{"id":1252,"date":"2015-08-27T09:54:09","date_gmt":"2015-08-27T00:54:09","guid":{"rendered":"http:\/\/www.roundown.net\/nyushi\/?p=1252"},"modified":"2021-09-09T06:05:42","modified_gmt":"2021-09-08T21:05:42","slug":"osr201501","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/osr201501\/","title":{"rendered":"\u962a\u5927\u7406\u7cfb2015\uff1a\u7b2c1\u554f"},"content":{"rendered":"
\n

\u81ea\u7136\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066\u95a2\u6570 \\(f _ n (x)\\) \u3092\r\n\\[\r\nf _ n (x) = \\dfrac{x}{n (1+x)} \\log \\left( 1 +\\dfrac{x}{n} \\right) \\quad ( x \\geqq 0 )\r\n\\]\r\n\u3067\u5b9a\u3081\u308b. \u4ee5\u4e0b\u306e\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n

    \r\n

  1. (1)<\/strong>\u3000\\(\\displaystyle\\int _ 0^n f(x) \\, dx \\leqq \\displaystyle\\int _ 0^1 \\log (1+x) \\, dx\\) \u3092\u793a\u305b.<\/p><\/li>\r\n

  2. (2)<\/strong>\u3000\u6570\u5217 \\(\\{ I _ n \\}\\) \u3092\r\n\\[\r\nI _ n = \\displaystyle\\int _ 0^n f(x) \\, dx\r\n\\]\r\n\u3067\u5b9a\u3081\u308b. \\(0 \\leqq x \\leqq 1\\) \u306e\u3068\u304d \\(\\log ( 1+x ) \\leqq \\log 2\\) \u3067\u3042\u308b\u3053\u3068\u3092\u7528\u3044\u3066\u6570\u5217 \\(\\{ I _ n \\}\\) \u304c\u53ce\u675f\u3059\u308b\u3053\u3068\u3092\u793a\u3057, \u305d\u306e\u6975\u9650\u5024\u3092\u6c42\u3081\u3088. \u305f\u3060\u3057, \\(\\displaystyle\\lim _ {x \\rightarrow \\infty} \\dfrac{\\log x}{x} = 0\\) \u3067\u3042\u308b\u3053\u3068\u306f\u7528\u3044\u3066\u3088\u3044.<\/p><\/li>\r\n<\/ol>\r\n

    \r\n\r\n

    \u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n

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

    \\(t = \\dfrac{x}{n}\\) \u3068\u304a\u304f\u3068\r\n\\[\r\ndt = \\dfrac{dx}{n} , \\ \\begin{array}{c|ccc} x & 0 & \\rightarrow & n \\\\ \\hline t & 0 & \\rightarrow & 1 \\end{array} \\ .\r\n\\]\r\n\u306a\u306e\u3067\r\n\\[\r\n\\displaystyle\\int _ 0^n f(x) \\, dx = \\displaystyle\\int _ 0^1 \\dfrac{nt}{1 +nt} \\log (1+t) \\, dt \\quad ... [1] \\ .\r\n\\]\r\n\\(0 \\leqq t \\leqq 1\\) \u306b\u304a\u3044\u3066\r\n\\[\r\n0 \\leqq \\dfrac{nt}{1 +nt} \\leqq 1 \\ .\r\n\\]\r\n\u306a\u306e\u3067\r\n\\[\r\n[1] \\leqq \\displaystyle\\int _ 0^1 \\log (1+t) \\, dt \\ .\r\n\\]\r\n\u3059\u306a\u308f\u3061\r\n\\[\r\n\\displaystyle\\int _ 0^n f(x) \\, dx \\leqq \\displaystyle\\int _ 0^1 \\log (1+x) \\, dx \\ .\r\n\\]\r\n

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

    \\(J _ n = \\displaystyle\\int _ 0^1 \\log (1+x) \\, dx -I _ n\\) \u3068\u304a\u304f\u3068, (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\r\nJ _ n \\geqq 0 \\quad ... [2] \\ .\r\n\\]\r\n[1] \u3068 \\(\\log (1+x) \\leqq \\log 2\\) \u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\nJ _ n & = \\displaystyle\\int _ 0^1 \\dfrac{1}{1 +nx} \\log (1+x) \\, dx \\\\\r\n& \\leqq \\log 2 \\displaystyle\\int _ 0^1 \\dfrac{1}{1 +nx} \\, dx \\\\\r\n& = \\log 2 \\left[ \\dfrac{\\log ( 1+nx )}{n} \\right] _ 0^1 \\\\\r\n& = \\log 2 \\cdot \\dfrac{\\log (1+n)}{n} \\quad ... [3] \\ .\r\n\\end{align}\\]\r\n\\(\\displaystyle\\lim _ {x \\rightarrow \\infty} \\dfrac{\\log x}{x} = 0\\) \u3092\u7528\u3044\u308c\u3070\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} \\log 2 \\cdot \\dfrac{\\log (1+n)}{n} = 0 \\ .\r\n\\]\r\n\u306a\u306e\u3067, [2] [3] \u3088\u308a\r\n\\[\\begin{align}\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} J _ n & = 0 \\\\\r\n\\text{\u2234} \\quad \\displaystyle\\lim _ {n \\rightarrow \\infty} I _ n & = \\displaystyle\\int _ 0^1 \\log (1+x) \\, dx \\ .\r\n\\end{align}\\]\r\n\u3088\u3063\u3066, \\(\\{ I _ n \\}\\) \u306f\u53ce\u675f\u3057, \u305d\u306e\u6975\u9650\u5024\u306f\r\n\\[\\begin{align}\r\nI _ n & = \\left[ (1+x) \\log (1+x) \\right] _ 0^1 -\\displaystyle\\int _ 0^1 (1+x) \\dfrac{1}{1+x} \\, dx \\\\\r\n& = 2 \\log 2 -[ x ] _ 0^1 \\\\\r\n& = \\underline{2 \\log 2 -1} \\ .\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\u81ea\u7136\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066\u95a2\u6570 \\(f _ n (x)\\) \u3092 \\[ f _ n (x) = \\dfrac{x}{n (1+x)} \\log \\left( 1 +\\dfrac{x}{n} \\right) \\quad ( … \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":[127],"tags":[142,137],"class_list":["post-1252","post","type-post","status-publish","format-standard","hentry","category-osaka_r_2015","tag-osaka_r","tag-137"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/1252","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=1252"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/1252\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=1252"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=1252"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=1252"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}