Lagrange basis polynomials

Remarks

Each $\ell_j$ has degree $n$, and $\ell_j(x_i) = 1$ if $i = j$ and $\ell_j(x_i) = 0$ if $i \ne j$. The $\ell_j$ form a basis of the space of polynomials of degree at most $n$, and every such $p$ satisfies $p = \sum_{j} p(x_j) \, \ell_j$. They also satisfy $\sum_j \ell_j = 1$, the case $p = 1$. With $\ell(x) = \prod_{k=0}^{n} (x - x_k)$, one has $\ell_j(x) = \dfrac{\ell(x)}{(x - x_j) \, \ell'(x_j)}$.

Example: for the nodes $0, 1, 2$ in $\R$, $\ell_1(x) = \frac{(x - 0)(x - 2)}{(1 - 0)(1 - 2)} = -x(x - 2)$. Check: $\ell_1(0) = 0$, $\ell_1(1) = 1$ and $\ell_1(2) = 0$.

Non-example: for the nodes $0, 0, 1$, the nodes are not distinct, and the factor $\frac{x - x_1}{x_0 - x_1}$ in $\ell_0$ has denominator $0$, so $\ell_0$ is not defined.