Math 381 Fall 2006 Homework # 11 Solutions
Due: Wednesday, December 6, 2006
1. Problem 10.2.2. We have
f (x) =
Z
∞
c(ω)e−iωx dω
−∞
and we want to show that if c(−ω) = c(ω) then f (x) is real. To show f (x)
is real we need to show f = f . If we take the complex conjugate of the
definition of f (x) above we obtain
Z ∞
f (x) =
c(ω)eiωx dω
−∞
Z ∞
=
c(−ω)eiωx dω.
−∞
Changing coordinates from ω = −t we get
Z ∞
Z −∞
−itx
c(t)e−itx dt = f (x)
c(t)e
− dt =
f (x) =
−∞
∞
as required.
2. 10.3.5. Let’s compute the inverse Fourier transform of eiωβ FT(f (x)) where
f (x) is a smooth function. We have
Z ∞
−1 iωβ
FT [e FT(f (x))] =
eiωβ FT(f (x))e−iωx dω
−∞
Z ∞
=
FT(f (x))e−iω(x−β) dω
−∞
Z ∞
=
FT(f (x))e−iωt dω
−∞
where we substituted t = x− β. Notice that the last line is just the inverse
Fourier transform of FT(f (x)) in the variable t. Since f (x) is smooth we
know FT−1 [FT(f (x))] = f (x). So we get that
FT−1 [eiωβ FT(f (x))] = f (t) = f (x − β).
This is called the shift theorem for Fourier transforms.
3. 10.3.8 The Fourier transform of f (x) is
Z ∞
1
F (ω) =
f (x)eiωx dx.
2π −∞
Differentiating both sides by ω we get
Z ∞
Z ∞
1
i
dF
=
f (x)ixeiωx dx =
xf (x)eiωx dx.
dω
2π −∞
2π −∞
Multiplying both sides by −i we get that −i dF
dω is the Fourier transform
of xf (x).
4. 10.3.11 (a) Suppose f (x) is a function with unit area. Then
Z ∞
Z ∞
1 x
f ( )dx =
f (t)dt = 1
α
−∞ α
−∞
where we used the substitution t = αx . Thus α1 f ( αx ) also has unit area.
R∞
(b) Suppose F (ω) = −∞ f (x)eiωx dx is the Fourier transform of f (x).
Then
Z ∞
Z ∞
dt
t
F (αω) =
f (x)eiωαx dx =
f ( )eiωt
α
α
−∞
−∞
where we used the substitution t = xα. Thus F (αω) is equal to the Fourier
transform of α1 f ( αt ) as required.
(c) Let f (x) be an arbitrary function with unit area and denote by F (ω) its
Fourier transform (as in part (b)). Consider g(x) = (1/α)f (x/α). Then
as α → 0 the function g(x) becomes sharply peaked around x = 0 while its
Fourier transform F (αω) becomes broadly spread. Conversely, if α → ∞
then g(x) becomes broadly spread while F (αω) becomes sharply peaked
around x = 0. This shows that the Fourier transform of a sharply peaked
function is broadly spread and vice versa.