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Three Chebyshev-root samples recover a quadratic by DCT-II

Example

Let $p (t) = 2 + T_{1} (t) - T_{2} (t)$ , where $T_{k}$ is the first-kind Chebyshev polynomial, and sample it at the three DCT-II nodes $ξ_{j} = \cos (π (j + 1 / 2) / 3)$ for $j = 0, 1, 2$ . The DCT-II output is $(6, 3 / 2, - 3 / 2)$ , and the inverse coefficient rule recovers $(2, 1, - 1)$ .

Facts & Assumptions

Given: The polynomial $p (t) = 2 + T_{1} (t) - T_{2} (t)$ and the samples $x_{j} = p (ξ_{j})$ at $ξ_{j} = \cos (π (j + 1 / 2) / 3)$ .

[F1]

At these DCT-II nodes, a polynomial $p = \sum_{k = 0}^{N - 1} a_{k} T_{k}$ has DCT values $X_{0} = N a_{0}$ and $X_{k} = (N / 2) a_{k}$ for $k \geq 1$ (DCT-II samples extract Chebyshev coefficients at first-kind roots).

[F2]

Chebyshev coefficient recovery from the DCT-II samples is $a_{0} = X_{0} / N$ and $a_{k} = 2 X_{k} / N$ for $1 \leq k < N$ (DCT-II samples extract Chebyshev coefficients at first-kind roots).

[F3]

$T_{0} (t) = 1$ , $T_{1} (t) = t$ , and $T_{2} (t) = 2 t^{2} - 1$ (Chebyshev polynomials of the first and second kinds by their three-term recurrences).

Verification

1.1givenF1F3algebra

By [F3], $p (t) = 2 + T_{1} (t) - T_{2} (t) = 2 + 1 \cdot T_{1} (t) - 1 \cdot T_{2} (t)$ has Chebyshev coefficient vector $(a_{0}, a_{1}, a_{2}) = (2, 1, - 1)$ and degree less than $3$ . Applying [F1] with $N = 3$ gives $X_{0} = 3 \cdot 2 = 6$ , $X_{1} = (3 / 2) \cdot 1 = 3 / 2$ , and $X_{2} = (3 / 2) (- 1) = - 3 / 2$ .

2.1step 1.1F2algebra∎

Applying [F2] gives $a_{0} = 6 / 3 = 2$ , $a_{1} = 2 (3 / 2) / 3 = 1$ , and $a_{2} = 2 (- 3 / 2) / 3 = - 1$ , recovering the original coefficients.

Depends on

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Sources

C. Sidney Burrus, Fast Fourier Transforms, Chapter 8 §8.3.2 (standard reference, not scraped)