Chapter Two
Fourier Series
1. Introduction. All of the eigenfunctions found in the last chapter for the eigenvalue
problem
φ′′(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 Pf 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 Pf is periodic on the whole line with
period p. Let us look at the eigenfunction expansions discussed in Chapter ST. If
MN = span{sinλnx}, λn = nπ/L, n = 1, 2, . . .
then Pf has period 2L. Moreover, since each sine function is odd it follows that Pf is an
odd 2L periodic function on (−∞,∞). If
MN = span{sinλnx}, λn =(π
2+ nπ
)/L, n = 0, 1, 2, . . .
then Pf 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
(π2
+ nπ)(L + ∆L)
L= sin
(π2
+ nπ)(L− ∆L)
Lfor ∆L > 0
1
so that this 4L periodic function is also even with respect to x = L. Similarly, if
MN = span{cosλnx}, λn =(π
2+ nπ
)/L, n = 0, 1,
then Pf 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
Pf(L+ ∆L) = −Pf(L− ∆L)
so that Pf also is odd with respect to x = L.
However, the orthogonal projection Pf into the span of eigenfunctions associated with
boundary conditions like
u(0) = 0, u′(L) = u(L)
is not periodic on (−∞,∞) because in general sin λmx and sinλnx have no common period.
Let us now turn to the eigenfunctions{cos mπx
L, sin nπx
L
}associated with
φ′′(x) = µφ(x)
φ(0) = φ(L)
φ′(0) = φ′(L).
The Sturm-Liouville theory assures that the orthogonal projection Pf 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 conver-
gence. 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 multiplica-
tion. (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) =
L∫
−L
f(x)g(x)dx,
the sequence
{1, cos
πx
L. sin
πx
L, cos 2
πx
L, sin 2
πx
L, cos 3
πx
L, sin 3
πx
L, . . .
}
is an orthogonal sequence. It is clear that the inner product (1, 1) = 2L and also easy to
verify that
(sinn
πx
L, sinn
πx
L
)=
L∫
−L
sin2 nπt
Ldt = L, and
(cos n
πx
L, cosn
πx
L
)=
L∫
−L
cos2 nπt
Ldt = L.
Thus for an integer N , the best least squares approximation Sn(x) of f ∈ X by a linear
combination of
{1, cos
πx
L. sin
πx
L, cos 2
πx
L, sin 2
πx
L, cos 3
πx
L, sin 3
πx
L, . . . , cosN
πx
L, sinN
πx
L
}
is given by
Sn(x) =a0
2+
N∑
n=1
(an cosn
πx
L+ bn sinn
πx
L
),
where
an =1
L
L∫
−L
f(t) cosnπt
Ldt, and bn =
1
L
L∫
−L
f(t) sinnπt
Ldt.
The seriesa0
2+
∞∑
n=1
(an cosn
πx
L+ bn sinn
πx
L
)
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
f ∼
a0
2+
∞∑
n=1
(an cosn
πx
L+ bn sinn
πx
L
).
The coefficients an and bn are called the Fourier coefficients of f .
3
Proposition 2.1. Suppose an and bn are the Fourier coefficients of f . Then the series∞∑
j=1
(a2k + b2k) converges.
Proof : Applying Bessell’s inequality to the orthonormal sequence
{1√2L,
1√L
cosπx
L,
1√L
sinπx
L,
1√L
cos 2πx
L,
1√L
sin 2πx
L,
1√L
cos 3πx
L,
1√L
sin 3πx
L, . . .
}
gives us
a20L
2+ L
∞∑
n=1
(a2n + b2n) ≤
2L∫
0
(f(x))2dx,
from which the proposition follows at once. The next proposition is an easy consequence
of Riemann’s Lemma.
Proposition 2.2. limn→∞
L∫−L
f(t) cosnπtLdt = lim
n→∞
L∫−L
f(t) sinnπtLdt = 0.
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 point-
wise 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
in the mean to the function f ifb∫
a
[fn(x) − f(x)]2 dx→ 0.
A few examples will help illuminate these definitions. For each positive integer n ≥ 2
and for x ∈ [0, 1], let
fn(x) =
n2x if 0 ≤ x ≤ 1/n−n2(x− 1
n) + n if 1/n < x ≤ 2/n
0 if 2/n < x ≤ 1.
Thus, fn looks like
4
(1/n,n)
2/n
It should be clear that for each x ∈ [0, 1], we have limn→∞
fn(x) = 0. Thus the sequence
(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 that1∫
0
(fn(x))2 dx =2
3n,
which tells us that (fn) does not converge in the mean to f .
