# Does this expression always vanish?

I have checked that the following expression

\begin{align} \sum_{i=1}^N\sum_{j=1\\ j\ne i}^N\frac{A_iA_j(A_i+A_j)}{(A_i-A_j)^3}\prod_{k=1\\ k\ne i\\ k\ne j}^N\frac{A_i A_k}{(A_i-A_k)^2} \end{align} is zero for $$N$$ up to 7. I would like a general proof that it is zero for $$N\in \mathbb Z_{+}$$. I tried induction but I didn't find any inductive structure that I could make use of.

This was previously asked on math stack exchange ( https://math.stackexchange.com/questions/4770113/prove-that-the-expression-is-identically-zero ) and since there was no answer, it was suggested that I post the question here.

• On MathOverflow, question titles that are commands (like "Prove that...") give the impression that the original source is a homework or textbook question. It's best to phrase the title as an actual question, not something that gives the impression you know the answer and you're challenging users here. Sep 23 at 4:54
• as a simple test, if I set $A_i=i$ for all $i$, the expression can be evaluated in closed form and gives 0. Sep 23 at 6:41
• Other observations: the expression can be written as $$(A_1\cdots A_n) \sum_{1\le i<j\le n} \frac{A_i+A_j}{(A_i-A_j)^3} \biggl( A_i^{n-2} \prod_{\substack{1\le k\le N\\k\ne i\\k\ne j}}\frac1{(A_i-A_k)^2} - A_j^{n-2} \prod_{\substack{1\le k\le N\\k\ne i\\k\ne j}}\frac1{(A_j-A_k)^2} \biggr).$$This makes it easy to see that the order of the pole at any $A_i=A_j$ is at most $2$ (multiplying through by $(A_i-A_j)^3$ and setting $A_i=A_j$ yields $0$). Sep 23 at 6:46
• It seems equivalent to \begin{align} \sum_{i=1}^N\sum_{j=1\\ j\ne i}^N\frac{A_i+A_j}{A_i-A_j}\prod_{k=1\\ k\ne i}^N\frac{A_i A_k}{(A_i-A_k)^2}=0. \end{align} Sep 23 at 6:59
• There should be some symmetric function identity hiding here. Sep 23 at 8:46

Recall that $$\sum_iA_i^{n-1}\prod_{j\neq i}\frac1{A_i-A_j}=1,$$ as follows from the Lagrange interpolation of $$x^{n-1}$$.
Now apply $$\prod_i \frac \partial{\partial A_i}$$. We get $$\sum_i\left((n-1)A_i^{n-2}-2A_i^{n-1}\sum_{j\neq i}\frac1{A_i-A_j}\right) \prod_{j\neq i}\frac1{(A_i-A_j)^2}\\ = - \sum_i A_i^{n-2}\left(\sum_{j\neq i}\frac{A_i+A_j}{A_i-A_j}\right) \prod_{j\neq i}\frac1{(A_i-A_j)^2}.$$ Multiplying by $$-\prod_iA_i$$, we get what we need.
• @IlyaBogdanov: Very nice answer and instructive approach. (+1) May I ask how you arrived at the expression with factor $n-1$? Is there a known formula, or is the expression tool based? Sep 26 at 19:39
• @epi163sqrt When dealing with the $i$th summand, you apply all $\partial/\partial x_j$ with $j\neq i$ first, that’s easy. Then it remains to apply $\partial/\partial x_i$, and, perhaps, it would be more clear if I wrote the result using the logarithmic derivative, as $\displaystyle \left(\frac{n-1}{A_i}-2\sum_{j\neq i}\frac1{A_i-A_j}\right) A_i^{n-1} \prod_{j\neq i}\frac1{(A_i-A_j)^2}$. Sep 27 at 8:43