{"id":841,"date":"2013-07-13T17:34:21","date_gmt":"2013-07-13T08:34:21","guid":{"rendered":"http:\/\/www.roundown.net\/nyushi\/?p=841"},"modified":"2021-09-14T08:29:51","modified_gmt":"2021-09-13T23:29:51","slug":"iks201303","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/iks201303\/","title":{"rendered":"\u533b\u79d1\u6b6f\u79d1\u59272013\uff1a\u7b2c3\u554f"},"content":{"rendered":"
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\\(m , n\\) \u3092\u81ea\u7136\u6570\u3068\u3057\u3066, \u95a2\u6570 \\(f(x) = x^m (1-x)^n\\) \u3092\u8003\u3048\u308b.\r\n\u3053\u306e\u3068\u304d\u4ee5\u4e0b\u306e\u5404\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n

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  1. (1)<\/strong>\u3000\\(0 \\leqq x \\leqq 1\\) \u306b\u304a\u3051\u308b \\(f(x)\\) \u306e\u6700\u5927\u5024\u3092 \\(m , n\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/p><\/li>\r\n

  2. (2)<\/strong>\u3000\u5b9a\u7a4d\u5206 \\(\\displaystyle\\int _ 0^1 f(x) \\, dx\\) \u3092 \\(m , n\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/p><\/li>\r\n

  3. (3)<\/strong>\u3000\\(a , b , c\\) \u3092\u5b9f\u6570\u3068\u3057\u3066, \u95a2\u6570 \\(g(x) = ax^2+bx+c\\) \u306e \\(0 \\leqq x \\leqq 1\\) \u306b\u304a\u3051\u308b\u6700\u5927\u5024\u3092 \\(M(a,b,c)\\) \u3068\u3059\u308b. \u6b21\u306e \\(2\\) \u6761\u4ef6 (i) , (ii) \u304c\u6210\u7acb\u3059\u308b\u3068\u304d, \\(M(a,b,c)\\) \u306e\u6700\u5c0f\u5024\u3092 \\(m , n\\) \u3092\u7528\u3044\u3066\u8868\u305b.<\/p><\/li>\r\n

  4. (i)\u3000\\(g(0) = g(1) = 0\\)<\/p><\/li>\r\n

  5. (ii)\u3000\\(0 \\lt x \\lt 1\\) \u306e\u3068\u304d \\(f(x) \\leqq g(x)\\)<\/p><\/li>\r\n

  6. (4)<\/strong>\u3000\\(m , n\\) \u304c \\(2\\) \u4ee5\u4e0a\u306e\u81ea\u7136\u6570\u3067 \\(m \\gt n\\) \u3067\u3042\u308b\u3068\u304d\r\n\\[\r\n\\dfrac{(m+n+1)!}{m! n!} \\gt \\dfrac{(m+n)^{m+n}}{m^m n^n} \\gt 2^{2n-1} \\ .\r\n\\]\r\n\u304c\u6210\u7acb\u3059\u308b\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n<\/ol>\r\n

    \r\n\r\n

    \u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n

    (1)<\/strong><\/p>\r\n

    \u6c42\u3081\u308b\u6700\u5927\u5024\u3092 \\(S(m,n)\\) \u3068\u304a\u304f.\r\n\\[\\begin{align}\r\nf'(x) & = mx^{m-1}(1-x)^{n} +x^{m} (-n)(1-x)^{n-1} \\\\\r\n& = \\left\\{ m(1-x) -nx \\right\\} x^{m-1}(1-x)^{n-1} \\\\\r\n& = \\left\\{ m-(m+n)x \\right\\} x^{m-1}(1-x)^{n-1} \\ .\r\n\\end{align}\\]\r\n\\(f'(x) = 0\\) \u3092\u3068\u304f\u3068\r\n\\[\r\nx = \\dfrac{m}{m+n} \\ .\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(0 \\leqq x \\leqq 1\\) \u306b\u304a\u3051\u308b \\(f(x)\\) \u306e\u5897\u6e1b\u306f\u4e0b\u8868\u306e\u3088\u3046\u306b\u306a\u308b.\r\n\\[\r\n\\begin{array}{c|ccccc} x & 0 & \\cdots & \\frac{m}{m+n} & \\cdots & 1 \\\\ \\hline f'(x) & & + & 0 & - & \\\\ \\hline f(x) & 0 & \\nearrow & \\text{\u6700\u5927} & \\searrow & 0 \\\\ \\end{array}\r\n\\]\r\n\u3088\u3063\u3066, \u6c42\u3081\u308b\u6700\u5927\u5024\u306f\r\n\\[\\begin{align}\r\nS(m,n) & = f \\left( \\dfrac{m}{m+n} \\right) \\\\\r\n& = \\left( \\dfrac{m}{m+n} \\right)^{m} \\left( \\dfrac{n}{m+n} \\right)^{n} \\\\\r\n& = \\underline{\\dfrac{m^m n^n}{(m+n)^{m+n}}} \\ .\r\n\\end{align}\\]\r\n

