LECTURE 11: HALF RANGE FOURIER SERIES AND CONVERGENCE OF FOURIER SERIES
MINGFENG ZHAO
July 23, 2015
Fourier series
Example 1. Assume that f (x) = x, 0 < x < 2 represents one full period of the function. Find the Fourier series.
By the assumption, we know that the period of f is 2, then the Fourier series of f is:
∞
f (x) =
a0 X
+
[an cos(nπx) + bn sin(nπx)] ,
2
n=1
where
an
=
2
Z
1
1
Z
f (x) cos(nπx) dx
∀n ≥ 0,
0
2
=
x cos(nπx) dx
0
=
bn
=
=

 2, if n = 0,
 0, if n ≥ 1
Z
1 2
f (x) sin(nπx) dx
1 0
Z 2
x sin(nπx)
0
=
−
2
,
nπ
∀n ≥ 1.
So we get
x=1−
∞
2 X sin(nπx)
,
π n=1
n
∀x ∈ (0, 2).
Even and odd functions
Definition 1. Let f be a function on the domain I which is symmetric with respect to 0 (that is, x ∈ I if and only if
−x ∈ I), then
1
2
MINGFENG ZHAO
I. we say that f is an even function if f (−x) = f (x) for all x ∈ I.
II. we say that f is an odd function if f (−x) = −f (x) for all x ∈ I.
Remark 1. It’s easy to see the following properties of even and odd functions:
• The sum (difference) and product (quotient) of two even functions are even.
• The sum (difference) of two odd functions is odd; the product (quotient) of two odd functions is even.
• The sum (difference) of an odd function and an even function is neither even nor odd; the product (quotient)
of two such functions is odd.
• Any function can be expressed as a sum of an even function and an odd function.
Lemma 1. For any L > 0, then


 0,
Z
f (x) dx =

−L
 2
Z
if f is odd in [−L, L],
L
L
f (x) dx, if f is even in [−L, L].
0
Proof. In fact, we have
Z
L
0
Z
f (x) dx
L
Z
=
f (x) dx +
−L
−L
f (x) dx
0
0
Z
Z
f (−x) · −dx +
=
f (x) dx
L
0
L
Z
L
=
Z
f (−x) dx +
0
L
f (x) dx
0
L
Z
=
[f (−x) + f (x)] dx
0
=


 0,
Z

 2
if f is odd in [−L, L],
L
f (x) dx, if f is even in [−L, L].
0
Remark 2. Let f be a function on [−L, L], then the Fourier series of f (x) is given by:
∞
f (x) =
nπ nπ i
a0 X h
+
an cos
x + bn sin
x ,
2
L
L
n=1
where
an
=
1
L
Z
L
f (x) cos
−L
nπ x dx
L
∀n ≥ 0,
LECTURE 11: HALF RANGE FOURIER SERIES AND CONVERGENCE OF FOURIER SERIES
bn
=
1
L
L
Z
f (x) sin
−L
nπ x dx
L
3
∀n ≥ 1.
Then
nπ nπ x is odd, and f (x) sin
x is even. By
L
L
• If f is an odd function on [−L, L], then for any n ≥ 0, f (x) cos
Lemma 1, then
an
bn
0 ∀n ≥ 0,
Z
nπ 2 L
f (x) sin
x dx,
=
L 0
L
=
∀n ≥ 1.
Then we get
f (x) =
∞
X
bn sin
n=1
nπ x ,
L
and
bn =
2
L
L
Z
f (x) sin
0
mπ x ,
L
∀n ≥ 1.
nπ nπ x is even, and f (x) sin
x is odd. By
L
L
• If f is an even function on [−L, L], then for any n ≥ 0, f (x) cos
Lemma 1, then
an
bn
=
=
2
L
L
Z
0,
f (x) cos
0
nπ x dx
L
∀n ≥ 0,
∀n ≥ 1.
Then we get
∞
f (x) =
nπ a0 X
+
an cos
x ,
2
L
n=1
and
an =
2
L
L
Z
f (x) cos
0
mπ x ,
L
∀n ≥ 0.
Half-range expansions
If we are given a function f on [0, L], and we want to represent f by a Fourier series with period 2L, then we
have two choices:

