Lagrange polynomial

Remarks

Each $\ell_j$ has degree $n$, with $\ell_j(x_j) = 1$ and $\ell_j(x_k) = 0$ for $k \ne j$. So $L$ lies in the polynomial ring $\R[x]$, has degree at most $n$, and satisfies $L(x_k) = y_k$ for all $k$. It is the unique polynomial with these properties: the difference of two of them has degree at most $n$ and $n + 1$ roots, so it is $0$. The same formula works over any field; this solves polynomial interpolation.

Example: nodes $0, 1, 2$ with values $1, 3, 7$. Here $\ell_0 = \frac{(x - 1)(x - 2)}{2}$, $\ell_1 = -x(x - 2)$ and $\ell_2 = \frac{x(x - 1)}{2}$. Then $L = \ell_0 + 3 \ell_1 + 7 \ell_2 = x^2 + x + 1$, and indeed $L(0) = 1$, $L(1) = 3$, $L(2) = 7$.

Non-example: if two nodes coincide, say $x_0 = x_1$, the factor $\frac{x - x_1}{x_0 - x_1}$ in $\ell_0$ has denominator $0$, so the formula is not defined.