Chebychev polynomial of $ (-x)$

Question

I know that the Chebyshev polynomials are defined as $T_n(x)=\cos(nx)$

How can I proove the following result:

$T_n(-x)= (-1)^n T_n(x)$

Answer 1

$$T_{n-1}(\cos(t))+T_{n+1}(\cos(t))=$$

$$\cos((n-1)t)+\cos((n+1)t)=$$

$$2\cos(nt)\cos(t)$$

Thus

$$T_{n-1}(x)+T_{n+1}(x)=2xT_n(x)$$

From here, you can prove it by induction.

Answer 2

Use induction:

$T_0(-x) = 1 = T_0(x)$, $T_1(-x) = -x = - T_1(x)$.

For $n \ge 1$, $T_{n+1}(x) = 2x T_n(x) - T_{n-1}(x)$, so if $n$ is even $T_{n+1}(-x) = -2x T_n(-x) - T_{n-1}(-x)=- (2x T_n(x) - T_{n-1}(x)) = -T_{n+1}(x)$ and similarly, if $n$ is odd, $T_{n+1}(-x) = -2x T_n(-x) - T_{n-1}(-x)=2x T_n(x) - T_{n-1}(x) = T_{n+1}(x)$.

Answer 3

Write $x = \cos t$. Then $T_n(x) = \cos nt$. Also, the equation you want to show ($T_n(-x)=(-1)^nT_n(x)$) is equivalent to $T_n(-\cos t)=(-1)^n \cos nt$, so this is what we show.

We have $$\begin{align*} T_n(-\cos t) &= T_n (\cos(t+\pi)) \tag{since $\cos(t+\pi)=-\cos t$} \\ &= \cos (n(t+\pi)) \tag{since $T_n(\cos \theta) = \cos n\theta$ for all $\theta$} \\ &= \cos(nt+n\pi)\\ &= \cos (nt)\cos(n\pi) - \sin(nt)\sin(n\pi) \tag{using $\cos(A+B)=\cos A\cos B -\sin A \sin B$} \\ &= (-1)^n\cos (nt), \tag{since $\cos (n\pi)=(-1)^n$ and $\sin(n\pi)=0$} \end{align*} $$ as required.