# Summation of Powers over Product of Differences/Proof 3

## Theorem

$\ds \sum_{j \mathop = 1}^n \begin{pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end{pmatrix} = \begin{cases} 0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \ds \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$

## Proof

### Definition

 $\ds \map p x$ $=$ $\ds \prod_{k \mathop = 1}^n \paren {x - x_k}$ $\ds \map {p_m} x$ $=$ $\ds \prod_{k \mathop = 1, \, k \mathop \ne m}^n \paren {x - x_k},\quad 1 \le m \le n$ $\ds A$ $=$ $\ds \begin{pmatrix} 1 & x_1 & \cdots & x_1^{n - 2} & x_1^{n - 1} \\  \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & \cdots & x_n^{n - 2} & x_n^{n - 1} \\ \end{pmatrix}$  where $\set {x_1, \ldots, x_n}$ is assumed to have distinct elements $\ds A_{\vec Y}$ $=$ $\ds \begin{pmatrix}  1 & x_1 & \cdots & x_1^{n - 2} & y_1 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & \cdots & x_n^{n -2} & y_n \\  \end{pmatrix}$ Vector $\vec Y$ has entries $y_1,\ldots,y_n$.

Symbol $\map {\mathbf {Cof } } {M, i, j}$ denotes cofactor $M_{i j}$ of matrix $M$

### Lemma 1

 $\ds \map \det A$ $=$ $\ds \map {p_m} {x_m} \map {\mathbf {Cof} } {A, m, n}$

Proof of Lemma 1

By Effect of Elementary Row Operations on Determinant, the determinant of $A$ is unchanged by adding a linear combination of the first $n - 1$ columns to the last column.

Let $\map f x$ be the monic polynomial $\map {p_m} x = x^{n - 1} +$ lower power terms.

Then:

 $\ds \det \begin {pmatrix}  1 & x_1 & \cdots & x_1^{n - 2} & x_1^{n - 1} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & \cdots & x_n^{n - 2} & x_n^{n - 1} \\ \end {pmatrix}$  $=$ $\ds \det \begin {pmatrix}  1 & x_1 & \cdots & x_1^{n - 2} & \map f {x_1} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & \cdots & x_n^{n - 2} & \map f {x_n} \\ \end {pmatrix}$ 

The last column on the right has all components zero except $m$th entry $\map {p_m} {x_m}$.

Apply cofactor expansion along column $n$.

$\Box$

### Lemma 2

 $\text {(1)}: \quad$ $\ds \map \det A$ $=$ $\ds \prod_{1 \mathop \le i \mathop < j \mathop \le n} \paren {x_j - x_i}$ $\text {(2)}: \quad$ $\ds \map {\mathbf {Cof} } {A, m, n}$ $=$ $\ds \paren {-1}^{m + n} \prod_{1 \mathop \le i \mathop < j \mathop \le n, \, i \mathop \ne m, \, j \mathop \ne m} \paren {x_j - x_i}$

Proof of Lemma 2:

Value of Vandermonde Determinant establishes $(1)$.

To prove (2), let:

$\set {y_1, \ldots, y_{n - 1} } = \set {x_1, \ldots, x_n} \setminus \set {x_m}$

then apply $(1)$ with $\set {x_1, \ldots, x_n}$ replaced by $\set {y_1, \ldots, y_{n - 1} }$.

$\Box$

Theorem Details:

The Lemmas change the Theorem's equation to:

 $\text {(3)}: \quad$ $\ds \sum_{m \mathop = 1}^n {x_m^r} \, \dfrac {\map {\mathbf {Cof} } {A, m, n} } {\map \det A}$ $=$ $\ds \begin{cases} 0 & : 0 \mathop \le r \mathop < n - 1 \\ 1 & : r = n - 1 \\ \sum_{j \mathop = 1}^n x_j & : r = n \\  \end{cases}$ 

Cofactor expansion changes identity $(3)$ to:

 $\text {(4)}: \quad$ $\ds \det \begin{pmatrix} 1 & x_1 & \cdots & x_1^{n - 2} & x_1^r \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & \cdots & x_n^{n - 2} & x_n^r \\  \end{pmatrix}$  $=$ $\ds \map \det A \begin{cases} 0 & : 0 \mathop \le r \mathop < n - 1 \\ 1 & : r = n - 1 \\ \sum_{j \mathop = 1}^n x_j & : r = n \\  \end{cases}$ 

The proof is divided into three cases.

Case 1: $0 \mathop \le r \mathop \le n - 2$:

The determinant on the left in $(4)$ is zero by Square Matrix with Duplicate Rows has Zero Determinant.

Case 2: $r = n - 1$:

The determinant on the left in $(4)$ is $\map \det A$.

Case 3: $r = n$:

Expand $\map p x = \prod_{k \mathop = 1}^n \paren {x - x_k}$ by Taylor's Theorem with remainder $R$ of degree $n - 2$:

$\map p x = x^n - \paren {x_1 + \cdots + x_n} x^{n - 1} + \map R x$

Let $x = x_k$ for $1 \mathop \le k \mathop \le n$.

Then $x$ is a root of $\map p x = \prod_{k \mathop = 1}^n \paren {x - x_k}$:

 $\ds 0$ $=$ $\ds \map p {x_k}$ $\ds$ $=$ $\ds x_k^n - \paren {x_1 + \cdots + x_n} x_k^{n - 1} + \map R {x_k}$

Let:

 $\ds c$ $=$ $\ds x_1 + \cdots + x_n$ $\ds \vec u$ $=$ $\ds \begin {pmatrix} x_1^n \\ \vdots \\ x_n^n \end{pmatrix}$ Column $n$ of the left side of $(4)$ $\ds \vec z$ $=$ $\ds c \begin {pmatrix} x_1^{n - 1} \\ \vdots \\ x_n^{n - 1} \end{pmatrix}$ Constant $c$ times column $n$ of $A$ $\ds \vec w$ $=$ $\ds \begin{pmatrix} -\map R {x_1} \\ \vdots \\ -\map R {x_n} \end{pmatrix}$

Then:

 $\text {(5)}: \quad$ $\ds \vec u$ $=$ $\ds \vec z + \vec w$ by equation $x_k^n = c x_k^{n - 1} - \map R {x_k}$ $\text {(6)}: \quad$ $\ds \map \det {A_{\vec z} }$ $=$ $\ds c \map \det A$ Effect of Elementary Row Operations on Determinant: factoring $c$ from last column of $A_{\vec z}$
 $\text {(7)}: \quad$ $\ds \det \paren { A_{\vec w} }$ $=$ $\ds 0$ Effect of Elementary Row Operations on Determinant: $\vec w$ is a linear combination of columns $1$ to $n - 1$

The left side of $(4)$:

 $\ds \map \det {A_{\vec u} }$ $=$ $\ds \map \det {A_{\vec z} } + \map \det {A_{\vec w} }$ by $(5)$ and Determinant as Sum of Determinants $\ds$ $=$ $\ds c \map \det A + 0$ by $(6)$ and $(7)$ $\ds$ $=$ $\ds \map \det A \displaystyle \sum_{i \mathop = 1}^n x_i$ definition of $c$

$\blacksquare$