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?
Best Answer
The answer is: $f_n(x)=(1-x^2)^n$.
You can prove that this is true by induction. If you subtract from the first row the second row times $x$, all the entries of the first line after the first one become $0$ (and the first one is $1-x^2$). Therefore, $f_n(x)=(1-x^2)f_{n-1}(x)$. Since $f_1(x)=1-x^2$, you're done.