\\(l _ 0\\) \u306f\u76f4\u7dda \\(3x+2y=1\\) \u3067\u3042\u308b.<\/p><\/li>\r\n
\u70b9 P \u304c \\(l _ n\\) \u4e0a\u3092\u52d5\u304f\u3068\u304d, \\(f( \\text{P} )\\) \u304c\u63cf\u304f\u76f4\u7dda\u3092 \\(l _ {n+1}\\) \u3068\u3059\u308b\uff08 \\(n =0, 1, 2, \\cdots\\) \uff09.<\/p><\/li>\r\n<\/ul>\r\n
- \r\n
(1)<\/strong>\u3000\\(a _ {n+1} , b _ {n+1}\\) \u3092 \\(a _ {n} , b _ {n}\\) \u3067\u8868\u305b.<\/p><\/li>\r\n
(2)<\/strong>\u3000\u4e0d\u7b49\u5f0f \\(a _ {n} x +b _ {n} y \\gt 1\\) \u304c\u5b9a\u3081\u308b\u9818\u57df\u3092 \\(D _ n\\) \u3068\u3059\u308b.\r\n\\(D _ 0 , D _ 1 , D _ 2 , \\cdots\\) \u3059\u3079\u3066\u306b\u542b\u307e\u308c\u308b\u3088\u3046\u306a\u70b9\u306e\u7bc4\u56f2\u3092\u56f3\u793a\u305b\u3088.<\/p><\/li>\r\n<\/ol>\r\n
\r\n\r\n\u3010 \u89e3 \u7b54 \u3011<\/h4>\r\n
(1)<\/strong><\/p>\r\n
\\(f\\) \u306b\u5bfe\u5fdc\u3059\u308b \\(2\\) \u6b21\u6b63\u65b9\u884c\u5217\u3092 \\(A\\) \u3068\u304a\u304f\u3068\r\n\\[\r\nA = \\left( \\begin{array}{cc} 3 & 1 \\\\ -2 & 0 \\end{array} \\right)\r\n\\]\r\n\\(l _ n : \\ a _ n x +b _ n y =1\\) \u4e0a\u306e \\(2\\) \u70b9 \\(\\left( \\dfrac{1}{a _ n} , 0 \\right) , \\left( 0 , \\dfrac{1}{b _ n} \\right)\\) \u306f,\r\n\\(f\\) \u306b\u3088\u3063\u3066\r\n\\[\\begin{align}\r\n\\left( \\begin{array}{cc} 3 & 1 \\\\ -2 & 0 \\end{array} \\right) \\left( \\begin{array}{c} \\dfrac{1}{a _ n} \\\\ 0 \\end{array} \\right) & = \\left( \\begin{array}{c} \\dfrac{3}{a _ n} \\\\ -\\dfrac{2}{a _ n} \\end{array} \\right) \\\\\r\n\\left( \\begin{array}{cc} 3 & 1 \\\\ -2 & 0 \\end{array} \\right) \\left( \\begin{array}{c} 0 \\\\ \\dfrac{1}{b _ n} \\end{array} \\right) & = \\left( \\begin{array}{c} \\dfrac{1}{b _ n} \\\\ 0 \\end{array} \\right)\r\n\\end{align}\\]\r\n\u306e \\(2\\) \u70b9\u306b\u79fb\u308b\u306e\u3067, \\(l _ {n+1}\\) \u306e\u5f0f\u306f\r\n\\[\\begin{align}\r\n\\left( \\dfrac{3}{a _ n} -\\dfrac{1}{b _ n} \\right) y -\\left( -\\dfrac{2}{a _ n} -0 \\right) x & = 0 \\\\\r\n2 b _ n x +( -a _ n +3 b _ n ) y & = 2 \\\\\r\n\\text{\u2234} \\quad b _ n x +\\dfrac{-a _ n +3 b _ n}{2} y & = 1\r\n\\end{align}\\]\r\n\u3088\u3063\u3066\r\n\\[\r\na _ {n+1} =\\underline{b _ n} , \\quad b _ {n+1} =\\underline{\\dfrac{-a _ n +3 b _ n}{2}}\r\n\\]\r\n
(2)<\/strong><\/p>\r\n
