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[数列] 两道数列题

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青青子衿 Post time 2013-11-15 21:04 |Read mode
$a_n^2=a_{n+1}+1$
$a_n^2=a_{n+1}+n$

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realnumber Post time 2013-11-16 09:24
本帖最后由 realnumber 于 2013-11-16 10:20 编辑 挺难,似乎没见过,把$a_1=?$打上,能说明题目来源最好了.
第一个,有n+1以上个1或无限.似乎符合
\[a_n=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+....}}}}\]
第2个似乎这样符合,应该不唯一吧,
\[a_n=\sqrt{n+\sqrt{n+1+\sqrt{n+2+\sqrt{n+3+....}}}}\]

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 Author| 青青子衿 Post time 2013-11-16 10:55
回复 2# realnumber
$a_{1}=k,a_n^2=a_{n+1}+1$
$a_{1}=k,a_n^2=a_{n+1}+n$

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战巡 Post time 2013-11-16 11:01
回复 3# 青青子衿


这类问题早有定论
只有极少数特殊情况有初等通项,大多数时候都极为恶心

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 Author| 青青子衿 Post time 2014-8-24 10:02
回复 2# realnumber
挺难,似乎没见过,把$a_1=?$打上,能说明题目来源最好了.
第一个,有n+1以上个1或无限.似乎符合
\[a_n=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+....}}}}\] ...
\[\color{red}{\varphi=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+....}}}}=\frac{\sqrt{5}+1}{2}}\]
\[\color{red}{\Phi=\Phi^{-1}-1=\varphi^{-1}=\varphi-1=\frac{1}{\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+....}}}}}=\frac{\sqrt{5}-1}{2}}\]
realnumber 发表于 2013-11-16 09:24

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isee Post time 2014-8-28 00:13
黄金分割数的倒数

连分数理论?有点兴趣,来源及背景是什么?

手机版|悠闲数学娱乐论坛(第3版)

2025-3-5 09:45 GMT+8

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