Next, for each integer n ≥ 1 and x ∈ [0, 1], let
fn(x) =
{n− n3x if 0 ≤ x ≤ 1/n3
0 if 1/n3 < x ≤ 1.
A picture:
1/n^3
n
5
It is clear that1∫0
(fn(x))2dx = 13n
, 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
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
b∫
a
|fn(x) − f(x)|2 dx <b∫
a
ε
b− adx = ε.
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.
Theorem 2.4. Consider the series of functions∞∑
n=0
gn(x) for x ∈ [a, b]. If for all n,
we have |gn(x)| ≤ Mn for all x ∈ [a, b] and if∞∑
n=0
Mn converges, then∞∑
n=0
gn(x) converges
uniformly and absolutely on [a, b].
Proof : First, we show absolute convergence. Let ε > 0. Choose N so thatn∑
k=m
Mk <
ε for n,m > N. Then for n,m > N,
n∑
k=m
|gk(x)| ≤n∑
k=m
Mk.
Thus∞∑
n=0
|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 thatn∑
k=m
Mk < ε/2 for all n ≥ N .
∣∣∣∣∣
n∑
k=0
gk(x) −G(x)
∣∣∣∣∣ ≤∣∣∣∣∣
m∑
k=0
gk(x) −G(x)
∣∣∣∣∣ +
∣∣∣∣∣
n∑
k=m
gk(x)
∣∣∣∣∣
≤∣∣∣∣∣
m∑
k=0
gk(x) −G(x)
∣∣∣∣∣ +n∑
k=m
Mk
6
3. Uniform convergence of Fourier series. Suppose f is a piecewise smooth con-
tinuous function of period 2L. We shall show that the Fourier series
f ∼
a0
2+
∞∑
n=1
(an cosn
πx
L+ bn sinn
πx
L
),
converges uniformly to some function g. Notice that this will not establish that the orthog-
onal sequence
{1, cos
πx
L. sin
πx
L, cos 2
πx
L, sin 2
πx
L, cos 3
πx
L, sin 3
πx
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 ′(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
an =1
L
L∫
−L
f(t) cosnπt
Ldt, bn =
1
L
L∫
−L
f(t) sinnπt
Ldt,
An =1
L
L∫
−L
g(t) cosnπt
Ldt, and Bn =
1
L
L∫
−L
g(t) sinnπt
Ldt
be the Fourier coefficients of f and g. Integration by parts tells us that for n ≥ 1,
An = nbn and Bn − nan.
Hence,n∑
j=1
(a2j + b2j)
1/2 =
n∑
j=1
1
j(A2
j +B2j )
1/2.
From the Cauchy-Schwartz inequality, we see that
n∑
j=1
(a2j + b2j)
1/2 =n∑
j=1
1
j(A2
j +B2j )
1/2 ≤[
n∑
j=1
1
j2
]1/2 [n∑
j=1
(A2j +B2
j )
]1/2
.
7
We know∞∑
j=1
1j2 converges and Proposition 1 tells us that
∞∑j=1
(A2j + B2
j ) converges. Hence
∞∑j=1
(a2j + b2j)
1/2 converges also. Observing that a cos θ+ b sin θ = (a2n + b2n)1/2 sin(θ+ϕ), where
tanϕ = a/b, we see that∣∣∣an cosn
πx
L+ bn sin n
πx
L
∣∣∣ ≤ (a2n + b2n)1/2
The Weierstrass M-test now tells us that the Fourier series
a0
2+
∞∑
n=1
(an cosn
πx
L+ bn sinn
πx
L
)
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
1
2+
n∑
k=1
cos kθ =sin(n + 1
2)θ
2 sin(θ/2).
Proof : Let
S =1
2+
n∑
k=1
cos kθ.
Then
2S sinθ
2= sin
θ
2+
n∑
k=1
2 cos kθ sinθ
2.
We remember that
2 cosα sin β = sin(α + β) − sin(α− β)
so the above equation becomes
2S sinθ
2= sin
θ
2+
n∑
k=1
[sin(k +
1
2)θ − sin(k − 1
2)θ
]
= sinθ
2+
n∑
k=1
sin(k +1
2)θ −
n∑
k=1
sin(k − 1
2)θ
= sinθ
2+
n∑
k=1
sin(k +1
2)θ −
n−1∑
k=0
sin(k +1
2)θ
= sin(n+1
2)θ.
8
Hence,
S =sin(n+ 1
2)θ
2 sin θ2
.
