Existence of Chebyshev Polynomials of the First Kind
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Theorem
There exists a Chebyshev polynomial of the first kind for all natural numbers $n$.
Proof
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For $n = 0$:
| \(\ds \map \cos {0 \theta}\) | \(=\) | \(\ds \map \cos 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
- $\map {T_0} x = 1$, $T_0 \in \Bbb P$
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For $n = 1$:
| \(\ds \map \cos {1 \theta}\) | \(=\) | \(\ds \map \cos \theta\) |
- $\map {T_1} x = x$, $T_1 \in \Bbb P$
Assume $\map {T_n} {\cos \theta} = \map \cos {n \theta}$.
| \(\ds \map {T_{n+1} } {\cos \theta}\) | \(=\) | \(\ds \map \cos { \paren {n+1} \theta}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \cos { n \theta + \theta}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \cos { n \theta} \map \cos \theta - \map \sin { n \theta} \map \sin \theta\) | Cosine of Sum | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \map \cos { n \theta} \map \cos \theta - \map \cos { n \theta} \map \cos \theta - \map \sin { n \theta} \map \sin \theta\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \map \cos { n \theta} \map \cos \theta - \paren { \map \cos { n \theta} \map \cos \theta + \map \sin { n \theta} \map \sin \theta }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \map \cos { n \theta} \map \cos \theta - \map \cos { n \theta - \theta }\) | Cosine of Difference | |||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \map \cos \theta \map \cos { n \theta} - \map \cos { \paren {n-1} \theta }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2 \map \cos \theta \map {T_n} {\cos \theta} - \map {T_{n-1} } {\cos \theta }\) | by hypothesis |
- $\map {T_{n+1} } x = 2x \map {T_n} x - \map {T_{n - 1} } x$, $T_{n+1} \in \Bbb P$
By the Second Principle of Mathematical Induction, $T_n \in \Bbb P$ for all natural numbers $n$.
$\blacksquare$