fourier series (by james stewart)

8/22/2019 Fourier Series (by James Stewart)

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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

Ϫ

f x dx 2 a0

x Ϫ

sin nx dx 0n

y

Ϫ

cos nx dx 1

nsin nx

Ϫ

1

n sin n Ϫ sin Ϫn 0

2 a0 ϩ

ϱ

n1

an y

Ϫ

cos nx dx ϩϱ

n1

bn y

Ϫ

sin nx dx

y

Ϫ

f x dx y

Ϫ

a0 dx ϩ y

Ϫ

ϱ

n1

an cos nx ϩ bn sin nx dx

cn f n a n!

f x cn x Ϫ a n

f bnan

Ϫ ഛ x ഛ f x a0 ϩ

ϱ

n1

an cos nx ϩ bn sin nx 2

Ϫ , f x

ϩ b1 sin x ϩ b2 sin 2 x ϩ b3 sin 3 x ϩ и и и

a0 ϩ a1 cos x ϩ a2 cos 2 x ϩ a3 cos 3 x ϩ и и и

f x a0 ϩ

ϱ

n1

an cos nx ϩ bn sin nx 1

f

1

S t e w a r t : C a l c u l u s ,

S i x t h E d i t i o n .

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© 2

0 0 8 B r o o k s / C o l e

. A l l r i g h t s r e s e r v e d

.



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 thatEquation 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

Ϫ

f x sin nx dx 6

Ϫ sin mx

n 1, 2, 3, . . .an 1

y

Ϫ

f x cos nx dx 5

nmam

y

Ϫ

f x cos mx dx am

am

y

Ϫ

cos nx cos mx dx 0

for n m

for n m

mny

Ϫ sin nx cos mx dx

0

a0y

Ϫ

cos mx dx ϩϱ

n1

an y

Ϫ

cos nx cos mx dx ϩϱ

n1

bn y

Ϫ

sin nx cos mx dx 4

y

Ϫ

f x cos mx dx y

Ϫ

a0 ϩ

ϱ

n1

an cos nx ϩ bn sin nx cos mx dx

Ϫ m ജ 1

mcos mx n ജ 1an

a0 1

2 y

Ϫ

f x dx 3

a0

2 ■ FOURIER SERIES



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8 B r o o k s / C o l e


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■ ■ Notice that is the average value

of over the interval . Ϫ , 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 calleda 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

Ϫ

f x dx 1

2 y0

Ϫ

0 dx ϩ1

2 y

01 dx 0 ϩ

1

2

1

2

0

F IGURE 1

Square-wave function

(a)

π 2π_π

1

y

x 0

π 2π_π

1

y

x

(b)

2 f

f x ϩ 2 f x f x 0

1

if Ϫ ഛ x Ͻ 0

if 0 ഛ x Ͻ

f

Ϫ

,

f

f x

f

bn 1

y

Ϫ

f x sin nx dx an 1

y

Ϫ

f x cos nx dx

a0 1

2 y

Ϫ

f x dx

bnan

a0 ϩ

ϱ

n1

an cos nx ϩ bn sin nx

f

Ϫ , f 7

FOURIER SERIES ■ 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

S n x 1

2ϩ

2

sin x ϩ2

3 sin 3 x ϩ и и и ϩ

2

n sin nx

1

2ϩ

ϱ

k 1

2

2k Ϫ 1

sin 2k Ϫ 1 x

k n 2k Ϫ 1

1

2 ϩ

2

sin x ϩ

2

3 sin 3 x ϩ

2

5 sin 5 x ϩ

2

7 sin 7 x ϩ и и и

ϩ2

sin x ϩ 0 sin 2 x ϩ2

3 sin 3 x ϩ 0 sin 4 x ϩ

2

5 sin 5 x ϩ и и и

1

2ϩ 0 ϩ 0 ϩ 0 ϩ и и и

ϩ b1 sin x ϩ b2 sin 2 x ϩ b3 sin 3 x ϩ и и и

a0 ϩ a1 cos x ϩ a2 cos 2 x ϩ a3 cos 3 x ϩ и и и

f

0

2

n

if n is even

if n is odd

Ϫ1

cos nx

n

0

Ϫ 1

n cos n Ϫ cos 0

bn 1

y

Ϫ

f x sin nx dx 1

y0

Ϫ

0 dx ϩ1

y

0sin x dx

0 ϩ1

sin nx

n

0

1

n sin n Ϫ sin 0 0

an 1

y

Ϫ

f x cos nx dx 1

y0

Ϫ

0 dx ϩ1

y

0cos nx dx

n ജ 1




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2 0 0



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■ ■ Note that equals 1 if is even

and if is odd.nϪ1

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 that this 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 andare 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

2ϩ

ϱ

k 1

2

2k Ϫ 1

sin 2k Ϫ 1 x f x 12

if n n

if x n

n1

2

f

f 0Ϫ lim x l0Ϫ

f x 0 f 0ϩ lim x l 0ϩ

f x 1

f

12 f x ϩ ϩ f x Ϫ

f x

f x f x Ϫ , f Ј

f 2

f 8

f aϪ lim x laϪ

f x f aϩ lim x laϩ

f x

f a Ϫ ,

f

f x x 0

f x S n x S n x n

F IGURE 2 Partial sums of the Fourier series for the square-wave function

x π_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¶




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2 0 0 8 B r o o k s / C o l e


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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?

F IGURE 3The triangular wave function

1 x 2_1

1

y

0

f x x

f x f x ϩ 2 f x Ϫ1 ഛ x ഛ 1

f x Խ x Խ

2 L

bn 1

L y L

Ϫ L

f x sinn x

L dx an

1

L yϪ L

L f x cos

n x

L dx

n ജ 1

a0 1

2 L y L

Ϫ L f x dx

a0 ϩ

ϱ

n1

an cosn x

L ϩ bn sin

n x

L

Ϫ L, L f 9

dt L dx t x L x Lt

bn 1

y

Ϫ

t t sin nt dt an 1

y

Ϫ

t t cos nt dt

a0 1

2 y

Ϫ

t t dt

a0 ϩ

ϱ

n1

an cos nt ϩ bn sin nt

tt Ϯ x Ϯ L2 t

t t f x f Lt

t x L

x f x ϩ 2 L f x 2 L f x 2 f




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. ©

2 0 0



.

■ ■ Notice that when these

formulas are the same as those in (7).

L




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 x f x 1

2Ϫ

ϱ

n1

4

2k Ϫ 1 2

2cos 2k Ϫ 1 x

1

2Ϫ

ϱ

n1

4

2k Ϫ 1 2

2cos 2k Ϫ 1 x

1

2Ϫ

4

2

cos x Ϫ4

9 2cos 3 x Ϫ

4

25 2cos 5 x Ϫ и и и

an 2

n2

2cos n Ϫ 1 0

Ϫ4

n2

2

if n is even

if n is odd

ncos n Ϫ1ncos n 1

1

2ϩ

ϱ

n1

2 cos n Ϫ 1

n2

2cos n x

bn y1

Ϫ1 Խ x Խ sin n x dx 0

y Խ x Խ sin n x

0 Ϫ2

n Ϫ

cos n x

n 1

0

2

n2

2cos n Ϫ 1

an 2x

n sin n x

0

1

Ϫ2

n y1

0sin n x dx

d v cos n x dx

u x y Խ x Խ cos n x

an y1

Ϫ1 Խ x Խ cos n x dx 2 y1

0 x cos n x dx

nജ

1

Ϫ14 x 2]Ϫ1

0ϩ

14 x 2]1

0 12

a0 1

2 y1

Ϫ1 Խ x Խ dx

1

2 y0

Ϫ1 Ϫ x dx ϩ

1

2 y1

0 x dx

L 1



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2 0 0 8 B r o o k s / C o l e


.

