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 f as an inﬁnite series of sine and cosine functions:

1 f ͑x͒ ෇ a0 ͚ ͑an cos nx bn sin nx͒ n෇1

෇ a0 a1 cos x a2 cos 2x a3 cos 3x

b1 sin x b2 sin 2x b3 sin 3x

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 function f ͑x͒ as its sum on the interval ͓, ͔ , that is,

2 f ͑x͒ ෇ a0 ͚ ͑an cos nx bn sin nx͒ x n෇1

Our aim is to find formulas for the coefficients an and bn in terms of f . Recall that for a ͸ n power series f ͑x͒ ෇ cn͑x a͒ we found a formula for the coefficients in terms of deriv- ͑n͒ atives: cn ෇ f ͑a͒͞n!. 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

y f ͑x͒ dx ෇ y a0 dx y ͚ ͑an cos nx bn sin nx͒ dx n෇1

෇ 2a0 ͚ an y cos nx dx ͚ bn y sin nx dx n෇1 n෇1 But 1 1 y cos nx dx ෇ sin nxͬ ෇ ͓sin n sin͑n͔͒ ෇ 0 n n

x ෇ because n is an integer. Similarly, sin nx dx 0 . So

Calculus, Stewart:

1 2 ■ FOURIER SERIES

and solving for a0 gives

■■ Notice that a0 is the average value 1 ͓ ͔ 3 a0 ෇ y f ͑x͒ dx of f over the interval , . 2

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 f x cos mx dx y a0 an cos nx bn sin nx cos mx dx n෇1

4 ෇ a0 y cos mx dx ͚ an y cos nx cos mx dx ͚ bn y sin nx cos mx dx n෇1 n෇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 ෇ 0 for all n and m

0 for n m y cos nx cos mx dx ෇ ͭ for n ෇ m

So the only nonzero term in (4) is am and we get

y f ͑x͒ cos mx dx ෇ am

Solving for am , and then replacing m by n , we have

1 5 an ෇ y f ͑x͒ cos nx dx n ෇ 1, 2, 3, . . .

Similarly, if we multiply both sides of Equation 2 by sin mx and integrate from to , we get

1 6 bn ෇ y f ͑x͒ sin nx dx n ෇ 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 ͓a, b͔ 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.) Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved.

Calculus, Stewart: FOURIER SERIES ■ 3

7 Definition Let f be a piecewise continuous function on ͓, ͔ . Then the Fourier series of f is the series

a0 ͚ ͑an cos nx bn sin nx͒ n෇1

where the coefﬁcients an and bn in this series are deﬁned by

1 a0 ෇ y f ͑x͒ dx 2

1 1 an ෇ y f ͑x͒ cos nx dx bn ෇ y f ͑x͒ sin nx dx

and are called the Fourier coefﬁcients of f .

Notice in Deﬁnition 7 that we are not saying f ͑x͒ 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 ͓, ͔ 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

0 if x 0 f ͑x͒ ෇ ͭ and f ͑x 2͒ ෇ f ͑x͒ 1 if 0 x

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 y and electromotive forces in an electric circuit 1 (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 _π 0 π2π x Figure 1(b), where you can see why it’s called a square wave. (a)

_π 0 π2π x FIGURE 1 Square-wave function (b)

SOLUTION Using the formulas for the Fourier coefﬁcients in Deﬁnition 7, we have

Calculus, Stewart: 4 ■ FOURIER SERIES

and, for n 1,

1 1 0 1 an ෇ y f ͑x͒ cos nx dx ෇ y 0 dx y cos nx dx 0

෇ 1 sin nxͬ ෇ 1 ͑ ͒ ෇ 0 sin n sin 0 0 n 0 n

1 1 0 1 bn ෇ y f ͑x͒ sin nx dx ෇ y 0 dx y sin x dx 0

෇ 1 cos nxͬ ෇ 1 ͑ ͒ cos n cos 0 n 0 n

0 if n is even ■■ Note that cos n equals 1 if n is even ෇ and 1 if n is odd. ͭ 2 if n is odd n

The Fourier series of f is therefore

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

b1 sin x b2 sin 2x b3 sin 3x

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

Since odd integers can be written as n ෇ 2k 1 , where k is an integer, we can write the Fourier series in sigma notation as

