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 On Sun, Aug 19, 2018 at 3:33 PM Adam P. Goucher <apgoucher@gmx.com> wrote:
The first product is, by definition, equal to the magnitude of the Vandermonde determinant (where omega is an Nth root of unity):
Prod[0 <= i < j < N] (omega^i - omega^j)
which is equal to the determinant of the Discrete Fourier Transform matrix M_ij := (omega^(ij)). Now, if we prepend the scaling factor of 1/sqrt(N), the DFT is unitary. Without this scaling factor, it therefore has a determinant of 1/sqrt(N)^N = N^(N/2).
Is there a more elementary way to see this (not relying on the Vandermonde determinant)?
Best wishes,
Adam P. Goucher
Sent: Sunday, August 19, 2018 at 10:42 PM From: "Mike Speciner" <ms@alum.mit.edu> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] Chord length product
If you inscribe a regular N-gon in a unit circle, then the product of all the vertex-to-vertex chord lengths is N**(N/2), and the product of all the fixed-vertex-to-other-vertex chord lengths is N. Is there some relatively intuitive reason that either of these should be true?
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