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[函数] 周期

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reny Posted 2013-12-6 12:58 |Read mode
Last edited by reny 2013-12-30 13:01定义在$R$上的函数$y=f(x)$,满足$f(x+2)=-\dfrac{1}{f(x)}$,问:$f(x)$有比4更小的正周期吗?
补充:定义在$R$上的函数$y=f(x)$,满足$f(x+2)=-f(x)$,问:$f(x)$也有比4更小的正周期吗?

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kuing Posted 2013-12-6 15:11

\begin{align*}
D_1&=\bigcup_{k\in\Bbb Z}\left[\frac43k,\frac43k+\frac23\right),\\
D_2&=\bigcup_{k\in\Bbb Z}\left[\frac43k+\frac23,\frac43k+\frac43\right),
\end{align*}
构造
\[f(x)=\begin{cases}
1, & x\in D_1,\\
-1,& x\in D_2,
\end{cases}\]
它满足条件,并且最小正周期为 $4/3$。

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 Author| reny Posted 2013-12-6 16:35
回复 2# kuing
厉害!感谢kk!

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isee Posted 2013-12-10 00:53
这个,我每次都说4是周期,是不是最小,从来只是一闪而过。


\begin{align*}
D_1&=\bigcup_{k\in\Bbb Z}\left[\frac43k,\frac43k+\frac23\right),\\
D_2&=\bigcup_{k ...
kuing 发表于 2013-12-6 15:11
一如继往的强


这个构造,也许从侧面说明,要使其为最小正整数周期,怕得附加严格的条件,并不容易。

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