I am trying to prove a something regarding Chebyshev polynomials. Given the polynomials $T_n(x), n = 0, 1, \ldots$ which are recursively defined by
$$\begin{cases} T_0(x) = 1\\ T_1(x) = x \\T_n(x) = 2x T_{nā1}(x) ā T_{nā2}(x), & \text{for } n \geq 2\end{cases}$$
I want to show that
For every $n$, $$T_n(x) = \cos(n \arccos(x))$$
Is there some type of proof I could use or is just plugging in values the only way?
Best Answer
Plugging in values will only prove finitely many instances.
This is a sequence of trigonometric identities. Since it's a definition by recursion, you do the proof by mathematical induction. It's obviously true if $n=0$ or $1$. So suppose it's true in the first $n$ cases, so you know $$ T_n(\cos\theta) = \cos(n\theta). $$ and similarly for $n-1$.
You want to prove $$ T_{n+1}(\cos\theta) = \cos((n+1)\theta). $$
So write $$ \begin{align} T_{n+1}(\cos\theta) & = 2(\cos\theta) T_n(\cos\theta) - T_{n-1}(\cos\theta) \\[8pt] & = 2(\cos\theta)(\cos(n\theta)) - \cos((n-1)\theta) \\[8pt] & = 2(\cos\theta)(\cos(n\theta)) - \Big( \cos(n\theta)\cos\theta + \sin(n\theta)\sin\theta \Big) \\[8pt] & = (\cos\theta)(\cos(n\theta)) -\sin(n\theta)\sin\theta \\[8pt] & = \cos((n+1)\theta). \end{align} $$
(Of course, you have to remember some basic trigonometric identities to follow this.)