    (2)<\/strong><\/p>\r\n

    \u6c42\u3081\u308b\u7a4d\u5206\u5024\u3092 \\(I(m,n)\\) \u3068\u304a\u304f.
    \r\n\u90e8\u5206\u7a4d\u5206\u3092\u7528\u3044\u308b\u3068\r\n\\[\\begin{align}\r\nI(m,n) & = \\left[ \\dfrac{x^{m+1}}{m+1} (1-x)^n \\right] _ 0^1 +\\dfrac{n}{m+1} \\displaystyle\\int _ 0^1 x^{m+1} (1-x)^{n-1} \\, dx \\\\\r\n& = \\dfrac{n}{m+1} I( m+1 , n-1 ) \\ .\r\n\\end{align}\\]\r\n\u3053\u308c\u3092\u7e70\u8fd4\u3057\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\nI(m,n) & = \\dfrac{n \\cdots 1}{(m+1) \\cdots (m+n)} I( m+n , 0 ) \\\\\r\n& = \\dfrac{m! n!}{(m+n)!} I( m+n , 0 ) \\ .\r\n\\end{align}\\]\r\n\u3053\u3053\u3067\r\n\\[\\begin{align}\r\nI( m+n , 0 ) & = \\displaystyle\\int _ 0^1 x^{m+n} \\, dx \\\\\r\n& = \\left[ \\dfrac{x^{m+n+1}}{m+n+1} \\right] _ 0^1 = \\dfrac{1}{m+n+1} \\ .\r\n\\end{align}\\]\r\n\u306a\u306e\u3067, \u6c42\u3081\u308b\u7a4d\u5206\u5024\u306f\r\n\\[\r\nI(m,n) = \\underline{\\dfrac{m! n!}{(m+n+1)!}} \\ .\r\n\\]\r\n

    (3)<\/strong><\/p>\r\n

    \u6761\u4ef6 (i) \u3088\u308a\r\n\\[\\begin{align}\r\ng(0) & = c = 0 \\\\\r\ng(1) & = a+b+c = 0 \\\\\r\n\\text{\u2234} \\quad b & = -a \\ .\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\\begin{align}\r\ng(x) & = ax^2-ax = -ax(1-x) \\\\\r\n& = a \\left( x-\\dfrac{1}{2} \\right)^2 -\\dfrac{a}{4} \\quad ... [1] \\ .\r\n\\end{align}\\]\r\n

