{"id":652,"date":"2013-02-24T13:43:32","date_gmt":"2013-02-24T04:43:32","guid":{"rendered":"http:\/\/www.roundown.net\/nyushi\/?p=652"},"modified":"2021-09-10T19:06:21","modified_gmt":"2021-09-10T10:06:21","slug":"osr200701","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/osr200701\/","title":{"rendered":"\u962a\u5927\u7406\u7cfb2007\uff1a\u7b2c1\u554f"},"content":{"rendered":"
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\\(n\\) \u3092\u81ea\u7136\u6570\u3068\u3059\u308b. \u95a2\u6570 \\(y = \\sqrt{x}\\) \u306e\u30b0\u30e9\u30d5\u3092 \\(C\\) \u3068\u3057, \\(C\\) \u4e0a\u306e \\(2\\) \u70b9 \\(( n , \\sqrt{n})\\) \u3068 \\(( n+1 , \\sqrt{n+1})\\) \u3092\u901a\u308b\u76f4\u7dda\u3092 \\(l\\) \u3068\u3059\u308b.\r\n\\(C\\) \u3068 \\(l\\) \u3067\u56f2\u307e\u308c\u305f\u90e8\u5206\u3092 \\(x\\) \u8ef8\u306e\u307e\u308f\u308a\u306b \\(1\\) \u56de\u8ee2\u3055\u305b\u3066\u3067\u304d\u308b\u7acb\u4f53\u306e\u4f53\u7a4d\u3092 \\(V\\) \u3068\u3059\u308b.\r\n\u3053\u306e\u3068\u304d \\(\\displaystyle\\lim _ {n \\rightarrow \\infty} n^a V = b\\) \u3092\u6e80\u305f\u3059\u6b63\u306e\u6570 \\(a , b\\) \u3092\u6c42\u3081\u3088.<\/p>\r\n

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\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n\"osaka_r_2007_01_01\"\r\n

\\(C\\) \u3068 \\(l\\) \u306b\u56f2\u307e\u308c\u305f\u90e8\u5206\u306f\u4e0a\u56f3\u659c\u7dda\u90e8\u306e\u3088\u3046\u306b\u306a\u308b\u306e\u3067\r\n\\[\\begin{align}\r\nV & = \\pi \\displaystyle\\int _ {n}^{n+1} \\left( \\sqrt{x} \\right)^2 \\, dx \\\\\r\n& \\qquad -\\dfrac{\\pi}{3} \\left( \\sqrt{n+1} \\right)^2 \\cdot \\dfrac{1 \\cdot \\sqrt{n+1}}{\\sqrt{n+1} -\\sqrt{n}} \\cdot \\left\\{ 1 -\\left( \\dfrac{\\sqrt{n}}{\\sqrt{n+1}} \\right)^3 \\right\\} \\\\\r\n& = \\pi \\left[ \\dfrac{x^2}{2} \\right] _ {n}^{n+1} -\\dfrac{\\pi}{3} \\cdot \\dfrac{\\left( \\sqrt{n+1} \\right)^3 -\\left( \\sqrt{n} \\right)^3}{\\sqrt{n+1} -\\sqrt{n}} \\\\\r\n& = \\dfrac{2n+1}{2} \\pi -\\dfrac{2n+1 + \\sqrt{n(n+1)}}{3} \\pi \\\\\r\n& = \\dfrac{\\pi}{6} \\left\\{ 2n+1 -2 \\sqrt{n(n+1)}\\right\\} \\ .\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\\begin{align}\r\nn^a V & = \\dfrac{n^a \\pi}{6} \\cdot \\dfrac{(2n+1)^2 -4 n(n+1)}{2n+1 +2 \\sqrt{n(n+1)}} \\\\\r\n& = \\dfrac{n^{a-1} \\pi}{6} \\cdot \\underline{\\dfrac{1}{2 +\\frac{1}{n} +2 \\sqrt{1 +\\frac{1}{n}}}} _ {[1]} \\ .\r\n\\end{align}\\]\r\n\u3053\u3053\u3067, \u4e0b\u7dda\u90e8 [1] \u306b\u3064\u3044\u3066\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} [1] = \\dfrac{1}{4} \\ .\r\n\\]\r\n\u306a\u306e\u3067, \\(\\displaystyle\\lim _ {n \\rightarrow \\infty} n^a V = b\\) \u3068\u306a\u308b\u306e\u306f, \\(n \\rightarrow \\infty\\) \u306e\u3068\u304d, \\(n^{a-1}\\) \u304c\u6b63\u306e\u5024\u306b\u53ce\u675f\u3059\u308b\u3068\u304d\u3067\u3042\u308b.<\/p>\r\n

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  1. 1*<\/strong>\u3000\\(0 \\lt a \\lt 1\\) \u306e\u3068\u304d\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} n^{a-1} = 0 \\ .\r\n\\]\r\n\u306a\u306e\u3067, \u4e0d\u9069.<\/p><\/li>\r\n

  2. 2*<\/strong>\u3000\\(a=1\\) \u306e\u3068\u304d\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} n^{a-1} = 1 \\ .\r\n\\]\r\n\u306a\u306e\u3067\r\n\\[\r\nb = \\dfrac{\\pi}{6} \\cdot \\dfrac{1}{4} = \\dfrac{\\pi}{24} \\ .\r\n\\]<\/li>\r\n

  3. 3*<\/strong>\u3000\\(a \\gt 1\\) \u306e\u3068\u304d\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} n^{a-1} = \\infty \\ .\r\n\\]\r\n\u306a\u306e\u3067, \u4e0d\u9069.<\/p><\/li>\r\n<\/ol>\r\n

    \u3088\u3063\u3066\r\n\\[\r\na = \\underline{1} , \\quad b = \\underline{\\dfrac{\\pi}{24}} \\ .\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\\(n\\) \u3092\u81ea\u7136\u6570\u3068\u3059\u308b. \u95a2\u6570 \\(y = \\sqrt{x}\\) \u306e\u30b0\u30e9\u30d5\u3092 \\(C\\) \u3068\u3057, \\(C\\) \u4e0a\u306e \\(2\\) \u70b9 \\(( n , \\sqrt{n})\\) \u3068 \\(( n+1 , \\sqrt{n+1} … \u7d9a\u304d\u3092\u8aad\u3080 →<\/span><\/a>","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false,"footnotes":""},"categories":[89],"tags":[142,109],"class_list":["post-652","post","type-post","status-publish","format-standard","hentry","category-osaka_r_2007","tag-osaka_r","tag-109"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/652","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/comments?post=652"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/652\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=652"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=652"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=652"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}