### y a0 dx y - Stewart Calculus

```FOURIER SERIES
When the French mathematician Joseph Fourier (1768–1830) was trying to solve a problem in heat conduction, he needed to express a function f as an infinite series of sine and
cosine functions:

f 共x兲 苷 a 0 1
n
cos nx bn sin nx兲
n苷1

The series in Equation 1 is called a trigonometric series or Fourier series and it turns
out that expressing a function as a Fourier series is sometimes more advantageous than
expanding it as a power series. In particular, astronomical phenomena are usually periodic,
as are heartbeats, tides, and vibrating strings, so it makes sense to express them in terms
of periodic functions.
We start by assuming that the trigonometric series converges and has a continuous function f 共x兲 as its sum on the interval 关, 兴, that is,
f 共x兲 苷 a 0 2

n
x cos nx bn sin nx兲
n苷1
Our aim is to find formulas for the coefficients a n and bn in terms of f . Recall that for a
power series f 共x兲 苷 冘 cn共x a兲 n we found a formula for the coefficients in terms of deriv共n兲
atives: cn 苷 f 共a兲兾n!. Here we use integrals.
If we integrate both sides of Equation 2 and assume that it’s permissible to integrate the
series term-by-term, we get
f 共x兲 dx 苷 y
y
a 0 dx y
cos nx bn sin nx兲 dx
n
But
y
n

1
sin nx
n
n
n苷1
cos nx dx 苷

n苷1

1

n
because n is an integer. Similarly, x sin nx dx 苷 0. So
y
f 共x兲 dx 苷 2 a0
1
2 ■ FOURIER SERIES
and solving for a0 gives
a0 苷
3
1
2
f 共x兲 dx
y
To determine an for n 1 we multiply both sides of Equation 2 by cos mx (where m is
an integer and m 1) and integrate term-by-term from to :
y
4
f 共x兲 cos mx dx 苷

y

a0 兺 共a
n

cos nx bn sin nx兲 cos mx dx
n苷1

cos mx dx n
n苷1
n
n苷1
We’ve seen that the first integral is 0. With the help of Formulas 81, 80, and 64 in the Table
of Integrals, it’s not hard to show that
y sin nx cos mx dx 苷 0
for all n and m
y cos nx cos mx dx 苷

0
for n 苷 m
for n 苷 m
So the only nonzero term in (4) is am and we get
y
f 共x兲 cos mx dx 苷 am
Solving for am , and then replacing m by n, we have
5
an 苷
1
y
f 共x兲 cos nx dx
n 苷 1, 2, 3, . . .
Similarly, if we multiply both sides of Equation 2 by sin mx and integrate from to ,
we get
6
bn 苷
1
y
f 共x兲 sin nx dx
n 苷 1, 2, 3, . . .
We have derived Formulas 3, 5, and 6 assuming f is a continuous function such that
Equation 2 holds and for which the term-by-term integration is legitimate. But we can still
consider the Fourier series of a wider class of functions: A piecewise continuous function
on 关a, b兴 is continuous except perhaps for a finite number of removable or jump discontinuities. (In other words, the function has no infinite discontinuities. See Section 2.5 for a
discussion of the different types of discontinuities.)
■ ■ Notice that a is the average value
0
of f over the interval 关, 兴.
FOURIER SERIES ■ 3
7 Definition Let f be a piecewise continuous function on 关, 兴. Then the
Fourier series of f is the series
a0 兺 共a
n
cos nx bn sin nx兲
n苷1
where the coefficients an and bn in this series are defined by
1
2
a0 苷
an 苷
1
y
y
f 共x兲 cos nx dx
f 共x兲 dx
1
bn 苷
y
f 共x兲 sin nx dx
and are called the Fourier coefficients of f .
Notice in Definition 7 that we are not saying f 共x兲 is equal to its Fourier series. Later we
will discuss conditions under which that is actually true. For now we are just saying that
associated with any piecewise continuous function f on 关, 兴 is a certain series called
a Fourier series.
EXAMPLE 1 Find the Fourier coefficients and Fourier series of the square-wave function
f defined by
f 共x兲 苷

0
1
if x 0
if 0 x f 共x 2兲 苷 f 共x兲
and
So f is periodic with period 2 and its graph is shown in Figure 1.
