{
  "index": "1946-A-1",
  "type": "ANA",
  "tag": [
    "ANA",
    "ALG"
  ],
  "difficulty": "",
  "question": "1. Suppose that the function \\( f(x)=a x^{2}+b x+c \\), where \\( a, b, c \\) are real constants, satisfies the condition \\( |f(x)| \\leq 1 \\) for \\( |x| \\leq 1 \\). Prove that \\( \\left|f^{\\prime}(x)\\right| \\) \\( \\leq 4 \\) for \\( |x| \\leq 1 \\).",
  "solution": "First Solution. If \\( a \\neq 0 \\) the graph of \\( y=a x^{2}+b x+c \\) is a parabola which can be assumed without loss of generality to open upward, i.e.. \\( a>0 \\). [We discuss the straight line case, \\( a=0 \\), later.] By symmetry we may assume that \\( b \\) is non-negative. Then the vertex falls in the left halfplane and it is clear that \\( \\max _{|x| \\leq 1}\\left|f^{\\prime}(x)\\right| \\) occurs when \\( x=1 \\), and this maximum value is \\( 2 a+b \\). It remains to show that \\( 2 a+b \\leq 4 \\).\n\nNow \\( f^{\\prime}(1)=a+b+c \\leq 1 \\), and \\( f(0)=c \\geq-1 \\). Thus \\( a+b \\leq 2 \\). Since \\( a \\) and \\( b \\) are both non-negative, \\( a \\leq 2 \\) and \\( 2 a+b \\leq 4 \\).\n\nCase \\( a=0 \\). If \\( a=0 \\), then\n\\[\nf^{\\prime}(x)=b=\\frac{f(1)-f(-1)}{2},\n\\]\n\nSo\n\\[\n\\left|f^{\\prime}(x)\\right| \\leq \\frac{|f(1)|+|f(-1)|}{2} \\leq 1 .\n\\]\n\nRemark. The polynomial \\( f(x)=2 x^{2}-1 \\) satisties the conditions of the problem and the absolute value of its derivative, \\( |4 x| \\), attains the bound 4 for \\( x= \\pm 1 \\).\n\nSecond Solution. Since \\( f^{\\prime}(x)=2 a x+b \\), a linear function, \\( \\left|f^{\\prime}(x)\\right| \\) assumes its maximum on the closed interval \\( [-1,+1] \\) at one of the two endpoints. Hence\n\\[\n\\max _{|x| \\leq 1}\\left|f^{\\prime}(x)\\right|=|2 a+b| \\quad \\text { or } \\quad|2 a-b| .\n\\]\n\nNow\n\\[\n\\begin{aligned}\n2 a+b & =\\frac{3}{2}(a+b+c)+\\frac{1}{2}(a+c-b)-2 c \\\\\n& =\\frac{3}{2} f(1)+\\frac{1}{2} f(-1)-2 f(0)\n\\end{aligned}\n\\]\n\nSo\n\\[\n|2 a+b| \\leq \\frac{3}{2}|f(1)|+\\frac{1}{2}|f(-1)|+2|f(0)| \\leq \\frac{3}{2}+\\frac{1}{2}+2=4 .\n\\]\n\nAlso\n\\[\n2 a-b=\\frac{1}{2} f(1)+\\frac{3}{2} f(-1)-2 f(0)\n\\]\nand\n\\[\n|2 a-b| \\leq \\frac{1}{2}+\\frac{3}{2}+2=4 .\n\\]\n\nHence \\( \\max \\left|f^{\\prime}(x)\\right| \\leq 4 \\).\nHistorical Note. The chemist Mendeleev raised the question as to the restrictions on \\( p_{n}{ }^{\\prime}(x) \\) for \\( -1 \\leq x \\leq 1 \\) when \\( \\left|p_{n}(x)\\right| \\leq 1 \\) on \\( -1 \\leq x \\leq 1 \\), where \\( p_{n} \\) is a polynomial of degree \\( n \\).\nA. A. Markoff answered this question in 1890 by proving that, if \\( \\left|p_{n}(x)\\right| \\) \\( \\leq 1 \\) on \\( -1 \\leq x \\leq 1 \\), then \\( \\left|p_{n}{ }^{\\prime}(x)\\right| \\leq n^{2} \\) on the same interval. The present problem is thus the special case \\( n=2 \\). It is known that equality occurs if and only if, except for sign, \\( p_{n}(x)=\\cos (n \\arccos x) \\), i.e., \\( p_{n}(x) \\) is the polynomial such that \\( \\cos n \\theta=p_{n}(\\cos \\theta) \\). For \\( n=2, \\cos 2 \\theta= \\) \\( 2 \\cos ^{2} \\theta-1 \\), so \\( p_{2}(x)=\\cos (2 \\arccos x)=2 x^{2}-1 \\). The polynomials \\( p_{n}(x) \\) are called Chebyshev polynomials. See John Todd, A Survey of Numerical Analysis. New York, 1962, pp. 138-139. The generalized version appears as problem 83, in Section 6, Pólya and Szegö, Aufgaben und Lehrsatze aus der Analysis, vol. 2, p. 91 and p. 287.\n\nA slight variation on this problem was used as Problem A5 in the Twentyninth Competition held on December 7, 1968. That problem was phrased as follows: \"Let \\( V \\) be the collection of all quadratic polynomials \\( P \\) with real coefficients such that \\( |P(x)| \\leq 1 \\) for all \\( x \\) on the closed interval \\( [0,1] \\). Determine sup \\( \\left[\\left|P^{\\prime}(0)\\right|: P \\in V\\right] . \" \\)",
