{
  "index": "1941-A-2",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "2. Find the \\( n \\)th derivative with respect to \\( x \\) of \\( \\int_{0}^{x}\\left[1+\\frac{(x-t)}{1!}+\\frac{(x-t)^{2}}{2!}+\\cdots+\\frac{(x-t)^{n-1}}{(n-1)!}\\right] e^{n t} d t \\).",
  "solution": "First Solution. Let\n\\[\n\\phi_{k}(x)=\\int_{0}^{x} \\frac{(x-t)^{k}}{k!} e^{m t} d t .\n\\]\n\nThen, for \\( k>0 \\),\n\\[\n\\phi_{k}{ }^{\\prime}(x)=\\phi_{k-1}(x) .\n\\]\n\nAlso\n\\[\n\\phi_{0}(x)=\\int_{0}^{x} e^{n t} d t=\\frac{e^{n x}-1}{n} .\n\\]\n\nTherefore\n\\[\n\\left(\\frac{d}{d x}\\right)^{n} \\phi_{k}(x)=\\left(\\frac{d}{d x}\\right)^{n-k} \\phi_{0}(x)=n^{n-k-1} e^{n x} \\text { for } n>k .\n\\]\n\nAccordingly, the \\( n \\)th derivative of the given function is\n\\[\n\\begin{aligned}\n\\left(\\frac{d}{d x}\\right)^{n} & {\\left[\\phi_{0}(x)+\\phi_{1}(x)+\\cdots+\\phi_{n-1}(x)\\right] } \\\\\n& =\\left[n^{n-1}+n^{n-2}+\\cdots+1\\right] e^{n x} \\\\\n& =\\left\\{\\begin{array}{l}\n\\frac{n^{n}-1}{n-1} e^{n x} \\text { for } n \\neq 1, \\\\\ne^{x} \\text { for } n=1\n\\end{array}\\right.\n\\end{aligned}\n\\]\n\nSecond Solution. Let\n\\[\nf(x)=\\int_{0}^{x}\\left(1+\\frac{(x-t)}{1!}+\\frac{(x-t)^{2}}{2!}+\\cdots+\\frac{(x-t)^{n-1}}{(n-1)!}\\right) e^{n t} d t .\n\\]\n\nSubstituting \\( t=x-u \\), we get \\( f(x)=e^{n x} \\psi(x) \\), where\n\\[\n\\psi(x)=\\int_{0}^{x}\\left(1+\\frac{u}{1!}+\\frac{u^{2}}{2!}+\\cdots+\\frac{u^{n-1}}{(n-1)!}\\right) e^{-n u} d u\n\\]\n\nThen\n\\[\n\\psi^{\\prime}(x)=\\left(1+\\frac{x}{1!}+\\frac{x^{2}}{2!}+\\cdots+\\frac{x^{n-1}}{(n-1)!}\\right) e^{-n x}\n\\]\nand\n\\[\nf^{\\prime}(x)-n f(x)=e^{n x} \\psi^{\\prime}(x),\n\\]\nwhere the right member is a polynomial of degree \\( n-1 \\).\nThe general solution of the differential equation (1) is\n\\[\nf(x)=C e^{n x}+P\n\\]\nwhere \\( P \\) is another polynomial of degree \\( \\boldsymbol{n}-1 \\). It follows that\n\\[\nf^{(n)}(x)=C n^{n} e^{n x},\n\\]\nand it remains to find \\( C \\).\nWe have\n\\[\n\\begin{aligned}\nC & =\\lim _{x \\rightarrow \\infty} f(x) e^{-n x}=\\lim _{x \\rightarrow \\infty} \\psi(x) \\\\\n& =\\int_{0}^{\\infty}\\left(1+\\frac{u}{1!}+\\frac{u^{2}}{2!}+\\cdots+\\frac{u^{n-1}}{(n-1)!}\\right) e^{-n u} d u \\\\\n& =\\frac{1}{n}+\\frac{1}{n^{2}}+\\cdots+\\frac{1}{n^{n}} .\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\nf^{(n)}(x) & =\\left(n^{n-1}+n^{n-2}+\\cdots+1\\right) e^{n x} \\\\\n& =\\left\\{\\begin{array}{ll}\n\\frac{n^{n}-1}{n-1} e^{n x}, \\quad \\text { if } n>1, \\\\\ne^{x}, \\quad \\text { if } n=1 .\n\\end{array}\\right.\n\\end{aligned}\n\\]",
  "vars": [
    "x",
    "t",
    "u",
    "k",
    "f",
    "\\\\phi_k",
    "\\\\psi"
  ],
  "params": [
    "n",
    "m",
    "C",
    "P"
  ],
  "sci_consts": [
    "e"
  ],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "abscissa",
        "t": "tempvar",
        "u": "auxvar",
        "k": "indexer",
        "f": "function",
        "\\phi_k": "phikern",
        "\\psi": "psifunc",
        "n": "degree",
        "m": "factor",
        "C": "ceevalue",
        "P": "polynom"
      },
      "question": "2. Find the \\( degree \\)th derivative with respect to \\( abscissa \\) of \\( \\int_{0}^{abscissa}\\left[1+\\frac{(abscissa-tempvar)}{1!}+\\frac{(abscissa-tempvar)^{2}}{2!}+\\cdots+\\frac{(abscissa-tempvar)^{degree-1}}{(degree-1)!}\\right] e^{degree tempvar} d tempvar \\).",