■ ■ Notice that is more easily calculated as

an area.

a0




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 airpressure 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 thatn A2n a2

n ϩ b2n

An s a2n ϩ b2

n

nP

an

cos

n t

Lϩ b

n

n t

L

n

P t a0 ϩ a1 cos t

Lϩ b1 sin

t

Lϩ a2 cos

2 t

L ϩ b2 sin

2 t

L ϩ и и и

F IGURE 4Waveforms (b) Violin(a) Flute

t t

Ϫ1 ഛ x ഛ 1Խ x Խ 1

2Ϫ

ϱ

k 1

4

2k Ϫ 1 2

2cos 2k Ϫ 1 x



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2 0 0



.




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 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.

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

n An2

P

A2n A2

n b2n

An Խ bn Խ

EXERCISES



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2 0 0 8 B r o o k s / C o l e


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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 ϩ 2 f t f t 0

E sin t

if Ϫ

ഛ t Ͻ 0

if 0 ഛ t Ͻ

t E sin t

Ϫ1 ഛ t ഛ 1 f t sin 3 t

f x ϩ 2 f x Ϫ1 ഛ x ഛ 1 f x 1 Ϫ x

f x ϩ 8 f x f x Ϫ x

0

if Ϫ4 ഛ x Ͻ 0

if 0 ഛ x Ͻ 4

f x ϩ 4 f x f x 0

1

0

if Ϫ2 ഛ x Ͻ 0

if 0 ഛ x Ͻ 1

if 1 ഛ x Ͻ 2

f x ϩ 4

f x f x 1

0

if Խ x

ԽϽ 1

if 1 ഛ Խ x Խ Ͻ 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 Ϫ1

1

0

if Ϫ ഛ x Ͻ Ϫ 2

if Ϫ 2 ഛ x Ͻ 0

if 0 ഛ x Ͻ

f x 0

cos x

if Ϫ ഛ x Ͻ 0

if 0 ഛ x Ͻ

f x x 2

f x x

f x 0

x

if Ϫ ഛ x Ͻ 0

if 0 ഛ x Ͻ

f x 1

Ϫ1

if Ϫ ഛ x Ͻ 0

if 0 ഛ x Ͻ

S 6S 4S 2 f

f x x f

f

2

f Ϫ , 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 8 f

1 Ϫ1

3ϩ

1

5Ϫ

1

7ϩ и и и

4

1 ϩ1

32ϩ

1

52ϩ

1

72ϩ и и и

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

ϱ

n1

1n2

2

6

x

x 2 1

3ϩ

ϱ

n1

Ϫ1 n 4

n2

2cos n x

Ϫ1 ഛ x ഛ 1

Ϫ2 ഛ x ഛ 2 f x e x

Ϫ1 ഛ x ഛ 1 f x x 3

f x x

1 Ϫ x

if Ϫ1 ഛ x Ͻ 0

if 0 ഛ x Ͻ 1

f x Ϫ1

3

if Ϫ4 ഛ x Ͻ 0

if 0 ഛ x Ͻ 4

f


T h o m s o n B r o o k s - C o l e c o p y r i g h t 2 0 0 8




SOLUTIONS

1. (a) a0 =1

2π

π

−πf (x) dx =

1

2π

0−π

dx − π

0dx

= 0.

an = 1π π−π

π−π f (x)cos nxdx = 1π

0−π cos nxdx − 1π

π0 cos nxdx = 0 [since cos nx is even].

bn =1

π

π

−π

π

−πf (x) sin nxdx =

1

π

0

−πsin nxdx −

1

π

π

0sin nxdx =

2

π

0

−πsin nxdx [since sin nx is odd]

= −2

nπ[1 − cos(−nπ)] =

0 if n even

−4

nπif n odd

(b) f (x) =∞k=0

−4

(2k + 1)πsin(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

−πf (x) dx =

1

2π

π

−πx dx = 0.

an =1

π π−π f (x)cos nxdx =

1

π π−π x cos nxdx = 0 [because x cos nx is odd]

bn =1

π

π

−πf (x)sin nxdx =

1

π

π

−πx sin nxdx =

2

π

π

0x sin nxdx [since x sin nx is odd]

