Summation of Powers over Product of Differences

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Theorem

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


Example

\(\ds \frac 1 {\paren {a - b} \paren {a - c} } + \frac 1 {\paren {b - a} \paren {b - c} } + \frac 1 {\paren {c - a} \paren {c - b} }\) \(=\) \(\ds 0\)
\(\ds \frac a {\paren {a - b} \paren {a - c} } + \frac b {\paren {b - a} \paren {b - c} } + \frac c {\paren {c - a} \paren {c - b} }\) \(=\) \(\ds 0\)
\(\ds \frac {a^2} {\paren {a - b} \paren {a - c} } + \frac {b^2} {\paren {b - a} \paren {b - c} } + \frac {c^2} {\paren {c - a} \paren {c - b} }\) \(=\) \(\ds 1\)
\(\ds \frac {a^3} {\paren {a - b} \paren {a - c} } + \frac {b^3} {\paren {b - a} \paren {b - c} } + \frac {c^3} {\paren {c - a} \paren {c - b} }\) \(=\) \(\ds a + b + c\)


Proof 1

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

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


Edge Cases

$\map P 0$ is a vacuous summation:

$\ds S_0 := \sum_{j \mathop = 1}^0 \paren {\dfrac { {x_j}^0} {\ds \prod_{\substack {1 \mathop \le k \mathop \le 0 \\ k \mathop \ne j} } \paren {x_j - x_k} } } = 0 = \sum_{j \mathop = 1}^0 x_0$

which is seen to hold.


$\map P 1$ is the case:

\(\ds \sum_{j \mathop = 1}^1 \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } }\) \(=\) \(\ds \dfrac { {x_1}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le 1 \\ k \mathop \ne 1} } \paren {x_1 - x_k} }\)
\(\ds \) \(=\) \(\ds \dfrac { {x_1}^r} 1\) as the product is vacuous
\(\ds \) \(=\) \(\ds {x_1}^r\) simplification
\(\ds \) \(=\) \(\ds \begin{cases} 1 & : r = 0 \\ x_1 & : r = 1 \end{cases}\)

This is also seen to hold.


Basis for the Induction

$\map P 2$ is the case where $n = 2$:

\(\ds S_2\) \(=\) \(\ds \sum_{j \mathop = 1}^2 \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le 2 \\ k \mathop \ne j} } \paren {x_j - x_k} } }\)
\(\ds \) \(=\) \(\ds \frac { {x_1}^r} {x_1 - x_2} + \frac { {x_2}^r} {x_2 - x_1}\)
\(\ds \) \(=\) \(\ds \frac { {x_1}^r - {x_2}^r} {x_1 - x_2}\)


When $0 \le r < n - 1$, it must be that $r = 0$:

\(\ds S_2\) \(=\) \(\ds \frac {1 - 1} {x_1 - x_2}\)
\(\ds \) \(=\) \(\ds 0\)


When $r = n - 1 = 1$:

\(\ds S_2\) \(=\) \(\ds \frac {x_1 - x_2} {x_1 - x_2}\)
\(\ds \) \(=\) \(\ds 1\)


When $r = n = 2$:

\(\ds S_2\) \(=\) \(\ds \frac { {x_1}^2 - {x_2}^2} {x_1 - x_2}\)
\(\ds \) \(=\) \(\ds \frac {\paren {x_1 + x_2} \paren {x_1 - x_2} } {x_1 - x_2}\)
\(\ds \) \(=\) \(\ds x_1 + x_2\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^2 x_j\)


Thus in all cases, $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $\map P m$ is true, where $m \ge 2$, then it logically follows that $\map P {m + 1}$ is true.


So this is the induction hypothesis:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:


For $n > 2$, let the formula be rewritten:

\(\ds S_n\) \(=\) \(\ds \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }\)
\(\ds \) \(=\) \(\ds \frac 1 {x_n - x_{n - 1} } \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r \paren {x_n - x_{n - 1} } } {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }\)
\(\ds \) \(=\) \(\ds \frac 1 {x_n - x_{n - 1} } \paren {\sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r \paren {x_j - x_n} } {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } - \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r \paren {x_j - x_{n - 1} } } {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } }\)


So:

\(\ds S_{n + 1}\) \(=\) \(\ds \frac 1 {x_{n + 1} - x_n} \paren {\sum_{j \mathop = 1}^{n + 1} \paren {\dfrac { {x_j}^r \paren {x_j - x_{n + 1} } } {\ds \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } } - \sum_{j \mathop = 1}^{n + 1} \paren {\dfrac { {x_j}^r \paren {x_j - x_n} } {\ds \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j} } \paren {x_j - x_k} } } }\)
\(\ds \) \(=\) \(\ds \frac 1 {x_{n + 1} - x_n} \paren {\sum_{\substack {1 \mathop \le j \mathop \le n + 1 \\ j \mathop \ne n} } \paren {\dfrac { {x_j}^r } {\ds \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j \\ k \mathop \ne n} } \paren {x_j - x_k} } } - \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } }\) as both parts are the original sum


When $r < n$, both parts are equal to $0$ or $1$ by the induction hypothesis.

