证明$f(x)=sinx^2$不是周期函数.
反证:假设是周期函数,周期为$T,T>0$.
$$f(0)=f(T)\Rightarrow sinT^2=0\Rightarrow T^2=k_1\pi,k_1\in N^{*}$$
$$f(\sqrt{2}T)=f(\sqrt{2}T+T)\Rightarrow sin2T^2=sin(\sqrt{2}T+T)^2$$
$$\Rightarrow sin2k_1\pi =s in(\sqrt{2}T+T)^2$$
$\Rightarrow(\sqrt{2}T+T)^2=k_2\pi,k_2\in N^{*}$
$\Rightarrow (\sqrt{2}+1)^2=\frac{k_2}{k_1}$,
等式左边为无理数$\ne$等式右边为有理数,矛盾,故假设不成立。$\therefore f(x)=sinx^2$不是周期函数.
评:此类证明非周期的题,套路基本都是反证,取一些特殊值,得出矛盾.
时间: 2024-10-20 15:19:07