\\(2\\) \u6b21\u306e\u6b63\u65b9\u884c\u5217 \\(A _ 0 , A _ 1 , A _ 2 , A _ 3 , \\cdots\\) \u3092\r\n\\[\r\nA _ 0 = O , \\quad A _ n = B +A _ {n-1} C \\quad ( n =1, 2, 3, \\cdots )\r\n\\]\r\n\u3067\u5b9a\u3081\u308b. \u305f\u3060\u3057, \\(O\\) \u306f \\(2\\) \u6b21\u306e\u96f6\u884c\u5217, \\(B\\) \u3068 \\(C\\) \u306f \\(2\\) \u6b21\u306e\u6b63\u65b9\u884c\u5217\u3068\u3059\u308b.<\/p>\r\n
(1)<\/strong>\u3000\\(A _ n ( E-C )\\) \u3092 \\(B\\) \u3068 \\(C\\) \u3092\u7528\u3044\u3066\u8868\u305b. \u3053\u3053\u3067 \\(E\\) \u306f \\(2\\) \u6b21\u306e\u5358\u4f4d\u884c\u5217\u3068\u3059\u308b.<\/p><\/li>\r\n (2)<\/strong>\u3000\\(B\\) \u3068 \\(C\\) \u3092\r\n\\[\r\nB = \\left( \\begin{array}{cc} 0 & 1 \\\\ 1 & 0 \\end{array} \\right) , \\quad C = \\left( \\begin{array}{cc} 1 & 3 \\\\ -1 & 1 \\end{array} \\right)\r\n\\]\r\n\u3068\u3059\u308b\u3068\u304d, \\(A _ {3n}\\) \u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n (1)<\/strong><\/p>\r\n \u6761\u4ef6\u306e\u5f0f\u306e\u4e21\u8fba\u306b, \u53f3\u304b\u3089 \\(C^{-n}\\) \u3092\u639b\u3051\u308b\u3068\r\n\\[\\begin{align}\r\nA _ n C^{-n} & = B C^{-n} +A _ {n-1} C^{-(n-1)} \\\\\r\n\\text{\u2234} \\quad A _ n C^{-n} - A _ {n-1} C^{-(n-1)} & = B C^{-n}\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(n \\geqq 1\\) \u306e\u3068\u304d\r\n\\[\\begin{align}\r\nA _ n C^{-n} & = A _ 0 + \\textstyle\\sum\\limits _ {k=1}^{n} B C^{-k} \\\\\r\n& = B \\left( C^{-1} +C^{-2} + \\cdots + C^{-n} \\right)\n\\end{align}\\]\r\n\u4e21\u8fba\u306b\u53f3\u304b\u3089 \\(C^n\\) \u3092\u639b\u3051\u308b\u3068\r\n\\[\r\nA _ n = B \\left( C^{n-1} +C^{n-2} + \\cdots + E \\right)\n\\]\r\n\u3088\u3063\u3066\r\n\\[\r\nA _ n ( E-C ) = \\underline{B \\left( E -C^n \\right)}\n\\]\r\n\u3053\u308c\u306f, \\(n=0\\) \u306e\u3068\u304d\u3082\u6e80\u305f\u3057\u3066\u3044\u308b.<\/p>\r\n (2)<\/strong><\/p>\r\n \\[\\begin{align}\r\nC^2 & = \\left( \\begin{array}{cc} 1 & 3 \\\\ -1 & 1 \\end{array} \\right) \\left( \\begin{array}{cc} 1 & 3 \\\\ -1 & 1 \\end{array} \\right) = -2\\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 1 \\end{array} \\right) , \\\\\r\nC^3 & = -2 \\left( \\begin{array}{cc} 1 & -3 \\\\ 1 & 1 \\end{array} \\right) \\left( \\begin{array}{cc} 1 & 3 \\\\ -1 & 1 \\end{array} \\right) = -8 E\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066\r\n\\[\r\nC^{3n} = (-8)^n E\n\\]\r\n\u307e\u305f\r\n\\[\r\n(E-C)^{-1} = \\left( \\begin{array}{cc} 0 & -3 \\\\ 1 & 0 \\end{array} \\right)^{-1} = \\dfrac{1}{3} \\left( \\begin{array}{cc} 0 & 3 \\\\ -1 & 0 \\end{array} \\right)\n\\]\r\n\u3088\u3063\u3066, (1)<\/strong> \u306e\u7d50\u679c\u304b\u3089\r\n\\[\\begin{align}\r\nA _ {3n} & = B \\left( E -C^{3n} \\right) (E-C)^{-1} \\\\\r\n& = \\dfrac{1 -(-8)^n}{3} \\left( \\begin{array}{cc} 0 & 1 \\\\ 1 & 0 \\end{array} \\right) \\left( \\begin{array}{cc} 0 & 3 \\\\ -1 & 0 \\end{array} \\right) \\\\\r\n& = \\underline{\\dfrac{1 -(-8)^n}{3} \\left( \\begin{array}{cc} -1 & 0 \\\\ 0 & 3 \\end{array} \\right)}\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\\(2\\) \u6b21\u306e\u6b63\u65b9\u884c\u5217 \\(A _ 0 , A _ 1 , A _ 2 , A _ 3 , \\cdots\\) \u3092 \\[ A _ 0 = O , \\quad A _ n = B +A _ {n-1} C \\quad ( … \u7d9a\u304d\u3092\u8aad\u3080
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n