1 2 ͚ sin͑2k 1͒x 2 k෇1 ͑2k 1͒

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

1 2 2 2 S ͑x͒ ෇ sin x sin 3x sin nx n 2 3 n

Calculus, Stewart: FOURIER SERIES ■ 5

y y y

1 1 1 S£ S∞ S¡

_π π x _π π x _π π x

y y y

1 1 1 S¶ S¡¡ S¡∞

_π π x _π π x _π π x

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

We see that, as n increases,Sn͑x͒ becomes a better approximation to the square-wave function. It appears that the graph of Sn͑x͒ is approaching the graph of f ͑x͒ , except where x ෇ 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 ͓, ͔ . At a number a where f has a jump discontinuity, the one-sided limits exist and we use the notation

f ͑a͒ ෇ lim f ͑x͒ f ͑a͒ ෇ lim f ͑x͒ x l a x l a

8 Fourier Convergence Theorem If f is a periodic function with period 2 and f and f are piecewise continuous on ͓, ͔ , then the Fourier series (7) is convergent. The sum of the Fourier series is equal to f ͑x͒ 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 f x f x

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 ͑0͒ ෇ lim f ͑x͒ ෇ 1 and f ͑0͒ ෇ lim f ͑x͒ ෇ 0 x l 0 x l 0 and similarly for the other points at which f is discontinuous. The average of these left and 1 right limits is 2 , so for any integer n the Fourier Convergence Theorem says that

1 2 f ͑x͒ if n n ͚ ͑ k ͒x ෇ ͭ ͑ ͒ sin 2 1 1 ෇ 2 k෇1 2k 1 2 if x n ෇

Calculus, Stewart: 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 ͑x͒ has period 2L , that is f ͑x 2L͒ ෇ f ͑x͒ for all x . If we let t ෇ x͞L and

t͑t͒ ෇ f ͑x͒ ෇ f ͑Lt͒͞

then, as you can verify,t has period 2 and x ෇ L corresponds to t ෇ . The Fourier series of t is

a0 ͚ ͑an cos nt bn sin nt͒ n෇1

where 1 a0 ෇ y t͑t͒ dt 2

1 1 an ෇ y t͑t͒ cos nt dt bn ෇ y t͑t͒ sin nt dt

If we now use the Substitution Rule with x ෇ Lt͞ , then t ෇ x͞L ,dt ෇ ͑͞L͒ dx , and we have the following

9 If f is a piecewise continuous function on ͓L, L͔ , its Fourier series is

ͫ ͩ n x ͪ ͩ n x ͪͬ a0 ͚ an cos bn sin n෇1 L L where ■■ ෇ Notice that when L these 1 L formulas are the same as those in (7). a0 ෇ y f ͑x͒ dx 2L L

and, for n 1,

L nx L ෇ 1 ͑ ͒ ͩ ͪ ෇ 1 ͑ ͒ ͩ n x ͪ an y f x cos dx bn y f x sin dx L L L L L L

Of course, the Fourier Convergence Theorem (8) is also valid for functions with period 2L.

EXAMPLE 2 Find the Fourier series of the triangular wave function deﬁned by f ͑x͒ ෇ Խ x Խ for 1 x 1 and f ͑x 2͒ ෇ f ͑x͒ for all x . (The graph of f is shown in Figure 3.) For which values of x is f ͑x͒ equal to the sum of its Fourier series? y

Calculus, Stewart: FOURIER SERIES ■ 7

SOLUTION We ﬁnd the Fourier coefﬁcients by putting L ෇ 1 in (9):

෇ 1 1 ෇ 1 0 ͑ ͒ 1 1 a0 2 y Խ x Խ dx 2 y x dx 2 y x dx 1 1 0 ■■ Notice that a0 is more easily calculated as an area. ෇ 1 2 0 1 2 1 ෇ 1 4 x ]1 4 x ] 0 2

and for n 1 ,

1 1 an ෇ y Խ x Խ cos͑nx͒ dx ෇ 2 y x cos͑nx͒ dx 1 0

because y ෇ Խ x Խ cos͑nx͒ is an even function. Here we integrate by parts with u ෇ x and dv ෇ cos͑nx͒ dx. Thus,

