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:

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

\(\text {(11)}: \quad\) \(\displaystyle \displaystyle \prod_{r \mathop = 1}^n \paren { x - x_r }\) \(=\) \(\displaystyle \sum_{i \mathop = 0}^n \paren {-1}^{n-i} {\map {e_{n-i} } X} x^i\) 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:

\(\text {(12)}: \quad\) \(\displaystyle \dfrac { D^m \map G {0} } { m! }\) \(=\) \(\displaystyle \map { e_m } X\)
\(\text {(13)}: \quad\) \(\displaystyle \dfrac { D^m \map F {0} } { m! }\) \(=\) \(\displaystyle \paren {-1}^m \map {p_{m+1} } X\)

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:

\(\text {(14)}: \quad\) \(\displaystyle D^m \map F z\) \(=\) \(\displaystyle \sum_{i \mathop = 1}^n \dfrac{ m!\paren {-1}^m x_i^{m+1} }{ \paren { 1 + x_i z }^{m+1} }\) 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

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

\(\text {(16)}: \quad\) \(\displaystyle k \, { \map {e_{k} } X }\) \(=\) \(\displaystyle \displaystyle \sum_{i \mathop = 1}^{k} \paren {-1}^{i-1} { \map {e_{k-i} } X } { \map {p_{i} } X }\) 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:

\(\text {(117)}: \quad\) \(\displaystyle 0\) \(=\) \(\displaystyle k \, { \map {e_{k} } X }\) 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.

$\blacksquare$