Download Math 304 Answers to Selected Problems 1 Section 5.5

Math 304 Answers to Selected Problems
1
Section 5.5
6. Let {u1 , u2 , u3 } be an orthonormal basis for an inner product space V
and let
u = u1 + 2u2 + 2u3
and
v = u1 + 7u3
Determine the value of each of the following.
(a) hu, vi
(b) kuk and kvk
(c) The angle θ between u and v
Answer:
(a) By Corollary 5.5.3,
hu, vi = 1(1) + 2(0) + 2(7) = 15
(b) By Corollary 5.5.4,
√
√
kuk =
12 + 22 + 22 = 9 = 3
√
√
√
12 + 7 2 = 50 = 5 2
kvk =
(c) The angle between u and v can be computed using the following
formula:
cos θ =
hu, vi
kukkvk
Thus,
cos θ =
Thus, θ =
π
.
4
1
15
1
√ =√
3·5 2
2
9. The set
S=
1
√ , cos x, cos 2x, cos 3x, cos 4x
2
is an orthonormal set of vectors in C[−π, π] with inner product defined
by (2).
(a) Use trigonometric identities to write the function sin4 x as a linear
combination of elements of S.
(b) Use part (a) and Theorem 5.5.2 to find the values of the following
integrals:
Rπ
(i) −π sin4 x cos x dx
Rπ
(ii) −π sin4 x cos 2x dx
Rπ
(iii) −π sin4 x cos 3x dx
Rπ
(iv) −π sin4 x cos 4x dx
Answer:
(a) We will use the trigonometric identity sin2 x =
1
2
− 21 cos 2x.
2
1 1
− cos 2x
2 2
1 1
1
− cos 2x + cos2 2x
4 2
4
1 1 1
1 1
− cos 2x +
+ cos 4x
4 2
4 2 2
3 1
1
− cos 2x + cos 4x
8 2
8
3
1
1
1
√
√
− cos 2x + cos 4x
2
8
4 2
2
4
sin x =
=
=
=
=
(b) Since S is an orthonormal set, Theorem 5.5.2 applies. Using part
(a) and the Theorem, we have that
2
1
4
sin x, √
2
4
sin x, cos x
4
sin x, cos 2x
4
sin x, cos 3x
4
sin x, cos 4x
=
=
=
=
=
3
√
4 2
0
1
−
2
0
1
8
We can use these to evaluate the integrals.
Z π
(i)
sin4 x cos x dx = π sin4 x, cos x = 0
−π
(ii)
Z
π
Z
π
π
sin4 x cos 2x dx = π sin4 x, cos 2x = −
2
−π
sin4 x cos 3x dx = π sin4 x, cos 3x = 0
(iii)
−π
Z
(iv)
π
π
sin4 x cos 4x dx = π sin4 x, cos 4x =
8
−π
21. Let





A=




1
2
1
2
1
2
1
2
− 12



− 12 



1 
2 

1
2
(a) Show that the column vectors of A form an orthonormal set in
R4 .
3
(b) Solve the least squares problem Ax = b for each of the following
choices of b.
(i) b = (4, 0, 0, 0)T
(ii) b = (1, 2, 3, 4)T
(iii) b = (1, 1, 2, 2)T
Answer:
(a) Let a1 and a2 denote the first and second column vectors of A,
respectively. To show that the column vectors of A form an orthonormal set in R4 , we just need to check that ha1 , a2 i = 0,
ka1 k = 1, and ka2 k = 1.
1
1
1
1
1
1
1
1
−
+
−
+
+
=0
ha1 , a2 i =
2
2
2
2
2
2
2
2
s 2 2 2
2
1
1
1
1
ka1 k =
+
+
+
=1
2
2
2
2
s 2 2 2
2
−1
1
1
−1
ka2 k =
+
+
+
=1
2
2
2
2
Thus, the column vectors of A form an orthonormal set in R4 .
(b) By Theorem 5.5.6, since the column vectors of A form an orthonormal set, the solution to the least squares problem is x̂ = AT b.
 
 1
 4
1
1
1
2
2
2
2
 0 
2




(i) x̂ =
 0  = −2
1
1
1
1
−2 −2 2 2
0
 
 1
 1
1
1
1
2
2
2
2
 2 
  = 5
(ii) x̂ = 
 3 
2
− 21 − 12 12 12
4
 
 1
 1
1
1
1
2
2
2
2
 1 
  = 3
(iii) x̂ = 
 2 
1
− 21 − 12 12 12
2
4
27. Given the vector space C[−1, 1] with inner product
Z
1
hf, gi =
f (x)g(x) dx
−1
and norm
kf k = (hf, f i)1/2
(a) Show that the vectors 1 and x are orthogonal.
(b) Compute k1k and kxk.
(c) Find the least squares approximation to x1/3 on [−1, 1] by a
linear function `(x) = c1 1 + c2 x.
(d) Sketch the graphs x1/3 and `(x) on [−1, 1].
Answer:
(a) The vectors 1 and x are orthogonal if h1, xi = 0.
Z
1
h1, xi =
x dx = 0
−1
The integral equals 0 because the function x is an odd function.
(b) First, we compute h1, 1i:
Z
1
h1, 1i =
1 dx = 2
−1
p
√
Thus, k1k = h1, 1i = 2.
We compute hx, xi:
1
1 3
2
hx, xi =
x dx = x
=
3
3
−1
−1
r
p
2
Thus, kxk = hx, xi =
.
3
(c) To find the least squares approximation to x1/3 on [−1, 1] by a linear function, we project x1/3 onto the subspace Span(1, x). The
formula to compute the projection requires an orthonormal basis
Z
1
2
5
for the subspace. Since 1 and x are orthogonal, we can find an
orthonormal basis
by k1k and kxk. Thus, the ortho( by dividing
√ )
3x
1
. The projection of x1/3 onto this
normal basis is √ , √
2
2
subspace is
*
√ +√
1
1
3x
3x
√ + x1/3 , √
√
x1/3 , √
2
2
2
2
We can compute each of these inner products:
4/3 1
Z 1 1/3
3x
x
1/3 1
√ dx =
√
x ,√
=
=0
2
2
4 2 −1
−1
*
" √
#1
√ + Z 1 √ 4/3
7/3
3x
3x
3x
3
√
√
x1/3 , √
=
dx =
=
2
2
7 2
−1
−1
√
6 3
√
7 2
Thus, the projection of x1/3 onto Span(1, x) is
*
√ +√
√
1
3x
3x
6 3
1
1/3
1/3
√ + x , √
√
= √
x ,√
2
2
2
2
7 2
9
x
=
7
√ !
3x
√
2
Thus, the least squares approximation to x1/3 by a linear function
9
is `(x) = x.
7
6