{"id":1987,"date":"2021-11-21T15:57:23","date_gmt":"2021-11-21T06:57:23","guid":{"rendered":"https:\/\/www.roundown.net\/nyushi\/?p=1987"},"modified":"2021-11-21T15:57:23","modified_gmt":"2021-11-21T06:57:23","slug":"osr202103","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/osr202103\/","title":{"rendered":"\u962a\u5927\u7406\u7cfb2021\uff1a\u7b2c3\u554f"},"content":{"rendered":"
\n

\\(n\\) \u3092 \u81ea\u7136\u6570\u3068\u3057, \\(t\\) \u3092 \\(t \\geqq 1\\) \u3092\u307f\u305f\u3059\u5b9f\u6570\u3068\u3059\u308b.<\/p>\r\n

    \r\n

  1. (1)<\/strong>\u3000\\(x \\geqq t\\) \u306e\u3068\u304d, \u4e0d\u7b49\u5f0f\r\n\\[\r\n-\\dfrac{(x-t)^2}{2} \\leqq \\log x -\\log t -\\dfrac{1}{t} (x-t) \\leqq 0\r\n\\]\r\n\u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n

  2. (2)<\/strong>\u3000\u4e0d\u7b49\u5f0f\r\n\\[\r\n-\\dfrac{1}{6 n^3} \\leqq \\displaystyle\\int_{t}^{t +\\frac{1}{n}} \\log x \\, dx -\\dfrac{1}{n} \\log t -\\dfrac{1}{2t n^2} \\leqq 0\r\n\\]\r\n\u304c\u6210\u308a\u7acb\u3064\u3053\u3068\u3092\u793a\u305b.<\/p><\/li>\r\n

  3. (3)<\/strong>\u3000\\(a_n = \\textstyle\\sum\\limits_{k=0}^{n-1} \\log \\left( 1 +\\dfrac{k}{n} \\right)\\) \u3068\u304a\u304f. \\(\\displaystyle\\lim_{n \\rightarrow \\infty} ( a_n -pn ) = q\\) \u3092\u307f\u305f\u3059\u3088\u3046\u306a\u5b9f\u6570 \\(p , q\\) \u306e\u5024\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n

    \r\n\r\n

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

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

    \\(u = \\dfrac{x}{t}\\) \u3068\u304a\u304f\u3068, \\(u \\geqq 1\\) .
    \r\n\u3053\u308c\u3092\u7528\u3044\u3066, \u793a\u3057\u305f\u3044\u4e0d\u7b49\u5f0f\u3092\u5909\u5f62\u3059\u308b\u3068\r\n\\[\r\n-\\dfrac{t^2 (u-1)^2}{2} \\leqq \\log u -u +1 \\leqq 0 \\quad ... [ \\text{A} ]\r\n\\]\r\n\u306a\u306e\u3067, \u3053\u308c\u3092\u793a\u305b\u3070\u3088\u3044.
    \r\n\\(f(u) = \\log u -u +1\\) \u3068\u304a\u304f\u3068\r\n\\[\r\nf'(u) = \\dfrac{1}{u} -1 \\leqq 0 \\quad ( \\ \\text{\u2235} \\ u \\geqq 1 \\ )\r\n\\]\r\n\u3086\u3048\u306b \\(f(u)\\) \u306f\u5358\u8abf\u6e1b\u5c11\u3059\u308b\u306e\u3067\r\n\\[\r\nf(u) \\leqq f(1) = 0 +1 -1 = 0 \\quad ... [1]\r\n\\]\r\n\\(g(u) = \\dfrac{t^2 (u-1)^2}{2} +\\log u -u +1\\) \u3068\u304a\u304f\u3068\r\n\\[\\begin{align}\r\ng'(u) & = t^2 (u-1) +\\dfrac{1}{u} -1 \\ , \\\\\r\ng''(u) & = t^2 -\\dfrac{1}{u^2} \\\\\r\n& = \\dfrac{(tu)^2 -1}{u^2} \\geqq 0 \\quad ( \\ \\text{\u2235} \\ t \\geqq 1 , u \\geqq 1 \\ )\r\n\\end{align}\\]\r\n\u3086\u3048\u306b \\(g'(x)\\) \u306f\u5358\u8abf\u5897\u52a0\u3057\r\n\\[\r\ng'(x) \\geqq g'(1) = 0 +1 -1 = 0\r\n\\]\r\n\u3055\u3089\u306b \\(g(x)\\) \u3082\u5358\u8abf\u5897\u52a0\u3057\r\n\\[\r\ng(x) \\geqq g(1) = 0 +0 +1 -1 = 0 \\quad ... [2]\r\n\\]\r\n\u3088\u3063\u3066, [1] [2] \u3088\u308a [A] \u304c\u793a\u3055\u308c\u3066, \u984c\u610f\u3082\u793a\u3055\u308c\u305f.<\/p>\r\n

