Consider a system of nodes: $$-1\le x_1<x_2<\ldots <x_n\le 1$$ Denote: $$w(x)=(x-x_1)(x-x_2)\ldots (x-x_n)$$ $$l_i(x)=w(x)/((x-x_i)w'(x_i)), \ \ \ i=1,2,\ldots , n$$ $$A_i(x)=(1-(w''(x_i)/w'(x_i))(x-x_i))l_i^2(x), \ \ \ i=1,\ldots,n$$
Now, he gives the following two identities: $$\sum_{i=1}^n A_i = 1$$ $$\sum_i x_i^kA_i(x)+k\sum_i x_i^{k-1}(x-x_i)l_i^2(x)=x^k, \ \ \ k=1,2$$
And to my question: How to prove the two last identities?
Thanks!
Hint: $g(x)=(x-x_i)\ell_i(x)^2$ is the unique polynomial of degree $<2n$ for which we have $g(x_j)=g'(x_j)=0$ for all $j\neq i$, $g(x_i)=0$ and $g'(x_i)=1$. Similarly, $A_i$ is the unique polynomial of degree $<2n$ for which we have $A_i(x_j)=A_i'(x_j)=0$ for all $j\neq i$, $A_i(x_i)=1$, $A_i'(x_i)=0$.