MA342A (Harmonic Analysis 1) Tutorial sheet 1
[October 15, 2015]
Name: Solutions
1. Express f (x) = 3 cos(4πx) + 7 cos(6πx) − sin(2πx) in terms of (i.e. as a linear combination of) the complex exponentials φn (x) = e2πinx (n ∈ Z).
Solution: Since eiθ = cos θ + i sin θ, e−iθ = cos θ − i sin θ, cos θ = (eiθ + e−iθ )/2 and
sin θ = (eiθ − e−iθ )/(2i). So cos(2nπx) = (e2nπix + e−2nπix )/2 = (φn (x) + φ−n (x))/2
and sin(2nπx) = (e2nπix − e−2nπix )/(2i) = (φn (x) − φ−n (x))/(2i).
3
1
1
3
f (x) = φ2 (x) + φ−2 (x) + 7φ3 (x) + 7φ−3 (x) − φ1 (x) + φ−1 (x)
2
2
2i
2i
2. Show that hφn , φn i = 1 and hφn , φm i = 0 if n 6= m.
Solution:
Z
1
hφn , φm i =
φn (x)φm (x) dx
0
Z
1
=
Z0 1
=
e2πinx e−2πimx dx
e2πi(n−m)x dx
0
If n = m, this is
R1
0
e0 dx =
R1
0
1 dx = 1.
If n 6= m
Z
1
e2πi(n−m)x dx
hφn , φm i =
0
Z
=
cos(2π(n − m)x) + i sin(2π(n − m)x) dx
(1)
Z 1
Z 1
cos(2π(n − m)x) dx + i
sin(2π(n − m)x) dx
0
0
1
1
1
−1
sin(2π(n − m)x)
+i
cos(2π(n − m)x)
2π(n − m)
2π(n − m)
x=0
x=0
1
i
1
sin(2π(n − m)x) −
cos(2π(n − m)x)
2π(n − m)
2π(n − m)
x=0
1
1
(i sin(2π(n − m)x) + cos(2π(n − m)x))
2πi(n − m)
x=0
1
1
e2πi(n−m)x
(2)
2πi(n − m)
x=0
1
1
e2πi(n−m) −
e0 = 0
2πi(n − m)
2πi(n − m)
0
=
=
=
=
=
=
1
In the steps (1) to (2), we have used the definition of the integral of a complex valued
function in terms of the integrals of the real and imaginary parts, but really you should
take this calculation as standard facts. So skip directly from the integral to (2).
3. Let f ∈ CP [0, 1] be given by f (x) = x for 0 ≤ x ≤ 1/2 and f (x) = 1 − x for
1/2 < x ≤ 1. Find the Fourier coefficients fˆ(n) (n ∈ Z).
Solution: If n 6= 0,
fˆ(n) =
Z
1
f (x)e−2πinx dx
0
Z
1/2
−2πinx
=
xe
Z
0
=
=
=
=
=
(1 − x)e−2πinx dx
1/2
1/2
−2πinx Z 1
e
e
xd
+
(1 − x) d
−2πin
−2πin
0
1/2
−2πinx 1/2 Z 1/2 −2πinx
e
e
x
−
dx
−2πin 0
−2πin
0
1
Z 1 −2πinx
e−2πinx
e
+ (1 − x)
−
(−1) dx
−2πin 1/2
1/2 −2πin
−2πinx 1/2
1
1 e−πin
e
+0+
−
2 2πin
2πin −2πin 0
−2πinx 1
−πin
1 e
1
e
+0 −
−
2 −2πin 2πin −2πin 1/2
1
(e−πin − 1 − (1 − e−πin ))
4π 2 n2
2
((−1)n − 1)
2 n2
4π
(
0
n even
2 2
1/(π n ) n odd
Z
=
1
dx +
−2πinx
For n = 0
fˆ(0) =
Z
1
Z
f (x) dx =
0
Z
1
(1 − x) dx
x dx +
0
1/2
1/2
1/2
= [x2 /2]0 + [x − x2 /2]11/2
1
1
1 1
1
=
−0+1− −( − )=
8
2
2 8
4
Richard M. Timoney
2