 f (x),
0<x<L
• Fourier sine series: We get an odd extension of f on [−L, L], that is, fo (x) =
,
 −f (−x), −L < x < 0
and fo (x + 2L) = fo (x) for all x ∈ R, then
∞
X
nπ fo (x) =
bn sin
x ,
L
n=1
and
2
bn =
L
Z
L
f (x) sin
0
mπ x ,
L
∀n ≥ 1.
4
MINGFENG ZHAO

 f (x),
0<x<L
• Fourier cosine series: We get an even extension of f on [−L, L], that is, fe (x) =
,
 f (−x), −L < x < 0
and fe (x + 2L) = fe (x) for all x ∈ R, then
∞
nπ a0 X
fe (x) =
+
x ,
an cos
2
L
n=1
2
an =
L
and
Z
L
f (x) cos
0
mπ x ,
L
∀n ≥ 0.
Example 2. Expand f (x) = x, 0 ≤ x ≤ 2 in a half-range: (a) Fourier sine series; (b) Fourier cosine series.

 x, 0 < x < 2
(a) Fourier sine series: The odd extension of f (x) is: fo (x) =
, and fo (x + 4) = fo (x) for all x ∈ R,
 x, −2 < x < 0
then
x=
∞
X
bn sin
n=1
nπ x ,
2
0 < x < 2,
where
bn
Z
2
nπ x dx
2
0
2
nπ x sin
x dx
2
0
2
2
Z
=
=
x sin
4(−1)n+1
.
nπ
=
Then
∞
nπ 4 X (−1)n+1
x=
sin
x , x ∈ (−2, 2).
π n=1
n
2

 x,
0<x<2
(a) Fourier cosine series: The even extension of f (x) is: fe (x) =
, and fe (x + 4) = fe (x) for all
 −x, −2 < x < 0
x ∈ R, then
∞
nπ a0 X
+
an cos
x , 0 < x < 2,
x=
2
2
n=1
where
an
2
2
Z
Z
2
nπ x dx
2
0
2
nπ =
x cos
x dx
2
0


if n = 0,
 2,
=
n

 4[(−1) − 1] , if n ≥ 1.
2
n π2
=
x cos
LECTURE 11: HALF RANGE FOURIER SERIES AND CONVERGENCE OF FOURIER SERIES
Then
∞
∞
nπ X
4 X (−1)n − 1
x = 1+
cos
x
π 2 n=1
n2
2
n=1
∞
(2k + 1)π
8 X
1
cos
= 1− 2
x
,
π
(2k + 1)2
2
∀x ∈ [−2, 2].
k=0
Complex form of Fourier series
Recall Euler’s identity eiθ = cos(θ) + i sin(θ), then we have
cos(θ) =
eiθ + e−iθ
,
2
and
sin(θ) =
eiθ − e−iθ
eiθ − e−iθ
= −i ·
.
2i
2
Let
∞
f (x) =
nπ i
nπ a0 X h
+
x + bn sin
x .
an cos
2
L
L
n=1
Then we have
f (x)
=
=
=
Let c0 =
a0
1
=
2
2L
Z
∞
nπ nπ i
a0 X h
+
an cos
x + bn sin
x
2
L
L
n=1
nπ
nπ
nπ
nπ
∞ a0 X
ei L x − e−i L x
ei L x + e−i L x
+
− ibn ·
an ·
2
2
2
n=1
∞
∞
a0 X an − ibn i nπ x X an + ibn −i nπ x
+
e L +
e L .
2
2
2
n=1
n=1
L
f (x) dx, for any n ≥ 1, let
−L
cn =
an − ibn
,
2
an + ibn
= cn .
2
and c−n =
Then we get
f (x) =
∞
X
cn ei
nπ
L x
.
n=−∞
Since
1
an =
L
Z
L
nπ f (x) cos
x dx,
L
−L
1
and bn =
L
Z
L
f (x) sin
−L
nπ x dx,
L
Then for all n ≥ 1, we have
cn
=
=
an − ibn
2
Z L
h
nπ nπ i
1
f (x) cos
x − i sin
x
dx
2L −L
L
L
∀n ≥ 1.
5
6
MINGFENG ZHAO
=
c−n
=
=
=
1
2L
L
Z
f (x)e−i
nπ
L x
dx
−L
an + ibn
2
Z L
h
nπ nπ i
1
f (x) cos
x + i sin
x
dx
2L −L
L
L
Z L
nπ
1
f (x)ei L x dx.
2L −L
In summary, we have
cm =
Z
1
2L
L
f (x)e−i
mπ
L x
dx,
∀m ∈ Z.
−L

