Existence of Chebyshev Polynomials of the Second Kind
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Theorem
There exists a Chebyshev polynomial of the second kind for all natural numbers $n$.
Proof
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For $n = 0$:
\(\ds \map \sin { \paren {0+1} \theta}\) | \(=\) | \(\ds \map \sin \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \cdot \sin \theta\) |
- $\map {U_0} x = 1$, $U_0 \in \Bbb P$
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For $n = 1$:
\(\ds \map \sin { \paren {1+1} \theta}\) | \(=\) | \(\ds \map \sin {2 \theta}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \sin \theta \cos \theta\) | Double Angle Formula for Sine | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \cos \theta \cdot \sin \theta\) |
- $\map {U_1} x = 2x$, $U_1 \in \Bbb P$
Assume $\map {U_n} {\cos \theta}\sin \theta = \map \sin { \paren {n+1} \theta }$.
\(\ds \map {U_{n+1} } {\cos \theta} \sin \theta\) | \(=\) | \(\ds \map \sin { \paren {n+1+1} \theta }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \sin { \paren {n+1} \theta + \theta }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \sin { \paren {n+1} \theta} \map \cos \theta + \map \cos { \paren {n+1} \theta } \map \sin \theta\) | Sine of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \map \sin { \paren {n+1} \theta} \map \cos \theta + \map \cos { \paren {n+1} \theta } \map \sin \theta - \map \sin { \paren {n+1} \theta} \map \cos \theta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \map \sin { \paren {n+1} \theta} \map \cos \theta - \paren { \map \sin { \paren {n+1} \theta} \map \cos \theta - \map \cos { \paren {n+1} \theta } \map \sin \theta }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \map \sin { \paren {n+1} \theta} \map \cos \theta - \map \sin { \paren {n+1} \theta - \theta }\) | Sine of Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \map \cos \theta \map \sin { \paren {n+1} \theta} - \map \sin { \paren {n-1+1} \theta }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \map \cos \theta \map {U_n} {\cos \theta} \map \sin \theta - \map {U_{n-1} } {\cos \theta} \sin \theta\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren { 2 \map \cos \theta \map {U_n} {\cos \theta} - \map {U_{n-1} } {\cos \theta} } \cdot \map \sin \theta\) |
- $\map {U_{n+1} } x = 2x \map {U_n} x - \map {U_{n - 1} } x$, $U_{n+1} \in \Bbb P$
By the Second Principle of Mathematical Induction, $U_n \in \Bbb P$ for all natural numbers $n$.
$\blacksquare$