Show that $\sum_{k=0}^n x_k^r l_k(x) = x^r$

Question

Let $n \in \Bbb N, x_k \in\Bbb R$ pair wise different grid points for $k = 0, \ ..., \ n$, and define the Lagrange polynomials

$l_k(x) = \prod_{j=0, j \neq k}^n {{x - x_j} \over {x_k - x_j}}$.

For $r = 0, \ ..., \ n$, Show that

$\sum_{k=0}^n x_k^r l_k(x) = x^r$.

How would I prove this statement? I know a proof for $\sum_{k=0}^n l_k(x) = 1$, but I don't know whether this one is similar to this one right here. It confuses me that, in the end, we have $x^r$, while we actually sum up different expressions. Does anyone have a hint for me?

Answer 1

The whole point of the definition of the Lagrange polynomials is that $\sum_{k=0}^n a_k l_k(x_j)=a_j$. So by taking $a_k=x_k^r$ you obtain a polynomial of degree at most $n$ with $p(x_j)=x_j^r$ for $j=0,1,\dots,n$. One such polynomial is clearly $x^r$ (note that the problem assumes $r \leq n$). Can there be another one? If not, then the Lagrange polynomial is just $x^r$ itself.