Prove that the expression is identically zero.

Question

I would like a proof 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 all $N\in \mathbb Z_{+}$. I have checked that it is zero for $N$ up to 7.

I obtained this expression at the next-to-leading order while expanding the Nekrasov partition function of an $\textsf{N}=2$ theory in the $\Omega$-background.

(Now crossposted at MathOverflow)

Answer 1

@IlyaBogdanov's answer to the problem posted at MathOverflow is as instructive as it is elegant. I have gone through his answer in detail to better understand what is going on. Here are some details that might also be of interest to other readers.

The claim is \begin{align*} \color{blue}{\sum_{i=1}^N\sum_{{j=1}\atop{j\ne i}}^N \frac{A_iA_j(A_i+A_j)}{(A_i-A_j)^3} \prod_{{k=1}\atop{{k\ne i}\atop{k\ne j}}}^N\frac{A_i A_k}{(A_i-A_k)^2}=0}\tag{1} \end{align*}

The first step is a small simplification that makes the structure of the expression a little clearer.

We obtain \begin{align*} \sum_{i=1}^N&\sum_{{j=1}\atop{j\ne i}}^N \frac{A_iA_j(A_i+A_j)}{(A_i-A_j)^3} \prod_{{k=1}\atop{{k\ne i}\atop{k\ne j}}}^N\frac{A_i A_k}{(A_i-A_k)^2}\\ &=\sum_{1\leq i\ne j\leq N}\frac{A_i+A_j}{A_i-A_j}\prod_{{k=1}\atop{k\ne i}}^N\frac{A_iA_k}{\left(A_i-A_k\right)^2}\\ &=\left(\prod_{k=1}^NA_k\right)\sum_{1\leq i\ne j\leq N}\frac{A_i+A_j}{A_i-A_j}\color{blue}{A_i^{N-2} \prod_{{k=1}\atop{k\ne i}}^N\frac{1}{\left(A_i-A_k\right)^2}}\tag{2} \end{align*}

A key step is to realise that the blue marked part in (2) is very close to a certain Lagrange polynomial. We consider the Lagrange polynomial of $x^{N-1}$ with nodes $A_1,\ldots,A_N$: \begin{align*} L_N(x)&=\sum_{i=1}^Ny_i\prod_{{k=1}\atop{k\ne i}}^N\frac{x-x_k}{x_i-x_k}\\ x^{N-1}&=\sum_{i=1}^NA_i^{N-1}\prod_{{k=1}\atop{k\ne i}}^N\frac{x-A_k}{A_i-A_k}\tag{3} \end{align*} Comparing the coefficient of $x^{N-1}$ in (3) we find the identity \begin{align*} \color{blue}{\sum_{i=1}^NA_i^{N-1}\prod_{{k=1}\atop{k\ne i}}^N\frac{1}{A_i-A_k}=1}\tag{4} \end{align*} Note the identity (4) looks very close to the blue marked part in (2). In fact, and this is the second important ingredient, we could apply a derivation to go from (4) to the blue marked part of (2). It turns out that the differential operator \begin{align*} \prod_{k=1}^N\frac{\partial}{\partial A_k} \end{align*} does the job. An important aspect here is also to apply the operators $\frac{\partial}{\partial A_k}$ in the right order to avoid getting complicated expressions. Here we apply them to (4) in the following order \begin{align*} \frac{\partial}{\partial A_i}\prod_{{k=1}\atop{k\ne i}}^N\frac{\partial}{\partial A_k}\tag{5} \end{align*} which enables us to use the $N-1$ factors $\prod_{{k=1}\atop{k\ne i}}^N\frac{\partial}{\partial A_k}$ in a single step as we will see soon.

