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\begin{center}
\textbf{Extra credit homework 7, due November 3} \\
\end{center}

I've included solutions for each part.

Let n be a non-negative integer.  The $n$th \emph{Chebyshev
polynomial} $T_{n}(x)$ is defined to be
\[
T_{n}(x) = \cos (n \arccos x).
\]

(a) What are the domain and range of this function?

\textbf{Solution:} Since $T_{n}(x)$ is the cosine of something, then
its range is $[-1,1]$.  Since you need to take the arccosine of $x$ to
define $T_{n}(x)$, then the only allowed values for $x$ are $[-1,1]$,
so this is the domain as well.

(b) Find simpler forms for $T_{0}(x)$, $T_{1}(x)$, $T_{2}(x)$, and
$T_{3}(x)$.

\textbf{Solution:}
I need some trig identities:
\begin{itemize}
\item [(1)] $\cos (y+z) = \cos y \cos z - \sin y \sin z$
\item [(2)] $\cos (2y) = 2 \cos^{2} y - 1$
\item [(3)] $\sin (y+z) = \sin y \cos z + \sin z \cos y$
\item [(4)] $\sin (2y) = 2 \sin y \cos y$
\item [(5)] $\sin y \sin z = 1/2 (\cos (y-z) - \cos (y+z))$
\item [(6)] $\sin y \cos z = 1/2 (\sin (y-z) + \sin (y+z))$
\end{itemize}

So we get:
\begin{align*}
T_{0} (x) &= \cos (0) = 1, \\
T_{1} (x) &= \cos (\arccos x) = x, \\
T_{2} (x) &= \cos (2 \arccos x) = 2 \cos^{2} (\arccos
x) - 1 = 2x^{2} - 1, \\
T_{3}(x) &= \cos (3 \arccos x) \\
     &= \cos (\arccos x) \cos (2 \arccos x) - \sin (\arccos x) \sin (2 \arccos x)\\
     &= x (2x^{2} - 1) - \sin (\arccos x) 2 \sin (\arccos x) \cos (\arccos x)\\
     &= 2x^{3} - x - 2 \sin^{2} (\arccos x) \cos (\arccos x)\\
     &= 2x^{3} - x - 2 (1 - \cos^{2} (\arccos x)) \cos (\arccos x)\\
     &= 2x^{3} - x - 2 (1 - x^{2}) x\\
     &= 4x^{3} - 3x.
\end{align*}

(c) Show that, when $n$ is at least 1, then
\[
T_{n+1}(x) = 2x T_{n}(x) - T_{n-1}(x).
\]

\textbf{Solution:}
Let $\theta = \arccos x$.
\begin{align*}
T_{n+1}(x) & = \cos((n+1) \theta) \\
     & = \cos (\theta + n\theta) \\ 
     & = \cos \theta \cos n\theta - \sin \theta \sin n\theta \\
         & \hspace{1.5in} \text{(using trig identity 1)}  \\
     & = x T_{n}(x) - \frac{1}{2} (\cos ((n-1) \theta) - \cos((n+1) \theta) \\
         & \hspace{1.5in} \text{(using trig identity 5)}  \\
     & = x T_{n}(x) - \frac{1}{2} (T_{n-1}(x) - T_{n+1}(x)) \\ 
     & = x T_{n}(x) - \frac{1}{2} T_{n-1}(x) + \frac{1}{2} T_{n+1}(x).
\end{align*}
Now solve for $T_{n+1}(x)$.

(d) Use part (c) to show that $T_{n}(x)$ is a polynomial of
degree $n$.

\textbf{Solution:} Well, by part (b), we know that this is correct
when $n$ is 0, 1, 2, or 3.  If we know that $T_{n}(x)$ is a polynomial
of degree $n$, and if $T_{n-1}(x)$ is a polynomial of degree $n-1$,
then by part (c), we can see that $T_{n+1}(x)$ is a polynomial of
degree $n+1$.  Starting with $n=3$, we see that $T_{4}(x)$ is a
polynomial of degree 4, $T_{5}(x)$ is a polynomial of degree 5, etc.
(By "etc.", I really mean that you should prove this by induction.)

(e) Use (b) and (c) to express $T_{4}$, $T_{5}$, and
$T_{6}$ as polynomials.

