TITLE: How to compute the determinant of this Toeplitz matrix?
QUESTION [11 upvotes]: Given a positive integer $n$, express$$
f_n(x) = \left|\begin{array}{c c c c c} 
1 & x & \cdots & x^{n - 1} & x^n\\
x & 1 & x & \cdots & x^{n - 1} \\
\vdots & x & \ddots & \ddots & \vdots\\
x^{n - 1} & \vdots & \ddots & 1 & x\\
x^n & x^{n - 1} & \cdots & x & 1
\end{array}\right|
$$
  as a polynomial of $x$.

I tried to find a recurrence relation of $\{f_n\}_{n \geqslant 1}$ using Laplace expansion, but there seems to be no patterns in the minors in the expansion. Is there a somewhat simple recurrence relation of $\{f_n\}_{n \geqslant 1}$ or these determinants can be computed with other methods?

REPLY [8 votes]: @José Carlos Santos I cannot compete with your straightforward proof.
Nevertheless, I thought it could be interesting to provide a (longer...) proof by using a rather peculiar property of the given matrix : the fact that its inverse is the following tridiagonal $(n+1) \times (n+1)$ matrix :
$$\dfrac{1}{1-x^2}T_n \ \ \ \ \text{where} \ \ \ \ T_n:=\begin{pmatrix} 
1&      -x&    0  & \cdots&       0&  0\\
-x& x^2 + 1&  -x &    \ddots&       0&       0\\
0&  -x& x^2 + 1&      \ddots& 0 &0\\
 0&0&  -x& \ddots&      -x &0\\
 0&0&  \ddots& \ddots&     x^2+1 &-x\\
  0&0&\cdots&  0&       -x&1\end{pmatrix}$$
with exceptional entries $1$ in $(1,1)$ and $(n+1,n+1)$.
Let $S_n$ be the initial matrix ; it is indeed easy to verify that $S_n T_n=(1-x^2)I_{n+1}$.
Let us prove now that, whatever $n$, the determinant of $T_n$ has the constant expression:
$$\tag{1}\det(T_n)=(1-x^2).$$
Let us multiply the first line of $\det(T_n)$ by $x$, then add this new line to the second line (this operation doesn't modify $\det(T_n)$). Laplace expansion along the first column gives:
$$x\det(T_n)=x\det(T_{n-1}).$$
Knowing that $\det(T_1)=1-x^2$ by direct computation, we have reached our objective (1).
As a consequence,
$$\det(S_n)=\left(\dfrac{1}{(1-x^2)^{n+1}}\det(T_n)\right)^{-1}=(1-x^2)^n.$$
Remark : There is a nice interpretation of the initial matrix $S_n$ as the covariance matrix of an autoregressive process $V_{n+1}=xV_n+aX$ with $X \sim N(0,1)$ (see for the continuous case (https://www.le.ac.uk/users/dsgp1/COURSES/ELOMET/LECTURE5.PDF)). The inverse of a covariance matrix is almost as significant as the initial covariance matrix. See (https://stats.stackexchange.com/q/73463) for its different uses/interpretations.