■ ■ Engineers use the square-wave function in
describing forces acting on a mechanical system
and electromotive forces in an electric circuit
(when a switch is turned on and off repeatedly).
Strictly speaking, the graph of f is as shown
in Figure 1(a), but it’s often represented as in
Figure 1(b), where you can see why it’s called a
square wave.
y
1
_π
0
π
2π
x
π
2π
x
(a)
y
1
_π
0
FIGURE 1
Square-wave function
(b)
SOLUTION Using the formulas for the Fourier coefficients in Definition 7, we have
a0 苷
1
2
y
f 共x兲 dx 苷
1
2
0
1
y 0 dx 2 y
0
1 dx 苷 0 1
1

2
2
4 ■ FOURIER SERIES
and, for n 1,
an 苷
1
y
1

Note that cos n equals 1 if n is even
and 1 if n is odd.
■ ■

1 sin nx
n

bn 苷
f 共x兲 cos nx dx 苷
y
0

0
1
0
y 0 dx y
0
cos nx dx
1

n
f 共x兲 sin nx dx 苷
1 cos nx
n
0
2
n

1
1
0
1
y 0 dx y
0
sin x dx
1

n
if n is even
if n is odd
The Fourier series of f is therefore
a 0 a1 cos x a2 cos 2x a3 cos 3x b1 sin x b2 sin 2x b3 sin 3x 苷
1
0 0 0 2

2
2
2
sin x 0 sin 2x sin 3x 0 sin 4x sin 5x 3
5
1
2
2
2
2
sin x sin 3x sin 5x sin 7x 2
3
5
7
1
2

sin共2k 1兲x
2 k苷1 共2k 1兲
In Example 1 we found the Fourier series of the square-wave function, but we don’t
know yet whether this function is equal to its Fourier series. Let’s investigate this question
graphically. Figure 2 shows the graphs of some of the partial sums
Sn共x兲 苷
1
2
2
2
sin x sin 3x sin nx
2
3
n
when n is odd, together with the graph of the square-wave function.
Since odd integers can be written as n 苷 2k 1, where k is an integer, we can write the
Fourier series in sigma notation as
FOURIER SERIES ■ 5
y
y
1
1
y
1
S£
S¡
_π
π
x
_π
y
x
π
_π
y
1
1
x
x
π
x
1
S¡¡
π
π
y
S¶
_π
S∞
_π
S¡∞
x
π
_π
FIGURE 2 Partial sums of the Fourier series for the square-wave function
We see that, as n increases, Sn共x兲 becomes a better approximation to the square-wave
function. It appears that the graph of Sn共x兲 is approaching the graph of f 共x兲, except where
x 苷 0 or x is an integer multiple of . In other words, it looks as if f is equal to the sum
of its Fourier series except at the points where f is discontinuous.
The following theorem, which we state without proof, says that this is typical of the
Fourier series of piecewise continuous functions. Recall that a piecewise continuous function has only a finite number of jump discontinuities on 关, 兴. At a number a where f
has a jump discontinuity, the one-sided limits exist and we use the notation
f 共a兲 苷 lim f 共x兲
f 共a兲 苷 lim f 共x兲
x l a
xla
8 Fourier Convergence Theorem If f is a periodic function with period 2 and f and
f are piecewise continuous on 关, 兴, then the Fourier series (7) is convergent.
The sum of the Fourier series is equal to f 共x兲 at all numbers x where f is continuous. At the numbers x where f is discontinuous, the sum of the Fourier series is
the average of the right and left limits, that is
1
2

If we apply the Fourier Convergence Theorem to the square-wave function f in
Example 1, we get what we guessed from the graphs. Observe that
f 共0兲 苷 lim f 共x兲 苷 1
xl0
and
f 共0兲 苷 lim f 共x兲 苷 0
x l 0
and similarly for the other points at which f is discontinuous. The average of these left and
right limits is 12 , so for any integer n the Fourier Convergence Theorem says that
1
2

sin共2k 1兲x 苷
2 k苷1 共2k 1兲
(Of course, this equation is obvious for x 苷 n.)

