Powers D.L. Boundary Value Problems and Partial Differential Equations
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88 Chapter 1 Fourier Series and Integrals
uous, it is certain that the differentiated series of f (x) will fail to converge, at
some points at least.
Example.
Let f be the function that is periodic with period 2π and has the formula
f (x) =|x|, −π<x <π.
This function is indeed continuous and sectionally smooth and is equal to its
Fourier series,
f (x ) =
π
2
−
4
π
cos(x) +
cos(3x)
9
+
cos(5x)
25
+···
.
According to Theorem 5, the differentiated series
4
π
sin(x) +
sin(3x)
3
+
sin(5x)
5
+···
converges to f
(x) at any point x where f
(x) exists. Now, the derivative of the
sawtooth function f (x) (see Fig. 10) is the square-wave function
f
(x) =
1, 0 < x <π,
−1, −π<x < 0
(9)
(see Fig. 9). Moreover, we know that the foregoing sine series is the Fourier
seriesofthesquarewavef
(x) andthatitconvergestothevaluesgivenby
Eq.(9),exceptatthepointswheref
(x) has a jump. These are precisely the
points where f
(x) does not exist.
Later on, it will frequently happen that we know a function only through its
Fourier series. Thus, it will be important to obtain properties of the function
by examining its coefficients, as the next theorem does.
Theorem 7. If f is periodic, w ith Fourier coefficients a
n
,b
n
,andiftheseries
∞
n=1
n
k
a
n
+
n
k
b
n
converges for some integer k ≥ 1, then f has continuous derivatives f
,...,f
(k)
whose Fourier series are differentiated series of f .
Example.
Consider the function defined by the series
f (x) =
∞
n=1
e
−nα
cos(nx),
1.5 Operations on Fourier Series 89
in which α is a positive parameter. For this function we have a
0
=0, a
n
=e
−nα
,
b
n
=0. By the integral test, the series
n
k
e
−nα
converges for any k. Therefore
f has derivatives of all orders. The Fourier series of f
and f
are
f
(x) =
∞
n=1
−ne
−nα
sin(nx),
f
(x) =
∞
n=1
−n
2
e
−nα
cos(nx).
EXERCISES
1. Evaluate the sum of the series
∞
n=1
1/n
2
by performing the integration
indicated in Eq. (5).
2. Sketch the graphs of the periodic extension of the function
f (x) =
π −x
2
, 0 < x < 2π,
and of its derivative f
(x) and of
F(x) =
x
0
f (t) dt.
3. Suppose that a function has the formula f (x) = x,0< x <π. What is its
derivative? Can the Fourier sine series of f be differentiated term by term?
What about the cosine series?
4. Verify Eqs. (6) and (7) by integration.
5. Suppose that a function f (x) is continuous and sectionally smooth in the
interval 0 < x < a. What additional conditions must f (x) satisfy in order
to guarantee that its sine series can be differentiated term by term? the
cosine series?
6. Is the derivative of a periodic function periodic? Is the integral of a peri-
odic function periodic?
7. It is known that the equality
ln
2cos
x
2
=
∞
n=1
(−1)
n+1
n
cos(nx)
is valid except when x is an odd multiple of π. Can the Fourier series be
differentiated term by term?
90 Chapter 1 Fourier Series and Integrals
8.
Use the series that follows, together with integration or differentiation, to
find a Fourier series for the function p(x) = x(π − x),0< x <π.
x = 2
∞
n=1
(−1)
n+1
n
sin(nx), 0 < x <π.
9. Let f (x) be an odd, periodic, sectionally smooth function with Fourier
sine coefficients b
1
, b
2
,.... Show that the function defined by
u(x, t) =
∞
n=1
b
n
e
−n
2
t
sin(nx), t ≥0,
has the following properties:
a.
∂
2
u
∂x
2
=
∞
n=1
−n
2
b
n
e
−n
2
t
sin(nx), t > 0;
b. u(0, t) = 0, u(π, t) = 0, t > 0;
c. u(x, 0) =
1
2
f (x+) +f (x−)
.
10. Let f be as in Exercise 9, but define u(x, y) by
u(x, y) =
∞
n=1
b
n
e
−ny
sin(nx), y > 0.
