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Math 415
Homework 7
10.2.21
f
(
x
) =
x
2
/
2
,

2
≤
x
≤
2;
f
(
x
+ 4) =
f
(
x
)
.
(a) Sketch the graph of the given function for three periods.
6
4
2
0
2
4
6
1
1
2
3
4
5
(b) Find the Fourier series.
a
0
=
1
2
Z
2

2
x
2
2
dx
=
1
12
x
3
±
±
±
±
2

2
=
4
3
.
a
m
=
1
4
Z
2

2
x
2
cos
mπx
2
=
²
1
2
mπ
x
2
sin
mπx
2
±
±
±
±
2

2

1
mπ
Z
2

2
x
sin
mπx
2
dx
=
²
1
2
mπ
x
2
sin
mπx
2
±
±
±
±
2

2
+
²
2
m
2
π
2
x
cos
mπx
2
±
±
±
±
2

2

2
m
2
π
2
Z
2

2
cos
mπx
2
dx
=
²
1
2
mπ
x
2
sin
mπx
2
±
±
±
±
2

2
+
²
2
m
2
π
2
x
cos
mπx
2
±
±
±
±
2

2

²
4
m
3
π
3
sin
mπx
2
±
±
±
±
2

2
=
0 +
8
m
2
π
2
(

1)
m
+ 0
.
Finally, since
f
is an even function and sine is an odd function
b
m
=
R
2

2
x
2
sin(
mπx/
2)
dx
is the integral
of an odd function over a symmetric domain, so is zero. Hence the Fourier series is
2
3
+
∞
X
m
=1
8(

1)
m
m
2
π
2
cos
mπx
2
.
(c) Plot
s
m
(
x
) versus
x
for
m
= 5, 10, and 20.
See ﬁgure 1.
(d) By the Fourier convergence theorem
s
m
→
f
as
m
→ ∞
since
f
has no discontinuities. But looking at
ﬁgure 1 we can see that it converges quite quickly everywhere but at
x
= 2 + 4
k
for any integer
k
.
1
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View Full Document01
2
1
2
Figure 1: The dotted line is a graph of
x
2
that I included for reference. If you above
x
= 2, the graph closest
to
y
=
x
2
is
s
20
, the graph furthest is
s
5
.
10.3.17
Assuming that
f
(
x
) =
a
0
2
+
∞
X
n
=1
±
a
n
cos
nπx
L
+
b
n
sin
nπx
L
²
,
(1)
Show that
1
L
Z
L

L
[
f
(
x
)]
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