(1)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\\begin{align}\r\n2 a _ {n+2} -3 a _ {n+1} +a _ n & = 0 \\\\\r\n\\text{\u2234} \\quad 2 a _ {n+2} -a _ {n+1} & = 2 a _ {n+1} -a _ n\r\n\\end{align}\\]\r\n\u3053\u308c\u3092\u7e70\u308a\u8fd4\u3057\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\n2 a _ {n+1} -a _ n = 2 b _ 1 -a _ 1 & = 2 \\cdot 2 -3 = 1 \\\\\r\n\\text{\u2234} \\quad a _ {n+1} -1 & = \\dfrac{1}{2} ( a _ n -1 )\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \u6570\u5217 \\(\\left\\{ a _ n -1 \\right\\}\\) \u306f, \u516c\u6bd4 \\(\\dfrac{1}{2}\\) , \u521d\u9805 \\(a _ 1 -1 =3-1 =2\\) \u306e\u7b49\u6bd4\u6570\u5217\u306a\u306e\u3067\r\n\\[\\begin{align}\r\na _ n -1 & = \\left( \\dfrac{1}{2} \\right)^{n-1} \\cdot 2 = 2^{2-n} \\\\\r\n\\text{\u2234} \\quad a _ n & = 2^{2-n} +1\r\n\\end{align}\\]\r\n\u518d\u3073 (1)<\/strong> \u306e\u7d50\u679c\u3092\u7528\u3044\u3066\r\n\\[\r\nb _ n = a _ {n+1} = 2^{1-n} +1\r\n\\]\r\n\\(l _ n\\) \u306e \\(x\\) \u5207\u7247 \\(\\dfrac{1}{a _ n}\\) , \\(y\\) \u5207\u7247 \\(\\dfrac{1}{b _ n}\\) \u306b\u7740\u76ee\u3059\u308b\u3068, \\(n \\geqq 0\\) \u306b\u5bfe\u3057\u3066\r\n\\[\\begin{align}\r\n\\dfrac{1}{3} \\leqq \\dfrac{1}{a _ n} & = \\dfrac{1}{2^{2-n} +1} \\lt 1 \\\\\r\n\\dfrac{1}{2} \\leqq \\dfrac{1}{b _ n} & = \\dfrac{1}{2^{1-n} +1} \\lt 1\r\n\\end{align}\\]\r\n\u307e\u305f, \\(l _ n\\) \u304c \\(n\\) \u306b\u3088\u3089\u305a\u901a\u308b\u70b9\u306b\u3064\u3044\u3066\u8003\u3048\u308b\u3068\r\n\\[\\begin{align}\r\n( 2^{2-n}+1 )x +( 2^{1-n}+1 )y & =1 \\\\\r\n(2x+y) 2^{1-n} +x+y-1 & = 0\r\n\\end{align}\\]\r\n\u306a\u306e\u3067\r\n\\[\\begin{align}\r\n\\left\\{ \\begin{array}{l} 2x+y=0 \\\\ x+y-1 =0 \\end{array} \\right. \\\\\r\n\\text{\u2234} \\quad (x,y) = ( -1 , 2 )\r\n\\end{align}\\]\r\n\u4ee5\u4e0a\u3088\u308a, \u6c42\u3081\u308b\u9818\u57df\u306f\u3059\u3079\u3066\u306e \\(l _ n\\) \uff08 \\(n =0, 1, 2, \\cdots\\) \uff09\u306b\u5bfe\u3057\u3066\u4e0a\u5074\u3068\u306a\u308b\u9818\u57df\u3067,\r\n\u4e0b\u56f3\u306e\u659c\u7dda\u90e8\uff08\u5b9f\u7dda\u306f\u542b\u307f, \u7834\u7dda\u3068\u25cb\u306f\u542b\u307e\u306a\u3044\uff09.<\/p>\r\n
![\"\"](\"\/\/www.roundown.net\/nyushi\/wp-content\/uploads\/tokyo_r_2008_01_01.png\" "\\"tokyo_r_2008_01_01\\"")
\r\n","protected":false},"excerpt":{"rendered":"\u5ea7\u6a19\u5e73\u9762\u306e\u70b9 \\((x,y)\\) \u3092 \\(( 3x+y , -2x )\\) \u3078\u79fb\u3059\u79fb\u52d5 \\(f\\) \u3092\u8003\u3048, \u70b9 P \u304c\u79fb\u308b\u5148\u3092 \\(f( \\text{P} )\\) \u3068\u8868\u3059. \\(f\\) \u3092\u7528\u3044\u3066\u76f4\u7dda \\(l _ 0 , … \u7d9a\u304d\u3092\u8aad\u3080