1 ෇ ͫ x ͑ ͒ͬ 2 y1 ͑ ͒ an 2 sin n x sin n x dx 0 n 0 n

1 2 cos͑nx͒ 2 ෇ 0 ͫ ͬ ෇ ͑cos n 1͒ 2 2 n n 0 n

Since y ෇ Խ x Խ sin͑nx͒ is an odd function, we see that

1 bn ෇ y Խ x Խ sin͑nx͒ dx ෇ 0 1

We could therefore write the series as

͑ ͒ 1 2 cos n 1 ͑ ͒ ͚ 2 2 cos n x 2 n෇1 n

But cos n ෇ 1 if n is even and cos n ෇ 1 if n is odd,so

0 if n is even ෇ 2 ͑ ͒ ෇ an 2 2 cos n 1 ͭ 4 n if n is odd n2 2

Therefore, the Fourier series is

1 4 4 4 cos͑x͒ cos͑3x͒ cos͑5x͒ 2 2 9 2 25 2

෇ 1 4 ͑͑ ͒ ͒ ͚ 2 2 cos 2k 1 x 2 n෇1 ͑2k 1͒

The triangular wave function is continuous everywhere and so, according to the Fourier Convergence Theorem, we have

Calculus, Stewart: 8 ■ FOURIER SERIES

In particular,

෇ 1 4 ͑͑ ͒ ͒ Խ x Խ ͚ 2 2 cos 2k 1 x for 1 x 1 2 k෇1 ͑2k 1͒

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 t

FIGURE 4 Waveforms(a) Flute (b) Violin

We gain insight into the differences between waveforms if we express them as sums of Fourier series:

t t 2t 2t P͑t͒ ෇ a a cosͩ ͪ b sinͩ ͪ a cosͩ ͪ b sinͩ ͪ 0 1 L 1 L 2 L 2 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,

nt nt a cosͩ ͪ b ͩ ͪ n L n L

is called the nth harmonic of P. The amplitude of the n th harmonic is

2 2 An ෇ san bn

Calculus, Stewart: FOURIER SERIES ■ 9

for a Fourier series with only sine terms, as in Example 1, the amplitude is An bn and 2 2 2 the energy is An bn .) The graph of the sequence An 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 2 that, for the flute,An 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.

A@n A@n

0 246810 n 0 246810 n FIGURE 5 Energy spectra (a) Flute (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 Find the Fourier series of the function. S Click here for solutions. 1 if Խ x Խ 1 7. f ͑x͒ ෇ ͭ f ͑x 4͒ ෇ f ͑x͒ 1–6 A function f is given on the interval ͓, ͔ and f is 0 if 1 Խ x Խ 2 periodic with period 2 . (a) Find the Fourier coefﬁcients of f . 0 if 2 x 0 ͑ ͒ (b) Find the Fourier series of f . For what values of x is f x 8. f ͑x͒ ෇ ͭ1 if 0 x 1 f ͑x 4͒ ෇ f ͑x͒ equal to its Fourier series? 0 if 1 x 2 ; (c) Graph f and the partial sums S2 ,S4 , and S6 of the Fourier series. x if 4 x 0 9. f ͑x͒ ෇ ͭ f ͑x 8͒ ෇ f ͑x͒ 1 if x 0 0 if 0 x 4 1. f ͑x͒ ෇ ͭ 1 if 0 x 10. f ͑x͒ ෇ 1 x, 1 x 1 f ͑x 2͒ ෇ f ͑x͒ 0 if x 0 2. f ͑x͒ ෇ ͭ x if 0 x 11. f ͑t͒ ෇ sin͑3t͒, 1 t 1

3. f ͑x͒ ෇ x ■■■■■■■■■■■■■

4. f ͑x͒ ෇ x 2 12. A voltage E sin t , where t represents time, is passed through a so-called half-wave rectiﬁer that clips the negative part of the 0 if x 0 wave. Find the Fourier series of the resulting periodic function 5. f ͑x͒ ෇ ͭ cos x if 0 x 0 if t 0 ͑ ͒ ෇ ͑ ͒͞ ෇ ͑ ͒ f t ͭ f t 2 f t ͞ 1 if x 2 E sin t if 0 t Sixth Edition. ISBN: 0495011606. © 2008 Brooks/Cole. All rights reserved. 6. f ͑x͒ ෇ ͭ1 if ͞2 x 0 0 if 0 x

