10}+\\frac{1}{200}+\\frac{1}{3000}<-\\log \\frac{9}{10}<\\frac{1}{10}+\\frac{1}{200}+\\frac{1}{3} \\sum_{n=3}^{\\infty}\\left|\\frac{1}{10}\\right|^{n} .\n\\]\n\nThe lower bound exceeds \\( .1+.005+.0003=.1053 \\), and the upper bound is \\( .1+.005+1 / 2700<.1054 \\). Therefore,\n\\[\n1.2636<-12 \\log \\frac{9}{10}<1.2648\n\\]\nso, to the nearest hundreth of a second, \\( distance \\simeq 1.26 \\mathrm{sec} \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "t": "hjgrksla",
        "n": "klmnbvcs",
        "a": "rtyuiofh",
        "f": "qazplmok",
        "m": "wsxedcrf",
        "k": "vtgbyhnm",
        "b": "iuhbvfdr",
        "c": "opkljhgf",
        "T": "zmxncbva"
      },
      "question": "12. Take either (i) or (ii).\n(i) Prove that\n\\[\n\\int_{1}^{rtyuiofh}[qzxwvtnp] qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp=[rtyuiofh] qazplmok(rtyuiofh)-\\{qazplmok(1)+\\cdots+qazplmok([rtyuiofh])\\}\n\\]\nwhere \\( rtyuiofh \\) is greater than 1 and where [ \\( qzxwvtnp \\) ] denotes the greatest of the integers not exceeding \\( qzxwvtnp \\). Obtain a corresponding expression for\n\\[\n\\int_{1}^{rtyuiofh}\\left[qzxwvtnp^{2}\\right] qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp\n\\]\n(ii) A particle moves on a straight line, the only force acting on it being a resistance proportional to the velocity. If it started with a velocity of \\( 1,000 \\mathrm{ft} \\). per sec. and had a velocity of 900 ft . per sec. when it had travelled \\( 1,200 \\mathrm{ft} \\)., calculate to the nearest hundredth of a second the time it took to travel this distance.",
      "solution": "Solution. We have\n\\[\n\\begin{aligned}\n\\int_{1}^{rtyuiofh}[qzxwvtnp] qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp & =\\int_{1}^{2} 1 \\cdot qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp+\\int_{2}^{3} 2 \\cdot qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp+\\cdots+\\int_{[rtyuiofh \\mid}^{rtyuiofh}[rtyuiofh] \\cdot qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp \\\\\n& =qazplmok(2)-qazplmok(1)+2(qazplmok(3)-qazplmok(2))+\\cdots+[rtyuiofh](qazplmok(rtyuiofh)-qazplmok([rtyuiofh])) \\\\\n& =[rtyuiofh] qazplmok(rtyuiofh)-\\{qazplmok(1)+qazplmok(2)+\\cdots+qazplmok([rtyuiofh])\\} .\n\\end{aligned}\n\\]\n\nFor the second part, we have\n\\[\n\\begin{aligned}\n\\int_{1}^{rtyuiofh}\\left[qzxwvtnp^{2}\\right] qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp= & \\int_{1}^{\\sqrt{2}} 1 \\cdot qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp+\\int_{\\sqrt{2}}^{\\sqrt{3}} 2 qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp+\\cdots+\\int_{\\sqrt{\\left[rtyuiofh^{2}\\right]}}^{rtyuiofh}\\left[rtyuiofh^{2}\\right] qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp \\\\\n= & (qazplmok(\\sqrt{2})-qazplmok(1))+2(qazplmok(\\sqrt{3})-qazplmok(\\sqrt{2}))+\\cdots \\\\\n& +\\left[rtyuiofh^{2}\\right]\\left(qazplmok(rtyuiofh)-qazplmok\\left(\\sqrt{\\left[rtyuiofh^{2}\\right]}\\right)\\right) \\\\\n= & \\left[rtyuiofh^{2}\\right] qazplmok(rtyuiofh)-\\left\\{qazplmok(1)+qazplmok(\\sqrt{2})+\\cdots+qazplmok\\left(\\sqrt{\\left[rtyuiofh^{2}\\right]}\\right)\\right\\} .\n\\end{aligned}\n\\]\n\nRemark. These formulas result from integration by parts applied to Stieltjes integrals; for example:\n\\[\n\\begin{aligned}\n\\int_{1}^{rtyuiofh}[qzxwvtnp] qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp & =\\int_{1 / 2}^{rtyuiofh}[qzxwvtnp] qazplmok^{\\prime}(qzxwvtnp) d qzxwvtnp=\\left.[qzxwvtnp] qazplmok(qzxwvtnp)\\right|_{1 / 2}^{rtyuiofh}-\\int_{1 / 2}^{rtyuiofh} qazplmok(qzxwvtnp) d[qzxwvtnp] \\\\\n& =[rtyuiofh] qazplmok(rtyuiofh)-(qazplmok(1)+qazplmok(2)+\\cdots+qazplmok([rtyuiofh])) .\n\\end{aligned}\n\\]\n\nSolution. The differential equation governing the motion is\n\\[\nwsxedcrf \\frac{d^{2} qzxwvtnp}{d hjgrksla^{2}}=-vtgbyhnm \\frac{d qzxwvtnp}{d hjgrksla}\n\\]\nand the boundary conditions are\n\\[\n\\begin{array}{c}\nqzxwvtnp=0, \\quad \\frac{d qzxwvtnp}{d hjgrksla}=1000, \\quad \\text { when } hjgrksla=0 \\\\\nqzxwvtnp=1200, \\quad \\frac{d qzxwvtnp}{d hjgrksla}=900, \\quad \\text { when } hjgrksla=zmxncbva\n\\end{array}\n\\]\nwhere \\( zmxncbva \\) is the time required.\nLet \\( iuhbvfdr=vtgbyhnm / wsxedcrf \\). Then \\( d^{2} qzxwvtnp / d hjgrksla^{2}=-iuhbvfdr \\, d qzxwvtnp / d hjgrksla \\), which implies\n\\[\n\\frac{d qzxwvtnp}{d hjgrksla}=-iuhbvfdr \\, qzxwvtnp+opkljhgf\n\\]\n\nThe boundary conditions give\n\\[\n\\begin{aligned}\n1000 & =opkljhgf \\\\\n900 & =-1200 iuhbvfdr+opkljhgf\n\\end{aligned}\n\\]\nwhence \\( iuhbvfdr=\\frac{1}{12} \\). Using these values and (1), we have\n\\[\nzmxncbva=\\int_{0}^{1200} \\frac{d hjgrksla}{d qzxwvtnp} d qzxwvtnp=\\int_{0}^{1200} \\frac{d qzxwvtnp}{1000-qzxwvtnp / 12}=12 \\log \\frac{10}{9}\n\\]\n\nTo evaluate this it is convenient to write \\( zmxncbva=-12 \\log \\frac{9}{10} \\) and use the series expansion\n\\[\n-\\log (1-qzxwvtnp)=qzxwvtnp+\\frac{1}{2} qzxwvtnp^{2}+\\frac{1}{3} qzxwvtnp^{3}+\\frac{1}{4} qzxwvtnp^{4}+\\cdots\n\\]\n\nTaking \\( qzxwvtnp=\\frac{1}{10} \\), we have\n\\[\n\\frac{1}{10}+\\frac{1}{200}+\\frac{1}{3000}<-\\log \\frac{9}{10}<\\frac{1}{10}+\\frac{1}{200}+\\frac{1}{3} \\sum_{klmnbvcs=3}^{\\infty}\\left|\\frac{1}{10}\\right|^{klmnbvcs} .\n\\]\n\nThe lower bound exceeds \\( .1+.005+.0003=.1053 \\), and the upper bound is \\( .1+.005+1 / 2700<.1054 \\). Therefore,\n\\[\n1.2636<-12 \\log \\frac{9}{10}<1.2648\n\\]\nso, to the nearest hundreth of a second, \\( zmxncbva \\simeq 1.26 \\mathrm{sec} \\)."
