证明xcosx无周期
假设\(xcos\,x\)有周期,依据周期函数的规律,可得
\[\begin{aligned}
xcos\,x & = (x+T)cos\,(x+T) \\
& = (x+T)cos\,xcos\,T - sin\,xsin\,T \\
& = xcos\,xcos\,T - xsin\,xsin\,T + Tcos\,xcos\,T - Tsin\,xsin\,T \\
\end{aligned}
\]
上式需要成立,则\(cos\,T = 1并且Tcos\,xcos\,T-Tsin\,xsin\,T-xsin\,xsin\,T=0\)
假设\(cos\,T=1\)成立,则\(sin\,T=1-cos^2\,T=0\),则\(Tcos\,xcos\,T-Tsin\,xsin\,T-xsin\,xsin\,T=Tcos\,x=0\)
\(Tcos\,x=0\)发现只有\(T=0\)时,\(Tcos\,x=0\)条件才成立,因此\(xcos\,x\)函数没有周期