Description
Unformatted Attachment Preview
Purchase answer to see full attachment
Explanation & Answer
View attached explanation and answer. Let me know if you have any questions.
5.1
2. ( x) = x2 , 0 x 1 = l
(a) The coefficients of the sine series
1
2
nx
2 x2
4
2
bn = ( x) sin
dx = 2 x sin ( nx ) dx = −
cos ( nx ) +
x cos ( nx ) dx
l 0
l
n
n 0
0
0
l
1
1
1
1
1
2
4x
4
2
4
=−
cos ( n ) + 2 2 sin ( nx ) − 2 2 sin ( nx ) dx = (−1) n +1
+ 3 3 cos ( nx ) .
n
n
n 0
n n
0
0
2
8
− 3
, n = 2k − 1
4
4
(2k − 1) (2k − 1)3
n +1 2
n
= (−1)
+ (−1) 3 3 − 3 3 =
n
n n
1
−
,
n = 2k
k
The Fourier sine series
+
sin ( 2 kx )
nx +
2
8
( x) ~ bn sin
=
− 3
sin ( (2k − 1) x ) −
3
l
(2k − 1)
k
n =1
k =1 ( 2k − 1)
(b) The coefficients of the cosine series
l
1
1
2
2 x3
2
a0 = ( x)dx = 2 x 2 dx =
=
l 0
3 0 3
0
1
2
nx
2x2
4
an = ( x) cos
dx = 2 x 2 cos ( nx ) dx =
sin ( nx ) −
0 x sin ( nx ) dx .
l 0
l
n
n
0
0
l
1
1
1
1
4x
4
4
= 2 2 cos ( nx ) + 2 2 cos ( nx ) dx = (−1) n 2 2
n
n 0
n
0
The Fourier cosine series
a +
nx 1 +
4
( x) ~ 0 + an cos
= + (−1) n 2 2 cos ( nx ) .
2 n =1
l
3 n =1
n
3. ( x) = x, x (0; l )
(a) The coefficients of the sine series
l
l
l
l
2
nx
2
nx
2x
nx
2
nx
bn = ( x) sin
dx = x sin
dx = −
cos
+
cos
dx
l 0
l
l 0
l
n
l 0 n 0
l
2l
2l
=−
cos ( n ) = (−1) n +1
n
n
The sum of the first three terms of the Fourier sine series
3
nx 2l x l
2 x 2l
3 x
.
S3 ( x) = bn sin
= sin
− sin
+ sin
l
l
l
3
l
n =1
(b) The coefficients of the cosine series
l
l
2
2
1 l
a0 = ( x)dx = xdx = x 2 = l
l 0
l 0
l 0
2
nx
2
nx
2x
nx
2
nx
an = ( x) cos
dx = x cos
dx =
sin
−
sin
dx .
l 0
l
l 0
l
n
l 0 n 0
l
l
l
l
l
2l
nx
2l
= 2 2 cos
= ( (−1) n − 1) 2 2
n
l 0
n
The sum of the first three nonzero terms of the Fourier cosine series
a
x
3 x l 4l
x 4l
3 x
C3 ( x) = 0 + a1 cos
+ a3 cos
= − 2 cos
− 2 cos
.
2
l
l
2
l 9
l
The graphs
l
.
5.
x
a0 +
x2
nx
= tdt and
(a) Let
be the Fourier cosine series for the function
+ an cos
2 0
2 n =1
l
the Fourier sine series from 3(a). Since
x +
x
+
a0 +
nx x 2 x
nt
nt
+ an cos
~
= tdt ~ bn sin
dt ~ bn sin
dt
2 n =1
l
2 0
l
l
n =1
0 n =1
0
lb
nt
= n − cos
l
n =1 n
+
+ lbn + lbn
nx 2l 2
+ −
= 2
~
cos
l
n =1 n
n =1 n
0
x
+
b sin
n =1
n
4l 2
+
(−1) n +1 2
2 l2 2
=
a
=
=
.
0
n2
4l 2
4l 2 3 12
n =1
+
6.
(a) The coefficients of the sine series for ( x) = x3 , x (0; l ) ,
l
2
nx
2 3
nx
2 x3
nx
6
nx
bn = ( x) sin
dx = x sin
dx = −
cos
+
x 2 cos
dx
l 0
l
l 0
l
n
l 0 n 0
l
l
1
l
l
2l 3
6 x 2l
nx
12l
nx
=−
cos ( n ) + 2 2 sin
− 2 2 x sin
dx
n
n
l 0 n 0
l
l
.
2
2 2
n +1 l
3
n +1 n − 6
(−1) n = 2l (−1)
3 n3
2l 3 12l
= (−1)
−
n 2 n2
The Fourier sine series
3( a )
n +1
+
( x) ~ bn sin
nx
+
= (−1) n +1 2l 3
2 n 2 − 6 nx
sin
3 n3
l
l
k =1
(b) The coefficients of the cosine series for ( x) = x 4 , x (0; l ) ,
n =1
l
l
l
2
2
2 x5
2l 4
a0 = ( x)dx = x 4 dx =
=
l 0
l 0
5l 0
5
l
2
nx
2 4
nx
2x4
nx
8
nx
an = ( x) cos
dx = x cos
dx =
sin
−
x3 sin
dx .
l 0
l
l 0
l
n
l 0 n 0
l
l
l
l
2 2
2 2
8 4
n +1 n − 6
4
n n −6
l
(
−
1)
=
8
l
(
−
1
)
n
3n3
4 n4
The Fourier cosine series
a +
nx l 4 + 4
2 n2 − 6
nx
( x) ~ 0 + an cos
= + 8l (−1) n
cos
.
4 4
2 n =1
l
5 n =1
n
l
6( a )
=−
l
2
(−1) n +1 +
nx
n 2l
+ (−1) 2 2 cos
2
n
n
l
n =1
n =1
+
(−1) n +1
. The series for a0 converges by the Alternate Test.
2 n =1 n 2
l
l
2 x2
1
1 l l2
(b) Since a0 = dx = x 2 dx = x3 = , we obtain
l 0 2
l0
3l 0 3
we get a0 =
nt
,
be
5.2
1. A function f is even (resp. odd) if f (− z ) = f ( z ) (resp. f (− z ) = − f ( z ) ) for all z
A function f has a period T if f ( z + T ) = f ( z ) for all z .
(a) The function is odd, sin ( −az ) = − sin(az ) , and periodic,
.
2 n
sin a z +
= sin ( az + 2 n ) = sin(az ) , n .
a
2
Each period is divided by
.
a
(b) The function is nor odd nor even and periodic,
e
2 in
a z +
a
...
Review
Review