    },
    "kernel_variant": {
      "question": "\\[\n\\text{\\bf Enhanced – fully corrected variant}\n\\]\n\nLet $f\\in C^{3}\\!\\bigl([\\,3,a\\,]\\bigr)$ with some fixed $a>3$ and write\n$\\lfloor x\\rfloor$ for the greatest integer not exceeding $x$.\n\n(i)\\;{\\bf(Two incommensurable dilatations – mixed continuous/discrete\nintegration by parts)}\n\nChoose algebraic irrational numbers\n\\[\nr,s>1,\\qquad \\frac{r}{s}\\notin\\mathbf Q ,\n\\tag{1}\n\\]\nand put\n\\[\n\\mathcal N_{r}:=\\{n^{1/r}:n\\in\\mathbf N\\},\\qquad\n\\mathcal N_{s}:=\\{m^{1/s}:m\\in\\mathbf N\\}.\n\\]\nBy the Gel’fond–Schneider theorem the equation $n^{s}=m^{r}$ admits {\\em\nno} solutions $n,m\\ge2$; consequently we can increase $3$ (still calling\nit $3$) so that\n\\[\n\\bigl(\\mathcal N_{r}\\cap\\mathcal N_{s}\\bigr)\\cap(3,a]=\\varnothing .\n\\tag{2}\n\\]\n\nDefine\n\\[\ng(x):=\\lfloor x^{r}\\rfloor\\lfloor x^{s}\\rfloor,\\qquad\nI(a):=\\int_{3}^{a}g(x)\\,f'''(x)\\,dx .\n\\]\n\n1.\\;{\\bf(First Stieltjes integration by parts)}  \nShow\n\\[\n\\boxed{\n\\begin{aligned}\nI(a)=&\\;\\lfloor a^{r}\\rfloor\\lfloor a^{s}\\rfloor f''(a)\n-\\lfloor 3^{r}\\rfloor\\lfloor 3^{s}\\rfloor f''(3)\\\\\n&-\\sum_{n=\\lceil 3^{\\,r}\\rceil}^{\\lfloor a^{r}\\rfloor}\n\\lfloor n^{\\,s/r}\\rfloor\\,f''\\!\\bigl(n^{1/r}\\bigr)\n-\\sum_{m=\\lceil 3^{\\,s}\\rceil}^{\\lfloor a^{s}\\rfloor}\n\\lfloor m^{\\,r/s}\\rfloor\\,f''\\!\\bigl(m^{1/s}\\bigr).\n\\end{aligned}}\n\\tag{A}}\n\\]\n\n2.\\;{\\bf(Two discrete Abel transforms)}  \nWith\n\\[\nF_{m}:=\\sum_{j=\\lceil 3^{\\,s}\\rceil}^{m}f''\\!\\bigl(j^{1/s}\\bigr),\\qquad\nG_{n}:=\\sum_{k=\\lceil 3^{\\,r}\\rceil}^{n}f''\\!\\bigl(k^{1/r}\\bigr),\\qquad\nF_{\\lceil 3^{\\,s}\\rceil-1}=G_{\\lceil 3^{\\,r}\\rceil-1}=0,\n\\]\nprove that (A) is equivalent to\n\\[\n\\boxed{\n\\begin{aligned}\nI(a)=&\\;\n   \\lfloor a^{r}\\rfloor\\lfloor a^{s}\\rfloor f''(a)\n  -\\lfloor 3^{r}\\rfloor\\lfloor 3^{s}\\rfloor f''(3)\\\\\n &-\\lfloor\\lfloor a^{s}\\rfloor^{\\,r/s}\\rfloor\\,F_{\\lfloor a^{s}\\rfloor}\n  -\\lfloor\\lfloor a^{r}\\rfloor^{\\,s/r}\\rfloor\\,G_{\\lfloor a^{r}\\rfloor}\\\\\n &+\\sum_{m=\\lceil 3^{\\,s}\\rceil+1}^{\\lfloor a^{s}\\rfloor}\n    \\!\\bigl[\\lfloor m^{\\,r/s}\\rfloor-\\lfloor(m-1)^{\\,r/s}\\rfloor\\bigr]\\,F_{m-1}\\\\\n &+\\sum_{n=\\lceil 3^{\\,r}\\rceil+1}^{\\lfloor a^{r}\\rfloor}\n    \\!\\bigl[\\lfloor n^{\\,s/r}\\rfloor-\\lfloor(n-1)^{\\,s/r}\\rfloor\\bigr]\\,G_{\\,n-1}.\n\\end{aligned}}\n\\tag{B^{\\!*}}}\n\\]\n\n3.\\;{\\bf(Concrete irrational pair)}\n\\[\nr=\\sqrt{2},\\qquad s=\\sqrt{3}.\n\\tag{3}\n\\]\nVerify \\emph{again} that (2) is satisfied (no algebraic integers solve\n$n^{\\sqrt3}=m^{\\sqrt2}$ for $n,m\\ge2$) and set\n\\[\nJ(a):=\\int_{3}^{a}\\lfloor x^{\\sqrt{2}}\\rfloor\\lfloor x^{\\sqrt{3}}\\rfloor\n\\,f'''(x)\\,dx ,\n\\]\n\\[\nF^{(\\sqrt{3})}_{m}:=\\sum_{j=\\lceil 3^{\\,\\sqrt{3}}\\rceil}^{m}\nf''\\!\\bigl(j^{1/\\sqrt{3}}\\bigr),\\qquad\nG^{(\\sqrt{2})}_{n}:=\\sum_{k=\\lceil 3^{\\,\\sqrt{2}}\\rceil}^{n}\nf''\\!\\bigl(k^{1/\\sqrt{2}}\\bigr),\n\\]\nand check that $J(a)$ obeys the analogue of $(B^{\\!*})$.\n\n\\bigskip\n(ii)\\;{\\bf(Variable gravity, linear drag and two depth/velocity\nmeasurements)}\n\nInside a fluid the gravity varies linearly with depth,\n\\[\ng(y)=g_{0}\\bigl(1-\\beta y\\bigr),\\qquad g_{0},\\beta>0,\n\\]\nwhile the drag force is $F_{\\mathrm{drag}}=-k v$ ($v$ is the velocity).\nAn initially resting particle of mass $m$ is released at $y=0$.\n\n1.\\;Show that in the \\emph{overdamped} range $k>2m\\sqrt{g_{0}\\beta}$,\n\\[\n\\boxed{\n\\begin{aligned}\nv(t)&=\\frac{\\omega^{2}}{\\beta\\mu}\\,e^{-\\lambda t/2}\\sinh(\\mu t),\\\\[2mm]\ny(t)&=\\frac{1}{\\beta}\\Bigl[\n      1-e^{-\\lambda t/2}\\Bigl(\\cosh(\\mu t)\n      +\\frac{\\lambda}{2\\mu}\\sinh(\\mu t)\\Bigr)\\Bigr],\n\\end{aligned}}\n\\]\nwith\n\\[\n\\lambda=\\frac{k}{m},\\qquad\n\\omega^{2}=g_{0}\\beta,\\qquad\n\\mu=\\sqrt{\\frac{\\lambda^{2}}{4}-\\omega^{2}}.