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I am trying to prove the following expression

$$\sum_{n=1}^{N}\prod_{m=1, m \neq n}^{N} \frac{1}{\frac{1}{x_m}-\frac{1}{x_n}}=0. $$ (1)

I have used an inductive approach to prove (1). In particular, when N= 2 or N=3, it is readily that (1) holds. However, when assuming that (1) holds for N = M, it becomes complicated to evaluate and confirm that (1) holds for N = M+1.

Does anyone have suggestions/advices for proving (1)?

Thank you.

PLe
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  • Hmm.... I might try to switch the product and summation. You might be able to do this directly (try some smaller examples and check) or you could turn the product into a sum by taking logs – Brevan Ellefsen May 10 '17 at 04:35
  • Thanks for your suggestions :) – PLe May 16 '17 at 16:19

2 Answers2

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This is $$\sum_{n=1}^N\prod_{m:m\ne n}\frac1{a_m-a_n}$$ where $a_m=1/x_m$. By Lagrange interpolation, $$1=\sum_{n=1}^N\prod_{m:m\ne n}\frac{x-a_n}{a_m-a_n}.$$ Compare coefficients of $x^{N-1}$ (as long as $N\ge2$).

ADDED IN EDIT

Lagrange interpolation finds the unique polynomial of degree at most $N-1$ with $f(a_1)=b_1,\ldots,f(a_N)=b_n$ with $a_1,\ldots,a_N$ distinct. It is $$\sum_{n=1}^N\prod_{m:m\ne n}b_n\frac{x-a_n}{a_m-a_n}.$$ The proof is straightforward: insert $a_n$ for $x$ and all but the $n$-th term vanishes, and that term reduces to $b_n$.

Angina Seng
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  • Thank you for your hint. I am trying to connect your guide to the solution of the problem. – PLe May 16 '17 at 16:30
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Let $z_n=\frac{1}{x_n}$ be arbitrary (different) complex numbers. Your sum is (up to the sign, $(-1)^{N-1}$) $$\sum^N_{n=1}\frac{1}{\omega'(z_n)},$$ where $\omega(z)=(z-z_1)\ldots(z-z_N)$. But that's the sum of the residues of $\frac{1}{\omega(z)}$, i.e. it's equal to $$\frac{1}{2\pi i}\oint_{|z|=R}\frac{dz}{\omega(z)},$$ if $R$ is large enough for the circle to contain all $z_n$. But the integral is of order $O(R^{-N+1})$ for large $R$, so the sum must be zero.