[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?
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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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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-- Forewarned is worth an octopus in the bush.
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
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
participants (4)
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Adam P. Goucher -
Michael Kleber -
Mike Speciner -
Veit Elser