chebyshev polynomials are orthogonal

[[concept]]

Theorem

Let $⟨ f, g ⟩ = \int_{- 1}^{1} f (x) g (x) \frac{1}{\sqrt{1 - x^{2}}} d x$

The Chebyshev Polynomials $T_{0}, T_{1}, \dots$ are orthogonal with respect to the inner product $⟨ \cdot, \cdot ⟩$ .

Proof

$\begin{aligned} \int_{- 1}^{1} T_{n} (x) T_{m} (x) \frac{1}{\sqrt{1 - x^{2}}} d x & ⟹ x = \cos θ, \frac{d x}{d θ} = \sin θ \\ = \int_{π}^{2} π T_{n} (\cos θ) T_{m} (\cos θ) \frac{d θ \sin θ}{\sqrt{1 - \cos^{2} θ}} \\ = \int_{π}^{2} π \cos (n θ) \cos (m θ) d θ since T_{n} = \cos (n θ) \\ = \int_{π}^{2} π \frac{e^{i n θ} + e^{- i n θ}}{2} + \frac{e^{i m θ} + e^{- i m θ}}{2} d θ \\ = \frac{1}{4} \int_{π}^{2 π} e^{i n θ} + e^{- i n θ} + e^{i m θ} + e^{- i m θ} d θ \\ = {\begin{cases} \frac{1}{4} \int_{π}^{2 π} 4 d θ = θ |_{π}^{2 π} = π if n = m = 0 \\ \frac{1}{4} \int_{π}^{2 π} 2 d θ + \frac{1}{4} \int_{π}^{2 π} 2 \cos ((n + m) θ) d θ = \frac{π}{2} + 0 = \frac{π}{2} if n = m \neq 0 \\ \frac{1}{4} \int_{π}^{2 π} 2 \cos ((n + m) θ) + 2 \cos ((n - m) θ) d θ = 0 if n \neq m \end{cases} \end{aligned}$

Mentions

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2025-02-17 graphs lecture 8