cos(2πkj/M)对k,j求和

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对于任意正整数$M$,\[\sum_{j=1}^{M/2}\sum_{k=1}^{M-1}\cos\left(\frac{2\pi}{M}kj\right)=-\left\lfloor\frac M2\right\rfloor\]
Source: MSE

证明

$M=1$时两边为0. 当$M>1$时, 等价于\[\sum_{j=1}^{M/2}\sum_{k=1}^{M}\cos\left(\frac{2\pi}{M}kj\right)=0\]
因为$1≤j<M$, 所以$M\nmid j$, 根据这帖式(1)\[\sum_{k=1}^M\cos\left(\frac{2\pi}{M}kj\right)=0\]