 −1, if −π ≤ x < 0,
Example 3. Let f (x) =
 1,
if 0 < x < π.
Then
cn
π
f (x)e−inx dx
−π
Z 0
Z π
1
−inx
−inx
−
e
dx +
e
dx .
2π
−π
0
=
Z
Z
1
2π
=
0
For
e−inx dx, then
−π
Z
0
e
−inx
Z
dx
0
einy · −dy
=
−π
π
Z π
=
Let y = −x, then dx = −dy
einy dy
0
Z
=
π
einx dx
Let y = x.
0
So we get
cn
=
=
=
=
Z π
Z π
1
−
einx dx +
e−inx dx
2π
0
0
Z π
1
[−einx + e−inx ] dx
2π 0
Z
−i π
sin(nx) dx
π 0


 0,
if n is even
−2i


, if n is odd.
nπ
LECTURE 11: HALF RANGE FOURIER SERIES AND CONVERGENCE OF FOURIER SERIES
7
So we get
∞
X
f (x) = −
k=−∞
2i
ei(2k+1)x .
(2k + 1)π
Convergence of Fourier series
Theorem 1. Suppose that f and f 0 are piecewise continuous on the interval [−L, L). Suppose that f is defined outside
the interval [−L, L) so that it is periodic with period 2L. Then f has a Fourier series:
∞
nπ i
nπ a0 X h
+
x + bn sin
x ,
an cos
2
L
L
n=1
f (x) =
where
an
=
bn
=
1
L
L
Z
nπ x dx ∀n ≥ 0,
L
−L
Z
nπ 1 L
x dx ∀n ≥ 1.
f (x) sin
L −L
L
f (x) cos
Moreover, the Fourier series converges to f (x) at all points where f is continuous, and it converges to
at all points where f is discontinuous.
Proof. For any N ≥ 1, let
N
SN (x) =
nπ nπ i
a0 X h
+
an cos
x + bn sin
x .
2
L
L
n=1
For any x ∈ R and fix it, for any n ∈ Z, then
Z x−L
Z L
nπ
y
−i nπ
L
dy =
f (x − z)e−i L (x−z) · −dz
f (y)e
−L
Let y = x − z
x+L
Z
x+L
f (x − z)ei
=
nπ
L z
· e−i
nπ
L x
dz
x−L
= e−i
nπ
L x
Z
x+L
f (x − z)ei
nπ
L z
dz
x−L
= e−i
nπ
L x
Z
L
f (x − z)ei
nπ
L z
dz
Since f has period 2L.
−L
By the computation in Complex form of Fourier series in Lecture 11, then
SN (x)
=
N
X
cn e i
nπ
L x
n=−N
=
N
X
n=−N
e
i nπ
L x
1
·
2L
Z
L
−L
f (y)e−i
nπ
L y
dy
f (x+) + f (x−)
2
8
MINGFENG ZHAO
1
2L
=
For
N
X
ei
nπ
L y
, if y = 0, then
n=−N
N
X
ei
nπ
L y
Z
"
L
f (x − y) ·
−L
#
N
X
e
= 2N + 1; if x 6= y, let θ =
ei
nπ
L z
dy.
n=−N
n=−N
N
X
i nπ
L y
N
X
=
n=−N
π
y, then
L
einθ
n=−N
=
e−iN θ − ei(N +1)θ
1 − eiθ
=
e−iN θ − ei(N +1)θ e−i 2
·
θ
1 − eiθ
e−i 2
θ
e−i
=
2N +1
θ
2
− ei
θ
2N +1
θ
2
θ
e−i 2 − ei 2
2N +1
2 θ
−2i sin θ2
−2i sin
=
2N +1
2 θ
sin θ2
sin
=
sin
=
(2N +1)π
y
2L
π
sin 2L
z
.
Let



2N + 1,
if z = 0,



DN (z) =
(2N +1)π

sin
z

2L


, if z 6= 0.

