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\vsize=9truein
\hsize=6.5truein
%\parindent=0.8truein
\def\C{{\bf C}}
\def\R{{\bf R}}
\def\N{{\bf N}}
\def\span{{\rm span}}
\font\sc=cmcsc10
\font\smalltt=cmtt10 scaled840
\font\smseven=cmr7 scaled840

%verbatim environment
\def\uncatcodespecials{\def\do##1{\catcode`##1=12 }\dospecials}
\newcount\lineno
%\def\listing#1{\par\begingroup\setupverbatim #1\endgroup}
\def\setupverbatimc{\parindent=1truecm\tt
	\smalltt\baselineskip=10pt plus 1pt minus 1pt
	\def\par{\leavevmode\endgraf}\catcode`\`=\active\obeylines
	\uncatcodespecials\obeyspaces
	\everypar{\global\advance\lineno by 1\llap{\smseven\the\lineno\ \ }}}
\def\setupverbatimb{\parindent=1truecm\global\lineno=0%
	\smalltt\baselineskip=10pt plus 1pt minus 1pt
	\def\par{\leavevmode\endgraf}\catcode`\`=\active\obeylines
	\uncatcodespecials\obeyspaces
	\everypar{\advance\lineno by 1{}}}
\def\setupverbatim{\global\lineno=0\setupverbatimc}

{\obeyspaces\global\let =\ }

\def\beginverbatim{\par\begingroup\setupverbatim\doverbatim}
{\catcode`\|=0\catcode`\\=12\catcode`|`=|active|gdef`{|relax|lq}%
|obeylines|gdef|doverbatim^^M#1\endverbatim{#1|endgroup}}

\def\beginverbatimb{\par\begingroup\setupverbatimb\doverbatim}
\def\listing#1{\medskip
	\begingroup\setupverbatim\input#1\endgroup\medskip
	\noindent\ignorespaces}
\def\listout#1{\medskip
	\begingroup\setupverbatimb\input#1\endgroup\medskip
	\noindent\ignorespaces}
\def\macinput#1{{\catcode`\$=10\catcode`@=3 \input #1 }}

\newcount\qnum\qnum=0
\def\qn{\advance\qnum by1\medskip\par\noindent\item{\the\qnum.}}

\input epsf

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\qn [Kincaid and Cheney Problem 6.8\#6] \ Show that the Hilbert matrix
with elements $a_{ij}=(i+j+1)^{-1}$ for $i,j=0,1,2,\ldots,n-1$ is a Gram
matrix for the functions $1,x,x^2,\ldots,x^{n-1}$.
\medskip\noindent
We define the inner product
$$
	\langle f,g\rangle = \int_0^1 f(x)g(x)\,dx$$
on the space of functions $L^2\big([0,1];\R\big)$ and note that
$$
	1,x,x^2,\ldots,x^{n-1}\in L^2\big([0,1];\R\big).$$
The corresponding Gram matrix has elements
$$
	\langle x^i,x^j\rangle = \int_0^1 x^{i+j}\,dx
	={1\over i+j+1} x^{i+j+1}\Big|_0^1= {1\over i+j+1}
$$
which are the entries of the Hilbert matrix.
\vfill\eject

\qn [Kincaid and Cheney Problem 6.8\#8] \ In the three-term
recurrence relation for the orthogonal polynomials, assume that the
inner product is
$$
	\langle f,g\rangle = \int_{-a}^a f(x)g(x)w(x)dx$$
where $w$ is an even function.  Prove that $a_n=0$ for all $n$.
Prove that $p_n$ is even if $n$ is even and that $p_n$ is
odd if $n$ is odd.
\medskip\noindent
By definition $p_0=1$ which is even.  Moreover
$$
	a_1={\langle xp_0,p_0\rangle\over \langle p_0,p_0\rangle }
		={\int_{-a}^a x\,dx\over \int_{-a}^a 1\,dx}=0
$$
and consequently $p_1=x-a_1=x$ is odd.  It further follows
that
$$
	a_2={\langle xp_1,p_1\rangle\over \langle p_1,p_1\rangle } =
		{\int_{-a}^a x^3\,dx\over \int_{-a}^a x^2\,dx} =0.$$
We now procede by induction which can be
stated as follows:  
\medskip
{\narrower\narrower\noindent
Suppose $p_k$ is even if $k$ is even and
that $p_k$ is odd if $k$ is odd for all $k<n$, then $p_n$ is even
if $n$ is even and $p_n$ is odd if $n$ is odd.\par}
\medskip\noindent
Notice no matter whether $k$ is even or odd that the
function $x\big(p_{k}(x)\big)^2$ is odd.  This can be seen
by the following two equalities:
\medskip\noindent
Case $k$ is even:
$$
	(-x)\big(p_k(-x)\big)^2=-x\big(p_k(x)\big)^2
$$
Case $k$ is odd:
$$
	(-x)\big(p_k(-x)\big)^2=-x\big(-p_k(x)\big)^2=-x\big(p_k(x)\big)^2
$$
\medskip\noindent
Therefore
$$
	a_{n}=
		{\langle xp_{n-1},p_{n-1}\rangle
			\over \langle p_{n-1},p_{n-1}\rangle } =
		{\int_{-a}^a x\big(p_{n-1}(x)\big)^2\,dx
			\over \int_{-a}^a \big(p_{n-1}(x)\big)^2\,dx} =0$$
and consequently
$$
	p_{n}(x)
		=(x-a_n)p_{n-1}(x)-b_np_{n-2}(x)
		xp_{n-1}(x)-b_np_{n-2}(x).
$$
We now consider the case when $n$ is even and $n$ odd separately.
\medskip\noindent
Case $n$ is odd:
By the induction hypothesis $p_{n-1}$ is even and $p_{n-1}$ is odd.
It follows that $xp_{n-1}(x)$ is odd and therefore $p_n(x)$, being the
sum of two odd functions, is again odd.
\medskip\noindent
Case $n$ is even:
By the induction hypothesis $p_{n-1}$ is odd and $p_{n-1}$ is even.
It follows that $xp_{n-1}(x)$ is even and therefore $p_n(x)$, being the
sum of two even functions, is again even.
\medskip\noindent
This completes the induction and the proof.
\vfill\eject

