Series de Fourier

Prévia do material em texto

FOURIER SERIES
When the French mathematician Joseph Fourier (1768–1830) was trying to solve a prob-
lem in heat conduction, he needed to express a function as an infinite series of sine and
cosine functions:
Earlier, Daniel Bernoulli and Leonard Euler had used such series while investigating prob-
lems concerning vibrating strings and astronomy.
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 func-
tion as its sum on the interval , that is,
Our aim is to find formulas for the coefficients and in terms of . Recall that for a
power series we found a formula for the coefficients in terms of deriv-
atives: . 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
But
because is an integer. Similarly, . So
y�
��
 f �x� dx � 2�a0
x�
��
 sin nx dx � 0n
y�
��
 cos nx dx �
1
n
 sin nx�
��
�
�
1
n
 �sin n� � sin��n��� � 0
 � 2�a0 � �
�
n�1
 an y�
��
 cos nx dx � �
�
n�1
 bn y�
��
 sin nx dx
y�
��
 f �x� dx � y�
��
 a0 dx � y�
��
 �
�
n�1
 �an cos nx � bn sin nx� dx
cn � f �n��a��n!
f �x� � 	 cn�x � a�n
fbnan
�� � x � �f �x� � a0 � �
�
n�1
 �an cos nx � bn sin nx�2
���, ��f �x�
� b1 sin x � b2 sin 2x � b3 sin 3x � � � �
� a0 � a1 cos x � a2 cos 2x � a3 cos 3x � � � �
f �x� � a0 � �
�
n�1
 �an cos nx � bn sin nx�1
f
1
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and solving for gives
To determine for we multiply both sides of Equation 2 by (where is
an integer and ) and integrate term-by-term from to :
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
for all and 
So the only nonzero term in (4) is and we get
Solving for , and then replacing by , we have
Similarly, if we multiply both sides of Equation 2 by and integrate from to ,
we get
We have derived Formulas 3, 5, and 6 assuming 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 is continuous except perhaps for a finite number of removable or jump disconti-
nuities. (In other words, the function has no infinite discontinuities. See Section 2.5 for a
discussion of the different types of discontinuities.)
�a, b�
f
n � 1, 2, 3, . . .bn �
1
�
 y�
��
 f �x� sin nx dx6
���sin mx
n � 1, 2, 3, . . .an �
1
�
 y�
��
 f �x� cos nx dx5
nmam
y�
��
 f �x� cos mx dx � am�
am�
y�
��
 cos nx cos mx dx � 
0
�
for n � m
for n � m
mny�
��
 sin nx cos mx dx � 0
� a0 y�
��
 cos mx dx � �
�
n�1
 
an y�
��
 cos nx cos mx dx � �
�
n�1
 bn y�
��
 sin nx cos mx dx4
y�
��
 f �x� cos mx dx � y�
��
 �a0 � ��
n�1
 �an cos nx � bn sin nx�� cos mx dx
���m � 1
mcos mxn � 1an
a0 �
1
2�
 y�
��
 f �x� dx3
a0
2 ■ FOUR IER SER I ES
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■ ■ Notice that is the average value
of over the interval .���, ��f
a0
Definition Let be a piecewise continuous function on . Then the
Fourier series of is the series
where the coefficients and in this series are defined by
and are called the Fourier coefficients of .
Notice in Definition 7 that we are not saying 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 on is a certain series called
a Fourier series.
EXAMPLE 1 Find the Fourier coefficients and Fourier series of the square-wave function
defined by
and
So is periodic with period and its graph is shown in Figure 1.
SOLUTION Using the formulas for the Fourier coefficients in Definition 7, we have
a0 �
1
2�
 y�
��
 f �x� dx � 1
2�
 y0
��
 0 dx �
1
2�
 y�
0
 1 dx � 0 �
1
2�
 ��� �
1
2
0
FIGURE 1
Square-wave function
(a)
π 2π_π
1
y
x0
π 2π_π
1
y
x
(b)
2�f
f �x � 2�� � f �x�f �x� � 
01 if �� � x 	 0if 0 � x 	 �
f
���, ��f
f �x�
f
bn �
1
�
 y�
��
 f �x� sin nx dxan � 1
�
 y�
��
 f �x� cos nx dx
a0 �
1
2�
 y�
��
 f �x� dx
bnan
a0 � �
�
n�1
 �an cos nx � bn sin nx�
f
���, ��f7
FOUR IER SER I ES ■ 3
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■ ■ 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 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.
f
and, for ,
The Fourier series of is therefore
Since odd integers can be written as , where is an integer, we can write the
Fourier series in sigma notation as
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
when is odd, together with the graph of the square-wave function.n
Sn�x� �
1
2
�
2
�
 sin x �
2
3�
 sin 3x � � � � �
2
n�
 sin nx
1
2
� �
�
k�1
 