    (1)<\/strong> \u3067\u8abf\u3079\u305f \\(f(x)\\) \u306e\u5897\u6e1b\u3088\u308a, \\(0 \\lt x \\lt 1\\) \u306b\u304a\u3044\u3066 \\(f(x) \\gt 0\\) \u3060\u304b\u3089, \u6761\u4ef6 (ii) \u3088\u308a, \\(g(x) \\gt 0\\) \u3067\u3042\u308b.
    \r\n\u4e00\u65b9, \\(g(x)\\) \u306f \\(2\\) \u6b21\u95a2\u6570\u3060\u304b\u3089, \u4e0a\u306b\u51f8\u3067\u3042\u308c\u3070\u3088\u304f\r\n\\[\r\na \\lt 0 \\ .\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066, [1] \u3088\u308a\r\n\\[\\begin{align}\r\nM(a,b,c) & = -\\dfrac{a}{4} \\\\\r\n\\text{\u2234} \\quad a & = -4 M(a,b,c) \\quad ... [2] \\ .\r\n\\end{align}\\]\r\n\u3055\u3089\u306b, \u6761\u4ef6 (ii) \u3088\u308a\r\n\\[\\begin{align}\r\ng(x) -f(x) & = x(1-x) \\left\\{ -a -x^{m-1} (1-x)^{n-1} \\right\\} \\geqq 0 \\\\\r\n\\text{\u2234} \\quad -a & \\geqq x^{m-1} (1-x)^{n-1} \\quad ( \\ \\text{\u2235} \\ 0 \\leqq x \\leqq 1 ) \\ .\r\n\\end{align}\\]\r\n\u3053\u308c\u304c \\(0 \\leqq x \\leqq 1\\) \u306b\u304a\u3044\u3066\u5e38\u306b\u6210\u7acb\u3059\u308b\u306e\u3067, \u53f3\u8fba\u306e\u6700\u5927\u5024\u3092 \\(A\\) \u3068\u304a\u3051\u3070, \\(-a \\geqq A\\) \u304c\u6210\u7acb\u3059\u308c\u3070\u3088\u3044.<\/p>\r\n

      \r\n

    1. 1*<\/strong>\u3000\\(n = 1\\) \u307e\u305f\u306f \\(m=1\\) \u306e\u3068\u304d
      \r\n\\(A = 1\\) \u3068\u306a\u308b\u306e\u3067, [2] \u3082\u7528\u3044\u3066\r\n\\[\\begin{align}\r\n4 M(a,b,c) & \\geqq 1 \\\\\r\n\\text{\u2234} \\quad M(a,b,c) & \\geqq \\dfrac{1}{4} \\ .\r\n\\end{align}\\]<\/li>\r\n

    2. 2*<\/strong>\u3000\\(m \\geqq 2\\) , \\(n \\geqq 2\\) \u306e\u3068\u304d
      \r\n(1)<\/strong> \u306e\u7d50\u679c\u3088\u308a, \\(A = S( m-1 , n-1 )\\) \u306a\u306e\u3067, [2] \u3082\u7528\u3044\u3066\r\n\\[\\begin{align}\r\n4 M(a,b,c) & \\geqq \\dfrac{(m-1)^{m-1} (n-1)^{n-1}}{(m+n-2)^{m+n-2}} \\\\\r\n\\text{\u2234} \\quad M(a,b,c) & \\geqq \\dfrac{(m-1)^{m-1} (n-1)^{n-1}}{4 (m+n-2)^{m+n-2}} \\ .\r\n\\end{align}\\]<\/li>\r\n<\/ol>\r\n

      \u4ee5\u4e0a\u3088\u308a, \u6c42\u3081\u308b\u6700\u5c0f\u5024\u306f\r\n\\[\r\n\\underline{\\left\\{ \\begin{array}{ll} \\dfrac{1}{4} & ( \\ m=1 \\ \\text{\u307e\u305f\u306f} \\ n=1 \\text{\u306e\u3068\u304d} ) \\\\ \\dfrac{(m-1)^{m-1} (n-1)^{n-1}}{4 (m+n-2)^{m+n-2}} & ( \\ m \\geqq 2 , \\ n \\geqq 2 \\text{\u306e\u3068\u304d} ) \\end{array} \\right.} \\ .\r\n\\]\r\n

      (4)<\/strong><\/p>\r\n

      \u793a\u3059\u3079\u304d\u4e0d\u7b49\u5f0f\u3092\u5909\u5f62\u3059\u308b\u3068\r\n\\[\r\nI(m,n) \\lt S(m,n) \\lt \\dfrac{1}{2^{2n-1}} \\quad ... [\\text{A}] \\ .\r\n\\]\r\n\u306a\u306e\u3067, [A] \u3092\u793a\u305b\u3070\u3088\u3044.<\/p>\r\n