Consider the Lagrange-interpolation-coefficients $\lambda_i$ ( $0 \leq i \leq n$) with nodes $x_0, x_1,…, x_n$.
Show: $$ \sum_{i=0}^n \lambda_i(x) = 1 $$ and
$$ \sum_{i=0}^n \lambda_i(0)x_i^{j} = 0 $$ for $j = 1,2,…,n$.
Remark: $\lambda_i(x)$ =$\prod_{j=0;i\neq j }^n \frac{x-x_j}{x_i-x_j}.$
My work :
So first claim was no big deal. Define $f(x) := \sum_{i=0}^n \lambda_i(x) $.
Then $f(x)$ has at most degree $n$, because the $\lambda_i$ has degree $n$. It holds: $f(x_i) = 1 $for $i=0,…,n$. Consider : $g(x) = f(x)-1$.
Then $g(x)$ has at most degree $n$ and $g(x_i) = 0$ for $i=0,…,n$. So $g(x)$ has $n+1$ roots. but $g(x)$ has at most degree $n$. We get that $g(x) = 0 $, which means $ \sum_{i=0}^n \lambda_i(x) = 1 $.
For the second claim we could consider the left side of the equation as interpolation-polynomial, that is evaluated at the point $0$. But this only an idea. Here I need your help.
Best Answer
The unique polynomial of degree $\leq n$ interpolates $(x_0,x_0^j),...,(x_n,x_n^j)$ is $f(x)=x^j$. Now consider Lagrange interpolation and put $x=0$.