2
�2k � 1��
 sin�2k � 1�x
kn � 2k � 1
�
1
2
�
2
�
 sin x �
2
3�
 sin 3x �
2
5� sin 5x �
2
7�
 sin 7x � � � �
�
2
�
 sin x � 0 sin 2x �
2
3�
 sin 3x � 0 sin 4x �
2
5� sin 5x � � � �
�
1
2
� 0 � 0 � 0 � � � �
� b1 sin x � b2 sin 2x � b3 sin 3x � � � �
a0 � a1 cos x � a2 cos 2x � a3 cos 3x � � � �
f
 � 
02
n�
if n is even
if n is odd
 � �
1
�
 
cos nx
n
�
0
�
� � 
1
n�
 �cos n� � cos 0�
bn �
1
�
 y�
��
 f �x� sin nx dx � 1
�
 y0
��
 0 dx �
1
�
 y�
0
 sin x dx
 � 0 �
1
�
 
sin nx
n
�
0
�
� 
1
n�
 �sin n� � sin 0� � 0
an �
1
�
 y�
��
 f �x� cos nx dx � 1
�
 y0
��
 0 dx �
1
�
 y�
0
 cos nx dx
n � 1
4 ■ FOUR IER SER I ES
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■ ■ Note that equals 1 if is even 
and if is odd.n�1
ncos n�
We see that, as increases, becomes a better approximation to the square-wave
function. It appears that the graph of is approaching the graph of , except where
or is an integer multiple of . In other words, it looks as if is equal to the sum
of its Fourier series except at the points where is discontinuous.
The following theorem, which we state without proof, says thatthis is typical of the
Fourier series of piecewise continuous functions. Recall that a piecewise continuous func-
tion has only a finite number of jump discontinuities on . At a number where 
has a jump discontinuity, the one-sided limits exist and we use the notation
Fourier Convergence Theorem If is a periodic function with period and and
are piecewise continuous on , then the Fourier series (7) is convergent.
The sum of the Fourier series is equal to at all numbers where is continu-
ous. At the numbers where is discontinuous, the sum of the Fourier series is
the average of the right and left limits, that is
If we apply the Fourier Convergence Theorem to the square-wave function in
Example 1, we get what we guessed from the graphs. Observe that
and
and similarly for the other points at which is discontinuous. The average of these left and
right limits is , so for any integer the Fourier Convergence Theorem says that
(Of course, this equation is obvious for .)x � n�
1
2
� �
�
k�1
 
