1. [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, …, n − 1 is a Gram matrix for the functions 1, x, x^{2}, …, x^{n−1}.

We define the inner product

⟨f, g⟩ = ∫_{0}^{1} f(x)g(x) dx

on the space of functions L^{2}([0, 1]; R) and note that

1, x, x^{2}, …, x^{n−1} ∈ L^{2}([0, 1]; R).

The corresponding Gram matrix has elements

⟨x^{i}, x^{j}⟩ = ∫_{0}^{1} x^{i+j} dx = \frac{1}{i + j + 1}x^{i+j+1}|_{0}^{1} = \frac{1}{i + j + 1}

which are the entries of the Hilbert matrix.

2. [Kincaid and Cheney Problem 6.8#8] In the three-term recurrence relation for the orthogonal polynomials, assume that the inner product is

⟨f, g⟩ = ∫_{−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.

By definition p_{0} = 1 which is even. Moreover

a_{1} = \frac{⟨xp_{0}, p_{0}⟩}{⟨p_{0}, p_{0}⟩} = \frac{∫_{−a}^{a} x dx}{∫_{−a}^{a} 1 dx} = 0

and consequently p_{1} = x − a_{1} = x is odd. It further follows that

a_{2} = \frac{⟨xp_{1}, p_{1}⟩}{⟨p_{1}, p_{1}⟩} = \frac{∫_{−a}^{a} x^{3} dx}{∫_{−a}^{a} x^{2} dx} = 0.

We now procede by induction which can be stated as follows:

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.

Notice no matter whether k is even or odd that the function x(p_{k}(x))^{2} is odd. This can be seen by the following two equalities:

Case k is even:

(−x)(p_{k}(−x))^{2} = −x(p_{k}(x))^{2}

Case k is odd:

(−x)(p_{k}(−x))^{2} = −x( − p_{k}(x))^{2} = −x(p_{k}(x))^{2}

Therefore

a_{n} = \frac{⟨xp_{n−1}, p_{n−1}⟩}{⟨p_{n−1}, p_{n−1}⟩} = \frac{∫_{−a}^{a} x(p_{n−1}(x))^{2} dx}{∫_{−a}^{a} (p_{n−1}(x))^{2} dx} = 0

and consequently

p_{n}(x) = (x − a_{n})p_{n−1}(x) − b_{n}p_{n−2}(x)xp_{n−1}(x) − b_{n}p_{n−2}(x).

We now consider the case when n is even and n odd separately.

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.

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.

This completes the induction and the proof.

3. [Kincaid and Cheney Problem 6.8#21] Derive these Legendre polynomials:

\begin{aligned}p_{3}(x) & = x^{3} − \frac{3}{5}x \\ p_{4}(x) & = x^{4} − \frac{6}{7}x^{2} + \frac{3}{35} \\ p_{5}(x) & = x^{5} − \frac{10}{9}x^{3} + \frac{5}{21}x \end{aligned}

I wrote a Maple script to implement the calculation given as Theorem 5 in Kincaid and Cheney on page 400. The script is

1 # Kincaid and Cheney Problem 6.8 # 21
2 # Written December 4 by Eric Olson for Math 761
3 restart;
4 kernel(printbytes=false):
5 l2prod:=(f,g)->int(f*g,x=-1..1);
6 p[0]:=1;
7 a[1]:=l2prod(x*p[0],p[0])/l2prod(p[0],p[0]);
8 p[1]:=x-a[1];
9 for n from 2 to 5 do
10  a[n]:=l2prod(x*p[n-1],p[n-1])/l2prod(p[n-1],p[n-1]);
11  b[n]:=l2prod(x*p[n-1],p[n-2])/l2prod(p[n-2],p[n-2]);
12  p[n]:=sort(collect((x-a[n])*p[n-1]-b[n]*p[n-2],x));
13 od;

and the output is

    |\^/|     Maple 9.5 (IBM INTEL LINUX)
._|\|   |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2004
 \  MAPLE  /  All rights reserved. Maple is a trademark of
 <________>  Waterloo Maple Inc.
      |       Type ? for help.
# Kincaid and Cheney Problem 6.8 # 21
# Written December 4 by Eric Olson for Math 761
> restart;
> kernel(printbytes=false):
> l2prod:=(f,g)->int(f*g,x=-1..1);
                                               1
                                              /
                                             |
                        l2prod := (f, g) ->  |   f g dx
                                             |
                                            /
                                              -1
> p[0]:=1;
                                   p[0] := 1
> a[1]:=l2prod(x*p[0],p[0])/l2prod(p[0],p[0]);
                                   a[1] := 0
> p[1]:=x-a[1];
                                   p[1] := x
> for n from 2 to 5 do

