\\(C _ 1\\) \u3068 \\(C _ 2\\) \u306f\u534a\u5f84 \\(1\\) \u306e\u5186\u3067, \u4e92\u3044\u306b\u5916\u63a5\u3059\u308b.<\/p><\/li>\r\n
\u6b63\u306e\u6574\u6570 \\(n\\) \u306b\u5bfe\u3057, \\(C _ {n+2}\\) \u306f \\(C _ n\\) \u3068 \\(C _ {n+1}\\) \u306b\u5916\u63a5\u3057, \\(C _ n\\) \u3068 \\(C _ {n+1}\\) \u306e\u5f27\u304a\u3088\u3073 \\(x\\) \u8ef8\u306b\u56f2\u307e\u308c\u308b\u90e8\u5206\u306b\u3042\u308b.<\/p><\/li>\r\n<\/ul>\r\n
\u5186 \\(C _ n\\) \u306e\u534a\u5f84\u3092 \\(r _ n\\) \u3068\u3059\u308b.<\/p>\r\n
- \r\n
(1)<\/strong>\u3000\u7b49\u5f0f \\(\\dfrac{1}{\\sqrt{r _ {n+2}}} = \\dfrac{1}{\\sqrt{r _ n}} +\\dfrac{1}{\\sqrt{r _ {n+1}}}\\) \u3092\u793a\u305b.<\/p><\/li>\r\n
(2)<\/strong>\u3000\u3059\u3079\u3066\u306e\u6b63\u306e\u6574\u6570 \\(n\\) \u306b\u5bfe\u3057\u3066 \\(\\dfrac{1}{\\sqrt{r _ n}} = s {\\alpha}^n +t {\\beta}^n\\) \u304c\u6210\u308a\u7acb\u3064\u3088\u3046\u306b, \\(n\\) \u306b\u3088\u3089\u306a\u3044\u5b9a\u6570 \\(\\alpha , \\beta , s , t\\) \u306e\u5024\u3092\u4e00\u7d44\u4e0e\u3048\u3088.<\/p><\/li>\r\n
(3)<\/strong>\u3000\\(n \\rightarrow \\infty\\) \u306e\u3068\u304d\u6570\u5217 \\(\\left\\{ \\dfrac{r _ n}{k^n} \\right\\}\\) \u304c\u6b63\u306e\u5024\u306b\u53ce\u675f\u3059\u308b\u3088\u3046\u306b\u5b9f\u6570 \\(k\\) \u306e\u5024\u3092\u5b9a\u3081, \u305d\u306e\u3068\u304d\u306e\u6975\u9650\u5024\u3092\u6c42\u3081\u3088.<\/p><\/li>\r\n<\/ol>\r\n
\r\n\u89e3\u7b54<\/h3>\r\n
(1)<\/strong><\/p>\r\n
![\"ngr20140301\"](\"\/\/www.roundown.net\/nyushi\/wp-content\/uploads\/ngr20140301.svg\")
\r\n\u5186 \\(C _ n\\) \u306e\u4e2d\u5fc3\u3092 \\(\\text{A} _ n\\) , \\(x\\) \u8ef8\u3068\u306e\u63a5\u70b9\u3092 \\(\\text{H} _ n\\) \u3068\u304a\u304f.\r\n\\[\\begin{align}\r\n\\text{H} _ n \\text{H} _ {n+1} & = \\sqrt{( r _ n +r _ {n+1})^2 -( r _ n -r _ {n+1})^2} \\\\\r\n& = 2 \\sqrt{r _ n r _ {n+1}} \\ .\r\n\\end{align}\\]\r\n\u540c\u69d8\u306b\r\n\\[\r\n\\text{H} _ {n+1} \\text{H} _ {n+2} = 2 \\sqrt{r _ {n+1} r _ {n+2}} , \\ \\text{H} _ n \\text{H} _ {n+2} = 2 \\sqrt{r _ n r _ {n+2}} \\ .\r\n\\]\r\n\\(\\text{H} _ n \\text{H} _ {n+1} = \\text{H} _ {n+1} \\text{H} _ {n+2} +\\text{H} _ n \\text{H} _ {n+2}\\) \u306a\u306e\u3067\r\n\\[\\begin{align}\r\n2 \\sqrt{r _ n r _ {n+1}} & = 2 \\sqrt{r _ {n+1} r _ {n+2}} +2 \\sqrt{r _ n r _ {n+2}} \\\\\r\n\\text{\u2234} \\quad \\dfrac{1}{\\sqrt{r _ {n+2}}} & = \\dfrac{1}{\\sqrt{r _ n}} +\\dfrac{1}{\\sqrt{r _ {n+1}}} \\ .\r\n\\end{align}\\]\r\n
(2)<\/strong><\/p>\r\n
