Thank you. These are all lovely elegant proofs (e.g., DFT Vandermonde determinant for all chords and polynomial ((x+1)**N-1)/x for chords from fixed vertex, perhaps also noting that the product of the roots of a polynomial is given by the constant term) of the facts I asked about, and I was actually aware of similar proofs, but I don't really find any intuition in these proofs. To me, they don't answer the question of "why should this be?" I was hoping for something perhaps geometric, although a physics interpretation would also be interesting. I admit that interpreting the product naturally as a hypervolume seems unlikely to provide much help. --ma On 20-Aug-18 07:39, Veit Elser wrote:
On Aug 20, 2018, at 1:08 AM, Michael Kleber <michael.kleber@gmail.com> wrote:
The vertices of the n-gon are the roots of x^n-1, so the vertices other than 1 are the roots of (x^n-1)/(x-1) = 1 + x + x^2 + ... + x^(n-1). This is the product of x-root for the n-1 other roots, so plug in x=1 and you get the product of the complex-number difference from 1 to each other root; the norm of the result is the product of the distances that you asked about. But evaluating the polynomial at x=1 you get n (and so no taking-the-norm step is needed).
--Michael I lost you somewhere in the middle. I would have said the chords you want are the roots of ((x+1)^n-1)/x = 1 + (x+1) + … +(x+1)^(n-1), and plugging in x=0 gives their product (which is the product of the distances up to a phase).
But is it intuitive? Was Mike looking for a physics interpretation?
-Veit _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun