8
Fourier Series
Our aim is to show that under reasonable assumptions a given 2π-periodic
function f can be represented as convergent series
∞
a0 X
(an cos nx + bn sin nx) .
+
2
n=1
f (x) =
(8.1)
By definition, the convergence of the series means that the sequence (sn (x))
of partial sums, defined by
n
sn (x) =
a0 X
+
(ak cos kx + bk sin kx) ,
2
k=1
converges at a given point x to f (x), sn (x) → f (x).
We start with some definitions. A real valued function defined on R is
said to be periodic with period L, or L-periodic, is f (x + L) = f (x) for
all x. In this case f is completely determined by its values on any interval
[a, a+L). A function f defined on [a, a+L) can be extended in a unique way
to a periodic function on R. Indeed, given x, there exists a unique integer
n so that x belongs to [a + nL, a + (n + 1)L) so that x − nL ∈ [a, a + P )
and we set f (x) = f (x − nL). A function f on [a, b] is piecewise continuous
if f is continuous except a finite numbers of points and at each such point
the one-sided limits f (x+ ) = limε→0− f (x + ε) and f (x− ) = limε→0− f (x +
ε) exist and are finite. We say that f is piecewise smooth (or piecewise
differentiable) on [a, b] if f and f ′ are piecewise continuous on [a, b]. If f
is piecewise continuous, then it is Riemann integrable over any bounded
interval contained in its domain. In addition, if f is L-periodic, then
Z L
Z a+L
f x) dx
f (x) dx =
0
a
for every a. In what follows we focus on functions which are 2π-periodic.
Assume
∞
a0 X
(an cos nx + bn sin nx) .
+
f (x) =
2
n=1
We also assume that the convergence is well-behaved so that we can integrate
term-by-term. We want to compute the coefficients ak and bk . We use
66
integral identities


0,
cos mx cos nx dx = 2π,

−π

π,
(
Z π
0,
sin mx sin nx dx =
π,
−π
Z π
cos mx sin nx dx = 0.
Z
π
m 6= n
m=n=0
m = n 6= 0
m 6= n
m = n 6= 0
−π
To compute a0 , we multiply both sides of (8.1) by the constant function 1
and integrate over [−π, π] and get
Z
Z π
Z π
Z
∞ X
a0 π
f (x) dx =
sin nx dx
cos nx dx + bn
1 dx +
an
2 −π
−π
−π
−π
n=1
a0
· (2π) = πa0
=
2
π
To compute am with m ≥ 1 we we multiply both sides of (8.1) by cos mx
and integrate over [−π, π] to get
Z π
Z
a0 π
f (x) cos mx dx =
cos mx dx
2 −π
−π
Z π
Z π
∞ X
sin nx cos mx dx
+
cos nx cos mx dx + bn
an
= am
−π
−π
n=1
Z
π
cos mx cos mx dx = πam .
−π
Similarly, multiplying both sides of (8.1) by sin mx and integrating over
[−π, π], we find that
Z π
f (x) sin mx dx = πbm .
−π
Summing up,
Z
1 π
an =
f (x) cos nx dx,
π −π
Z
1 π
f (x) sin nx dx,
bn =
π −π
67
n≥0
(8.2)
n ≥ 1.
The coefficients an and bn are called the Fourier coefficients of f and the
series
∞
a0 X
[an cos nx + bn sin nx]
+
2
n=1
the Fourier series of f . We denote this fact by
∞
f (x) ∼
a0 X
[an cos nx + bn sin nx] .
+
2
n=1
The symbol ∼ should be read as f “has Fourier serier”.
Example 8.1. Let f be a 2π-periodic function given by f (x) = x for x ∈ (−π, π].
Then integrating by parts
π
Z
Z π
1 π
1 x sin nx
1
an =
−
x cos nx dx =
sin nx dx = 0
π −π
π
n
nπ −π
−π
and
1
bn =
π
Z π
1
1 h x cos nx iπ
−
+
x sin nx dx =
cos nx dx
π
n
nπ −π
−π
−π
Z
π
=−
2(−1)n+1
2 cos nπ
=
.
n
n
Consequently,
f (x) ∼ 2
X (−1)n+1
sin nx.
