[[concept]]Theorem
Let
The Chebyshev Polynomials are orthogonal with respect to the inner product .
Proof
\int_{-1}^1 T_{n}(x)T_{m}(x) \frac{1}{\sqrt{ 1-x^2 }} \, dx &\implies x=\cos \theta,\;\;\; \frac{dx}{d\theta}=\sin \theta\\ &= \int_{\pi}^2\pi T_{n}(\cos\theta) T_{m}(\cos\theta) \frac{d\theta \sin\theta}{\sqrt{ 1-\cos^2\theta }} \\ &= \int_{\pi}^2\pi \cos(n\theta) \cos(m\theta) d\theta \;\; \text{ since } T_{n} = \cos(n\theta) \\ &= \int_{\pi}^2\pi \frac{e^{i n\theta} + e^{- i n \theta}}{2} + \frac{e^{i m\theta} + e^{- i m \theta}}{2} \, d\theta \\ &= \frac{1}{4} \int_{\pi}^{2\pi} e^{i n\theta} + e^{- i n \theta} + e^{i m\theta} + e^{- i m \theta}\, d\theta \\ &= \begin{cases} \frac{1}{4} \int_{\pi}^{2\pi} 4 d\theta = \theta |_{\pi}^{2\pi} = \pi \;\;\;\text{ if } n=m=0\\ \frac{1}{4} \int_{\pi}^{2\pi} 2 d\theta + \frac{1}{4}\int_{\pi}^{2\pi}2 \cos((n+m)\theta) d\theta = \frac{\pi}{2} + 0 = \frac{\pi}{2}\;\;\; \text{ if } n=m\neq 0 \\ \frac{1}{4} \int_{\pi}^{2\pi}2\cos((n+m)\theta) + 2\cos((n-m)\theta)d\theta = 0\;\; \; \text{ if } n \neq m \end{cases} \end{aligned}$$ ^proof
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