%%  chebyshev.tex
%%  Approximation Theory: 
%%  Exercises on Chebyshev Polynomials
%%  Neal L. Carothers
%%  Bowling Green State University
%%  Bowling Green, Ohio  43403
%%  carother@math.bgsu.edu
%%  http://www.bgsu.edu/~carother/


\input 680-setup.tex

\chaptertitle = {Chebyshev Polynomials}

\centerline{\hfil\tf Chebyshev Polynomials\hfil}
\vskip-\baselineskip
\line{\sc Math 680 \hfil 7/11/94}

\noindent
We've shown that $\cos n\theta$ and $\sin(n+1)\theta/\sin\theta$
can be written as algebraic
polynomials of degree $n$ in $\cos\theta$; we use this
observation to define the {\sl Chebyshev polynomials}.
The Chebyshev polynomials of the {\sl first kind},  
$(T_n(x))$, are defined by  $T_n(\cos\theta)=\cos n\theta$, 
for $n=0,1,2,\ldots$, while 
the Chebyshev polynomials of the {\sl second kind}, $(U_n(x))$, 
are defined by $U_n(\cos\theta)=\sin(n+1)\theta/\sin\theta$
for $n=0,1,2,\ldots$.

{\baselineskip=24 true pt
\prob{\cmark}{%
Establish the following properties of $T_n(x)$.
\itemitem{(i)}
$T_0(x) = 1$, $T_1(x) = x$, and 
$T_n(x) = 2xT_{n-1}(x)-T_{n-2}(x)$ for $n\ge 2$.
\itemitem{(ii)}
$T_n(x)$  is a polynomial of degree $n$ having
leading coefficient  $2^{n-1}$ for  $n \ge 1$, 
and containing only even (resp., odd) powers of 
$x$  if  $n$  is even (resp., odd).
\itemitem{(iii)}
$|T_n(x)| \le 1$  for  $-1 \le x \le 1$; when
does equality occur?  Where are the zeros of 
$T_n(x)$\/?  Show that between two consecutive
zeros of  $T_n(x)$  there is exactly one zero
of  $T_{n-1}(x)$.  Can  $T_n(x)$  and 
$T_{n-1}(x)$ have a common zero?  
\itemitem{(iv)}
$|T_n'(x)|\le n^2$ for $-1\le x\le1$, and $|T_n'(\pm 1)|=n^2$.
\itemitem{(v)}
$T_m(x)+T_n(x)={1\over2}\bigl[T_{m+n}(x)+T_{m-n}(x)\bigr]$ for $m>n$.
\itemitem{(vi)}
$T_m(T_n(x)) = T_{mn}(x)$.
\itemitem{(vii)}
Evaluate 
$\displaystyle\int_{-1}^1T_n(x)\,T_m(x)\,{{dx}\over{\sqrt{1-x^2}}}$\/.
\itemitem{(viii)}
Show that  $T_n$  is a solution to 
$(1-x^2)y{''} - xy' + n^2y = 0$.
\itemitem{(ix)}
$T_n(x) = {1\over2}\bigl[(x+\sqrt{x^2-1}\,)^n+(x-\sqrt{x^2-1}\,)^n\bigr]$  
for any  $x$, real or complex.
\itemitem{(x)}
${\rm Re}\left(\sum_{n=0}^\infty t^ne^{in\theta}\right)
	= \sum_{n=0}^\infty t^n\cos n\theta = 
	\displaystyle{{1-t\cos\theta}\over{1-2t\cos\theta +t^2}}$\ 
for  $-1 < t < 1$; that is,
$\sum_{n=0}^\infty t^nT_n(x)= 
	\displaystyle{{1-tx}\over{1-2tx +t^2}}$\ 
(this is a {\sl generating function\/} for  $T_n$; 
it's closely related to the Poisson kernel).
\itemitem{(xi)}
Find analogues of  (i)--(x)  (if possible) for  $U_n(x)$.
}\par}

\prob{\cmark}{%
Show that every  $p \in {\cal P}_n$  has a unique
representation as  $p=a_0 + a_1T_1 +\cdots + a_nT_n$.  Find
this representation in the case  $p(x) = x^n$.
}

\prob{\cmark}{%
The polynomial of degree  $n$  having leading
coefficient  $1$  and deviating least from  $0$  on 
$[-1,1\,]$  is given by  $T_n(x)/2^{n-1}$.  On an
arbitrary interval $[\,a,b\,]$ we would instead take 
$$\displaystyle{{{(b-a)^n}\over{2^{2n-1}}}\ 
	T_n\!\left({{2x-b-a}\over{b-a}}\right)}.$$
Is this solution unique?  Explain.
}

\prob{}{%
If  $p$  is a polynomial on  $[\,a,b\,]$  of
degree  $n$  having leading coefficient  $a_n > 0$,
then  $\Vert�p\Vert\ge a_n(b-a)^n/2^{2n-1}$.  
If  $b - a \ge 4$, then no polynomial of degree exactly  
$n$  with {\sl integer\/} coefficients can satisfy  
$\Vert p\Vert< 2$  (compare this with problem 9 on the
``Uniform Approximation by Polynomials'' handout).
}

\prob{}{%
Given $p\in{\cal P}_n$, show that 
$|p(x)|\le\Vert p\Vert\,|T_n(x)|$ for $|x|>1$.
}

\prob{}{%
If $p\in{\cal P}_n$ with $\Vert p\Vert=1$ on $[-1,1\,]$,
and if $|p(x_i)|=1$ at $n+1$ distinct point $x_0,\ldots,x_n$
in $[-1,1\,]$, show that either $p=\pm1$, or else
$p=\pm T_n$.  [Hint: One approach is to compare the polynomials 
$1-p^2$ and $(1-x^2)(p\,')^2$.]
}

\prob{}{%
Compute $T_n^{(k)}(1)$ for $k=0,1,\ldots,n$, where $T_n^{(k)}$
is the $k$-th derivative of $T_n$.
For $x\ge1$ and $k=0,1,\ldots,n$, show that $T_n^{(k)}(x)>0$.
}



\bye


%%  end of chebyshev.tex

