# Newton's Identities/Proof 2

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## Theorem

Let $X$ be a set of $n$ numbers $\set {x_1, x_2, \ldots, x_n}$.

Define:

\(\ds {\mathbf S}_m\) | \(=\) | \(\ds \set { \paren {j_1,\ldots,j_m} : 1 \le j_1 \lt \cdots \lt j_m \le n}\) | $1 \le m \le n$ | |||||||||||

\(\ds \map {e_m} {X}\) | \(=\) | \(\ds \begin{cases} 1 & m = 0\\ \ds \sum_{ {\mathbf S}_m } x_{j_1} \cdots x_{j_m} & 1 \leq m \leq n \\ 0 & m \gt n \\ \end{cases}\) | elementary symmetric function | |||||||||||

\(\ds \map {p_k} X\) | \(=\) | \(\ds \begin{cases} n & k = 0 \\ \ds \sum_{i \mathop = 1}^n x_i^k & k \ge 1 \end {cases}\) | power sums |

Then **Newton's Identities** are:

\(\text {(1)}: \quad\) | \(\ds k \map {e_k} X\) | \(=\) | \(\ds \ds \sum_{i \mathop = 1}^k \paren {-1}^{i - 1} \map {e_{k - i} } X \map {p_i} X\) | for $1 \leq k \leq n$ | ||||||||||

\(\text {(2)}: \quad\) | \(\ds 0\) | \(=\) | \(\ds \ds \sum_{i \mathop = k - n}^k \paren {-1}^{i - 1} \map {e_{k - i} } X \map {p_i} X\) | for $1 \leq n \lt k$ |

## Proof

### Outline

Calculus is used to prove identities (1) and (2) in a single effort.

The tools are Viète's Formulas, the calculus derivative of powers $x^n$ and logarithm $\ln \size x$, Maclaurin series expansion coefficients, mathematical induction, and Leibniz's Rule: One Variable.

### Lemma 1

\(\ds \prod_{r \mathop = 1}^n \paren {1 + x_r z}\) | \(=\) | \(\ds \sum_{m \mathop = 0}^n {\map {e_m} X} z^m\) |

**Proof of Lemma 1**

Begin with:

\(\text {(11)}: \quad\) | \(\ds \prod_{r \mathop = 1}^n \paren {x - x_r}\) | \(=\) | \(\ds \sum_{i \mathop = 0}^n \paren {-1}^{n - i} {\map {e_{n - i} } X} x^i\) | Viète's Formulas |

Change variables in $(11)$: $x = -1/z$.

Details: Generating Function for Elementary Symmetric Function.

$\Box$

### Lemma 2

Denote by $D^k \map f z$ the $k$th calculus derivative of $\map f z$.

Let:

\(\ds \map G z\) | \(=\) | \(\ds \prod_{k \mathop = 1}^n \paren {1 + x_k z}\) | as in Lemma 1 | |||||||||||

\(\ds \map F z\) | \(=\) | \(\ds \dfrac {\map {DQ} z} {\map Q z}\) | the calculus derivative of $\ln \size {\map Q z}$ |

Then:

\(\text {(12)}: \quad\) | \(\ds \dfrac {D^m \map G 0} {m!}\) | \(=\) | \(\ds \map {e_m } X\) | |||||||||||

\(\text {(13)}: \quad\) | \(\ds \dfrac {D^m \map F 0} {m!}\) | \(=\) | \(\ds \paren {-1}^m \map {p_{m + 1} } X\) |

**Proof of Lemma 2**

\(\ds \map G z\) | \(=\) | \(\ds \sum_{m \mathop = 0}^n {\map {e_m} X} z^m\) | by Lemma 1 |

Then identity $(12)$ holds by Maclaurin series expansion applied to polynomial $G$.

Identity $(13)$ will be proved after mathematical induction establishes $(14)$ *infra*.

Let $\map {\mathbf P} m$ be the statement:

\(\text {(14)}: \quad\) | \(\ds D^m \map F z\) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \dfrac{ m!\paren {-1}^m x_i^{m+1} }{ \paren { 1 + x_i z }^{m+1} }\) | for $m \ge 0$ |

Basis for the induction: $m=0$

By calculus and the definition of $G$:

\(\ds \map F z\) | \(=\) | \(\ds \dfrac { D \map Q z}{\map Q z}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \dfrac { x_i }{ 1 + x_i z }\) |

Then $\map {\mathbf P} 0$ is true.

