{"id":204,"date":"2011-12-02T23:17:34","date_gmt":"2011-12-02T14:17:34","guid":{"rendered":"http:\/\/roundown.main.jp\/nyushi\/?p=204"},"modified":"2021-09-16T06:27:18","modified_gmt":"2021-09-15T21:27:18","slug":"ngr200903","status":"publish","type":"post","link":"https:\/\/www.roundown.net\/nyushi\/ngr200903\/","title":{"rendered":"\u540d\u53e4\u5c4b\u5927\u7406\u7cfb2009\uff1a\u7b2c3\u554f"},"content":{"rendered":"
\n

\u884c\u5217 \\(A = \\dfrac{1}{2} \\left( \\begin{array}{cc} 0 & -1 \\\\ 1 & -1 \\end{array} \\right)\\) \u306b\u5bfe\u3057\u3066, \u5ea7\u6a19\u7a7a\u9593\u306e\u70b9 \\(\\text{P} _ n\\) \u306e\u5ea7\u6a19 \\(( a _ n , b _ n , c _ n ) \\ ( n = 1, 2, 3, \\cdots )\\) \u3092, \\(( a _ 1 , b _ 1 , c _ 1 ) = ( 1, 0, 0 )\\) .\r\n\\[\r\n\\left( \\begin{array}{c} a _ {n+1} \\\\ b _ {n+1} \\end{array} \\right) = A \\left( \\begin{array}{c} a _ n \\\\ b _ n \\end{array} \\right) , \\ c _ {n+1} = c _ n +\\sqrt{a _ n b _ n} \\quad ( n= 1, 2, 3, \\cdots )\n\\]\r\n\u3067\u5b9a\u3081\u308b.<\/p>\r\n

    \r\n

  1. (1)<\/strong>\u3000\\(A^3\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n

  2. (2)<\/strong>\u3000\u70b9 \\(\\text{P} _ 2 , \\text{P} _ 3 , \\text{P} _ 4\\) \u306e\u5ea7\u6a19\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n

  3. (3)<\/strong>\u3000\u70b9 \\(\\text{P} _ n\\) \u306e\u5ea7\u6a19\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

    \\[\\begin{align}\r\nA^2 & = \\dfrac{1}{4} \\left( \\begin{array}{cc} 0 & -1 \\\\ 1 & -1 \\end{array} \\right) \\left( \\begin{array}{cc} 0 & -1 \\\\ 1 & -1 \\end{array} \\right) = \\dfrac{1}{4} \\left( \\begin{array}{cc} -1 & 1 \\\\ -1 & 0 \\end{array} \\right) , \\\\\r\nA^3 & = \\dfrac{1}{8} \\left( \\begin{array}{cc} -1 & 1 \\\\ -1 & 0 \\end{array} \\right) \\left( \\begin{array}{cc} 0 & -1 \\\\ 1 & -1 \\end{array} \\right) = \\underline{\\dfrac{1}{8} \\left( \\begin{array}{cc} 1 & 0 \\\\ 0 & 1 \\end{array} \\right)}\n\\end{align}\\]\r\n

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

    \\[\\begin{align}\r\n\\left( \\begin{array}{c} a _ 2 \\\\ b _ 2 \\end{array} \\right) & = A \\left( \\begin{array}{c} a _ 1 \\\\ b _ 1 \\end{array} \\right) \\\\\r\n& = \\dfrac{1}{2} \\left( \\begin{array}{cc} 0 & -1 \\\\ 1 & -1 \\end{array} \\right) \\left( \\begin{array}{c} 1 \\\\ 0 \\end{array} \\right) = \\dfrac{1}{2} \\left( \\begin{array}{c} 0 \\\\ 1 \\end{array} \\right) , \\\\\r\nc _ 2 & = c _ 1 +\\sqrt{a _ 1 b _ 1} = 0\n\\end{align}\\]\r\n\u3088\u3063\u3066, \\(\\text{P} _ 2 \\ \\underline{\\left( 0 , \\dfrac{1}{2} , 0 \\right)}\\) .\r\n\\[\\begin{align}\r\n\\left( \\begin{array}{c} a _ 3 \\\\ b _ 3 \\end{array} \\right) & = A^2 \\left( \\begin{array}{c} a _ 1 \\\\ b _ 1 \\end{array} \\right) \\\\\r\n& = \\dfrac{1}{4} \\left( \\begin{array}{cc} -1 & 1 \\\\ -1 & 0 \\end{array} \\right) \\left( \\begin{array}{c} 1 \\\\ 0 \\end{array} \\right) = \\dfrac{1}{4} \\left( \\begin{array}{c} -1 \\\\ -1 \\end{array} \\right) , \\\\\r\nc _ 3 & = c _ 2 +\\sqrt{a _ 2 b _ 2} = 0\n\\end{align}\\]\r\n\u3088\u3063\u3066, \\(\\text{P} _ 3 \\ \\underline{\\left( -\\dfrac{1}{4} , -\\dfrac{1}{4} , 0 \\right)}\\) .\r\n\\[\\begin{align}\r\n\\left( \\begin{array}{c} a _ 4 \\\\ b _ 4 \\end{array} \\right) & = A^3 \\left( \\begin{array}{c} a _ 1 \\\\ b _ 1 \\end{array} \\right) \\\\\r\n& = \\dfrac{1}{8} \\left( \\begin{array}{cc} 1 & 0 \\\\ 0 & 1 \\end{array} \\right) \\left( \\begin{array}{c} 1 \\\\ 0 \\end{array} \\right) = \\dfrac{1}{8} \\left( \\begin{array}{c} 1 \\\\ 0 \\end{array} \\right) , \\\\\r\nc _ 4 & = c _ 3 +\\sqrt{a _ 3 b _ 3} = \\dfrac{1}{4}\n\\end{align}\\]\r\n\u3088\u3063\u3066, \\(\\text{P} _ 4 \\ \\underline{\\left( \\dfrac{1}{8} , 0 , \\dfrac{1}{4} \\right)}\\) .<\/p>\r\n