      "solution": "First Solution. Let\n\\[\nphikern_{indexer}(abscissa)=\\int_{0}^{abscissa} \\frac{(abscissa-tempvar)^{indexer}}{indexer!} e^{factor tempvar} d tempvar .\n\\]\n\nThen, for \\( indexer>0 \\),\n\\[\nphikern_{indexer}{ }^{\\prime}(abscissa)=phikern_{indexer-1}(abscissa) .\n\\]\n\nAlso\n\\[\nphikern_{0}(abscissa)=\\int_{0}^{abscissa} e^{degree tempvar} d tempvar=\\frac{e^{degree abscissa}-1}{degree} .\n\\]\n\nTherefore\n\\[\n\\left(\\frac{d}{d abscissa}\\right)^{degree} phikern_{indexer}(abscissa)=\\left(\\frac{d}{d abscissa}\\right)^{degree-indexer} phikern_{0}(abscissa)=degree^{degree-indexer-1} e^{degree abscissa} \\text { for } degree>indexer .\n\\]\n\nAccordingly, the \\( degree \\)th derivative of the given function is\n\\[\n\\begin{aligned}\n\\left(\\frac{d}{d abscissa}\\right)^{degree} & {\\left[phikern_{0}(abscissa)+phikern_{1}(abscissa)+\\cdots+phikern_{degree-1}(abscissa)\\right] } \\\\\n& =\\left[degree^{degree-1}+degree^{degree-2}+\\cdots+1\\right] e^{degree abscissa} \\\\\n& =\\left\\{\\begin{array}{l}\n\\frac{degree^{degree}-1}{degree-1} e^{degree abscissa} \\text { for } degree \\neq 1, \\\\\ne^{abscissa} \\text { for } degree=1\n\\end{array}\\right.\n\\end{aligned}\n\\]\n\nSecond Solution. Let\n\\[\nfunction(abscissa)=\\int_{0}^{abscissa}\\left(1+\\frac{(abscissa-tempvar)}{1!}+\\frac{(abscissa-tempvar)^{2}}{2!}+\\cdots+\\frac{(abscissa-tempvar)^{degree-1}}{(degree-1)!}\\right) e^{degree tempvar} d tempvar .\n\\]\n\nSubstituting \\( tempvar=abscissa-auxvar \\), we get \\( function(abscissa)=e^{degree abscissa} psifunc(abscissa) \\), where\n\\[\npsifunc(abscissa)=\\int_{0}^{abscissa}\\left(1+\\frac{auxvar}{1!}+\\frac{auxvar^{2}}{2!}+\\cdots+\\frac{auxvar^{degree-1}}{(degree-1)!}\\right) e^{-degree auxvar} d auxvar\n\\]\n\nThen\n\\[\npsifunc^{\\prime}(abscissa)=\\left(1+\\frac{abscissa}{1!}+\\frac{abscissa^{2}}{2!}+\\cdots+\\frac{abscissa^{degree-1}}{(degree-1)!}\\right) e^{-degree abscissa}\n\\]\nand\n\\[\nfunction^{\\prime}(abscissa)-degree function(abscissa)=e^{degree abscissa} psifunc^{\\prime}(abscissa),\n\\]\nwhere the right member is a polynomial of degree \\( degree-1 \\).\nThe general solution of the differential equation (1) is\n\\[\nfunction(abscissa)=ceevalue e^{degree abscissa}+polynom\n\\]\nwhere \\( polynom \\) is another polynomial of degree \\( \\boldsymbol{degree}-1 \\). It follows that\n\\[\nfunction^{(degree)}(abscissa)=ceevalue degree^{degree} e^{degree abscissa},\n\\]\nand it remains to find \\( ceevalue \\).\nWe have\n\\[\n\\begin{aligned}\nceevalue & =\\lim _{abscissa \\rightarrow \\infty} function(abscissa) e^{-degree abscissa}=\\lim _{abscissa \\rightarrow \\infty} psifunc(abscissa) \\\\\n& =\\int_{0}^{\\infty}\\left(1+\\frac{auxvar}{1!}+\\frac{auxvar^{2}}{2!}+\\cdots+\\frac{auxvar^{degree-1}}{(degree-1)!}\\right) e^{-degree auxvar} d auxvar \\\\\n& =\\frac{1}{degree}+\\frac{1}{degree^{2}}+\\cdots+\\frac{1}{degree^{degree}} .\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\nfunction^{(degree)}(abscissa) & =\\left(degree^{degree-1}+degree^{degree-2}+\\cdots+1\\right) e^{degree abscissa} \\\\\n& =\\left\\{\\begin{array}{ll}\n\\frac{degree^{degree}-1}{degree-1} e^{degree abscissa}, \\quad \\text { if } degree>1, \\\\\ne^{abscissa}, \\quad \\text { if } degree=1 .\n\\end{array}\\right.\n\\end{aligned}\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "labyrinth",
        "t": "mosaicron",
        "u": "parallax",
        "k": "driftwood",
        "f": "tapestry",
        "\\\\phi_k": "farthingale",
        "\\\\psi": "chandelier",
        "n": "quagmire",
        "m": "flagstaff",
        "C": "whispering",
        "P": "breadcrumb"
      },
      "question": "Problem:\n<<<\n2. Find the \\( quagmire \\)th derivative with respect to \\( labyrinth \\) of \\( \\int_{0}^{labyrinth}\\left[1+\\frac{(labyrinth-mosaicron)}{1!}+\\frac{(labyrinth-mosaicron)^{2}}{2!}+\\cdots+\\frac{(labyrinth-mosaicron)^{quagmire-1}}{(quagmire-1)!}\\right] e^{quagmire mosaicron} d mosaicron \\).\n>>>",
