{
  "index": "1978-A-3",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "Problem A-3\nLet \\( p(x)=2+4 x+3 x^{2}+5 x^{3}+3 x^{4}+4 x^{5}+2 x^{6} \\). For \\( k \\) with \\( 0<k<5 \\), define\n\\[\nI_{k}=\\int_{0}^{\\infty} \\frac{x^{k}}{p(x)} d x\n\\]\n\nFor which \\( k \\) is \\( I_{k} \\) smallest?",
  "solution": "A-3.\nSince the integral converges for \\( -1<k<5 \\), one can consider \\( I_{k} \\) to be defined on this open interval. Letting \\( x=1 / t \\), one finds that\n\\[\nI_{k}=\\int_{\\infty}^{0} \\frac{t^{-k}}{t^{-6} p(t)}\\left(\\frac{-d t}{t^{2}}\\right)=\\int_{0}^{\\infty} \\frac{t^{4-k} d t}{p(t)}=I_{4-k}\n\\]\n\nThen\n\\[\nI_{k}=\\left(I_{k}+I_{4-k}\\right) / 2=\\int_{0}^{\\infty} \\frac{\\left[\\left(x^{k}+x^{4-k}\\right) / 2\\right] d x}{p(x)} \\geqslant \\int_{0}^{\\infty} \\frac{x^{2} d x}{p(x)}=I_{2}\n\\]\nsince \\( \\left(x^{k}+x^{4-k}\\right) / 2 \\geqslant \\sqrt{x^{k} \\cdot x^{4-k}}=x^{2} \\) by the Arithmetic Mean-Geometric Mean Inequality. Thus \\( I_{k} \\) is smallest for \\( k=2 \\).",
  "vars": [
    "x",
    "t"
  ],
  "params": [
    "k",
    "p",
    "I_k",
    "I_4-k",
    "I_2"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variate",
        "t": "auxvar",
        "k": "exponent",
        "p": "polynomf",
        "I_k": "integralexp",
        "I_4-k": "integralcmp",
        "I_2": "integralref"
      },
      "question": "Problem A-3\nLet \\( polynomf(variate)=2+4 variate+3 variate^{2}+5 variate^{3}+3 variate^{4}+4 variate^{5}+2 variate^{6} \\). For \\( exponent \\) with \\( 0<exponent<5 \\), define\n\\[\nintegralexp=\\int_{0}^{\\infty} \\frac{variate^{exponent}}{polynomf(variate)} d variate\n\\]\n\nFor which \\( exponent \\) is \\( integralexp \\) smallest?",
      "solution": "A-3.\nSince the integral converges for \\( -1<exponent<5 \\), one can consider \\( integralexp \\) to be defined on this open interval. Letting \\( variate=1 / auxvar \\), one finds that\n\\[\nintegralexp=\\int_{\\infty}^{0} \\frac{auxvar^{-exponent}}{auxvar^{-6} polynomf(auxvar)}\\left(\\frac{-d auxvar}{auxvar^{2}}\\right)=\\int_{0}^{\\infty} \\frac{auxvar^{4-exponent} d auxvar}{polynomf(auxvar)}=integralcmp\n\\]\n\nThen\n\\[\nintegralexp=\\left(integralexp+integralcmp\\right) / 2=\\int_{0}^{\\infty} \\frac{\\left[\\left(variate^{exponent}+variate^{4-exponent}\\right) / 2\\right] d variate}{polynomf(variate)} \\geqslant \\int_{0}^{\\infty} \\frac{variate^{2} d variate}{polynomf(variate)}=integralref\n\\]\nsince \\( \\left(variate^{exponent}+variate^{4-exponent}\\right) / 2 \\geqslant \\sqrt{variate^{exponent} \\cdot variate^{4-exponent}}=variate^{2} \\) by the Arithmetic Mean-Geometric Mean Inequality. Thus \\( integralexp \\) is smallest for \\( exponent=2 \\)."
