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Let $n \in \mathbb{N}$ and $a_0, a_1, \dots, a_n$ be positive integers such that $a_i\neq a_j$ for $i \neq j$. Prove that

$$\sum_{0\leq k \leq n} \prod_{\substack{0\leq i \leq n\\i\neq k}} \frac{1}{a_i-a_k} = 0$$


The original problem was to evaluate the determinant of the matrix below: $$ A =\, \begin{bmatrix} a_0!&(a_0+1)!&\cdots&(a_0+n)!\\ a_1!&(a_1+1)!&&\vdots\\ \vdots&&\ddots&\vdots\\ a_n!&\cdots&\cdots&(a_n+n)! \end{bmatrix} $$ and the determinant should be equal to $$ \prod_{0\leq i \leq n}a_i! \prod_{0\leq i<j\leq n}(a_j-a_i)\,. $$ To prove this, I tried cofactor expansion and mathematical induction respect to $n$; \begin{align*} \mathrm{det}\,A &= \sum_{0\leq k \leq n} (-1)^k a_k! \prod_{\substack{0\leq i \leq n\\i\neq k}}(a_i+1)!\prod_{\substack{0\leq i<j\leq n\\i,j\neq k}}(a_j-a_i)\\ &= \prod_{0\leq i \leq n}a_i! \sum_{0\leq k \leq n} (-1)^k \prod_{\substack{0\leq i \leq n\\i\neq k}}(a_i+1)\prod_{\substack{0\leq i<j\leq n\\i,j\neq k}}(a_j-a_i)\\ &= \prod_{0\leq i \leq n}a_i! \prod_{0\leq i<j\leq n}(a_j-a_i) \sum_{0\leq k \leq n}\prod_{\substack{0\leq i \leq n\\i\neq k}}\frac{a_i+1}{a_i-a_k} \end{align*} and now it's done if the sum of the most right hand side is equal to $1$. I applied induction again to prove this; Let $b_i=a_i+1$ then the sum is \begin{align*} \sum_{0\leq k \leq n}\prod_{\substack{0\leq i \leq n\\i\neq k}}\frac{b_i}{b_i-b_k} &= \sum_{0\leq k < n}\prod_{\substack{0\leq i \leq n\\i\neq k}}\frac{b_i}{b_i-b_k} + \prod_{\substack{0\leq i \leq n\\i\neq n}}\frac{b_i}{b_i-b_n}\\ &= \sum_{0\leq k < n}\frac{b_n}{b_n-b_k}\prod_{\substack{0\leq i < n\\i\neq k}}\frac{b_i}{b_i-b_k} + \prod_{\substack{0\leq i \leq n\\i\neq n}}\frac{b_i}{b_i-b_n}\\ &= \sum_{0\leq k < n}\left(1+\frac{b_k}{b_n-b_k}\right)\prod_{\substack{0\leq i < n\\i\neq k}}\frac{b_i}{b_i-b_k} + \prod_{\substack{0\leq i \leq n\\i\neq n}}\frac{b_i}{b_i-b_n} \end{align*} from the induction hypothesis, \begin{align*} &= 1 + \sum_{0\leq k < n}\frac{b_k}{b_n-b_k}\prod_{\substack{0\leq i < n\\i\neq k}}\frac{b_i}{b_i-b_k} + \prod_{\substack{0\leq i \leq n\\i\neq n}}\frac{b_i}{b_i-b_n}\\ &= 1 + \prod_{0\leq i<n} b_i \left(\sum_{0\leq k < n}\frac{1}{b_n-b_k}\prod_{\substack{0\leq i < n\\i\neq k}}\frac{1}{b_i-b_k} + \prod_{\substack{0\leq i \leq n\\i\neq n}}\frac{1}{b_i-b_n}\right)\\ &= 1 + \prod_{0\leq i<n} b_i \left(\sum_{0\leq k < n}\prod_{\substack{0\leq i \leq n\\i\neq k}}\frac{1}{b_i-b_k} + \prod_{\substack{0\leq i \leq n\\i\neq n}}\frac{1}{b_i-b_n}\right)\\ &= 1 + \prod_{0\leq i<n} b_i\sum_{0\leq k\leq n}\prod_{\substack{0\leq i \leq n\\i\neq k}}\frac{1}{b_i-b_k}\,. \end{align*} It'll be finally proven if the sum of most right hand side is $0$, and this is identical to the left hand side of the very first problem, except for $a_i$ is now $b_i$.

Later, I managed to prove the original problem in the different way, so the first problem is also proven. However I'm looking for more 'direct' proof not through a such tricky route. Does anyone have idea for this? Thank you.

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  • $\begingroup$ Are you acquainted with Cauchy matrices? $\endgroup$ Jul 29, 2020 at 17:12
  • $\begingroup$ Indeed closely related (see the connection between my answer below and this one). $\endgroup$
    – metamorphy
    Aug 5, 2020 at 6:00

1 Answer 1

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In fact the same holds with the $a$'s being (distinct) elements of an abritrary field.

To prove it, we may assume $n>0$. Consider the partial fraction expansion $$\prod_{j=0}^n\frac{1}{x-a_j}=\sum_{i=0}^n\frac{A_i}{x-a_i},\quad A_i=\prod_{j\neq i}\frac{1}{a_i-a_j}$$ ($A_i$ are found by multiplying the identity by $x-a_i$, and letting $x=a_i$ then).

To get $\sum_i A_i$ out of it, an idea would be to multiply the identity by $x$ and take $x\to\infty$. But, in an abstract field, we can't do that. Instead, we put $x=1/z$ and divide the result by $z$: $$z^n\prod_{i=0}^n\frac{1}{1-a_i z}=\sum_{i=0}^n\frac{1}{1-a_i z}\prod_{j\neq i}\frac{1}{a_i-a_j}.$$ And it remains to put $z=0$.

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