{"id":1873,"date":"2021-09-12T09:07:50","date_gmt":"2021-09-12T00:07:50","guid":{"rendered":"https:\/\/www.roundown.net\/nyushi\/?p=1873"},"modified":"2021-09-12T09:07:50","modified_gmt":"2021-09-12T00:07:50","slug":"iks201603","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/iks201603\/","title":{"rendered":"\u533b\u79d1\u6b6f\u79d1\u59272016\uff1a\u7b2c3\u554f"},"content":{"rendered":"<hr \/>\n<p>\u95a2\u6570 \\(f(x) = \\langle \\&#33; \\langle x \\rangle \\&#33; \\rangle -2 \\langle \\&#33; \\langle x-1 \\rangle \\&#33; \\rangle +\\langle \\&#33; \\langle x-2 \\rangle \\&#33; \\rangle\\) \u3092\u8003\u3048\u308b. \u3053\u3053\u3067, \u5b9f\u6570 \\(u\\) \u306b\u5bfe\u3057\u3066 \\(\\langle \\&#33; \\langle u \\rangle \\&#33; \\rangle = \\dfrac{u +|u|}{2}\\) \u3068\u3059\u308b. \u3053\u306e\u3068\u304d\u4ee5\u4e0b\u306e\u5404\u554f\u3044\u306b\u7b54\u3048\u3088.<\/p>\r\n<ol>\r\n<li><p><strong>(1)<\/strong>\u3000\\(f(x)\\) \u306e\u30b0\u30e9\u30d5\u3092\u304b\u3051.<\/p><\/li>\r\n<li><p><strong>(2)<\/strong>\u3000\\(g(x) = \\displaystyle\\int _ {0}^{1} f(x-t) \\, dt\\) \u3068\u304a\u304f\u3068\u304d, \\(g(x)\\) \u306e\u6700\u5927\u5024\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<li><p><strong>(3)<\/strong>\u3000<strong>(2)<\/strong> \u306e \\(g(x)\\) \u306b\u5bfe\u3057\u3066, \\(p(s) = \\displaystyle\\int _ {0}^{3} (x-s)^2 g(x) \\, dx\\) \u3068\u304a\u304f\u3068\u304d, \\(p(s)\\) \u306e\u6700\u5c0f\u5024\u3092\u6c42\u3081\u3088.<\/p><\/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>\\[\\begin{align}\r\nf(x) & = \\dfrac{x +|x|}{2} -2 \\cdot \\dfrac{x-1 +| x-1 |}{2} +\\dfrac{x-2 +| x-2 |}{2} \\\\\r\n& = \\dfrac{1}{2} \\left( |x| -2 | x-1 | +| x-2 | \\right) \\ .\r\n\\end{align}\\]\r\n<ul>\r\n<li><p>\\(x \\lt 0\\) \u306e\u3068\u304d\r\n\\[\r\nf(x) = \\dfrac{1}{2} \\left\\{ -x +2 (x-1) -(x-2) \\right\\} = 0 \\ .\r\n\\]<\/li>\r\n<li><p>\\(0 \\leqq x \\lt 1\\) \u306e\u3068\u304d\r\n\\[\r\nf(x) = \\dfrac{1}{2} \\left\\{ x +2 (x-1) -(x-2) \\right\\} = x \\ .\r\n\\]<\/li>\r\n<li><p>\\(1 \\leqq x \\lt 2\\) \u306e\u3068\u304d\r\n\\[\r\nf(x) = \\dfrac{1}{2} \\left\\{ x -2 (x-1) -(x-2) \\right\\} = 2-x \\ .\r\n\\]<\/li>\r\n<li><p>\\(x \\geqq 2\\) \u306e\u3068\u304d\r\n\\[\r\nf(x) = \\dfrac{1}{2} \\left\\{ x -2 (x-1) +(x-2) \\right\\} = 0 \\ .\r\n\\]<\/li>\r\n<\/ul>\r\n<p>\u3088\u3063\u3066, \\(f(x)\\) \u306e\u30b0\u30e9\u30d5\u306f, \u4e0b\u56f3\u306e\u901a\u308a.<\/p>\r\n<img decoding=\"async\" src=\"\/\/www.roundown.net\/nyushi\/wp-content\/uploads\/ika20160301.svg\" alt=\"\" class=\"aligncenter size-full\" \/>\r\n<p><strong>(2)<\/strong><\/p>\r\n<p>\\(u = x-t\\) \u3068\u304a\u304f\u3068\r\n\\[\r\ndt = -du , \\quad \\begin{array}{c|c} t & 0 \\rightarrow 1 \\\\ \\hline u & x \\rightarrow x-1\\end{array} \\ .\r\n\\]\r\n\u3086\u3048\u306b\r\n\\[\r\ng(x) = \\displaystyle\\int _ {x-1}^{x} f(u) \\, du \\quad ... [1] \\ .\r\n\\]\r\n<strong>(1)<\/strong> \u3067\u6c42\u3081\u305f\u30b0\u30e9\u30d5\u306e\u5f62\u72b6\u3088\u308a, \\(1 \\leqq x \\lt 2\\) \u3067 \\(g(x)\\) \u306f\u6700\u5927\u3068\u306a\u308b.<br \/>\r\n\u3053\u306e\u533a\u9593\u3067\u306f\r\n\\[\\begin{align}\r\ng(x) & = \\displaystyle\\int _ {x-1}^{1} u \\, du +\\displaystyle\\int_{1}^{x} ( 2-u ) \\, du \\\\\r\n& = \\left[ \\dfrac{u^2}{2} \\right] _ {x-1}^{1} +\\left[ 2u -\\dfrac{u^2}{2} \\right]_{1}^{x} \\\\\r\n& = \\dfrac{1}{2} -\\dfrac{(x-1)^2}{2} +2x -\\dfrac{x^2}{2} -\\dfrac{3}{2} \\\\\r\n& = -x^2 +3x -\\dfrac{3}{2} \\\\\r\n& = -\\left( x -\\dfrac{3}{2} \\right)^2 +\\dfrac{3}{4} \\ .\r\n\\end{align}\\]\r\n\u3088\u3063\u3066, \\(g(x)\\) \u306e\u6700\u5927\u5024\u306f\r\n\\[\r\ng \\left( \\dfrac{3}{2} \\right) = \\underline{\\dfrac{3}{4}} \\ .\r\n\\]\r\n<p><strong>(3)<\/strong><\/p>\r\n[1] \u3088\u308a<\/p>\r\n<ul>\r\n<li><p>\\(0 \\leqq x \\lt 1\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\ng(x) & = \\displaystyle\\int _ {0}^{x} u \\, du \\\\\r\n& = \\left[ \\dfrac{u^2}{2} \\right] _ {0}^{x} = \\dfrac{x^2}{2} \\ .\r\n\\end{align}\\]<\/li>\r\n<li><p>\\(2 \\leqq x \\lt 3\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\ng(x) & = \\displaystyle\\int _ {x-1}^{2} (2-u) \\, du \\\\\r\n& = \\left[ 2u -\\dfrac{u^2}{2} \\right] _ {x-1}^{2} \\\\\r\n& = \\dfrac{1}{2} ( x-3 )^2 \\ .\r\n\\end{align}\\]<\/li>\r\n<\/ul>\r\n<p>\\(A = \\displaystyle\\int_{0}^{3} g(x) \\, dx\\) , \\(B = \\displaystyle\\int_{0}^{3} x g(x) \\, dx\\) , \\(C = \\displaystyle\\int_{0}^{3} x^2 g(x) \\, dx\\) \u3068\u304a\u304f\u3068\r\n\\[\\begin{align}\r\np(s) & = A s^2 -2B s +C \\\\\r\n& = A \\left( s -\\dfrac{B}{A} \\right)^2 -\\dfrac{B^2}{A} +C\r\n\\end{align}\\]\r\n\u306a\u306e\u3067, \\(p(s)\\) \u306e\u6700\u5c0f\u5024\u306f, \\(-\\dfrac{B^2}{A} +C\\) \u3068\u8868\u305b\u308b.\r\n\\[\\begin{align}\r\nA & = \\displaystyle\\int _ {0}^{1} \\dfrac{x^2}{2} \\, dx +\\displaystyle\\int _ {1}^{2} \\left( -x^2 +3x -\\dfrac{3}{2} \\right) \\, dx +\\displaystyle\\int _ {2}^{3} \\dfrac{1}{2} (x-3)^2 \\, dx \\\\\r\n& = \\left[ \\dfrac{x^3}{6} \\right] _ {0}^{1} +\\left[ -\\dfrac{x^3}{3} +\\dfrac{3x^2}{2} -\\dfrac{3x}{2} \\right] _ {1}^{2} +\\left[ \\dfrac{(x-3)^3}{6} \\right] _ {2}^{3} \\\\\r\n& = \\dfrac{1}{6} +\\dfrac{1}{3} -\\left( -\\dfrac{1}{3} \\right) -\\left( -\\dfrac{1}{6} \\right) = 1 \\ . \\\\\r\nB & = \\displaystyle\\int _ {0}^{1} \\dfrac{x^3}{2} \\, dx +\\displaystyle\\int_ {1}^{2} \\left( -x^3 +3x^2 -\\dfrac{3x}{2} \\right) \\, dx +\\displaystyle\\int_ {2}^{3} \\dfrac{1}{2} x (x-3)^2 \\, dx \\\\\r\n& = \\left[ \\dfrac{x^4}{8} \\right] _ {0}^{1} +\\left[ -\\dfrac{x^4}{4} +x^3 -\\dfrac{3x^2}{4} \\right] _ {1}^{2} +\\left[ \\dfrac{(x-3)^4}{8} +\\dfrac{(x-3)^3}{2} \\right] _ {2}^{3} \\\\\r\n& = \\dfrac{1}{8} +1 -0 -\\left( -\\dfrac{3}{8} \\right) = \\dfrac{3}{2} \\ . \\\\\r\nC & = \\displaystyle\\int _ {0}^{1} \\dfrac{x^4}{2} \\, dx +\\displaystyle\\int _ {1}^{2} \\left( -x^4 +3x^3 -\\dfrac{3x^2}{2} \\right) \\, dx +\\displaystyle\\int _ {2}^{3} \\dfrac{1}{2} x^2 (x-3)^2 \\, dx \\\\\r\n& = \\left[ \\dfrac{x^5}{10} \\right] _ {0}^{1} +\\left[ -\\dfrac{x^5}{5} +\\dfrac{3x^4}{4} -\\dfrac{x^3}{2} \\right] _ {1}^{2} +\\left[ \\dfrac{(x-3)^5}{10} +\\dfrac{3 (x-3)^4}{4} +\\dfrac{3 (x-3)^3}{2} \\right] _ {2}^{3} \\\\\r\n& = \\dfrac{1}{10} +\\dfrac{8}{5} -\\dfrac{1}{20} -\\left( -\\dfrac{17}{20} \\right) = \\dfrac{5}{2} \\ .\r\n\\end{align}\\]\r\n\u3088\u3063\u3066, \u6c42\u3081\u308b\u6700\u5c0f\u5024\u306f\r\n\\[\r\n-\\left( \\dfrac{3}{2} \\right)^2 +\\dfrac{5}{2} = \\underline{\\dfrac{1}{4}} \\ .\r\n\\]\r\n","protected":false},"excerpt":{"rendered":"\u95a2\u6570 \\(f(x) = \\langle \\&#33; \\langle x \\rangle \\&#33; \\rangle -2 \\langle \\&#33; \\langle x-1 \\rangle \\&#33; \\rang &hellip; <a href=\"https:\/\/www.roundown.net\/nyushi\/iks201603\/\">\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":[154],"tags":[145,162],"class_list":["post-1873","post","type-post","status-publish","format-standard","hentry","category-ikashika_2016","tag-ikashika","tag-162"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/1873","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=1873"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/1873\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=1873"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=1873"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=1873"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}