Show that u(x, y) has these properties:
a.
∂
2
u
∂x
2
=
∞
n=1
−n
2
b
n
e
−ny
sin(nx), y > 0;
b. u(0, y) = 0, u(π, y) = 0, y > 0;
c. u(x, 0) =
1
2
f (x+) +f (x−)
.
1.6 Mean Error a nd Convergence in Mean
While we can study the behavior of infinite series, we must almost always use
finite series in practice. Fortunately, Fourier series have some properties that
make them very useful in this setting. Before going on to these properties, we
shall develop a useful formula.
Suppose f is a function defined in the interval −a < x < a, for which
a
−a
f (x )
2
dx
1.6 Mean Error and Convergence in Mean 91
is a finite number. Let
f (x) ∼ a
0
+
∞
n=1
a
n
cos
n πx
a
+b
n
sin
n πx
a
and let g(x) have a finite Fourier series
g(x) = A
0
+
N
1
A
n
cos
n πx
a
+B
n
sin
n πx
a
.
Then we may perform the following operations:
a
−a
f (x )g(x) dx =
a
−a
f (x)
A
0
+
N
1
A
n
cos
n πx
a
+B
n
sin
n πx
a
dx
= A
0
a
−a
f (x ) dx +
N
1
A
n
a
−a
f (x) cos
n πx
a
dx
+
N
1
B
n
a
−a
f (x) sin
n πx
a
dx.
We recognize the integrals as multiples of the Fourier coefficients of f and
rewrite
1
a
a
−a
f (x )g(x) dx = 2a
0
A
0
+
N
1
(a
n
A
n
+b
n
B
n
). (1)
Now suppose we wish to approximate f (x) by a finite Fourier series. The
difficulty here is deciding what “approximate” means. Of the many ways we
can measure approximation, the one that is easiest to use is the following:
E
N
=
a
−a
f (x ) −g(x)
2
dx. (2)
(Here g is the function with a Fourier series containing terms up to and in-
cluding cos(Nπx/a).) Clearly, E
N
can never be negative, and if f and g are
“close,” then E
N
will be small. Thus our problem is to choose the coefficients
of g so as to minimize E
N
.(WeassumeN fixed.)
To c o m p u t e E
N
, we first expand the integrand:
E
N
=
a
−a
f
2
(x) dx − 2
a
−a
f (x)g(x) dx +
a
−a
g
2
(x) dx. (3)
The first integral has nothing to do with g; the other two integrals clearly de-
pend on the choice of g and can be manipulated so as to minimize E
N
.We
92 Chapter 1 Fourier Series and Integrals
already have an expression for the middle integral. The last one can be found
by replacing f with g in Eq. (1):
a
−a
g
2
(x) dx = a
2A
2
0
+
N
1
A
2
n
+B
2
n
. (4)
Now we have a formula for E
N
in terms of the variables A
0
, A
n
, B
n
:
E
N
=
a
−a
f
2
(x) dx − 2a
2A
0
a
0
+
N
1
A
n
a
n
+B
n
b
n
+a
2A
2
0
+
N
1
A
2
n
+B
2
n
. (5)
The error E
N
takes its minimum value when all of the partial derivatives
with respect to the variables are zero. We must then solve the equations
∂E
N
∂A
0
=−4aa
0
+4aA
0
=0,
∂E
N
∂A
n
=−2aa
n
+2aA
n
=0,
∂E
N
∂B
n
=−2ab
n
+2aB
n
=0.
These equations require that A
0
= a
0
, A
n
= a
n
, B
n
= b
n
.Thusg should be
chosen to be the truncated Fourier series of f ,
g(x) = a
0
+
N
n=1
a
n
cos
n πx
a
+b
n
sin
n πx
a
in order to minimize E
N
.
