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In Problems from the book by Andreescu, there's the following problem :

Let $x_0,\ldots,x_n$ be distinct complex numbers.

  1. Prove $\displaystyle \sum_{k=0}^n\dfrac{x_k^{n+1}}{\prod_{j\neq k}(x_k-x_j)}=\sum_{k=0}^nx_k$

  2. Evaluate $\displaystyle \sum_{k=0}^n\dfrac{x_k^{n+2}}{\prod_{j\neq k}(x_k-x_j)}$

My try:

  1. Set $\Pi= \prod_{i=0}^nx_i$ and $f:x\to \dfrac{(-1)^n}{\Pi}x^{n+2}$

    • Setting $P$ as the Lagrange polynomial for the function $f$ and points $x_0,\ldots,x_n$ yields $$P(X)=\sum_{k=0}^n \dfrac{(-1)^n}{\Pi}x_k^{n+2} \prod_{j\neq k }\frac{X-x_j}{x_k-x_j}$$ and $$P(0)= \sum_{k=0}^n\dfrac{x_k^{n+1}}{\prod_{j\neq k}(x_k-x_j)} $$

It remains to prove that $P(0)=\sum_{k=0}^nx_k$, which would be true if hypothetically $P=\sum_{0}^nx_kX^k$

I don't know if indeed $P=\sum_{0}^nx_kX^k$ ...

  • Besides, noting that $f$ is a polynomial with degree $n+2$, which matches with $P$ at $n+1$ points, there exists some $z$ such that $$f(X)-P(X)= \dfrac{(-1)^n}{\Pi}(X-x_0)\ldots(X-x_n)(X-z)$$

Hence $-P(0)=f(0)-P(0)=z$.

How to prove that $-z=\sum_{k=0}^nx_k$ ?

EDIT: Surprisingly (at least for me), this sum amounts to this one

$2.$ I haven't thought much about it, but I don't see any trivial connection with $1.$

Community
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Gabriel Romon
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1 Answers1

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Put $\displaystyle R(x)=\frac{x^{n+1}}{\prod_{k=0}^n (x-x_k)}$.

We have: $$R(x)=1+\sum_{k=0}^n \frac{b_k}{x-x_k}$$ with $$b_k=\lim_{x\to x_k}(x-x_k)R(x)=\frac{x_k^{n+1}}{\prod_{j\not = k}(x_k-x_j)}$$

Hence $$R(x)-1=\frac{x^{n+1}-\prod_{k=0}^n (x-x_k)}{\prod_{k=0}^n (x-x_k)}=\frac{x^{n}(\sum_{k=0}^n x_k)+...}{\prod_{k=0}^n (x-x_k)}=\sum_{k=0}^n \frac{b_k}{x-x_k}$$

Now you multiply this equality by $x$, and let $x\to +\infty$ and you are done.

Same method for the second point.

Kelenner
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