\\(a _ n = \\dfrac{1}{\\sqrt{r _ n}}\\) \u3068\u304a\u304f\u3068, (1)<\/strong> \u306e\u7d50\u679c\u3088\u308a\r\n\\[\r\na _ {n+2} -a _ {n+1} -a _ n = 0 \\quad ... [1] \\ .\r\n\\]\r\n\u65b9\u7a0b\u5f0f \\(x^2 -x -1 = 0\\) \u306e\u89e3\u3092 \\(p , q \\ ( p \\gt q )\\) \u3068\u304a\u304f\u3068\r\n\\[\r\np = \\dfrac{1 +\\sqrt{5}}{2} , \\ q = \\dfrac{1 -\\sqrt{5}}{2} \\ .\r\n\\]\r\n\u307e\u305f, \u89e3\u3068\u4fc2\u6570\u306e\u95a2\u4fc2\u3088\u308a \\(p+q = 1\\) , \\(pq = -1\\) \u306a\u306e\u3067, [1] \u3088\u308a\r\n\\[\\begin{align}\r\na _ {n+2} -p a _ {n+1} & = q \\left( a _ {n+1} -p a _ n \\right) , \\\\\r\na _ {n+2} -q a _ {n+1} & = p \\left( a _ {n+1} -q a _ n \\right) \\ .\r\n\\end{align}\\]\r\n\u3057\u305f\u304c\u3063\u3066, \\(a _ 1 = a _ 2 = 1\\) \u3067\u3042\u308b\u3053\u3068\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\na _ {n+1} -p a _ n & = q^{n-1} \\left( a _ 2 -p a _ 1 \\right) = q^{n-1} (1-p) = q^n, \\\\\r\na _ {n+1} -q a _ n & = p^{n-1} \\left( a _ 2 -q a _ 1 \\right) = p^{n-1} (1-q) = p^n \\ .\r\n\\end{align}\\]\r\n\u8fba\u3005\u3092\u5f15\u304f\u3068\r\n\\[\\begin{align}\r\n(p-q) a _ n & = p^n -q^n \\\\\r\n\\text{\u2234} \\quad a _ n & = \\dfrac{1}{p-q} \\left( p^n -q^n \\right) \\ .\r\n\\end{align}\\]\r\n\\(p-q = \\sqrt{5}\\) \u306a\u306e\u3067, \u6c42\u3081\u308b\u7d44\u5408\u305b\u306e \\(1\\) \u3064\u306f\r\n\\[\r\n\\alpha = \\underline{\\dfrac{1 +\\sqrt{5}}{2}} , \\ \\beta = \\underline{\\dfrac{1 -\\sqrt{5}}{2}} , \\ s = t = \\underline{\\dfrac{1}{\\sqrt{5}}} \\ .\r\n\\]\r\n
(3)<\/strong><\/p>\r\n
(2)<\/strong> \u306e\u7d50\u679c\u3092\u7528\u3044\u308c\u3070\r\n\\[\\begin{align}\r\n\\dfrac{r _ n}{k^n} & = \\dfrac{1}{k^n \\left( s {\\alpha}^n +t {\\beta}^n \\right)^2} \\\\\r\n& = \\underline{\\dfrac{1}{\\left( k {\\alpha}^2 \\right)^n}} _ {[2]} \\cdot \\underline{\\dfrac{1}{\\left\\{s +t \\left( \\frac{\\beta}{\\alpha} \\right)^n \\right\\}^2}} _ {[3]} \\ .\r\n\\end{align}\\]\r\n\u4e0b\u7dda\u90e8 [3] \u306b\u3064\u3044\u3066, \\(\\left| \\dfrac{\\beta}{\\alpha} \\right| \\lt 1\\) \u306a\u306e\u3067, \\(n \\rightarrow \\infty\\) \u306e\u3068\u304d\r\n\\[\r\n[3] \\rightarrow \\dfrac{1}{s^2} = \\dfrac{1}{5} \\ .\r\n\\]\r\n\u3057\u305f\u304c\u3063\u3066, \u4e0b\u7dda\u90e8 [2] \u304c, \u6b63\u306e\u5024\u306b\u53ce\u675f\u3059\u308b\u6761\u4ef6\u3092\u8003\u3048\u308c\u3070\u3088\u3044.
\r\n\u3088\u3063\u3066, \u6c42\u3081\u308b \\(k\\) \u306e\u5024\u306f\r\n\\[\\begin{align}\r\nk {\\alpha}^2 & = 1 \\\\\r\n\\text{\u2234} \\quad k & = \\dfrac{1}{{\\alpha}^2} = \\underline{\\dfrac{3 -\\sqrt{5}}{2}} \\ .\r\n\\end{align}\\]\r\n\u307e\u305f, \u3053\u306e\u3068\u304d\r\n\\[\\begin{align}\r\n\\displaystyle\\lim _ {n \\rightarrow \\infty} \\dfrac{r _ n}{k^n} = \\underline{\\dfrac{1}{5}} \\ .\r\n\\end{align}\\]\r\n","protected":false},"excerpt":{"rendered":"\\(xy\\) \u5e73\u9762\u306e \\(y \\geqq 0\\) \u306e\u90e8\u5206\u306b\u3042\u308a, \\(x\\) \u8ef8\u306b\u63a5\u3059\u308b\u5186\u306e\u5217 \\(C _ 1 , C _ 2 , C _ 3 , \\cdots\\) \u3092\u6b21\u306e\u3088\u3046\u306b\u5b9a\u3081\u308b. \\(C _ 1\\) \u3068 \\(C … \u7d9a\u304d\u3092\u8aad\u3080