Chapter Two
Fourier Series
1. Introduction. All of the eigenfunctions found in the last chapter for the eigenvalue
problem
φ00 (x) = µφ(x)
subject to the various boundary conditions were combinations of sines and cosines and thus
periodic functions. We recall that a function f defined on (−∞, ∞) is periodic with period
p if
f (x + p) = f (x)
for all x ∈ (−∞, ∞).
Since f (x + mp) = f (x + (m − 1)p) = · · · = f (p) it follows that any periodic function with
period p also has period mp for any integer m. For example, the functions sin(mωx) and
cos(mωx) have period p = 2π/mω since
sin(mω(x + p)) = sin(mωx + 2π).
Note that this identity implies that 2π/ω is the period common for all these functions for
m = 1, 2, . . . .
It now follows that the orthogonal projection P f of a square integrable function f defined
on [0, L] into the span of trigonometric functions is automatically defined on (−∞, ∞), and
if all of these functions have a common period p then P f is periodic on the whole line with
period p. Let us look at the eigenfunction expansions discussed in Chapter ST. If
MN = span{sin λn x},
λn = nπ/L,
n = 1, 2, . . .
then P f has period 2L. Moreover, since each sine function is odd it follows that P f is an
odd 2L periodic function on (−∞, ∞). If
MN = span{sin λn x},
λn =
π
2
+ nπ /L,
n = 0, 1, 2, . . .
then P f is an odd periodic function with period 4L. Moreover, it follows from the addition
formula
sin(x + y) = sin x cos y + cos x sin y
that
sin
π
2
+ nπ (L + ∆L)
= sin
L
π
2
+ nπ (L − ∆L)
L
1
for ∆L > 0
so that this 4L periodic function is also even with respect to x = L. Similarly, if
MN = span{cos λn x},
λn =
π
2
+ nπ /L,
n = 0, 1,
then P f is an even approximation of f defined on (−∞, ∞) with period 4L. Moreover, the
addition formula
cos(x + y) = cos x cos y − sin x sin y
shows that
P f (L + ∆L) = −P f (L − ∆L)
so that P f also is odd with respect to x = L.
However, the orthogonal projection P f into the span of eigenfunctions associated with
boundary conditions like
u(0) = 0,
u0 (L) = u(L)
is not periodic on (−∞, ∞) because in general sin λm x and sin λn x have no common period.
nπx
Let us now turn to the eigenfunctions cos mπx
associated with
,
sin
L
L
φ00 (x) = µφ(x)
φ(0) = φ(L)
φ0 (0) = φ0 (L).
The Sturm-Liouville theory assures that the orthogonal projection P f of a function f in
L2 [0, L] converges in the mean square sense to f . However, for these functions a great
deal more is known about their approximating properties than just mean square convergence. These periodic functions are the basis of Fourier series approximations to continuous
and discontinuous periodic functions which are widely used in signal recognition and data
compression and, of course in diffusion and vibration studies. We shall give next a fairly
self-contained outline of the mathematics of Fourier series.
Let X be the linear space of all continuous, piecewise smooth functions that are periodic
with period 2L for some L > 0 with the usual definitions of addition and scalar multiplication. (The function f is piecewise continuous if it has at most a finite number of jump
discontinuities on any finite interval. It is piecewise smooth if it is piecewise continuous
2
and on every finite interval has a piecewise continuous derivative except at possible a finite
number of points.) It is easy to verify that with inner product
(f, g) =
ZL
f (x)g(x)dx,
−L
the sequence
n
1, cos
o
πx
πx
πx
πx
πx
πx
. sin
, cos 2 , sin 2 , cos 3 , sin 3 , . . .
L
L
L
L
L
L
is an orthogonal sequence. It is clear that the inner product (1, 1) = 2L and also easy to
verify that
πx πx
=
sin n , sin n
L
L
ZL
−L
ZL
πx
πx πt
πt
sin n dt = L, and cos n , cos n
= cos2 n dt = L.