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

    (1)<\/strong> \u3067\u793a\u3057\u305f\u5f0f\u306e\u5404\u8fba\u3092 \\(x\\) \u306b\u3064\u3044\u3066 \\(t \\rightarrow t +\\dfrac{1}{n}\\) \u3067\u7a4d\u5206\u3059\u308b.
    \r\n\u5404\u9805\u306b\u3064\u3044\u3066\r\n\\[\\begin{align}\r\n\\displaystyle\\int _ {t}^{t +\\frac{1}{n}} \\dfrac{(x-t)^2}{2} \\, dx & = \\left[ \\dfrac{(x-t)^3}{6} \\right] _ {t}^{t +\\frac{1}{n}} = \\dfrac{1}{6n^3} \\\\\r\n\\displaystyle\\int _ {t}^{t +\\frac{1}{n}} \\log t \\, dx & = \\dfrac{1}{n} \\log t \\\\\r\n\\displaystyle\\int _ {t}^{t +\\frac{1}{n}} \\dfrac{1}{t} (x-t) \\, dx & = \\left[ \\dfrac{(x-t)^2}{2t} \\right] _ {t}^{t +\\frac{1}{n}} = \\dfrac{1}{2n^2}\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\r\n-\\dfrac{1}{6 n^3} \\leqq \\displaystyle\\int_{t}^{t +\\frac{1}{n}} \\log x \\, dx -\\dfrac{1}{n} \\log t -\\dfrac{1}{2t n^2} \\leqq 0\r\n\\]\r\n

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

    (2)<\/strong> \u3067\u793a\u3057\u305f\u5f0f\u306b, \\(t = 1 , 1+\\dfrac{1}{n} , \\cdots , 1 +\\dfrac{n-1}{n}\\) \u3092\u4ee3\u5165\u3057\u3066\u3067\u304d\u308b \\(n\\) \u500b\u306e\u4e0d\u7b49\u5f0f\u3092\u8fba\u3005\u52a0\u3048\u3066, \\(n\\) \u500d\u3059\u308b\u3068\r\n\\[\\begin{gather}\r\n-\\dfrac{1}{6n} \\leqq n \\displaystyle\\int _ {1}^{2} \\log x \\, dx -a_n -\\dfrac{1}{2n} \\textstyle\\sum\\limits _ {k=0}^{n-1} \\dfrac{1}{1 +\\dfrac{k}{n}} \\leqq 0 \\\\\r\n\\text{\u2234} \\quad -\\dfrac{1}{2n} \\textstyle\\sum\\limits _ {k=0}^{n-1} \\dfrac{1}{1 +\\dfrac{k}{n}} \\leqq a_n -n \\underline{\\displaystyle\\int _ {1}^{2} \\log x \\, dx} _ {[3]} \\leqq -\\dfrac{1}{2n} \\textstyle\\sum\\limits _ {k=0}^{n-1} \\dfrac{1}{1 +\\dfrac{k}{n}} +\\dfrac{1}{6n} \\quad ... [4]\r\n\\end{gather}\\]\r\n\u3053\u3053\u3067\r\n\\[\r\n[3] = \\left[ x \\log x -x \\right] _ {1}^{2} = 2 \\log 2 -1\r\n\\]\r\n[4] \u306b\u304a\u3044\u3066 \\(n \\rightarrow \\infty\\) \u3068\u3059\u308c\u3070\r\n\\[\\begin{align}\r\n( \\text{\u7b2c 1 \u8fba} ) & \\longrightarrow -\\dfrac{1}{2} \\displaystyle\\int _ {0}^{1} \\dfrac{1}{1+x} \\, dx \\\\\r\n& = -\\dfrac{1}{2} \\left[ \\log (1+x) \\right] _ {0}^{1} = -\\dfrac{1}{2} \\log 2 \\ , \\\\\r\n( \\text{\u7b2c 3 \u8fba} ) & \\longrightarrow -\\dfrac{1}{2} \\log 2\r\n\\end{align}\\]\r\n\u306a\u306e\u3067, \u306f\u3055\u307f\u3046\u3061\u306e\u539f\u7406\u3088\u308a\r\n\\[\r\n( \\text{\u7b2c 2 \u8fba} ) = a_n -( 2 \\log 2 -1 ) n \\longrightarrow -\\dfrac{1}{2} \\log 2\r\n\\]\r\n\u3088\u3063\u3066\r\n\\[\r\np = \\underline{2 \\log 2 -1} \\ , \\ q = \\underline{-\\dfrac{1}{2} \\log 2}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\\(n\\) \u3092 \u81ea\u7136\u6570\u3068\u3057, \\(t\\) \u3092 \\(t \\geqq 1\\) \u3092\u307f\u305f\u3059\u5b9f\u6570\u3068\u3059\u308b. (1)\u3000\\(x \\geqq t\\) \u306e\u3068\u304d, \u4e0d\u7b49\u5f0f \\[ -\\dfrac{(x-t)^2}{2} \\leqq \\log x … \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":[172],"tags":[142,165],"class_list":["post-1987","post","type-post","status-publish","format-standard","hentry","category-osaka_r_2021","tag-osaka_r","tag-165"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/1987","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=1987"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/1987\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=1987"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=1987"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=1987"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}