{
  "index": "1992-B-4",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "having no nonconstant factor in common with $x^3 - x$. Let\n\\[\n\\frac{d^{1992}}{dx^{1992}} \\left( \\frac{p(x)}{x^3 - x} \\right) =\n\\frac{f(x)}{g(x)}\n\\]\nfor polynomials $f(x)$ and $g(x)$. Find the smallest possible degree of $f(x)$.",
  "solution": "Solution. By the Division Algorithm, we can write \\( p(x)=\\left(x^{3}-x\\right) q(x)+r(x) \\), where \\( q(x) \\) and \\( r(x) \\) are polynomials, the degree of \\( r(x) \\) is at most 2 , and the degree of \\( q(x) \\) is less than 1989. Then\n\\[\n\\frac{d^{1992}}{d x^{1992}}\\left(\\frac{p(x)}{x^{3}-x}\\right)=\\frac{d^{1992}}{d x^{1992}}\\left(\\frac{r(x)}{x^{3}-x}\\right)\n\\]\n\nWrite \\( r(x) /\\left(x^{3}-x\\right) \\) in the form\n\\[\n\\frac{A}{x-1}+\\frac{B}{x}+\\frac{C}{x+1} .\n\\]\n\nBecause \\( p(x) \\) and \\( x^{3}-x \\) have no common factor, neither do \\( r(x) \\) and \\( x^{3}-x \\), and thus \\( A, B \\), and \\( C \\) are nonzero. Thus\n\\[\n\\begin{array}{l}\n\\frac{d^{1992}}{d x^{1992}}\\left(\\frac{r(x)}{x^{3}-x}\\right) \\\\\n\\quad=1992!\\left(\\frac{A}{(x-1)^{1993}}+\\frac{B}{x^{1993}}+\\frac{C}{(x+1)^{1993}}\\right) \\\\\n\\quad=1992!\\left(\\frac{A x^{1993}(x+1)^{1993}+B(x-1)^{1993}(x+1)^{1993}+C(x-1)^{1993} x^{1993}}{\\left(x^{3}-x\\right)^{1993}}\\right)\n\\end{array}\n\\]\n\nSince \\( A, B, C \\) are nonzero, the numerator and denominator have no common factor. Expanding the numerator yields\n\\[\n(A+B+C) x^{3986}+1993(A-C) x^{3985}+1993(996 A-B+996 C) x^{3984}+\\cdots\n\\]\n\nFrom \\( A=C=1, B=-2 \\) we see that the degree can be as low as 3984. A lower degree would imply \\( A+B+C=0, A-C=0,996 A-B+996 C=0 \\), implying that \\( A=B=C=0 \\), a contradiction.\n\nExpressing \\( \\frac{d^{1992}}{d x^{1992}}\\left(\\frac{p(x)}{x^{3}-x}\\right) \\) in other than lowest terms can only increase the degree of the numerator. (Nowhere in the problem does it say that \\( f(x) / g(x) \\) is to be in lowest terms.)",
  "vars": [
    "x",
    "p",
    "q",
    "r",
    "f",
    "g"
  ],
  "params": [
    "A",
    "B",
    "C"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "unknown",
        "p": "polyfunc",
        "q": "quotient",
        "r": "remainder",
        "f": "numerator",
        "g": "denominator",
        "A": "constanta",
        "B": "constantb",
        "C": "constantc"
      },
      "question": "having no nonconstant factor in common with $unknown^3 - unknown$. Let\n\\[\n\\frac{d^{1992}}{d unknown^{1992}} \\left( \\frac{polyfunc(unknown)}{unknown^3 - unknown} \\right) =\n\\frac{numerator(unknown)}{denominator(unknown)}\n\\]\nfor polynomials $numerator(unknown)$ and $denominator(unknown)$. Find the smallest possible degree of $numerator(unknown)$.",
      "solution": "Solution. By the Division Algorithm, we can write \\( polyfunc(unknown)=\\left(unknown^{3}-unknown\\right) quotient(unknown)+remainder(unknown) \\), where \\( quotient(unknown) \\) and \\( remainder(unknown) \\) are polynomials, the degree of \\( remainder(unknown) \\) is at most 2, and the degree of \\( quotient(unknown) \\) is less than 1989. Then\n\\[\n\\frac{d^{1992}}{d unknown^{1992}}\\left(\\frac{polyfunc(unknown)}{unknown^{3}-unknown}\\right)=\\frac{d^{1992}}{d unknown^{1992}}\\left(\\frac{remainder(unknown)}{unknown^{3}-unknown}\\right)\n\\]\n\nWrite \\( remainder(unknown) /\\left(unknown^{3}-unknown\\right) \\) in the form\n\\[\n\\frac{constanta}{unknown-1}+\\frac{constantb}{unknown}+\\frac{constantc}{unknown+1} .\n\\]\n\nBecause \\( polyfunc(unknown) \\) and \\( unknown^{3}-unknown \\) have no common factor, neither do \\( remainder(unknown) \\) and \\( unknown^{3}-unknown \\), and thus constanta, constantb, and constantc are nonzero. Thus\n\\[\n\\begin{array}{l}\n\\frac{d^{1992}}{d unknown^{1992}}\\left(\\frac{remainder(unknown)}{unknown^{3}-unknown}\\right) \\\\\n\\quad=1992!\\left(\\frac{constanta}{(unknown-1)^{1993}}+\\frac{constantb}{unknown^{1993}}+\\frac{constantc}{(unknown+1)^{1993}}\\right) \\\\\n\\quad=1992!\\left(\\frac{constanta\\, unknown^{1993}(unknown+1)^{1993}+constantb(unknown-1)^{1993}(unknown+1)^{1993}+constantc(unknown-1)^{1993} unknown^{1993}}{\\left(unknown^{3}-unknown\\right)^{1993}}\\right)\n\\end{array}\n\\]\n\nSince constanta, constantb, constantc are nonzero, the numerator and denominator have no common factor. Expanding the numerator yields\n\\[\n(constanta+constantb+constantc) \\, unknown^{3986}+1993(constanta-constantc) \\, unknown^{3985}+1993(996\\, constanta-constantb+996\\, constantc) \\, unknown^{3984}+\\cdots\n\\]\n\nFrom constanta=constantc=1, constantb=-2 we see that the degree can be as low as 3984. A lower degree would imply constanta+constantb+constantc=0, constanta-constantc=0, 996\\, constanta-constantb+996\\, constantc=0, implying that constanta=constantb=constantc=0, a contradiction.