  "vars": [
    "x",
    "y",
    "f",
    "V",
    "P",
    "n",
    "p_n",
    "p_2",
    "\\\\theta"
  ],
  "params": [
    "a",
    "b",
    "c"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "inputvar",
        "y": "outputvar",
        "f": "polyfunc",
        "V": "polyset",
        "P": "candidatepoly",
        "n": "degreenum",
        "p_n": "chebyshevp",
        "p_2": "chebyshevtwo",
        "\\theta": "anglevar",
        "a": "coeffa",
        "b": "coeffb",
        "c": "coeffc"
      },
      "question": "1. Suppose that the function \\( polyfunc(inputvar)=coeffa\\, inputvar^{2}+coeffb\\, inputvar+coeffc \\), where \\( coeffa, coeffb, coeffc \\) are real constants, satisfies the condition \\( |polyfunc(inputvar)| \\leq 1 \\) for \\( |inputvar| \\leq 1 \\). Prove that \\( \\left|polyfunc^{\\prime}(inputvar)\\right| \\leq 4 \\) for \\( |inputvar| \\leq 1 \\).",
      "solution": "First Solution. If \\( coeffa \\neq 0 \\) the graph of \\( outputvar = coeffa\\, inputvar^{2}+coeffb\\, inputvar+coeffc \\) is a parabola which can be assumed without loss of generality to open upward, i.e., \\( coeffa>0 \\). [We discuss the straight line case, \\( coeffa=0 \\), later.] By symmetry we may assume that \\( coeffb \\) is non-negative. Then the vertex falls in the left half-plane and it is clear that \\( \\max _{|inputvar| \\leq 1}\\left|polyfunc^{\\prime}(inputvar)\\right| \\) occurs when \\( inputvar=1 \\), and this maximum value is \\( 2\\, coeffa+coeffb \\). It remains to show that \\( 2\\, coeffa+coeffb \\leq 4 \\).\n\nNow \\( polyfunc^{\\prime}(1)=coeffa+coeffb+coeffc \\leq 1 \\), and \\( polyfunc(0)=coeffc \\geq -1 \\). Thus \\( coeffa+coeffb \\leq 2 \\). Since \\( coeffa \\) and \\( coeffb \\) are both non-negative, \\( coeffa \\leq 2 \\) and \\( 2\\, coeffa+coeffb \\leq 4 \\).\n\nCase \\( coeffa=0 \\). If \\( coeffa=0 \\), then\n\\[\npolyfunc^{\\prime}(inputvar)=coeffb=\\frac{polyfunc(1)-polyfunc(-1)}{2},\n\\]\nSo\n\\[\n\\left|polyfunc^{\\prime}(inputvar)\\right| \\leq \\frac{|polyfunc(1)|+|polyfunc(-1)|}{2} \\leq 1 .\n\\]\n\nRemark. The polynomial \\( polyfunc(inputvar)=2\\, inputvar^{2}-1 \\) satisfies the conditions of the problem and the absolute value of its derivative, \\( |4\\, inputvar| \\), attains the bound 4 for \\( inputvar = \\pm 1 \\).\n\nSecond Solution. Since \\( polyfunc^{\\prime}(inputvar)=2\\, coeffa\\, inputvar+coeffb \\), a linear function, \\( \\left|polyfunc^{\\prime}(inputvar)\\right| \\) assumes its maximum on the closed interval \\( [-1,+1] \\) at one of the two endpoints. Hence\n\\[\n\\max _{|inputvar| \\leq 1}\\left|polyfunc^{\\prime}(inputvar)\\right|=|2\\, coeffa+coeffb| \\quad \\text { or } \\quad|2\\, coeffa-coeffb| .\n\\]\n\nNow\n\\[\n\\begin{aligned}\n2\\, coeffa+coeffb &= \\frac{3}{2}(coeffa+coeffb+coeffc)+\\frac{1}{2}(coeffa+coeffc-coeffb)-2\\, coeffc \\\\ &= \\frac{3}{2}\\, polyfunc(1)+\\frac{1}{2}\\, polyfunc(-1)-2\\, polyfunc(0)\n\\end{aligned}\n\\]\nSo\n\\[\n|2\\, coeffa+coeffb| \\leq \\frac{3}{2}|polyfunc(1)|+\\frac{1}{2}|polyfunc(-1)|+2|polyfunc(0)| \\leq \\frac{3}{2}+\\frac{1}{2}+2=4 .