6.2 Power sum equality

Lemma 6.2.1

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Let \(\xi _k = \cos ((2k + 1)\pi /(2N))\) for \(k = 0, \ldots , N-1\) denote the roots of \(T_N(x)\). Then for any \(1 \leq j {\lt} N\),

\begin{equation} \sum _{k=0}^{N-1} \xi _k^j = \begin{cases} \frac{N}{2^j} \binom {j}{j/2} & \text{if } j \text{ is even}, \\ 0 & \text{if } j \text{ is odd.} \end{cases} \end{equation}

(See Lemma 6.1 in ../paper/chebyshev_circles.pdfthe paper)

Proof ▶

The proof parallels that of Theorem 4.3.1. Expand \(\xi _k^j = \cos ^j((2k+1)\pi /(2N))\) using the binomial formula:

\begin{equation} \cos ^j\theta = \frac{1}{2^j}\sum _{\ell =0}^j \binom {j}{\ell } \mathrm{e}^{i(j-2\ell )\theta }. \end{equation}

Summing over \(k\) and using Lemmas 3.2.1 and 3.2.2, only the term with \(j - 2\ell = 0\) survives (when \(j\) is even). This gives the stated formula.

Theorem 6.2.2 Power sum equality

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For \(1 \leq j {\lt} N\),

\begin{equation} \sum _{k=0}^{N-1} \cos ^j\left(\theta + \frac{2\pi k}{N}\right) = \sum _{k=0}^{N-1} \cos ^j\left(\frac{(2k + 1)\pi }{2N}\right). \end{equation}

(See Theorem 6.2 in ../paper/chebyshev_circles.pdfthe paper)

Proof ▶

Both sides are independent of \(\theta \) by Theorem 4.3.1. Evaluating at any particular value of \(\theta \) establishes equality. The formalization uses \(\theta = \pi /(2N)\) as the bridge: the rotated roots at this angle coincide with a permutation of the Chebyshev roots, making the equality manifest.