Induction step $\map {\mathbf P} m$ implies $\map {\mathbf P} {m+1}$:

\(\ds D^{m + 1} \map F z\) | \(=\) | \(\ds \map D {\sum_{i \mathop = 1}^n \dfrac {m! \paren {-1}^m x_i^{m + 1} } {\paren {1 + x_i z}^{m + 1} } }\) | by induction hypothesis $\map {\mathbf P} m$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^n \dfrac{ m! \paren {-1}^m x_i^{m + 1} \paren {-1} \paren {m + 1} x_i} {\paren {1 + x_i z}^{m + 2} }\) | Power Rule for Derivatives: $\dfrac {\d u^{-n} } {\d z } = \paren {-n} u^{-n - 1} \dfrac {\d u} {\d z}$ |

Simplify to prove $\map {\mathbf P} {m + 1}$ is true.

The induction is complete.

To prove equation (13), first let $z = 0$ in equation (14).

Divide by $m!$ to isolate $\map {p_{m + 1} } X$, which proves (13).

$\Box$

### Lemma 3

\(\text {(15)}: \quad\) | \(\ds \paren {m + 1} \map {e_{m + 1} } X\) | \(=\) | \(\ds \sum_{r \mathop = 0}^m \paren {-1}^r {\map {e_{m - r} } X} {\map {p_{r + 1} } X}\) | for $m \ge 0$ |

**Proof of Lemma 3**

Begin with $D \map G z = {\map F z} {\map G z}$ and differentiate $m$ times on variable $z$:

\(\ds D^{m + 1} \map G z\) | \(=\) | \(\ds \sum_{r \mathop = 0}^m {\dbinom m r} D^r {\map F z} D^{m - r} {\map G z}\) | Leibniz's Rule/One Variable | |||||||||||

\(\ds D^{m + 1} {\map G 0}\) | \(=\) | \(\ds \sum_{r \mathop = 0}^m \dbinom m r r! \paren {-1}^r {\map {p_{r + 1} } X} \paren {m - r}! \map {e_{m-r} } X\) | Evaluate at $ z = 0$ and use equations (12) and (13) in Lemma 2 | |||||||||||

\(\ds \paren {m + 1} {\map {e_m} X}\) | \(=\) | \(\ds \sum_{r \mathop = 0}^m \paren {-1}^r {\map {e_{m-r} } X} {\map {p_{r+1} } X}\) | Use (12), then collect factorials and simplify |

$\Box$

**Proof of the Theorem**

To prove (1), start with equation (15) in Lemma 3.

Change indices via equations $m + 1 = k$, $r + 1 = i$.

The summation is from $i = 0 + 1$ to $i = m + 1$, which gives range $i = 1$ to $k$.

Subscript $m - r$ equals $k - 1- i + 1$, which simplifies to $k - i$.

Then:

\(\text {(16)}: \quad\) | \(\ds k \map {e_k} X\) | \(=\) | \(\ds \sum_{i \mathop = 1}^{k} \paren {-1}^{i - 1} {\map {e_{k - i} } X} {\map {p_i} X}\) | for $k \ge 1$, which is equation (1) |

To prove (2), assume $k > n \ge 1$ and $X = \set {x_1, \ldots, x_n}$.

Equation (16) implies:

\(\text {(117)}: \quad\) | \(\ds 0\) | \(=\) | \(\ds k \map {e_{k} } X\) | because $ \map {e_j} X = 0$ for $j = n + 1, \ldots, k$. | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds 0\) | \(=\) | \(\ds \sum_{i \mathop = 1}^k \paren {-1}^{i - 1} {\map {e_{k - i} } X} {\map {p_i} X}\) | by (16) for $k \ge 1$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds 0\) | \(=\) | \(\ds \sum_{i \mathop = k - n}^k \paren {-1}^{i - 1} {\map {e_{k - i} } X} {\map {p_i} X}\) | because $ \map {e_{k - i} } X = 0$ when $n + 1 \le k - i \le k$ |

Then (2) holds.

$\blacksquare$