Math 222 - Selected Homework Solutions from
Chapter 5.3
Instructor - Al Boggess
Fall 1998
Section 5.3
6. To show the second property, we have
Zb
hf; gi =
a
Zb
f (x)g(x) dx
=
g(x)f (x) dx
a
= hg; f i
To show the third property, we have
hf + g; hi =
Zb
a
(f + g)h dx
Zb
Zb
= fh dx + gh dx
a
a
= hf; hi + hg; hi
9. To show cos(mx) is orthogonal to sin(nx), we must show
Z
,
sin(nx) cos(mx) = 0
If n = m, then we use the identity
sin nx cos nx = (1=2) sin(2nx)
and
Z
,
sin(2nx) = ,2n1 cos(2nx)j, = 0
1
If n 6= m, then we use the identity
sin nx cos mx = (1=2) (sin(m + n)x + sin(n , m)x)
which is shown by expanding the right side using the sum and dierence formulas for sine. The integral of the right side is
Z
(1=2) (sin(m + n)x + sin(n , m)x) dx = 2(m,+1 n) cos(m+n)x+ 2(n,,1 m) cos(n,m)xj,
,
which equals zero since cos is periodic with period 2.
Each of these functions are unit vectors since
Z
1
2
sin2 nx dx
jj sin(nx)jj =
Z,
= 1 (1=2)(1 , cos(2nx)) dx
,
1
= [(1=2)(x) , (1=4n) sin(2nx)]j
,
= 1
and
jj cos(mx)jj2 = 1 1
Z
Z
,
cos2 mx dx
= 1 (1=2)(1 + cos(2mx)) dx
,
1
= [(1=2)(x) + (1=4m) sin(2mx)]j
,
= 1
From the Pythagorean Theorem,
p the distance between any two unit
vectors that are orthogonal is 2.
P
13. Dene jjxjj = i jxi j for x = (x1 ; : : : ; xn ). For the rst property of
norm, we have jjxjj 0 since jjxjj is a sum of nonnegative numbers.
If jjxjj = 0, then each jxi j is zero and so x = (x1 ; : : : ; xn ) is the zero
vector. For the second property, we have
X
X
jjxjj = jxij = jj jxij = jjjjxjj
i
i
For the third property, we have
X
X
X
jjx + yjj = jxi + yij jxij + jyij
i
i
2
i
The right side is just jjxjj + jjyjj and so
jjx + yjj jjxjj + jjyjj
as desired.
24. We have
jju + vjj2 + jju , vjj2 = (u + v) (u + v) + (u , v) (u , v)
= jjujj2 + 2u v + jjvjj2 + jjujj2 , 2u v + jjvjj2
= 2jjujj2 + 2jjvjj2
The geometric interpretation is that the sum of the squares of the
lengths of the diagonals of the parallelogram formed by u and v is
twice the sum of the squares of the lengths of the two sides of the
parallelogram.
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