论文标题
一种新颖的永久身份，具有应用
A novel permanent identity with applications
论文作者
论文摘要
令$ n $为一个正整数，并定义有理函数$ s（x_1，\ ldots，x_ {2n}）$作为矩阵$ [x_ {x {j，k}] _ {1 \ le j，k \ le 2n} $的永久$ x_ {j，k} = \ begin {cases}（x_j+x_k）/（x_j-j-x_k）＆\ text {if} \ j \ not = k，\\ 1＆\ \\ 1＆\ text {if}具有以下结果：如果变量之一$ x_1，\ ldots，x_ {2n} $为零，则为$ s（x_1，\ ldots，x_ {2n}）$ nishes $消失，I.E. τ(j)\not=j}^{2n}\frac{x_j+x_{τ(j)}}{x_j-x_{τ(j)}}=0,$$ where we view an empty product $\prod_{i\in\emptyset}a_i$ as $1$.作为一个应用程序，我们表明，如果$ζ$是unity的原始$ 2N $ - $ \ sum_ {τ\ in s_ {2n}}} \ prod_ {j = 1 \ atop τ（j）\ not = j}^{2n} \ frac {1+ζ^{j-τ（j）}}} {1-ζ^{j-ζ^{j-τ（j）}} =（（2n-1）！太阳。
Let $n$ be a positive integer, and define the rational function $S(x_1,\ldots,x_{2n})$ as the permanent of the matrix $[x_{j,k}]_{1\le j,k\le 2n}$, where $$x_{j,k}=\begin{cases}(x_j+x_k)/(x_j-x_k)&\text{if}\ j\not=k,\\1&\text{if}\ j=k.\end{cases}$$ We give an explicit formula for $S(x_1,\ldots,x_{2n})$ which has the following consequence: If one of the variables $x_1,\ldots,x_{2n}$ takes zero, then $S(x_1,\ldots,x_{2n})$ vanishes, i.e., $$\sum_{τ\in S_{2n}}\prod_{j=1\atop τ(j)\not=j}^{2n}\frac{x_j+x_{τ(j)}}{x_j-x_{τ(j)}}=0,$$ where we view an empty product $\prod_{i\in\emptyset}a_i$ as $1$. As an application, we show that if $ζ$ is a primitive $2n$-th root of unity then $$\sum_{τ\in S_{2n}}\prod_{j=1\atop τ(j)\not=j}^{2n}\frac{1+ζ^{j-τ(j)}}{1-ζ^{j-τ(j)}}=((2n-1)!!)^2$$ as conjectured by Z.-W. Sun.
