Let $\nu$ be the minimum integer which satisfies $2\sin(\frac{\pi \nu}{2(N+1)})>\tau$, for $N$ an integer, and $\tau$ an arbitrary positive number. Since the LHS is bounded and RHS is not, there might be cases where no such value of $\nu$ satisfying the inequality does exist. In that case, we will take $\nu=0$. Now consider the polynomial:($a=\tau^2-2)$ $$\sum_{i=1}^{\nu-1}ac_{i}^2-\sum_{i=\nu}^{N}ac_{i}^2+2\sum_{i=1}^{\nu-2}c_{i}c_{i+1}+2ic_{\nu}c_{\nu-1}-2\sum_{i=\nu}^{N}c_{i}c_{i+1}$$
Suppose I want to express the above as $c^{T}Ac$ with $c^{T}=\begin{matrix} (c_{1} & c_{2} ... &c_{N}) \end{matrix}$
I want to calculate the NxN determinant of A:
$$ \begin{matrix} a & 1 & 0 & 0 .....&0 \\ 1 & a & 1 & 0.......&0 \\ 0 & 1 & a &1....... &0 \\ 0 & 0 & 1 &a.....\\ .\\ .\\ 0.\\\ \end{matrix} $$ So we have repeating blocks of $\begin{matrix} a & 1\\ 1 & a\\ \end{matrix}$ until the $(\nu-2)^{th}$ row and similar repeating blocks of $\begin{matrix} -a & -1\\ -1 & -a\\ \end{matrix}$ from the $(\nu+1)^{th}$ row and a block $\begin{matrix} a & i\\ i & -a\\ \end{matrix}$ in the $(\nu-1)^{th}$ and $(\nu)^{th}$ rows. I could not write down the matrix here, so I would be obliged greatly if you check it yourself. As you can guess I am doing to this calculate $\int_{-\infty}^{\infty}dc e^{-c^{T}Ac}$
$$D_n=\begin{vmatrix} a & 1 & 0 & \cdots & 0 & 0 & 0\\ 1 & a & 1 & \cdots & 0 & 0 & 0\\ 0 & 1 & a & \cdots & 0 & 0 & 0\\ \cdots & \cdots & \cdots & \cdots &\cdots & \cdots & \cdots\\ 0 & 0 & 0 & \cdots & a & 1 & 0\\ 0 & 0 & 0 & \cdots & 1 & a & 1\\ 0 & 0 & 0 &\cdots & 0 & 1 & a\end{vmatrix}$$
$$D_1=a,\\D_2={{a}^{2}}-1,\\D_3={{a}^{3}}-2 a,\\D_4={{a}^{4}}-3 {{a}^{2}}+1,\\D_5={{a}^{5}}-4 {{a}^{3}}+3a,\\ \dots\\ D_n=aD_{n-1}-D_{n-2}$$ Determinants $D_n$ is Chebyshev polynomials of the second kind: $$D_n=U_n\left(\frac{a}2\right)$$ https://en.wikipedia.org/wiki/Chebyshev_polynomials
http://mathworld.wolfram.com/ChebyshevPolynomialoftheSecondKind.html