\u5ea7\u6a19\u5e73\u9762\u4e0a\u306b, \u30d9\u30af\u30c8\u30eb \\(\\overrightarrow{u _ 1} , \\overrightarrow{u _ 2} , \\cdots , \\overrightarrow{u _ n}\\) \u304c\u3042\u308b.\r\n\u5ea7\u6a19\u5e73\u9762\u4e0a\u306e\u30d9\u30af\u30c8\u30eb \\(\\overrightarrow{p}\\) \u306e\u3046\u3061, \\(\\textstyle\\sum\\limits _ {k=1}^n k \\left| \\overrightarrow{p} -\\overrightarrow{u _ k} \\right|^2\\) \u3092\u6700\u5c0f\u306b\u3059\u308b\u3082\u306e\u3092 \\(\\overrightarrow{v}\\) \u3068\u3057, \u305d\u306e\u3068\u304d\u306e\u6700\u5c0f\u5024\u3092 \\(m\\) \u3068\u3059\u308b. \u6b21\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n
\u3000\u4ee5\u4e0b, \\(\\overrightarrow{u _ k} = \\left( \\cos \\dfrac{k \\alpha}{n} , \\sin \\dfrac{k \\alpha}{n} \\right) \\ ( k = 1, 2, \\cdots , n )\\) \u306e\u5834\u5408\u3092\u8003\u3048\u308b, \u305f\u3060\u3057, \\(\\alpha\\) \u306f\u6b63\u306e\u5b9a\u6570\u3068\u3059\u308b.<\/p>\r\n
(2)<\/strong>\u3000\\(\\displaystyle\\lim _ {n \\rightarrow \\infty} \\left| \\overrightarrow{v} \\right|^2\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n (3)<\/strong>\u3000\\(\\displaystyle\\lim _ {n \\rightarrow \\infty} \\dfrac{m}{n^2}\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \\[\\begin{align}\r\n\\textstyle\\sum\\limits _ {k=1}^n k \\left| \\overrightarrow{p} -\\overrightarrow{u _ k} \\right|^2 & = \\textstyle\\sum\\limits _ {k=1}^n k \\left( \\left| \\overrightarrow{p} \\right|^2 -2 \\overrightarrow{p} \\cdot \\overrightarrow{u _ k} + \\left| \\overrightarrow{u _ k} \\right|^2 \\right) \\\\\r\n& = \\dfrac{n(n+1)}{2} \\left| \\overrightarrow{p} \\right|^2 -2 \\left( \\textstyle\\sum\\limits _ {k=1}^n k \\overrightarrow{u _ k} \\right) \\cdot \\overrightarrow{p} +\\textstyle\\sum\\limits _ {k=1}^n k \\left| \\overrightarrow{u _ k} \\right|^2 \\\\\r\n& = \\dfrac{n(n+1)}{2} \\left| \\overrightarrow{p} -\\dfrac{2}{n(n+1)} \\textstyle\\sum\\limits _ {k=1}^n k \\overrightarrow{u _ k} \\right|^2 \\\\\r\n& \\qquad +\\textstyle\\sum\\limits _ {k=1}^n k \\left| \\overrightarrow{u _ k} \\right|^2 -\\dfrac{2}{n(n+1)} \\left| \\textstyle\\sum\\limits _ {k=1}^n k \\overrightarrow{u _ k} \\right|^2 \\quad ... [1]\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\r\n\\overrightarrow{v} = \\underline{\\dfrac{2}{n(n+1)} \\textstyle\\sum\\limits _ {k=1}^n k \\overrightarrow{u _ k}}\r\n\\]\r\n (2)<\/strong><\/p>\r\n (1)<\/strong> \u306e\u7d50\u679c\u3068\u6761\u4ef6\u3088\u308a\r\n\\[\\begin{align}\r\n\\left| \\overrightarrow{v} \\right|^2 & = \\dfrac{4}{n^2 (n+1)^2} \\left\\{ \\left( \\textstyle\\sum\\limits _ {k=1}^n k \\cos \\dfrac{k \\alpha}{n} \\right)^2 +\\left( \\textstyle\\sum\\limits _ {k=1}^n k \\sin \\dfrac{k \\alpha}{n} \\right)^2 \\right\\} \\\\\r\n& = \\underline{\\dfrac{4}{\\left( 1 +\\frac{1}{n} \\right)^2}} _ {[2]} \\left\\{ \\left( \\underline{\\dfrac{1}{n} \\textstyle\\sum\\limits _ {k=1}^n \\dfrac{k}{n} \\cos \\dfrac{k \\alpha}{n}} _ {[3]} \\right)^2 +\\left( \\underline{\\dfrac{1}{n} \\textstyle\\sum\\limits _ {k=1}^n \\dfrac{k}{n} \\sin \\dfrac{k \\alpha}{n}} _ {[4]} \\right)^2 \\right\\}\r\n\\end{align}\\]\r\n\u3053\u3053\u3067, [2] \uff5e [4] \u306b\u3064\u3044\u3066, \\(n \\rightarrow \\infty\\) \u3068\u3059\u308c\u3070\r\n\\[\\begin{align}\r\n[2] & \\rightarrow 4 , \\\\\r\n[3] & \\rightarrow \\displaystyle\\int _ 0^1 x \\cos \\alpha x \\, dx \\\\\r\n& = \\left[ \\dfrac{x \\sin \\alpha x}{\\alpha} \\right] _ 0^1 -\\dfrac{1}{\\alpha} \\displaystyle\\int _ 0^1 \\sin \\alpha x \\, dx \\\\\r\n& = \\dfrac{\\sin \\alpha}{\\alpha} +\\dfrac{1}{\\alpha} \\left[ \\dfrac{\\cos \\alpha x}{\\alpha} \\right] _ 0^1 \\\\\r\n& = \\dfrac{\\alpha \\sin \\alpha +\\cos \\alpha -1}{\\alpha^2} , \\\\\r\n[4] & \\rightarrow \\displaystyle\\int _ 0^1 x \\sin \\alpha x \\, dx \\\\\r\n& = \\left[ -\\dfrac{x \\cos \\alpha x}{\\alpha} \\right] _ 0^1 +\\dfrac{1}{\\alpha} \\displaystyle\\int _ 0^1 \\cos \\alpha x \\, dx \\\\\r\n& = -\\dfrac{\\cos \\alpha}{\\alpha} +\\dfrac{1}{\\alpha} \\left[ \\dfrac{\\sin \\alpha x}{\\alpha} \\right] _ 0^1 \\\\\r\n& = \\dfrac{-\\alpha \\cos \\alpha +\\sin \\alpha}{\\alpha^2}\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\\begin{align}\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} \\left| \\overrightarrow{v} \\right|^2 & = 4 \\left\\{ \\left( \\dfrac{\\alpha \\sin \\alpha +\\cos \\alpha -1}{\\alpha^2} \\right)^2 +\\left( \\dfrac{-\\alpha \\cos \\alpha +\\sin \\alpha}{\\alpha^2} \\right)^2 \\right\\} \\\\\r\n& = \\underline{\\dfrac{4}{\\alpha^4} \\left( \\alpha^2 -2 \\alpha \\sin \\alpha -2 \\cos \\alpha +2 \\right)}\r\n\\end{align}\\]\r\n (3)<\/strong><\/p>\r\n (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\\begin{align}\r\n\\left| \\overrightarrow{v} \\right| & = \\dfrac{2}{n(n+1)} \\left| \\textstyle\\sum\\limits _ {k=1}^n k \\overrightarrow{u _ k} \\right| \\\\\r\n\\text{\u2234} \\quad \\left| \\textstyle\\sum\\limits _ {k=1}^n k \\overrightarrow{u _ k} \\right| & = \\dfrac{n(n+1)}{2} \\left| \\overrightarrow{v} \\right|\r\n\\end{align}\\]\r\n\u3053\u308c\u3092\u7528\u3044\u308c\u3070, [1] \u3088\u308a\r\n\\[\\begin{align}\r\nm & = \\textstyle\\sum\\limits _ {k=1}^n k \\left| \\overrightarrow{u _ k} \\right|^2 -\\dfrac{2}{n(n+1)} \\left| \\textstyle\\sum\\limits _ {k=1}^n k \\overrightarrow{u _ k} \\right|^2 \\\\\r\n& = \\dfrac{n(n+1)}{2} \\left( 1 -\\left| \\overrightarrow{v} \\right|^2 \\right)\r\n\\end{align}\\]\r\n\u3088\u3063\u3066, (2)<\/strong> \u306e\u7d50\u679c\u3082\u7528\u3044\u3066\r\n\\[\\begin{align}\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} \\dfrac{m}{n^2} & = \\displaystyle\\lim _ {n \\rightarrow \\infty} \\dfrac{1 +\\frac{1}{n}}{2} \\left( 1 -\\left| \\overrightarrow{v} \\right|^2 \\right) \\\\\r\n& = \\underline{\\dfrac{\\alpha^4 -4 \\alpha^2 +8 \\alpha \\sin \\alpha +8 \\cos \\alpha -8}{2 \\alpha^4}}\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u5ea7\u6a19\u5e73\u9762\u4e0a\u306b, \u30d9\u30af\u30c8\u30eb \\(\\overrightarrow{u _ 1} , \\overrightarrow{u _ 2} , \\cdots , \\overrightarrow{u _ n}\\) \u304c\u3042\u308b. \u5ea7\u6a19\u5e73\u9762\u4e0a\u306e … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n