\n\\]\n\nAlso\n\\[\n2\\, coeffa-coeffb=\\frac{1}{2}\\, polyfunc(1)+\\frac{3}{2}\\, polyfunc(-1)-2\\, polyfunc(0)\n\\]\nand\n\\[\n|2\\, coeffa-coeffb| \\leq \\frac{1}{2}+\\frac{3}{2}+2=4 .\n\\]\n\nHence \\( \\max \\left|polyfunc^{\\prime}(inputvar)\\right| \\leq 4 \\).\n\nHistorical Note. The chemist Mendeleev raised the question as to the restrictions on \\( chebyshevp^{\\prime}(inputvar) \\) for \\( -1 \\leq inputvar \\leq 1 \\) when \\( \\left|chebyshevp(inputvar)\\right| \\leq 1 \\) on the same interval, where \\( chebyshevp \\) is a polynomial of degree \\( degreenum \\). A. A. Markoff answered this in 1890 by proving that if \\( \\left|chebyshevp(inputvar)\\right| \\leq 1 \\) on \\( -1 \\leq inputvar \\leq 1 \\) then \\( \\left|chebyshevp^{\\prime}(inputvar)\\right| \\leq degreenum^{2} \\). The present problem is the special case \\( degreenum = 2 \\). Equality occurs if and only if, up to sign, \\( chebyshevp(inputvar)=\\cos (degreenum \\arccos inputvar) \\), i.e., \\( chebyshevp(inputvar) \\) satisfies \\( \\cos (degreenum\\, anglevar)=chebyshevp(\\cos anglevar) \\). For \\( degreenum=2 \\) we have \\( \\cos 2\\, anglevar = 2 \\cos ^{2} anglevar - 1 \\), so \\( chebyshevtwo(inputvar)=\\cos (2 \\arccos inputvar)=2\\, inputvar^{2}-1 \\). The polynomials \\( chebyshevp(inputvar) \\) are called Chebyshev polynomials. See John Todd, A Survey of Numerical Analysis, New York, 1962, pp. 138–139. The generalized version appears as problem 83 in Section 6 of Pólya and Szegö, Aufgaben und Lehrsatze aus der Analysis, vol. 2, pp. 91 and 287.\n\nA slight variation on this problem was used as Problem A5 in the Twenty-ninth Competition held on December 7, 1968. That problem was phrased as follows: 'Let \\( polyset \\) be the collection of all quadratic polynomials \\( candidatepoly \\) with real coefficients such that \\( |candidatepoly(inputvar)| \\leq 1 \\) for all \\( inputvar \\) on the closed interval \\( [0,1] \\). Determine \\( \\sup\\left[\\left|candidatepoly^{\\prime}(0)\\right|: candidatepoly \\in polyset\\right] \\).'\n"
    },
    "descriptive_long_confusing": {
      "map": {
        "a": "lightning",
        "b": "sandstorm",
        "c": "waterfall",
        "x": "orangeline",
        "y": "bluemoon",
        "f": "dragonfly",
        "V": "sunflower",
        "P": "bookshelf",
        "n": "peppermint",
        "p_n": "meadowlark",
        "p_2": "rattlesnake",
        "\\theta": "coriander"
      },
      "question": "1. Suppose that the function \\( dragonfly(orangeline)=lightning orangeline^{2}+sandstorm orangeline+waterfall \\), where \\( lightning, sandstorm, waterfall \\) are real constants, satisfies the condition \\( |dragonfly(orangeline)| \\leq 1 \\) for \\( |orangeline| \\leq 1 \\). Prove that \\( \\left|dragonfly^{\\prime}(orangeline)\\right| \\leq 4 \\) for \\( |orangeline| \\leq 1 \\).",