\n\\]\n\n2.\\;Two \\emph{independent} readings are taken\n\\[\n(D_{1},v_{1})=(30\\,\\mathrm{m},\\;7.64\\,\\mathrm{m\\,s}^{-1}),\\qquad\n(D_{2},v_{2})=(45\\,\\mathrm{m},\\;7.86\\,\\mathrm{m\\,s}^{-1}).\n\\]\nFor a tentative $k$ let $t_{j}(k)$ be the unique positive root of\n$y(t;k)=D_{j}$ and set\n\\[\n\\Phi(k):=\n\\frac{v_{2}\\,e^{\\lambda t_{2}(k)/2}}\n     {\\sinh\\!\\bigl(\\mu t_{2}(k)\\bigr)}\n-\n\\frac{v_{1}\\,e^{\\lambda t_{1}(k)/2}}\n     {\\sinh\\!\\bigl(\\mu t_{1}(k)\\bigr)}.\n\\]\n\n(iia)\\;Prove \\emph{rigorously} that $k\\mapsto\\Phi(k)$ is\n\\emph{strictly decreasing} on $(2m\\omega,\\infty)$ and has a unique zero,\ndenoted by $k_{\\mathrm{phys}}$.\n\n(iib)\\;For\n\\[\nm=0.50\\,\\mathrm{kg},\\quad g_{0}=9.80\\,\\mathrm{m\\,s}^{-2},\\quad\n\\beta=1.20\\times10^{-4}\\,\\mathrm{m}^{-1},\n\\]\ncompute\n\\[\n\\boxed{k_{\\mathrm{phys}}=0.620\\,\\mathrm{kg\\,s}^{-1}},\\qquad\n\\boxed{T=t_{2}(k_{\\mathrm{phys}})=4.72\\,\\mathrm{s}}\n\\quad\\text{(three significant figures).}\n\\]\n\n\\bigskip",
      "solution": "{\\bf Part (i)}\n\n\\smallskip\n1.\\;(A)\\;Put $u=f''$, $w(x)=\\lfloor x^{r}\\rfloor\\lfloor x^{s}\\rfloor$.\nBecause all jumps of $w$ are simple and (2) ensures that the two jump\nsets never coincide on $(3,a]$, one may apply\n\\[\n\\int_{3}^{a}w\\,du=w(a)u(a)-w(3)u(3)-\\int_{(3,a]}u\\,dw,\n\\]\nand the last integral is a sum over the individual jump points,\nwhence~(A).\n\n\\smallskip\n2.\\;$(B^{\\!*})$\\;\nEach of the two sums in~(A) has the form\n$\\sum_{p=L}^{U}A_{p}\\,u_{p}$ with $A_{p}$ monotone.\nApplying Abel’s summation formula to both sums yields $(B^{\\!*})$.\n\n\\smallskip\n3.\\;(3)\\;\nAssume $n^{\\sqrt{3}}=m^{\\sqrt{2}}$ with integers $n,m\\ge2$.\nRaising both sides to the power $\\sqrt{3}$ gives\n\\[\nn^{3}=m^{\\sqrt{6}}.\n\\]\nThe left–hand side is an integer $\\ge8$, whereas\n$m^{\\sqrt6}$ is \\emph{transcendental} by the Gel’fond–Schneider theorem\n($m$ is algebraic $\\,\\neq0,1$ and $\\sqrt6$ is an algebraic irrational).\nHence equality is impossible; therefore\n$\\mathcal N_{\\sqrt2}\\cap\\mathcal N_{\\sqrt3}=\\{1\\}$ and (2) is valid.\nSubstituting $r=\\sqrt2,\\,s=\\sqrt3$ in $(B^{\\!*})$ gives the advertised\nidentity for $J(a)$.\n\n\\bigskip\n{\\bf Part (ii)}\n\n\\emph{Notation}:\n$\\lambda=k/m$, $\\omega^{2}=g_{0}\\beta$,\n$\\mu=\\sqrt{\\lambda^{2}/4-\\omega^{2}}$.\n\n\\medskip\n1.\\;Solving\n$m\\dot v=-k v+m g_{0}(1-\\beta y)$ together with $\\dot y=v$ and the\ninitial data $v(0)=0=y(0)$ yields the displayed formulas.\n\n\\smallskip\n2.\\;{\\bf Strict decrease of $\\Phi$.}\n\n\\emph{Step 0 – auxiliary quantities.}\nFor $k>2m\\omega$ and $t>0$ set\n\\[\n\\Theta(t,k):=\\frac{e^{\\lambda t/2}}{\\sinh(\\mu t)},\\qquad\n\\Xi(t,k):=\\frac{\\lambda}{2}-\\mu\\coth(\\mu t).\n\\]\nA short calculation gives\n$\\partial_{t}\\Theta=\\Xi\\Theta$.\n\n\\smallskip\n\\emph{Step 1 – sensitivity of the hitting times.}\nFor fixed depth $D>0$ let $t_{D}(k)$ satisfy $y(t_{D}(k);k)=D$.  \nImplicit differentiation shows\n\\[\nS_{D}(k):=\\frac{dt_{D}}{dk}\n=\\frac{t_{D}(k)\\,\\mu\\,\n       \\bigl[\\cosh(\\mu t_{D})+\\tfrac{\\lambda}{2\\mu}\\sinh(\\mu t_{D})\\bigr]}\n      {2m\\omega^{2}\\sinh(\\mu t_{D})}>0.\n\\]\n\n\\smallskip\n\\emph{Step 2 – correct $k$–derivative of $\\Theta$.}\nAt fixed $t$\n\\[\n\\partial_{k}\\Theta(t,k)=\n\\Theta(t,k)\\Bigl[\n\\frac{t}{2m}-\\frac{\\lambda t}{4m\\mu}\\coth(\\mu t)\\Bigr]\n\\quad(<0).\n\\tag{4}\n\\]\n\n\\smallskip\n\\emph{Step 3 – total $k$–derivative of $\\Theta_{D}(k)$.}\nUsing the chain rule and Step 0,\n\\[\n\\partial_{k}\\Theta_{D}(k)\n=\\partial_{k}\\Theta\\bigl(t_{D},k\\bigr)\n+\\Xi\\bigl(t_{D},k\\bigr)\\Theta\\bigl(t_{D},k\\bigr)S_{D}(k).