6.3 Completion of the proof

We now have all the ingredients to prove the main theorem.

Proof of Theorem 1.0.4 ▶

Let \(r_k(\theta ) = \cos (\theta + 2\pi k/N)\) for \(k = 0, \ldots , N-1\) be the roots of \(P_N(x; \theta )\), and let \(\xi _k = \cos ((2k + 1)\pi /(2N))\) be the roots of \(T_N(x)\).

By Theorem 6.2.2, the power sums of \(\{ r_k(\theta )\} \) and \(\{ \xi _k\} \) are equal for all \(j\) with \(1 \leq j {\lt} N\):

\begin{equation} \sum _{i=0}^{N-1} r_i(\theta )^j = \sum _{i=0}^{N-1} \xi _i^j. \end{equation}

By Newton’s identities (Theorem 2.3.1), the elementary symmetric polynomials \(e_k\) for \(k = 1, \ldots , N\) are determined recursively from the power sums \(p_1, \ldots , p_k\). Since the power sums match for \(j {\lt} N\), and the recursion for \(e_k\) with \(k \leq N\) only depends on \(p_1, \ldots , p_k {\lt} p_N\), the elementary symmetric polynomials match:

\begin{equation} e_k(r_0(\theta ), \ldots , r_{N-1}(\theta )) = e_k(\xi _0, \ldots , \xi _{N-1}), \quad k = 1, \ldots , N. \end{equation}

The monic polynomial \(P_N(x; \theta )\) has roots \(r_k(\theta )\) and can be written as

\begin{equation} P_N(x; \theta ) = \prod _{k=0}^{N-1}(x - r_k(\theta )) = \sum _{k=0}^N (-1)^k e_k \, x^{N-k}, \end{equation}

where \(e_k = e_k(r_0(\theta ), \ldots , r_{N-1}(\theta ))\) are the elementary symmetric polynomials.

Similarly, the Chebyshev polynomial \(T_N(x)\) has leading coefficient \(2^{N-1}\) and roots \(\xi _k\), so

\begin{equation} T_N(x) = 2^{N-1} \prod _{k=0}^{N-1}(x - \xi _k). \end{equation}

By Vieta’s formulas, the coefficient of \(x^{N-k}\) in \(P_N(x; \theta )\) is \((-1)^k e_k(r_0, \ldots , r_{N-1})\), and the coefficient of \(x^{N-k}\) in \(T_N(x)/2^{N-1}\) is \((-1)^k e_k(\xi _0, \ldots , \xi _{N-1})\).

Since the elementary symmetric polynomials match for \(k = 1, \ldots , N\), we have \(P_N(x; \theta )\) and \(T_N(x)/2^{N-1}\) have the same coefficients for all degrees \(k = 1, \ldots , N\). They may differ only in the constant term (degree 0).

Multiplying both by \(2^{N-1}\), the scaled polynomial

\begin{equation} S_N(x; \theta ) = 2^{N-1} P_N(x; \theta ) \end{equation}

and \(T_N(x)\) have the same coefficients for degrees \(k = 1, \ldots , N\). They may differ in the constant term.

Hence

\begin{equation} S_N(x; \theta ) = T_N(x) + c(\theta ), \end{equation}

where \(c(\theta )\) is given explicitly by ?? as the difference in constant terms.

Moreover, since the coefficients of \(x^k\) for \(k \geq 1\) match exactly, we have coefficient-wise equality:

\begin{equation} [x^k] S_N(x; \theta ) = [x^k] T_N(x) \quad \text{for all } k \geq 1. \end{equation}

This completes the proof of Theorem 1.0.4.