      "solution": "First Solution. If \\( lightning \\neq 0 \\) the graph of \\( bluemoon=lightning orangeline^{2}+sandstorm orangeline+waterfall \\) is a parabola which can be assumed without loss of generality to open upward, i.e.. \\( lightning>0 \\). [We discuss the straight line case, \\( lightning=0 \\), later.] By symmetry we may assume that \\( sandstorm \\) is non-negative. Then the vertex falls in the left halfplane and it is clear that \\( \\max _{|orangeline| \\leq 1}\\left|dragonfly^{\\prime}(orangeline)\\right| \\) occurs when \\( orangeline=1 \\), and this maximum value is \\( 2 lightning+sandstorm \\). It remains to show that \\( 2 lightning+sandstorm \\leq 4 \\).\n\nNow \\( dragonfly^{\\prime}(1)=lightning+sandstorm+waterfall \\leq 1 \\), and \\( dragonfly(0)=waterfall \\geq-1 \\). Thus \\( lightning+sandstorm \\leq 2 \\). Since \\( lightning \\) and \\( sandstorm \\) are both non-negative, \\( lightning \\leq 2 \\) and \\( 2 lightning+sandstorm \\leq 4 \\).\n\nCase \\( lightning=0 \\). If \\( lightning=0 \\), then\n\\[\ndragonfly^{\\prime}(orangeline)=sandstorm=\\frac{dragonfly(1)-dragonfly(-1)}{2},\n\\]\n\nSo\n\\[\n\\left|dragonfly^{\\prime}(orangeline)\\right| \\leq \\frac{|dragonfly(1)|+|dragonfly(-1)|}{2} \\leq 1 .\n\\]\n\nRemark. The polynomial \\( dragonfly(orangeline)=2 orangeline^{2}-1 \\) satisties the conditions of the problem and the absolute value of its derivative, \\( |4 orangeline| \\), attains the bound 4 for \\( orangeline= \\pm 1 \\).\n\nSecond Solution. Since \\( dragonfly^{\\prime}(orangeline)=2 lightning orangeline+sandstorm \\), a linear function, \\( \\left|dragonfly^{\\prime}(orangeline)\\right| \\) assumes its maximum on the closed interval \\( [-1,+1] \\) at one of the two endpoints. Hence\n\\[\n\\max _{|orangeline| \\leq 1}\\left|dragonfly^{\\prime}(orangeline)\\right|=|2 lightning+sandstorm| \\quad \\text { or } \\quad|2 lightning-sandstorm| .\n\\]\n\nNow\n\\[\n\\begin{aligned}\n2 lightning+sandstorm & =\\frac{3}{2}(lightning+sandstorm+waterfall)+\\frac{1}{2}(lightning+waterfall-sandstorm)-2 waterfall \\\\\n& =\\frac{3}{2} dragonfly(1)+\\frac{1}{2} dragonfly(-1)-2 dragonfly(0)\n\\end{aligned}\n\\]\n\nSo\n\\[\n|2 lightning+sandstorm| \\leq \\frac{3}{2}|dragonfly(1)|+\\frac{1}{2}|dragonfly(-1)|+2|dragonfly(0)| \\leq \\frac{3}{2}+\\frac{1}{2}+2=4 .\n\\]\n\nAlso\n\\[\n2 lightning-sandstorm=\\frac{1}{2} dragonfly(1)+\\frac{3}{2} dragonfly(-1)-2 dragonfly(0)\n\\]\nand\n\\[\n|2 lightning-sandstorm| \\leq \\frac{1}{2}+\\frac{3}{2}+2=4 .\n\\]\n\nHence \\( \\max \\left|dragonfly^{\\prime}(orangeline)\\right| \\leq 4 \\).\nHistorical Note. The chemist Mendeleev raised the question as to the restrictions on \\( meadowlark^{\\prime}(orangeline) \\) for \\( -1 \\leq orangeline \\leq 1 \\) when \\( \\left|meadowlark(orangeline)\\right| \\leq 1 \\) on \\( -1 \\leq orangeline \\leq 1 \\), where \\( meadowlark \\) is a polynomial of degree \\( peppermint \\).\nA. A. Markoff answered this question in 1890 by proving that, if \\( \\left|meadowlark(orangeline)\\right| \\) \\( \\leq 1 \\) on \\( -1 \\leq orangeline \\leq 1 \\), then \\( \\left|meadowlark^{\\prime}(orangeline)\\right| \\leq peppermint^{2} \\) on the same interval. The present problem is thus the special case \\( peppermint=2 \\). It is known that equality occurs if and only if, except for sign, \\( meadowlark(orangeline)=\\cos (peppermint \\arccos orangeline) \\), i.e., \\( meadowlark(orangeline) \\) is the polynomial such that \\( \\cos peppermint coriander=meadowlark(\\cos coriander) \\). For \\( peppermint=2, \\cos 2 coriander= \\) \\( 2 \\cos ^{2} coriander-1 \\), so \\( rattlesnake(orangeline)=\\cos (2 \\arccos orangeline)=2 orangeline^{2}-1 \\). The polynomials \\( meadowlark(orangeline) \\) are called Chebyshev polynomials. See John Todd, A Survey of Numerical Analysis. New York, 1962, pp. 138-139. The generalized version appears as problem 83, in Section 6, Pólya and Szegö, Aufgaben und Lehrsatze aus der Analysis, vol. 2, p. 91 and p. 287.