$x_0, x_1, ... x_n$ are distinct points and $A(x)=\prod_{j=0}^n{(x-x_j)}$, $A_k(x)=\prod_{j=0, j\neq k}^n{(x-x_j)}$, $L_k(x)=\frac{A_k(x)}{A_k(x_k)}$. Prove each of the following:
a) $\sum_{k=0}^{n}{L_k(x)}$=1 and $\sum_{k=0}^{n}{\frac{A_k^\prime(x)}{A^\prime(x_k)}}$=0 for all x
b)$\sum_{k=0}^{n}{\frac{1}{A^\prime(x_k)}}$=0. [Hint:Use part a with suitable values of x]
Have solved part (a) - need help with part (b).
Note first that $A(x)=A_k(x)(x-x_k)$, so differentiating both sides and putting $x=x_k$ we get
$A'(x_k)=A_k(x_k)$.
Thus $\displaystyle\sum\limits_{k=0}^{n}{\dfrac{1}{A'(x_k)}} = \displaystyle\sum\limits_{k=0}^{n}{\dfrac{1}{A_k(x_k)}}$ is to be found.
Now $\displaystyle\sum\limits_{k=0}^{n}{L_k(x)}= \displaystyle\sum\limits_{k=0}^{n}{\dfrac{A_k(x)}{A_k(x_k)}}$
Since $\displaystyle\sum\limits_{k=0}^{n}{L_k(x)}=1$ as a polynomial in $x$, we can look at the coefficient of $x^n$ in this polynomial and equate it to $0$.
Each $A_k(x)=\displaystyle \prod\limits_{\substack{j=0\\ j\neq k}}^{n}{(x-x_j)}$ is a monic polynomial of degree $n$ so coefficient of $x^n$ in $A_k(x)$ is $1$.
Thus coefficient of $x^n$ in $\displaystyle\sum\limits_{k=0}^{n}{L_k(x)}$ is $\displaystyle\sum\limits_{k=0}^{n}{\dfrac{1}{A_k(x_k)}}$ and is thus equal to $0$.
