3
$\begingroup$

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 $1.$

$\endgroup$
  • $\begingroup$ here is the answer to the first question given by....yourself :) $\endgroup$ – Petite Etincelle Aug 8 '14 at 11:38
  • $\begingroup$ @LiuGang Yes, you're right. That is very odd, it didn't occur to me at anytime that both questions were related... Do you know how to deal with the second one ? (I think this is an overkill) $\endgroup$ – Gabriel Romon Aug 8 '14 at 12:32
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Yes, a standard method, I should have done this. $\endgroup$ – Gabriel Romon Aug 8 '14 at 16:32
  • $\begingroup$ Now that I look back it, it's a very tricky proof... Well done. $\endgroup$ – Gabriel Romon Aug 8 '14 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.