      "solution": "Solution:\n<<<\nFirst Solution. Let\n\\[\nfarthingale_{driftwood}(labyrinth)=\\int_{0}^{labyrinth} \\frac{(labyrinth-mosaicron)^{driftwood}}{driftwood!} e^{flagstaff mosaicron} d mosaicron .\n\\]\n\nThen, for \\( driftwood>0 \\),\n\\[\nfarthingale_{driftwood}{ }^{\\prime}(labyrinth)=farthingale_{driftwood-1}(labyrinth) .\n\\]\n\nAlso\n\\[\nfarthingale_{0}(labyrinth)=\\int_{0}^{labyrinth} e^{quagmire mosaicron} d mosaicron=\\frac{e^{quagmire labyrinth}-1}{quagmire} .\n\\]\n\nTherefore\n\\[\n\\left(\\frac{d}{d labyrinth}\\right)^{quagmire} farthingale_{driftwood}(labyrinth)=\\left(\\frac{d}{d labyrinth}\\right)^{quagmire-driftwood} farthingale_{0}(labyrinth)=quagmire^{quagmire-driftwood-1} e^{quagmire labyrinth} \\text { for } quagmire>driftwood .\n\\]\n\nAccordingly, the \\( quagmire \\)th derivative of the given function is\n\\[\n\\begin{aligned}\n\\left(\\frac{d}{d labyrinth}\\right)^{quagmire} & {\\left[farthingale_{0}(labyrinth)+farthingale_{1}(labyrinth)+\\cdots+farthingale_{quagmire-1}(labyrinth)\\right] } \\\\\n& =\\left[quagmire^{quagmire-1}+quagmire^{quagmire-2}+\\cdots+1\\right] e^{quagmire labyrinth} \\\\\n& =\\left\\{\\begin{array}{l}\n\\frac{quagmire^{quagmire}-1}{quagmire-1} e^{quagmire labyrinth} \\text { for } quagmire \\neq 1, \\\\\ne^{labyrinth} \\text { for } quagmire=1\n\\end{array}\\right.\n\\end{aligned}\n\\]\n\nSecond Solution. Let\n\\[\ntapestry(labyrinth)=\\int_{0}^{labyrinth}\\left(1+\\frac{(labyrinth-mosaicron)}{1!}+\\frac{(labyrinth-mosaicron)^{2}}{2!}+\\cdots+\\frac{(labyrinth-mosaicron)^{quagmire-1}}{(quagmire-1)!}\\right) e^{quagmire mosaicron} d mosaicron .\n\\]\n\nSubstituting \\( mosaicron=labyrinth-parallax \\), we get \\( tapestry(labyrinth)=e^{quagmire labyrinth} chandelier(labyrinth) \\), where\n\\[\nchandelier(labyrinth)=\\int_{0}^{labyrinth}\\left(1+\\frac{parallax}{1!}+\\frac{parallax^{2}}{2!}+\\cdots+\\frac{parallax^{quagmire-1}}{(quagmire-1)!}\\right) e^{-quagmire parallax} d parallax\n\\]\n\nThen\n\\[\nchandelier^{\\prime}(labyrinth)=\\left(1+\\frac{labyrinth}{1!}+\\frac{labyrinth^{2}}{2!}+\\cdots+\\frac{labyrinth^{quagmire-1}}{(quagmire-1)!}\\right) e^{-quagmire labyrinth}\n\\]\nand\n\\[\ntapestry^{\\prime}(labyrinth)-quagmire tapestry(labyrinth)=e^{quagmire labyrinth} chandelier^{\\prime}(labyrinth),\n\\]\nwhere the right member is a polynomial of degree \\( quagmire-1 \\).\nThe general solution of the differential equation (1) is\n\\[\ntapestry(labyrinth)=whispering e^{quagmire labyrinth}+breadcrumb\n\\]\nwhere \\( breadcrumb \\) is another polynomial of degree \\( \\boldsymbol{quagmire}-1 \\). It follows that\n\\[\ntapestry^{(quagmire)}(labyrinth)=whispering \\, quagmire^{quagmire} e^{quagmire labyrinth},\n\\]\nand it remains to find \\( whispering \\).\nWe have\n\\[\n\\begin{aligned}\nwhispering & =\\lim _{labyrinth \\rightarrow \\infty} tapestry(labyrinth) e^{-quagmire labyrinth}=\\lim _{labyrinth \\rightarrow \\infty} chandelier(labyrinth) \\\\\n& =\\int_{0}^{\\infty}\\left(1+\\frac{parallax}{1!}+\\frac{parallax^{2}}{2!}+\\cdots+\\frac{parallax^{quagmire-1}}{(quagmire-1)!}\\right) e^{-quagmire \\, parallax} d parallax \\\\\n& =\\frac{1}{quagmire}+\\frac{1}{quagmire^{2}}+\\cdots+\\frac{1}{quagmire^{quagmire}} .\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\ntapestry^{(quagmire)}(labyrinth) & =\\left(quagmire^{quagmire-1}+quagmire^{quagmire-2}+\\cdots+1\\right) e^{quagmire labyrinth} \\\\\n& =\\left\\{\\begin{array}{ll}\n\\frac{quagmire^{quagmire}-1}{quagmire-1} e^{quagmire labyrinth}, \\quad \\text { if } quagmire>1, \\\\\ne^{labyrinth}, \\quad \\text { if } quagmire=1 .\n\\end{array}\\right.\n\\end{aligned}\n\\]\n>>>"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "ycoordinate",
        "t": "staticspace",
        "u": "fixpoint",
        "k": "floatindex",
        "f": "constantval",
        "\\phi_k": "failurefunc",
        "\\psi": "clarityfunc",
        "n": "nothingness",
        "m": "emptiness",
        "C": "variable",
        "P": "transcendent"
      },