\qn [Kincaid and Cheney Problem 6.8\#21] \ Derive these Legendre
polynomials:
$$\eqalign{
	p_3(x)&=\textstyle x^3-{3\over 5}x\cr
	p_4(x)&=\textstyle x^4-{6\over 7}x^2+{3\over 35}\cr
	p_5(x)&=\textstyle x^5-{10\over 9}x^3+{5\over 21}x\cr}$$
\medskip\noindent
I wrote a Maple script to implement the calculation given as
Theorem 5 in Kincaid and Cheney on page 400.  The script is
\listing{legendre.mpl}
and the output is
\listout{legendre.out}
\vfill\eject

\qn [Kincaid and Cheney Problem 6.9\#2] \ Find the best approximation
of $\sqrt x$ by a first-degree polynomial on the interval $[0,1]$.
\medskip\noindent
In light of Corollary 2 in Kincaid and Cheney page 408 and Example 1
on the preceding page we solve the following system of equations:
$$\eqalign{
	g(0)-f(0)=\delta\cr
	g(\xi)-f(\xi)=-\delta\cr
	g(1)-f(1)=\delta\cr
	g'(\xi)-f'(\xi)=0\cr}$$
where $g(x)=ax+b$ and $f(x)=\sqrt x$.  The Maple
script
\listing{prob692.mpl}
solves these equations.  The best approximation is
$$
	f(x)=x+1/8$$
which has a graph 

\centerline{\epsfxsize=2.5truein\epsfbox{graph692.eps}}

\noindent The Maple output follows:
\listout{prob692.out}
\vfill\eject

\qn [Kincaid and Cheney Problem 6.9\#3] \ Show that the subspaces 
in $C[0,1]$ spanned by these sets are Haar subspaces:
$$
A=\{1,x^2,x^3\},\qquad
B=\{1,e^x,e^{2x}\},\qquad
C=\{(x+2)^{-1},(x+3)^{-1},(x+4)^{-1}\}.$$
\medskip\noindent
Let $a+bx^2+cx^3$ be in the span of $A$.  Claim that this element
has at most two roots in the interval $[-1,1]$.
if $c=0$ then $a+bx^2$ clearly has at most two roots.  If $c\ne 1$
we may consider the polynmial $\alpha+\beta x+x^3$ were
$\alpha=a/c$ and $\beta=b/c$.
Suppose for contradiction there were three distinct roots 
$0\le x_1<x_2<x_3\le 1$ such that
$$\eqalign{
	\alpha+\beta x^2+x^3&=(x-x_1)(x-x_2)(x-x_3)\cr
	&=
	x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_1x_3+x_2x_3)x-x_1x_2x_3.}$$
Equating coefficients we obtain
$0=x_1x_2+x_1x_3+x_2x_3\ge x_2 x_3>0$ which is a contradiction.
\medskip\noindent
Let $a+be^{x}+ce^{2x}$ be in the span of $B$.  Writing $w=e^x$ we
may consider the polynomial
$a+bw+cw^2$ which has at most $2$ roots $w_1$ and $w_2$.  Since
the function $x\to e^x$ is injective on $[0,1]$ there are at most
two numbers $x_1$ and $x_2$ such that $w_1=e^{x_1}$ and $w_2=e^{x_2}$.
It follows that $a+be^x+ce^{2x}=0$ has at most $2$ solutions.
\medskip\noindent
Let $a(x+2)^{-1}+b(x+3)^{-1}+c(x+4)^{-1}$ be in the span of $C$.  
Finding a common denominator shows this expression is equal to
$$
	{a(x+3)(x+4)+b(x+2)(x+4)+c(x+2)(x+3)\over (x+1)(x+2)(x+3)}
	={p(x)\over (x+1)(x+2)(x+3)}
$$
for some polynomial $p(x)$ of degree less than or equal $2$.
Since $p(x)$ has at most $2$ roots than $a(x+2)^{-1}+b(x+3)^{-1}+c(x+4)^{-1}=0$
has at most $2$ solutions.
\vfill\eject

