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If $\lambda_1,\dots,\lambda_n$ are distinct positive real numbers, then $$\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1.$$ This identity follows from a probability calculation that you can find at the top of page 311 in the 10th edition of Introduction to Probability Models by Sheldon Ross.

Is there a slick or obvious explanation for this identity?

This question is sort of similar to my previous problem; clearly algebra is not my strong suit!

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up vote 25 down vote accepted

It's the Lagrange interpolation polynomial for the constant function $1$ evaluated at $0$: $$\sum_{i=1}^n 1 \prod_{j\neq i} {{\lambda_j-0}\over \lambda_j-\lambda_i}=1$$

In general, you have that the polynomial below interpolates the data points $(\lambda_i,y_i$): $$\sum_{i=1}^n y_i \prod_{j\neq i} {{\lambda_j-x}\over \lambda_j-\lambda_i}$$

See http://en.wikipedia.org/wiki/Lagrange_polynomial

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+1: You could add some more info so that this can be accepted as an answer. –  user17762 Mar 2 '11 at 2:57
    
@Sivaram: done, thanks for the nudge. –  lhf Mar 2 '11 at 3:05
    
Ah, i knew this, forgot the name. I studied this in numerical analysis. –  anonymous Mar 2 '11 at 3:43
    
Thanks. This is exactly what I was looking for. –  Byron Schmuland Mar 2 '11 at 4:31
7  
I guess to rephrase: if $P(x) = \sum \prod (x-\lambda_j)/(\lambda_i - \lambda_j)$, then we can easily see that $P(\lambda_i) = 1$. Since $P(x)$ is a polynomial of degree $n-1$, it is identically $1$. What we have here is $P(0)$. –  Aryabhata Mar 2 '11 at 4:43
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