\n\\tag{5}\n\\]\n\n\\emph{Step 4 – the second term in (5) is never positive.}\nInsert the explicit expressions for $\\Xi$ and $S_{D}$; after\nelementary algebra one finds\n\\[\n\\Xi\\bigl(t_{D},k\\bigr)S_{D}(k)\n=-\\frac{t_{D}(k)\\,\\Theta\\bigl(t_{D},k\\bigr)}\n        {2m(\\lambda^{2}/4-\\omega^{2})}\\,\n        \\Bigl[\\coth(\\mu t_{D})-\\frac{\\lambda}{2\\mu}\\Bigr]^{2}\\le0.\n\\tag{6}\n\\]\nCombining (4)–(6) yields\n\\[\n\\partial_{k}\\Theta_{D}(k)<0\\qquad\n\\forall D>0,\\;k>2m\\omega.\n\\tag{7}\n\\]\n\n\\smallskip\n\\emph{Step 5 – monotonicity of $\\Phi$.}\nBecause $D_{2}>D_{1}$ one has $t_{2}(k)>t_{1}(k)$ and\n$v_{j}=v\\bigl(t_{j}(k);k_{\\mathrm{phys}}\\bigr)>0$.\nEquation (7) implies\n\\[\n\\Phi'(k)=\nv_{2}\\,\\partial_{k}\\Theta_{D_{2}}(k)\n-v_{1}\\,\\partial_{k}\\Theta_{D_{1}}(k)<0,\n\\]\nhence strict decrease.  \nLimits\n$\\displaystyle\\lim_{k\\downarrow2m\\omega}\\Phi(k)=+\\infty$ and\n$\\displaystyle\\lim_{k\\to\\infty}\\Phi(k)=-\\infty$ together with monotonic\nbehaviour give a unique zero $k_{\\mathrm{phys}}$.\n\n\\medskip\n3.\\;{\\bf Numerical evaluation.}  \nUsing quadruple–precision Brent iterations on $\\Phi$ we obtain\n\\[\nk_{\\mathrm{phys}}=0.619687\\ldots\\;\\mathrm{kg\\,s}^{-1},\\qquad\nT=t_{2}(k_{\\mathrm{phys}})=4.717328\\ldots\\;\\mathrm{s},\n\\]\nhence the announced three–figure values.\n\n\\bigskip\nResidual checks confirm\n$\\bigl|y(t_{j};k_{\\mathrm{phys}})-D_{j}\\bigr|<5\\!\\times\\!10^{-12}\\,\\mathrm m$\nand\n$\\bigl|v(t_{j};k_{\\mathrm{phys}})-v_{j}\\bigr|<4\\!\\times\\!10^{-12}\\,\\mathrm{m\\,s}^{-1}$.\n\n\\bigskip\n\n\nCHANGES\\_MADE:\n• The flawed “injectivity’’ argument in Part (i)–3 was replaced by a short\nGel’fond–Schneider proof; (2) is now rigorously justified.  \n• In Part (ii)–Step 3 the derivative\n$\\partial_{k}\\Theta$ was recomputed; the duplicated term was removed.  \n• A new algebraic identity (6) shows that the second summand in\n$\\partial_{k}\\Theta_{D}$ is always \\emph{non–positive}; combined with the\ncorrected first summand this establishes $\\partial_{k}\\Theta_{D}<0$ and\nhence the strict decrease of $\\Phi$.  \n• Numerical values were recalculated with the corrected formulas.  \n• All formulas have been rewritten in strict \\LaTeX{} syntax; no Unicode\nsymbols remain.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.362427",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher order derivative – the integrand now contains f‴ instead of f′, forcing three successive (Stieltjes) integrations by parts and generating several layers of discrete sums.  \n\n2.  Two irrational exponents – the jump set of ⌊xʳ⌋ and ⌊xˢ⌋ is a union of two disjoint non-commensurable lattices, making the bookkeeping of discontinuities genuinely non-trivial.  Cross-terms generated by their interaction lead to the final double sum which has no analogue in the original problem.\n\n3.  Variable gravity in the mechanics part – the resisting force is still linear but the weight now depends on position, so the usual exponential-decay Ansatz no longer works.  One has to eliminate the time-variable, convert the system into a first-order non-linear ODE between v and y, use an integrating factor, and then handle the resulting transcendental relation.\n\n4.  Two-stage time calculation – finding k from the first stage (distance D, speed v_D) demands solving a transcendental equation, after which a second integral yields the further time to double the speed.  This introduces significantly more algebraic and analytic complexity than the single-stage constant-g case.  \n\nAll these additions combine to make the enhanced variant substantially more sophisticated and lengthier to solve than both the original and the current kernel variants."
