Newton's Identities/Proof 2

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 in One Variable.

Lemma 1
Proof of Lemma 1

Begin with:

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

Details: Generating Function for Elementary Symmetric Function.

Lemma 2
Denote by $D^k \map f z$ the $k$th calculus derivative of $\map f z$.

Let:

Then:

Proof of Lemma 2

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:

Basis for the induction: $m=0$

By calculus and the definition of $G$:

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

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

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).

Lemma 3
Proof of Lemma 3

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

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:

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

Equation (16) implies:

Then (2) holds.