π
sin 2L
z
Notice that
Z
L
Z
Dn (z) dz
L
=
−L
N
X
ei
nπ
L z
dz
−L n=−N
Z
L
=
dz +
−L
=
X
2L +
1≤|n|≤N
X
1≤|n|≤N
=
2L,
Z
Z
L
ei
nπ
L z
dz
−L
L
h
nπ nπ i
cos
z + i sin
z
dz
L
L
−L
By Lemma 3 and Lemma 4 in Lecture 10.
LECTURE 11: HALF RANGE FOURIER SERIES AND CONVERGENCE OF FOURIER SERIES
9
So we get
|f (x) − SN (x)| =
=
=
≤
Let F (z) =
Z L
1
f (x − z)DN (z) dz f (x) −
2L −L
Z L
1 Z L
1
f (x)DN (z) dz −
f (x − z)DN (z) dz 2L −L
2L −L
1 Z L
[f (x) − f (x − z)]DN (z) dz 2L −L
Z L
1
|f (x) − f (x − z)||DN (z)| dz.
2L −L
f (x) − f (x − z)
for z 6=, then
z
1
|f (x) − SN (x)| ≤
2L
Z
L
|F (z)| · |zDN (z)| dz.
−L
If f is differentiable at x, one can use the Riemann’s Lemma to prove that |f (x) − SN (x)| → 0, as N → ∞.
Remark 3. Riemann’s Lemma states that: for any integrable function f on [−L, L], then
Z
L
f (x)einx dx = 0.
lim
n→∞
−L

 0, if −L < x < 0,
Example 4. Let L > 0 and f (x) =
, and let f be defined outside (−L, L) so that f (x + 2L) =
 L, if 0 < x < L
f (x) for all x. Find the Fourier series for f and determine where it converges.
It’s easy to see that f satisfies the conditions in Theorem 1, then f has a Fourier series:
∞
nπ nπ i
a0 X h
f (x) =
+
an cos
x + bn sin
x ,
2
L
L
n=1
where
a0
=
=
an
1
L
Z
1
L
Z
=
1
=
1
L
L
f (x) dx
−L
L
L dx
0
Z
L
f (x) cos
−L
nπ x dx
L
10
MINGFENG ZHAO
Z
nπ 1 L
L cos
x dx
L 0
L
Z L
nπ =
cos
x dx
L
0
nπ L
L
=
sin
x nπ
L
=
0
=
=
bn
=
=
=
=
L
sin(nπ)
nπ
0 ∀n ≥ 1,
Z
nπ 1 L
f (x) sin
x dx
L −L
L
Z
nπ 1 L
L sin
x dx
L 0
L
Z L
nπ x dx
sin
L
0
nπ L
L
−
cos
x nπ
L
0
L
[cos(nπ) − 1]
nπ
1 − (−1)n
, ∀n ≥ 1.
nπ
= −
=
So we get
∞
f (x)
=
=
nπ L X 1 − (−1)n
+
sin
x
2 n=1
nπ
L
∞
L
2X 1
(2k + 1)π
+
sin
x .
2
π
2k + 1
L
k=0
Since f is continuous in (−L, 0) and (0, L), then
∞
L
2X 1
(2k + 1)π
f (x) = +
sin
x ,
2
π
2k + 1
L
∀x 6= ZL.
k=0
Notice that f can not be continuous at x = ZL, but we have
f (0−) + f (0+)
f (L−) + f (L+)
f (mL−) + f (mL+)
L
=
=
= ,
2
2
2
2
On the other hand, we know that
∞
L
2X 1
(2k + 1)π
L
+
sin
· (mL) = ,
2
π
2k + 1
L
2
k=0
∀m ∈ Z.
∀m ∈ Z.
LECTURE 11: HALF RANGE FOURIER SERIES AND CONVERGENCE OF FOURIER SERIES
11
Gibbs phenomenon
Theorem 2. Let f be a piecewise continuously differentiable function which is periodic with period 2L > 0. Suppose
that f is not continuous at x0 but f (x0 +) and f (x0 −) exists such that
f (x0 +) − f (x0 −) = α 6= 0.
For the Fourier series of f :
∞
f (x) =
nπ i
nπ a0 X h
+
an cos
x + bn sin
x ,
2
L
L
n=1
where
Let SN (x) =
an
=
bn
=
1
L
Z
L
nπ x dx ∀n ≥ 0,
L
−L
Z
nπ 1 L
x dx ∀n ≥ 1.
f (x) sin
L −L
L
f (x) cos
N
−1 h
nπ i
nπ X
a0
+
x + bn sin
x , then there exists a positive constant C > 0 such that
an cos
2
L
L
n=1
L
lim SN x0 +
= f (x0 +) + C · α
N →∞
N
L
= f (x0 −) − C · α
lim SN x0 −
N →∞
N