\qn [Kincaid and Cheney Problem 6.9\#4] \ Show that the subspaces 
in $C[-1,1]$ spanned by these sets are Haar subspaces:
$$
A=\{1,x^2,x^3\},\qquad
B=\{|x|,|x-1|\},\qquad
C=\{e^x,x+1\}.$$
\medskip\noindent
We consider again the equation $x_1x_2+x_1x_3+x_2x_3=0$ from part A
of the previous problem and this equation has solutions such that
$-1\le x_1<x_2<x_3\le 1$.  In particular, if $x_2=1/2$ and $x_3=1/3$,
then
$$
	x_1={-x_2x_3\over x_2+x_3}=-{1\over 5}\in [-1,1].
$$
Therefore the span of $A$ is not a Haar subspace.
\medskip\noindent
Consider the function $f(x)=2|x|-|x-1|$ in the span of $B$.  Clearly
$f(1)=2-1=0$ and $f(1/3)=2/3-|-2/3|=0$ therefore $B$ is not a Haar
subspace.
\medskip\noindent
Consider the function $f(x)=1+x-{4\over 5}e^x$.  Since
$$
	\eqalign{
		f(-1)=-{4\over 5}e^{-1}<0\cr
		f(0)=1-{4\over 5}>0\cr
		f(1)=2-{4\over 5}e<4-{4\over 5}(2.7)<0,\cr}
$$
then by the intermediate value property of continuous functions
there must be points $x_1$ and $x_2$ such that
$$-1<x_1<0<x_2<1\qquad\hbox{and}\qquad f(x_1)=f(x_2)=0.$$
Therefore $C$ is not a Haar subspace.
\vfill\eject

\qn [Kincaid and Cheney Problem 6.9\#8] \ Prove the quadratic
polynomial of best approximation to the function $\cosh x$ on
the interval $[-1,1]$ is $a+bx^2$ where $b=\cosh 1 - 1$
and $a$ is obtained by simultaneously solving
for $a$ and $t$ in the system
$$\left\{\eqalign{
	2a&=1+\cosh t - t^2b\cr
	\sinh t&=2 t b.\cr
}\right.$$
\medskip\noindent
Define $f(x)=\cosh x$.
First we claim the quadratic polynomial that best approximates $f$
on the interval $[-1,1]$ is of the form $a+bx^2$.
Let $F=\{a+bx^2: a,b\in\R\}$, $G=\{cx:x\in R\}$ and 
	$P_2=\{a+cx+bx^2:a,b,c\in\R\}$.
Let $a$, $b$ and $c$ be chosen so that
$$
	\big\|f-(a+cx^2+bx^2)\big\|=\min \big\{ \|f-g\|:g\in P_2\big\}.
$$
Define $f_2(x)=\cosh x -(a+bx^2)$.  Since $f_2(x)$ is even
then $x\in{\rm crit}(f_2)$ implies $-x\in{\rm crit}(f_2)$
and moreover $f_2(x)$ and $f_2(-x)$ have the same sign.
Thus, there is no function in $G$ that has
the same signs as $f_2$ on ${\rm crit}(f_2)$.
It follows that Kolmogorov's Characterization
Theorem from Kincaid and Cheney page 407 implies 
$\|f_2\|={\rm dist}(f_2,G)$.
In particular
$$
	\big\|f_2-(a+cx^2+bx^2)\big\|={\rm dist}(f_2,G)=\|f_2\|
$$
and therefore there is a function of the form $a+bx^2$ such
that
$$
	\big\|f-(a+bx^2)\big\|=\min \big\{ \|f-g\|:g\in P_2\big\}.
$$
Now make the change of variables $y=\sqrt x$.  To obtain the
following equivalent minimization problem:  Find 
$a+by$ that 
best approximates the function $\cosh \sqrt y$ on the
interval $[0,1]$.
\medskip\noindent
This problem is in the form covered by Corollary 2 from Kincaid
and Cheney page 408 so we may obtain the solution by solving the
system of equations:
$$\eqalign{
    g_3(0)-f_3(0)&=\delta\cr
    g_3(\xi)-f_3(\xi)&=-\delta\cr
    g_3(1)-f_3(1)&=\delta\cr
    g_3'(\xi)-f_3'(\xi)&=0\cr}$$
where $g_3(y)=a+by$ and $f_3(y)=\cosh \sqrt y$.  Simplifying obtains
$$\eqalign{
    a-1&=\delta\cr
    a+b\xi-\cosh\sqrt\xi&=-\delta\cr
    a+b-\cosh 1&=\delta\cr
    2b \sqrt\xi-\sinh\sqrt \xi&=0\cr}$$
Elimination of $\delta$ from the 2nd and 3rd equation, setting
$t=\sqrt \xi$ and futher simplification obtains the desired result
$$\eqalign{
	2a&=1+\cosh t-t^2b\cr
	b&=\cosh 1-1\cr
	\sinh t&=2b t.\cr}
$$
\vfill
\end