n
n≥1
The series converges for every x in view of the following Dirichlet’s test. If (an ) is
a decreasing
sequence converging to 0 and
P
P the sequence (sn ) of partial sums of 1the
series
bn is bounded, then the series
an bn converges. In our case, an = n is
decreasing and converges to 0, (−1)n sin nx = sin n(x + π), and
n
(n + 1)ϑ
nϑ
ϑ X
sin kϑ = sin
· sin
sin ·
2
2
2
k=1
which implies that
if ϑ 6= 2πk for all k ∈ Z.
n
X
1
sin kϑ ≤ sin ϑ2 k=1
Example 8.2. Consider f be a 2π-periodic function defined by f (x) = |x| for
x ∈ [−π, π]. Note that f is even. Since the product of even function with the odd
function is odd, it follows that
Z π
f (x) sin mxdx = 0.
−π
68
Hence bn = 0 for all n ≥ 1. To compute bn note that the product of an even
function with an even function is even so that
Z
Z
Z
1 π
2 π
2 π
an =
f (x) cos nx dx =
f (x) cos nxdx =
x cos nx dx.
π −π
π 0
π 0
If n = 0, then
a0 =
2
π
Z
π
x dx = π,
0
and if n ≥ 1, then integrating by parts, one finds that
π
Z
Z π
2
2 x sin nx
2 π
−
x cos nx dx =
sin nx dx
π 0
π
n
nπ 0
0
π
2
2
= 2 cos nx = 2 [(−1)n − 1] .
n π
n π
0
Hence an = 0 if n is even and an = − n24π when n is odd, and hence
f (x) ∼
4 X cos(2n − 1)x
π
−
.
2
π
(2n − 1)2
n≥1
Since the series
series converges.
P
1
n2
converges, the M-Weierstarss test implies that the above
In the above examples we have used the following identities,
( RL
Z L
2 0 g(x) dx
if g is even
g(x) dx =
0
if g is odd.
−L
Using that the sum of two even functions is even and the sum of two odd
functions is odd, that the product of two even functions or two odd functions
is even, and the product of an even function and an odd function is odd, we
have the following proposition.
Proposition 8.3. (i) If f is even, then its Fourier sine coefficients bn
are equal to 0 and f is represented by Fourier cosine series
∞
f (x) ∼
where
2
an =
π
Z
a0 X
an cos nx
+
2
n=1
π
f (x) cos nx dx
−π
69
for n ≥ 0.
(ii) If f is odd, then its Fourier cosine coefficients an are equal to 0 and
f is represented by Fourier sine series
f (x) ∼
∞
X
an sin nx
n=1
where
8.0.1
2
bn =
π
Z
π
f (x) sin nx dx
for n ≥ 1.
−π
Convergence Theorem
Theorem 8.4. Assume that f is a 2π-periodic function which is piecewise
smooth. Then
n
a0 X
1
−
+
[ak cos kx + bk sin kx]
+
f (x ) + f (x ) =
2
2
k=1
where f (x± ) = limy→x± f (y). In particular, if f is continuous at x, then
n
a0 X
f (x) =
+
[ak cos kx + bk sin kx] .
2
k=1
Example 8.5. Consider the function f from Example 8.1. Then f is smooth at
all points except points mπ where m is odd. The one sided limits at these points
are
f (mπ − ) =
lim f (x) = π
x→mπ −
and
f (mπ + ) =
lim f (x) = −π
x→mπ +
provided that m is odd. Consequently, the Fourier series converges of f converges
to f at every point except points mπ with m odd. At these points the Fourier series
converges to 21 [f (mπ − ) + f (mπ + )] = 0. In particular,
∞
x X (−1)n+1
=
sin nx,
2 n=1
n
for x ∈ (−π, π).
Taking x = π/2, we find that
∞
X
(−1)n−1
π
=
.
4
2n − 1
n=1
70
Example 8.6. Consider the function 2π-periodic function f given by f (x) = |x|
for x ∈ (−π, π). (See Example 8.2.) Then f is continuous at every point and it is
smooth except points mπ with m odd. Hence the Fourier series of f converges to
f at every point. In particular,
|x| =
∞
π
4 X cos(2n − 1)x
−
2
π n=1 (2n − 1)2
for x ∈ [−π, π].
Substituting x = 0, we find that
∞
X
1
π2
.
=
8
(2n
+
1)2
n=0
Set S =
P∞
1
n=1 n2 .