>     a[n]:=l2prod(x*p[n-1],p[n-1])/l2prod(p[n-1],p[n-1]);
>     b[n]:=l2prod(x*p[n-1],p[n-2])/l2prod(p[n-2],p[n-2]);
>     p[n]:=sort(collect((x-a[n])*p[n-1]-b[n]*p[n-2],x));
> od;
                                   a[2] := 0
                                  b[2] := 1/3
                                         2
                                p[2] := x  - 1/3
                                   a[3] := 0
                                  b[3] := 4/15
                                        3
                               p[3] := x  - 3/5 x
                                   a[4] := 0
                                  b[4] := 9/35
                                    4        2
                           p[4] := x  - 6/7 x  + 3/35
                                   a[5] := 0
                                           16
                                   b[5] := --
                                           63
                                  5         3
                         p[5] := x  - 10/9 x  + 5/21 x
> quit
bytes used=753184, alloc=655240, time=0.10

4. [Kincaid and Cheney Problem 6.9#2] Find the best approximation of \sqrt{x} by a firstdegree polynomial on the interval [0, 1].

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:

g(0) − f(0) = δ
g(ξ) − f(ξ) = −δ
g(1) − f(1) = δ
g′(ξ) − f′(ξ) = 0

where g(x) = ax + b and f(x) = \sqrt{x}. The Maple script

1 # Kincaid and Cheney Problem 6.9 # 2
2 # Written December 4 by Eric Olson for Math 761
3 restart;
4 kernel(printbytes=false):
5 eq1:=g(0)-f(0)=delta;
6 eq2:=g(xi)-f(xi)=-delta;
7 eq3:=g(1)-f(1)=delta;
8 eq4:=D(g)(xi)-D(f)(xi)=0;
9 g:=x->a*x+b;
10 f:=sqrt;
11 eqns:={eq1,eq2,eq3,eq4};
12 S:=solve(eqns,{a,b,xi,delta});
13 g1:=subs(S,g(x));

solves these equations. The best approximation is

f(x) = x + 1/8

which has a graph

The Maple output follows:

      |\^/|     Maple 9.5 (IBM INTEL LINUX)
  ._|\|   |/|_. Copyright (c) Maplesoft, a division of Waterloo Maple Inc. 2004
   \  MAPLE  /  All rights reserved. Maple is a trademark of
   <________>  Waterloo Maple Inc.
        |       Type ? for help.

# Kincaid and Cheney Problem 6.9 # 2
# Written December 4 by Eric Olson for Math 761
> restart;
> kernel(printbytes=false):
> eq1:=g(0)-f(0)=delta;
                           eq1 := g(0) - f(0) = delta
> eq2:=g(xi)-f(xi)=-delta;
                         eq2 := g(xi) - f(xi) = -delta
> eq3:=g(1)-f(1)=delta;
                           eq3 := g(1) - f(1) = delta
> eq4:=D(g)(xi)-D(f)(xi)=0;
                         eq4 := D(g)(xi) - D(f)(xi) = 0
> g:=x->a*x+b;
                               g := x -> a x + b
> f:=sqrt;
                                   f := sqrt
> eqns:={eq1,eq2,eq3,eq4};
eqns :=
                             1/2                                     1
    {b = delta, a xi + b - xi    = -delta, a + b - 1 = delta, a - ------- = 0}
                                                                      1/2
                                                                  2 xi
> S:=solve(eqns,{a,b,xi,delta});
                  S := {b = 1/8, a = 1, xi = 1/4, delta = 1/8}
> g1:=subs(S,g(x));
                                 g1 := x + 1/8
> quit
bytes used=1419008, alloc=1179432, time=0.14

5. [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}},  B = {1, e^{x}, e^{2x}},  C = {(x + 2)^{−1}, (x + 3)^{−1}, (x + 4)^{−1}}.

