Prove that for all $x$, $\sum_{j=0}^{n}ℓ_j(x)=1$

Question

Prove that for all $x$, $\sum_{j=0}^{n}ℓ_j(x)=1$

$ℓ_j(x)$ are obtained by means of Lagrange interpolation and are defined as $ℓ_j(x)=\prod_{i=0, i\neq j}^{n}\frac{x-x_i}{x_j-x_i}$ where $x_i, i=0,1,...,n$ are the nodes given to interpolate, with which I have to prove that $\sum_{j=0}^{n}\prod_{i=0, i\neq j}^{n}\frac{x-x_i}{x_j-x_i}=1$ but I do not know how to do this, could someone help me please? Thank you very much.

Answer 1

Consider the polynomial $$ Q(x) \stackrel{\rm def}{=} \sum_{j=0}^n \ell_j(x) -1\,. $$ which is a polynomial of degree at most $n$ as every $\ell_j$ is.

It is easy to check that, for any $j$ and $i$, $$\ell_j(x_i) = \begin{cases} 0&\text{ if } i\neq j\\ 1&\text{ if } i=j \end{cases}$$ and therefore, we have, for every $0\leq i\leq n$, $$ Q(x_i) = \sum_{j=0}^n \ell_j(x_i) - 1 = \ell_i(x_i) - 1 = 1-1=0 $$ and since $Q$ is a polynomial of degree at most $n$ with $n+1$ roots, it has to be identically zero.

Answer 2

$ \sum_{j=0}^{n}\ell_j(x)= \sum_{j=0}^{n} 1 \ell_j(x)$ is the expression of Lagrange interpolation for the constant function $1$.