{"id":94,"date":"2011-11-27T16:17:30","date_gmt":"2011-11-27T07:17:30","guid":{"rendered":"http:\/\/roundown.main.jp\/nyushi\/?p=94"},"modified":"2021-03-15T19:54:39","modified_gmt":"2021-03-15T10:54:39","slug":"tkr201002","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/tkr201002\/","title":{"rendered":"\u6771\u5927\u7406\u7cfb2010\uff1a\u7b2c2\u554f"},"content":{"rendered":"<ol>\r\n<li><strong>(1)<\/strong>\u3000\u3059\u3079\u3066\u306e\u81ea\u7136\u6570 \\(k\\) \u306b\u5bfe\u3057\u3066, \u6b21\u306e\u4e0d\u7b49\u5f0f\u3092\u793a\u305b.\r\n\\[\r\n\\dfrac{1}{2(k+1)} \\lt \\int _ 0^1 \\dfrac{1-x}{k+x} dx \\lt \\dfrac{1}{2k}\r\n\\]<\/li>\r\n<li><strong>(2)<\/strong>\u3000\\(m \\gt n\\) \u3067\u3042\u308b\u3088\u3046\u306a\u3059\u3079\u3066\u306e\u81ea\u7136\u6570 \\(m , n\\) \u306b\u5bfe\u3057\u3066, \u6b21\u306e\u4e0d\u7b49\u5f0f\u3092\u793a\u305b.\r\n\\[\r\n\\dfrac{m-n}{2(m+1)(n+1)} \\lt \\log \\dfrac{m}{n} -\\textstyle\\sum\\limits _ {k=n+1}^m \\dfrac{1}{k} \\lt \\dfrac{m-n}{2mn}\r\n\\]<\/li>\r\n<\/ol>\r\n<hr>\r\n<!--more-->\r\n<h4>\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n<p><strong>(1)<\/strong><\/p>\r\n<p>\\(0 \\lt x \\lt 1\\) \u306b\u304a\u3044\u3066, \\(k \\lt k+x \\lt k+1\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\n\\dfrac{1}{k+1} & \\lt \\dfrac{1}{k+x} \\lt \\dfrac{1}{k} \\\\\r\n\\dfrac{1-x}{k+1} & \\lt \\dfrac{1-x}{k+x} \\lt \\dfrac{1-x}{k} \\quad ( \\ \\text{\u2235} \\ 1-x \\gt 0 \\ ) \\\\\r\n\\text{\u2234} \\quad \\displaystyle\\int _ 0^1 \\dfrac{1-x}{k+1} \\, dx & \\lt \\displaystyle\\int _ 0^1 \\dfrac{1-x}{k+x} \\, dx \\lt \\int _ 0^1 \\dfrac{1-x}{k} \\, dx\r\n\\end{align}\\]\r\n\\(\\displaystyle\\int _ 0^1 (1-x) \\, dx = \\dfrac{1}{2} \\cdot 1 \\cdot 1 = \\dfrac{1}{2}\\) \u306a\u306e\u3067\r\n\\[\r\n\\dfrac{1}{2(k+1)} \\lt \\int _ 0^1 \\dfrac{1-x}{k+x} dx \\lt \\dfrac{1}{2k}\r\n\\]\r\n<p><strong>(2)<\/strong><\/p>\r\n<p>\\[\\begin{align}\r\n\\displaystyle\\int _ 0^1 \\dfrac{1-x}{k+x} \\, dx & = \\int _ 0^1 \\left( \\dfrac{k+1}{k+x} -1 \\right) \\, dx \\\\\r\n& = \\left[ (k+1) \\log (k+x) -x \\right] _ 0^1 \\\\\r\n& = (k+1) \\left\\{ \\log (k+1) -\\log k \\right\\} -1\r\n\\end{align}\\]\r\n<p><strong>(1)<\/strong> \u306e\u7d50\u679c\u306b\u3053\u308c\u3092\u7528\u3044\u3066, \u8fba\u3005\u3092 \\(k+1\\) \u3067\u5272\u308b\u3068\r\n\\[\r\n\\dfrac{1}{2(k+1)^2} \\lt \\left\\{ \\log (k+1) -\\log k \\right\\} -\\dfrac{1}{k+1} \\lt \\dfrac{1}{2k(k+1)}\r\n\\]\r\n\u3053\u3053\u3067\r\n\\[\\begin{align}\r\n\\dfrac{1}{2(k+1)^2} & \\gt \\dfrac{1}{2(k+1)(k+2)} = \\dfrac{1}{2} \\left( \\dfrac{1}{k+1} - \\dfrac{1}{k+2} \\right) , \\\\\r\n\\dfrac{1}{2k(k+1)} & = \\dfrac{1}{2} \\left( \\dfrac{1}{k} - \\dfrac{1}{k+1} \\right)\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\r\n\\dfrac{1}{2} \\left( \\dfrac{1}{k+1} - \\dfrac{1}{k+2} \\right) \\lt \\left\\{ \\log (k+1) -\\log k \\right\\} -\\dfrac{1}{k+1} \\lt \\dfrac{1}{2} \\left( \\dfrac{1}{k} - \\dfrac{1}{k+1} \\right)\r\n\\]\r\n\u3053\u306e\u5f0f\u306b \\(k = n , \\cdots , m-1\\) \u3092\u4ee3\u5165\u3057\u305f\u5f0f\u3092\u8fba\u3005\u52a0\u3048\u308b\u3068,\r\n\\[\\begin{align}\r\n\\textstyle\\sum\\limits _ {k=n}^{m-1} \\left( \\dfrac{1}{k+1} - \\dfrac{1}{k+2} \\right) & = \\left( \\dfrac{1}{n+1} - \\dfrac{1}{n+2} \\right) + \\cdots +\\left( \\dfrac{1}{m} - \\dfrac{1}{m+1} \\right) \\\\\r\n& = \\dfrac{1}{n+1} - \\dfrac{1}{m+1} = \\dfrac{m-n}{(n+1)(m+1)} , \\\\\r\n\\textstyle\\sum\\limits _ {k=n}^{m-1} \\left( \\dfrac{1}{k} - \\dfrac{1}{k+1} \\right) & = \\dfrac{1}{n} - \\dfrac{1}{m} = \\dfrac{m-n}{nm} , \\\\\r\n\\textstyle\\sum\\limits _ {k=n}^{m-1} \\left\\{ \\log (k+1) -\\log k \\right\\} & = \\log m -\\log n = \\log \\dfrac{m}{n} , \\\\\r\n\\textstyle\\sum\\limits _ {k=n}^{m-1} \\dfrac{1}{k+1} & = \\textstyle\\sum\\limits _ {k=n+1}^m \\dfrac{1}{k}\r\n\\end{align}\\]\r\n\u4ee5\u4e0a\u3088\u308a\r\n\\[\r\n\\dfrac{m-n}{2(m+1)(n+1)} \\lt \\log \\dfrac{m}{n} - \\textstyle\\sum\\limits _ {k=n+1}^m \\dfrac{1}{k} \\lt \\dfrac{m-n}{2mn}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"(1)\u3000\u3059\u3079\u3066\u306e\u81ea\u7136\u6570 \\(k\\) \u306b\u5bfe\u3057\u3066, \u6b21\u306e\u4e0d\u7b49\u5f0f\u3092\u793a\u305b. \\[ \\dfrac{1}{2(k+1)} \\lt \\int _ 0^1 \\dfrac{1-x}{k+x} dx \\lt \\dfrac{1}{2k} \\] &hellip; <a href=\"https:\/\/www.roundown.net\/nyushi\/tkr201002\/\">\u7d9a\u304d\u3092\u8aad\u3080 <span class=\"meta-nav\">&rarr;<\/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":[19],"tags":[139,14],"class_list":["post-94","post","type-post","status-publish","format-standard","hentry","category-tokyo_r_2010","tag-tokyo_r","tag-14"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/94","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=94"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/94\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=94"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=94"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=94"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}