      "question": "2. Find the \\( nothingness \\)th derivative with respect to \\( ycoordinate \\) of \\( \\int_{0}^{ycoordinate}\\left[1+\\frac{(ycoordinate-staticspace)}{1!}+\\frac{(ycoordinate-staticspace)^{2}}{2!}+\\cdots+\\frac{(ycoordinate-staticspace)^{nothingness-1}}{(nothingness-1)!}\\right] e^{nothingness\\,staticspace} d\\,staticspace \\).",
      "solution": "First Solution. Let\n\\[\nfailurefunc_{floatindex}(ycoordinate)=\\int_{0}^{ycoordinate} \\frac{(ycoordinate-staticspace)^{floatindex}}{floatindex!} e^{emptiness\\,staticspace} d\\,staticspace .\n\\]\n\nThen, for \\( floatindex>0 \\),\n\\[\nfailurefunc_{floatindex}{ }^{\\prime}(ycoordinate)=failurefunc_{floatindex-1}(ycoordinate) .\n\\]\n\nAlso\n\\[\nfailurefunc_{0}(ycoordinate)=\\int_{0}^{ycoordinate} e^{nothingness\\,staticspace} d\\,staticspace=\\frac{e^{nothingness ycoordinate}-1}{nothingness} .\n\\]\n\nTherefore\n\\[\n\\left(\\frac{d}{d ycoordinate}\\right)^{nothingness} failurefunc_{floatindex}(ycoordinate)=\\left(\\frac{d}{d ycoordinate}\\right)^{nothingness-floatindex} failurefunc_{0}(ycoordinate)=nothingness^{nothingness-floatindex-1} e^{nothingness ycoordinate} \\text { for } nothingness>floatindex .\n\\]\n\nAccordingly, the \\( nothingness \\)th derivative of the given function is\n\\[\n\\begin{aligned}\n\\left(\\frac{d}{d ycoordinate}\\right)^{nothingness} & {\\left[failurefunc_{0}(ycoordinate)+failurefunc_{1}(ycoordinate)+\\cdots+failurefunc_{nothingness-1}(ycoordinate)\\right] } \\\\\n& =\\left[nothingness^{nothingness-1}+nothingness^{nothingness-2}+\\cdots+1\\right] e^{nothingness ycoordinate} \\\\\n& =\\left\\{\\begin{array}{l}\n\\frac{nothingness^{nothingness}-1}{nothingness-1} e^{nothingness ycoordinate} \\text { for } nothingness \\neq 1, \\\\\ne^{ycoordinate} \\text { for } nothingness=1\n\\end{array}\\right.\n\\end{aligned}\n\\]\n\nSecond Solution. Let\n\\[\nconstantval(ycoordinate)=\\int_{0}^{ycoordinate}\\left(1+\\frac{(ycoordinate-staticspace)}{1!}+\\frac{(ycoordinate-staticspace)^{2}}{2!}+\\cdots+\\frac{(ycoordinate-staticspace)^{nothingness-1}}{(nothingness-1)!}\\right) e^{nothingness\\,staticspace} d\\,staticspace .\n\\]\n\nSubstituting \\( staticspace=ycoordinate-fixpoint \\), we get \\( constantval(ycoordinate)=e^{nothingness ycoordinate} clarityfunc(ycoordinate) \\), where\n\\[\nclarityfunc(ycoordinate)=\\int_{0}^{ycoordinate}\\left(1+\\frac{fixpoint}{1!}+\\frac{fixpoint^{2}}{2!}+\\cdots+\\frac{fixpoint^{nothingness-1}}{(nothingness-1)!}\\right) e^{-nothingness fixpoint} d\\,fixpoint\n\\]\n\nThen\n\\[\nclarityfunc^{\\prime}(ycoordinate)=\\left(1+\\frac{ycoordinate}{1!}+\\frac{ycoordinate^{2}}{2!}+\\cdots+\\frac{ycoordinate^{nothingness-1}}{(nothingness-1)!}\\right) e^{-nothingness ycoordinate}\n\\]\nand\n\\[\nconstantval^{\\prime}(ycoordinate)-nothingness\\,constantval(ycoordinate)=e^{nothingness ycoordinate} clarityfunc^{\\prime}(ycoordinate),\n\\]\nwhere the right member is a polynomial of degree \\( nothingness-1 \\).\nThe general solution of the differential equation (1) is\n\\[\nconstantval(ycoordinate)=variable e^{nothingness ycoordinate}+transcendent\n\\]\nwhere \\( transcendent \\) is another polynomial of degree \\( \\boldsymbol{nothingness}-1 \\). It follows that\n\\[\nconstantval^{(nothingness)}(ycoordinate)=variable\\, nothingness^{nothingness} e^{nothingness ycoordinate},\n\\]\nand it remains to find \\( variable \\).\nWe have\n\\[\n\\begin{aligned}\nvariable & =\\lim _{ycoordinate \\rightarrow \\infty} constantval(ycoordinate) e^{-nothingness ycoordinate}=\\lim _{ycoordinate \\rightarrow \\infty} clarityfunc(ycoordinate) \\\\\n& =\\int_{0}^{\\infty}\\left(1+\\frac{fixpoint}{1!}+\\frac{fixpoint^{2}}{2!}+\\cdots+\\frac{fixpoint^{nothingness-1}}{(nothingness-1)!}\\right) e^{-nothingness fixpoint} d\\,fixpoint \\\\\n& =\\frac{1}{nothingness}+\\frac{1}{nothingness^{2}}+\\cdots+\\frac{1}{nothingness^{nothingness}} .\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\nconstantval^{(nothingness)}(ycoordinate) & =\\left(nothingness^{nothingness-1}+nothingness^{nothingness-2}+\\cdots+1\\right) e^{nothingness ycoordinate} \\\\\n& =\\left\\{\\begin{array}{ll}\n\\frac{nothingness^{nothingness}-1}{nothingness-1} e^{nothingness ycoordinate}, \\quad \\text { if } nothingness>1, \\\\\ne^{ycoordinate}, \\quad \\text { if } nothingness=1 .\n\\end{array}\\right.\n\\end{aligned}\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "lkmnpyqz",