\textbf{Solution:} To get $T_{4}$, we apply (c) with
$n=3$:
\begin{align*}
T_{4}(x) & = 2x T_{3}(x) - T_{2}(x) \\ 
      & = 2x (4x^{3} - 3x) - (2x^{2} - 1) \\ 
      & = 8x^{4} - 8x^{2} - 1.
\end{align*}
Similarly,
\begin{gather*}
T_{5}(x) = 16x^{5} - 20x^{3} + 5x, \\
T_{6}(x) = 32x^{6} - 48x^{4} + 18x^{2} - 1.
\end{gather*}

(f) What are the zeros of $T_{n}$?  Where are its local maxima and
minima?

\textbf{Solution:} $T_{n} (x) = \cos (n \arccos x)$, and $\cos
(\theta) = 0$ when $\theta=\pm \pi/2$, $\pm 3\pi/2$, $\pm 5\pi/2$,
\dots.  So $T_{n} (x) = 0$ when $n \arccos x = \pm \pi/2$, $\pm
3\pi/2$, $\pm 5\pi/2$, \dots.  In other words, $T_{n} (x) = 0$ when
\begin{gather*}
  \arccos x = \pm \pi/2n, \pm 3\pi/2n, \pm 5\pi/2n, \dots,
     \hspace{0.5in} \text{or} \\
  x = \cos(\pi/2n), \cos(3\pi/2n), \cos(5\pi/2n), \dots 
\end{gather*}
Since $T_{n} (x)$ is a polynomial of degree $n$, then it should only
have $n$ zeros; in fact, the terms in my list of zeros start to repeat
after a while, so the zeros are actually:
\[
  x = \cos(\pi/2n), \cos(3\pi/2n), \cos(5\pi/2n), ..., \cos((2n-1)\pi/2n).
\]

To find the local maxima and minima, we differentiate $T_{n}(x)$ and
set the derivative equal to zero:
\[
  T_{n}'(x) = \frac{n \sin (n \arccos x)} {\sqrt{1-x^{2}}} = 0.
\]
This is zero only when $\sin (n \arccos x) = 0$, which happens when 
\begin{gather*}
  n \arccos x = 0, \pi, 2\pi, 3\pi, \dots, \hspace{0.5in} \text{or} \\
  \arccos x = 0, \frac{\pi}{n}, \frac{2\pi}{n}, \frac{3\pi}{n}...,  \hspace{0.5in} \text{or} \\
  x = \cos(0)=1, \cos(\frac{\pi}{n}), \cos(\frac{2\pi}{n}), ...,
\cos(\frac{(n-1)\pi}{n}),
\cos(\frac{n\pi}{n}) = -1.
\end{gather*}

(g) What is the integral of $T_{n}(x)$ with respect to $x$, as $x$
goes from $-1$ to 1?

\textbf{Solution:} 
Let $u = \arccos x$, so $\cos u = x$, so $-\sin u \ du = dx$.
\begin{align*}
\int_{-1}^{1} \cos(n \arccos x) dx & = - \int_{\pi}^{0} \cos (nu) \sin u
\ du \\
   & = \frac{1}{2} \int_{0}^{\pi} (\sin((1-n) u) + \sin((1+n) u)) du \\
   & \hspace{2in}  \text{(by trig identity 6)} \\
   & = \frac{1}{2} \left(\frac{1}{n-1} \cos((n-1) u) - \frac{1}{n+1}
\cos((n+1) u) \right)_{0}^{\pi}\\
   & = 
\begin{cases}
\frac{1}{2} \left(-\frac{2}{n-1} + \frac{2}{n+1} \right), & \text{$n$
even}, \\ 
0, & \text{$n$ odd}
\end{cases} \\
   & = 
\begin{cases}
-\frac{2}{n^{2}-1}, & \text{$n$ even}, \\ 
0, & \text{$n$ odd}.
\end{cases}
\end{align*}

(h) You can define Chebyshev polynomials $T_{a}$ for any number $a$,
not just a non-negative integer.  What can you tell me about $T_{a}$
in general?  What about $T_{a}$ for specific values of $a$ (for
instance, $a=\frac{1}{2}$ or $a=\frac{1}{3}$)?

\textbf{Solution:} This is an open-ended question; I wasn't looking
for anything in particular.  For example, though, $T_{\frac{1}{2}}(x)
= \cos (\frac{1}{2} \arccos x)$, and you might try to simplify that
using a half-angle formula.

(i) What are these Chebyshev polynomials good for, anyway?

\textbf{Solution:} If you want to know, then go to the library
and do some research.  When you get to the library, you may need to
know how to spell Chebyshev: 

For what it's worth, ``Chebyshev'' is sometimes transliterated as
``Tschebyscheff'' or ``\u{C}ebysev'' or in various other ways.  See
\emph{The Thread} by Philip Davis for more information (on the
transliteration problem).
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