Prove $\sum_{i=0}^n \prod_{j=0,\ i\neq j}^n \frac{x-x_j}{x_i-x_j}=1$

Question

Let $$ l_i(x)=\prod_{j=0,\ i\neq j}^n \frac{x-x_j}{x_i-x_j} $$ where $x_0,...x_n\in \mathbb{R}$ and $\forall_{i,j}\ i\neq j \implies x_i\neq x_j$

Show that: $$ \forall_{x\in \mathbb {R}}\ \sum_{i=0}^n l_i(x)=1 $$

Answer 1

It suffices to note that the polynomial $p(x) = \left(\sum_{i=0}^n l_i(x)\right) - 1$ has degree at most $n$, but the equation $p(x) = 0$ has at least $n+1$ solutions at $x = x_0,x_1,\dots,x_n$.

Answer 2

Actually to prove that you can use the fact that :

Given 2 polynoms $P,Q$ of degree $n$:

$$ P=Q $$ if and only if $$ \exists(x_1,...,x_{n+1})\in R^{n+1}, \forall i \in [1,n+1], P(x_i)=Q(x_i) $$

So here you have to show that the polynom $ \sum l_i $ is equal to the polynom constant to $1$.

For this, evaluate $\sum_i l_i$ in each $x_i$ (hopefully you have (n+1) values: the $x_i$ ... and conclude.