# Double summation problem

I'm reading a research paper, and I'm nott sure how certain identities are derived. I've listed them below:

$$\sum_{i=1}^n\sum_{i\neq j}^{n}\frac{1}{z_i-z_j}=0\\$$ $$\sum_{i=1}^n\sum_{i\neq j}^{n}\frac{z_i}{z_i-z_j}=\frac{1}{2}n(n-1)$$ $$\sum_{i=1}^n\sum_{i\neq j}^{n}\frac{z_i^2}{z_i-z_j}=(n-1)\sum_{i=1}^n z_i$$ $$\sum_{i=1}^n\sum_{i\neq j}^{n}\frac{z_i^3}{z_i-z_j}=(n-1)\sum_{i=1}^n z_i^2 +\sum_{i<j}^n z_i z_j$$

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First one, by symmetry (flip $i, j,$ which sends the sum to minus itself. Second one, by symmetry, noting that $\frac{z_i}{z_i - z_j} = 1 + \frac{z_j}{z_i - z_j}.$ Third an fourth, using the previous ones and long division.

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Third and forth are also by symmetry. – Calvin Lin Jan 6 '14 at 1:35
@Calvin Lin -- long division then symmetry (the same idea as before). – Igor Rivin Jan 6 '14 at 1:39
It wasn't obvious to me what you meant by "using the previous ones", since it's not immediate how to convert the square/cube term into the linear/constant version. – Calvin Lin Jan 6 '14 at 1:40
The square thing reduces to (basically) sum of $z_i z_j/(z_i - z_j),$ so an identical argument to that in (1) works. – Igor Rivin Jan 6 '14 at 1:42
I'm not certain how $\frac{z_i z_j} { z_i - z_j}$ comes in. We must be thinking of something different. – Calvin Lin Jan 6 '14 at 1:44

Try and change the order of summation over the single term $\sum_{i < j }$. This is the "symmetry" that is reference by Igor.

Do you see why the first summation is equal to

$$\sum_{i< j } \frac{1-1} { x_i - x_j } = 0 ?$$

Do you see why the second summation is equal to

$$\sum_{i<j} \frac{x_i - x_j} { x_i - x_j} = \sum_{i<j} 1 = { n \choose 2 } ?$$

Do you see why the third summation is equal to

$$\sum_{i<j} \frac{x_i^2 - x_j^2 } { x_i - x_j} = \sum_{i<j} (x_i + x_j) = (n-1) \sum x_i?$$

Do you see why the fourth summation is equal to

$$\sum_{i<j} \frac{x_i^3 - x_j^3} { x_i - x_j} = \sum_{i<j} (x_i^2 + x_ix_j+ x_j^2) ?$$

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I'm sorry but it is not clear to me why you used $\sum_{i<j}$. Could you please elaborate? – Millardo Peacecraft Jan 6 '14 at 1:55
@MillardoPeacecraft It is a "change of variables" that converts the summation, and makes it much easier to deal with. – Calvin Lin Jan 6 '14 at 5:25
Also, where did the $\sum_{i=1}^n go?$ – Millardo Peacecraft Jan 6 '14 at 17:04