\\(x\\) \u306e \\(3\\) \u6b21\u95a2\u6570 \\(f(x) = ax^3+bx^2+cx+d\\) \u304c, \\(3\\) \u3064\u306e\u6761\u4ef6\r\n\\[\r\nf(1) = 1 , \\ f(-1) = -1 , \\ \\displaystyle\\int _ {-1}^{1} ( bx^2+cx+d ) \\, dx = 1\r\n\\]\r\n\u3092\u5168\u3066\u6e80\u305f\u3057\u3066\u3044\u308b\u3068\u3059\u308b. \u3053\u306e\u3088\u3046\u306a \\(f(x)\\) \u306e\u4e2d\u3067\u5b9a\u7a4d\u5206\r\n\\[\r\nI = \\displaystyle\\int _ {-1}^{\\frac{1}{2}} \\left\\{ f''(x) \\right\\}^2 \\, dx\r\n\\]\r\n\u3092\u6700\u5c0f\u306b\u3059\u308b\u3082\u306e\u3092\u6c42\u3081, \u305d\u306e\u3068\u304d\u306e \\(I\\) \u306e\u5024\u3092\u6c42\u3081\u3088. \u305f\u3060\u3057, \\(f''(x)\\) \u306f \\(f'(x)\\) \u306e\u5c0e\u95a2\u6570\u3092\u8868\u3059.<\/p>\r\n
\u6761\u4ef6\u3088\u308a\r\n\\[\\begin{align}\r\nf(1) & = a+b+c+d = 1 , \\\\\r\nf(-1) & = -a+b-c+d = -1\r\n\\end{align}\\]\r\n\\(2\\) \u5f0f\u3092\u52a0\u6e1b\u3059\u308b\u3068\r\n\\[\\begin{align}\r\nb+d = 0 & , \\ a+c = 1 \\\\\r\n\\text{\u2234} \\quad d = -b & , \\ c = 1-a \\quad ... [1]\r\n\\end{align}\\]\r\n\u3055\u3089\u306b\r\n\\[\\begin{align}\r\n\\displaystyle\\int _ {-1}^{1} ( bx^2+cx+d ) \\, dx & = 2 \\left[ \\dfrac{bx^3}{3} +dx \\right] _ {0}^{1} \\\\\r\n& = \\dfrac{2b}{3} +2d \\\\\r\n& = -\\dfrac{4b}{3} \\quad ( \\ \\text{\u2235} \\ [1] \\ )\r\n\\end{align}\\]\r\n\u306a\u306e\u3067, \u6761\u4ef6\u3088\u308a\r\n\\[\\begin{align}\r\n-\\dfrac{4b}{3} & = 1 \\\\\r\n\\text{\u2234} \\quad b & = -\\dfrac{3}{4}\r\n\\end{align}\\]\r\n[1] \u3088\u308a\r\n\\[\r\nd = \\dfrac{3}{4}\r\n\\]\r\n\u3053\u308c\u3089\u3092\u7528\u3044\u308b\u3068\r\n\\[\\begin{align}\r\nf'(x) & = 3ax^2 +2bx +c , \\\\\r\nf''(x) & = 6ax +2b = \\dfrac{3}{2} ( 4ax-1 )\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\\begin{align}\r\nI & = \\dfrac{9}{4} \\displaystyle\\int _ {-1}^{\\frac{1}{2}} ( 4ax-1 )^2 \\, dx \\\\\r\n& = \\dfrac{9}{4} \\left[ \\dfrac{16a^2x^3}{3} -4ax^2 +x \\right] _ {-1}^{\\frac{1}{2}} \\\\\r\n& = \\dfrac{9}{4} \\left\\{ \\left( \\dfrac{2a^2}{3} -a +\\dfrac{1}{2} \\right) -\\left( -\\dfrac{16a^2}{3} -4a -1 \\right) \\right\\} \\\\\r\n& = \\dfrac{9}{4} \\left( 6a^2 +3a +\\dfrac{3}{2} \\right) \\\\\r\n& = \\dfrac{9}{4} \\left\\{ 6 \\left( a +\\dfrac{1}{4} \\right)^2 +\\dfrac{9}{8} \\right\\} \\\\\r\n& \\geqq \\dfrac{9}{4} \\cdot \\dfrac{9}{8} = \\dfrac{81}{32}\r\n\\end{align}\\]\r\n\u7b49\u53f7\u6210\u7acb\u306f, \\(a = -\\dfrac{1}{4}\\) \u306e\u3068\u304d\u3067\u3042\u308b.
\r\n[1] \u3088\u308a, \\(c = \\dfrac{5}{4}\\) .
\r\n\u4ee5\u4e0a\u3088\u308a, \\(I\\) \u306f\r\n\\[\r\nf(x) = \\underline{-\\dfrac{1}{4}x^3 -\\dfrac{3}{4}x^2 +\\dfrac{5}{4}x +\\dfrac{3}{4}}\r\n\\]\r\n\u306e\u3068\u304d, \u6700\u5c0f\u5024 \\(\\underline{\\dfrac{81}{32}}\\) \u3092\u3068\u308b.<\/p>\r\n","protected":false},"excerpt":{"rendered":"\\(x\\) \u306e \\(3\\) \u6b21\u95a2\u6570 \\(f(x) = ax^3+bx^2+cx+d\\) \u304c, \\(3\\) \u3064\u306e\u6761\u4ef6 \\[ f(1) = 1 , \\ f(-1) = -1 , \\ \\displaystyle\\int _ { … \u7d9a\u304d\u3092\u8aad\u3080