Math 241
Quiz 1
Name:
Section:
Please note:
There is one question on the back, don’t miss it!
You may consult to your own homework, which is the only valid reference for this quiz.
Calculator is not allowed.
Please show your work also for multiple choice questions.
Problem 1 (Spring 11 Final, Problem
If H(x) = −1 for −π ≤ x < 0 and H(x)
periodic, then when we expand H(x) in a
sum c−2 + c−1 + c0 equals
a) 2πi
b) 3πi
c) πi
d) 2i
π
Solution: We have
c0 =
1
2π
II)
= 1 for 0 ≤ x < π andPif we extend H to be 2π
ikx , we find the
complex Fourier Series ∞
k=−∞ ck e
e) −1.
Z
π
H(x)dx = 0,
−π
and
1
ck =
2π
Z
π
H(x)e−ikx dx
−π
Z 0
Z π
1
1
−ikx
=
−e
dx +
e−ikx d
2π −π
2π 0
1 −ikx 0
1 −ikx π
=
e
e
−
2πik
2πik
−π
0
1 − (−1)n i
=−
.
kπ
Therefore
c−2 + c−1 + c0 = 0 +
2i
2i
+0= .
π
π
So we choose d.
Remark: Here we should treat c0 separately, because when computing ck we use integration
by parts, and that usually doesn’t take care of the c0 term.
Problem 2 (Textbook 12.3:35)
Expand the function f (x) = x2 , 0 < x < 2π
in a Fourier series.
Solution: Let’s expand it into full Fourier series,
∞
a0 X
f (x) =
+
an cos nx + bn sin nx,
2
n=1
where
1
a0 =
π
Z
Z
2π
0
x2 dx =
8π 2
,
3
2π
1
x2 cos nxdx
π 0
2π 1 Z 2π 2x
1 21
=
x sin nx −
sin nxdx
π
n
π 0 n
0
Z
1 2π 2x
sin nxdx
=0−
π 0 n
2π 1 Z 2π 2
1 2x
=0+
cos
nx
cos nxdx
−
π n2
π 0 n2
0
2π
1 2x
=
cos
nx
−0
π n2
0
4
= 2,
n
an =
Z
1 2π 2
bn =
x sin nxdx
π 0
2π 1 Z 2π 2x
1
1
=
− x2 cos nx +
cos nxdx
π
n
π 0 n
0
Z
4π
1 2π 2x
=−
+
cos nxdx
n
π 0 n
2π 1 Z 2π 2
4π
1 2x
=−
+
sin nx −
sin nxdx
2
n
π n
π 0 n2
0
4π
=− .
n
Therefore
f (x) =
∞ X
4π 2
1
π
+4
cos
nx
−
sin
nx
.
3
n2
n
n=1
Remark: We can also extend f to be an even function or an odd function, and expand it into
half range Fourier series. Again, we treat a0 separately, and don’t forget in the series, the first
2
2
term is a0 /2, so in this problem it is 4π3 rather than 8π3 .