Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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

share|cite|improve this question

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.

share|cite|improve this answer
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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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