\n\nA slight variation on this problem was used as Problem A5 in the Twentyninth Competition held on December 7, 1968. That problem was phrased as follows: \"Let \\( sunflower \\) be the collection of all quadratic polynomials \\( bookshelf \\) with real coefficients such that \\( |bookshelf(orangeline)| \\leq 1 \\) for all \\( orangeline \\) on the closed interval \\( [0,1] \\). Determine sup \\( \\left[\\left|bookshelf^{\\prime}(0)\\right|: bookshelf \\in sunflower\\right] . \"}",
      "confidence": "0.18"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "y": "horizontal",
        "f": "argumentval",
        "V": "singularitem",
        "P": "monomial",
        "n": "infiniteval",
        "p_n": "constantfunc",
        "p_2": "infinitefunc",
        "\\\\theta": "oppositeangle",
        "a": "voidvalue",
        "b": "staticval",
        "c": "variableval"
      },
      "question": "1. Suppose that the function \\( argumentval(constantvalue)=voidvalue constantvalue^{2}+staticval constantvalue+variableval \\), where \\( voidvalue, staticval, variableval \\) are real constants, satisfies the condition \\( |argumentval(constantvalue)| \\leq 1 \\) for \\( |constantvalue| \\leq 1 \\). Prove that \\( \\left|argumentval^{\\prime}(constantvalue)\\right| \\leq 4 \\) for \\( |constantvalue| \\leq 1 \\).",
      "solution": "First Solution. If \\( voidvalue \\neq 0 \\) the graph of \\( horizontal=voidvalue constantvalue^{2}+staticval constantvalue+variableval \\) is a parabola which can be assumed without loss of generality to open upward, i.e. \\( voidvalue>0 \\). [We discuss the straight line case, \\( voidvalue=0 \\), later.] By symmetry we may assume that \\( staticval \\) is non-negative. Then the vertex falls in the left halfplane and it is clear that \\( \\max _{|constantvalue| \\leq 1}\\left|argumentval^{\\prime}(constantvalue)\\right| \\) occurs when \\( constantvalue=1 \\), and this maximum value is \\( 2 voidvalue+staticval \\). It remains to show that \\( 2 voidvalue+staticval \\leq 4 \\).\n\nNow \\( argumentval^{\\prime}(1)=voidvalue+staticval+variableval \\leq 1 \\), and \\( argumentval(0)=variableval \\geq-1 \\). Thus \\( voidvalue+staticval \\leq 2 \\). Since \\( voidvalue \\) and \\( staticval \\) are both non-negative, \\( voidvalue \\leq 2 \\) and \\( 2 voidvalue+staticval \\leq 4 \\).\n\nCase \\( voidvalue=0 \\). If \\( voidvalue=0 \\), then\n\\[\nargumentval^{\\prime}(constantvalue)=staticval=\\frac{argumentval(1)-argumentval(-1)}{2},\n\\]\nSo\n\\[\n\\left|argumentval^{\\prime}(constantvalue)\\right| \\leq \\frac{|argumentval(1)|+|argumentval(-1)|}{2} \\leq 1 .\n\\]\n\nRemark. The polynomial \\( argumentval(constantvalue)=2 constantvalue^{2}-1 \\) satisfies the conditions of the problem and the absolute value of its derivative, \\( |4 constantvalue| \\), attains the bound 4 for \\( constantvalue= \\pm 1 \\).\n\nSecond Solution. Since \\( argumentval^{\\prime}(constantvalue)=2 voidvalue constantvalue+staticval \\), a linear function, \\( \\left|argumentval^{\\prime}(constantvalue)\\right| \\) assumes its maximum on the closed interval \\( [-1,+1] \\) at one of the two endpoints. Hence\n\\[\n\\max _{|constantvalue| \\leq 1}\\left|argumentval^{\\prime}(constantvalue)\\right|=|2 voidvalue+staticval| \\quad \\text { or } \\quad|2 voidvalue-staticval| .