Problem 1.
(1). Since
f
(
x
) = 2
x
for 0
< x < π
,
f
(
x
)
∼
X
n
≥
1
A
n
sin(
nx
)
.
Then by deﬁnition, we have
A
n
=
2
π
Z
π
0
2
x
sin(
nx
)
dx
=

4
nπ
cos(
nx
)
x

π
0
+
Z
π
0
4
nπ
cos(
nx
)
dx
=

4
n
(

1)
n
+
4
n
2
π
sin(
nx
)

π
0
= (

1)
n
+1
4
n
.
(2). Let
˜
f
be the odd and 2
π
periodic extension of
f
. By convergence theo
rem, the Fourier sine series
S
[
f
] of
f
equals to
˜
f
at the point of continuity, and
˜
f
(
x

)+
˜
f
(
x
+)
2
at the point of discontinuity. In conclusion, we know that for
S
[
f
](
x
) =
0
x
=

3
π
2(
x
+ 2
π
)

3
π < x <

π
0
x
=

π
2
x

π < x < π
0
x
=
π
2(
x

2
π
)
π < x <
3
π
0
x
= 3
π
(3). If we use partial sum of eigenfunctions to approximate a function, we know
that we have Gibbs phenomenon (roughly 9% overshoot) at the point of disconti
nuity.
Here the only points of discontinuity are
x
k
=
π
+2
kπ
, at
x
k
,
f
(
x
k
+)

f
(
x
k

) =
4
π
, the overshoot is roughly 9%
×
4
π
= 0
.
36
π
.
(4). If we studied the Fourier cosine series instead, we need to consider
ˆ
f
which is
the even and 2
π
periodic extension of
f
. We can know that
ˆ
f
is in fact a continuous
function, then by convergence theorem, the Fourier cosine series
C
[
f
](
x
) of
f
equals
to
ˆ
f
.
Problem 2.
By the method of separation of variables, we assume ﬁrst that
u
(
x, t
) =
φ
(
x
)
h
(
t
)
.
Then by the equation, we know that
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 Spring '08
 GOLDBERG
 Math, Fourier Series, Cos, Partial differential equation, Joseph Fourier

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