I have the $N$x$N$ matrix below where $N$ is a power of 2 (usually 64 or 256) and $\omega = 2\pi/N$.  What is its largest eigenvalue?

$\begin{bmatrix}
2 & 1 & 0 & 0 & \cdots & 0 & 0 & 1\\
1 & 2\cos(\omega) & 1 & 0 & \cdots & 0 & 0 & 0\\
0 & 1 & 2\cos(2\omega) & 1 & \cdots & 0 & 0 & 0\\
0 & 0 & 1 & 2\cos(3\omega) & \cdots & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 2\cos((N-3)\omega) & 1 & 0\\
0 & 0 & 0 & 0 & \cdots & 1 & 2\cos((N-2)\omega) & 1\\
1 & 0 & 0 & 0 & \cdots & 0 & 1 & 2\cos((N-1)\omega)
\end{bmatrix}$

This is exactly the $S$ matrix [here](https://www.cs.princeton.edu/~ken/Eigenvectors82.pdf) (which explains how this eigenvector is also the simplest eigenvector of the Discrete Fourier Transform); I am wondering if our analysis of this eigenvector has improved since 1982.

For large $N$, I know this eigenvalue tends to 4 and its eigenvector tends to a Gaussian, and I can numerically find the roots of the characteristic polynomial for more precision (e.g., this eigenvalue is 3.903025 for $N$=64), but is there a faster more-accurate method?  Is there a closed-form solution?

To reiterate, the helpful points here are that $N$ is a power of 2 and I am only concerned with the highest eigenvalue.