 −1, −π ≤ x < 0,
Example 5. Let f (x) =
, then f is odd, and the Fourier series of f is:
 1,
0≤x<π
∞
f (x) =
a0 X
[an cos(nx) + bn sin(nπ)] .
+
2
n=1
Then we have
an
=
=
bn
=
Z
1
1
f (x) cos(nx) dx
−1
for all n ≥ 0,
0,
1
1
Z
1
f (x) sin(nx) dx
−1
Z
=
1
2
1
sin(nx) dx
0
12
MINGFENG ZHAO
=
2[1 − (−1)n ]
,
nπ
for all n ≥ 1.
So we have
f (x)
=
∞
2 X 1 − (−1)n
sin(nx)
π n=1
n
=
∞
4X 1
sin[(2k + 1)x],
π
2k + 1
∀x ∈ (−π, 0)
[
(0, π).
k=0
For any N ≥ 1, let
SN (x) =
N −1
4 X sin[(2k + 1)x]
·
.
π
2k + 1
k=0
Then
SN
2π
2N
=
N −1
2π
4 X sin (2k + 1) 2N
·
π
2k + 1
k=0
=
=
=
N −1
4 2π X sin 2k+1
2N · 2π
·
·
2k+1
π 2N
2N · 2π
k=0
N −1
2 X sin 2k+1
2N · 2π
·
2k+1
N
2N · 2π
k=0
N −1
1 2π X sin 2k+1
2N · 2π
·
·
.
2k+1
π N
2N · 2π
k=0
It’s easy to see that SN
2π
2N
is exactly the midpoint Riemann sum of
sin(x)
with N intervals over [0, 2π], which
x
implies that
lim SN
N →∞
Z
2π
For
π
2π
2N
1
=
π
Z
2π
sin(x)
1
dx =
x
π
0
Z
0
π
sin(x)
dx +
x
Z
2π
π
sin(x)
dx .
x
sin(x)
dx, we have
x
Z
2π
π
π
Z
sin(x)
dx
x
sin(y + π)
dy
y+π
=
0
π
sin(y)
dy
y+π
π
0
sin(x)
dx
x+π
Z
= −
0
Z
= −
Let x = y + π
Let y = x.
Then
lim SN
N →∞
2π
2N
1
=
π
Z
0
π
1
1
sin(x) ·
−
x x+π
Z
dx =
0
π
sin(x)
dx > 0.
x(x + π)
LECTURE 11: HALF RANGE FOURIER SERIES AND CONVERGENCE OF FOURIER SERIES
13
Since SN (x) is odd, then
lim SN
N →∞
2π
−
2N
Z
=−
0
π
sin(x)
dx < 0.
x(x + π)
Notice that f (0+) − f (0−) = 2.
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca