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A trigonometric identity for a circulant matrix
David V. Ingerman

Abstract. We conjecture a trigonometric identity that arises in the inverse problem for discrete Laplacian for radially symmetric graphs. The results were tested on open source mathematical software Sage. The identities are self-contained and connect such important mathematical objects as golden ratio, Pascal triangle and characteristic polynomials of circulant matrices.

1. Main theorem and corollary
In this paper we will conjecture trigonometric identities that arise in the inverse problems with radial symmetry. The identity and the corollary are self-contained and are of interest to general mathematics.

Conjecture 1.1

 * $$\prod_{k=1}^n(x-4\sin^2 \frac{k\pi}{2n+1}) = x^n+(2n+1)\sum_{k=0}^{n-1}(-1)^{k}{2n-k-1 \choose 2n-2k-1}\frac{x^{k}}{k+1} $$

Corollary 1.2

 * $$4\sum_{k = 1}^n \sin^2 \frac{k\pi}{2n+1}=\prod_{k = 1}^n (4\sin^2 \frac{k\pi}{2n+1})=(2n+1) $$

Conjecture 1.3

 * $$\prod_{k=1}^n(\cot^2\frac{k\pi}{2n+1}-x)=\sum_0^n\frac{(-1)^k}{2n+1-2k}{2n \choose 2k}x^k $$

Corollary 1.4

 * $$\prod_{k=1}^n\cot^2\frac{k\pi}{2n+1}=\frac{1}{2n+1}$$

2. Outline of proofs
The proof of corollaries follows directly from conjectures by matching the coefficients. We will prove the identity 1.1 by constructing the circulant matrix with eigenvalues that are equal to the terms in the sums and products of the expressions in our conjecture and corollary.

Since
 * $$\sin^2\theta=\frac{1 - \cos 2\theta}{2} $$

We can rewrite the expression in the following way
 * $$\prod_{k=1}^n(x-4\sin^2 \frac{k\pi}{2n+1})=\prod_{k=1}^{n}(x - 2(1-\frac{\cos 2k\pi}{2n+1}))=\prod_{k=1}^{n}(2\cos\frac{2k\pi}{2n+1}-(2-x))$$

Therefore,


 * $$\prod_{k=1}^n(x-4\sin^2 \frac{k\pi}{2n+1})=\prod_{k=1}^{n}(2\cos\frac{2k\pi}{2n+1}-z)$$

where $$z = 2-x$$

Since
 * $$\cos 0 = 1 $$ and $$\cos x$$ is an even periodic function

we can complete the product to the product over all $$(2n+1)$$ roots of unity:

$$\prod_{k=1}^{n}(2\cos\frac{2k\pi}{2n+1}-z) = \sqrt{\frac{1}{2-z}\prod_{k = 0}^{2n}(2\cos\frac{2k\pi}{2n+1}-z)}=\sqrt{\frac{1}{2-z}\det(C-zI)}$$

where $$C$$ is the following symmetric 2-band matrix



C=\begin{bmatrix} 0 & 1 & \dots & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ \vdots & 0 & \ddots & \vdots & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & \dots & 1 & 0 \\ \end{bmatrix}

$$

since taking the discrete Fourier transform we can find the the eigenvalues of this matrix to be $$2\cos \frac{2k\pi}{2n+1}$$. Matrix $$C$$ is an instance of a circulant matrix.

Expanding the determinant leads to

Lemma 2.1

 * $$\det(C-zI)=\det

\begin{bmatrix} -z & 1 & \dots & 0 & 1 \\ 1 & -z & 1 & 0 & 0 \\ \vdots & 0 & \ddots & \vdots & 0 \\ 0 & 0 & 1 & -z & 1 \\ 1 & 0 & \dots & 1 & -z \\ \end{bmatrix} $$ $$ =-z^{2n+1}+Tr(C)z^{2n}+\dots+\det(C)=-z^{2n+1}+0z^{2n}+\dots+2= $$

$$ =-z^{2n+1}+(2n+1)\sum_{k=0}^{n-1}(-1)^{k}{2n-k-1 \choose 2n-2k-1} z^{2k+1}+2=(2-z)(\sum^n_{l=0}a_l z^l)^2. $$