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "sandwich",
        "t": "playhouse",
        "k": "pineapple",
        "p": "horseshoe",
        "I_k": "tangerine",
        "I_4-k": "blueberry",
        "I_2": "raspberry"
      },
      "question": "Problem A-3\nLet \\( horseshoe(sandwich)=2+4 sandwich+3 sandwich^{2}+5 sandwich^{3}+3 sandwich^{4}+4 sandwich^{5}+2 sandwich^{6} \\). For \\( pineapple \\) with \\( 0<pineapple<5 \\), define\n\\[\ntangerine=\\int_{0}^{\\infty} \\frac{sandwich^{pineapple}}{horseshoe(sandwich)} d sandwich\n\\]\n\nFor which \\( pineapple \\) is \\( tangerine \\) smallest?",
      "solution": "A-3.\nSince the integral converges for \\( -1<pineapple<5 \\), one can consider \\( tangerine \\) to be defined on this open interval. Letting \\( sandwich=1 / playhouse \\), one finds that\n\\[\ntangerine=\\int_{\\infty}^{0} \\frac{playhouse^{-pineapple}}{playhouse^{-6} horseshoe(playhouse)}\\left(\\frac{-d playhouse}{playhouse^{2}}\\right)=\\int_{0}^{\\infty} \\frac{playhouse^{4-pineapple} d playhouse}{horseshoe(playhouse)}=blueberry\n\\]\nThen\n\\[\ntangerine=\\left(tangerine+blueberry\\right) / 2=\\int_{0}^{\\infty} \\frac{\\left[\\left(sandwich^{pineapple}+sandwich^{4-pineapple}\\right) / 2\\right] d sandwich}{horseshoe(sandwich)} \\geqslant \\int_{0}^{\\infty} \\frac{sandwich^{2} d sandwich}{horseshoe(sandwich)}=raspberry\n\\]\nsince \\( \\left(sandwich^{pineapple}+sandwich^{4-pineapple}\\right) / 2 \\geqslant \\sqrt{sandwich^{pineapple} \\cdot sandwich^{4-pineapple}}=sandwich^{2} \\) by the Arithmetic Mean-Geometric Mean Inequality. Thus \\( tangerine \\) is smallest for \\( pineapple=2 \\)."
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantval",
        "t": "timeless",
        "k": "unchanged",
        "p": "constantfn",
        "I_k": "sumdiscrete",
        "I_4-k": "summirror",
        "I_2": "sumcenter"
      },
      "question": "Problem A-3\nLet \\( constantfn(constantval)=2+4 constantval+3 constantval^{2}+5 constantval^{3}+3 constantval^{4}+4 constantval^{5}+2 constantval^{6} \\). For \\( unchanged \\) with \\( 0<unchanged<5 \\), define\n\\[\nsumdiscrete=\\int_{0}^{\\infty} \\frac{constantval^{unchanged}}{constantfn(constantval)} d constantval\n\\]\n\nFor which \\( unchanged \\) is \\( sumdiscrete \\) smallest?",
      "solution": "A-3.\nSince the integral converges for \\( -1<unchanged<5 \\), one can consider \\( sumdiscrete \\) to be defined on this open interval. Letting \\( constantval=1 / timeless \\), one finds that\n\\[\nsumdiscrete=\\int_{\\infty}^{0} \\frac{timeless^{-unchanged}}{timeless^{-6} constantfn(timeless)}\\left(\\frac{-d timeless}{timeless^{2}}\\right)=\\int_{0}^{\\infty} \\frac{timeless^{4-unchanged} d timeless}{constantfn(timeless)}=summirror\n\\]\n\nThen\n\\[\nsumdiscrete=\\left(sumdiscrete+summirror\\right) / 2=\\int_{0}^{\\infty} \\frac{\\left[\\left(constantval^{unchanged}+constantval^{4-unchanged}\\right) / 2\\right] d constantval}{constantfn(constantval)} \\geqslant \\int_{0}^{\\infty} \\frac{constantval^{2} d constantval}{constantfn(constantval)}=sumcenter\n\\]\nsince \\( \\left(constantval^{unchanged}+constantval^{4-unchanged}\\right) / 2 \\geqslant \\sqrt{constantval^{unchanged} \\cdot constantval^{4-unchanged}}=constantval^{2} \\) by the Arithmetic Mean-Geometric Mean Inequality. Thus \\( sumdiscrete \\) is smallest for \\( unchanged=2 \\)."