L
L
L
L
2
−L
Thus for an integer N , the best least squares approximation Sn (x) of f ∈ X by a linear
combination of
n
πx
πx
πx
πx
πx
πx
πx
πx o
1, cos
. sin
, cos 2 , sin 2 , cos 3 , sin 3 , . . . , cos N , sin N
L
L
L
L
L
L
L
L
is given by
a0 X πx
πx an cos n
Sn (x) =
+
+ bn sin n
,
2
L
L
n=1
N
where
1
an =
L
ZL
1
πt
f (t) cos n dt, and bn =
L
L
−L
The series
ZL
f (t) sin n
πt
dt.
L
−L
a0 X πx
πx +
an cos n
+ bn sin n
2
L
L
n=1
∞
is called the Fourier series of f , named for the French mathematician Joseph Fourier
(1768-1830), who introduced it in his solutions of the heat equation−more of this later. The
fact that a series is the Fourier series of a function f is usually indicated by the notation
πx
πx a0 X +
an cos n
+ bn sin n
f∼
.
2
L
L
n=1
∞
The coefficients an and bn are called the Fourier coefficients of f .
3
∞
P
j=1
Proposition 2.1. Suppose an and bn are the Fourier coefficients of f . Then the series
(a2k + b2k ) converges.
Proof : Applying Bessell’s inequality to the orthonormal sequence
1
1
πx 1
πx 1
πx 1
πx 1
πx 1
πx
√ , √ cos
, √ sin
, √ cos 2 , √ sin 2 , √ cos 3 , √ sin 3 , . . .
L
L
L
L
L
L
2L L
L
L
L
L
L
gives us
Z
∞
X
a20 L
2
2
+L
(an + bn ) ≤ (f (x))2 dx,
2
n=1
2L
0
from which the proposition follows at once.
of Riemann’s Lemma.
Proposition 2.2.
lim
RL
n→∞ −L
The next proposition is an easy consequence
dt = lim
f (t) cos n πt
L
RL
n→∞ −L
f (t) sin n πt
dt = 0.
L
Is the given orthogonal sequence is an approximating basis? That is a good question
which we shall address in this chapter. We shall also investigate not only mean convergence
of this series in several different subspaces of X, but also other types of convergence.
2. Convergence. Before we come to grips with convergence questions regarding the
Fourier series, a brief review of the convergence of sequences of functions is in order. We
shall look at three different types of convergence. Let us recall the definitions.
Definition. A sequence (fn ) of functions all with domain D is said to converge pointwise to the function f if for each x ∈ D it is true that fn (x) → f (x).
Definition. A sequence (fn ) of functions all with domain D is said to converge uni-
formly to the function f if given an ε > 0, there is an integer N so that |fn (x) − f (x)| < ε
for all n ≥ N and all x ∈ D.
Definition. A sequence (fn ) of functions all with domain D = [a, b] is said to converge
Rb
in the mean to the function f if [fn (x) − f (x)]2 dx → 0.
a
A few examples will help illuminate these definitions. For each positive integer n ≥ 2
and for x ∈ [0, 1], let
Thus, fn looks like


n2 x
if 0 ≤ x ≤ 1/n
1
2
−n (x − n ) + n if 1/n < x ≤ 2/n .
fn (x) =

0
if 2/n < x ≤ 1
4
(1/n,n)
2/n
It should be clear that for each x ∈ [0, 1], we have lim fn (x) = 0. Thus the sequence
n→∞
(fn ) converges pointwise to the function f (x) = 0. Note that this sequence does not, however,
converge uniformly to f (x) = 0. In fact, for every n, there is an x ∈ [0, 1] such that
|fn (x) − f (x)| ≥ n.
Next, note that
Z1
2
(fn (x))2 dx = n,
3
0
which tells us that (fn ) does not converge in the mean to f .
Next, for each integer n ≥ 1 and x ∈ [0, 1], let
n − n3 x if 0 ≤ x ≤ 1/n3
fn (x) =
.