\n\nExpressing \\( \\frac{d^{1992}}{d unknown^{1992}}\\left(\\frac{polyfunc(unknown)}{unknown^{3}-unknown}\\right) \\) in other than lowest terms can only increase the degree of the numerator. (Nowhere in the problem does it say that \\( numerator(unknown) / denominator(unknown) \\) is to be in lowest terms.)"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "blackbird",
        "p": "teacupful",
        "q": "driftwood",
        "r": "signboard",
        "f": "moonlight",
        "g": "stargazer",
        "A": "crossroads",
        "B": "sandstone",
        "C": "horsewhip"
      },
      "question": "having no nonconstant factor in common with $blackbird^3 - blackbird$. Let\n\\[\n\\frac{d^{1992}}{d blackbird^{1992}} \\left( \\frac{teacupful(blackbird)}{blackbird^3 - blackbird} \\right) =\n\\frac{moonlight(blackbird)}{stargazer(blackbird)}\n\\]\nfor polynomials $moonlight(blackbird)$ and $stargazer(blackbird)$. Find the smallest possible degree of $moonlight(blackbird)$.",
      "solution": "Solution. By the Division Algorithm, we can write \\( teacupful(blackbird)=\\left(blackbird^{3}-blackbird\\right) driftwood(blackbird)+signboard(blackbird) \\), where \\( driftwood(blackbird) \\) and \\( signboard(blackbird) \\) are polynomials, the degree of \\( signboard(blackbird) \\) is at most 2 , and the degree of \\( driftwood(blackbird) \\) is less than 1989. Then\n\\[\n\\frac{d^{1992}}{d blackbird^{1992}}\\left(\\frac{teacupful(blackbird)}{blackbird^{3}-blackbird}\\right)=\\frac{d^{1992}}{d blackbird^{1992}}\\left(\\frac{signboard(blackbird)}{blackbird^{3}-blackbird}\\right)\n\\]\n\nWrite \\( signboard(blackbird) /\\left(blackbird^{3}-blackbird\\right) \\) in the form\n\\[\n\\frac{crossroads}{blackbird-1}+\\frac{sandstone}{blackbird}+\\frac{horsewhip}{blackbird+1} .\n\\]\n\nBecause \\( teacupful(blackbird) \\) and \\( blackbird^{3}-blackbird \\) have no common factor, neither do \\( signboard(blackbird) \\) and \\( blackbird^{3}-blackbird \\), and thus \\( crossroads, sandstone \\), and \\( horsewhip \\) are nonzero. Thus\n\\[\n\\begin{array}{l}\n\\frac{d^{1992}}{d blackbird^{1992}}\\left(\\frac{signboard(blackbird)}{blackbird^{3}-blackbird}\\right) \\\\\n\\quad=1992!\\left(\\frac{crossroads}{(blackbird-1)^{1993}}+\\frac{sandstone}{blackbird^{1993}}+\\frac{horsewhip}{(blackbird+1)^{1993}}\\right) \\\\\n\\quad=1992!\\left(\\frac{crossroads\\; blackbird^{1993}(blackbird+1)^{1993}+sandstone(blackbird-1)^{1993}(blackbird+1)^{1993}+horsewhip(blackbird-1)^{1993} blackbird^{1993}}{\\left(blackbird^{3}-blackbird\\right)^{1993}}\\right)\n\\end{array}\n\\]\n\nSince \\( crossroads, sandstone, horsewhip \\) are nonzero, the numerator and denominator have no common factor. Expanding the numerator yields\n\\[\n(crossroads+sandstone+horsewhip) blackbird^{3986}+1993(crossroads-horsewhip) blackbird^{3985}+1993(996\\;crossroads-sandstone+996\\;horsewhip) blackbird^{3984}+\\cdots\n\\]\n\nFrom \\( crossroads=horsewhip=1, sandstone=-2 \\) we see that the degree can be as low as 3984. A lower degree would imply \\( crossroads+sandstone+horsewhip=0, crossroads-horsewhip=0,996\\;crossroads-sandstone+996\\;horsewhip=0 \\), implying that \\( crossroads=sandstone=horsewhip=0 \\), a contradiction.\n\nExpressing \\( \\frac{d^{1992}}{d blackbird^{1992}}\\left(\\frac{teacupful(blackbird)}{blackbird^{3}-blackbird}\\right) \\) in other than lowest terms can only increase the degree of the numerator. (Nowhere in the problem does it say that \\( moonlight(blackbird) / stargazer(blackbird) \\) is to be in lowest terms.)"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "immutablevalue",
        "p": "irrationalfun",
        "q": "productpoly",
        "r": "completepart",
        "f": "denominator",
        "g": "numerator",
        "A": "flexible",
        "B": "changing",
        "C": "unstable"
      },
      "question": "having no nonconstant factor in common with $immutablevalue^3 - immutablevalue$. Let\n\\[\n\\frac{d^{1992}}{d immutablevalue^{1992}} \\left( \\frac{irrationalfun(immutablevalue)}{immutablevalue^3 - immutablevalue} \\right) =\n\\frac{denominator(immutablevalue)}{numerator(immutablevalue)}\n\\]\nfor polynomials $denominator(immutablevalue)$ and $numerator(immutablevalue)$. Find the smallest possible degree of $denominator(immutablevalue)$.",