2
�2k � 1��
 sin�2k � 1�x �
 f �x�1
2
if n � n�
if x � n�
n
1
2
f
f �0�� � lim 
xl 0�
 f �x� � 0f �0�� � lim 
xl 0�
 f �x� � 1
f
1
2 � f �x�� � f �x���
fx
fxf �x�
���, ��f 
f2�f8
f �a�� � lim 
xl a�
 f �x�f �a�� � lim 
xl a�
 f �x�
fa���, ��
f
f�xx � 0
f �x�Sn�x�
Sn�x�n
FIGURE 2 Partial sums of the Fourier series for the square-wave function
xπ_π
1
y
S¡
π x_π
1
y
S£
π x_π
1
y
S∞
π x_π
S¡∞
1
y
π x_π
1
y
S¡¡
π x_π
1
y
S¶
FOUR IER SER I ES ■ 5
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FUNCTIONS WITH PERIOD 2L
If a function has period other than , we can find its Fourier series by making a change
of variable. Suppose has period , that is for all . If we let
and
then, as you can verify, has period and corresponds to . The Fourier
series of is
where
If we now use the Substitution Rule with , then , , and
we have the following
If is a piecewise continuous function on , its Fourier series is
where
and, for ,
Of course, the Fourier Convergence Theorem (8) is also valid for functions with 
period .
EXAMPLE 2 Find the Fourier series of the triangular wave function defined by 
for and for all . (The graph of is shown in Figure 3.)
For which values of is equal to the sum of its Fourier series?
FIGURE 3
The triangular wave function
1 x2_1
1
y
0
f �x�x
fxf �x � 2� � f �x��1 � x � 1
f �x� � � x �
2L
bn �
1
L
 yL
�L
 f �x� sin
n�xL � dxan � 1L y�LL f �x� cos
n�xL � dx
n � 1
a0 �
1
2L
 yL
�L
 f �x� dx
a0 � �
�
n�1
 �an cos
n�xL � � bn sin
n�xL ��
��L, L�f9
dt � ���L� dxt � �x�Lx � Lt��
bn �
1
�
 y�
��
 t�t� sin nt dtan �
1
�
 y�
��
 t�t� cos nt dt
a0 �
1
2�
 y�
��
 t�t� dt
a0 � �
�
n�1
 �an cos nt � bn sin nt�
t
t � ��x � �L2�t
t�t� � f �x� � f �Lt���
t � �x�L
xf �x � 2L� � f �x�2Lf �x�
2�f
6 ■ FOUR IER SER I ES
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■ ■ Notice that when these
formulas are the same as those in (7).
L � �
FOUR IER SER I ES ■ 7
SOLUTION We find the Fourier coefficients by putting in (9):
and for ,
because is an even function. Here we integrate by parts with 
and . Thus,
Since is an odd function, we see that
We could therefore write the series as
But if is even and if is odd, so
Therefore, the Fourier series is
The triangular wave function is continuous everywhere and so, according to the Fourier
Convergence Theorem, we have
for all xf �x� � 1
2
� �
�
n�1
 
4
�2k � 1�2� 2
cos��2k � 1��x�
�
1
2
� �
�
n�1
 
4
�2k � 1�2� 2
cos��2k � 1��x�
1
2
�
4
� 2
cos��x� �
4
9� 2
cos�3�x� �
4
25� 2 cos�5�x� � � � �
an �
2
n2� 2
�cos n� � 1� � 
0� 4
n2� 2
if n is even
if n is odd
ncos n� � �1ncos n� � 1
1
2
� �
�
n�1
 
2�cos n� � 1�
n2� 2
cos�n�x�
bn � y1
�1
 � x � sin�n�x� dx � 0
y � � x � sin�n�x�
 � 0 �
2
n�
 �� cos�n�x�
n�
�1
0
�
2
n2� 2
�cos n� � 1�
an � 2� x
n�
 sin�n�x��
0
1
�
2
n�
 y1
0
 sin�n�x� dx
dv � cos�n�x� dx
u � xy � � x � cos�n�x�
an � y1
�1
 � x � cos�n�x� dx � 2 y10 x cos�n�x� dx
n � 1
 � �14 x
2]�10 � 14 x 2]10 � 12
a0 �
1
2 y1
�1
 � x � dx � 12 y0
�1
 ��x� dx � 12 y10 x dx
L � 1
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■ ■ Notice that is more easily calculated as
an area.
a0
8 ■ FOUR IER SER I ES
In particular,
for 
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.
We gain insight into the differences between waveforms if we express them as sums of
Fourier series:
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 th term of the Fourier series, that is,
is called the nth harmonic of . The amplitude of the th harmonic is
and its square, , is sometimes called energy of the th harmonic. (Notice thatnA2n � a2n � b2n
An � sa2n � b2n
nP
an cos
n� tL � � bn
n� tL �
n
P�t� � a0 � a1 cos
� tL � � b1 sin
� tL � � a2 cos
2� tL � � b2 sin
2� tL � � � � �
FIGURE 4
Waveforms (b) Violin(a) Flute
tt
�1 � x � 1� x � � 12 � �
�
k�1
 