0
if 1/n3 < x ≤ 1
A picture:
n
1/n^3
5
R1
It is clear that (fn (x))2 dx =
0
1
,
3n
and so our sequence converges in the mean to f (x) = 0.
Clearly it does not converge pointwise or uniformly.
We see next that uniform convergence is the nicest of the three.
Theorem 2.3. Suppose the sequence (fn ) of functions with domain [a, b] converges
uniformly to the function f . Then (fn ) converges pointwise to f and also converges in the
mean to f .
Proof : It is obvious that the sequence converges to f pointwise. For convergence in the
p
mean, let ε > 0 and choose n sufficiently large to insure that |fn (x) − f (x)| < ε/(b − a)
for all x ∈ [a, b]. Then
Zb
|fn (x) − f (x)|2 dx <
a
Zb
a
ε
dx = ε.
b−a
Hence, (fn ) converges in the mean to f .
We shall make use of the so-called Weierstrass M-test, named for the German
mathematician Karl Weierstrass (1815-1897), given in the following theorem.
∞
P
Theorem 2.4. Consider the series of functions
gn (x) for x ∈ [a, b]. If for all n,
we have |gn (x)| ≤ Mn for all x ∈ [a, b] and if
∞
P
n=0
Mn converges, then
n=0
uniformly and absolutely on [a, b].
∞
P
gn (x) converges
n=0
Proof : First, we show absolute convergence. Let ε > 0. Choose N so that
k=m
Thus
∞
P
n=0
n
P
k=m
Mk <
k=m
ε for n, m > N. Then for n, m > N,
n
X
n
P
|gk (x)| ≤
n
X
Mk .
k=m
|gn (x)| converges to a limit function G(x).
Now, to see the convergence to G(x) is uniform, let ε > 0 and choose N so that
Mk < ε/2 for all n ≥ N .
n
n
m
X
X
X
gk (x) − G(x) + gk (x)
gk (x) − G(x) ≤ k=m
k=0
k=0
n
m
X
X
≤ gk (x) − G(x) +
Mk
k=0
6
k=m
3. Uniform convergence of Fourier series. Suppose f is a piecewise smooth continuous function of period 2L. We shall show that the Fourier series
πx
πx a0 X +
an cos n
+ bn sin n
,
f∼
2
L
L
n=1
∞
converges uniformly to some function g. Notice that this will not establish that the orthogonal sequence
n
o
πx
πx
πx
πx
πx
πx
. sin
, cos 2 , sin 2 , cos 3 , sin 3 , . . .
1, cos
L
L
L
L
L
L
in the space of continuous piecewise smooth functions with the usual inner product is an
approximating basis because it does not say that the function g is the function f with which
we started. This is in fact the case, but a proof of must wait until later.
Theorem 2.5. If f is a piecewise smooth continuous function, the Fourier series of f
converges uniformly.
Proof : Let g(x) = f 0 (x) at all points where f has a derivative. Define g(x) = 0 at
the other points—recall there are only a finite number of them on any any bounded interval.
Let
1
an =
L
ZL
1
πt
f (t) cos n dt, bn =
L
L
−L
1
An =
L
ZL
ZL
f (t) sin n
πt
dt,
L
−L
πt
1
g(t) cos n dt, and Bn =
L
L
−L
ZL
g(t) sin n
πt
dt
L
−L
be the Fourier coefficients of f and g. Integration by parts tells us that for n ≥ 1,
An = nbn and Bn − nan .
Hence,
n
X
(a2j
j=1
+
b2j )1/2
=
n
X
1
j=1
j
(A2j + Bj2 )1/2 .
From the Cauchy-Schwartz inequality, we see that
n
X
j=1
(a2j + b2j )1/2 =
n
X
1
j=1
j
(A2j + Bj2 )1/2
7
"
n
X
1
≤
j2
j=1
#1/2 "
n
X
j=1
(A2j + Bj2 )
#1/2
.
We know
∞
P
∞
P
j=1
1
j2
converges and Proposition 1 tells us that
∞
P
j=1
(A2j + Bj2 ) converges. Hence
(a2j + b2j )1/2 converges also. Observing that a cos θ + b sin θ = (a2n + b2n )1/2 sin(θ + ϕ), where
j=1
tan ϕ = a/b, we see that
πx πx
+ bn sin n ≤ (a2n + b2n )1/2
an cos n
L
L
The Weierstrass M-test now tells us that the Fourier series
∞
πx
πx a0 X +
an cos n
+ bn sin n
2
L
L
n=1
converges uniformly and absolutely.
It is worth repeating what this theorem does not say. It does not tell us that the
uniform limit of the series is the function f .
4. Pointwise convergence of Fourier series. A few preliminary results are in order
first.
Proposition 2.6. If θ is not an integral multiple of 2π, then
n
sin(n + 21 )θ
1 X
+
cos kθ =
.
2 k=1
2 sin(θ/2)
Proof : Let
Then
n
1 X
cos kθ.
S= +
2 k=1
n
We remember that
θ
θ X
θ
2S sin = sin +
2 cos kθ sin .
2
2 k=1
2
2 cos α sin β = sin(α + β) − sin(α − β)
so the above equation becomes
n θ X
1
1
θ
= sin +
sin(k + )θ − sin(k − )θ
2S sin
2
2 k=1
2
2
n
n
X
θ X
1
1
= sin +
sin(k + )θ −
sin(k − )θ
2 k=1
2
2
k=1
n
n−1
k=1
k=0
X
1
1
θ X
sin(k + )θ −
sin(k + )θ
= sin +
2
2
2
1
= sin(n + )θ.
2
8
Hence,
S=
sin(n + 21 )θ
.
2 sin θ2
sin(n+ 12 )θ
θ
θ→2kπ 2 sin 2
Proposition 2.7. If k is an integer, then lim
Definition. The function
n
1 X
cos kθ =
Dn (θ) = +
2 k=1
(
sin(n+ 21 )θ
2 sin θ2
n + 12
= n + 12 .
if θ 6= 2kπ
if θ = 2kπ
is continuous and periodic with period 2π. (This function is called the Dirichlet kernel,
named for the German mathematician Lejeune Dirichlet (1805-1859). )
Integrate the formula in Proposition 2.6 to get
R2π
Proposition 2.8. Dn (θ)dθ = π.
0
Notation. The left and right hand limits of f at x are denoted by f (x−) and f (x+),
respectively. In other words
f (x−) = lim f (x − h), and f (x+) = lim f (x + h), where h > 0.
h→0
h→0
Theorem 2.9. Let f be a piecewise smooth function of period 2π. The the Fourier series
of f converges pointwise to the function
f (x+) + f (x−)
fe(x) =
.
2
Proof : Let
a0 X
+
(an cos nx + bn sin nx) .
f∼
2
n=1
∞
Note that because f is piecewise smooth f and fe differ at at most a finite number of points
on any bounded interval. Thus the Fourier series for f and fe are the same.
Now then,
1
an cos nx + bn sin nx =
π
Z2π
1
π
Z2π
0
=
0
Hence,
fe(t) [cos nt cos nx + sin nt sin nx] dt
fe(t) cos n(x − t)dt.
N
a0 X
1
Sn (x) =
+
(an cos nx + bn sin nx) =
2
π
n=1
9
Z2π
0
"
#
N
X
1
fe(t)
+
cos n(x − t) dt,
2 n=0
and so from Proposition 2.6, we have
N
or,
a0 X
1
SN (x) =
+
(an cos nx + bn sin nx) =
2
π
n=1
1
SN (x) =
π
Z2π
0
We also have
1
SN (x) = −
π
1
fe(t)DN (x − t)dt =
π
Z2π
0
Z2π
0
sin(N + 12 )(x − t)
fe(t)
dt,
2 sin((x − t)/2)
fe(x − t)DN (t)dt.
−2π
Z
Z2π
1
fe(x + t)DN (−t)dt =
fe(x + t)DN (t)dt.
π
0
0
(Remember that the Dirichlet kernel is an even function and DN and fe both have period
2π.)
Next from Proposition 2.8,
1
SN (x) − fe(x) = SN (x) − fe(x)
π
=
1
π
Z2π
0
and
1
SN (x) − fe(x) =
π
Adding these two, we have
1
2[SN (x) − fe(x)] =
π
Z2π h
0
h
Z2π
DN (t)dt
0
i
fe(x − t) − fe(x) DN (t)dt,
i
e
e
f (x + t) − f (x) DN (t)dt.
Z2π
[fe(x − t) + fe(x + t) − 2fe(x)]DN (t)dt.
0
Next observe that the integrand in this expression is an even function. Thus,
Zπ
1
[fe(x − t) + fe(x + t) − 2fe(x)]DN (t)dt
SN (x) − fe(x) =
π
0
Next, note that 2fe(x) = f (x+) + f (x−), and so we have
Zπ
1
SN (x) − fe(x) =
[fe(x − t) + fe(x + t) − f (x+) − f (x−)]DN (t)dt
π
0
1
=
π
Zπ
0
1
[fe(x − t) − f (x−)]DN (t)dt +
π
10
Zπ
0
[fe(x + t) − f (x+)]DN (t)dt.
We shall show that
lim
N →∞
by showing that
lim
N →∞
Zπ
0
SN (x) − fe(x) = 0
[fe(x − t) − f (x−)]DN (t)dt = lim
N →∞
Zπ
0
[fe(x + t) − f (x+)]DN (t)dt = 0.
First, look at [fe(x − t) − f (x−)]DN (t).Observe that the function F defined by
F (t) =
is perfectly nice at t = 0:
lim+
t→0
Thus,
t
fe(x − t) − f (x−)
t
2 sin t/2
t
fe(x − t) − f (x−)
= f 0 (x−) and lim+
= 1.
t→0 2 sin t/2
t
Fe(t) =
(
fe(x−t)−f (x−)
t
t
2 sin t/2
0
f (x−)
0<t≤π
t=0
is piecewise smooth on [0, π].
Now,
fe(x − t) − f (x−)
t
1
sin(N + )t
t
2 sin t/2
2
= Fe(t) cos t/2 sin N t + Fe(t) sin t/2 cos N t.
[fe(x − t) − f (x−)]DN (t) =
Hence,
Zπ
0
[fe(x − t) − f (x−)]DN (t)dt =
Finally, observe that
Zπ
0
Fe(t) cos t/2 sin N tdt +
Zπ
0
Fe(t) sin t/2 cos N tdt.
{sin t, sin 2t, sin 3t, . . .} and {cos t, cos 2t, cos 3t, . . .}
are orthogonal sequences with the inner product (u, v) =
lim
u(t)v(t)dt.The Riemann lemma
0
then tells us that
N →∞
Rπ
Zπ
0
Fe(t) cos t/2 sin N tdt = lim
N →∞
11
Zπ
0
Fe(t) sin t/2 cos N tdt = 0.
Thus we have finally shown that
lim
N →∞
Zπ
0
[fe(x−t)−f (x−)]DN (t)dt = lim
N →∞
Zπ
0
The demonstration that
lim
N →∞
Zπ
0
Fe(t) cos t/2 sin N tdt+ lim
N →∞
Zπ
0
Fe(t) sin t/2 cos N tdt = 0
[fe(x + t) − f (x+)]DN (t)dt = 0
is almost the same and is omitted. This at last proves the theorem.
This result allows use to tidy up the loose ends of Theorem 2.5. Recall that this one
told us that the Fourier series of a continuous piecewise differentiable function f converges
uniformly to something but did not say what. Now we know from the just proved Theorem
2.9 that the something to which the series converges must indeed be f . This is worth stating
explicitly.
