P7487 「Stoi2031」兰亭序 的式子推导
∏x1=1n∏x2=1n...∏xtn(1+e2iπx1x2...xtn)\prod_{x_1=1}^n\prod_{x_2=1}^n...\prod_{x_t}^n(1+e^{\frac{2i\pi x_1 x_2...x_t}{n}})∏x1=1n∏x2=1n...∏xtn(1+en2iπx1x2...xt)
=∏x1=1n∏x2=1n...∏xtn(1+(−1)2x1x2...xtn)=\prod_{x_1=1}^n\prod_{x_2=1}^n...\prod_{x_t}^n(1+(-1)^{\frac{2 x_1 x_2...x_t}{n}})=∏x1=1n∏x2=1n...∏xtn(1+(−1)n2x1x2...xt)
=∏x1=1n∏x2=1n...∏xtn(1+1x1x2...xtn)=\prod_{x_1=1}^n\prod_{x_2=1}^n...\prod_{x_t}^n(1+1^{\frac{x_1 x_2...x_t}{n}})=∏x1=1n∏x2=1n...∏xtn(1+1nx1x2...xt)
=∏x1=1n∏x2=1n...∏xtn2=\prod_{x_1=1}^n\prod_{x_2=1}^n...\prod_{x_t}^n2=∏x1=1n∏x2=1n...∏xtn2