4
�2k � 1�2� 2
cos��2k � 1��x�
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FOUR IER SER I ES ■ 9
for a Fourier series with only sine terms, as in Example 1, the amplitude is and
the energy is .) The graph of the sequence is called the energy spectrum of
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, tends to diminish rapidly as increases whereas, for the violin, the
higher harmonics are fairly strong. This accounts for therelative 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.
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).
FIGURE 5
Energy spectra
n2 4 6 8 10
(b) Violin
0
A@n
0 n2 4 6 8 10
(a) Flute
A@n
nAn2
P
�A2n�A2n � b2n
An � � bn �
EXERCISES
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7–11 Find the Fourier series of the function.
7.
8.
9.
10. ,
11. ,
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
12. A voltage , where represents time, is passed through a
so-called half-wave rectifier that clips the negative part of the
wave. Find the Fourier series of the resulting periodic function
f �t � 2���� � f �t�f �t� � 
0E sin �t if �
�
�
� t 	 0
if 0 � t 	 �
�
tE sin �t
�1 � t � 1f �t� � sin�3�t�
f �x � 2� � f �x��1 � x � 1f �x� � 1 � x
f �x � 8� � f �x�f �x� � 
�x0 if �4 � x 	 0if 0 � x 	 4
f �x � 4� � f �x�f �x� � 
01
0
 if �2 � x 	 0
 if 0 � x 	 1
 if 1 � x 	 2
f �x � 4� � f �x�f �x� � 
10 if � x � 	 1if 1 � � x � 	 21–6 A function is given on the interval and is 
periodic with period .
(a) Find the Fourier coefficients of .
(b) Find the Fourier series of . For what values of is 
equal to its Fourier series?
; (c) Graph and the partial sums , , and of the Fourier
series.
1.
2.
3.
4.
5.
6.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
f �x� � 
�11
0
 if �� � x 	 ���2
 if ���2 � x 	 0
 if 0 � x 	 �
f �x� � 
0
cos x
if �� � x 	 0
if 0 � x 	 �
f �x� � x 2
f �x� � x
f �x� � 
0
x
if �� � x 	 0
if 0 � x 	 �
f �x� � 
1
�1
if �� � x 	 0
if 0 � x 	 �
S6S4S2f
f �x�xf
f
2�
f���, ��f
Click here for solutions.S
18. Use the result of Example 2 to show that
19. Use the result of Example 1 to show that
20. Use the given graph of and Simpson’s Rule with to
estimate the Fourier coefficients . Then use
them to graph the second partial sum of the Fourier series and
compare with the graph of .
x
y
1
0.25
f
a0, a1, a2, b1, and b2
n � 8f
1 �
1
3
�
1
5 �
1
7
� � � � �
�
4
1 �
1
32
�
1
52 �
1
72
� � � � �
� 2
8
13–16 Sketch the graph of the sum of the Fourier series of 
without actually calculating the Fourier series.
13.
14.
15. ,
16. ,
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
17. (a) Show that, if , then
(b) By substituting a specific value of , show that
�
�
n�1
 