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "t": "hjgrksla",
        "k": "mnbvcxzq",
        "p": "lkjhgfds",
        "I_k": "poiuytre",
        "I_4-k": "qazxswed",
        "I_2": "plmoknij"
      },
      "question": "Problem A-3\nLet \\( lkjhgfds(qzxwvtnp)=2+4 qzxwvtnp+3 qzxwvtnp^{2}+5 qzxwvtnp^{3}+3 qzxwvtnp^{4}+4 qzxwvtnp^{5}+2 qzxwvtnp^{6} \\). For \\( mnbvcxzq \\) with \\( 0<mnbvcxzq<5 \\), define\n\\[\npoiuytre=\\int_{0}^{\\infty} \\frac{qzxwvtnp^{mnbvcxzq}}{lkjhgfds(qzxwvtnp)} d qzxwvtnp\n\\]\n\nFor which \\( mnbvcxzq \\) is \\( poiuytre \\) smallest?",
      "solution": "A-3.\nSince the integral converges for \\( -1<mnbvcxzq<5 \\), one can consider \\( poiuytre \\) to be defined on this open interval. Letting \\( qzxwvtnp=1 / hjgrksla \\), one finds that\n\\[\npoiuytre = \\int_{\\infty}^{0} \\frac{hjgrksla^{-mnbvcxzq}}{hjgrksla^{-6} lkjhgfds(hjgrksla)}\\left(\\frac{-d hjgrksla}{hjgrksla^{2}}\\right)=\\int_{0}^{\\infty} \\frac{hjgrksla^{4-mnbvcxzq} d hjgrksla}{lkjhgfds(hjgrksla)}=qazxswed\n\\]\n\nThen\n\\[\npoiuytre = \\left(poiuytre + qazxswed\\right) / 2 = \\int_{0}^{\\infty} \\frac{\\left[\\left(qzxwvtnp^{mnbvcxzq}+qzxwvtnp^{4-mnbvcxzq}\\right) / 2\\right] d qzxwvtnp}{lkjhgfds(qzxwvtnp)} \\geqslant \\int_{0}^{\\infty} \\frac{qzxwvtnp^{2} d qzxwvtnp}{lkjhgfds(qzxwvtnp)} = plmoknij\n\\]\nsince \\( \\left(qzxwvtnp^{mnbvcxzq}+qzxwvtnp^{4-mnbvcxzq}\\right) / 2 \\geqslant \\sqrt{qzxwvtnp^{mnbvcxzq} \\cdot qzxwvtnp^{4-mnbvcxzq}}=qzxwvtnp^{2} \\) by the Arithmetic Mean-Geometric Mean Inequality. Thus \\( poiuytre \\) is smallest for \\( mnbvcxzq=2 \\)."