f 共x兲 if n 苷 n
1
if x 苷 n
2
6 ■ FOURIER SERIES
FUNCTIONS WITH PERIOD 2L
If a function f has period other than 2, we can find its Fourier series by making a change
of variable. Suppose f 共x兲 has period 2L, that is f 共x 2L兲 苷 f 共x兲 for all x. If we let
t 苷 x兾L and
t共t兲 苷 f 共x兲 苷 f 共Lt兾兲
then, as you can verify, t has period 2 and x 苷 L corresponds to t 苷 . The Fourier
series of t is

a0 n
cos nt bn sin nt兲
n苷1
where
a0 苷
an 苷
1
1
2
y t共t兲 dt
bn 苷
y t共t兲 cos nt dt
1
y t共t兲 sin nt dt
If we now use the Substitution Rule with x 苷 Lt兾, then t 苷 x兾L, dt 苷 共兾L兲 dx, and
we have the following
9
If f is a piecewise continuous function on 关L, L兴, its Fourier series is

a0 冋 冉 冊
a n cos
n苷1
n x
L

bn sin
n x
L
where
1
2L
a0 苷
and, for n 1,
an 苷
1
L
y
L
L

f 共x兲 cos
n x
L
y
L
L
f 共x兲 dx
bn 苷
dx
1
L
y
L
L

f 共x兲 sin
n x
L
dx
Of course, the Fourier Convergence Theorem (8) is also valid for functions with
period 2L.
ⱍ ⱍ
EXAMPLE 2 Find the Fourier series of the triangular wave function defined by f 共x兲 苷 x
for 1 x 1 and f 共x 2兲 苷 f 共x兲 for all x. (The graph of f is shown in Figure 3.)
For which values of x is f 共x兲 equal to the sum of its Fourier series?
y
1
FIGURE 3
The triangular wave function
_1
0
1
2
x
Notice that when L 苷 these
formulas are the same as those in (7).
■ ■
FOURIER SERIES ■ 7
SOLUTION We find the Fourier coefficients by putting L 苷 1 in (9):
a 0 苷 12 y
1
1
■ ■ Notice that a is more easily calculated as
0
an area.
ⱍ x ⱍ dx 苷 y
1
2
]
0
1

14 x 2
0
1
1
]
1
0

0

and for n 1,
an 苷 y
1
1
ⱍ x ⱍ cos共n x兲 dx 苷 2 y
1
0
x cos共n x兲 dx
ⱍ ⱍ
because y 苷 x cos共n x兲 is an even function. Here we integrate by parts with u 苷 x
and dv 苷 cos共n x兲 dx. Thus,

an 苷 2

x
sin共n x兲
n
2

n

1
0
2
n
y

cos共n x兲
n
1
0
1
0
sin共n x兲 dx
2

n2 2
ⱍ ⱍ
Since y 苷 x sin共n x兲 is an odd function, we see that
bn 苷 y
1
1
ⱍ x ⱍ sin共n x兲 dx 苷 0
We could therefore write the series as
1
2共cos n 1兲

cos共n x兲
2 n苷1
n2 2
But cos n 苷 1 if n is even and cos n 苷 1 if n is odd, so
an 苷

0
2

n
1兲

4
n2 2
2 2
n
if n is even
if n is odd
Therefore, the Fourier series is
1
4
4
4
2 cos共 x兲 cos共5 x兲 2 cos共3 x兲 2
9
25 2

1
4

cos共共2k 1兲 x兲
2 n苷1 共2k 1兲2 2
The triangular wave function is continuous everywhere and so, according to the Fourier
Convergence Theorem, we have
f 共x兲 苷
1
4

cos共共2k 1兲 x兲
2 n苷1 共2k 1兲2 2
for all x
8 ■ FOURIER SERIES
In particular,
1
ⱍxⱍ 苷 2 兺
k苷1
4
cos共共2k 1兲 x兲

for 1 x 1
FOURIER SERIES AND MUSIC
One of the main uses of Fourier series is in solving some of the differential equations that
arise in mathematical physics, such as the wave equation and the heat equation. (This is
covered in more advanced courses.) Here we explain briefly how Fourier series play a role
in the analysis and synthesis of musical sounds.
We hear a sound when our eardrums vibrate because of variations in air pressure. If a
guitar string is plucked, or a bow is drawn across a violin string, or a piano string is struck,
the string starts to vibrate. These vibrations are amplified and transmitted to the air. The
resulting air pressure fluctuations arrive at our eardrums and are converted into electrical
impulses that are processed by the brain. How is it, then, that we can distinguish between
a note of a given pitch produced by two different musical instruments?
