Find the zeros, critical points, and critical values of the Chebyshev polynomial \(T_n(z)\) for \(n\ge1\).
Solution
Recall that \(T_n\) is the degree-\(n\) polynomial satisfying \[ T_n(\cos\theta)=\cos(n\theta). \]
1. The zeros of \(T_n\) are\[ \boxed{ \cos\frac{(2k-1)\pi}{2n}, \qquad k=1,\ldots,n.} \]
For \[ \theta_k=\frac{(2k-1)\pi}{2n}, \] we have \[ T_n(\cos\theta_k) =\cos(n\theta_k) =\cos\frac{(2k-1)\pi}{2} =0. \] The numbers \(\theta_k\) lie strictly between \(0\) and \(\pi\) and are distinct, so their cosines are distinct. Since \(T_n\) has degree \(n\), these are all of its zeros.
2. For \(0<\theta<\pi\),\[ T_n'(\cos\theta) =n\frac{\sin(n\theta)}{\sin\theta}. \]
Differentiate \[ T_n(\cos\theta)=\cos(n\theta) \] with respect to \(\theta\). This gives \[ -\sin\theta\,T_n'(\cos\theta) =-n\sin(n\theta). \] Since \(\sin\theta\neq0\) on \((0,\pi)\), division by \(\sin\theta\) yields the formula.
3. For \(n\ge2\), the critical points are\[ \boxed{ \cos\frac{k\pi}{n}, \qquad k=1,\ldots,n-1.} \]For \(n=1\) there are no critical points.
By 2, if \[ \theta_k=\frac{k\pi}{n}, \qquad 1\le k\le n-1, \] then \[ T_n'(\cos\theta_k) =n\frac{\sin(k\pi)}{\sin(k\pi/n)} =0. \] These \(n-1\) points are distinct because cosine is strictly decreasing on \([0,\pi]\). The derivative \(T_n'\) has degree \(n-1\), so these are all of its zeros. When \(n=1\), \(T_1(z)=z\) and \(T_1'(z)=1\), so there are none.
4. At the critical point \(\cos(k\pi/n)\),\[ T_n\left(\cos\frac{k\pi}{n}\right)=(-1)^k. \]
Using the defining cosine identity, \[ T_n\left(\cos\frac{k\pi}{n}\right) =\cos(k\pi) =(-1)^k. \]
5. Hence the critical values are\[ \boxed{\{-1,1\}} \]for \(n\ge3\), while for \(n=2\) the only critical value is \(-1\), and for\(n=1\) there are no critical values.
For \(n\ge3\), the indices \(k=1,2\) already produce the two values \(-1\) and \(1\), and 4 shows that no other value occurs. For \(n=2\), only \(k=1\) is available, giving \(-1\). The case \(n=1\) follows from 3.