Chebyshev nodes

Remarks

Example: for $n = 3$ the nodes are $\cos(\pi/6), \cos(\pi/2), \cos(5\pi/6)$, that is $\sqrt{3}/2$, $0$ and $-\sqrt{3}/2$. Check: $T_3(x) = 4x^3 - 3x = x(4x^2 - 3)$ vanishes exactly there.

Non-example: the equally spaced points $-1, 0, 1$ are not the Chebyshev nodes of order $3$, since $T_3(1) = 1 \ne 0$.

The node polynomial is $\prod_{k=1}^{n} (x - x_k) = 2^{1-n} T_n(x)$, whose maximum on $[-1, 1]$ is $2^{1-n}$, the least possible for a monic polynomial of degree $n$. Hence, if $f$ has $n$ continuous derivatives and $p$ interpolates $f$ at these nodes, then $\max_{[-1, 1]} \abs{f - p} \le \frac{\max \abs{f^{(n)}}}{2^{n-1} n!}$. The points $\cos(k\pi/n)$, $k = 0, \dots, n$, where $T_n = \pm 1$, are the Chebyshev extreme points, also called Chebyshev–Lobatto points.