Now that we know which choice of A’s and B’s minimizes E
N
,wecancom-
pute that minimum value. After some algebra, we see that
min(E
N
) =
a
−a
f
2
(x) dx − a
2a
2
0
+
N
1
a
2
n
+b
2
n
. (6)
Eventhisminimumerrormustbegreaterthanorequaltozero,andthuswe
have the Bessel inequality
1
a
a
−a
f
2
(x) dx ≥ 2a
2
0
+
N
1
a
2
n
+b
2
n
. (7)
1.6 Mean Error and Convergence in Mean 93
This inequality is valid for any N and therefore is also valid in the limit as N
tends to infinity. The actual fact is that, in the limit, the inequality becomes
Parseval’ s equality:
1
a
a
−a
f
2
(x) dx = 2a
2
0
+
∞
1
a
2
n
+b
2
n
. (8)
Another very important consequence of Bessel’s inequality is that the two
series
a
2
n
and
b
2
n
must converge if the left-hand side of Eqs. (7) and (8) is
finite. Thus, the numbers a
n
and b
n
must tend to 0 as n tends to infinity.
By comparing Eqs. (6) and (8), we get a different expression for the mini-
mum error:
min(E
N
) =a
∞
N+1
a
2
n
+b
2
n
.
This quantity decreases steadily to zero as N increases. Since min(E
N
) is, ac-
cording to Eq. (2), a mean deviation between f and the truncated Fourier se-
ries of f , we often say, “The Fourier series of f converges to f in the mean.”
(Another kind of convergence!)
Summary
If f (x) has been defined in the interval −a < x < a and if
a
−a
f
2
(x) dx
is finite, then:
1. Among all finite series of the form
g(x) = A
0
+
N
1
A
n
cos
n πx
a
+B
n
sin
n πx
a
the one that best approximates f in the sense of the error described by
Eq. (2) is the truncated Fourier series of f :
a
0
+
N
1
a
n
cos
n πx
a
+b
n
sin
n πx
a
.
2.
1
a
a
−a
f
2
(x) dx = 2a
2
0
+
∞
1
a
2
n
+b
2
n
.
94 Chapter 1 Fourier Series and Integrals
3. a
n
=
1
a
a
−a
f (x ) cos
n πx
a
dx → 0asn →∞;
b
n
=
1
a
a
−a
f (x) sin
n πx
a
dx → 0asn →∞.
4. The Fourier series of f converges to f in the sense of the mean.
Properties 2 and 3 are very useful for checking computed values of Fourier
coefficients.
EXERCISES
1. Use properties of Fourier series to evaluate the definite integral
1
π
π
−π
ln
2cos
x
2
2
dx.
(Hint: See Section 10, Eq. (4), and Section 5, Eq. (5).)
2. Verify Parseval’s equality for these functions:
a. f (x) = x, −1 < x < 1;
b. f (x) = sin(x ), −π<x <π.
3. What can be said about the behavior of the Fourier coefficients of the fol-
lowing functions as n →∞?
a. f (x) =|x |
1/2
, −1 < x < 1;
b. f (x) =|x |
−1/2
, −1 < x < 1.
4. How do we know that E
N
has a minimum and not a maximum?
5. If a function f defined on the interval −a < x < a has Fourier coefficients
a
n
=0, b
n
=
1
√
n
,
what can you say about
a
−a
f
2
(x) dx?
6. Show that, as n →∞, the Fourier sine coefficients of the function
f (x ) =
1
x
, −π<x <π,
tend to a nonzero constant. (Since this is an odd function, we can take the
cosine coefficients to be zero, although strictly speaking they do not exist.)
1.7 Proof of Convergence 95
Use the fact that
∞
0
sin(t)
t
dt =
π
2
.
1.7 Proof of Convergence
In this section we prove the Fourier convergence theorem stated in Section 3.
Most of the proof requires nothing more than simple calculus, but there are
three technical points that we state here.
Lemma 1. For all N = 1, 2,...,
1
π
π
−π
1
2
+
N
n=1
cos(ny)
dy =1.
Lemma 2. For all N = 1, 2,...,
1
2
+
N
n=1
cos(ny) =
sin
(N +
1
2
)y
2sin(
1
2
y)
.
Lemma 3. If φ(y) is sectionally continuous, −π<y <π,thenitsFouriercoef-
ficients tend to 0 with n:
lim
n→∞
1
π
π
−π
φ(y) cos(ny) dy = 0,
lim
n→∞
1
π
π
−π
φ(y) sin(ny) dy = 0.
