Permanent of a matrix A = $\|a_{i,j}\|_{i,j=1}^{n}$ is defined as $$ \mathrm{Perm}(A) = \sum\limits_{\sigma \in S_{n}} a_{1,\sigma_{1}},\ldots,a_{n,\sigma_{n}} $$ Is there some representation for permanent of Vandermonde's matrix similar to its determinant? Is there some numerical methods for computing Vandermonde's permanent that uses the particularity of Vandermonde's matrix?
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No, but i have to evaluate it using computer. In general, I can use Ryser's formula ($O(n2^n)$ operations), but it doesn't use the particularity of Wandermonde's matrix. – Appliqué Oct 26 '11 at 20:29
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1An apropos paper. (At least you can bound your permanent...) – J. M. ain't a mathematician Oct 26 '11 at 23:13
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@J.M.ain'tamathematician Fixed link: https://www.emis.de/journals/AMEN/2006/050915-3.pdf Jia-jin Wen and Zhi-hua Zhang Vandermonde type determinants and inequalities Appl. Math. E-Notes, 6(2006), 211-218 – Jairo Bochi Mar 21 '24 at 12:47
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Muirhead inequality will also give bounds to the "Vandermonde-permanent". – Jairo Bochi Mar 21 '24 at 12:48