      "solution": "Solution. By the Division Algorithm, we can write \\( irrationalfun(immutablevalue)=\\left(immutablevalue^{3}-immutablevalue\\right) productpoly(immutablevalue)+completepart(immutablevalue) \\), where \\( productpoly(immutablevalue) \\) and \\( completepart(immutablevalue) \\) are polynomials, the degree of \\( completepart(immutablevalue) \\) is at most 2 , and the degree of \\( productpoly(immutablevalue) \\) is less than 1989. Then\n\\[\n\\frac{d^{1992}}{d immutablevalue^{1992}}\\left(\\frac{irrationalfun(immutablevalue)}{immutablevalue^{3}-immutablevalue}\\right)=\\frac{d^{1992}}{d immutablevalue^{1992}}\\left(\\frac{completepart(immutablevalue)}{immutablevalue^{3}-immutablevalue}\\right)\n\\]\n\nWrite \\( completepart(immutablevalue) /\\left(immutablevalue^{3}-immutablevalue\\right) \\) in the form\n\\[\n\\frac{flexible}{immutablevalue-1}+\\frac{changing}{immutablevalue}+\\frac{unstable}{immutablevalue+1} .\n\\]\n\nBecause \\( irrationalfun(immutablevalue) \\) and \\( immutablevalue^{3}-immutablevalue \\) have no common factor, neither do \\( completepart(immutablevalue) \\) and \\( immutablevalue^{3}-immutablevalue \\), and thus \\( flexible, changing \\), and \\( unstable \\) are nonzero. Thus\n\\[\n\\begin{array}{l}\n\\frac{d^{1992}}{d immutablevalue^{1992}}\\left(\\frac{completepart(immutablevalue)}{immutablevalue^{3}-immutablevalue}\\right) \\\\\n\\quad=1992!\\left(\\frac{flexible}{(immutablevalue-1)^{1993}}+\\frac{changing}{immutablevalue^{1993}}+\\frac{unstable}{(immutablevalue+1)^{1993}}\\right) \\\\\n\\quad=1992!\\left(\\frac{flexible \\, immutablevalue^{1993}(immutablevalue+1)^{1993}+changing(immutablevalue-1)^{1993}(immutablevalue+1)^{1993}+unstable(immutablevalue-1)^{1993} immutablevalue^{1993}}{\\left(immutablevalue^{3}-immutablevalue\\right)^{1993}}\\right)\n\\end{array}\n\\]\n\nSince \\( flexible, changing, unstable \\) are nonzero, the numerator and denominator have no common factor. Expanding the numerator yields\n\\[\n(flexible+changing+unstable) immutablevalue^{3986}+1993(flexible-unstable) immutablevalue^{3985}+1993(996 flexible-changing+996 unstable) immutablevalue^{3984}+\\cdots\n\\]\n\nFrom \\( flexible=unstable=1, changing=-2 \\) we see that the degree can be as low as 3984. A lower degree would imply \\( flexible+changing+unstable=0, flexible-unstable=0,996 flexible-changing+996 unstable=0 \\), implying that \\( flexible=changing=unstable=0 \\), a contradiction.\n\nExpressing \\( \\frac{d^{1992}}{d immutablevalue^{1992}}\\left(\\frac{irrationalfun(immutablevalue)}{immutablevalue^{3}-immutablevalue}\\right) \\) in other than lowest terms can only increase the degree of the numerator. (Nowhere in the problem does it say that \\( denominator(immutablevalue) / numerator(immutablevalue) \\) is to be in lowest terms.)"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "p": "hjgrksla",
        "q": "xmptovli",
        "r": "beznadey",
        "f": "vbrclnue",
        "g": "tkhsmpqo",
        "A": "lkdjfhre",
        "B": "cpowmizt",
        "C": "qabnduso"
      },
      "question": "having no nonconstant factor in common with $qzxwvtnp^3 - qzxwvtnp$. Let\n\\[\n\\frac{d^{1992}}{d qzxwvtnp^{1992}} \\left( \\frac{hjgrksla(qzxwvtnp)}{qzxwvtnp^3 - qzxwvtnp} \\right) =\n\\frac{vbrclnue(qzxwvtnp)}{tkhsmpqo(qzxwvtnp)}\n\\]\nfor polynomials $vbrclnue(qzxwvtnp)$ and $tkhsmpqo(qzxwvtnp)$. Find the smallest possible degree of $vbrclnue(qzxwvtnp)$.",
      "solution": "Solution. By the Division Algorithm, we can write \\( hjgrksla(qzxwvtnp)=\\left(qzxwvtnp^{3}-qzxwvtnp\\right) xmptovli(qzxwvtnp)+beznadey(qzxwvtnp) \\), where \\( xmptovli(qzxwvtnp) \\) and \\( beznadey(qzxwvtnp) \\) are polynomials, the degree of \\( beznadey(qzxwvtnp) \\) is at most 2 , and the degree of \\( xmptovli(qzxwvtnp) \\) is less than 1989. Then\n\\[\n\\frac{d^{1992}}{d qzxwvtnp^{1992}}\\left(\\frac{hjgrksla(qzxwvtnp)}{qzxwvtnp^{3}-qzxwvtnp}\\right)=\\frac{d^{1992}}{d qzxwvtnp^{1992}}\\left(\\frac{beznadey(qzxwvtnp)}{qzxwvtnp^{3}-qzxwvtnp}\\right)\n\\]\n\nWrite \\( beznadey(qzxwvtnp) /\\left(qzxwvtnp^{3}-qzxwvtnp\\right) \\) in the form\n\\[\n\\frac{lkdjfhre}{qzxwvtnp-1}+\\frac{cpowmizt}{qzxwvtnp}+\\frac{qabnduso}{qzxwvtnp+1} .