Corollary 2.10. If f is continuous, then the series converges uniformly to f .
Remark. Although the theorem refers to functions having period 2π, it applies as well
to other periodic functions; the choice of 2π is simply a computational convenience. Given
a function f having period 2L, apply Theorem 2.9 to the function F (x) = f (xL/π).
Example. Let us find the Fourier series for the function f (x) = x on the interval [−π, π] :
1
ak =
π
Zπ
1
f (t) cos ktdt =
π
−π
bk
1
=
π
Zπ
Zπ
t cos ktdt = 0,
−π
1
f (t) sin ktdt =
π
−π
Zπ
−π
2
2
= − cos kπ = (−1)k+1
k
k
The Fourier series is thus
2
π
1
t sin ktdt = 2 (sin kt − kt cos kt)
k
−π
∞
X
(−1)k+1
k=1
k
sin kx
Now, what does the limit of this series look like? We simply apply Theorem 2.9 to
the periodic extension of f . This results in
12
3
2
1
–5
5
x
10
–1
–2
–3
Another example. We shall find the Fourier series of the function g(x) = |x| on the
interval [−1, 1].
a0 =
Z1
|t| dt = 1
−1
an =
Z1
|t| cos nπtdt = 2
−1
Z1
t cos nπtdt = 2
0
bn =
Z1
−1 + (−1)n
for n ≥ 1;
n2 π 2
|t| sin nπtdt = 0.
−1
Life can be made a bit simpler by noting that for n ≥ 1
0 if n even
an =
.
−4
if n odd
n2 π 2
Then letting n = 2k + 1, we have the Fourier series
∞
1
4 X
1
− 2
cos(2k + 1)πx.
2 π k=0 (2k + 1)2
Now, where does this series converge and what does the limit look like? Again, we
simply look at the periodic extension of f . A picture:
1
0.8
0.6
0.4
0.2
–4
–2
0
13
2 x
4
5. Fourier series approximation. If f is continuous, then we know its Fourier series
converges uniformly to f . This means that for ”large” N , the graph of the partial sum
N
P
a0
+
an cos n πx
+ bn sin n πx
will look pretty much like the graph of f . This, of course,
2
L
L
n=1
cannot be so if f is not continuous; for any N , the partial sum is a continuous function.
Look at an example. The Fourier series for the periodic extension of f defined by
−π/2 for − π < x < 0
f (x) =
π/2
0<x<π
is easily found to be
2
∞
X
sin(2n − 1)x
2n − 1
n=1
Pictures of SN (x) = 2
N
P
n=1
N =8:
sin(2n−1)x
2n−1
.
for
1.5
1
0.5
–4
–2
2
4
x
–0.5
–1
–1.5
N = 20 :
1.5
1
0.5
–4
–2
2
–0.5
–1
–1.5
14
x
4
N = 30 :
1.5
1
0.5
–4
–2
2
x
4
–0.5
–1
–1.5
One might hope that the limiting graph is simply the graph of f with vertical lines
at the points of discontinuity. It is, as the pictures seem to indicate, not to be so. It turns
out that we inevitably have a situation like that indicated in the examples of Section 2 in
which the sequence of functions SN (x) converges pointwise to the limit f , but no matter
how large N is, there will be values of x at which the difference |SN (x) − f (x)| is larger than
some fixed value. Let us see precisely what we are talking about. First with this specific
example, and then in general.