Proposition 2.7. If k is an integer, then limθ→2kπ
sin(n+ 1
2)θ
2 sin θ
2
= n+ 12.
Definition. The function
Dn(θ) =1
2+
n∑
k=1
cos kθ =
{sin(n+ 1
2)θ
2 sin θ
2
if θ 6= 2kπ
n+ 12
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
Proposition 2.8.2π∫0
Dn(θ)dθ = π.
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−) = limh→0
f(x− h), and f(x+) = limh→0
f(x + h), where 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
f̃(x) =f(x+) + f(x−)
2.
Proof : Let
f ∼
a0
2+
∞∑
n=1
(an cosnx + bn sinnx) .
Note that because f is piecewise smooth f and f̃ differ at at most a finite number of points
on any bounded interval. Thus the Fourier series for f and f̃ are the same.
Now then,
an cos nx+ bn sin nx =1
π
2π∫
0
f̃(t) [cosnt cosnx + sinnt sin nx] dt
=1
π
2π∫
0
f̃(t) cosn(x− t)dt.
Hence,
Sn(x) =a0
2+
N∑
n=1
(an cos nx+ bn sin nx) =1
π
2π∫
0
f̃(t)
[1
2+
N∑
n=0
cosn(x− t)
]dt,
9
and so from Proposition 2.6, we have
SN(x) =a0
2+
N∑
n=1
(an cosnx + bn sinnx) =1
π
2π∫
0
f̃(t)sin(N + 1
2)(x− t)
2 sin((x− t)/2)dt,
or,
SN(x) =1
π
2π∫
0
f̃(t)DN(x− t)dt =1
π
2π∫
0
f̃(x− t)DN (t)dt.
We also have
SN(x) = − 1
π
−2π∫
0
f̃(x + t)DN(−t)dt =1
π
2π∫
0
f̃(x + t)DN(t)dt.
(Remember that the Dirichlet kernel is an even function and DN and f̃ both have period
2π.)
Next from Proposition 2.8,
SN(x) − f̃(x) = SN(x) − f̃(x)1
π
2π∫
0
DN(t)dt
=1
π
2π∫
0
[f̃(x− t) − f̃(x)
]DN (t)dt,
and
SN(x) − f̃(x) =1
π
2π∫
0
[f̃(x + t) − f̃(x)
]DN(t)dt.
Adding these two, we have
2[SN(x) − f̃(x)] =1
π
2π∫
0
[f̃(x− t) + f̃(x+ t) − 2f̃(x)]DN(t)dt.
Next observe that the integrand in this expression is an even function. Thus,
SN(x) − f̃(x) =1
π
π∫
0
[f̃(x− t) + f̃(x+ t) − 2f̃(x)]DN(t)dt
Next, note that 2f̃(x) = f(x+) + f(x−), and so we have
SN(x) − f̃(x) =1
π
π∫
0
[f̃(x− t) + f̃(x + t) − f(x+) − f(x−)]DN (t)dt
=1
π
π∫
0
[f̃(x− t) − f(x−)]DN (t)dt+1
π
π∫
0
[f̃(x+ t) − f(x+)]DN (t)dt.
10
We shall show that
limN→∞
(SN(x) − f̃(x)
)= 0
by showing that
limN→∞
π∫
0
[f̃(x− t) − f(x−)]DN (t)dt = limN→∞
π∫
0
[f̃(x+ t) − f(x+)]DN(t)dt = 0.
First, look at [f̃(x− t) − f(x−)]DN (t).Observe that the function F defined by
F (t) =f̃(x− t) − f(x−)
t
t
2 sin t/2
is perfectly nice at t = 0:
limt→0+
f̃(x− t) − f(x−)
t= f ′(x−) and lim
t→0+
t
2 sin t/2= 1.
Thus,
F̃ (t) =
{f̃(x−t)−f(x−)
tt
2 sin t/20 < t ≤ π
f ′(x−) t = 0
is piecewise smooth on [0, π].
Now,
[f̃(x− t) − f(x−)]DN (t) =f̃(x− t) − f(x−)
t
t
2 sin t/2sin(N +
1
2)t
= F̃ (t) cos t/2 sinNt + F̃ (t) sin t/2 cosNt.
Hence,
π∫
0
[f̃(x− t) − f(x−)]DN (t)dt =
π∫
0
F̃ (t) cos t/2 sinNtdt +
π∫
0
F̃ (t) sin t/2 cosNtdt.
