Talk:Sum of Elements in Inverse of Cauchy Matrix

Statement errors:

The statement of the theorem needs the assumption:


 * Distinct values $\set {x_1, \ldots, x_n, -y_1, \ldots, -y_n}$ required (Did Knuth mention this?).

There is a Querty keyboard proximity typo in the last row of matrix $C_n$: letter $m$ should be letter $n$.

Proposal: A proof using Vandermonde Matrix Identity for Cauchy Matrix was finished October 2019. However, the proof uses the statement below. I would like to keep Knuth's statement with fewest corrections (errors cited above). The matrix $C_{-}$ would appear only in the proof. The other results in the new statement would be the initial text for a Corollary, added after the Knuth theorem and its proof are stable. The Corollary would need no proof.--Gbgustafson (talk) 06:13, 3 November 2019 (EST)

Theorem for October 2019 Proof
Define Cauchy matrices of order $n$:


 * $C_+ = \begin{pmatrix}

\dfrac 1 {x_1 + y_1} & \dfrac 1 {x_1 + y_2 } & \cdots & \dfrac 1 {x_1 + y_n} \\ \dfrac 1 {x_2 + y_1} & \dfrac 1 {x_2 + y_2 } & \cdots & \dfrac 1 {x_2 + y_n} \\ \vdots & \vdots & \ddots & \vdots \\ \dfrac 1 {x_n + y_1} & \dfrac 1 {x_n + y_2 } & \cdots & \dfrac 1 {x_n + y_n} \\ \end{pmatrix}$ Distinct values $\set {x_1, \ldots, x_n, -y_1, \ldots, -y_n}$ required.


 * $C_- = \begin{pmatrix}

\dfrac 1 {x_1 - y_1} & \dfrac 1 {x_1 - y_2 } & \cdots & \dfrac 1 {x_1 - y_n} \\ \dfrac 1 {x_2 - y_1} & \dfrac 1 {x_2 - y_2 } & \cdots & \dfrac 1 {x_2 - y_n} \\ \vdots & \vdots & \ddots & \vdots \\ \dfrac 1 {x_n - y_1} & \dfrac 1 {x_n - y_2 } & \cdots & \dfrac 1 {x_n - y_n} \\ \end{pmatrix}$ Distinct values $\set {x_1, \ldots, x_n, y_1, \ldots, y_n}$ required.

Then:


 * The sum of elements in the inverse matrix $\paren { b_{ij} }$ of $C_+$ is:


 * $\ds \sum_{1 \mathop \le i, \ j \mathop \le n} b_{i j} = \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k$


 * The sum of elements in the inverse matrix $\paren { c_{ij} }$ of $C_-$ is:


 * $\ds \sum_{1 \mathop \le i, \ j \mathop \le n} c_{i j} = \sum_{k \mathop = 1}^n x_k - \sum_{k \mathop = 1}^n y_k$

Summations
Inverse of Cauchy Matrix applied respectively to $C_+$ and $C_-$ implies:


 * $\ds \sum_{ i \mathop = 1}^n \sum_{ j \mathop = 1}^n

\dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_j + y_k} \paren {x_k + y_i} } {\ds \paren {x_j + y_i} \paren { \prod_{k \mathop = 1, k \mathop \ne j }^n \paren {x_j - x_k} } \paren { \prod_{k \mathop = 1, k \mathop \ne i }^n \paren {y_i - y_k} } } = \sum_{k \mathop = 1}^n x_k + \sum_{k \mathop = 1}^n y_k$


 * $\ds \sum_{ i \mathop = 1}^n \sum_{ j \mathop = 1}^n

\dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_j - y_k} \paren {x_k - y_i} } {\ds \paren {x_j - y_i} \paren { \prod_{k \mathop = 1, k \mathop \ne j }^n \paren {x_j - x_k} } \paren { \prod_{k \mathop = 1, k \mathop \ne i }^n \paren {y_k - y_i} } } = \sum_{k \mathop = 1}^n x_k - \sum_{k \mathop = 1}^n y_k$