\n\\]\n\nBecause \\( hjgrksla(qzxwvtnp) \\) and \\( qzxwvtnp^{3}-qzxwvtnp \\) have no common factor, neither do \\( beznadey(qzxwvtnp) \\) and \\( qzxwvtnp^{3}-qzxwvtnp \\), and thus \\( lkdjfhre, cpowmizt \\), and \\( qabnduso \\) are nonzero. Thus\n\\[\n\\begin{array}{l}\n\\frac{d^{1992}}{d qzxwvtnp^{1992}}\\left(\\frac{beznadey(qzxwvtnp)}{qzxwvtnp^{3}-qzxwvtnp}\\right) \\\\\n\\quad=1992!\\left(\\frac{lkdjfhre}{(qzxwvtnp-1)^{1993}}+\\frac{cpowmizt}{qzxwvtnp^{1993}}+\\frac{qabnduso}{(qzxwvtnp+1)^{1993}}\\right) \\\\\n\\quad=1992!\\left(\\frac{lkdjfhre \\, qzxwvtnp^{1993}(qzxwvtnp+1)^{1993}+cpowmizt(qzxwvtnp-1)^{1993}(qzxwvtnp+1)^{1993}+qabnduso(qzxwvtnp-1)^{1993} qzxwvtnp^{1993}}{\\left(qzxwvtnp^{3}-qzxwvtnp\\right)^{1993}}\\right)\n\\end{array}\n\\]\n\nSince \\( lkdjfhre, cpowmizt, qabnduso \\) are nonzero, the numerator and denominator have no common factor. Expanding the numerator yields\n\\[\n(lkdjfhre+cpowmizt+qabnduso) qzxwvtnp^{3986}+1993(lkdjfhre-qabnduso) qzxwvtnp^{3985}+1993(996 lkdjfhre-cpowmizt+996 qabnduso) qzxwvtnp^{3984}+\\cdots\n\\]\n\nFrom \\( lkdjfhre=qabnduso=1, cpowmizt=-2 \\) we see that the degree can be as low as 3984. A lower degree would imply \\( lkdjfhre+cpowmizt+qabnduso=0, lkdjfhre-qabnduso=0,996 lkdjfhre-cpowmizt+996 qabnduso=0 \\), implying that \\( lkdjfhre=cpowmizt=qabnduso=0 \\), a contradiction.\n\nExpressing \\( \\frac{d^{1992}}{d qzxwvtnp^{1992}}\\left(\\frac{hjgrksla(qzxwvtnp)}{qzxwvtnp^{3}-qzxwvtnp}\\right) \\) in other than lowest terms can only increase the degree of the numerator. (Nowhere in the problem does it say that \\( vbrclnue(qzxwvtnp) / tkhsmpqo(qzxwvtnp) \\) is to be in lowest terms.)"
    },
    "kernel_variant": {
      "question": "Let\n\\[\nD(x)=x\\,(x-1)\\,(x-2)^{2}(x+2)^{3},\\qquad n=3000 .\n\\]\n\nFor every non–zero polynomial \\(p(x)\\in\\mathbb C[x]\\) that is coprime to \\(D\\),\ndefine\n\\[\n\\frac{d^{\\,n}}{dx^{\\,n}}\\!\\Bigl(\\tfrac{p(x)}{D(x)}\\Bigr)=\\frac{f(x)}{g(x)},\n\\qquad f(x),g(x)\\in\\mathbb C[x],\n\\]\nwhere the fraction on the right is \\emph{any} one that is obtained\ndirectly from differentiating; common factors of \\(f\\) and \\(g\\)\n\\emph{need not} be cancelled.\n\nDetermine\n\\[\n\\boxed{\\displaystyle\\min_{p}\\deg f(x)}\n\\]\nand prove that this minimum is achieved for at least one admissible\npolynomial \\(p(x)\\).\n\n--------------------------------------------------------------------",
      "solution": "Throughout we work over \\(\\mathbb C\\).\n\n0.  Pre–computation, notation  \n    \\[\n        m_{0}=m_{1}=1,\\;m_{2}=2,\\;m_{-2}=3,\n        \\qquad \\mathcal P:=\\{(a,s)\\mid a\\in\\{0,1,2,-2\\},\\,1\\le s\\le m_{a}\\}.\n    \\]\n    Hence \\(\\#\\mathcal P=7\\).\n\n====================================================================\n1.  Reduction to \\(\\deg p\\le 6\\).\n\nBecause \\(\\deg D=7\\) we may write \\(p=qD+r\\) with\n\\(\\deg r\\le 6\\).  The summand \\(qD\\) differentiates to an honest\npolynomial, hence it disappears from any fractional representation, so\n\\[\n   \\frac{d^{n}}{dx^{n}}\\Bigl(\\tfrac{p}{D}\\Bigr)=\n   \\frac{d^{n}}{dx^{n}}\\Bigl(\\tfrac{r}{D}\\Bigr).\n\\]\nConsequently\n\\[\n   \\min_{p\\ \\mathrm{cpr}\\,D}\\deg f=\n   \\min_{\\substack{p\\ \\mathrm{cpr}\\,D\\\\ \\deg p\\le 6}}\\deg f.\n\\tag{1.1}\n\\]\n\n====================================================================\n2.  Partial fractions and a \\(7\\)-dimensional residue isomorphism.\n\nFor \\(\\deg p\\le 6\\) there is a unique decomposition\n\\[\n   \\frac{p(x)}{D(x)}=\n   \\sum_{(a,s)\\in\\mathcal P}\\frac{\\lambda_{a,s}(p)}{(x-a)^s},\n   \\qquad \\lambda_{a,s}(p)\\in\\mathbb C .\n\\tag{2.1}\n\\]\nThe residue map  \n\\[\n   \\mathscr R:\\mathbb C[x]_{\\le 6}\\longrightarrow\\mathbb C^{7},\\qquad\n   p\\longmapsto\\bigl(\\lambda_{a,s}(p)\\bigr)_{(a,s)\\in\\mathcal P}\n\\]\nis an isomorphism (both spaces have dimension \\(7\\) and the images of\n\\(1,x,\\dots ,x^{6}\\) are independent).  In particular, every choice of\nresidues can be realised by an admissible \\(p\\).\n\n====================================================================\n3.  The \\(n\\)-th derivative.\n\nFor \\(k\\ge 1\\)\n\\[\n   \\frac{d^{\\,n}}{dx^{\\,n}}\\!\\Bigl(\\frac1{(x-a)^{k}}\\Bigr)\n   =(-1)^{n}\\frac{(k+n-1)!}{(k-1)!}\\,\n     \\frac1{(x-a)^{k+n}} .\n\\]\nIntroduce  \n\\[\n   P_{n}(x):=x^{\\,n+1}(x-1)^{\\,n+1}(x-2)^{\\,n+2}(x+2)^{\\,n+3},\n   \\qquad \\deg P_{n}=4n+7,\n\\]\nand put\n\\[\n   M_{a,s}(x):=\\frac{P_{n}(x)}{(x-a)^{\\,n+s}},\\qquad\n   \\deg M_{a,s}=3n+7-s .\n\\]\nWith  \n\\[\n   \\Lambda_{a,s}:=(-1)^{n}\\frac{(n+s-1)!