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Trigonometric formulas for Chebyshev cardinal polynomials

Statement

Let $n \geq 1$ , $0 \leq θ \leq π$ , and choose $j$ so $θ \neq θ_{j}^{R, n}$ . Then $∣ ℓ_{j, n}^{R} (\cos θ) ∣ = \frac{∣ \cos ((n + 1) θ) ∣ \sin θ_{j}^{R, n}}{(n + 1) ∣ \cos θ - \cos θ_{j}^{R, n} ∣} .$ For the Lobatto family, put $c_{j} = 1 / 2$ for $j = 0, n$ and $c_{j} = 1$ otherwise. If $θ \neq θ_{j}^{L, n}$ , then $∣ ℓ_{j, n}^{L} (\cos θ) ∣ = \frac{c_{j} ∣ \sin θ \sin (n θ) ∣}{n ∣ \cos θ - \cos θ_{j}^{L, n} ∣} .$

Facts & Assumptions

Given: $n \geq 1$ , $0 \leq θ \leq π$ , and one of the node families.

[F1]

Best uniform polynomial error and Chebyshev interpolation operators defines the nodes and cardinal polynomials.

[F2]

Chebyshev cardinal interpolation reproduces polynomials and its norm is the cardinal sum: these nodes are distinct and their cardinal polynomials are the unique degree-at-most- $n$ polynomials with the cardinal values.

[F3]

$T_{n} (\cos θ) = \cos (n θ)$ and $U_{n} (\cos θ) \sin θ = \sin ((n + 1) θ)$ for every $n \in N$ : $T_{m} (\cos θ) = \cos (m θ)$ and $U_{m} (\cos θ) \sin θ = \sin ((m + 1) θ)$ .

[F4]

Chebyshev polynomials of the first and second kinds by their three-term recurrences and Degrees and leading coefficients of the Chebyshev polynomials: $T_{m}$ and $U_{m}$ have degree $m$ for $m \geq 1$ ; $U_{0} = 1$ has degree $0$ .

[F5]

The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , The derivatives of sine and cosine are cosine and minus sine, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , and For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term: polynomial functions and sine/cosine are differentiable, and the chain and product rules compute their derivatives.

[F6]

A real polynomial vanishing at $a$ is divisible by $x - a$ : a polynomial that vanishes at $a$ factors as $(x - a) q (x)$ .

[F7]

Signs, monotonicity intervals, and ranges of sine and cosine, Quarter-turn values and shifts by pi/2 and pi, and The derivatives of sine and cosine are cosine and minus sine: $\sin θ > 0$ for $0 < θ < π$ , while $\sin 0 = \sin π = 0$ and $\cos π = - 1$ .

Proof

1.1givenF3F4F5F7algebra

Put $N = n + 1$ and $θ_{j} = (2 j + 1) π / (2 N)$ . By [F3], $T_{N} (\cos θ_{j}) = 0$ . Differentiating $T_{N} (\cos θ) = \cos (N θ)$ using [F4]–[F5] gives $\sin θ T_{N}^{'} (\cos θ) = N \sin (N θ)$ , hence $T_{N}^{'} (\cos θ_{j}) = N (- 1)^{j} / \sin θ_{j} \neq 0$ by [F7].

1.2givenF3F4F5F7algebra

Set $ω_{n} (x) = (x^{2} - 1) U_{n - 1} (x)$ . By [F3], $ω_{n} (\cos θ) = - \sin θ \sin (n θ)$ , so it vanishes at every Lobatto node; its degree is $n + 1$ by [F4]. For an interior node $θ_{j} = j π / n$ , differentiating this identity gives $- \sin θ_{j} ω_{n}^{'} (x_{j}) = - n \sin θ_{j} (- 1)^{j}$ , hence $ω_{n}^{'} (x_{j}) = n (- 1)^{j}$ by [F7].

1.3givenF3F4F5F7algebra

Differentiating $U_{n - 1} (\cos θ) \sin θ = \sin (n θ)$ at $θ = 0, π$ gives $U_{n - 1} (1) = n$ and $U_{n - 1} (- 1) = (- 1)^{n - 1} n$ by [F7]. The product rule then gives $ω_{n}^{'} (1) = 2 n$ and $ω_{n}^{'} (- 1) = 2 n (- 1)^{n}$ , so at every Lobatto node $ω_{n}^{'} (x_{j}) = n (- 1)^{j} / c_{j}$ .

2.1givenF1F2F3F4F5F6step 1.1algebra

By [F6], $T_{N} (x) / (x - x_{j})$ is a polynomial. Writing $T_{N} = (x - x_{j}) Q_{j}$ and differentiating gives $Q_{j} (x_{j}) = T_{N}^{'} (x_{j})$ ; at every other node $Q_{j}$ vanishes. Therefore $Q_{j} / T_{N}^{'} (x_{j})$ is a degree-at-most- $n$ cardinal polynomial, which equals $ℓ_{j, n}^{R}$ by uniqueness [F2]. Substitution of $x = \cos θ$ and [F3] gives the first formula.

3.1givenF1F2F3F4F6step 1.2step 1.3algebra∎

By [F6], write $ω_{n} (x) = (x - x_{j}) Q_{j} (x)$ with $Q_{j}$ a polynomial. Differentiating this factorization gives $Q_{j} (x_{j}) = ω_{n}^{'} (x_{j})$ ; at every other Lobatto node $Q_{j}$ vanishes because $ω_{n}$ does. Thus $Q_{j} / ω_{n}^{'} (x_{j})$ is a degree-at-most- $n$ cardinal polynomial, hence $ℓ_{j, n}^{L}$ by [F2]. Substitution of $ω_{n} (\cos θ) = - \sin θ \sin (n θ)$ and $∣ ω_{n}^{'} (x_{j}) ∣ = n / c_{j}$ gives the second formula.

Depends on

Used by

Chebyshev-root and Lobatto interpolation norms have logarithmic upper bounds Theorem

Dependency tree · two levels

50 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Lloyd N. Trefethen, Approximation Theory and Approximation Practice, Chs. 2–5 (standard reference, not scraped)
Tobin A. Driscoll and Richard J. Braun, Fundamentals of Numerical Computation, Polynomial interpolation (standard reference, not scraped)