CHAPTER 11 SOLUTIONS
n=1 bn
1. The half range Fourier sine series is given by f (x)
L
2
L
2
=
L
bn =
=
f (x) sin(
0
L
sin(
0
2
L
=
nx
)dx
L
nx
)dx
L
L
L
nx
cos
n
L
0
0
2
2
[(1)n 1] =
[1 (1)n ] =
n
n
This can be written neatly as:
The Fourier series is
4
CHAPTER 9 SOLUTIONS
For each of these questions, note the following results.
nx
L
nx
dx =
x sin
L
n
L
x cos
a
b
b
L
nx
=
x sin
n
L
b
b
a
n=1 bn
1. The function f is odd, so f (x)
L
1
L
x sin
L
2L
x
L
n=1
a0
2
2. The function g is even, so g(x)
2
L
+
b
a
CHAPTER 13 SOLUTIONS
1.
1
1
1
1
1
1
2. (a) f is linear. To see this let (x, y) R2 and R. Then f (x, y) =
f (x, y) = (2x + 4y, x 3y)(2x + 4y, x 3y) = f (x, y).
Next let (x, y) and (x , y ) R2 .
Then f (x, y) + (x , y ) = f (x + x , y + y ) = (2(x + x ) + 4
CHAPTER 10 SOLUTIONS
1. The function is piecewise dierentiable with a bounded derivative, so we plug in x =
to obtain
2
2
cos(k)
=
,
4
6
k2
k=1
so
2 2
2
=
=
4
6
12
k=1
2
(1)k+1
.
k2
2. Notice that f is odd, so that the Fourier series involves only sine te