Past exam of the numerical analysis course of the University of Oxford / 2023 / 2 / b

The monomial $x^k$ lies in ${\Pi }_n$ and already fits the data. Uniqueness of polynomial interpolation therefore givesApply this result with a nonnegative integer $m\le n$. Its Lagrange interpolation representation is the polynomial identityThe numerator in each summand is a monic polynomial of degree $n$, so the coefficient of $x^n$ on the right is exactly ${\sum }_i{x}_i^m/{\prod }_{j\ne i}(x_i-x_j)$. The coefficient on the left is zero for $m<n$ and one for $m=n$. Comparing coefficients proves the Lagrange leading-coefficient identityThe intended exponent range is $m\ge 0$: arbitrary negative exponents would not give this polynomial identity. For $m=0$, $x_i^0$ denotes the value of the constant polynomial $1$, including at a zero interpolation node.