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Feb 25, 2015 at 23:58 vote accept Liliam
Feb 21, 2015 at 18:52 comment added Alex Degtyarev @JasonStarr: thanks, this is nice. For proof, I guess, this is the interpolation by Newton "monomials" rather than conventional ones, right?
Feb 21, 2015 at 17:53 comment added Jason Starr @AlexDegtyarev. For elements $a_0,\dots,a_{n-1}$ in a commutative ring $R$, the Vandermonde determinant $\text{det}(a_i^j)_{0\leq i,j\leq n-1}$ equals $0!\cdot 1!\cdots (n-1)!$ times the determinant $\text{det}(\binom{a_i}{j})_{0\leq i,j\leq n-1}$. Thus, the determinant above is, up to a product of factorials, a Vandermonde determinant.
Feb 21, 2015 at 17:01 answer added Chris McDaniel timeline score: 6
Feb 21, 2015 at 17:01 review Close votes
Feb 21, 2015 at 17:43
Feb 21, 2015 at 16:59 history edited Stefan Kohl CC BY-SA 3.0
Tried to improve the formulation of the question.
Feb 21, 2015 at 16:50 comment added Alex Degtyarev @JasonStarr How do you reduce this to Vandermonde? Just curious :)
Feb 21, 2015 at 16:40 comment added Liliam Yes, but Vandermonde determinant didn't solve the problem. I found a paper that solves the problem but to a much more general case. I would like a more simpler solution, I believe it exists.
Feb 21, 2015 at 16:33 review First posts
Feb 21, 2015 at 16:50
Feb 21, 2015 at 16:32 comment added Jason Starr Have you searched for "Vandermonde determinant"?
Feb 21, 2015 at 16:30 history asked Liliam CC BY-SA 3.0