}{(s-1)!}\\lambda_{a,s}(p)\n\\]\nwe obtain  \n\\[\n   \\frac{d^{\\,n}}{dx^{\\,n}}\\!\\Bigl(\\tfrac{p}{D}\\Bigr)\n   =\\frac{N(x)}{P_{n}(x)},\\qquad\n   N(x):=\\sum_{(a,s)\\in\\mathcal P}\\Lambda_{a,s}\\,M_{a,s}(x).\n\\tag{3.1}\n\\]\nHence the numerators lie in the \\(7\\)-dimensional space\n\\[\n   \\mathcal S_{n}:=\\operatorname{Span}\\{M_{a,s}\\mid(a,s)\\in\\mathcal P\\}.\n\\]\n\n====================================================================\n4.  The first seven leading–coefficient functionals are independent.\n\nFor \\(0\\le t\\le 6\\) define\n\\[\n   L_{t}(R):=[x^{\\,3n+6-t}]\\,R(x)\\qquad(R\\in\\mathcal S_{n}).\n\\]\nThus \\(L_{0}\\) extracts the top coefficient, \\(L_{1}\\) the next one,\nand so on.  We claim:\n\nLemma 4.1.  \nThe linear functionals \\(L_{0},L_{1},\\dots ,L_{6}\\) are linearly\nindependent on \\(\\mathcal S_{n}\\).\n\nProof.  Order the seven basis elements as  \n\\[\n   M_{0,1},\\  M_{1,1},\\  M_{2,1},\\  M_{-2,1},\\  \n   M_{2,2},\\  M_{-2,2},\\  M_{-2,3}.\n\\]\nTaking the \\((t,j)\\)-entry to be\n\\([x^{3n+6-t}]M_{j}(x)\\) gives a \\(7\\times7\\) matrix \\(\\mathcal A\\).\nA direct degree inspection shows that \\(\\mathcal A\\) is upper block–\ntriangular with non–zero diagonal, hence \\(\\det\\mathcal A\\neq0\\).  \\(\\square\\)\n\n====================================================================\n5.  Optimal lower bound for \\(\\deg N\\).\n\nSuppose \\(N\\in\\mathcal S_{n}\\setminus\\{0\\}\\).\nIf the top \\(k\\;(0\\le k\\le 7)\\) coefficients of \\(N\\) vanish, then\n\\(L_{0}(N)=\\dots=L_{k-1}(N)=0\\).  \nBy Lemma 4.1 at most six of the \\(L_{t}\\) can vanish simultaneously;\nhence \\(k\\le6\\).  Therefore\n\\[\n    \\deg N\\ge(3n+6)-6=3n .\n\\tag{5.1}\n\\]\nBecause \\(f=N\\) (we never cancel common factors),  \n\\[\n     \\boxed{\\deg f\\ge 3n}.\n\\]\n\n====================================================================\n6.  A degree–\\(3n\\) numerator whose \\emph{four} simple residues are\nnon–zero.\n\nThe six homogeneous linear conditions\n\\[\n   L_{0}(N)=\\dots=L_{5}(N)=0\n\\tag{6.1}\n\\]\ndefine a codimension–\\(6\\) subspace of \\(\\mathcal S_{n}\\); by\n\\(\\dim\\mathcal S_{n}=7\\) that subspace is a \\emph{line}, say\n\\(\\mathcal L\\subset\\mathcal S_{n}\\setminus\\{0\\}\\).\nEvery non–zero \\(N\\in\\mathcal L\\) automatically satisfies \\(\\deg N\\le\n3n\\) by construction, while (5.1) gives \\(\\deg N\\ge3n\\); hence\n\\[\n       \\deg N=3n\\qquad\\forall\\,N\\in\\mathcal L\\setminus\\{0\\}.\n\\tag{6.2}\n\\]\n\nWe must still find an \\(N\\in\\mathcal L\\) that corresponds to a\n\\emph{coprime} polynomial \\(p\\).\nRecall (Section 2) that for a simple pole \\(a\\in\\{0,1,2,-2\\}\\)\n\\[\n   \\lambda_{a,1}(p)=\\frac{p(a)}{D^{\\prime}(a)};\n\\]\nhence \\(p(a)\\neq0\\iff\\lambda_{a,1}(p)\\neq0\\).\nConsequently we have to show that \\emph{all four simple residues\n\\(\\lambda_{0,1},\\lambda_{1,1},\\lambda_{2,1},\\lambda_{-2,1}\\)}\nare non–zero for a suitable element of \\(\\mathcal L\\).\n\nTo that end let\n\\[\n   R_{a}:\\mathcal S_{n}\\longrightarrow\\mathbb C,\\qquad\n   R_{a}(N):=\\text{ the coefficient of }M_{a,1}\\text{ in }N\n\\]\n(so \\(R_{a}(N)=\\Lambda_{a,1}\\), hence\n\\(R_{a}(N)=0\\iff\\lambda_{a,1}(p)=0\\)).\nEach \\(R_{a}\\) is a non–trivial linear functional on \\(\\mathcal\nS_{n}\\).  Together with the six functionals \\(L_{0},\\dots,L_{5}\\) they\nform \\(7\\) linear conditions on the \\(7\\)-dimensional space\n\\(\\mathcal S_{n}\\).  Because the \\(L_{t}\\) are independent\n(Lemma 4.1), the family\n\\(\\{L_{0},\\dots,L_{5},R_{a}\\}\\) is independent as well:\nindeed, the matrix \\(\\mathcal A\\) used in the proof of Lemma 4.1\nalready contains the column\ncorresponding to \\(R_{a}\\) (it is the \\emph{unit} vector which picks\nout the coefficient of \\(M_{a,1}\\)), hence adding the row of \\(R_{a}\\)\nsimply appends a non–zero coordinate.\n\nTherefore\n\\[\n   \\dim\\bigl(\\ker L_{0}\\cap\\dots\\cap\\ker L_{5}\\cap\\ker R_{a}\\bigr)=0\n\\qquad\\bigl(a=0,1,2,-2\\bigr).\n\\]\nIn words: the line \\(\\mathcal L\\) is \\emph{not} contained in\n\\(\\ker R_{a}\\) for any of the four simple poles.  Consequently\n\\[\n   R_{a}\\!\\restriction_{\\mathcal L}\\not\\equiv0\\quad\n   (a=0,1,2,-2).\n\\]\nBecause \\(\\mathcal L\\) is one–dimensional, every non–zero \\(N\\in\\mathcal\nL\\) may be written as \\(N=t\\,N_{0}\\;(t\\in\\mathbb C^{\\times})\\) for a\nfixed generator \\(N_{0}\\).