In Exercises 1 and 2 of this section, you are asked to verify Lemmas 1 and 2
(also see Miscellaneous Exercise 17 at the end of this chapter). Lemma 3 was
proved in Section 6.
The theorem we are going to prove is restated here for easy reference. Period
2π is used for typographic convenience; we have seen that any other period
can be obtained by a simple change of variables.
Theorem. If f (x) is sectionally smooth and periodic with period 2π , then the
Fourier series corresponding to f converges at every x, and the sum of the series is
a
0
+
∞
n=1
a
n
cos(nx) + b
n
sin(nx) =
1
2
f (x +) +f (x−)
. (1)
96 Chapter 1 Fourier Series and Integrals
Proof: Let the point x be chosen; it is to remain fixed. To begin with, we
assume that f is continuous at x, so the sum of the series should be f (x).
Another way to say this is that
lim
N→∞
S
N
(x) −f (x) = 0,
where S
N
is the partial sum of the Fourier series of f ,
S
N
(x) = a
0
+
N
n=1
a
n
cos(nx) + b
n
sin(nx). (2)
Of course, the a’s and b’s are the Fourier coefficients of f ,
a
0
=
1
2π
π
−π
f (z) dz,
a
n
=
1
π
π
−π
f (z) cos(nz) dz,
b
n
=
1
2π
π
−π
f (z) sin(nz) dz.
(3)
The integrals have z as their variable of integration, but that does not affect
their value.
Part 1. Transformation of S
N
(x).
In order to show a relationship between S
N
(x) and f , we replace the co-
efficients in Eq. (2) by the integrals that define them and use elementary
algebra on the results:
S
N
(x) =
1
2π
π
−π
f (z) dx +
N
n=1
1
π
π
−π
f (z) cos(nz) dz cos(nx)
+
1
π
π
−π
f (z) sin(nz) dz sin(nx)
(4)
=
1
2π
π
−π
f (z) dx +
N
n=1
1
π
π
−π
f (z) cos(nz) cos(nx) dz
+
1
π
π
−π
f (z) sin(nz) sin(nx) dz
(5)
=
1
2π
π
−π
f (z) dx +
N
n=1
1
π
π
−π
f (z)
cos(nz) cos(nx)
+sin(nz) sin(nx)
dz
(6)
1.7 Proof of Convergence 97
=
1
π
π
−π
f (z)
1
2
+
N
n=1
cos(nz) cos(nx) +sin(nz) sin(nx)
dz
(7)
=
1
π
π
−π
f (z)
1
2
+
N
n=1
cos
n(z −x)
dz. (8)
In this very compact formula for S
N
(x), we now change the variable of in-
tegration from z to y = z − x:
S
N
(x) =
1
π
π+x
−π+x
f (x +y)
1
2
+
N
n=1
cos(ny)
dy. (9)
Note that both factors in the integrand are periodic with period 2π.The
interval of integration can be any interval of length 2π with no change in
the result. (See Exercise 5 of Section 1.) Therefore,
S
N
(x) =
1
π
π
−π
f (x +y)
1
2
+
N
n=1
cos(ny)
dy. (10)
Part 2. Expression for S
N
(x) −f (x).
Since we must show that the difference S
N
(x) − f (x) goes to 0, we need to
have f (x) in a form compatible with that for S
N
(x).Recallthatx is fixed
(although arbitrary), so f (x) is to be thought of as a number. Lemma 1
suggests the appropriate form,
f (x) = f (x) ·
1
π
π
−π
1
2
+
N
n=1
cos(ny)
dy
=
1
π
π
−π
f (x)
1
2
+
N
n=1
cos(ny)
dy. (11)
Now, using Eq. (10) to represent S
N
(x),wehave
S
N
(x) −f (x) =
1
π
π
−π
f (x +y) −f (x)
1
2
+
N
n=1
cos(ny)
dy. (12)
Part 3. The limit.
ThenextstepistouseLemma2toreplacethesuminEq.(12).Theresult
is
S
N
(x) −f (x) =
1
π
π
−π
f (x +y) −f (x)
sin
(N +
1
2
)y
2sin(
1
2
y)
dy. (13)