Proposition 2.11. Let
g(x) =
−π/2 for − π < x < 0
,
π/2
0<x<π
and let
Sn (x) = 2
n
X
sin(2k − 1)x
2k − 1
k=1
be the nth partial sum of the Fourier series for g.Then for given ε > 0, there is an integer N
such that
Zπ
sin
t
π
Sn ( ) −
< ε for all n > N.
dt
2n
t
0
Proof : First, observe that
n
X
d
sin 2nx
Sn (x) = 2
cos(2k − 1)x =
dx
sin x
k=1
Thus,
Sn (x) =
Zx
sin 2nt
dt, and so
sin t
0
Sn (x) −
Zx
0
sin 2nt
dt =
t
Zx
0
15
1
1
sin 2nt
−
dt
sin t t
Now, note that
1
1
−
lim
=0
t→0 sin t
t
and so there is a number K so that sin 2nt sin1 t − 1t ≤ K for all t ∈ [−π, π]. Thus
x
Z
Zx
Sn (x) − sin 2nt dt = sin 2nt 1 − 1 dt ≤ Kx
t
sin t t
0
0
Let ε > 0 be given and choose δ so that
Zx
sin
2nt
< ε for 0 < x < δ.
Sn (x) −
dt
t
0
Next, observe that
Zx
sin 2nt
dt =
t
0
so that we have
Z2nx
sin t
dt
t
0
Z2nx
sin
t
< ε for 0 < x < δ.
Sn (x) −
dt
t
0
Choose N large enough to ensure that π/2N < δ. Then for n > N , letting x = π/2n, we
have
Zπ
Z2nx
sin
t
sin
t
π
= Sn ( ) −
< ε,
Sn (x) −
dt
dt
t
2n
t
0
0
and the proposition is proved.
R∞
Rπ
π
Comment. Note that sint t dt = π/2. Thus Sn ( 2n
)−
0
Hence,
0
sin t
dt
t
π
= Sn ( 2n
)−
π
2
+
R∞ sin t
π
t
dt.
Z∞
h
sin t
π
πi
π
π i
lim Sn ( ) −
= lim Sn ( ) − g( ) = −
dt ≈ 0.281 14.
n→∞
n→∞
2n
2
2n
2n
t
h
π
You can see this in the graphs pictured above.
Let us reflect on what all this tells us. We know that for fixed x 6= 0 in the interval
(−π, π), the sequence (sn (x)) converges to π/2. Thus we can have Sn (x) as close to π/2 ≈
1. 570 8 as we wish by choosing n large enough. There is, on the other hand, an x (viz.,
Rπ
x = π/2n) such that Sn (x) is as close to sint t dt ≈ 1. 851 9 as we wish. As we shall see
0
in the sequel, this behavior is not restricted just to this particular function, but is seen at
16
any point at which the piecewise smooth function f fails to be continuous. This behavior is
called the Gibbs Phenomenon, in honor of the American physicist, J. Willard Gibbs
(1839-1903).
Now we seen about the general case.
Theorem 2.12.
Suppose f is piecewise smooth and continuous everywhere on the
interval [−π, π] except at x = 0.Let fn (x) be the nth partial sum of the Fourier series for f .
Then
h
π
π i
−1
lim fn ( ) − f ( ) = [f (0+) − f (0−)]
n→∞
2n
2n
π
Proof : Let ψ(x) = f (x) − 12 [f (0+) + f (0−)] −
the function defined in Proposition 2.11.Then
1
[f (0+) + f (0−)] −
2
1
ψ(0−) = f (0−) − [f (0+) + f (0−)] +
2
ψ(0+) = f (0+) −
1
π
Z∞
sin t
dt
t
π
[f (0+) − f (0−)] g(x), where g is
1
π
[f (0+) − f (0−)] = 0, and
π
2
1
π
[f (0+) − f (0−)] = 0.
π
2
Thus ψ is continuous at 0 and hence everywhere on the given interval. In particular, the
sequence (ψn (x)) of partial sums of its Fourier series converges uniformly to ψ.Hence,
π i
π
lim ψn ( ) − ψ( ) = 0.
n→∞
2n
2n
h
Now
π
π
π
1
π
π
π
ψn ( ) − ψ( ) = fn ( ) − f ( ) − [f (0+) − f (0−)] gn ( ) −
;
2n
2n
2n
2n
π
2n
2
and so
lim
n→∞
Thus,
π
π
1
π
π
fn ( ) − f ( ) − [f (0+) − f (0−)] gn ( ) −
2n
2n
π
2n
2
h
π i
π
lim fn ( ) − f ( ) =
n→∞
2n
2n
= 0.
