Chebyshev Polynomials

Question

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?

Show that $\cos (n \arccos (x))$ satisfies the recursive definition. — Qiaochu Yuan, Sep 10 '12 at 01:43
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 01:48
I am getting: 2xcos(narccosx-arccosc) - cos(narcosx - 2arccosx), is there an angle property I need to use? — Samuel Gregory, Sep 10 '12 at 02:01
You can solve Chebyshev's equation using the change of variables: $x=\cos t$ — jinawee, Mar 29 '14 at 20:48

score 9 · Accepted Answer · answered Sep 10 '12 at 02:27

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.)

@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

user2662833 · Answer 2 · 2018-01-04T01:56:51.180

I had to answer this question even though its years late :P

$$ \cos\left((n+1)\theta\right) = \cos(n\theta + \theta) = \cos(n\theta)\cos(\theta) - \sin(n\theta) \sin(\theta) \\ \cos\left((n-1)\theta\right) = \cos(n\theta - \theta) = \cos(n\theta)\cos(\theta) + \sin(n\theta) \sin(\theta) $$

Now if you add the left and right sides of this equation you get:

$$ \cos((n+1)\theta) + \cos((n-1)\theta) = 2 \cos(n\theta) \cos(\theta) $$

If you let $x = \cos(\theta)$ and $T_n(x) = \cos(n\theta)$, you get:

$$ T_{n+1} + T_{n-1} = 2 x T_{n} $$

So we have:

$$ T_{n+1} = 2 x T_{n} -T_{n-1} $$

Which is equivalent to what you asked :)

Chebyshev Polynomials

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