We apply the differential operator (5) to (4) and obtain \begin{align*} \color{blue}{0}&=\left(\prod_{k=1}^{N}\frac{\partial}{\partial A_k}\right) \left(\sum_{i=1}^NA_i^{N-1}\prod_{{j=1}\atop{j\ne i}}^N\frac{1}{A_i-A_j}\right)\\ &=\sum_{i=1}^N\frac{\partial}{\partial A_i} \left(A_i^{N-1}\left(\left(\prod_{{k=1}\atop{k\ne i}}^N\frac{\partial}{\partial A_k}\right) \prod_{{j=1}\atop{j\ne i}}^N\frac{1}{A_i-A_j}\right)\right)\tag{6}\\ &=\sum_{i=1}^N\frac{\partial}{\partial A_i} \left(A_i^{N-1}\prod_{{j=1}\atop{j\ne i}}^N\frac{1}{\left(A_i-A_j\right)^2}\right)\tag{7}\\ &=\sum_{i=1}^N\left((N-1)A_i^{N-2}\prod_{{j=1}\atop{j\ne i}}^N\frac{1}{\left(A_i-A_j\right)^2}\right.\\ &\qquad\qquad\quad\left.+A_i^{N-1}\frac{\partial}{\partial A_i} \left(\prod_{{j=1}\atop{j\ne i}}^N\frac{1}{\left(A_i-A_j\right)^2}\right)\right)\tag{8}\\ &=\sum_{i=1}^N\left((N-1)A_i^{N-2}\prod_{{j=1}\atop{j\ne i}}^N\frac{1}{\left(A_i-A_j\right)^2}\right.\\ &\qquad\qquad\left.+A_i^{N-1}\sum_{{k=1}\atop{k\ne i}}^N\left(\frac{\partial}{\partial A_i}\,\frac{1}{\left(A_i-A_k\right)^2}\right) \prod_{{j=1}\atop{{j\ne i}\atop{j\ne k}}}^N\frac{1}{\left(A_i-A_j\right)^2}\right)\tag{9}\\ &=\sum_{i=1}^N\left((N-1)A_i^{N-2}-2A_i^{N-1} \sum_{{k=1}\atop{k\ne i}}^N\frac{1}{A_i-A_k}\right) \prod_{{j=1}\atop{j\ne i}}^N\frac{1}{\left(A_i-A_j\right)^2}\tag{10}\\ &=\sum_{i=1}^N\left(\sum_{{k=1}\atop{k\ne i}}^N\frac{A_i-A_k}{A_i-A_k}A_i^{N-2}-2A_i^{N-1} \sum_{{k=1}\atop{k\ne i}}^N\frac{1}{A_i-A_k}\right) \prod_{{j=1}\atop{j\ne i}}^N\frac{1}{\left(A_i-A_j\right)^2}\tag{11}\\ &\,\,\color{blue}{=-\sum_{i=1}^N\sum_{{k=1}\atop{k\ne i}}^N\frac{A_i+A_k}{A_i-A_k}A^{N-2} \prod_{{j=1}\atop{j\ne i}}^N\frac{1}{\left(A_i-A_j\right)^2}}\tag{12}\\ \end{align*} and finally multiplying (12) by $\left(-\prod_{k=1}^NA_k\right)$ so that (2) and so also the claim (1) follows.

Comment:

• In (6) we order the differential operators as stated in (5) and use the linearity of the operators.

• In (7) we apply the product of the $N-1$ operators in one step, noting that each of the operators modifies exactly one factor and takes the other factors as constant.

• In (8) we use the product rule of differentiation: $\left(fg\right)^{\prime}=f^{\prime}g+fg^{\prime}$.

• In (9) we use the generalized product rule of differentiation: \begin{align*} \left(f_1\cdots f_N\right)^{\prime}=\sum_{i=1}^Nf_i^{\prime}\prod_{{j=1}\atop{j\ne i}}^Nf_k \end{align*}

• In (10) we differentiate \begin{align*} \frac{\partial}{\partial A_i}\frac{1}{\left(A_i-A_k\right)^2}=\frac{-2}{\left(A_i-A_k\right)^3} \end{align*} and factor out the product after small rearrangements.

• In (11) we use the identity: \begin{align*} N-1=\sum_{{k=1}\atop{k\ne i}}^N\frac{A_i-A_k}{A_i-A_k} \end{align*}

• In (12) we make some final simplifications.