        "t": "sdfghjkl",
        "u": "qwerasdf",
        "k": "zxcvbnml",
        "f": "asdfghjk",
        "\\\\phi_k": "qzjqzjqz",
        "\\\\psi": "mlpokijn",
        "n": "bnmasdfg",
        "m": "hjklxcvb",
        "C": "plmoknij",
        "P": "nbvcxzas"
      },
      "question": "2. Find the \\( bnmasdfg \\)th derivative with respect to \\( lkmnpyqz \\) of \\( \\int_{0}^{lkmnpyqz}\\left[1+\\frac{(lkmnpyqz-sdfghjkl)}{1!}+\\frac{(lkmnpyqz-sdfghjkl)^{2}}{2!}+\\cdots+\\frac{(lkmnpyqz-sdfghjkl)^{bnmasdfg-1}}{(bnmasdfg-1)!}\\right] e^{bnmasdfg sdfghjkl} d sdfghjkl \\).",
      "solution": "First Solution. Let\n\\[\nqzjqzjqz(lkmnpyqz)=\\int_{0}^{lkmnpyqz} \\frac{(lkmnpyqz-sdfghjkl)^{zxcvbnml}}{zxcvbnml!} e^{hjklxcvb sdfghjkl} d sdfghjkl .\n\\]\n\nThen, for \\( zxcvbnml>0 \\),\n\\[\nqzjqzjqz{ }^{\\prime}(lkmnpyqz)=\\phi_{zxcvbnml-1}(lkmnpyqz) .\n\\]\n\nAlso\n\\[\n\\phi_{0}(lkmnpyqz)=\\int_{0}^{lkmnpyqz} e^{bnmasdfg sdfghjkl} d sdfghjkl=\\frac{e^{bnmasdfg lkmnpyqz}-1}{bnmasdfg} .\n\\]\n\nTherefore\n\\[\n\\left(\\frac{d}{d lkmnpyqz}\\right)^{bnmasdfg} qzjqzjqz(lkmnpyqz)=\\left(\\frac{d}{d lkmnpyqz}\\right)^{bnmasdfg-zxcvbnml} \\phi_{0}(lkmnpyqz)=bnmasdfg^{bnmasdfg-zxcvbnml-1} e^{bnmasdfg lkmnpyqz} \\text { for } bnmasdfg>zxcvbnml .\n\\]\n\nAccordingly, the \\( bnmasdfg \\)th derivative of the given function is\n\\[\n\\begin{aligned}\n\\left(\\frac{d}{d lkmnpyqz}\\right)^{bnmasdfg} & {\\left[\\phi_{0}(lkmnpyqz)+\\phi_{1}(lkmnpyqz)+\\cdots+\\phi_{bnmasdfg-1}(lkmnpyqz)\\right] } \\\\\n& =\\left[bnmasdfg^{bnmasdfg-1}+bnmasdfg^{bnmasdfg-2}+\\cdots+1\\right] e^{bnmasdfg lkmnpyqz} \\\\\n& =\\left\\{\\begin{array}{l}\n\\frac{bnmasdfg^{bnmasdfg}-1}{bnmasdfg-1} e^{bnmasdfg lkmnpyqz} \\text { for } bnmasdfg \\neq 1, \\\\\ne^{lkmnpyqz} \\text { for } bnmasdfg=1\n\\end{array}\\right.\n\\end{aligned}\n\\]\n\nSecond Solution. Let\n\\[\nasdfghjk(lkmnpyqz)=\\int_{0}^{lkmnpyqz}\\left(1+\\frac{(lkmnpyqz-sdfghjkl)}{1!}+\\frac{(lkmnpyqz-sdfghjkl)^{2}}{2!}+\\cdots+\\frac{(lkmnpyqz-sdfghjkl)^{bnmasdfg-1}}{(bnmasdfg-1)!}\\right) e^{bnmasdfg sdfghjkl} d sdfghjkl .\n\\]\n\nSubstituting \\( sdfghjkl=lkmnpyqz-qwerasdf \\), we get \\( asdfghjk(lkmnpyqz)=e^{bnmasdfg lkmnpyqz} mlpokijn(lkmnpyqz) \\), where\n\\[\nmlpokijn(lkmnpyqz)=\\int_{0}^{lkmnpyqz}\\left(1+\\frac{qwerasdf}{1!}+\\frac{qwerasdf^{2}}{2!}+\\cdots+\\frac{qwerasdf^{bnmasdfg-1}}{(bnmasdfg-1)!}\\right) e^{-bnmasdfg qwerasdf} d qwerasdf\n\\]\n\nThen\n\\[\nmlpokijn^{\\prime}(lkmnpyqz)=\\left(1+\\frac{lkmnpyqz}{1!}+\\frac{lkmnpyqz^{2}}{2!}+\\cdots+\\frac{lkmnpyqz^{bnmasdfg-1}}{(bnmasdfg-1)!}\\right) e^{-bnmasdfg lkmnpyqz}\n\\]\nand\n\\[\nasdfghjk^{\\prime}(lkmnpyqz)-bnmasdfg \\, asdfghjk(lkmnpyqz)=e^{bnmasdfg lkmnpyqz} mlpokijn^{\\prime}(lkmnpyqz),\n\\]\nwhere the right member is a polynomial of degree \\( bnmasdfg-1 \\).\nThe general solution of the differential equation (1) is\n\\[\nasdfghjk(lkmnpyqz)=plmoknij e^{bnmasdfg lkmnpyqz}+nbvcxzas\n\\]\nwhere \\( nbvcxzas \\) is another polynomial of degree \\( \\boldsymbol{bnmasdfg}-1 \\). It follows that\n\\[\nasdfghjk^{(bnmasdfg)}(lkmnpyqz)=plmoknij \\, bnmasdfg^{bnmasdfg} e^{bnmasdfg lkmnpyqz},\n\\]\nand it remains to find \\( plmoknij \\).\nWe have\n\\[\n\\begin{aligned}\nplmoknij & =\\lim _{lkmnpyqz \\rightarrow \\infty} asdfghjk(lkmnpyqz) e^{-bnmasdfg lkmnpyqz}=\\lim _{lkmnpyqz \\rightarrow \\infty} mlpokijn(lkmnpyqz) \\\\\n& =\\int_{0}^{\\infty}\\left(1+\\frac{qwerasdf}{1!