Finally, observe that
{sin t, sin 2t, sin 3t, . . .} and {cos t, cos 2t, cos 3t, . . .}
are orthogonal sequences with the inner product (u, v) =π∫0
u(t)v(t)dt.The Riemann lemma
then tells us that
limN→∞
π∫
0
F̃ (t) cos t/2 sinNtdt = limN→∞
π∫
0
F̃ (t) sin t/2 cosNtdt = 0.
11
Thus we have finally shown that
limN→∞
π∫
0
[f̃(x−t)−f(x−)]DN (t)dt = limN→∞
π∫
0
F̃ (t) cos t/2 sinNtdt+ limN→∞
π∫
0
F̃ (t) sin t/2 cosNtdt = 0
The demonstration that
limN→∞
π∫
0
[f̃(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] :
ak =1
π
π∫
−π
f(t) cos ktdt =1
π
π∫
−π
t cos ktdt = 0,
bk =1
π
π∫
−π
f(t) sin ktdt =1
π
π∫
−π
t sin ktdt =1
k2(sin kt− kt cos kt)
∣∣∣∣π
−π
= −2
kcos kπ = (−1)k+1 2
k
The Fourier series is thus
2
∞∑
k=1
(−1)k+1
ksin 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
1
2
3
–5 5 10x
Another example. We shall find the Fourier series of the function g(x) = |x| on the
interval [−1, 1].
a0 =
1∫
−1
|t| dt = 1
an =
1∫
−1
|t| cosnπtdt = 2
1∫
0
t cos nπtdt = 2−1 + (−1)n
n2π2for n ≥ 1;
bn =
1∫
−1
|t| sinnπtdt = 0.
Life can be made a bit simpler by noting that for n ≥ 1
an =
{0 if n even−4
n2π2 if n odd.
Then letting n = 2k + 1, we have the Fourier series
1
2− 4
π2
∞∑
k=0
1
(2k + 1)2cos(2k + 1)πx.
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:
0
0.2
0.4
0.6
0.8
1
–4 –2 2 4x
13
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
a0
2+
N∑n=1
(an cosnπx
L+ bn sin nπx
L
)will look pretty much like the graph of f . This, of course,
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
f(x) =
{−π/2 for − π < x < 0π/2 0 < x < π
is easily found to be
2∞∑
n=1
sin(2n− 1)x
2n− 1.
Pictures of SN (x) = 2N∑
n=1
sin(2n−1)x2n−1
for
N = 8 :
–1.5
–1
–0.5
0.5
1
1.5
–4 –2 2 4x
N = 20 :
–1.5
–1
–0.5
0.5
1
1.5
–4 –2 2 4x
14
N = 30 :
–1.5
–1
–0.5
0.5
1
1.5
–4 –2 2 4x
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 < 0π/2 0 < x < π
,
and let
Sn(x) = 2
n∑
k=1
sin(2k − 1)x
2k − 1
be the nth partial sum of the Fourier series for g.Then for given ε > 0, there is an integer N
such that ∣∣∣∣∣∣Sn(
π
2n) −
π∫
0
sin t
tdt
∣∣∣∣∣∣< ε for all n > N.
Proof : First, observe that
d
dxSn(x) = 2
n∑
k=1
cos(2k − 1)x =sin 2nx
sin x
Thus,
Sn(x) =
x∫
0
sin 2nt
sin tdt, and so
Sn(x) −x∫
0
sin 2nt
tdt =
x∫
0
sin 2nt
[1
sin t− 1
t
]dt
15
Now, note that
limt→0
[1
sin t− 1
t
]= 0
and so there is a number K so that∣∣sin 2nt
[1
sin t− 1
t
]∣∣ ≤ K for all t ∈ [−π, π]. Thus
∣∣∣∣∣∣Sn(x) −
x∫
0
sin 2nt
tdt
∣∣∣∣∣∣=
∣∣∣∣∣∣
x∫
0
sin 2nt
[1
sin t− 1
t
]dt
∣∣∣∣∣∣≤ Kx
Let ε > 0 be given and choose δ so that∣∣∣∣∣∣Sn(x) −
x∫
0
sin 2nt
tdt
∣∣∣∣∣∣< ε for 0 < x < δ.
Next, observe thatx∫
0
sin 2nt
tdt =
2nx∫
0
sin t
tdt
so that we have ∣∣∣∣∣∣Sn(x) −
2nx∫
0
sin t
tdt
∣∣∣∣∣∣< ε for 0 < x < δ.