    },
    "kernel_variant": {
      "question": "Let  \n  p(x)=2+5x+4x^{2}+7x^{3}+4x^{4}+5x^{5}+2x^{6}.  \n\nFor every real parameter α>0 and every real k with −1<k<6α−1 define  \n  I_{k}(α)=∫_{0}^{∞} \\dfrac{x^{k}}{p(x)^{α}}\\;dx.  \n\n(a)  Show that the integral converges precisely for  \n  −1<k<6α−1.  \n\n(b)  Prove the reflection identity  \n  I_{k}(α)=I_{6α−2−k}(α).  \n\n(c)  Assume from now on α>1/3 so that the open interval  \n  (0,6α−2) is non-empty.  \n  Fix such an α and determine all k∈(0,6α−2) at which I_{k}(α) attains its minimum value.  \n\n(d)  Describe explicitly how the minimising k depends on α, and list every α>1/3 for which this minimiser is an integer.  \n\n(e)  Analyse the complementary regime 0<α≤1/3.  Prove that for these α the set  \n  {I_{k}(α):k∈(0,6α−2)}  \n  is empty, hence has infimum +∞, and consequently no minimum can be realised at either endpoint of the (degenerate) interval.",
      "solution": "Throughout denote d:=6 (the degree of p).  By inspection p has palindromic coefficients, whence  \n  x^{d}\\,p(1/x)=p(x)  for all x>0.  (0.1)\n\nStep (a) – Domain of convergence.  \nNear the origin p(x)=p(0)+O(x)=2+O(x), so  \n  x^{k}/p(x)^{α}=x^{k}/2^{α}+O(x^{k+1}).  \nHence ∫_{0}^{1} converges iff k>−1.\n\nFor large x one has p(x)=2x^{6}+O(x^{5}).  Therefore  \n  p(x)^{α}=2^{α}x^{6α}\\bigl(1+O(x^{−1})\\bigr),  \nand  \n  x^{k}/p(x)^{α}=2^{−α}x^{k-6α}\\bigl(1+o(1)\\bigr).  \nThe tail integral ∫_{1}^{∞} converges iff k−6α<−1, i.e. k<6α−1.\n\nCombining the two local conditions gives the announced strip  \n  −1<k<6α−1. ∎\n\n\n\nStep (b) – Reflection identity.  \nSubstitute x=1/t.  Using (0.1) and dx=−t^{−2}dt,\n\nI_{k}(α)=∫_{∞}^{0}\\frac{t^{−k}}{t^{−6α}p(t)^{α}}\\,(−t^{−2})dt\n        =∫_{0}^{∞}\\frac{t^{6α−2−k}}{p(t)^{α}}\\,dt\n        =I_{6α−2−k}(α). ∎\n\n\n\nStep (c) – The minimiser for α>1/3.\n\n(i) Symmetry.  \nIdentity (b) shows that k↦I_{k}(α) is symmetric about  \n  k_{0}:=(6α−2)/2=3α−1.  (3.1)\n\n(ii) Strict log-convexity.  \nLet k_{1},k_{2}∈(−1,6α−1), k_{1}≠k_{2}, and θ∈(0,1); set k_{θ}=(1−θ)k_{1}+θk_{2}.  \nBy Hölder’s inequality,\n\nI_{k_{θ}}(α)=∫_{0}^{∞}\\!\\frac{x^{k_{θ}}}{p(x)^{α}}dx\n          ≤\\Bigl(∫_{0}^{∞}\\!\\frac{x^{k_{1}}}{p(x)^{α}}dx\\Bigr)^{1−θ}\n           \\Bigl(∫_{0}^{∞}\\!\\frac{x^{k_{2}}}{p(x)^{α}}dx\\Bigr)^{θ}\n          =I_{k_{1}}(α)^{1−θ}I_{k_{2}}(α)^{θ},\n\nwith strict inequality when k_{1}≠k_{2}.  Hence ln I_{k}(α) is strictly convex, and the map k↦I_{k}(α)=exp(ln I_{k}(α)) is strictly convex as well.\n\n(iii) Uniqueness of the minimum.  \nA strictly convex function symmetric about k_{0} has exactly one minimiser, namely k_{0}.  For α>1/3 we have\n\n−1<3α−1<6α−1 and 0<3α−1<6α−2,\n\nso k_{0} lies inside both the convergence strip and the interval demanded in the question.  Consequently  \n\n  k_{min}(α)=3α−1  for every α>1/3. ∎\n\n\n\nStep (d) – Dependence on α and integral minimisers.  \nThe minimiser is the linear function k_{min}(α)=3α−1.  It is integral iff 3α−1∈ℤ, i.e.\n\n  α=(m+1)/3 with m∈ℤ.\n\nImposing α>1/3 forces m≥1, whence the complete list\n\n  α=2/3,1,4/3,5/3,… (m=1,2,3,4,…)  \n\nwith corresponding integer minima k_{min}=1,2,3,4,… ∎\n\n\n\nStep (e) – The range 0<α≤1/3.\n\nFor 0<α<1/3 we have 6α−2<0, so the open interval (0,6α−2) is empty.  \nFor α=1/3 one obtains 6α−2=0, and the open interval (0,0) is again empty.\n\nTherefore in either case the set of admissible indices\n\n  S_{α}:=(0,6α−2)\n\nis empty, and consequently\n\n  inf {I_{k}(α):k∈S_{α}} = inf ∅ = +∞.\n\nBecause this infimum is infinite, it is certainly not realised by any k, let alone by an endpoint such as k=0.  In particular, although I_{0}(1/3)=∫_{0}^{∞}p(x)^{−1/3}dx is finite, the value k=0 does not belong to S_{α} and hence plays no rôle in the minimisation problem that was posed.\n\nSummary for 0<α≤1/3: the minimisation problem over (0,6α−2) has no solution, and the infimum over that interval equals +∞, so it is not attained at either endpoint. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.635717",
        "was_fixed": false,
        "difficulty_analysis": "1. Extra parameter α introduces a two-variable family of integrals; the solver must work uniformly in α.\n2. Denominator raised to an arbitrary power demands careful convergence analysis and produces the general reflection k↦6α−2−k, rather than the fixed reflection in the original problem.\n3. Proving strict convexity of k↦I_{k}(α) by Hölder’s inequality is a deeper analytical step absent from the original A-3 (which relied only on AM–GM for two points).\n4. Part (d) mixes discrete (integrality) and continuous (real α) aspects, adding a number-theoretic flavour.\n5. Altogether the solution requires: asymptotic analysis, substitution symmetry, Hölder’s inequality, convexity theory, and a final diophantine argument—far more varied and sophisticated than in the original problem."
      }
    },
    "original_kernel_variant": {
      "question": "Let  \n  p(x)=2+5x+4x^{2}+7x^{3}+4x^{4}+5x^{5}+2x^{6}.  \n\nFor every real parameter α>0 and every real k with −1<k<6α−1 define  \n  I_{k}(α)=∫_{0}^{∞} \\dfrac{x^{k}}{p(x)^{α}}\\;dx.  \n\n(a)  Show that the integral converges precisely for  \n  −1<k<6α−1.  \n\n(b)  Prove the reflection identity  \n  I_{k}(α)=I_{6α−2−k}(α).  \n\n(c)  Assume from now on α>1/3 so that the open interval  \n  (0,6α−2) is non-empty.  \n  Fix such an α and determine all k∈(0,6α−2) at which I_{k}(α) attains its minimum value.  \n\n(d)  Describe explicitly how the minimising k depends on α, and list every α>1/3 for which this minimiser is an integer.  \n\n(e)  Analyse the complementary regime 0<α≤1/3.  Prove that for these α the set  \n  {I_{k}(α):k∈(0,6α−2)}  \n  is empty, hence has infimum +∞, and consequently no minimum can be realised at either endpoint of the (degenerate) interval.",
      "solution": "Throughout denote d:=6 (the degree of p).  By inspection p has palindromic coefficients, whence  \n  x^{d}\\,p(1/x)=p(x)  for all x>0.  (0.1)\n\nStep (a) – Domain of convergence.  \nNear the origin p(x)=p(0)+O(x)=2+O(x), so  \n  x^{k}/p(x)^{α}=x^{k}/2^{α}+O(x^{k+1}).  \nHence ∫_{0}^{1} converges iff k>−1.\n\nFor large x one has p(x)=2x^{6}+O(x^{5}).  Therefore  \n  p(x)^{α}=2^{α}x^{6α}\\bigl(1+O(x^{−1})\\bigr),  \nand  \n  x^{k}/p(x)^{α}=2^{−α}x^{k-6α}\\bigl(1+o(1)\\bigr).  \nThe tail integral ∫_{1}^{∞} converges iff k−6α<−1, i.e. k<6α−1.\n\nCombining the two local conditions gives the announced strip  \n  −1<k<6α−1. ∎\n\n\n\nStep (b) – Reflection identity.  \nSubstitute x=1/t.  Using (0.1) and dx=−t^{−2}dt,\n\nI_{k}(α)=∫_{∞}^{0}\\frac{t^{−k}}{t^{−6α}p(t)^{α}}\\,(−t^{−2})dt\n        =∫_{0}^{∞}\\frac{t^{6α−2−k}}{p(t)^{α}}\\,dt\n        =I_{6α−2−k}(α). ∎\n\n\n\nStep (c) – The minimiser for α>1/3.\n\n(i) Symmetry.  \nIdentity (b) shows that k↦I_{k}(α) is symmetric about  \n  k_{0}:=(6α−2)/2=3α−1.  (3.1)\n\n(ii) Strict log-convexity.  \nLet k_{1},k_{2}∈(−1,6α−1), k_{1}≠k_{2}, and θ∈(0,1); set k_{θ}=(1−θ)k_{1}+θk_{2}.  \nBy Hölder’s inequality,\n\nI_{k_{θ}}(α)=∫_{0}^{∞}\\!\\frac{x^{k_{θ}}}{p(x)^{α}}dx\n          ≤\\Bigl(∫_{0}^{∞}\\!\\frac{x^{k_{1}}}{p(x)^{α}}dx\\Bigr)^{1−θ}\n           \\Bigl(∫_{0}^{∞}\\!\\frac{x^{k_{2}}}{p(x)^{α}}dx\\Bigr)^{θ}\n          =I_{k_{1}}(α)^{1−θ}I_{k_{2}}(α)^{θ},\n\nwith strict inequality when k_{1}≠k_{2}.  Hence ln I_{k}(α) is strictly convex, and the map k↦I_{k}(α)=exp(ln I_{k}(α)) is strictly convex as well.\n\n(iii) Uniqueness of the minimum.  \nA strictly convex function symmetric about k_{0} has exactly one minimiser, namely k_{0}.  For α>1/3 we have\n\n−1<3α−1<6α−1 and 0<3α−1<6α−2,\n\nso k_{0} lies inside both the convergence strip and the interval demanded in the question.  Consequently  \n\n  k_{min}(α)=3α−1  for every α>1/3. ∎\n\n\n\nStep (d) – Dependence on α and integral minimisers.  \nThe minimiser is the linear function k_{min}(α)=3α−1.  It is integral iff 3α−1∈ℤ, i.e.\n\n  α=(m+1)/3 with m∈ℤ.\n\nImposing α>1/3 forces m≥1, whence the complete list\n\n  α=2/3,1,4/3,5/3,… (m=1,2,3,4,…)  \n\nwith corresponding integer minima k_{min}=1,2,3,4,… ∎\n\n\n\nStep (e) – The range 0<α≤1/3.\n\nFor 0<α<1/3 we have 6α−2<0, so the open interval (0,6α−2) is empty.  \nFor α=1/3 one obtains 6α−2=0, and the open interval (0,0) is again empty.\n\nTherefore in either case the set of admissible indices\n\n  S_{α}:=(0,6α−2)\n\nis empty, and consequently\n\n  inf {I_{k}(α):k∈S_{α}} = inf ∅ = +∞.\n\nBecause this infimum is infinite, it is certainly not realised by any k, let alone by an endpoint such as k=0.  In particular, although I_{0}(1/3)=∫_{0}^{∞}p(x)^{−1/3}dx is finite, the value k=0 does not belong to S_{α} and hence plays no rôle in the minimisation problem that was posed.\n\nSummary for 0<α≤1/3: the minimisation problem over (0,6α−2) has no solution, and the infimum over that interval equals +∞, so it is not attained at either endpoint. ∎",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.505949",
        "was_fixed": false,
        "difficulty_analysis": "1. Extra parameter α introduces a two-variable family of integrals; the solver must work uniformly in α.\n2. Denominator raised to an arbitrary power demands careful convergence analysis and produces the general reflection k↦6α−2−k, rather than the fixed reflection in the original problem.\n3. Proving strict convexity of k↦I_{k}(α) by Hölder’s inequality is a deeper analytical step absent from the original A-3 (which relied only on AM–GM for two points).\n4. Part (d) mixes discrete (integrality) and continuous (real α) aspects, adding a number-theoretic flavour.\n5. Altogether the solution requires: asymptotic analysis, substitution symmetry, Hölder’s inequality, convexity theory, and a final diophantine argument—far more varied and sophisticated than in the original problem."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}