# Chebyshev Polynomials

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 $$\begin{cases} T_0(x) = 1\\ T_1(x) = x \\T_n(x) = 2x T_{n−1}(x) − T_{n−2}(x), & \text{for } n \geq 2\end{cases}$$

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?

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Show that $\cos (n \arccos (x))$ satisfies the recursive definition. –  Qiaochu Yuan Sep 10 '12 at 1:43
So do I set it equal to the recursive definition? –  Samuel Gregory Sep 10 '12 at 1:47
No, you show that it satisfies the recursive definition. In other words, set $P_n(x) = \cos (n \arccos x)$ and show that $P_0(x) = 1, P_1(x) = x, P_n(x) = 2x P_{n-1}(x) - P_{n-2}(x)$. By induction, it then follows that $T_n(x) = P_n(x)$. (You already need induction to show that a recursive definition is actually a definition of anything so this should not be too surprising.) –  Qiaochu Yuan Sep 10 '12 at 1:48
Should that be P(n-2)? –  Samuel Gregory Sep 10 '12 at 1:53
Yes, sorry about that. –  Qiaochu Yuan Sep 10 '12 at 1:54

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.)

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I was able to follow, thank you! –  Samuel Gregory Sep 10 '12 at 2:52
Shouldnt it be Tn(theta) and not cos(theta)? –  Samuel Gregory Sep 10 '12 at 3:13
@SamuelGregory : No, the argument above won't work in that case. Just let $x=\cos\theta$, so $\theta=\arccos x$ and $T_n(x)=T_n(\cos\theta)$. –  Michael Hardy Sep 10 '12 at 17:11