The discrete Fourier transform's Gaussian-like eigenvector

Question

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 (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.

Answer 1

A quick numerical investigation for $N$ up to 512 leads to a conjectured exact large-$N$ expansion of the largest eigenvalue, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} - c_5 \frac{\pi^5}{N^5} + \ldots. $$ The coefficient $c_5^{-1} = -114.63(1)$ is negative and seems to be a more complicated expression.

Added 16.07.22, 18:00 CEST:

Extending the high-precision calculation to $N=1024$, the next two constants seem to be $c_5=-67/(2^6 5!)$ and $c_6=653/(2^7 6!)$. These terms should be verified with a precision higher than 30 digits.

Edited 17.07.22, 15:00 CEST:

I did this verification with 100 digits precision, verifying $c_{5,6}$ and extending the series to 10th order. Therefore, $$ \lambda_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4} + \frac{67\pi^5}{7680 N^5} + \frac{653\pi^6}{92160 N^6} +\frac{32519 \pi ^7}{5160960 N^7} +\frac{135001 \pi ^8}{20643840 N^8} +\frac{45750727 \pi ^9}{5945425920 N^9} +\frac{1198585643 \pi ^{10}}{118908518400 N^{10}} +\ldots. $$ Note that the denominators of $c_n$ are divisible by $n!$.

All constants were successively determined by a least-square fit of a 20th order polynomial in $\pi/N$ to $λ_0(N)$. I did not find a known function with this series expansion.

For reference, I'll append the 30 digits results: \begin{array}{ll} N & λ_0(N)\\ 128 & 3.95121320281088603898478521135 \\ 256 & 3.97553152996970070083424756460 \\ 384 & 3.98367098180919839228844082991 \\ 512 & 3.98774696887736782094115794509 \\ 640 & 3.99019456589827273460033342852 \\ 768 & 3.99182713285240169577614505117 \\ 896 & 3.99299366147114034625800714548 \\ 1024 & 3.99386878184121881247367446866 \\ \end{array}