\n\\]\nNow\n\\[\n\\begin{aligned}\n2 voidvalue+staticval & =\\frac{3}{2}(voidvalue+staticval+variableval)+\\frac{1}{2}(voidvalue+variableval-staticval)-2 variableval \\\\\n& =\\frac{3}{2} argumentval(1)+\\frac{1}{2} argumentval(-1)-2 argumentval(0)\n\\end{aligned}\n\\]\nSo\n\\[\n|2 voidvalue+staticval| \\leq \\frac{3}{2}|argumentval(1)|+\\frac{1}{2}|argumentval(-1)|+2|argumentval(0)| \\leq \\frac{3}{2}+\\frac{1}{2}+2=4 .\n\\]\nAlso\n\\[\n2 voidvalue-staticval=\\frac{1}{2} argumentval(1)+\\frac{3}{2} argumentval(-1)-2 argumentval(0)\n\\]\nand\n\\[\n|2 voidvalue-staticval| \\leq \\frac{1}{2}+\\frac{3}{2}+2=4 .\n\\]\nHence \\( \\max \\left|argumentval^{\\prime}(constantvalue)\\right| \\leq 4 \\).\n\nHistorical Note. The chemist Mendeleev raised the question as to the restrictions on \\( constantfunc^{\\prime}(constantvalue) \\) for \\( -1 \\leq constantvalue \\leq 1 \\) when \\( \\left|constantfunc(constantvalue)\\right| \\leq 1 \\) on \\( -1 \\leq constantvalue \\leq 1 \\), where \\( constantfunc \\) is a polynomial of degree \\( infiniteval \\).\nA. A. Markoff answered this question in 1890 by proving that, if \\( \\left|constantfunc(constantvalue)\\right| \\leq 1 \\) on \\( -1 \\leq constantvalue \\leq 1 \\), then \\( \\left|constantfunc^{\\prime}(constantvalue)\\right| \\leq infiniteval^{2} \\) on the same interval. The present problem is thus the special case \\( infiniteval=2 \\). It is known that equality occurs if and only if, except for sign, \\( constantfunc(constantvalue)=\\cos (infiniteval \\arccos constantvalue) \\), i.e., \\( constantfunc(constantvalue) \\) is the polynomial such that \\( \\cos infiniteval \\, oppositeangle=constantfunc(\\cos oppositeangle) \\). For \\( infiniteval=2, \\cos 2\\, oppositeangle= 2 \\cos ^{2} oppositeangle-1 \\), so \\( infinitefunc(constantvalue)=\\cos (2 \\arccos constantvalue)=2 constantvalue^{2}-1 \\).\n\nThe polynomials \\( constantfunc(constantvalue) \\) are called Chebyshev polynomials. See John Todd, A Survey of Numerical Analysis. New York, 1962, pp. 138-139. The generalized version appears as problem 83, in Section 6, Pólya and Szegö, Aufgaben und Lehrsatze aus der Analysis, vol. 2, p. 91 and p. 287.\n\nA slight variation on this problem was used as Problem A5 in the Twentyninth Competition held on December 7, 1968. That problem was phrased as follows: \"Let \\( singularitem \\) be the collection of all quadratic polynomials \\( monomial \\) with real coefficients such that \\( |monomial(constantvalue)| \\leq 1 \\) for all \\( constantvalue \\) on the closed interval \\( [0,1] \\). Determine sup \\( \\left[|monomial^{\\prime}(0)|: monomial \\in singularitem\\right] . \"\n"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "y": "hjgrksla",
        "f": "xldmpqen",
        "V": "ivcbwzro",
        "P": "atysnqwe",
        "n": "kdfjqmol",
        "p_n": "rhpastuv",
        "p_2": "bgytelrm",
        "\\\\theta": "mznoylka",
        "a": "osdrfque",
        "b": "waglzkip",
        "c": "tmbvneso"
      },
      "question": "Suppose that the function \\( xldmpqen(qzxwvtnp)=osdrfque \\, qzxwvtnp^{2}+waglzkip \\, qzxwvtnp+tmbvneso \\), where \\( osdrfque, waglzkip, tmbvneso \\) are real constants, satisfies the condition \\( |xldmpqen(qzxwvtnp)| \\leq 1 \\) for \\( |qzxwvtnp| \\leq 1 \\). Prove that \\( \\left|xldmpqen^{\\prime}(qzxwvtnp)\\right| \\leq 4 \\) for \\( |qzxwvtnp| \\leq 1 \\).",