1
π
π
[f (0+) − f (0−)] gn ( ) −
n→∞ π
2n
2
Z∞
−1
sin t
= [f (0+) − f (0−)]
dt
π
t
lim
π
The theorem shows that the Gibbs Phenomenon ”overshoot” at the jump
R∞ sin t
dt ≈ 8. 949 0 × 10−2 times the total jump of f .
discontinuity is −1
π
t
Comment.
π
17
6. Cosine and sine series. In our discussion of orthogonal collections of eigenfunctions,
we also considered the approximation of functions by a series of cosine functions and sine
functions. We shall now look at these in more detail. Mercifully, most of the results are easy
consequence of what we have done for the Fourier series.
Let X be the linear space of all continuous, piecewise smooth functions that are
periodic with period L for some L > 0. Then we know the collection
n
o
πx
πx
πx
1, cos
. cos 2 , cos 3 , . . .
L
L
L
is an approximating base that is orthogonal with respect to the inner product (f, g) =
RL
f (t)g(t)dt. For f X, we thus have mean convergence to f of the series
0
πx
A0 X
An cos n , where
+
2
L
n=1
∞
(f, cos n πx
)
2
L
An =
πx
πx =
(cos n L , cos n L )
L
Now, let fe be defined by
fe(x) =
ZL
f (t) cos n
πt
dt.
L
0
f (−x) for − L < x < 0
.
f (x)
for 0 ≤ x < L
Observe that fe is an even function.
fe.):
Now, compute the Fourier series for fe (more precisely. for the periodic extension of
1
an =
L
ZL
−L
2
πt
fe(t) cos n dt = an =
L
L
ZL
f (t) cos n
πt
dt = An ,
L
0
since the integrand is an even function. And
1
bn =
L
ZL
−L
πt
fe(t) sin n dt = 0,
L
since the integrand is an odd function. In other words, the above series
πx
A0 X
+
An cos n
2
L
n=1
∞
18
is “really” the Fourier series for the extension fe. This means we know all about it! The series
is usually called the Fourier cosine series for f .
Example. Let f be defined by f (x) = x for 0 ≤ x ≤ π. Then ff
, the even extension of
f looks like
The cosine series is
and it converges to this:
The collection
∞
π 2 X (−1)k − 1
c(x) = +
cos kx;
2 π k=1
k2
n
o
πx
πx
πx
sin
, sin 2 , sin 3 , . . .
L
L
L
is also an orthogonal approximating basis for the space X. It is easy to show that the
Fourier sine series for a function f
∞
X
Bn sin n
n=1
19
πx
,
L
where
2
Bn =
L
ZL
f (t) sin n
πt
dt
L
0
is the Fourier series of the odd periodic extension fb of f :
−f (−x) −L < x < 0
b
.
f (x) =
f (x)
0≤x<L
Example. Let f be as in the previous example. Then fb looks like
3
2
1
–3
–2
0
–1
1
x
2
3
–1
–2
–3
Here is the sine series:
s(x) = 2
∞
X
(−1)k+1
k
k=1
The limit of this series is thus
sin kx.
3
2
1
–8
–6
–4
–2
2
4 x
6
8
–1
–2
–3
Note that the cosine series for f (x) = x on the interval (0, L) converges uniformly
to f , while the sine series approximation will be plagued with the Gibbs phenomenon. Let’s
take a look. First, a picture of the partial sum of the first 10 terms of the cosine series:
20
3
2.5
2
1.5
1
0.5
0
0.5
1
1.5
x
2
2.5
3
3
2.5
2
1.5
1
0.5
0
0.5
1
1.5
x
2
2.5
3
Although on the interval (0, π) both the cosine and the sine series converge pointwise
to f (x) = x, the convergence of the sine series is much nastier because the odd extension of
f is not continuous.
21