设对任意的x满足f(x+w)=-f(x)或对任意的x满足f(x+w)=1/f(x) (w>0).
证明:f(x)是以2w为周期的周期函数.
參考答案:f(x+w)=-f(x)
=>
f(x+2w)=f(x+w+w)=-f(x+w)=f(x)
f(x+w)=1/f(x)
=>
f(x+2w)=f(x+w+w)=1/f(x+w)=f(x)
设对任意的x满足f(x+w)=-f(x)或对任意的x满足f(x+w)=1/f(x) (w>0).
证明:f(x)是以2w为周期的周期函数.
參考答案:f(x+w)=-f(x)
=>
f(x+2w)=f(x+w+w)=-f(x+w)=f(x)
f(x+w)=1/f(x)
=>
f(x+2w)=f(x+w+w)=1/f(x+w)=f(x)