$\sum_{i=0}^{n}{\prod_{j=0\\j\not=i}^{n}\frac{x-x_j}{x_i-x_j}}=1$

Question

can you help me to show this identy: $$\sum_{i=0}^{n}{\prod_{j=0\\j\not=i}^{n}\frac{x-x_j}{x_i-x_j}}=1,$$ I try to expand it but I only have that is equal to $$\frac{n\prod_{j=0}^{n}(x-x_j)}{\sum_{i=0}^{n}{(x-x_i)\prod_{j=0\\j\not=i}^{n}(x_i-x_j)}}$$

Answer 1

The product $\displaystyle\prod_{j=0,\ j\neq i}^n\frac{x-x_j}{x_i-x_j}$ is equal to $0$ if $j\neq i$ (because $x_j-x_j=0$) and it is equal to $1$ otherwise. So$$\sum_{j=0}^n\prod_{j=0,\ j\neq i}^n\frac{x-x_j}{x_i-x_j}=1,$$since it is the sum of $n$ $0$'s and one $1$.

Answer 2

We assume that all $x_i$ are distinct, otherwise the product has ill-defined denominators. Let

$$p(x) = \sum_{i=0}^{n}{\prod_{\substack{j=0\\j\not=i}}^{n}\frac{x-x_j}{x_i-x_j}}.$$

Each term in the product is an $n$ degree polynomial in $x$, so $p$ is the sum of $n$ degree polynomials and hence of degree $\leqslant n$.

Now, notice that for each $i=0, \dots, n$ we have

$$p(x_i) = \prod_{\substack{j=0\\j\not=i}}^{n}\frac{x_i-x_j}{x_i-x_j} = 1,$$

because all other terms in the outer sum involve a product in which one of the factors has a vanishing numerator $(x_i - x_i)$.

Then $p$ is a polynomial with degree $\leqslant n$ that assumes the value $1$ at least $n+1$ times, and hence must be constant and equal to $1$ everywhere.