Choose N large enough to ensure that π/2N < δ. Then for n > N , letting x = π/2n, we
have ∣∣∣∣∣∣Sn(x) −
2nx∫
0
sin t
tdt
∣∣∣∣∣∣=
∣∣∣∣∣∣Sn(
π
2n) −
π∫
0
sin t
tdt
∣∣∣∣∣∣< ε,
and the proposition is proved.
Comment. Note that∞∫0
sin ttdt = π/2. Thus Sn( π
2n) −
π∫0
sin ttdt = Sn(
π2n
) − π2
+∞∫π
sin ttdt.
Hence,
limn→∞
[Sn(
π
2n) − π
2
]= lim
n→∞
[Sn(
π
2n) − g(
π
2n)]
= −∞∫
π
sin t
tdt ≈ 0.281 14.
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.,
x = π/2n) such that Sn(x) is as close toπ∫0
sin ttdt ≈ 1. 851 9 as we wish. As we shall see
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
limn→∞
[fn(
π
2n) − f(
π
2n)]
= [f(0+) − f(0−)]−1
π
∞∫
π
sin t
tdt
Proof : Let ψ(x) = f(x) − 12[f(0+) + f(0−)] − 1
π[f(0+) − f(0−)] g(x), where g is
the function defined in Proposition 2.11.Then
ψ(0+) = f(0+) − 1
2[f(0+) + f(0−)] − 1
π[f(0+) − f(0−)]
π
2= 0, and
ψ(0−) = f(0−) − 1
2[f(0+) + f(0−)] +
1
π[f(0+) − f(0−)]
π
2= 0.
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,
limn→∞
[ψn(
π
2n) − ψ(
π
2n)]
= 0.
Now
ψn(π
2n) − ψ(
π
2n) = fn(
π
2n) − f(
π
2n) − 1
π[f(0+) − f(0−)]
(gn(
π
2n) − π
2
);
and so
limn→∞
{fn(
π
2n) − f(
π
2n) − 1
π[f(0+) − f(0−)]
(gn(
π
2n) − π
2
)}= 0.
Thus,
limn→∞
[fn(
π
2n) − f(
π
2n)]
= limn→∞
1
π[f(0+) − f(0−)]
(gn(
π
2n) − π
2
)
= [f(0+) − f(0−)]−1
π
∞∫
π
sin t
tdt
Comment. The theorem shows that the Gibbs Phenomenon ”overshoot” at the jump
discontinuity is −1π
∞∫π
sin ttdt ≈ 8. 949 0× 10−2 times the total jump of f .
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
{1, cos
πx
L. cos 2
πx
L, cos 3
πx
L, . . .
}
is an approximating base that is orthogonal with respect to the inner product (f, g) =L∫0
f(t)g(t)dt. For f ε X, we thus have mean convergence to f of the series
A0
2+
∞∑
n=1
An cosnπx
L, where
An =(f, cos nπx
L)
(cosnπxL, cosnπx
L)
=2
L
L∫
0
f(t) cosnπt
Ldt.
Now, let f̃ be defined by
f̃(x) =
{f(−x) for − L < x < 0f(x) for 0 ≤ x < L
.
Observe that f̃ is an even function.
Now, compute the Fourier series for f̃ (more precisely. for the periodic extension of
f̃ .):
an =1
L
L∫
−L
f̃(t) cosnπt
Ldt = an =
2
L
L∫
0
f(t) cosnπt
Ldt = An,
since the integrand is an even function. And
bn =1
L
L∫
−L
f̃(t) sin nπt
Ldt = 0,
since the integrand is an odd function. In other words, the above series
A0
2+
∞∑
n=1
An cosnπx
L
18
is “really” the Fourier series for the extension f̃ . 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 f̃ , the even extension of
f looks like
The cosine series is
c(x) =π
2+
2
π
∞∑
k=1
(−1)k − 1
k2cos kx;
and it converges to this:
The collection {sin
πx
L, sin 2
πx
L, sin 3
πx
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
∞∑
n=1
Bn sinnπx
L,
19
where
Bn =2
L
L∫
0
f(t) sinnπt
Ldt
is the Fourier series of the odd periodic extension f̂ of f :
f̂(x) =
{−f(−x) −L < x < 0f(x) 0 ≤ x < L
.
Example. Let f be as in the previous example. Then f̂ looks like
–3
–2
–1
0
1
2
3
–3 –2 –1 1 2 3x
Here is the sine series:
s(x) = 2∞∑
k=1
(−1)k+1
ksin kx.
The limit of this series is thus
–3
–2
–1
1
2
3
–8 –6 –4 –2 2 4 6 8x
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