(1)<\/strong>\u3000\\(1 \\leqq k \\leqq n\\) \u3092\u6e80\u305f\u3059\u81ea\u7136\u6570 \\(k\\) \u306b\u5bfe\u3057\u3066\r\n\\[\r\n\\displaystyle\\int _ {\\frac{k-1}{2n} \\pi}^{\\frac{k}{2n} \\pi} \\sin 2nt \\cos t \\, dt\r\n= (-1)^{k+1} \\dfrac{2n}{4n^2-1} \\left( \\cos \\dfrac{k}{2n} \\pi +\\cos \\dfrac{k-1}{2n} \\pi \\right)\r\n\\]\r\n\u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n (2)<\/strong>\u3000\u5a92\u4ecb\u5909\u6570 \\(t\\) \u306b\u3088\u3063\u3066\r\n\\[\r\nx = \\sin t , \\ y = \\sin 2nt \\ \uff08 0 \\leqq t \\leqq \\pi \uff09\r\n\\]\r\n\u3068\u8868\u3055\u308c\u308b\u66f2\u7dda \\(C _ n\\) \u3067\u56f2\u307e\u308c\u305f\u90e8\u5206\u306e\u9762\u7a4d \\(S _ n\\) \u3092\u6c42\u3081\u3088. \u305f\u3060\u3057\u5fc5\u8981\u306a\u3089\r\n\\[\r\n\\textstyle\\sum\\limits _ {k=1}^{n-1} \\cos \\dfrac{k}{2n} \\pi = \\dfrac{1}{2} \\left( \\dfrac{1}{\\tan \\frac{\\pi}{4n}} -1 \\right) \\quad ( n \\geqq 2 )\r\n\\]\r\n\u3092\u7528\u3044\u3066\u3082\u3088\u3044.<\/p><\/li>\r\n (3)<\/strong>\u3000\u6975\u9650\u5024 \\(\\displaystyle\\lim _ {n \\rightarrow \\infty} S _ n\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\(f(x) = \\sin 2nx \\cos x\\) , \\(F(x) = \\displaystyle\\int f(x) \\, dx\\) \u3068\u304a\u304f.\r\n\\[\r\nf(x) = \\dfrac{1}{2} \\left\\{ \\sin (2n+1)x +\\sin (2n-1)x \\right\\}\r\n\\]\r\n\u306a\u306e\u3067\r\n\\[\r\nF(x) = -\\dfrac{1}{2} \\left\\{ \\dfrac{\\cos (2n+1)x}{2n+1} +\\dfrac{\\cos (2n-1)x}{2n-1}\\right\\}+C \\quad ... [1]\r\n\\]\r\n\u305f\u3060\u3057, \\(C\\) \u306f\u7a4d\u5206\u5b9a\u6570. (2)<\/strong><\/p>\r\n \\(x(t) = \\sin t\\) , \\(y(t) = \\sin 2nt \\ ( 0 \\leqq t \\leqq \\pi )\\) \u3068\u304a\u304f.\r\n\\[\\begin{align}\r\nx( \\pi -t ) & = \\sin t = x(t) , \\\\\r\ny( \\pi -t ) & = \\sin (-2nt) = -\\sin 2nt = -y(t) \\\\\r\n\\end{align}\\]\r\n\u306a\u306e\u3067, \\(C\\) \u306f \\(x\\) \u8ef8\u306b\u3064\u3044\u3066\u5bfe\u79f0\u3067\u3042\u308b. (3)<\/strong><\/p>\r\n \\[\\begin{align}\r\nS & = \\dfrac{16n^2 \\cos \\frac{\\pi}{4n}}{\\pi ( 4n^2-1 )} \\cdot \\dfrac{1}{\\frac{4n}{\\pi} \\sin \\frac{\\pi}{4n}} \\\\\r\n& = \\dfrac{16 \\cos \\frac{\\pi}{4n}}{\\pi \\left( 4-\\frac{1}{n^2} \\right)} \\cdot \\dfrac{1}{\\frac{4n}{\\pi} \\sin \\frac{\\pi}{4n}} \\\\\r\n& \\rightarrow \\dfrac{16 \\cdot 1}{\\pi (4-0)} \\cdot \\dfrac{1}{1} \\quad ( n \\rightarrow \\infty \\text{\u306e\u3068\u304d} ) \\\\\r\n& = \\dfrac{4}{\\pi}\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} S _ n = \\underline{\\dfrac{4}{\\pi}}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\\(n\\) \u306f\u81ea\u7136\u6570\u3068\u3059\u308b. (1)\u3000\\(1 \\leqq k \\leqq n\\) \u3092\u6e80\u305f\u3059\u81ea\u7136\u6570 \\(k\\) \u306b\u5bfe\u3057\u3066 \\[ \\displaystyle\\int _ {\\frac{k-1}{2n} \\pi}^{\\frac … \u7d9a\u304d\u3092\u8aad\u3080 ![\"tsukuba_r_2013_02_01\"](\"\/\/www.roundown.net\/nyushi\/wp-content\/uploads\/tsukuba_r_2013_02_01.png\")
\r\n
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n
\r\n\u7d9a\u3044\u3066, \\(\\cos k \\pi = (-1)^k\\) , \\(\\sin k \\pi = 0\\) \u3067\u3042\u308b\u3053\u3068\u3092\u7528\u3044\u308c\u3070, \u52a0\u6cd5\u5b9a\u7406\u3088\u308a\r\n\\[\\begin{align}\r\n\\cos \\left( k \\pi +\\dfrac{k}{m} \\pi \\right) & = \\cos k \\pi \\cos \\dfrac{k}{m} \\pi \\\\\r\n& = (-1)^k \\cos \\dfrac{k}{m} \\pi \\quad ... [2] , \\\\\r\n\\cos \\left( k \\pi -\\dfrac{k}{m} \\pi \\right) & = \\cos k \\pi \\cos \\left( -\\dfrac{k}{m} \\pi \\right) \\\\\r\n& = (-1)^{k+1} \\cos \\dfrac{k}{m} \\pi \\quad ... [3]\r\n\\end{align}\\]\r\n\u305f\u3060\u3057, \u3053\u3053\u3067 \\(m , k\\) \u306f\u81ea\u7136\u6570\u3067 \\(1 \\leqq k \\leqq m\\) \u3092\u307f\u305f\u3057\u3066\u3044\u308b.
