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:
| \(\displaystyle {\mathbf S}_m\) | \(=\) | \(\displaystyle \set { \paren {j_1,\ldots,j_m} : 1 \le j_1 \lt \cdots \lt j_m \le n}\) | $1 \le m \le n$ | ||||||||||
| \(\displaystyle \map {e_m} {X}\) | \(=\) | \(\displaystyle \begin{cases} 1 & m = 0\\ \displaystyle \sum_{ {\mathbf S}_m } x_{j_1} \cdots x_{j_m} & 1 \leq m \leq n \\ 0 & m \gt n \\ \end{cases}\) | elementary symmetric function | ||||||||||
| \(\displaystyle \map {p_k} X\) | \(=\) | \(\displaystyle \begin{cases} \displaystyle n & k = 0 \\ \displaystyle \sum_{i \mathop = 1}^n x_i^k & k \ge 1 \\ \end{cases}\) | power sums |
Then Newton's Identities are:
| \((1):\quad\) | \(\displaystyle k \, \map {e_k} X\) | \(=\) | \(\displaystyle \displaystyle \sum_{i \mathop = 1}^k \paren {-1}^{i-1} \map {e_{k-i} } X \map {p_i} X\) | for $1 \leq k \leq n$ | |||||||||
| \((2):\quad\) | \(\displaystyle 0\) | \(=\) | \(\displaystyle \displaystyle \sum_{i \mathop = k-n}^k \paren {-1}^{i-1} \map {e_{k-i} } X \map {p_i} X\) | for $1 \leq n \lt k$ |
Proof
Outline
Calculus is used to prove identities (1) and (2) in a single effort.
The tools are Viete's Formulas, the calculus derivative of powers $x^n$ and logarithm $\ln \vert x \vert$, Maclaurin series expansion coefficients, mathematical induction, and Leibniz's Rule: One Variable.
Lemma 1
| \(\displaystyle \prod_{r \mathop = 1}^n \paren { 1 + x_r z }\) | \(=\) | \(\displaystyle \sum_{m \mathop = 0}^n {\map {e_m} X} z^m\) |
Proof of Lemma 1
Begin with:
| \((11):\quad\) | \(\displaystyle \displaystyle \prod_{r \mathop = 1}^n \paren { x - x_r }\) | \(=\) | \(\displaystyle \sum_{i \mathop = 0}^n \paren {-1}^{n-i} {\map {e_{n-i} } X} x^i\) | Viete'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:
| \(\displaystyle \map G z\) | \(=\) | \(\displaystyle \displaystyle \prod_{k \mathop = 1}^n \paren {1 + x_k z}\) | as in Lemma 1 | ||||||||||
| \(\displaystyle \map F z\) | \(=\) | \(\displaystyle \dfrac { \map {DQ} z }{\map Q z}\) | the calculus derivative of $\ln \vert { \map Q z } \vert$ |
Then:
| \((12):\quad\) | \(\displaystyle \dfrac { D^m \map G {0} } { m! }\) | \(=\) | \(\displaystyle \map { e_m } X\) | ||||||||||
| \((13):\quad\) | \(\displaystyle \dfrac { D^m \map F {0} } { m! }\) | \(=\) | \(\displaystyle \paren {-1}^m \map {p_{m+1} } X\) |
Proof of Lemma 2
| \(\displaystyle \map G z\) | \(=\) | \(\displaystyle \sum_{m \mathop = 0}^n {\map {e_m} X} z^m\) | 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:
| \((14):\quad\) | \(\displaystyle D^m \map F z\) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^n \dfrac{ m!\paren {-1}^m x_i^{m+1} }{ \paren { 1 + x_i z }^{m+1} }\) | for $m \ge 0$ |
Basis for the induction: $m=0$
By calculus and the definition of $G$:
| \(\displaystyle \map F z\) | \(=\) | \(\displaystyle \dfrac { D \map Q z}{\map Q z}\) | |||||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle \) | \(=\) | \(\displaystyle \displaystyle \sum_{i \mathop = 1}^n \dfrac { x_i }{ 1 + x_i z }\) |
Then $\map {\mathbf P} 0$ is true.
Induction step $\map {\mathbf P} m$ implies $\map {\mathbf P} {m+1}$:
| \(\displaystyle D^{m+1} \map F z\) | \(=\) | \(\displaystyle D \paren {\sum_{i \mathop = 1}^n \dfrac{ m!\paren {-1}^m x_i^{m+1} }{ \paren { 1 + x_i z }^{m+1} } }\) | by induction hypothesis $\map {\mathbf P} m$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^n \dfrac{ m! \paren {-1}^m x_i^{m+1} \, \paren {-1} \paren {m+1} \, x_i}{ \paren { 1 + x_i z }^{m+2} }\) | by calculus power rule $\dfrac { \d u^{-n} }{ \d z } = \paren { -n } u^{-n-1} \dfrac { \d u }{ \d z }$ |
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
| \((15):\quad\) | \(\displaystyle \paren {m+1} \map {e_{m+1} } X\) | \(=\) | \(\displaystyle \displaystyle \sum_{r \mathop = 0}^m \paren {-1}^r { \map {e_{m-r} } X } { \map {p_{r+1} } X }\) | for $m \geq 0$ |
Proof of Lemma 3
Begin with $D \map G z = {\map F z} {\map G z}$ and differentiate $m$ times on variable $z$:
| \(\displaystyle D^{m+1} \map G z\) | \(=\) | \(\displaystyle \displaystyle \sum_{r \mathop = 0}^m {\dbinom {m} {r} } D^r {\map F z} \, D^{m-r} {\map G z}\) | Leibniz's Rule/One Variable | ||||||||||
| \(\displaystyle D^{m+1} {\map G 0}\) | \(=\) | \(\displaystyle \displaystyle \sum_{r \mathop = 0}^m \dbinom {m} {r} r! \, \paren {-1}^r {\map { p_{r+1} } X} \, \paren {m-r}! \, {\map {e_{m-r} } X}\) | Evaluate at $ z = 0$ and use equations (12) and (13) in Lemma 2. | ||||||||||
| \(\displaystyle \paren {m+1} {\map {e_m} X}\) | \(=\) | \(\displaystyle \displaystyle \sum_{r \mathop = 0}^m \paren {-1}^r {\map {e_{m-r} } X} \, {\map {p_{r+1} } X}\) | 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:
| \((16):\quad\) | \(\displaystyle k \, { \map {e_{k} } X }\) | \(=\) | \(\displaystyle \displaystyle \sum_{i \mathop = 1}^{k} \paren {-1}^{i-1} { \map {e_{k-i} } X } { \map {p_{i} } X }\) | for $k \geq 1$, which is equation (1) |
To prove (2), assume $k > n \ge 1$ and $X = \set {x_1,\ldots,x_n}$.
Equation (16) implies:
| \((117):\quad\) | \(\displaystyle 0\) | \(=\) | \(\displaystyle k \, { \map {e_{k} } X }\) | because $ \map {e_{j} } X = 0$ for $j=n+1,\ldots,k$. | |||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle 0\) | \(=\) | \(\displaystyle \sum_{i \mathop = 1}^{k} \paren {-1}^{i-1} { \map {e_{k-i} } X } { \map {p_{i} } X }\) | by (16) for $k \geq 1$ | |||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle 0\) | \(=\) | \(\displaystyle \sum_{i \mathop = k-n}^{k} \paren {-1}^{i-1} { \map {e_{k-i} } X } { \map {p_{i} } X }\) | because $ \map {e_{k-i} } X = 0$ when $n+1 \le k-i \le k$ |
Then (2) holds.
$\blacksquare$