Then we note that
∞
∞
X
S
1
1X 1
=
=
2
4
4 n=1 n2
(2n)
n=1
and hence
∞
∞
∞
X
X
X
1
1
1
π2
S
3
−
=
=
S=S− =
.
2
2
2
4
4
n
(2n)
(2n + 1)
8
n=1
n=0
n=1
From this we conclude that
S=
∞
X
1
π2
=
.
2
n
6
n=1
Integration of Fourier series
We start with the following observation.
R x Assume that f is 2π-periodic,
piecewise continuous,
and let F (x) = 0 f (y) dy. Then F is 2π periodic if
Rπ
and only if −π f (y)dy = 0. Indeed, we have
F (x + 2π) − F (x) =
Z
x+2π
f (y)dy =
x
Z
π
f (y)dy.
−π
This means that
R π the constant term in the Fourier series of f is equal to 0.
1
Since a20 = 2π
−π f (y) dy, the number a0 /2 is the mean of the function f
over the interval [−π.π].
Theorem 8.7. Assume that f is 2π-periodic and piecewise continuous and
its mean is equal to 0. Then its Fourier series
X
f (x) ∼
[an cos nx + bn sin nx]
n≥1
71
can be integrated term by term and produce the Fourier series
Z x
X bn
an
sin nx
f (y) dy ∼ C0 +
− cos nx +
F (x) =
n
n
0
n≥1
Rπ
1
where the constant C0 = 2π
−π F (y) dy.
Example 8.8. Consider a 2π-periodic function f given by f (x) = x on (−π, π].
(See Example 8.1.) Since f is odd, its mean value over [−π, π] is equal to 0. Its
Fourier series is given by
X (−1)n−1
sin nx,
for x ∈ (−π, π].
x=2
n
n≥1
Integrating term by term we find that
X (−1)n−1
x2
π2
cos nx
=
−2
2
6
n2
for x ∈ (−π, π]
n≥1
where π 2 /6 =
1
2π
Rπ
−π
(x2 /2) dx.
Differentiation of Fourier series
Proposition 8.9. Assume that f is 2π-periodic, continuous, and piecewise
smooth. Abbreviate by an , bn the Fourier coefficients of f and by a′n , b′n the
Fourier coefficients of f ′ . Then
a′n = nbn
and
b′n = −nan .
Proof. Integrating by parts,
a′n =
1
π
Z
π
f ′ (x) cos nx dx =
−π
Similarly, b′n = −nan .
Z
iπ
1h
n π
f (x) cos nx
+
f (x) sin nx dx = nbn .
π
π −π
−π
As a consequence, we get:
Theorem 8.10. Let f be 2π-periodic, continuous, and piecewise smooth.
In addition, assume that f ′ is piecewise smooth. If
∞
then
a0 X
f (x) =
[an cos nx + bn sin nx] ,
+
2
n=1
∞
i X
1h ′ −
′ +
[nbn cos nx − nan sin nx] .
f (x ) + f (x ) =
2
n=1
In particular, at points x at which f ′ (x) exists, the series converges to f ′ (x).
72
Proof. By Proposition 8.9,
f ′ (x) ∼
∞
X
[nbn cos nx − nan sin nx] ,
n=1
and since f ′ is piecewise smooth, the theorem follows from Theorem 8.4.
Example 8.11. Recall the function 2π-periodic function f given by f (x) = |x|
for x ∈ (−π, π). (See Example 8.2 and Example 8.6.) The function f is continuous
and piecewise smooth, and
|x| =
∞
4 X cos(2n − 1)x
π
,
−
2
π n=1 (2n − 1)2
x ∈ [−π, π].
The derivative of f is given by
d
|x| =
dx
(
1,
−1,
x > 0,
x < 0.
In view of Theorem 8.10,
∞
4 X sin(2n − 1)x
=
π n=1 (2n − 1)
(
1,
−1,
0 < x < π,
−π < x < 0.
At x = 0, the series converges to 12 [f ′ (x− ) + f ′ (x+ )] = 12 [−1 + 1] = 0.
8.0.2
Absolute and Uniform Convergence
Theorem 8.12. Let f be 2π-periodic, continuous, and piecewise smooth.
Then its Fourier series converges absolutely and uniformly.
Proof. We prove the result under additional assumption that f ′ is piecewise smooth.