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

    (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a, \\(A^3 = \\dfrac{1}{8} E\\) \u306a\u306e\u3067<\/p>\r\n

      \r\n

    1. 1*<\/strong>\u3000\\(n = 3m-2 \\ ( m = 1, 2, \\cdots )\\) \u306e\u3068\u304d\r\n\\[\r\n\\left( \\begin{array}{c} a _ {3m-2} \\\\ b _ {3m-2} \\end{array} \\right) =\\dfrac{1}{8^{m-1}} \\left( \\begin{array}{c} a _ 1 \\\\ b _ 1 \\end{array} \\right) =\\dfrac{1}{8^{m-1}} \\left( \\begin{array}{c} 1 \\\\ 0 \\end{array} \\right)\n\\]<\/li>\r\n

    2. 2*<\/strong>\u3000\\(n = 3m-1 \\ (m = 0, 1, 2, \\cdots )\\) \u306e\u3068\u304d\r\n\\[\r\n\\left( \\begin{array}{c} a _ {3m-1} \\\\ b _ {3m-1} \\end{array} \\right) =\\dfrac{1}{8^{m-1}} \\left( \\begin{array}{c} a _ 2 \\\\ b _ 2 \\end{array} \\right) =\\dfrac{1}{8^{m-1}} \\left( \\begin{array}{c} 0 \\\\ \\dfrac{1}{2} \\end{array} \\right)\n\\]<\/li>\r\n

    3. 3*<\/strong>\u3000\\(n = 3m \\ ( m=1, 2, \\cdots )\\) \u306e\u3068\u304d\r\n\\[\r\n\\left( \\begin{array}{c} a _ {3m} \\\\ b _ {3m} \\end{array} \\right) =\\dfrac{1}{8^{m-1}} \\left( \\begin{array}{c} a _ 3 \\\\ b _ 3 \\end{array} \\right) = -\\dfrac{1}{8^{m-1}} \\left( \\begin{array}{c} \\dfrac{1}{4} \\\\ \\dfrac{1}{4} \\end{array} \\right)\n\\]<\/li>\r\n<\/ol>\r\n

      \u307e\u305f\r\n\\[\\begin{align}\r\nc _ {3m+3} & = c _ {3m+2} +\\sqrt{a _ {3m+2} b _ {3m+2}} = c _ {3m+2} \\\\\r\n& = c _ {3m+1} +\\sqrt{a _ {3m+1} b _ {3m+1}} = c _ {3m+1} \\\\\r\n& = c _ {3m} +\\sqrt{a _ {3m} b _ {3m}} = c _ {3m} +\\dfrac{1}{4 \\cdot 8^{m-1}}\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(m \\geqq 2\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\nc _ {3m} & = c _ 3 +\\textstyle\\sum\\limits _ {k=1}^{m-1} \\dfrac{1}{4 \\cdot 8^{k-1}} \\\\\r\n& = \\dfrac{1}{4} \\cdot \\dfrac{1 -\\frac{1}{8^{m-1}}}{1 -\\frac{1}{8}} \\\\\r\n& = \\dfrac{2}{7} \\left( 1 -\\dfrac{1}{8^{m-1}} \\right)\n\\end{align}\\]\r\n\\(c _ 3 = 0\\) \u306a\u306e\u3067, \\(m = 1\\) \u306e\u3068\u304d\u3082\u6e80\u305f\u3057\u3066\u3044\u308b.
      \r\n\u4ee5\u4e0a\u3088\u308a, \\(m= 1, 2, \\cdots\\) \u306b\u5bfe\u3057\u3066 \\(\\text{P} _ n\\) \u306e\u5ea7\u6a19\u306f\r\n\\[\r\n\\underline{\\left\\{ \\begin{array}{ll} \\left( \\dfrac{1}{8^{m-1}} , 0 , \\dfrac{2}{7} \\left( 1 -\\dfrac{1}{8^{m-1}} \\right) \\right) & \\left( n=3m-2 \\text{\u306e\u3068\u304d} \\right) \\\\ \\left( 0 , \\dfrac{1}{2 \\cdot 8^{m-1}} , \\dfrac{2}{7} \\left( 1 -\\dfrac{1}{8^{m-1}} \\right) \\right) & \\left( n=3m-1 \\text{\u306e\u3068\u304d} \\right) \\\\ \\left( -\\dfrac{1}{4 \\cdot 8^{m-1}} , -\\dfrac{1}{4 \\cdot 8^{m-1}} , \\dfrac{2}{7} \\left( 1 -\\dfrac{1}{8^{m-1}} \\right) \\right) & \\ \\left( n=3m \\text{\u306e\u3068\u304d}\\right) \\end{array} \\right.}\n\\]\r\n","protected":false},"excerpt":{"rendered":"\u884c\u5217 \\(A = \\dfrac{1}{2} \\left( \\begin{array}{cc} 0 & -1 \\\\ 1 & -1 \\end{array} \\right)\\) \u306b\u5bfe\u3057\u3066, \u5ea7\u6a19\u7a7a\u9593\u306e\u70b9 \\(\\text{P} … \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":[33],"tags":[143,15],"class_list":["post-204","post","type-post","status-publish","format-standard","hentry","category-nagoya_r_2009","tag-nagoya_r","tag-15"],"_links":{"self":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/204","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=204"}],"version-history":[{"count":0,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/posts\/204\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/media?parent=204"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/categories?post=204"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.roundown.net\/nyushi\/wp-json\/wp\/v2\/tags?post=204"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}