Trigonometric interpolation

Remarks

Existence and uniqueness: the conditions form a square linear system in $c_{-n}, \ldots, c_n$. Let $p$ have degree at most $n$ and vanish at all $x_j$. With $z = e^{ix}$, the function $z^n p$ is an ordinary polynomial of degree at most $2n$ in $z$. It has the $2n + 1$ distinct roots $e^{ix_j}$, so $p = 0$. Hence the system has a unique solution. If all $y_j$ are real, then $c_{-k} = \overline{c_k}$ and $p$ is real-valued.

For equally spaced nodes $x_j = 2\pi j / (2n+1)$, the coefficients are $$c_k = \frac{1}{2n+1} \sum_{j=0}^{2n} y_j e^{-ik x_j},$$ a discrete Fourier transform of the data.

Example: $n = 1$, nodes $0, 2\pi/3, 4\pi/3$ and values $1, 0, 0$. The formula gives $c_0 = c_1 = c_{-1} = 1/3$, so $p(x) = \frac{1}{3} + \frac{2}{3} \cos x$. Check: $p(0) = 1$ and $p(2\pi/3) = p(4\pi/3) = \frac{1}{3} - \frac{1}{3} = 0$.

Non-example: the nodes $0$ and $2\pi$ are equal modulo $2\pi$, so they are not allowed. The data $p(0) = 1$, $p(2\pi) = 0$ cannot be met by any $2\pi$-periodic function.

With an even number $2n$ of nodes one uses trigonometric polynomials of degree $n$ with a constraint on the top frequency, such as a cosine term only.