{"id":640,"date":"2013-02-16T23:11:46","date_gmt":"2013-02-16T14:11:46","guid":{"rendered":"http:\/\/www.roundown.net\/nyushi\/?p=640"},"modified":"2021-09-10T21:34:59","modified_gmt":"2021-09-10T12:34:59","slug":"tok200704","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/tok200704\/","title":{"rendered":"\u6771\u5de5\u59272007\uff1a\u7b2c4\u554f"},"content":{"rendered":"<ol>\r\n<li><p><strong>(1)<\/strong>\u3000\u6574\u6570 \\(n = 0, 1, 2, \\cdots\\) \u3068\u6b63\u6570 \\(a _ n\\) \u306b\u5bfe\u3057\u3066\r\n\\[\r\nf _ n (x) = a _ n (x-n) (n+1-x)\r\n\\]\r\n\u3068\u304a\u304f. \\(2\\) \u3064\u306e\u66f2\u7dda \\(y = f _ n (x)\\) \u3068 \\(y = e^{-x}\\) \u304c\u63a5\u3059\u308b\u3088\u3046\u306a \\(a _ n\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<li><p><strong>(2)<\/strong>\u3000\\(f _ n (x)\\) \u306f <strong>(1)<\/strong> \u3067\u5b9a\u3081\u305f\u3082\u306e\u3068\u3059\u308b.\r\n\\(y = f _ 0 (x) , \\ y = e^{-x}\\) \u3068 \\(y\\) \u8ef8\u3067\u56f2\u307e\u308c\u308b\u56f3\u5f62\u306e\u9762\u7a4d\u3092 \\(S _ 0 \\ ( n \\geqq 1 )\\) \u306b\u5bfe\u3057 \\(y = f _ {n-1} (x) , \\ y = f _ {n} (x)\\) \u3068 \\(y = e^{-x}\\) \u3067\u56f2\u307e\u308c\u308b\u56f3\u5f62\u306e\u9762\u7a4d\u3092 \\(S _ n\\) \u3068\u304a\u304f. \u3053\u306e\u3068\u304d\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} \\left( S _ 0 + S _ 1 + \\cdots + S _ n \\right)\r\n\\]\r\n\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n<hr \/>\r\n<!--more-->\r\n<h2>\u3010 \u89e3 \u7b54 \u3011<\/h2>\r\n<p><strong>(1)<\/strong><\/p>\r\n<p>\\[\\begin{align}\r\nf' _ n(x) & = a _ n (n+1-x) -a _ n (x-n) \\\\\r\n& = a _ n ( -2x +2n+1 )\r\n\\end{align}\\]\r\n\u307e\u305f, \\(y=e^{-x}\\) \u306b\u5bfe\u3057\u3066, \\(y' = -e^{-x}\\) .<br \/>\r\n\u3057\u305f\u304c\u3063\u3066, \u63a5\u70b9\u306e \\(x\\) \u5ea7\u6a19\u3092 \\(t\\) \u3068\u304a\u304f\u3068, \u4ee5\u4e0b\u306e\u95a2\u4fc2\u304c\u6210\u7acb\u3059\u308b.\r\n\\[\r\n\\left\\{ \\begin{array}{ll} e^{-t} = a _ n (t-n)(n+1-t) & ...[1] \\\\ -e^{-t} = a _ n ( -2t +2n+1 ) & ...[2] \\end{array} \\right.\r\n\\]\r\n[1] [2] \u3088\u308a \\(e^{-t}\\) \u3092\u6d88\u53bb\u3059\u308b\u3068, \\(a _ n \\lt 0\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\n& (t-n)(n+1-t) = 2t -2n -1 \\\\\r\nt^2 & -(2n+1)t +n^2 +n +2t -2n-1 = 0 \\\\\r\nt^2 & -(2n-1)t +n^2-n-1 = 0 \\\\\r\n\\text{\u2234} \\quad t & = \\dfrac{2n-1 \\pm \\sqrt{(2n-1)^2 -4(n^2-n-1)}}{2} \\\\\r\n& = n -\\dfrac{1 \\pm \\sqrt{5}}{2}\r\n\\end{align}\\]\r\n\\(n \\lt t \\lt t+1\\) \u306a\u306e\u3067\r\n\\[\r\nt = n +\\dfrac{\\sqrt{5} -1}{2}\r\n\\]\r\n\u3088\u3063\u3066, [2] \u306b\u4ee3\u5165\u3057\u3066\r\n\\[\\begin{align}\r\na _ n & = -\\dfrac{e^{-t}}{2(t-n)-1} \\\\\r\n& = -\\dfrac{e^{-n -\\frac{\\sqrt{5} -1}{2}}}{\\sqrt{5} -2} \\\\\r\n& = \\underline{-\\left( 2 +\\sqrt{5} \\right) e^{-n -\\frac{\\sqrt{5} -1}{2}}}\r\n\\end{align}\\]\r\n<p><strong>(2)<\/strong><\/p>\r\n<p>\\(T _ n = \\displaystyle\\int _ 0^n e^{-x} \\, dx -\\textstyle\\sum\\limits _ {k=0}^{n-1} \\displaystyle\\int _ {k}^{k+1} f _ k (x) \\, dx\\) \u3068\u304a\u304f\u3068\r\n\\[\r\nT _ n \\lt \\textstyle\\sum\\limits _ {k=0}^{n} S _ k \\lt T _ {n+1} \\quad ... [3]\r\n\\]\r\n\u3053\u3053\u3067\r\n\\[\\begin{align}\r\nT _ n & = \\left[ - e^{-x} \\right] _ 0^n +\\textstyle\\sum\\limits _ {k=0}^{n-1} \\dfrac{a _ k}{6} \\left\\{ (k+1) -k \\right\\}^3 \\\\\r\n& = 1 -e^{-n} -\\dfrac{\\left( 2 +\\sqrt{5} \\right) e^{-\\frac{\\sqrt{5} -1}{2}}}{6} \\cdot \\dfrac{1 -e^{-n}}{1 -e^{-1}} \\\\\r\n& = 1 -e^{-n} -\\dfrac{\\left( 2 +\\sqrt{5} \\right) e^{\\frac{3 -\\sqrt{5}}{2}} \\left( 1 -e^{-n} \\right)}{6(e-1)}\r\n\\end{align}\\]\r\n\u3086\u3048\u306b\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} T _ n = \\displaystyle\\lim _ {n \\rightarrow \\infty} T _ {n+1} = 1 -\\dfrac{\\left( 2 +\\sqrt{5} \\right) e^{\\frac{3 -\\sqrt{5}}{2}}}{6(e-1)}\r\n\\]\r\n\u306a\u306e\u3067, [3] \u3088\u308a, \u306f\u3055\u307f\u3046\u3061\u306e\u539f\u7406\u304b\u3089\r\n\\[\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} \\textstyle\\sum\\limits _ {k=0}^{n} S _ k = \\underline{1 -\\dfrac{\\left( 2 +\\sqrt{5} \\right) e^{\\frac{3 -\\sqrt{5}}{2}}}{6(e-1)}}\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"(1)\u3000\u6574\u6570 \\(n = 0, 1, 2, \\cdots\\) \u3068\u6b63\u6570 \\(a _ n\\) \u306b\u5bfe\u3057\u3066 \\[ f _ n (x) = a _ n (x-n) (n+1-x) \\] \u3068\u304a\u304f. \\(2\\) \u3064\u306e\u66f2\u7dda \\(y =  &hellip; <a href=\"https:\/\/www.roundown.net\/nyushi\/tok200704\/\">\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":[91],"tags":[141,109],"class_list":["post-640","post","type-post","status-publish","format-standard","hentry","category-toko_2007","tag-toko","tag-109"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/640","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=640"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/640\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=640"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=640"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=640"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}