\nFor such a parametrisation\n\\(R_{a}(tN_{0})=t\\,R_{a}(N_{0})\\); hence once\n\\(R_{a}(N_{0})\\neq0\\) the same holds for \\emph{every} non–zero\nmultiple.  Since none of the four $R_a$ vanish identically on\n\\(\\mathcal L\\), we can choose the generator \\(N_{0}\\) so that\n\\[\n   R_{0}(N_{0}),\\,R_{1}(N_{0}),\\,R_{2}(N_{0}),\\,R_{-2}(N_{0})\\neq0.\n\\tag{6.3}\n\\]\n(Equivalently: starting from any \\(\\widehat N\\in\\mathcal L\\setminus\\{0\\}\\),\ndivide by those $R_a(\\widehat N)$ that are non–zero and take a small\ngeneric linear combination; only finitely many choices give one of the\nfour coordinates equal to \\(0\\).)\n\nNow lift \\(N_{0}\\) back to a polynomial of degree at most\n\\(6\\):\n\\[\n      p^{*}(x):=\\sum_{(a,s)\\in\\mathcal P}\n                 \\frac{\\lambda_{a,s}(p^{*})\\,D(x)}{(x-a)^{s}},\n                 \\qquad \\mathscr R(p^{*})=\\bigl(\\lambda_{a,s}(p^{*})\\bigr)\n                 =\\bigl((-1)^{n}\\tfrac{(s-1)!}{(n+s-1)!}\\Lambda_{a,s}(N_{0})\\bigr).\n\\]\nBecause of (6.3) the four simple residues of \\(p^{*}\\) are non–zero,\nand therefore\n\\[\n   p^{*}(0)\\neq0,\\quad p^{*}(1)\\neq0,\\quad\n   p^{*}(2)\\neq0,\\quad p^{*}(-2)\\neq0.\n\\]\nHence \\(\\gcd\\bigl(p^{*},D\\bigr)=1\\), i.e.\\ \\(p^{*}\\) is \\emph{coprime}\nto \\(D\\).\nFor this admissible polynomial the associated numerator equals \\(N_{0}\\)\nand satisfies \\(\\deg f=3n\\).\n\n====================================================================\n7.  Final answer.\n\nCombining the universal lower bound (5.1) with\nthe explicit construction in Section 6 we conclude\n\\[\n   \\boxed{\\displaystyle\\min_{p}\\deg f(x)=3n=9000}.\n\\]\nThe minimum is achieved, for instance, by the polynomial \\(p^{*}\\)\nconstructed in Section 6.\n\\(\\hfill\\blacksquare\\)\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.727015",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher multiplicities and more poles  \n    • The denominator now has four distinct factors with total multiplicity 7, compared with three simple factors in the original.  \n    • Repeated factors force the use of higher-order partial fractions and factorial constants from formula (3).\n\n2.  Larger derivative order  \n    • Order 3000 instead of 1992 adds computational weight and magnifies every exponent appearing in the denominator.\n\n3.  Linear-algebra dimension argument  \n    • Seven independent parameters give a six-dimensional space of admissible relations; a careful rank argument is needed to see exactly how many top coefficients can be killed, rather than the three-coefficient cancellation in the original.\n\n4.  Coprimality constraints intertwined with optimisation  \n    • Some coefficients (A,B, etc.) are compelled to be non-zero, so the optimisation is a constrained one; showing that the constraints are compatible with the six equations requires an extra geometric argument (avoiding finitely many hyperplanes).\n\n5.  Existence argument for a witnessing p(x)  \n    • Constructing (rather than guessing) a polynomial that attains the bound uses the explicit structure of the solution space of the linear system, something unnecessary in the basic problem.\n\nAll these extra layers—higher multiplicities, a larger derivative order, a higher-dimensional parameter space, constrained linear algebra, and an existence construction—make the enhanced variant markedly deeper and technically heavier than both the original and the intermediate kernel version."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n\\[\nD(x)=x\\,(x-1)\\,(x-2)^{2}(x+2)^{3},\\qquad\\;n=3000 .\n\\]\n\nFor every non–zero polynomial \\(p(x)\\in\\mathbb{C}[x]\\) that is coprime to \\(D\\), i.e.  \n\\[\n\\gcd\\bigl(p(x),D(x)\\bigr)=1 ,\n\\]\ndefine  \n\\[\nF(x)\\;=\\;\\frac{d^{\\,n}}{dx^{\\,n}}\\!\\Bigl(\\tfrac{p(x)}{D(x)}\\Bigr)\n      \\;=\\;\\frac{f(x)}{g(x)},\\qquad f(x),g(x)\\in\\mathbb{C}[x],\n\\]\nwhere \\(\\tfrac{f}{g}\\) is \\emph{any} polynomial representation obtained\nafter differentiating; common factors of \\(f\\) and \\(g\\) do \\emph{not}\nhave to be cancelled.\n\nDetermine  \n\\[\n\\boxed{\\displaystyle \n\\min_{\\,p}\\deg f(x)}\n\\]\nand prove that this minimum is attained by at least one admissible\npolynomial \\(p(x)\\).\n\n--------------------------------------------------------------------",
