Timeline for About a circular variant of Vandermonde matrix
Current License: CC BY-SA 4.0
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| Mar 14 at 7:17 | comment | added | Vladimir Dotsenko | Moreover, the answer obtained by @FredHucht admits an immediate proof. Subtract from the first line $x_1$ times the second, develop along the first row, and do the same with the remaining matrix. Every time you earn $1-x_1x_2\cdots x_n$ as a factor. | |
| Mar 13 at 21:37 | comment | added | Fred Hucht |
A quick Mathematica calculation (using Mat[n_]:=Table[Product[Subscript[x,Mod[k,n,1]], {k,i,If[j<i,j+n,j] - 1}], {i,n}, {j,n}]) reveals that $$ \det(M_n) = (1-x_1 x_2 \ldots x_n)^{n-1}\,. $$
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| Mar 13 at 20:20 | history | edited | lntk | CC BY-SA 4.0 |
deleted 72 characters in body
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| S Mar 13 at 20:20 | review | First questions | |||
| Mar 13 at 22:08 | |||||
| S Mar 13 at 20:20 | history | asked | lntk | CC BY-SA 4.0 |