The graphs in Figure 4 show these fluctuations (deviations from average air pressure)
for a flute and a violin playing the same sustained note D (294 vibrations per second) as
functions of time. Such graphs are called waveforms and we see that the variations in air
pressure are quite different from each other. In particular, the violin waveform is more
complex than that of the flute.
t
(a) Flute
(b) Violin
We gain insight into the differences between waveforms if we express them as sums of
Fourier series:

P共t兲 苷 a 0 a1 cos
t
L

b1 sin
t
L

a2 cos
2 t
L

b2 sin
2 t
L
In doing so, we are expressing the sound as a sum of simple pure sounds. The difference
in sounds between two instruments can be attributed to the relative sizes of the Fourier
coefficients of the respective waveforms.
The n th term of the Fourier series, that is,

a n cos
n t
L
bn
n t
L
is called the nth harmonic of P. The amplitude of the n th harmonic is
A n 苷 sa 2n b2n
and its square, A2n 苷 a 2n b2n , is sometimes called energy of the n th harmonic. (Notice that
FIGURE 4
Waveforms
t
FOURIER SERIES ■ 9
for a Fourier series with only sine terms, as in Example 1, the amplitude is A n bn and
the energy is A2n b 2n.) The graph of the sequence A2n is called the energy spectrum of
P and shows at a glance the relative sizes of the harmonics.
Figure 5 shows the energy spectra for the flute and violin waveforms in Figure 4. Notice
that, for the flute, A2n tends to diminish rapidly as n increases whereas, for the violin, the
higher harmonics are fairly strong. This accounts for the relative simplicity of the flute
waveform in Figure 4 and the fact that the flute produces relatively pure sounds when
compared with the more complex violin tones.
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0
FIGURE 5
Energy spectra
2
4
6
8
0
n
10
2
4
(a) Flute
6
8
n
10
(b) Violin
In addition to analyzing the sounds of conventional musical instruments, Fourier series
enable us to synthesize sounds. The idea behind music synthesizers is that we can combine
various pure tones (harmonics) to create a richer sound through emphasizing certain
harmonics by assigning larger Fourier coefficients (and therefore higher corresponding
energies).
EXERCISES
7–11
S

1
1
if x 1
if 1 x 2
f 共x 4兲 苷 f 共x兲
0
1
0
if 2 x 0
if 0 x 1
if 1 x 2
f 共x 4兲 苷 f 共x兲
ⱍ ⱍ

x if 4 x 0
0
if 0 x 4
9. f 共x兲 苷
if x 0
if 0 x ⱍ ⱍ
1
0
10. f 共x兲 苷 1 x,
11. f 共t兲 苷 sin共3 t兲,
1 x 1
f 共x 8兲 苷 f 共x兲
f 共x 2兲 苷 f 共x兲
1 t 1
3. f 共x兲 苷 x
■
4. f 共x兲 苷 x 2
12. A voltage E sin t, where t represents time, is passed through a
5. f 共x兲 苷

■
■
0
if x 0
cos x if 0 x ■
■
■
■
■
■

0
f 共t兲 苷
if x 兾2
if 兾2 x 0
if 0 x ■
■
■
■
■
■
■
■
■
■
so-called half-wave rectifier that clips the negative part of the
wave. Find the Fourier series of the resulting periodic function

1
6. f 共x兲 苷 1
0
■

8. f 共x兲 苷
if x 0
if 0 x 0
2. f 共x兲 苷
x

7. f 共x兲 苷
A function f is given on the interval 关, 兴 and f is
periodic with period 2.
(a) Find the Fourier coefficients of f .
(b) Find the Fourier series of f . For what values of x is f 共x兲
equal to its Fourier series?
; (c) Graph f and the partial sums S2, S4, and S6 of the Fourier
series.
1–6
1. f 共x兲 苷
Find the Fourier series of the function.
■
■
■
■
■
E sin t
t0
if 0 t if f 共t 2兾兲 苷 f 共t兲
■
10 ■ FOURIER SERIES
18. Use the result of Example 2 to show that
13–16
Sketch the graph of the sum of the Fourier series of f
without actually calculating the Fourier series.