      "solution": "Throughout we work over \\(\\mathbb{C}\\).\n\n0.  Notation once and for all  \n    \\[\n        m_{0}=m_{1}=1,\\;m_{2}=2,\\;m_{-2}=3,\n        \\qquad \\mathcal P:=\\{(a,s)\\;|\\;a\\in\\{0,1,2,-2\\},\\;1\\le s\\le m_{a}\\}.\n    \\]\n    Thus \\(\\#\\mathcal P=7\\).\n\n====================================================================\n1.  Reduction to \\(\\deg p\\le 6\\).\n\nBecause \\(\\deg D=7\\) we may write \\(p=qD+r\\) with\n\\(\\deg r\\le 6\\).  The summand \\(qD\\) differentiates to a polynomial and\nhence disappears in the fraction, so\n\\[\n   \\frac{d^{n}}{dx^{n}}\\Bigl(\\tfrac{p}{D}\\Bigr)\n   =\n   \\frac{d^{n}}{dx^{n}}\\Bigl(\\tfrac{r}{D}\\Bigr).\n\\]\nHence\n\\[\n   \\min_{p\\;\\text{coprime to }D}\\deg f\n   =\\;\n   \\min_{\\substack{p\\;\\text{coprime to }D\\\\ \\deg p\\le 6}}\\deg f.\n\\tag{1.1}\n\\]\n\n====================================================================\n2.  Partial fractions and the residue isomorphism.\n\nFor \\(\\deg p\\le 6\\) there is a unique decomposition\n\\[\n   \\frac{p(x)}{D(x)}\n   =\n   \\sum_{(a,s)\\in\\mathcal P}\\frac{\\lambda_{a,s}(p)}{(x-a)^s},\n   \\qquad\n   \\lambda_{a,s}(p)\\in\\mathbb{C}.\n\\tag{2.1}\n\\]\nDefine the residue map\n\\[\n   \\mathscr R:\\mathbb{C}[x]_{\\le 6}\\longrightarrow\\mathbb{C}^{7},\n   \\qquad\n   p\\longmapsto\\bigl(\\lambda_{a,s}(p)\\bigr)_{(a,s)\\in\\mathcal P}.\n\\]\nBecause \\(\\dim_{\\mathbb C}\\mathbb{C}[x]_{\\le 6}=7=\\dim_{\\mathbb C}\\mathbb{C}^{7}\\)\nand the images \\(\\mathscr R(1),\\dots,\\mathscr R(x^6)\\) are easily seen\nto be linearly independent, \\(\\mathscr R\\) is an isomorphism.  In\nparticular every vector of residues may be realised by a polynomial of\ndegree \\(\\le 6\\).\n\n====================================================================\n3.  The \\(n\\)-th derivative.\n\nFor \\(k\\ge 1\\)\n\\[\n   \\frac{d^{\\,n}}{dx^{\\,n}}\\!\\Bigl(\\frac{1}{(x-a)^{k}}\\Bigr)\n   =\n   (-1)^{n}\\frac{(k+n-1)!}{(k-1)!}\\,\n   \\frac{1}{(x-a)^{k+n}} .\n\\]\nTherefore, with\n\\[\n   P_{n}(x):=x^{\\,n+1}(x-1)^{\\,n+1}(x-2)^{\\,n+2}(x+2)^{\\,n+3},\n   \\qquad \\deg P_{n}=4n+7,\n\\]\none obtains\n\\[\n   F(x)=\\frac{N(x)}{P_{n}(x)},\\quad\n   N(x):=\\sum_{(a,s)\\in\\mathcal P}\n          \\Lambda_{a,s}\\,M_{a,s}(x),\n\\tag{3.1}\n\\]\nwhere\n\\[\n   \\Lambda_{a,s}:=(-1)^{n}\\frac{(n+s-1)!}{(s-1)!}\\lambda_{a,s}(p),\n   \\qquad\n   M_{a,s}(x):=\\frac{P_{n}(x)}{(x-a)^{\\,n+s}}.\n\\]\nThe seven polynomials \\(\\{M_{a,s}\\}\\) are linearly independent (each is\ncharacterised by its different pole structure) and\n\\[\n   \\deg M_{a,s}=3n+7-s.\n\\]\nConsequently the numerators that can occur are\n\\[\n   \\mathcal S_{n}:=\\operatorname{Span}_{\\mathbb C}\\{M_{a,s}\\}\n   \\quad (\\dim_{\\mathbb C}\\mathcal S_{n}=7).\n\\]\n\n====================================================================\n4.  Seven relevant invariants of a numerator.\n\nFor \\(R(x)\\in\\mathcal{S}_{n}\\) put  \n\n(i) leading–coefficient functionals  \n\\[\n   L_{r}(R):=[x^{\\,3n+6-r}]\\,R(x), \\qquad 0\\le r\\le 4;\n\\]\n\n(ii) vanishing orders at the critical points  \n\\[\n   u:=\\nu_{x=2}(R),\\qquad v:=\\nu_{x=-2}(R),\n\\]\nso \\(0\\le u\\le 1,\\;0\\le v\\le 2\\);\n\n(iii) their numerical values  \n\\[\n   V(R):=R(2),\\quad W_{0}(R):=R(-2),\\quad W_{1}(R):=R'(-2).\n\\]\n\nBecause \\(\\deg R\\le 3n+6\\), the five coefficients\n\\(L_{0},\\dots,L_{4}\\) together with \\(V,W_{0},W_{1}\\) give eight linear\nfunctionals on \\(\\mathcal{S}_{n}\\).\n\n--------------------------------------------------------------------\nLemma 4.1  (No simultaneous annihilation by the eight functionals).  \nIf \\(R\\in\\mathcal S_{n}\\) satisfies\n\\[\n   L_{0}(R)=\\dots=L_{4}(R)=V(R)=W_{0}(R)=W_{1}(R)=0,\n\\]\nthen \\(R=0\\).\n\nProof.  \nThe vanishing of \\(V,W_{0},W_{1}\\) means that\n\\((x-2)(x+2)^{2}\\mid R\\); write \\(R=(x-2)(x+2)^{2}Q\\).\nThen \\(\\deg Q\\le 3n+3\\).\nThe five equalities \\(L_{0}=\\dots=L_{4}=0\\) say that the\nfive leading coefficients of \\(R\\) are \\(0\\), hence\n\\(\\deg R\\le 3n+1\\) and consequently \\(\\deg Q\\le 3n-2\\).\n\nBut\n\\[\n   \\mathcal T:=\n   \\bigl\\{R\\in\\mathcal S_{n}\\,\\bigm|\\, (x-2)(x+2)^{2}\\mid R\\bigr\\}\n\\]\nis a linear subspace of \\(\\mathcal S_{n}\\) of dimension at most\n\\(7-3=4\\) (three independent conditions).\nInside this \\(4\\)-dimensional space the five\nconditions \\(L_{0}=\\dots=L_{4}=0\\) force \\(R=0\\).\n\\(\\square\\)\n\n--------------------------------------------------------------------\nCorollary 4.2.  \nFor every integer \\(n\\ge 0\\) (hence in particular for \\(n=3000\\))\nthe matrix  \n\\[\n   \\mathcal M(n):=\n   \\bigl(\\,L_{0},L_{1},L_{2},L_{3},L_{4},V,W_{0},W_{1}\\bigr)^{\\!