Thus either:

$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {1 - 1} = 0$

or:

$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {0 - 0} = 0$

and so $\map P {m + 1}$ holds for $r < n$.


When $r = n$:

\(\ds \sum_{\substack {1 \mathop \le j \mathop \le n + 1 \\ j \mathop \ne n} } \paren {\dfrac { {x_j}^r } {\ds \prod_{\substack {1 \mathop \le k \mathop \le n + 1 \\ k \mathop \ne j \\ k \mathop \ne n} } \paren {x_j - x_k} } }\) \(=\) \(\ds \sum_{\substack {1 \mathop \le j \mathop \le n + 1 \\ j \mathop \ne n} } x_j\)
\(\ds \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }\) \(=\) \(\ds \sum_{j \mathop = 1}^n x_j\)

Thus:

$S_{n + 1} = \dfrac 1 {x_{n + 1} - x_n} \paren {x_{n + 1} - x_n} = 1$

and so $\map P {m + 1}$ holds for $r = n$.


Now:

\(\ds \sum_{j \mathop = 1}^n \paren {\dfrac {\ds \prod_{k \mathop = 1}^n \paren {x_j - x_k} } {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }\) \(=\) \(\ds 0\)
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^n - \paren {x_1 + \cdots + x_n} {x_j}^{n + 1} + \map P {x_j} } {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }\) where $\map P {x_j}$ is a polynomial of degree $n - 2$
\(\ds \leadsto \ \ \) \(\ds \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^n} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } }\) \(=\) \(\ds \sum_{l \mathop = 1}^n \paren {\sum_{j \mathop = 1}^n x_l \paren {\dfrac { {x_j}^{n - 1} } {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } }\)
\(\ds \) \(=\) \(\ds \sum_{l \mathop = 1}^n x_l\)

Thus $\map P {m + 1}$ holds for $r = n + 1$.


So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

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

$\blacksquare$


Proof 2

By the Residue Theorem:

$\ds \sum_{j \mathop = 1}^n \begin {pmatrix} {\dfrac { {x_j}^r} {\ds \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } \end{pmatrix} = \dfrac 1 {2 \pi i} \int \limits_{\size z \mathop = R} \dfrac {z^r \rd z} {\paren {z - z_1} \cdots \paren {z - z_n} }$

where $R > \size {z_1}, \ldots, \size {z_n}$.


The Laurent series of the integrand converges uniformly on $\size z = R$:

\(\ds \) \(\) \(\ds z^{r - n} \paren {\dfrac 1 {1 - x_1 / z} } \cdots \paren {\dfrac 1 {1 - x_n / z} }\)
\(\ds \) \(=\) \(\ds z^{r - n} + \paren {x_1 + \cdots x_n} z^{r - n - 1} + \paren {x_1^2 + x_1 x_2 + \cdots} z^{r - n - 2} + \cdots\)

On integrating term my term, everything vanishes except the coefficient of $z^{-1}$.


Thus:

$\ds \sum_{\substack {j_1 \mathop + \mathop \cdots \mathop + j_n \mathop = r \mathop - n \mathop + 1 \\ j_1, \mathop \ldots j_n \mathop \ge 0} } {x_1}^{j_1} {x_2}^{j_2} \cdots {x_n}^{j_n}$

$\blacksquare$


Proof 3

Definition

\(\ds \map p x\) \(=\) \(\ds \prod_{k \mathop = 1}^n \paren {x - x_k}\)
\(\ds \map {p_m} x\) \(=\) \(\ds \prod_{k \mathop = 1, \, k \mathop \ne m}^n \paren {x - x_k},\quad 1 \le m \le n\)
\(\ds A\) \(=\) \(\ds \begin{pmatrix} 1 & x_1 & \cdots & x_1^{n - 2} & x_1^{n - 1} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 1 & x_n & \cdots & x_n^{n - 2} & x_n^{n - 1} \\ \end{pmatrix}\) Set $\set {x_1, \ldots, x_n}$ 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 $c$

$\blacksquare$


Historical Note

The result Summation of Powers over Product of Differences was discussed by Leonhard Paul Euler in a letter to Christian Goldbach in $1762$.

He subsequently published it in his Institutiones Calculi Integralis, Volume 2 of $1769$.

The proof involving complex analysis was devised in $1857$ by James Joseph Sylvester.


Sources