1
n2
�
� 2
6
x
x 2 �
1
3
� �
�
n�1
��1�n 
4
n2� 2
cos�n�x�
�1 � x � 1
�2 � x � 2f �x� � e x
�1 � x � 1f �x� � x 3
f �x� � 	x1 � x if �1 � x � 0if 0 � x � 1
f �x� � 	�13 if �4 � x � 0if 0 � x � 4
f
10 ■ FOUR IER SER I ES
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FOUR IER SER I ES ■ 11
SOLUTIONS
1. (a) a0 =
1
2π
∫
π
−π
f(x) dx =
1
2π
(∫ 0
−π
dx−
∫
π
0
dx
)
= 0.
an =
1
π
∫
π
−π
∫
π
−π
f(x) cosnxdx =
1
π
∫ 0
−π
cosnxdx−
1
π
∫
π
0
cosnxdx = 0 [since cosnx is even].
bn =
1
π
∫
π
−π
∫
π
−π
f(x) sinnxdx =
1
π
∫ 0
−π
sinnxdx−
1
π
∫
π
0
sinnxdx =
2
π
∫ 0
−π
sinnxdx [since sinnx is odd]
= −
2
nπ
[1− cos(−nπ)] =


0 if n even
−
4
nπ
if n odd
(b) f(x) =
∞∑
k=0
−
4
(2k + 1)π
sin(2k + 1)x when−π < x < 0 and 0 < x < π.
(c)
2.51.250-1.25-2.5
1
0.5
0
-0.5
-1
x
y
x
y
3. (a) a0 =
1
2π
∫
π
−π
f(x) dx =
1
2π
∫
π
−π
xdx = 0.
an =
1
π
∫
π
−π
f (x) cosnxdx =
1
π
∫
π
−π
x cosnxdx = 0 [because x cosnx is odd]
bn =
1
π
∫
π
−π
f (x) sinnxdx =
1
π
∫
π
−π
x sinnxdx =
2
π
∫
π
0
x sinnxdx [since x sinnx is odd]
= −
2
n
cosnπ [using integration by parts] =
{
−(2/n) if n even
(2/n) if n odd
(b) f(x) =
∞∑
n=1
(−1)n+1
2
n
sinnx
when−π < x < π.
(c)
2.51.250-1.25-2.5
2.5
1.25
0
-1.25
-2.5
x
y
x
y
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12 ■ FOUR IER SER I ES
5. (a) a0 =
1
2π
∫
π
−π
f(x) dx =
1
2π
∫
π
0
cosxdx = 0
an =
1
π
∫
π
−π
f(x) cosnxdx =
1
π
∫
π
0
cosx cosnxdx =
{
1
2
if n = 1
0 if n �= 1
[by symmetry about x = π
2
]
bn =
1
π
∫
π
−π
f(x) sinnxdx =
1
π
∫
π
0
cosx sinnxdx
=


2n
π(n2 − 1)
if n even
0 if n odd
[
using an integral table,
and simplified using the addition formula for cos(a+ b)
]
(b) f (x) = 1
2
cosx+
∞∑
k=1
4k
π (4k2 − 1)
sin(2k) when−π < x < 0, 0 < x < π.
(c)
2.51.250-1.25-2.5
1
0.5
0
-0.5
-1
x
y
x
y
7. Use f (x) =


0 if − 2 ≤ x ≤ −1
1 if − 1 < x < 1,
0 if 1 ≤ x ≤ 2
L = 2.
a0 =
1
2L
∫
L
−L
f(x) dx = 1
4
∫ 1
−1
dx = 1
2
an =
1
L
∫
L
−L
f(x) cos
(nπx
L
)
dx = 1
2
∫ 1
−1
cos
(nπx
2
)
dx =
2
nπ
sin
(π
2
n
)
=