Conjecture 2.2
$$ \sum^n_{l=0}a_l z^l =(-1)^n\sum_{l=0}^{[n/2]}(-1)^{l+1}({n-l \choose l}z^{n-2l}+{n-l-1 \choose l} z^{n-2l-1}) $$

where $$a_l$$'s are binomial coefficients forming a Pascal triangle

Pluginning back $$x=2-z$$ we get our main conjecture:


 * $$\prod_{k=1}^n(x-4\sin^2 \frac{k\pi}{2n+1})=

\sqrt{\frac{1}{2-z}(2-z)(\sum_{l=0}a_l z^l)^2}=x^n+\sum_{k=0}^{n-1}(-1)^{k}({2n-k \choose k + 1}+{2n-k-1 \choose k}) x^{k}$$.

3. Golden ratio
We will show the calculations more explicitly for the "pentagon" case that gives a connection of the main identity and corollary to properties of golden ratio. The golden ratio is the algebraic irrational number


 * $$\varphi = 2\cos\frac{2\pi}{5}+1 = \frac{1+\sqrt{5}}{2}\approx 1.61803\,39887\ldots\,$$

Since
 * $$\sin^2\theta=\frac{1 - \cos 2\theta}{2} $$

we can rewrite the expression in the following way
 * $$\prod_{k=1}^2(x-4\sin^2 \frac{k\pi}{5})=\prod_{k=1}^{2}(x - 2(1-\frac{\cos 2k\pi}{5}))=\prod_{k=1}^{2}(x-2+2\cos\frac{2k\pi}{5})$$

Therefore,


 * $$\prod_{k=1}^2(x-4\sin^2 \frac{k\pi}{5})=\prod_{k=1}^{2}(2\frac{\cos 2k\pi} {5}-z)$$

where $$z = 2-x$$

Since
 * $$\cos 0 = 1 $$

We can complete the product in the following way:

$$\prod_{1}^{2}(2\frac{\cos 2k\pi}{5}-z) = \sqrt{\frac{1}{2-z}\prod_{0}^{5}( 2\frac{\cos 2k\pi}{5}-z)}=\sqrt{\frac{1}{2-z}\det(C-zI)}$$

where $$C$$ is the circulant matix



C= \begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 \\ \end{bmatrix}. $$

Lemma 3.1
The eigenvalues of $$C$$ are $$2\cos\frac{2k\pi}{5}, k = 0,1,2,3,4$$

Proof



C= \begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ w  \\ w^2 \\ w^3 \\ w^4 \\ \end{bmatrix}= (w+w^{-1})

\begin{bmatrix} 1 \\ w  \\ w^2 \\ w^3 \\ w^4 \\ \end{bmatrix} $$

where $$w = e^{\frac{2k\pi}{5}i}$$

Expanding the determinant leads to
 * $$\det

\begin{bmatrix} -z & 1 & 0 & 0 & 1 \\ 1 & -z & 1 & 0 & 0 \\ 0 & 1 & -z & 1 & 0 \\ 0 & 0 & 1 & -z & 1 \\ 1 & 0 & 0 & 1 & -z \\ \end{bmatrix} =-z^5+5z^3-5z+2=(2-z)(z^2+z-1)^2=(2-z)(z+1/\varphi)^2(z-\varphi)^2. $$

Pluginning back

$$x=2-z$$

we get:


 * $$\prod_{k=1}^2(x-4\sin^2 \frac{k\pi}{5})=\sqrt{\frac{1}{2-z}(2-z)(z^2+z-1)^2}=

(z^2+z-1)=x^2-5x+5$$

The proof of the corollary follows directly by matching the coefficients.

We deduce

Corollary 3.2

 * $$4\sin^2\frac{2\pi}{5}+4\sin^2\frac{\pi}{5} = 4\sin^2\frac{2\pi}{5}*4\sin^2\frac{\pi}{5} = 5$$