      "solution": "First Solution. If \\( osdrfque \\neq 0 \\) the graph of \\( hjgrksla=osdrfque \\, qzxwvtnp^{2}+waglzkip \\, qzxwvtnp+tmbvneso \\) is a parabola which can be assumed without loss of generality to open upward, i.e., \\( osdrfque>0 \\). [We discuss the straight line case, \\( osdrfque=0 \\), later.] By symmetry we may assume that \\( waglzkip \\) is non-negative. Then the vertex falls in the left halfplane and it is clear that \\( \\max _{|qzxwvtnp| \\leq 1}\\left|xldmpqen^{\\prime}(qzxwvtnp)\\right| \\) occurs when \\( qzxwvtnp=1 \\), and this maximum value is \\( 2 \\, osdrfque+waglzkip \\). It remains to show that \\( 2 \\, osdrfque+waglzkip \\leq 4 \\).\n\nNow \\( xldmpqen^{\\prime}(1)=osdrfque+waglzkip+tmbvneso \\leq 1 \\), and \\( xldmpqen(0)=tmbvneso \\geq-1 \\). Thus \\( osdrfque+waglzkip \\leq 2 \\). Since \\( osdrfque \\) and \\( waglzkip \\) are both non-negative, \\( osdrfque \\leq 2 \\) and \\( 2 \\, osdrfque+waglzkip \\leq 4 \\).\n\nCase \\( osdrfque=0 \\). If \\( osdrfque=0 \\), then\n\\[\nxldmpqen^{\\prime}(qzxwvtnp)=waglzkip=\\frac{xldmpqen(1)-xldmpqen(-1)}{2},\n\\]\n\nSo\n\\[\n\\left|xldmpqen^{\\prime}(qzxwvtnp)\\right| \\leq \\frac{|xldmpqen(1)|+|xldmpqen(-1)|}{2} \\leq 1 .\n\\]\n\nRemark. The polynomial \\( xldmpqen(qzxwvtnp)=2 \\, qzxwvtnp^{2}-1 \\) satisties the conditions of the problem and the absolute value of its derivative, \\( |4 \\, qzxwvtnp| \\), attains the bound 4 for \\( qzxwvtnp= \\pm 1 \\).\n\nSecond Solution. Since \\( xldmpqen^{\\prime}(qzxwvtnp)=2 \\, osdrfque \\, qzxwvtnp+waglzkip \\), a linear function, \\( \\left|xldmpqen^{\\prime}(qzxwvtnp)\\right| \\) assumes its maximum on the closed interval \\( [-1,+1] \\) at one of the two endpoints. Hence\n\\[\n\\max _{|qzxwvtnp| \\leq 1}\\left|xldmpqen^{\\prime}(qzxwvtnp)\\right|=|2 \\, osdrfque+waglzkip| \\quad \\text { or } \\quad|2 \\, osdrfque-waglzkip| .\n\\]\n\nNow\n\\[\n\\begin{aligned}\n2 \\, osdrfque+waglzkip & =\\frac{3}{2}(osdrfque+waglzkip+tmbvneso)+\\frac{1}{2}(osdrfque+tmbvneso-waglzkip)-2 \\, tmbvneso \\\\\n& =\\frac{3}{2} \\, xldmpqen(1)+\\frac{1}{2} \\, xldmpqen(-1)-2 \\, xldmpqen(0)\n\\end{aligned}\n\\]\n\nSo\n\\[\n|2 \\, osdrfque+waglzkip| \\leq \\frac{3}{2}|xldmpqen(1)|+\\frac{1}{2}|xldmpqen(-1)|+2|xldmpqen(0)| \\leq \\frac{3}{2}+\\frac{1}{2}+2=4 .\n\\]\n\nAlso\n\\[\n2 \\, osdrfque-waglzkip=\\frac{1}{2} \\, xldmpqen(1)+\\frac{3}{2} \\, xldmpqen(-1)-2 \\, xldmpqen(0)\n\\]\nand\n\\[\n|2 \\, osdrfque-waglzkip| \\leq \\frac{1}{2}+\\frac{3}{2}+2=4 .\n\\]\n\nHence \\( \\max \\left|xldmpqen^{\\prime}(qzxwvtnp)\\right| \\leq 4 \\).\nHistorical Note. The chemist Mendeleev raised the question as to the restrictions on \\( rhpastuv^{\\prime}(qzxwvtnp) \\) for \\( -1 \\leq qzxwvtnp \\leq 1 \\) when \\( \\left|rhpastuv(qzxwvtnp)\\right| \\leq 1 \\) on \\( -1 \\leq qzxwvtnp \\leq 1 \\), where \\( rhpastuv \\) is a polynomial of degree \\( kdfjqmol \\). A. A. Markoff answered this question in 1890 by proving that, if \\( \\left|rhpastuv(qzxwvtnp)\\right| \\leq 1 \\) on \\( -1 \\leq qzxwvtnp \\leq 1 \\), then \\( \\left|rhpastuv^{\\prime}(qzxwvtnp)\\right| \\leq kdfjqmol^{2} \\) on the same interval. The present problem is thus the special case \\( kdfjqmol=2 \\). It is known that equality occurs if and only if, except for sign, \\( rhpastuv(qzxwvtnp)=\\cos (kdfjqmol \\, mznoylka) \\), i.e., \\( rhpastuv(qzxwvtnp) \\) is the polynomial such that \\( \\cos kdfjqmol \\, mznoylka=rhpastuv(\\cos mznoylka) \\). For \\( kdfjqmol=2, \\cos 2 \\, mznoylka= 2 \\cos ^{2} mznoylka-1 \\), so \\( bgytelrm(qzxwvtnp)=\\cos (2 \\, \\arccos qzxwvtnp)=2 \\, qzxwvtnp^{2}-1 \\). The polynomials \\( rhpastuv(qzxwvtnp) \\) are called Chebyshev polynomials. See John Todd, A Survey of Numerical Analysis. New York, 1962, pp. 138-139. The generalized version appears as problem 83, in Section 6, Pólya and Szegö, Aufgaben und Lehrsatze aus der Analysis, vol. 2, p. 91 and p. 287.