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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.

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    $\begingroup$ 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. $\endgroup$
    – David Roberts ♦
    Sep 23 at 4:54
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    $\begingroup$ as a simple test, if I set $A_i=i$ for all $i$, the expression can be evaluated in closed form and gives 0. $\endgroup$
    – Carlo Beenakker
    Sep 23 at 6:41
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    $\begingroup$ 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 $0$). $\endgroup$
    – Greg Martin
    Sep 23 at 6:46
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    $\begingroup$ 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} $\endgroup$
    – T. Amdeberhan
    Sep 23 at 6:59
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    $\begingroup$ There should be some symmetric function identity hiding here. $\endgroup$
    – Per Alexandersson
    Sep 23 at 8:46

1 Answer 1

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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.

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    $\begingroup$ Thanks a lot for the answer. $\endgroup$
    – Silly Point
    Sep 24 at 3:17
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    $\begingroup$ @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? $\endgroup$
    – Markus Scheuer
    Sep 26 at 19:39
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    $\begingroup$ @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 $\displaystyle \left(\frac{n-1}{A_i}-2\sum_{j\neq i}\frac1{A_i-A_j}\right) A_i^{n-1} \prod_{j\neq i}\frac1{(A_i-A_j)^2}$. $\endgroup$
    – Ilya Bogdanov
    Sep 27 at 8:43
  • $\begingroup$ @IlyaBogdanov: I see. Many thanks for your reply and thank you very much for this elegant answer. I have taken the liberty of writing a slightly more detailed version of your response to MSE. $\endgroup$
    – Markus Scheuer
    Sep 30 at 6:51

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