1
13. f x 3
x
1x
14. f x 1 x 1
16. f x e ,
2 x 2
■
■
■
■
■
■
1
1
1
1
3
5
7
4
20. Use the given graph of f and Simpson’s Rule with n 8 to
■
■
■
■
17. (a) Show that, if 1 x 1, then
x2 1
1
2
1
2 2 2
3
5
7
8
19. Use the result of Example 1 to show that
if 1 x 0
if 0 x 1
15. f x x 3,
x
1
if 4 x 0
if 0 x 4
■
■
■
estimate the Fourier coefficients a 0, a1, a 2, b1, and b2. Then use
them to graph the second partial sum of the Fourier series and
compare with the graph of f .
y
1
4
1 n 2 2 cosn x
3
n
n1
1
(b) By substituting a specific value of x, show that
n1
1
2
n2
6
0.25
x
FOURIER SERIES ■ 11
SOLUTIONS
π 1 π
1 0
f (x) dx =
dx − 0 dx = 0.
−π
−π
2π
2π
1 π π
1 0
1 π
f(x) cos nx dx =
cos nx dx −
cos nx dx = 0 [since cos nx is even].
an =
π −π −π
π −π
π 0
1 π π
1 0
1 π
2 0
f (x) sin nx dx =
sin nx dx −
sin nx dx =
sin nx dx [since sin nx is odd]
bn =
π −π −π
π −π
π 0
π −π

if n even
 0
2
= − [1 − cos(−nπ)] =
− 4
nπ
if n odd
nπ
1. (a) a0 =
(b) f (x) =
∞
k=0
−
4
sin(2k + 1)x when −π < x < 0 and 0 < x < π.
(2k + 1)π
(c)
y
1
0.5
0
-2.5
-1.25
0
1.25
2.5
x
-0.5
-1
1 π
1 π
f (x) dx =
x dx = 0.
−π
2π
2π −π
1 π
1 π
an =
f (x) cos nx dx =
x cos nx dx = 0 [because x cos nx is odd]
π −π
π −π
1 π
1 π
2 π
bn =
f (x) sin nx dx =
x sin nx dx =
x sin nx dx [since x sin nx is odd]
−π
−π
π
π
π 0
−(2/n) if n even
2
= − cos nπ [using integration by parts] =
n
(2/n)
if n odd
3. (a) a0 =
(b) f (x) =
∞
n=1
(−1)n+1
2
sin nx
n
y
(c)
2.5
when −π < x < π.
1.25
0
-2.5
-1.25
0
1.25
2.5
x
-1.25
-2.5
12 ■ FOURIER SERIES
1 π
1 π
f (x) dx =
cos x dx = 0
2π −π
2π 0
1
if n = 1
1 π
1 π
2
an =
f (x) cos nx dx =
cos x cos nx dx =
−π
0
π
π
0 if n = 1
1 π
1 π
f (x) sin nx dx =
cos x sin nx dx
bn =
π −π
π 0

2n

if n even
using an integral table,
2 − 1)
π(n
=

and simplified using the addition formula for cos(a + b)
0
if n odd
5. (a) a0 =
(b) f (x) =
1
2
cos x +
∞
k=1
π
]
2
4k
sin(2k) when −π < x < 0, 0 < x < π.
π (4k2 − 1)
y
(c)
1
0.5
0
-2.5
-1.25
0
1.25
2.5
x
-0.5
-1

0


7. Use f (x) = 1


0
a0 =
if − 2 ≤ x ≤ −1
if − 1 < x < 1,
L = 2.
if 1 ≤ x ≤ 2
1 L
f (x) dx =
2L −L
1
4
1
−1
dx =
nπx 1 L
dx =
an =
f
(x)
cos
L −L
L
1
2
nπx 1 L
dx =
f (x) sin
−L
L
L
1
2
bn =
Fourier Series:
1
2
1
2
1
2

0

π 
nπx 2
2/nπ
dx =
sin n =
cos
−1

2
nπ
2

−2/nπ
1
1
−1
sin
nπx 2
if n even
if n = 4n + 1
if n = 4n + 3
dx = 0
πx 2
3πx
2
5πx
2
−
cos
+
cos
−···
+ cos
π
2
3π
2
5π
2
π
π
∞
2
2
sin (4k + 1) −
sin (4k + 3)
+
2
(4k + 3) π
2
k=1 (4k + 1) π
y
1
0.75
0.5
0.25
0
-5
-2.5
0
2.5
5
x
FOURIER SERIES ■ 13
9. Use f (x) =
−x
if − 4 ≤ x < 0
0
if 0 ≤ x ≤ 4
0
a0 =
1 L
f (x) dx =
2L −L
an =
nπx 1 L
dx =
f
(x)
cos
L −L
L
0
1
8
−4
, L = 4.
