# Existence of Chebyshev Polynomials of the First Kind

## Theorem

There exists a Chebyshev polynomial of the first kind for all natural numbers $n$.

## Proof

For $n = 0$:

 $\ds \map \cos {0 \theta}$ $=$ $\ds \map \cos 0$ $\ds$ $=$ $\ds 1$
$\map {T_0} x = 1$, $T_0 \in \Bbb P$

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 $n$.

$\blacksquare$