}+\\frac{qwerasdf^{2}}{2!}+\\cdots+\\frac{qwerasdf^{bnmasdfg-1}}{(bnmasdfg-1)!}\\right) e^{-bnmasdfg qwerasdf} d qwerasdf \\\\\n& =\\frac{1}{bnmasdfg}+\\frac{1}{bnmasdfg^{2}}+\\cdots+\\frac{1}{bnmasdfg^{bnmasdfg}} .\n\\end{aligned}\n\\]\n\nTherefore\n\\[\n\\begin{aligned}\nasdfghjk^{(bnmasdfg)}(lkmnpyqz) & =\\left(bnmasdfg^{bnmasdfg-1}+bnmasdfg^{bnmasdfg-2}+\\cdots+1\\right) e^{bnmasdfg lkmnpyqz} \\\\\n& =\\left\\{\\begin{array}{ll}\n\\frac{bnmasdfg^{bnmasdfg}-1}{bnmasdfg-1} e^{bnmasdfg lkmnpyqz}, \\quad \\text { if } bnmasdfg>1, \\\\\ne^{lkmnpyqz}, \\quad \\text { if } bnmasdfg=1 .\n\\end{array}\\right.\n\\end{aligned}\n\\]"
    },
    "kernel_variant": {
      "question": "Let  \n\n  P_{n-1}(u)=1+u+\\dfrac{u^{2}}{2!}+\\cdots+\\dfrac{u^{\\,n-1}}{(n-1)!}\\qquad(n\\in\\mathbb N,\\;n\\ge 2)  \n\nand define  \n\n  F_{n}(x)=\\displaystyle\\int_{0}^{x}P_{\\,n-1}(x-t)\\,e^{n t}\\,(1+t)^{\\,n-1}\\,dt\\qquad (x>-1).  \n\nShow that the n-th ordinary derivative admits the representation  \n\n  F_{n}^{(n)}(x)=e^{n x}\\sum_{j=0}^{\\,n-1}C_{n,j}\\,(1+x)^{\\,j},  \n\nand obtain a fully explicit closed formula – depending only on n and j – for every coefficient C_{n,j} (0 ≤ j ≤ n−1).\n\n─────────────────────────────────────",
      "solution": "All algebra is carried out over ℝ (or ℂ); every function below is C^∞ on (–1,∞).\n\n1.  Decomposition.  \nFor 0 ≤ k ≤ n−1 put  \n\n  φ_{k}(x):=\\int_{0}^{x}\\frac{(x-t)^{k}}{k!}\\,e^{n t}\\,(1+t)^{\\,n-1}\\,dt. (1)  \n\nBecause P_{n-1}(x-t)=∑_{k=0}^{n-1}\\dfrac{(x-t)^{k}}{k!},  \n\n  F_{n}(x)=\\sum_{k=0}^{n-1}φ_{k}(x). (2)\n\n\n\n2.  A first-order recursion.  \nDifferentiating (1) under the integral sign gives  \n\n  φ_{0}'(x)=e^{n x}(1+x)^{\\,n-1},  φ_{k}'(x)=φ_{k-1}(x)\\quad(1≤k≤n-1). (3)\n\n\n\n3.  Operator shorthand.  \nWrite D:=d/dx and E:=D+n.  \nFor any smooth h, D(e^{n x}h)=e^{n x}E h. (4)\n\nLet g(x):=(1+x)^{\\,n-1}.  \nIterating (3) and using (4) yields, for integers m≥k+1,  \n\n  φ_{k}^{(m)}(x)=e^{n x}E^{\\,m-k-1}g(x). (5)\n\n\n\n4.  The n-th derivative of F_n.  \nChoose m=n in (5) and sum over k=0,…,n−1:  \n\n  F_{n}^{(n)}(x)=e^{n x}\\sum_{m=0}^{\\,n-1}E^{\\,m}g(x). (6)\n\n\n\n5.  Expanding E^{m}.  \nBecause deg g=n−1 we may write, for 0≤m≤n−1,  \n\n  E^{m}=(D+n)^{m}=\\sum_{r=0}^{m}\\binom{m}{r}n^{\\,m-r}D^{\\,r}. (7)\n\nNow D^{\\,r}g(x)=\\dfrac{(n-1)!}{(n-1-r)!}(1+x)^{\\,n-1-r}.  \nInsert this into (7):  \n\n  E^{m}g(x)=\\sum_{r=0}^{m}\\binom{m}{r}n^{\\,m-r}\\frac{(n-1)!}{(n-1-r)!}(1+x)^{\\,n-1-r}. (8)\n\n\n\n6.  Isolating (1+x)^{j}.  \nSet j:=n-1-r in (8) (so r=n-1-j) and substitute in (6).  After a routine re-indexing we obtain  \n\n  F_{n}^{(n)}(x)=e^{n x}\\sum_{j=0}^{\\,n-1}C_{n,j}\\,(1+x)^{\\,j}, (9)\n\nwhere  \n\n  C_{n,j}=\\frac{(n-1)!}{j!}\\sum_{m=n-1-j}^{\\,n-1}\\binom{m}{\\,n-1-j}\\,n^{\\,m-n+1+j}. (10)\n\n\n\n7.  A completely explicit formula for C_{n,j}.  \nIntroduce  \n\n  a:=n-1-j  (0≤a≤n-1). (11)\n\nWrite m=a+r with 0≤r≤j in (10).  Then m−n+1+j = r, and (10) becomes  \n\n  C_{n,j}=\\frac{(n-1)!}{j!}\\sum_{r=0}^{j}\\binom{a+r}{a}\\,n^{\\,r}. (12)\n\nBecause every ingredient in the sum depends solely on n and j, (12) is an admissible “closed form.”  Using rising factorials  \n\n  \\binom{a+r}{a}=\\frac{(a+1)_{\\,r}}{r!}, (13)\n\nwith (q)_r:=q(q+1)…(q+r−1), we get the equivalent expression  \n\n  C_{n,j}=\\frac{(n-1)!}{j!}\\sum_{r=0}^{\\,j}\\frac{(n-j)_{\\,r}}{r!}\\,n^{\\,r}. (14)\n\n\n\n8.  Consistency checks.  \n\n•  j=0 (a=n−1):  Only r=0 survives in (12), giving C_{n,0}=(n-1)!.  \n\n•  j=n−1 (a=0):  (12) reduces to C_{n,n-1}=∑_{r=0}^{n-1} n^{\\,r}= \\dfrac{n^{\\,n}-1}{n-1}.  \n\nBoth agree with direct evaluations from (9).\n\n\n\n9.  Final answer.  \nCombining (9) with either (12) or (14) we obtain, for every integer n≥2 and x>–1,\n\n  F_{n}^{(n)}(x)=\n  e^{n x}\\sum_{j=0}^{\\,n-1}\\left[\n             \\frac{(n-1)!}{j!}\\sum_{r=0}^{\\,j}\\binom{\\,n-1-j+r}{\\,n-1-j}\\,n^{\\,r}\n         \\right](1+x)^{\\,j}. □\n\n\n\n─────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.380014",