= −2

ncos nπ [using integration by parts] =

−(2/n) if n even

(2/n) if n odd

(b) f (x) =∞n=1

(−1)n+12

nsin nx

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





5. (a) a0 =1

2π

π

−πf (x) dx =

1

2π

π

0cos x dx = 0

an =1

π

π

−πf (x) cos nxdx =

1

π

π

0cos x cos nxdx =

1

2if n = 1

0 if n = 1[by symmetry about x = π

2]

bn =1

π π

−πf (x) sin nxdx =

1

π π

0cos x sin nxdx

=

2nπ(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

2cos x +

∞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

−Lf (x) cos

nπx

L

dx = 1

2

1−1

cosnπx

2

dx =

2

nπsinπ

2n

=

0 if n even

2/nπ if n = 4n + 1

−2/nπ if n = 4n + 3

bn =1

L

L

−Lf (x)sin

nπx

L

dx = 1

2

1

−1sinnπx

2

dx = 0

Fourier Series: 1

2+

2

πcosπx

2

−

2

3πcos

3πx

2

+

2

5πcos

5πx

2

− · · ·

1

2+∞

k=12

(4k + 1) πsin

π

2(4k + 1)

−

2

(4k + 3) πsin

π

2(4k + 3)

52.50-2.5-5

1

0.75

0.5

0.25

0

x

y

x

y





9. Use f (x) =

−x if − 4 ≤ x < 0

0 if 0 ≤ x ≤ 4, L = 4.

a0 =1

2L

L

−Lf (x) dx = 1

8

0

−4−x dx = 1

an = 1L L−L f (x) cos

nπxL

dx = 1

4

0−4 −x cos

nπx4

dx = 4

(nπ)2(cos(nπ) − 1) =

0 if n is even

−8/(nπ)2 if n is odd

bn =1

L

L

−Lf (x) sin

nπx

L

dx = 1

4

0

−4−x sin

nπx

4

dx =

4

nπcos(nπ) =

4/nπ if n is even

−4/nπ if n is odd

Fourier Series:

1 +∞k=1

−4

(2k − 1)πsinπ

4(2k − 1)x

−

8

(2k − 1)2π2cosπ

4(2k − 1) x

+

4

(2k)πsinπ

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 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

2L L

−Lf (x) dx = 1

2 1

−1sin(3πx) dx = 0

an =1

L

1

−1f (x)cos

nπx

L

dx =

1

−1sin(3πx) cos(nπx) dx

= 0 [applying change of variables to a formula in the section]

bn =1

L

L

−Lf (x) sin

nπx

L

dx =

1

−1sin(3πx) sin(nπx) dx

=

6sin nπ

π (−9 + n2)if n = 3

1 if n = 3

[using integral table and addition formula =

0 if n = 3

1 if n = 3





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

−Lf (x) dx = 1

2

1−1

x2 dx = 1

3

an =1

L 1

−1f (x) cos

nπx

L

dx =

1

−1x2 cos(nπx) dx =

4

(nπ)2cos nπ =

4

(nπ)2if n even

−4

(nπ)2

if n odd

bn =1

L

L

−Lf (x)sin

nπx

L

dx =

1

−1x2 sin(nπx) dx = 0 because x2 sin(nπx) is odd.

So we have x2 = 1

3+∞n=1

(−1)n4

(nπ)2cos(nπx) for −1 ≤ x ≤ 1.

(b) We let x = 1 in the above to obtain

1 = 1

3+∞n=1

(−1)n4

(nπ)2cos(nπ) 2

3=∞n=1

4

n2π2π2

6=∞n=1

1

n2





19. Example 1 says that, for 0 ≤ x < π , 1 = 1

2+∞

k=1

2

(2k − 1)πsin((2k − 1)x).

Let x = π

2to obtain

1 = 1

2+∞

k=12

(2k − 1)πsinπ

2(2k − 1)

π4

=∞k=1

1(2k − 1)

sin((2k − 1))

π

4= 1 − 1

3+ 1

5− 1

7+ · · ·


fourier series (by james stewart)

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