−x dx = 1
1
4
nπx 4
dx =
−x
cos
(cos (nπ) − 1) =
−4
4
(nπ)2
0
if n is even
2
−8/(nπ)
if n is odd
nπx 1 L
bn =
f(x) sin
dx =
−L
L
L
1
4
nπx 4
cos (nπ) =
−x sin
dx =
−4
4
nπ
0
4/nπ
if n is even
−4/nπ
if n is odd
Fourier Series:
π
π
π
∞
8
4
4
1+
sin (2k − 1)x −
(2k − 1) x +
sin (2k)x
−
cos
2
2
(2k − 1)π
4
(2k − 1) π
4
(2k)π
4
k=1
y
4
3
2
1
0
-5
-2.5
0
2.5
5
x
11. Use f (x) = {sin(3πt) if − 1 ≤ t ≤ 1 , L = 1.
Note: This can be done instantly if one observes that the period of sin(3πt) is 23 , and the period of f (x) = 2 which
is an integer multiple of 23 . Therefore f (x) is the same as sin(3πt) for all t, and its Fourier series is therefore
sin(3πt).
We can get this result using the standard coefficient formulas: a0 =
1 L
f (x) dx =
2L −L
1
2
1
−1
sin(3πx) dx = 0
nπx 1
1 1
dx = −1 sin(3πx) cos(nπx) dx
f
(x)
cos
L −1
L
= 0 [applying change of variables to a formula in the section]
an =
nπx 1
1 L
dx = −1 sin(3πx) sin(nπx) dx
f (x) sin
−L
L
L

sin nπ
6
if n = 3
0
2)
π
(−9
+
n
=
[using integral table and addition formula =

1
1
if n = 3
bn =
if n = 3
if n = 3
14 ■ FOURIER SERIES
Fourier Series: sin(3πx)
y
1
0.5
0
-5
-2.5
0
2.5
5
x
-0.5
-1
y
13.

3




 −1

3




−1
3
2
if −5 ≤ x < −4
if −4 ≤ x < 0
1
if 0 ≤ x < 4
if 4 ≤ x < 5
0
-5
-2.5
0
2.5
5
x
-1
15.
y
1
0.5
0
-1.25
0
1.25
2.5
3.75
x
-0.5
-1
17. (a) We find the Fourier series for f (x) = {x2
1 L
f (x) dx =
a0 =
2L −L
1
2
1
−1
x2 dx =
if −1 ≤ x ≤ 1, L = 1
1
3

4

if n even

2

(nπ)
1 1
nπx
4
1
2
dx = −1 x cos(nπx) dx =
an =
f (x) cos
cos nπ =

4
L −1
L
(nπ)2

−
if n odd
(nπ)2
nπx 1
1 L
dx = −1 x2 sin(nπx) dx = 0 because x2 sin(nπx) is odd.
f (x) sin
bn =
−L
L
L
So we have x2 =
1
3
+
∞
n=1
(−1)n
4
cos(nπx) for −1 ≤ x ≤ 1.
(nπ)2
(b) We let x = 1 in the above to obtain
1=
1
3
+
∞
n=1
(−1)n
4
cos(nπ)
(nπ)2
2
3
=
∞
4
n=1
n2 π2
∞ 1
π2
=
2
6
n=1 n
FOURIER SERIES ■ 15
19. Example 1 says that, for 0 ≤ x < π, 1 =
Let x =
π
2
1
2
+
1
2
+
∞
k=1
2
sin((2k − 1)x).
(2k − 1)π
to obtain
1=
π
2
sin (2k − 1)
2
k=1 (2k − 1)π
∞
∞
π
1
=
sin((2k − 1))
4
k=1 (2k − 1)