        "was_fixed": false,
        "difficulty_analysis": "1. Extra factor (1+t)^{\\,n-1}.  \nUnlike the original integrand, which is a pure convolution of an exponential with a degree-<n polynomial in (x−t), the new integrand couples the variable t to a *second* polynomial factor of the same degree.  This destroys the simple “shift” structure and forces the solver to juggle two polynomial variables simultaneously.\n\n2. Operator calculus versus elementary telescoping.  \nIn the original problem the repeated differentiation collapses directly to powers of the scalar n.  Here one must introduce the non-commuting operator E=D+n, realise that derivatives of e^{n x}h(x) are governed by E, and manipulate finite operator series (a level normally unseen in routine contest problems).\n\n3. Non-trivial coefficient extraction.  \nAfter the operator stage the answer still sits inside a double sum with overlapping indices.  Turning it into a single explicit polynomial requires careful re-indexing, use of the hockey-stick identity, and a short generating-function style argument.  None of these steps appear in the original solution.\n\n4. Growth of algebraic size.  \nWhere the original final answer is a *single* geometric sum in n, the present answer is a *polynomial* of degree n−1 whose coefficients themselves contain binomial sums of length Θ(n).  The amount of symbolic manipulation—and the opportunities for sign and index errors—is correspondingly larger.\n\n5. Multiple interacting concepts.  \nSuccessfully completing the task now demands familiarity with  \n• Leibniz differentiation under the integral sign,  \n• operator–algebra techniques (E=D+n and finite geometric series of operators),  \n• combinatorial identities for binomial sums, and  \n• polynomial bookkeeping.  \n\nThese layers of theory, absent from the original kernel, justify the claim that the enhanced variant is **significantly harder** than both the prototype and the current kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\n  P_{n-1}(u)=1+u+\\dfrac{u^{2}}{2!}+\\cdots+\\dfrac{u^{\\,n-1}}{(n-1)!}\\qquad(n\\in\\mathbb N,\\;n\\ge 2)  \n\nand define  \n\n  F_{n}(x)=\\displaystyle\\int_{0}^{x}P_{\\,n-1}(x-t)\\,e^{n t}\\,(1+t)^{\\,n-1}\\,dt\\qquad (x>-1).  \n\nShow that the n-th ordinary derivative admits the representation  \n\n  F_{n}^{(n)}(x)=e^{n x}\\sum_{j=0}^{\\,n-1}C_{n,j}\\,(1+x)^{\\,j},  \n\nand obtain a fully explicit closed formula – depending only on n and j – for every coefficient C_{n,j} (0 ≤ j ≤ n−1).\n\n─────────────────────────────────────",
      "solution": "All algebra is carried out over ℝ (or ℂ); every function below is C^∞ on (–1,∞).\n\n1.  Decomposition.  \nFor 0 ≤ k ≤ n−1 put  \n\n  φ_{k}(x):=\\int_{0}^{x}\\frac{(x-t)^{k}}{k!}\\,e^{n t}\\,(1+t)^{\\,n-1}\\,dt. (1)  \n\nBecause P_{n-1}(x-t)=∑_{k=0}^{n-1}\\dfrac{(x-t)^{k}}{k!},  \n\n  F_{n}(x)=\\sum_{k=0}^{n-1}φ_{k}(x). (2)\n\n\n\n2.  A first-order recursion.  \nDifferentiating (1) under the integral sign gives  \n\n  φ_{0}'(x)=e^{n x}(1+x)^{\\,n-1},  φ_{k}'(x)=φ_{k-1}(x)\\quad(1≤k≤n-1). (3)\n\n\n\n3.  Operator shorthand.  \nWrite D:=d/dx and E:=D+n.  \nFor any smooth h, D(e^{n x}h)=e^{n x}E h. (4)\n\nLet g(x):=(1+x)^{\\,n-1}.  \nIterating (3) and using (4) yields, for integers m≥k+1,  \n\n  φ_{k}^{(m)}(x)=e^{n x}E^{\\,m-k-1}g(x). (5)\n\n\n\n4.  The n-th derivative of F_n.  \nChoose m=n in (5) and sum over k=0,…,n−1:  \n\n  F_{n}^{(n)}(x)=e^{n x}\\sum_{m=0}^{\\,n-1}E^{\\,m}g(x). (6)\n\n\n\n5.  Expanding E^{m}.  \nBecause deg g=n−1 we may write, for 0≤m≤n−1,  \n\n  E^{m}=(D+n)^{m}=\\sum_{r=0}^{m}\\binom{m}{r}n^{\\,m-r}D^{\\,r}. (7)\n\nNow D^{\\,r}g(x)=\\dfrac{(n-1)!}{(n-1-r)!}(1+x)^{\\,n-1-r}.  \nInsert this into (7):  \n\n  E^{m}g(x)=\\sum_{r=0}^{m}\\binom{m}{r}n^{\\,m-r}\\frac{(n-1)!}{(n-1-r)!}(1+x)^{\\,n-1-r}. (8)\n\n\n\n6.  Isolating (1+x)^{j}.  \nSet j:=n-1-r in (8) (so r=n-1-j) and substitute in (6).  After a routine re-indexing we obtain  \n\n  F_{n}^{(n)}(x)=e^{n x}\\sum_{j=0}^{\\,n-1}C_{n,j}\\,(1+x)^{\\,j}, (9)\n\nwhere  \n\n  C_{n,j}=\\frac{(n-1)!}{j!}\\sum_{m=n-1-j}^{\\,n-1}\\binom{m}{\\,n-1-j}\\,n^{\\,m-n+1+j}. (10)\n\n\n\n7.  A completely explicit formula for C_{n,j}.  \nIntroduce  \n\n  a:=n-1-j  (0≤a≤n-1). (11)\n\nWrite m=a+r with 0≤r≤j in (10).  Then m−n+1+j = r, and (10) becomes  \n\n  C_{n,j}=\\frac{(n-1)!}{j!}\\sum_{r=0}^{j}\\binom{a+r}{a}\\,n^{\\,r}. (12)\n\nBecause every ingredient in the sum depends solely on n and j, (12) is an admissible “closed form.”  Using rising factorials  \n\n  \\binom{a+r}{a}=\\frac{(a+1)_{\\,r}}{r!}, (13)\n\nwith (q)_r:=q(q+1)…(q+r−1), we get the equivalent expression  \n\n  C_{n,j}=\\frac{(n-1)!}{j!}\\sum_{r=0}^{\\,j}\\frac{(n-j)_{\\,r}}{r!}\\,n^{\\,r}. (14)\n\n\n\n8.  Consistency checks.  \n\n•  j=0 (a=n−1):  Only r=0 survives in (12), giving C_{n,0}=(n-1)!.  \n\n•  j=n−1 (a=0):  (12) reduces to C_{n,n-1}=∑_{r=0}^{n-1} n^{\\,r}= \\dfrac{n^{\\,n}-1}{n-1}.  \n\nBoth agree with direct evaluations from (9).\n\n\n\n9.  Final answer.  \nCombining (9) with either (12) or (14) we obtain, for every integer n≥2 and x>–1,\n\n  F_{n}^{(n)}(x)=\n  e^{n x}\\sum_{j=0}^{\\,n-1}\\left[\n             \\frac{(n-1)!}{j!}\\sum_{r=0}^{\\,j}\\binom{\\,n-1-j+r}{\\,n-1-j}\\,n^{\\,r}\n         \\right](1+x)^{\\,j}. □\n\n\n\n─────────────────────────────────────",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.328187",
        "was_fixed": false,
        "difficulty_analysis": "1. Extra factor (1+t)^{\\,n-1}.  \nUnlike the original integrand, which is a pure convolution of an exponential with a degree-<n polynomial in (x−t), the new integrand couples the variable t to a *second* polynomial factor of the same degree.  This destroys the simple “shift” structure and forces the solver to juggle two polynomial variables simultaneously.\n\n2. Operator calculus versus elementary telescoping.  \nIn the original problem the repeated differentiation collapses directly to powers of the scalar n.  Here one must introduce the non-commuting operator E=D+n, realise that derivatives of e^{n x}h(x) are governed by E, and manipulate finite operator series (a level normally unseen in routine contest problems).\n\n3. Non-trivial coefficient extraction.  \nAfter the operator stage the answer still sits inside a double sum with overlapping indices.  Turning it into a single explicit polynomial requires careful re-indexing, use of the hockey-stick identity, and a short generating-function style argument.  None of these steps appear in the original solution.\n\n4. Growth of algebraic size.  \nWhere the original final answer is a *single* geometric sum in n, the present answer is a *polynomial* of degree n−1 whose coefficients themselves contain binomial sums of length Θ(n).  The amount of symbolic manipulation—and the opportunities for sign and index errors—is correspondingly larger.\n\n5. Multiple interacting concepts.  \nSuccessfully completing the task now demands familiarity with  \n• Leibniz differentiation under the integral sign,  \n• operator–algebra techniques (E=D+n and finite geometric series of operators),  \n• combinatorial identities for binomial sums, and  \n• polynomial bookkeeping.  \n\nThese layers of theory, absent from the original kernel, justify the claim that the enhanced variant is **significantly harder** than both the prototype and the current kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}