*}\n   \\quad\\text{restricted to }\\mathcal S_{n}\n\\]\nhas rank \\(7\\).  \nEquivalently, there is \\emph{no} non–zero \\(R\\in\\mathcal S_{n}\\) that\nannihilates the eight functionals, while the collection of all eight\nfunctionals is linearly dependent (because \\(\\dim\\mathcal S_{n}=7\\)).\n\n====================================================================\n5.  A universal lower bound.\n\nTake an arbitrary non–zero \\(N\\in\\mathcal S_{n}\\) and put  \n\n\\[\n  t:=\\#\\{\\,r\\mid 0\\le r\\le 4,\\;L_{r}(N)=0\\},\\qquad\n  u:=\\nu_{2}(N),\\qquad\n  v:=\\nu_{-2}(N)\\;(0\\le u\\le1,\\;0\\le v\\le2).\n\\]\nNecessarily \\(t+u+v\\le 7\\); otherwise the eight functionals would all\nvanish, contradicting Lemma 4.1.\n\nNow  \n\\[\n   \\deg N = 3n+6-t,\\qquad\n   \\deg f = \\deg N - (u+v) = 3n+6-(t+u+v).\n\\]\nBecause \\(t+u+v\\le 7\\),\n\\[\n   \\boxed{\\deg f\\ge 3n+6-7=3n-1}.\n\\tag{5.1}\n\\]\n\n====================================================================\n6.  Existence of an extremal numerator.\n\nBecause the rank of the \\(8\\times7\\) matrix in Corollary 4.2 is \\(7\\),\n\\emph{exactly one} (up to scaling) linear relation exists among the\neight functionals.  Consequently at least one of the seven–element\nsubsets is linearly \\emph{dependent}.  Choose such a subset as\n\\[\n   \\{L_{0},L_{1},L_{2},L_{3},V,W_{0},W_{1}\\},\n\\]\nso there exists a non–zero \\(N_{*}\\in\\mathcal S_{n}\\) satisfying  \n\\[\n      L_{0}(N_{*})=\\dots=L_{3}(N_{*})=0,\\quad\n      V(N_{*})=W_{0}(N_{*})=W_{1}(N_{*})=0 .\n\\tag{6.1}\n\\]\nFor this \\(N_{*}\\) we have  \n\\[\n   t=4,\\;u=1,\\;v=2,\\qquad t+u+v=7,\n\\]\nhence, by (5.1),\n\\[\n   \\deg f_{*}=3n+6-7=3n-1,\n   \\qquad\n   \\deg N_{*}=3n+2 .\n\\tag{6.2}\n\\]\nBecause \\(\\mathscr R\\) is an isomorphism, the coefficients of \\(N_{*}\\)\nprovide residues \\(\\lambda_{a,s}^{*}\\).  Setting\n\\[\n    p^{*}(x):=\\sum_{(a,s)\\in\\mathcal P}\n               \\frac{\\lambda_{a,s}^{*}\\,D(x)}{(x-a)^{s}}\n\\]\ngives a polynomial of degree at most \\(6\\); its four simple residues\n(with \\(s=1\\)) are non–zero, so \\(p^{*}\\) is coprime to \\(D\\).  By\nconstruction the associated numerator after the \\(n\\)-th derivative is\nexactly \\(f_{*}\\) with degree \\(3n-1\\).\n\n====================================================================\n7.  Final answer.\n\nCombining the lower bound (5.1) with the example (6.2) we conclude  \n\\[\n   \\boxed{\\displaystyle\\min_{\\,p}\\deg f(x)=3n-1=8999}.\n\\]\nThe minimum is achieved, for instance, by the explicitly constructed\npolynomial \\(p^{*}\\).\n\\(\\hfill\\blacksquare\\)\n\n--------------------------------------------------------------------",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.564860",
        "was_fixed": false,
        "difficulty_analysis": "1.  Higher multiplicities and more poles  \n    • The denominator now has four distinct factors with total multiplicity 7, compared with three simple factors in the original.  \n    • Repeated factors force the use of higher-order partial fractions and factorial constants from formula (3).\n\n2.  Larger derivative order  \n    • Order 3000 instead of 1992 adds computational weight and magnifies every exponent appearing in the denominator.\n\n3.  Linear-algebra dimension argument  \n    • Seven independent parameters give a six-dimensional space of admissible relations; a careful rank argument is needed to see exactly how many top coefficients can be killed, rather than the three-coefficient cancellation in the original.\n\n4.  Coprimality constraints intertwined with optimisation  \n    • Some coefficients (A,B, etc.) are compelled to be non-zero, so the optimisation is a constrained one; showing that the constraints are compatible with the six equations requires an extra geometric argument (avoiding finitely many hyperplanes).\n\n5.  Existence argument for a witnessing p(x)  \n    • Constructing (rather than guessing) a polynomial that attains the bound uses the explicit structure of the solution space of the linear system, something unnecessary in the basic problem.\n\nAll these extra layers—higher multiplicities, a larger derivative order, a higher-dimensional parameter space, constrained linear algebra, and an existence construction—make the enhanced variant markedly deeper and technically heavier than both the original and the intermediate kernel version."
      }
    }
  },
  "checked": true,
  "problem_type": "calculation"
}