# Summation of Powers over Product of Differences/Proof 1

## 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

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:

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

$\blacksquare$

## 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.