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[函数] $\cos(π(2j^2-n)/4n)$求和

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hbghlyj posted 2023-4-28 02:27 |Read mode
$$\sqrt{2n}=\sum _{j=0}^{2 n-1} \cos \left(\frac{\pi  \left(2 j^2-n\right)}{4 n}\right)$$与$$0=\sum _{j=0}^{2 n-1} \sin\left(\frac{\pi  \left(2 j^2-n\right)}{4 n}\right)$$

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original poster hbghlyj posted 2023-5-6 19:13
顶啊

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