\r\n\u3057\u305f\u304c\u3063\u3066, [1] \uff5e [3] \u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\n\\displaystyle\\int _ {\\frac{k-1}{2n} \\pi}^{\\frac{k}{2n} \\pi} & \\sin 2nt \\cos t \\, dt \\\\\r\n& = F \\left( \\dfrac{k \\pi}{2n} \\right) -F \\left( \\dfrac{(k-1) \\pi}{2n} \\right) \\\\\r\n& = -\\dfrac{1}{2} \\left\\{ \\dfrac{(-1)^k \\cos \\frac{k \\pi}{2n}}{2n+1} +\\dfrac{(-1)^{k+1} \\cos \\frac{k \\pi}{2n}}{2n-1} \\right\\} \\\\\r\n& \\hspace{2em} +\\dfrac{1}{2} \\left\\{ \\dfrac{(-1)^k \\cos \\frac{(k-1) \\pi}{2n}}{2n+1} +\\dfrac{(-1)^{k+1} \\cos \\frac{(k-1) \\pi}{2n}}{2n-1} \\right\\} \\\\\r\n& = \\dfrac{(-1)^{k+1}}{2} \\left( \\dfrac{1}{2n+1} +\\dfrac{1}{2n-1} \\right) \\left( \\cos \\dfrac{k}{2n} \\pi +\\cos \\dfrac{k-1}{2n} \\pi \\right) \\\\\r\n& = (-1)^{k+1} \\dfrac{2n}{4n^2-1} \\left( \\cos \\dfrac{k}{2n} \\pi +\\cos \\dfrac{k-1}{2n} \\pi \\right)\r\n\\end{align}\\]\r\n
\r\n\u3055\u3089\u306b \\(0 \\leqq t \\leqq \\dfrac{\\pi}{2}\\) \u306b\u304a\u3044\u3066\r\n\\[\r\nx'(t) = \\cos t \\geqq 0\r\n\\]\r\n\u306a\u306e\u3067, \\(C\\) \u306e\u3046\u3061 \\(0 \\leqq t \\leqq \\dfrac{\\pi}{2}\\) \u306e\u90e8\u5206\u304c\u8868\u3059\u66f2\u7dda \\(C'\\) \u3068\u304a\u304f\u3068, \\(C'\\) \u306f\u81ea\u8eab\u3068\u306e\u4ea4\u70b9\u3092\u3082\u305f\u305a, \\(x\\) \u8ef8\u304c\u56f2\u3080\u90e8\u5206\u306e\u9762\u7a4d\u3092 \\(S'\\) \u3068\u304a\u3051\u3070\r\n\\[\r\nS = 2 S' \\quad ... [4]\r\n\\]\r\n\\(y(t) = 0\\) \u3092\u3068\u304f\u3068\r\n\\[\r\nt = \\dfrac{k \\pi}{2n} \\quad ( k = 0 , 1 , \\cdots , n )\r\n\\]\r\n\u306a\u306e\u3067, (1)<\/strong> \u306e\u7d50\u679c\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\nS' & = \\textstyle\\sum\\limits _ {k=1}^n \\displaystyle\\int _ {\\sin \\frac{k-1}{2n} \\pi}^{\\sin \\frac{k}{2n} \\pi} |y| \\, dx \\\\\r\n& = \\textstyle\\sum\\limits _ {k=1}^n \\displaystyle\\int _ {\\frac{k-1}{2n} \\pi}^{\\frac{k}{2n} \\pi} | y(t) | x'(t) \\, dt \\\\\r\n& = \\textstyle\\sum\\limits _ {k=1}^n \\displaystyle\\int _ {\\frac{k-1}{2n} \\pi}^{\\frac{k}{2n} \\pi} \\left| \\sin 2nt \\cos t \\right| \\, dt \\ \uff08 \u2235 \\cos t \\geqq 0 \uff09\\\\\r\n& = \\textstyle\\sum\\limits _ {k=1}^n \\dfrac{2n}{4n^2-1} \\left( \\cos \\dfrac{k}{2n} \\pi +\\cos \\dfrac{k-1}{2n} \\pi \\right) \\\\\r\n& = \\dfrac{2n}{4n^2-1} \\left( \\cos 0 +2 \\textstyle\\sum\\limits _ {k=1}^{n-1} \\cos \\dfrac{k}{2n} \\pi +\\cos \\dfrac{\\pi}{2} \\right) \\\\\r\n& = \\dfrac{2n}{4n^2-1} \\left( 1 +\\dfrac{1}{\\tan \\frac{\\pi}{4n}} -1 +0 \\right) \\\\\r\n& = \\dfrac{2n}{( 4n^2-1 ) \\tan \\frac{\\pi}{4n}}\r\n\\end{align}\\]\r\n\u3088\u3063\u3066, [4] \u3088\u308a\r\n\\[\r\nS = 2S' = \\underline{\\dfrac{4n}{( 4n^2-1 ) \\tan \\frac{\\pi}{4n}}}\r\n\\]\r\n