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 ≠ 1 we may consider the polynmial α + βx + x^{3} were α = a/c and β = b/c. Suppose for contradiction there were three distinct roots 0 ≤ x_{1} < x_{2} < x_{3} ≤ 1 such that

\begin{aligned}α + βx^{2} + x^{3} & = (x − x_{1})(x − x_{2})(x − x_{3}) \\ & = x^{3} − (x_{1} + x_{2} + x_{3})x^{2} + (x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3})x − x_{1}x_{2}x_{3}. \end{aligned}

Equating coefficients we obtain 0 = x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3} ≥ x_{2}x_{3} > 0 which is a contradiction.

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 → 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.

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

\frac{a(x + 3)(x + 4) + b(x + 2)(x + 4) + c(x + 2)(x + 3)}{(x + 1)(x + 2)(x + 3)} = \frac{p(x)}{(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.

6. [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}},  B = {|x|, |x − 1|},  C = {e^{x}, x + 1}.

We consider again the equation x_{1}x_{2} + x_{1}x_{3} + x_{2}x_{3} = 0 from part A of the previous problem and this equation has solutions such that −1 ≤ x_{1} < x_{2} < x_{3} ≤ 1. In particular, if x_{2} = 1/2 and x_{3} = 1/3, then

x_{1} = \frac{−x_{2}x_{3}}{x_{2} + x_{3}} = −\frac{1}{5} ∈ [−1, 1].

Therefore the span of A is not a Haar subspace.

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.

Consider the function f(x) = 1 + x − \frac{4}{5}e^{x}. Since

f(−1) = −\frac{4}{5}e^{−1} < 0

f(0) = 1 − \frac{4}{5} > 0

f(1) = 2 − \frac{4}{5}e < 4 − \frac{4}{5}(2.7) < 0,

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  and  f(x_{1}) = f(x_{2}) = 0.

Therefore C is not a Haar subspace.

7. [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

\{ \begin{matrix}2a & = 1 + cosh t − t^{2}b \\ sinh t & = 2tb. \end{matrix}

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 ∈ R}, G = {cx : x ∈ R} and P_{2} = {a + cx + bx^{2} : a, b, c ∈ R}. Let a, b and c be chosen so that

∥f − (a + cx^{2} + bx^{2})∥ = min {∥f − g∥ : g ∈ P_{2}}.

Define f_{2}(x) = cosh x−(a+bx^{2}). Since f_{2}(x) is even then x ∈ crit(f_{2}) implies −x ∈ 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 crit(f_{2}). It follows that Kolmogorov’s Characterization Theorem from Kincaid and Cheney page 407 implies ∥f_{2}∥ = dist(f_{2}, G). In particular

∥f_{2} − (a + cx^{2} + bx^{2})∥ = dist(f_{2}, G) = ∥f_{2}∥

and therefore there is a function of the form a + bx^{2} such that

∥f − (a + bx^{2})∥ = min {∥f − g∥ : g ∈ P_{2}}.

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].

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:

\begin{aligned}g_{3}(0) − f_{3}(0) & = δ \\ g_{3}(ξ) − f_{3}(ξ) & = −δ \\ g_{3}(1) − f_{3}(1) & = δ \\ g′_{3}(ξ) − f′_{3}(ξ) & = 0 \end{aligned}

where g_{3}(y) = a + by and f_{3}(y) = cosh \sqrt{y}. Simplifying obtains

a − 1 = δ
a + bξ − cosh \sqrt{ξ} = −δ
a + b − cosh 1 = δ
2b\sqrt{ξ} − sinh \sqrt{ξ} = 0

Elimination of δ from the 2nd and 3rd equation, setting t = \sqrt{ξ} and futher simplification obtains the desired result

2a = 1 + cosh t − t^{2}b
b = cosh 1 − 1
sinh t = 2bt.