引用:
原帖由 cplee8tcfsh 於 2012-5-19 07:16 AM 發表
(1)\( a_{n+1} = \sqrt{2+\sqrt{a_n}} \geq 0 \) 知數列 \( a_n \) 有下界 0
(2)
\( a^2_{n+1} = 2+ \sqrt{a_n} \)\( a^2_{n+2} = 2+ \sqrt{a_{n+1}} \)
\( a^2_{n+2} - a^2_{n+1} = \sqrt{a_{n+1}} -\sqrt{a_{n}}\) ...
(1)
\( a_{n+1} = \sqrt{2+\sqrt{a_n}} \geq 0 \)
知數列 \( a_n \) 有下界 0
\( a^2_{n+1} = 2+ \sqrt{a_n} \)
\( a^2_{n+2} = 2+ \sqrt{a_{n+1}} \)
\( a^2_{n+2} - a^2_{n+1} = \sqrt{a_{n+1}} -\sqrt{a_{n}} \)
\( (a_{n+2} - a_{n+1})(a_{n+2} + a_{n+1}) = \frac{a_{n+1} -a_{n}}{\sqrt{a_{n+1}} +\sqrt{a_{n}}} \)
知 數列 遞增
因遞增有上界 故數列收斂
令數列極限值為x
得 \( x=\sqrt{2+\sqrt{x}} \) 整理得 \( x^4 -4 x^2 - x +4 =0 \)
一元四次方程懶得自己解,呼叫Alpha
解得 \( x = \frac{1}{3} ( -1 + \sqrt[3]{ \frac{79}{2} -\frac{3}{2} \sqrt{249} } + \sqrt[3]{ \frac{79}{2} +\frac{3}{2} \sqrt{249} } ) \approx 1.83118 \)
原帖由 cplee8tcfsh 於 2012-5-19 07:16 AM 發表
(1)\( a_{n+1} = \sqrt{2+\sqrt{a_n}} \geq 0 \) 知數列 \( a_n \) 有下界 0
(2)
\( a^2_{n+1} = 2+ \sqrt{a_n} \)\( a^2_{n+2} = 2+ \sqrt{a_{n+1}} \)
\( a^2_{n+2} - a^2_{n+1} = \sqrt{a_{n+1}} -\sqrt{a_{n}}\) ...
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