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10
Chapter 2
Fourier Series
Solutions to Exercises 2.3
1.
(a) and (b) Since
f
is odd, all the
a
n
’s are zero and
b
n
=
2
p
±
p
0
sin
nπ
p
dx
=

2
nπ
cos
nπ
p
²
²
²
π
0
=

2
nπ
³
(

1)
n

1
´
=
µ
0i
f
n
is even,
4
nπ
if
n
is odd
.
Thus the Fourier series is
4
π
∞
¶
k
=0
1
(2
k
+1)
sin
(2
k
π
p
x
. At the points of discon
tinuity, the Fourier series converges to the average value of the function. In this
case, the average value is 0 (as can be seen from the graph.
5.
(a) and (b) The function is even. It is also continuous for all
x
. All the
b
n
s are
0. Also, by computing the area between the graph of
f
and the
x
axis, from
x
=0
to
x
=
p
, we see that
a
0
= 0. Now, using integration by parts, we obtain
a
n
=
2
p
±
p
0

·
2
c
p
¸
(
x

p/
2) cos
nπ
p
xdx
=

4
c
p
2
±
p
0
u
¹
º»
¼
(
x

p/
2)
v
±
¹
º»
¼
cos
nπ
p
=

4
c
p
2
=0
¹
º»
¼
p
nπ
(
x

p/
2) sin
nπ
p
x
²
²
²
p
x
=0

p
nπ
±
p
0
sin
nπ
p
=

4
c
p
2
p
2
n
2
π
2
cos
nπ
p
x
²
²
²
p
x
=0
=
4
c
n
2
π
2
(1

cos
nπ
)
=
½
f
n
is even
,
8
c
n
2
π
2
if
n
is odd
.
Thus the Fourier series is
f
(
x
)=
8
c
π
2
∞
¶
k
=0
cos
¾
(2
k
π
p
x
¿
(2
k
2
.
9.
The function is even; so all the
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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