f x a 0 n 1 a 0 a 1 cos x a 2 cos 2x a 3 cos 3x b 1 sin x b 2 sin 2x b 3 sin 3x a n cos nx b n sin nx n 1 f x dx y

||||

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

 fx dx  2a⁰

x

 sin nx dx 0 n

y

 cos nx dx 1

n sin nx



 2a⁰_n1



y

 cos nx dx_n1



y

sin nx dx

y

 fx dx 

y

 a0 dx

y

_n



cn ff^nx an!

  x   fx  a⁰_n1



2

, 

fx

 b¹ sin x b² sin 2x b³ sin 3x   

 a⁰ a¹ cos x a² cos 2x a³ cos 3x    fx  a⁰_n1



1

f

1

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

 fx sin nx dx

6





sin mx

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



y

 fx cos nx dx

5

n m am

y

 fx cos mx dx  a^m am

y

 cos nx cos mx dx

m

y

 sin nx cos mx dx 0

 a⁰

y

 cos mx dx_n



y

 cos nx cos mx dx _n



y

sin nx cos mx dx

4

y

 fx cos mx dx 

y



^







ma 1n n 1 cos mx m

a0 1 2

y

 fx dx

3

a0

|||| 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

 fx dx  1 2

y

 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

x 0

π 2π

_π 1 y

x

(b)

2

f

fx  2  f x

fx 

f

, 

f fx

f

bn 1



y

 fx sin nx dx an 1



y

 fx cos nx dx a0 1

2

y

 fx dx bn

an

a0_n1



f

, 

f

7

|||| 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 Snx  1

2  2

 sin x 2

3 sin 3x     2 n sin nx 1

2 _k1



k n 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



 1

 cos nx

n



bn 1



y

 fx sin nx dx  1



y

 0 dx 1



y

0 sin x dx  0  1

 sin nx

n





y

 fx cos nx dx  1



y

 0 dx 1



y

0 cos nx dx n 1

|||| Note that equals 1 if is even and if 1 nis odd.

n cos 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 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 n

2

f

f0^  lim

x l 0^ fx  0 f0^  lim

x l 0^ fx  1

f

1

2  f x^  f x^

f x

f x

fx

, 

f8 f 2 f

fa^  lim

x l a^ fx

fa^  lim

x l a^ fx

f

,  a f

 f x

x 0 n SnxSnx fx

FIGURE 2 Partial sums of the Fourier series for the square-wave function π x

_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¶

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 2 x

_1 1 y

0

fx

x

f x

fx  2  f x

1  x  1 fx 

2L

bn 1 L

y

L fx sin



an 1 L

y

L

L fx cos



n 1

a0 1 2L

y

L fx dx

a0_n







f

9

dt L dx txL

x Lt

bn 1



y

 tt sin nt dt an 1



y

 tt cos nt dt a0 1

2

y

 tt dt a0_n1



t

t  x L

2

t

tt  f x  f Lt

txL f fx 22L fx  2L  f x x

|||| 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 fx  1

2 _n1



 1

2 _n1



1 2  4

²cosx  4

9²cos3x  4

25²cos5x     an 2

n²²cos n  1 

cos n  1 n

1

2 _n1



n²² cosnx

bn

y

1

 0  2

n



an 2



^y

d^v cosnx dxy

an

y

1

y

0 x cosnx dx n 1

 ¹4x²

]

0 ¹4x²

]

a0¹2

y

1

^y

y

0 x dx L 1

|||| 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 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, A²n a²ⁿ b²ⁿ, is sometimes called energy of the th harmonic. (Notice thatn Ans_a²_{n b}²_n

n P

an cos





n

Pt  a⁰ a¹ cos









FIGURE 4

Waveforms (a) Flute (b) Violin

t t

1  x  1

^

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

2 4 6 8 10 n

(b) Violin 0

A@n

0 2 4 6 8 10 n

(a) Flute A@n

n An2

P

A²ⁿ

A²n b²ⁿ An

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

ft  2  f t

ft 

t E sin t

■

■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

1  t  1 ft  sin3t

fx  2  f x

1  x  1 fx  1  x

fx  8  f x

fx 

fx  4  f x

fx 

fx 

1–6 |||| 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.

■

■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

fx 

fx  x² fx  x

fx 

fx ^f

fx

x f

f

2 ,  f

f

|||| ^Exercises

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 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



n1

1 n²  ²

6 x x² 1

3 _n



1  x  1

■

■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■ ■■

2  x  2 fx  e^x

1  x  1 fx  x³

fx 

fx 

f