Show that ${x_0^k\over \prod_{j\neq0} (x_0-x_j)}+{x_1^k\over \prod_{j\neq1} (x_1-x_j)}+...+{x_n^k\over \prod_{j\neq n} (x_n-x_j)}=0$

Question

Show that $${x_0^k\over \prod_{j\neq0} (x_0-x_j)}+{x_1^k\over \prod_{j\neq1} (x_1-x_j)}+...+{x_n^k\over \prod_{j\neq n} (x_n-x_j)}=0$$ For $k=0,1,...n-1$ and for $k=n$ $${x_0^n\over \prod_{j\neq0} (x_0-x_j)}+{x_1^n\over \prod_{j\neq1} (x_1-x_j)}+...+{x_n^n\over \prod_{j\neq n} (x_n-x_j)}=1$$ I know this identity $$x_0^kL_0(x)+x_1^kL_1(x)+...+x^k_nL_n(x)=x^k$$ Somehow I should be able to compare the coefficient of $x^n$ to the one above and get the result. Let $P_n(x)$ interpolates $f(x)$ at n+1 nodes. So I can prove that $$P_n(x)=f(x)$$ and the identity is shown.

Answer 1

Incomplete answer. There is this beatiful lemma, that any polynomial of degree $n$ with more than $n$ roots is identically zero.

Now, given that $L_j(x)=\prod\limits_{k=0,\\k\ne j}^n\frac{x-x_k}{x_j-x_k}$ is of degree $n$, then $$P(x)=x^k-\sum\limits_{j=0}^nx_j^kL_j(x)$$ is of degree at most $n$ for $k\in\{1,2,...,n\}$, but it has $n+1$ roots $\{x_0,x_1,...,x_{n}\}$. Thus $$x^k-\sum\limits_{j=0}^nx_j^kL_j(x)\equiv 0 \Rightarrow x^k=\sum\limits_{j=0}^nx_j^kL_j(x)$$ and $$\sum\limits_{j=0}^nx_j^kL_j(0)=0 \iff \sum\limits_{j=0}^nx_j^k\prod\limits_{k=0,\\k\ne j}^n\frac{0-x_k}{x_j-x_k}=0 \iff\\ (-1)^n\left(\prod\limits_{k=0}^n x_k\right)\left(\sum\limits_{j=0}^n\frac{x_j^{k-1}}{\prod\limits_{k=0,\\k\ne j}^n(x_j-x_k)}\right)=0$$ or, assuming none of $x_k=0$ $$\sum\limits_{j=0}^n\frac{x_j^{k-1}}{\prod\limits_{k=0,\\k\ne j}^n(x_j-x_k)}=0, k\in\{1,2,...,n\} \iff \\ \sum\limits_{j=0}^n\frac{x_j^{k}}{\prod\limits_{k=0,\\k\ne j}^n(x_j-x_k)}=0, k\in\{0,2,...,n-1\}$$

Answer 2

If $R>\max|x_j|$ is sufficiently large radius and $q(z)=\prod_{j=0}^n(z-x_j)$, then the denominators in the given formula are the derivative values $q'(x_j)$ and by the residue theorem of complex analysis $$ \frac1{2\pi i}\int_{|z|=R}\frac{z^k}{q(z)}dz=\sum_{j=0}^n\mathrm{Res}\left(\frac{z^k}{q(z)}, x_j\right)=\sum_{j=0}^n\frac{x_j^k}{q'(x_j)}. $$ On the other hand, for $R\to\infty$ the left side converges to (and is in consequence already equal to it for all $R>\max|x_j|$) $$ \frac1{2\pi i}\int_{|z|=1}\frac{R^{k+1}z^k}{R^{n+1}z^{n+1}+O(R^{n})}dz \to \lim_{R\to\infty}R^{k-n}\cdot\frac1{2\pi i}\int_{|z|=1}\frac{dz}{z^{n+1-k}} = \begin{cases}0,& k<n,\\1,&k=n.\end{cases} $$