0 if n even
2/nπ if n = 4n+ 1
−2/nπ if n = 4n+ 3
bn =
1
L
∫
L
−L
f (x) sin
(nπx
L
)
dx = 1
2
∫ 1
−1
sin
(nπx
2
)
dx = 0
Fourier Series: 1
2
+
2
π
cos
(πx
2
)
−
2
3π
cos
(
3πx
2
)
+
2
5π
cos
(
5πx
2
)
− · · ·
1
2
+
∞∑
k=1
2
(4k + 1)π
sin
(π
2
(4k + 1)
)
−
2
(4k + 3)π
sin
(π
2
(4k + 3)
)
52.50-2.5-5
1
0.75
0.5
0.25
0
x
y
x
y
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FOUR IER SER I ES ■ 13
9. Use f(x) =
{
−x if − 4 ≤ x < 0
0 if 0 ≤ x ≤ 4
, L = 4.
a0 =
1
2L
∫
L
−L
f(x) dx = 1
8
∫ 0
−4
−xdx = 1
an =
1
L
∫
L
−L
f(x) cos
(nπx
L
)
dx = 1
4
∫ 0
−4
−x cos
(nπx
4
)
dx =
4
(nπ)2
(cos (nπ)− 1) =
{
0 if n is even
−8/(nπ)2 if n is odd
bn =
1
L
∫
L
−L
f(x) sin
(nπx
L
)
dx = 1
4
∫ 0
−4
−x sin
(nπx
4
)
dx =
4
nπ
cos (nπ) =
{
4/nπ if n is even
−4/nπ if n is odd
Fourier Series:
1 +
∞∑
k=1
−
4
(2k − 1)π
sin
(π
4
(2k − 1)x
)
−
8
(2k − 1)2π2
cos
(π
4
(2k − 1)x
)
+
4
(2k)π
sin
(π
4
(2k)x
)
52.50-2.5-5
4
3
2
1
0
x
y
x
y
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 2
3
, and the period of f(x) = 2 which
is an integer multiple of 2
3
. 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
2L
∫
L
−L
f(x) dx = 1
2
∫ 1
−1
sin(3πx) dx = 0
an =
1
L
∫ 1
−1
f(x) cos
(nπx
L
)
dx =
∫ 1
−1
sin(3πx) cos(nπx) dx
= 0 [applying change of variables to a formula in the section]
bn =
1
L
∫
L
−L
f(x) sin
(nπx
L
)
dx =
∫ 1
−1
sin(3πx) sin(nπx) dx
=


6
sinnπ
π (−9 + n2)
if n �= 3
1 if n = 3
[using integral table and addition formula =
{
0 if n �= 3
1 if n = 3
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14 ■ FOUR IER SER I ES
Fourier Series: sin(3πx)
52.50-2.5-5
1
0.5
0
-0.5
-1
x
y
x
y
13.


3 if −5 ≤ x < −4
−1 if −4 ≤ x < 0
3 if 0 ≤ x < 4
−1 if 4 ≤ x < 5
52.50-2.5-5
3
2
1
0
-1
x
y
x
y
15.
3.752.51.250-1.25
1
0.5
0
-0.5
-1
x
y
x
y
17. (a) We find the Fourier series for f(x) = {x2 if −1 ≤ x ≤ 1, L = 1
a0 =
1
2L
∫
L
−L
f(x) dx = 1
2
∫ 1
−1
x2 dx = 1
3
an =
1
L
∫
1
−1
f(x) cos
(nπx
L
)
dx =
∫
1
−1
x2 cos(nπx) dx =
4
(nπ)2
cosnπ =


4
(nπ)2
if n even
−
4
(nπ)2
if n odd
bn =
1
L
∫
L
−L
f (x) sin
(nπx
L
)
dx =
∫ 1
−1
x2 sin(nπx) dx = 0 because x2 sin(nπx) is odd.
So we have x2 = 1
3
+
∞∑
n=1
(−1)n
4
(nπ)2
cos(nπx) for −1 ≤ x ≤ 1.
(b) We let x = 1 in the above to obtain
1 = 1
3
+
∞∑
n=1
(−1)n
4
(nπ)2
cos(nπ) 2
3
=
∞∑
n=1
4
n2π2
π2
6
=
∞∑
n=1
1
n2
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FOUR IER SER I ES ■ 15
19. Example 1 says that, for 0 ≤ x < π, 1 = 1
2
+
∑
∞
k=1
2
(2k − 1)π
sin((2k − 1)x).
Let x = π
2
to obtain
1 = 1
2
+
∞∑
k=1
2
(2k − 1)π
sin
(π
2
(2k − 1)
)
π
4
=
∞∑
k=1
1
(2k − 1)
sin((2k − 1))
π
4
= 1− 1
3
+ 1
5
− 1
7
+ · · ·
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