This means that f ′′ is piecewise continuous. Abbreviate by an , bn the Fourier coefficients of f and by a′n , b′n the Fourier coefficients of f ′ . Since f is continuous,
∞
f (x) =
a0 X
[an cos nx + bn sin nx] .
+
2
n=1
To prove the absolute convergence it suffices to show that |an | , |bnP
| ≤ M/n2 for
some constant M independent of n. Then, by the comparison test, n≥1 |an | and
P
n≥1 |bn | converge which imply the absolute convergence of the Fourier series.
Also, the Weierstrass M-test implies the uniform convergence. By Proposition 8.9,
an = −b′n /n. That is,
Z
Z π
1
1 π
f (x) cos nx dx = −
f ′ (x) sin nx dx.
an =
π −π
nπ −π
73
By assumption, f ′ is piecewise continuous and so f ′′ is continuous except a finite
number of point at which it has a jump. If (a, b) is an interval on which f ′′ is
continuous, then
1
nπ
b
Z b
1
1
′
f
(x)
cos
nx
−
f ′′ (x) cos nx dx
2π
n2 π
n
−π
a
a
Z b
1
1
−
+
f ′′ (x) cos nx dx.
= 2 f (b ) cos nx − f (a ) cos nx + 2
n π
n π a
Z
π
f ′ (x) sin nx dx =
Since |f ′ (x)| , |f ′′ (x)| ≤ K for some constant K and all x (at points x where f ′ and
f ′′ has a jump discontinuity, f ′ (x) one has to take left- and hand-side derivatives).
So,
Z
K(2 + (b − a)
1 π ′
2K(1 + π
4K
≤
≤ 2.
f (x) sin nx dx ≤
2
2
nπ −π
n π
n π
n
Rπ ′
′′
Since f has a finite number of discontinuities, the integral −π f (x) sin nx dx can
Rb
be written as the finite number, say m, of integrals a f ′ (x) sin nx dx over intervals
on which f ′′ is continuous. Then
Z
4Km
1 π ′
|an | =
.
f (x) sin nx dx ≤
nπ −π
n2
Similarly, |bn | ≤
4Km
n2 .
Theorem 8.13. Let f be 2π-periodic. Assume that f is of class C k−1 and
that f (k−1) is piecewise smooth. If the Fourier coefficients satisfy
|an | ≤
C
nk+α
and
|bn | ≤
C
nk+α
for some α > 1 and C > 0, then f is of class C k .
(k)
Proof. Let a(j)
series of f (j) . Then using
n and bn are coefficients of the Fourier
(j) (j) Proposition 8.9 an = nj |an | if j is even and an = nk |bn | if j is odd. Similarly,
(k)
for bn . Observe, using the assumption, that if j ≤ k, then
X
X
n j an ≤ C
n≥1
n≥1
1
nk−j+α
≤C
X 1
nα
n≥1
P
for j ≤ k. By the Weierstrass M-test, the series n≥1 nj an converges absolutely
P
and uniformly for j ≤ k. Similarly, the series n≥1 nj bn converges absolutely
P
j)
j)
and uniformly for j ≤ k Hence, f (j) = n≥1 [an cos nx + bn sin nx] for j ≤ k and
f (k) is continuous.
74
8.1
Changing a scale
So far we have considered 2π-periodic functions and their Fourier series over
the interval [−π,
π]. Now, given a function f defined on [−L, L], we define
L
F (y) = f π y . Then F is defined on [−π, π] and its Fourier series is given
by
a0 X
an cos ny + bn sin ny
F (y) ∼
+
2
n≥1
where
an =
1
π
Z
π
F (y) cos ny dy
and
bn =
−π
1
π
Z
π
F (y) sin ny dy.
(8.3)
−π
Since f (x) = F ( Lπ x) for x ∈ [−L, L], we deduce that
f (x) ∼
a0 X
nπ
nπ an cos
+
x + bn sin
x
2
L
L
n≥1
The coefficients an and bn can be computed by integrating (8.3) by substitution x = Lπ y. We get
Z
Z
L
1 π
1 π
y cos ny dy
F (y) cos ny dy =
f
an =
π −π
π −π
π
(8.4)
Z
1 L
nπx
=
dx
f (x) cos
L −L
L
and similarly
bn =
1
π
Z
π
F (y) sin ny dy =
−π
75
1
L
Z
L
−L
f (x) cos
nπx
dx.
L
(8.5)