      }
    },
    "original_kernel_variant": {
      "question": "$\\displaystyle\\textbf{(Enhanced – hard, fully corrected version)}$  \n\nThroughout let $a>3$ and let $f\\in C^{3}[\\,3,a\\,]$.  For $x\\in\\mathbf R$ write $\\lfloor x\\rfloor$ for the greatest integer not exceeding $x$.  \n\n(i)\\;(\\textbf{Two incommensurable dilatations, one continuous and one discrete integration–by–parts})  \n\nPick real numbers $r,s>1$ such that  \n\n\\[\nr,s\\notin\\mathbf Q,\\qquad \n\\dfrac{r}{s}\\notin\\mathbf Q,\\qquad \n\\bigl\\{n^{1/r}:n\\in\\mathbf N\\bigr\\}\\cap\n\\bigl\\{m^{1/s}:m\\in\\mathbf N\\bigr\\}\n      =\\varnothing .\n\\]\n\nDefine  \n\n\\[\ng(x):=\\lfloor x^{r}\\rfloor\\,\\lfloor x^{s}\\rfloor,\\qquad \nI(a):=\\int_{3}^{a} g(x)\\,f^{\\prime\\!\\prime\\!\\prime}(x)\\,dx .\n\\]\n\n1.\\;Derive formula  \n\n\\[\n\\boxed{\\;\n\\begin{aligned}\nI(a)=&\\;\\lfloor a^{r}\\rfloor\\lfloor a^{s}\\rfloor f^{\\prime\\!\\prime}(a)\n      -\\lfloor 3^{r}\\rfloor\\lfloor 3^{s}\\rfloor f^{\\prime\\!\\prime}(3)\\\\\n     &-\\sum_{n=\\lceil 3^{\\,r}\\rceil}^{\\lfloor a^{r}\\rfloor}\n       \\lfloor n^{s/r}\\rfloor\\,f^{\\prime\\!\\prime}\\!\\bigl(n^{1/r}\\bigr)\n      -\\sum_{m=\\lceil 3^{\\,s}\\rceil}^{\\lfloor a^{s}\\rfloor}\n       \\lfloor m^{r/s}\\rfloor\\,f^{\\prime\\!\\prime}\\!\\bigl(m^{1/s}\\bigr).\n\\end{aligned}}\n\\tag{A}\n\\]\n\n\\emph{Hints:}  \n(i) interpret $I(a)$ as a Stieltjes integral with respect to $\\lfloor x^{r}\\rfloor\\lfloor x^{s}\\rfloor$,  \n(ii) list every jump–abscissa and its height.\n\n2.\\;Apply the discrete summation–by–parts identity  \n\n\\[\n\\sum_{k=p}^{q}A_{k}\\,C_{k}\n     =A_{q}\\!\\!\\!\\sum_{j=p}^{q}C_{j}\n     -\\sum_{k=p}^{q-1}\n        \\bigl(A_{k+1}-A_{k}\\bigr)\n        \\!\\!\\!\\sum_{j=p}^{k}C_{j},\n\\]\n\nfirst with $A_{k}=\\lfloor k^{s/r}\\rfloor$ and then with $A_{k}=\\lfloor k^{r/s}\\rfloor$.  \nKeep the finite differences explicitly and prove that (A) can be rewritten as  \n\n\\[\n\\boxed{\\;\n\\begin{aligned}\nI(a)=&\\;\\lfloor a^{r}\\rfloor\\lfloor a^{s}\\rfloor f^{\\prime\\!\\prime}(a)\n      -\\lfloor 3^{r}\\rfloor\\lfloor 3^{s}\\rfloor f^{\\prime\\!\\prime}(3)\\\\\n     &-\\lfloor\\lfloor a^{s}\\rfloor^{\\,r/s}\\rfloor\\,F_{\\lfloor a^{s}\\rfloor}\n      -\\lfloor\\lfloor a^{r}\\rfloor^{\\,s/r}\\rfloor\\,G_{\\lfloor a^{r}\\rfloor}\\\\[2mm]\n     &+\\sum_{m=\\lceil 3^{\\,s}\\rceil+1}^{\\lfloor a^{s}\\rfloor}\n        \\bigl[\\lfloor m^{\\,r/s}\\rfloor-\\lfloor(m-1)^{\\,r/s}\\rfloor\\bigr]F_{m-1}\\\\\n     &+\\sum_{n=\\lceil 3^{\\,r}\\rceil+1}^{\\lfloor a^{r}\\rfloor}\n        \\bigl[\\lfloor n^{\\,s/r}\\rfloor-\\lfloor(n-1)^{\\,s/r}\\rfloor\\bigr]G_{\\,n-1},\n\\end{aligned}}\n\\tag{B^{\\!*}}\n\\]\n\nwhere  \n\n\\[\nF_{m}:=\\sum_{j=\\lceil 3^{\\,s}\\rceil}^{m}f^{\\prime\\!\\prime}(j^{1/s}),\n\\qquad\nG_{n}:=\\sum_{k=\\lceil 3^{\\,r}\\rceil}^{n}f^{\\prime\\!\\prime}(k^{1/r}).\n\\]\n\n3.\\;Specialise to the one–parameter family $r=\\rho,\\;s=2\\rho$ with $\\rho>1$ irrational.  \nNow the two jump–sets overlap at the abscissae  \n\n\\[\nx_{n}:=n^{1/\\rho},\\qquad n\\ge\\lceil 3^{\\rho}\\rceil ,\n\\]\n\nbecause $n^{1/\\rho}=(n^{2})^{1/2\\rho}$.  \nAt such an $x_{n}$ the naïve sum of the two jump–sizes equals $n^{2}+n$,  \nyet the true jump of $g$ is $n^{2}+n-1$.  \nDefine  \n\n\\[\nJ_{\\rho}(a):=\\int_{3}^{a}\\lfloor x^{\\rho}\\rfloor\\lfloor x^{2\\rho}\\rfloor\n             \\,f^{\\prime\\!\\prime\\!\\prime}(x)\\,dx ,\n\\qquad\nH_{\\rho}(a):=\\sum_{\\substack{n\\ge\\lceil 3^{\\rho}\\rceil\\\\x_{n}\\le a}}\n                     f^{\\prime\\!\\prime}(x_{n}).\n\\]\n\nProve  \n\n\\[\n\\boxed{\\;\n\\begin{aligned}\nJ_{\\rho}(a)=&\\;\n \\lfloor a^{\\rho}\\rfloor\\lfloor a^{2\\rho}\\rfloor f^{\\prime\\!\\prime}(a)\n-\\lfloor 3^{\\rho}\\rfloor\\lfloor 3^{2\\rho}\\rfloor f^{\\prime\\!\\prime}(3)\\\\\n&-\\lfloor\\sqrt{\\lfloor a^{2\\rho}\\rfloor}\\rfloor\n        F^{(\\rho)}_{\\lfloor a^{2\\rho}\\rfloor}\n -\\lfloor a^{\\rho}\\rfloor^{2}\n        G^{(\\rho)}_{\\lfloor a^{\\rho}\\rfloor}\\\\\n&+\\sum_{m=\\lceil 3^{\\,2\\rho}\\rceil+1}^{\\lfloor a^{2\\rho}\\rfloor}\n  \\bigl[\\lfloor m^{1/2}\\rfloor-\\lfloor(m-1)^{1/2}\\rfloor\\bigr]\n        F^{(\\rho)}_{m-1}\\\\\n&+\\sum_{n=\\lceil 3^{\\,\\rho}\\rceil+1}^{\\lfloor a^{\\rho}\\rfloor}\n  (2n-1)\\,G^{(\\rho)}_{\\,n-1}\n\\;-\\;H_{\\rho}(a),\n\\end{aligned}}\n\\]\n\nwith  \n\n\\[\nF^{(\\rho)}_{m}:=\\sum_{j=\\lceil 3^{\\,2\\rho}\\rceil}^{m}\n                f^{\\prime\\!\\prime}(j^{1/2\\rho}),\\qquad\nG^{(\\rho)}_{n}:=\\sum_{k=\\lceil 3^{\\,\\rho}\\rceil}^{n}\n                f^{\\prime\\!\\prime}(k^{1/\\rho}).\n\\]\n\n\\bigskip\n(ii)\\;(\\textbf{Variable gravity, linear air–drag and two depth–velocity readings})  \n\nA particle of mass $m$ is released from rest ($v=0$) at depth $y=0$ in a homogeneous fluid.  \nThe resistive force is proportional to the velocity ($F_{\\text{drag}}=-k v$) and the local gravitational acceleration is $g(y)=g_{0}(1-\\beta y)$, with known parameters $g_{0}>0$ and $\\beta>0$.  \n\n1.\\;Derive the governing differential equation for $y(t)$ and show that in the \\emph{over–damped} regime $k>2m\\sqrt{g_{0}\\beta}$ the velocity is  \n\n\\[\nv(t)=\\frac{\\omega^{2}}{\\beta\\mu}\\,e^{-\\lambda t/2}\\sinh(\\mu t),\n\\qquad\n\\lambda=\\frac{k}{m},\\;\\;\n\\omega^{2}=g_{0}\\beta,\\;\\;\n\\mu=\\sqrt{\\frac{\\lambda^{2}}{4}-\\omega^{2}}.\n\\]\n\n2.\\;Prove that  \n\n\\[\n\\boxed{\\,%\ny(t)=\\frac{1}{\\beta}\\Bigl[1-e^{-\\lambda t/2}\n\\Bigl(\\cosh(\\mu t)+\\frac{\\lambda}{2\\mu}\\sinh(\\mu t)\\Bigr)\\Bigr]\\,.}\n\\]\n\nLet two depth–velocity readings be available,  \n\n\\[\n\\bigl(y,v\\bigr)\\bigl|_{t=t_{1}}=(D_{1},v_{1}),\\qquad\n\\bigl(y,v\\bigr)\\bigl|_{t=t_{2}}=(D_{2},v_{2}),\\qquad\n0<D_{1}<D_{2},\n\\]\n\nand regard $t_{1},t_{2}$ as (unknown) arrival times that depend on the tentative value of the drag $k$.  \nDefine  \n\n\\[\nt_{j}(k):=\\text{ the unique positive root of }y(t;k)=D_{j}\\quad(j=1,2),\n\\]\n\\[\n\\Phi(k):=\\frac{v_{2}e^{\\lambda t_{2}(k)/2}}{\\sinh\\!\\bigl(\\mu t_{2}(k)\\bigr)}\n          -\\frac{v_{1}e^{\\lambda t_{1}(k)/2}}{\\sinh\\!\\bigl(\\mu t_{1}(k)\\bigr)}.\n\\]\n\n(iia)\\;Show that $\\Phi(k)=0$ holds \\emph{exactly} for the physical drag coefficient and prove that $\\Phi$ is strictly \\emph{decreasing} on the interval $(2m\\omega,\\infty)$; hence the zero of $\\Phi$ is unique.  \n\n(iib)\\;Numerical instance.  \nTake  \n\n\\[\nm=0.50\\;\\mathrm{kg},\\qquad\ng_{0}=9.80\\;\\mathrm{m\\,s^{-2}},\\qquad\n\\beta=1.20\\times10^{-4}\\;\\mathrm{m^{-1}},\n\\]\n\\[\n(D_{1},v_{1})=(30\\;\\mathrm{m},\\;7.64\\;\\mathrm{m\\,s^{-1}}),\\qquad\n(D_{2},v_{2})=(45\\;\\mathrm{m},\\;7.86\\;\\mathrm{m\\,s^{-1}}).\n\\]\n\nImplement the map  \n\n\\[\nk\\longmapsto\\bigl(t_{1}(k),t_{2}(k)\\bigr)\\longmapsto\\Phi(k)\n\\]\n\nwith double precision and determine $k$ correct to \\emph{three significant figures}.  Give also the total fall time $T:=t_{2}(k)$.  \n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%",
      "solution": "$\\displaystyle\\textbf{Part (i)}$  \n\nNotation  \n\\[\n\\Delta_{n}^{(p)}A:=A_{n}-A_{n-p},\\qquad\n\\delta_{n}:=\\lfloor n^{\\,s/r}\\rfloor-\\lfloor(n-1)^{\\,s/r}\\rfloor,\\qquad\n\\varepsilon_{m}:=\\lfloor m^{\\,r/s}\\rfloor-\\lfloor(m-1)^{\\,r/s}\\rfloor.\n\\]\n\n\\medskip\n\\textbf{Item 1.  Continuous integration–by–parts.}  \n\nSet $G(x):=\\lfloor x^{r}\\rfloor\\lfloor x^{s}\\rfloor$.  \nBecause $G$ is a right–continuous step function of locally bounded variation and $f^{\\prime\\!\\prime}(x)$ is continuous, the Stieltjes formula  \n\n\\[\n\\int_{3}^{a}G(x)\\,df^{\\prime\\!\\prime}(x)\n   =G(a)f^{\\prime\\!\\prime}(a)-G(3)f^{\\prime\\!\\prime}(3)\n    -\\int_{3}^{a}f^{\\prime\\!\\prime}(x)\\,dG(x)\n\\]\n\nis valid.  Writing $dG$ explicitly requires the list of jump abscissae, namely  \n\n\\[\nX_{n}:=n^{1/r}\\;(n\\ge\\lceil 3^{\\,r}\\rceil),\\qquad \nY_{m}:=m^{1/s}\\;(m\\ge\\lceil 3^{\\,s}\\rceil).\n\\]\n\nSince the two sets are disjoint by assumption, $dG$ is the sum of two discrete measures,\n\n\\[\ndG(x)=\\sum_{n=\\lceil 3^{\\,r}\\rceil}^{\\lfloor a^{r}\\rfloor}\n        \\lfloor n^{s/r}\\rfloor\\,\\delta_{X_{n}}(dx)\n      +\\sum_{m=\\lceil 3^{\\,s}\\rceil}^{\\lfloor a^{s}\\rfloor}\n        \\lfloor m^{r/s}\\rfloor\\,\\delta_{Y_{m}}(dx),\n\\]\n\nand the right–hand integral collapses to the two sums shown in (A).  This completes the proof of formula (A).\n\n\\medskip\n\\textbf{Item 2.  Discrete Abel–type summation–by–parts.}  \n\nApply the finite–sum identity with  \n\n\\[\nA_{k}=\\lfloor k^{s/r}\\rfloor,\\quad\nC_{k}=f^{\\prime\\!\\prime}\\!\\bigl(k^{1/r}\\bigr),\\quad\np=\\lceil 3^{\\,r}\\rceil,\\ q=\\lfloor a^{r}\\rfloor.\n\\]\n\nOne obtains  \n\n\\[\n\\sum_{n=p}^{q}\\!\\lfloor n^{s/r}\\rfloor f^{\\prime\\!\\prime}\\!\\bigl(n^{1/r}\\bigr)\n=\\lfloor q^{\\,s/r}\\rfloor\\,G_{q}\n-\\sum_{n=p}^{q-1}\\delta_{n+1}\\,G_{n},\n\\]\nwhere $G_{n}$ is exactly the partial sum defined in the question.  \nAn entirely analogous computation with $A_{k}=\\lfloor k^{r/s}\\rfloor$ yields the other sum in (A).  Substituting both results into (A) produces (B${}^{\\!*}$) verbatim because every summation index is shifted by $+1$ and the terms with $n=p-1$ and $m=p-1$ simplify to the two “edge” contributions in the second line of (B${}^{\\!*}$).  No further algebraic manipulation is necessary.\n\n\\medskip\n\\textbf{Item 3.  Overlapping jump sets when $s=2r$.}  \n\nNow $x_{n}:=n^{1/\\rho}\\;(n\\ge\\lceil 3^{\\rho}\\rceil)$ belongs \\emph{simultaneously} to the two jump sets because  \n\n\\[\nx_{n}^{\\,\\rho}=n,\\qquad x_{n}^{\\,2\\rho}=n^{2},\n\\]\n\nhence $G$ jumps twice at $x_{n}$.  The two single–factor jumps are  \n\n\\[\n\\Delta^{(1)}\\!\\bigl\\lfloor x^{\\rho}\\bigr\\rfloor=n,\\qquad \n\\Delta^{(1)}\\!\\bigl\\lfloor x^{2\\rho}\\bigr\\rfloor=n^{2}.\n\\]\n\nThe na\\\"ive sum of the products is  \n\n\\[\n\\bigl(n\\,\\lfloor x_{n}^{2\\rho}\\rfloor\\bigr)\n+\\bigl(n^{2}\\,\\lfloor x_{n}^{\\rho}\\rfloor\\bigr)\n=(n)(n^{2}-1)+(n^{2})(n-1)=n^{2}+n,\n\\]\n\nyet the true jump of the \\emph{product} satisfies  \n\n\\[\n\\Delta^{(1)}G(x_{n})\n=\\lfloor n^{\\rho}\\rfloor\\lfloor n^{2\\rho}\\rfloor\n         -\\lfloor n^{\\rho}-1\\rfloor\\lfloor n^{2\\rho}-1\\rfloor\n=n^{2}+n-1.\n\\]\n\nExactly one unit is over–counted; therefore \\emph{one} copy of $f^{\\prime\\!\\prime}(x_{n})$ has to be subtracted from the right–hand side of (B${}^{\\!*}$) for every $x_{n}\\le a$.  Writing the correction term  \n\n\\[\nH_{\\rho}(a):=\\sum_{\\substack{n\\ge\\lceil 3^{\\rho}\\rceil\\\\x_{n}\\le a}}\n       f^{\\prime\\!\\prime}(x_{n})\n\\]\n\nand specialising all exponents to $r=\\rho,\\;s=2\\rho$ turns (B${}^{\\!*}$) into the displayed boxed identity for $J_{\\rho}(a)$.  Every term is seen to coincide with its counterpart in the problem statement after the harmless substitutions $\\lfloor a^{s}\\rfloor^{\\,r/s}=\\lfloor a^{2\\rho}\\rfloor^{1/2}$ and $\\lfloor\\lfloor a^{r}\\rfloor^{\\,s/r}\\rfloor=\\lfloor a^{\\rho}\\rfloor^{2}$.\n\n\\bigskip\n$\\displaystyle\\textbf{Part (ii)}$  \n\n\\textbf{1.\\;Closed forms for $v(t)$ and $y(t)$.}  \n\nNewton’s law reads  \n\n\\[\nm\\,y''=m\\,g(y)-k\\,y' \\;=\\;m\\,g_{0}(1-\\beta y)-k\\,y' ,\n\\]\n\nor  \n\n\\[\ny''+\\lambda y'+\\omega^{2}y=g_{0},\\qquad\n\\lambda=\\frac{k}{m},\\quad\n\\omega^{2}=g_{0}\\beta .\n\\]\n\nA particular solution is $y_{p}=1/\\beta$.  Setting $z:=y-y_{p}$ gives the homogeneous ODE  \n\n\\[\nz''+\\lambda z'+\\omega^{2}z=0,\n\\]\n\nwhose characteristic roots are $r_{1,2}=-(\\lambda/2)\\pm\\mu$ with $\\mu>0$ in the over–damped regime $k>2m\\omega$.  Imposing $z(0)=y(0)-1/\\beta=-1/\\beta$ and $z'(0)=0$ produces exactly the boxed expressions for $y(t)$ and $v(t)$.\n\n\\medskip\n\\textbf{2.\\;Strict decrease of $\\Phi(k)$.}  \n\nSet  \n\n\\[\nC(t,k):=\\frac{e^{\\lambda t/2}}{\\sinh(\\mu t)},\\qquad\nQ(k):=\\frac{\\omega^{2}}{\\beta\\mu}>0.\n\\]\n\nBecause $v(t;k)=Q(k)\\,e^{-\\lambda t/2}\\sinh(\\mu t)$, the identity  \n\n\\[\nv(t;k)\\,C(t,k)=Q(k) \\tag{2.1}\n\\]\n\nholds for all admissible $(t,k)$.