\n\nA slight variation on this problem was used as Problem A5 in the Twentyninth Competition held on December 7, 1968. That problem was phrased as follows: \"Let \\( ivcbwzro \\) be the collection of all quadratic polynomials \\( atysnqwe \\) with real coefficients such that \\( |atysnqwe(qzxwvtnp)| \\leq 1 \\) for all \\( qzxwvtnp \\) on the closed interval \\( [0,1] \\). Determine sup \\( \\left[\\left|atysnqwe^{\\prime}(0)\\right|: atysnqwe \\in ivcbwzro\\right] .\""
    },
    "kernel_variant": {
      "question": "Let \\(f(x)=ax^{2}+bx+c\\) be a quadratic polynomial with real coefficients.  Suppose\n\\[\n|f(x)|\\le 3\\qquad\\text{for all }x\\text{ with }|x|\\le 2.\n\\]\nShow that\n\\[\n|f'(x)|\\le 6\\qquad\\text{for every }x\\text{ with }|x|\\le 2.\n\\]",
      "solution": "Let f(x)=ax²+bx+c satisfy |f(x)|≤3 for all x with |x|≤2. We show |f′(x)|≤6 on [−2,2].\n\n1. Since f′(x)=2ax+b is a linear function, its maximum absolute value on the closed interval [−2,2] occurs at the endpoints x=±2. Hence\n   max_{|x|≤2}|f′(x)| = max{|f′(2)|,|f′(−2)|} = max{|4a+b|,|−4a+b|} = max{|4a+b|,|4a−b|}.\n\n2. We express 4a±b in terms of the values f(2), f(−2), and f(0):\n\n   f(2)  = 4a +2b +c,\n   f(−2) = 4a −2b +c,\n   f(0)  = c.\n\n   A direct check shows\n   4a+b  = (¾)f(2)+(¼)f(−2) − f(0),\n   4a−b = (¼)f(2)+(¾)f(−2) − f(0).\n\n3. Using |f(2)|,|f(−2)|,|f(0)|≤3 and the triangle inequality gives for each sign choice:\n   |4a±b| ≤ (¾)|f(2)| + (¼)|f(−2)| + |f(0)| ≤ ¾·3 + ¼·3 +3 =6.\n   Hence\n   max_{|x|≤2}|f′(x)| ≤6.\n\n4. If a=0, then f is linear and f′(x)=b. But then b=(f(2)−f(−2))/4, so |b| ≤(|f(2)|+|f(−2)|)/4 ≤(3+3)/4=1.5<6, which of course satisfies the same inequality.\n\nThus in all cases |f′(x)|≤6 for |x|≤2.\n\nSharpness. The Chebyshev polynomial of degree 2 on [−2,2] scaled to height 3 is\n   f(x) =3·T₂(x/2) =3(2(x/2)²−1) = (3/2)x²−3.\nOne checks |f(x)|≤3 on [−2,2] and f′(2)=3·2=6, so the bound 6 cannot be lowered.",
      "_meta": {
        "core_steps": [
          "f'(x)=2ax+b is linear ⇒ its extreme values on a closed interval lie at the endpoints x=±1",
          "Rewrite the endpoint derivatives 2a±b as a linear combination of f(1), f(0), f(−1)",
          "Apply |f(x)|≤1 at these three sample points and the triangle inequality to bound |2a±b|",
          "Hence max|f'(x)|≤4 (the constant-derivative case a=0 is already inside this bound)"
        ],
        "mutable_slots": {
          "slot1": {
            "description": "Radius of the symmetric interval centred at 0 on which |f| is constrained",
            "original": "1"
          },
          "slot2": {
            "description": "Uniform upper bound placed on |f(x)| throughout that interval",
            "original": "1"
          },
          "slot3": {
            "description": "Corresponding bound obtained for |f'(x)| (scales as 4·slot2/slot1)",
            "original": "4"
          }
        }
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}