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i found a equation that holds for any natural number of n and any $x_i \ne x_j$ as follows:

$\sum\limits_{i = 1}^{n } {\prod\limits_{\substack{j = 1\\j \ne i}}^{n } {\frac{{x_i }}{{x_i - x_j }}} } = 1$

when n=2, it is given by

$\frac{x_1}{x_1-x_2}+\frac{x_2}{x_2-x_1}=\frac{x_1 - x_2}{x_1 - x_2} = 1$

when n=3, it is given by

$\frac{x_1^2}{(x_1-x_2)(x_1-x_3)}+\frac{x_2^2}{(x_2-x_1)(x_2-x_3)}+\frac{x_3^2}{(x_3-x_1)(x_3-x_2)}=1$

But, how can I prove for general $n$?

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    Thanks for your comments. Acctually i cound't dare to try the n=4 case. Even i tried the n=3 case with the help of mathematica. :)2012-11-20

2 Answers 2

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Consider the Lagrange interpolation of the polynomial $f(x)=x^{n-1}$ with interpolation points $x_1,\ldots,x_n$. We have: $x^{n-1}=\sum_{i=1}^{n}x_i^{n-1}\prod_{j\neq i}\frac{x-x_j}{x_i-x_j},$ so, by comparing the leading coefficients of RHS and LHS, the result follows.

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    No$i$don't. beacause less or equal to $n$th polynomial is unique... so, $a_k = (\sum \prod ~~ )x^{k}$ for k are zero from the origin....2012-11-23
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For $n>1$, the partial fractional decomposition of $ \frac1z\prod_{j=1}^{n-1}\frac{z}{z-x_j} =\frac{z^{n-2}}{\prod\limits_{j=1}^{n-1}(z-x_j)} =\sum_{i=1}^{n-1}\frac{A_i}{z-x_i}\tag{1} $ using the Heaviside method yields $ A_i=\frac{x_i^{n-2}}{\prod\limits_{\substack{j=1\\j\ne i}}^{n-1}(x_i-x_j)} =\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\tag{2} $ Combine $(1)$ and $(2)$: $ \begin{align} \prod_{j=1}^{n-1}\frac{z}{z-x_j} &=\sum_{i=1}^{n-1}A_i\frac{z}{z-x_i}\\ &=\sum_{i=1}^{n-1}\frac{z}{z-x_i}\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\\ &=\sum_{i=1}^{n-1}\left(1-\frac{x_i}{x_i-z}\right)\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\tag{3} \end{align} $ Set $z=x_n$ in $(3)$: $ \prod_{j=1}^{n-1}\frac{x_n}{x_n-x_j} =\sum_{i=1}^{n-1}\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j} -\sum_{i=1}^{n-1}\prod_{\substack{j=1\\j\ne i}}^n\frac{x_i}{x_i-x_j}\tag{4} $ Add the second term of the right side of $(4)$ to both sides: $ \sum_{i=1}^n\prod_{\substack{j=1\\j\ne i}}^n\frac{x_i}{x_i-x_j} =\sum_{i=1}^{n-1}\prod_{\substack{j=1\\j\ne i}}^{n-1}\frac{x_i}{x_i-x_j}\tag{5} $ Noting that the case $n=1$ follows vacuously, using $(5)$ and induction proves that $ \sum_{i=1}^n\prod_{\substack{j=1\\j\ne i}}^n\frac{x_i}{x_i-x_j}=1\tag{6} $

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    Thank you for your answer. It's relatively straitforward to follow up.2012-11-23