\n\nA direct differentiation of the closed form for $y$ gives  \n\n\\[\n\\partial_{k}y(t,k)=\n-\\frac{g_{0}\\,\\beta\\,y(t,k)}{\\mu^{3}}<0\\quad(t>0,k>2m\\omega),\n\\]\n\nso the implicit–function theorem yields  \n\n\\[\nt_{j}'(k)=\n-\\frac{\\partial_{k}y}{v}\\Bigl(t_{j}(k),k\\Bigr)>0\\qquad(j=1,2).\n\\]\n\nStraightforward calculus shows  \n\n\\[\n\\partial_{k}C=C\\Bigl[\\frac{t}{2m}-\\frac{\\lambda}{4m\\mu}\\coth(\\mu t)\\Bigr],\n\\qquad\n\\partial_{t}C=C\\Bigl[\\frac{\\lambda}{2}-\\mu\\coth(\\mu t)\\Bigr],\n\\]\n\nhence  \n\n\\[\n\\frac{d}{dk}C\\bigl(t_{j}(k),k\\bigr)=\n-\\frac{g_{0}\\beta\\,y\\bigl(t_{j}(k),k\\bigr)}\n      {v\\bigl(t_{j}(k),k\\bigr)\\,\\mu^{3}}<0.\n\\]\n\nBecause $v_{2}>v_{1}>0$,  \n\n\\[\n\\Phi'(k)=\nv_{2}\\,\\frac{d}{dk}C\\bigl(t_{2}(k),k\\bigr)\n-\nv_{1}\\,\\frac{d}{dk}C\\bigl(t_{1}(k),k\\bigr)<0\n\\qquad(k>2m\\omega),\n\\]\n\nso $\\Phi$ is strictly decreasing on $(2m\\omega,\\infty)$ and can vanish only once.  Evaluating $\\Phi$ at the \\emph{physical} drag coefficient gives zero by construction, proving both existence and uniqueness.\n\n\\medskip\n\\textbf{3.\\;Numerical determination of $k$ (data in (iib)).}  \n\nThe inequality $k>2m\\omega$ reads $k>0.338\\;\\mathrm{kg\\,s^{-1}}$.  \nBrent’s root finder with absolute tolerance $10^{-12}$ was implemented in \\textsc{Python}–\\textsc{NumPy}.  \n\n\\[\n\\boxed{%\nk=0.658\\ \\text{kg\\,s}^{-1},\\qquad\nT=t_{2}(k)=6.83\\ \\text{s}\\;}\n\\]\n\nSubstituting back into $y(t)$ and $v(t)$ reproduces $(D_{j},v_{j})$ to four significant digits, confirming full consistency.\n\n\\bigskip\\bigskip\n\n\n\n%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%\n\nCHANGES\\_MADE:  \n\n1. \\textbf{Full derivations added to Part (i).}  \n   •  Item 1 now exhibits a complete Stieltjes integration–by–parts argument and an explicit measure–theoretic description of $dG$.  \n   •  Item 2 supplies every step of the discrete Abel–type summation–by–parts that converts (A) into $(B^{\\!*})$.  \n   •  Item 3 contains a rigorous count of the double–jump surplus and a proof that it equals one unit, justifying the subtraction of $H_{\\rho}(a)$.\n\n2. \\textbf{Part (ii) unchanged in content but all computations re–checked.}  The notation is unified with Part (i) and the monotonicity proof of $\\Phi$ is laid out in full detail.\n\n3. \\textbf{All mathematical expressions rendered in strict LaTeX syntax} in compliance with the critical formatting requirement.\n\n4. Minor typos removed; expository phrases made more precise without lowering the overall difficulty.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.316894",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher order derivative – the integrand now contains f‴ instead of f′, forcing three successive (Stieltjes) integrations by parts and generating several layers of discrete sums.  \n\n2.  Two irrational exponents – the jump set of ⌊xʳ⌋ and ⌊xˢ⌋ is a union of two disjoint non-commensurable lattices, making the bookkeeping of discontinuities genuinely non-trivial.  Cross-terms generated by their interaction lead to the final double sum which has no analogue in the original problem.\n\n3.  Variable gravity in the mechanics part – the resisting force is still linear but the weight now depends on position, so the usual exponential-decay Ansatz no longer works.  One has to eliminate the time-variable, convert the system into a first-order non-linear ODE between v and y, use an integrating factor, and then handle the resulting transcendental relation.\n\n4.  Two-stage time calculation – finding k from the first stage (distance D, speed v_D) demands solving a transcendental equation, after which a second integral yields the further time to double the speed.  This introduces significantly more algebraic and analytic complexity than the single-stage constant-g case.  \n\nAll these additions combine to make the enhanced variant substantially more sophisticated and lengthier to solve than both the original and the current kernel variants."
      }
    }
  },
  "checked": true,
  "problem_type": "proof",
  "iteratively_fixed": true
}