Selected Solutions for Homework 6
Math 2280
Nov. 23, 2003
1. (Problem 7.1.7) We want to nd the Fourier cosine and sine representation for the function
<x<1
f (x) = 10 0otherwise.
Recall the the representation is given by
f (x) =
where
1
Z
0
[A(!) cos(x!) + B (!) sin(x!)]d!;
Z 1
1
A(!) = f (y) cos(y!)dy;
Z 1
1
B (!) = f (y) sin(y!)dy:
?1
First we compute A:
?1
1
Z
A(!) = 1
f (y) cos(y!)dy = 1
?1
sin
!
= ! :
Next we compute B :
1
Z
B (!) = 1
f (y) sin(y!)dy = 1
?1
1
?
cos(!)
=
! :
Putting this all together, we have
f (x) =
Z
0
1
Z 1
0
Z 1
0
1 sin(y!)j1
cos(y!)dy = !
0
1 cos(y!)j1
sin(y!)dy = ? !
0
!?1 [sin(!) cos(x!) + (1 ? cos(!)) sin(x!)]d!:
2. (Problem 7.1.13)
(a) We know from Example 1 of the text that if
f (x) =
then
f (x) = 2
1 jxj 1
0 jxj > 1
1 sin(!) cos(!x)
Z
!
0
d!:
Evaluating this at x = 1, which a point of discontinuity of f , we see
2 Z 1 sin ! cos ! d! = 1 (f (1 ) + f (1 )) = 1 (1 + 0) = 1 ;
?
0
!
2 +
2
2
or
= Z 1 sin ! cos ! d!:
4
!
0
(b) Now we integrate
Z
0
by parts. Let u = sin2 ! and dv = d!=!2. Then
1 sin2 !
!2 d!
du = 2 sin ! cos !d!;
so
Z
0
v = ? !1 ;
1
Z 1
2
sin
!
2 cos ! sin ! d! = 2 Z 1 cos ! sin ! d! = :
+
d!
=
?
!2
! 0 0
!
!
2
0
1 sin2 !
3. (Problem 7.2.1) We want to compute the Fourier transform of
f (x) =
We have
f^(!) = p1
Z
1
e?ix! f (x)dx = p1
1 jxj < 2
0 jxj 2:
Z 2
1p e?ix! 2
?2
2 ?1
2 ?2
?i! 2
p1 (e2i! ? e?2i! ) = p1 (cos(2!) + i sin(2!) ? cos(2!) + i sin(2!))
=
i! 2
i! 2
p
sin(2
!
2
sin(2
!
)
p = 2= ! ) :
=
! 2
e?ix! dx =
4. (Problem 7.2.4) We want to compute the Fourier transform of the function
f (x) =
We have
x jxj > 1
0 jxj 1:
1
Z 1
x e?ix! x=1 + 1 Z 1 e?ix! dx]
e?ix! f (x)dx = p1
xe?ix! dx = p1 [ ? i!
x=?1 i! ?1
2 ?1
2 ?1
2
i!
?
i!
?
i!
1 [? 2 cos ! + e ? ei! ] = p1 [ 2i cos(!) + 2 sin(!) ]
p
= p1 [? ei! ? ei! + !12 e?ix! xx=1
]
=
=?1
i!
!2
!
!2
2p
2
2
2
= i!p 2 [! cos(!) + sin(!)]
f^(!) = p1
Z
5. (Problem 7.2.10)
We start by recalling the formulas for the Fourier transform and inverse Fourier transform:
Z 1
Z 1
1
1
?
ixy
^
f (x) = p
e f (y)dy; f (x) = p
eixy f (y)dy:
2 ?1
2 ?1
(a) f^(x) = f(?x)
This is clear from the formulas above. Indeed,
Z 1
Z 1
ei(?x)y f (y)dy = p1
e?ixy f (y)dy = f^(x):
f(?x) = p1
2 ?1
2 ?1
(b) F 2 (f )(x) = f (?x).
Applying the Fourier transform to both sides of the equation of part (a), we have
F (f )(x) = F (F ? (f ))(?x) = f (?x):
2
1
(c) By denition, f is even if and only if f (?x) = f (x). However, F 2 (f )(x) = f (?x), so f is even if and only
if F 2 (f )(x) = f (x). The argument for odd functions is similar.
(d) F 4 (f ) = f .
This time we just apply part (b) twice. Indeed,
F (f )(x) = F (F (f ))(x) = f (?(?x)) = f (x):
4
2
2