Calculus,

■■■■■■■■■■■■■ Stewart: 10 ■ FOURIER SERIES

13–16 Sketch the graph of the sum of the Fourier series of f 18. Use the result of Example 2 to show that without actually calculating the Fourier series. 1 1 1 2 1 1 if 4 x 0 2 2 2 13. f x 3 5 7 8 3 if 0 x 4 19. Use the result of Example 1 to show that x if 1 x 0 14. f x 1 1 1 1 x if 0 x 1 1 3 5 7 4

3 15. f x x , 1 x 1 20. Use the given graph of f and Simpson’s Rule with n 8 to

estimate the Fourier coefﬁcients a0, a1, a2, b1, and b2 . Then use x 16. f x e , 2 x 2 them to graph the second partial sum of the Fourier series and compare with the graph of f . ■■■■■■■■■■■■■ y 17. (a) Show that, if 1 x 1 , then

1 4 2 n x 1 2 2 cos n x 3 n1 n 1

(b) By substituting a speciﬁc value of x , show that 0.25 x

SOLUTIONS

1 1 1. a = π f(x) dx = 0 dx − π dx =0 (a) 0 2π −π 2π −π 0 . 1 1 1 a = π π f(x)cosnx dx = 0 cos nx dx − π cos nx dx =0 cos nx n π −π −π π −π π 0 [since is even]. 1 1 1 2 b = π π f(x)sinnx dx = 0 sin nx dx − π sin nx dx = 0 sin nx dx sin nx n π −π −π π −π π 0 π −π [since is odd]  0 n 2  if even = − [1 − cos(−nπ)] = 4 nπ − n nπ if odd ∞ 4 (b) f(x)= − sin(2k +1)x when −π

0.5

0 -2.5 -1.25 0 1.25 2.5

x -0.5

-1

1 1 3. a = π f(x) dx = π xdx=0 (a) 0 2π −π 2π −π . 1 1 a = π f (x)cosnx dx = π x cos nx dx =0 x cos nx n π −π π −π [because is odd] 1 1 2 b = π f (x)sinnx dx = π x sin nx dx = π x sin nx dx x sin nx n π −π π −π π 0 [since is odd] 2 −(2/n) if n even = − cos nπ [using integration by parts] = n (2/n) if n odd

∞ 2 y (b) f(x)= (−1)n+1 sin nx (c) n=1 n 2.5 when −π

1.25

0 -2.5 -1.25 0 1.25 2.5

-1.25

1 1 5. a = π f(x) dx = π cos xdx=0 (a) 0 −π 0 2π 2π 1 1 1 if n =1 a = π f(x)cosnx dx = π cos x cos nx dx = 2 x = π ] n −π 0 [bysymmetryabout 2 π π 0 if n =1 1 1 b = π f(x)sinnx dx = π cos x sin nx dx n π −π π 0   2n if n even usinganintegraltable, = π(n2 − 1)  and simpliﬁed using the addition formula for cos(a + b) 0 if n odd

∞ 4k f (x)= 1 cos x + sin(2k) −π

0.5

0 -2.5 -1.25 0 1.25 2.5

x -0.5

-1

 0 − 2 ≤ x ≤−1  if 7. Use f (x)= 1 if − 1

1 a = L f(x) dx = 1 1 dx = 1 0 2L −L 4 −1 2   0 if n even  1 L nπx 1 1 nπx 2 π an = f(x)cos dx = cos dx = sin n = 2/nπ if n =4n +1 L −L L 2 −1 2 nπ 2  −2/nπ if n =4n +3 1 nπx nπx b = L f (x)sin dx = 1 1 sin dx =0 n L −L L 2 −1 2 2 πx 2 3πx 2 5πx 1 + cos − cos + cos −··· Fourier Series: 2 π 2 3π 2 5π 2 ∞ 2 π 2 π 1 + sin (4k +1) − sin (4k +3) 2 k=1 (4k +1)π 2 (4k +3)π 2

y 1

0.75

0.5

0.25

0 -5 -2.5 0 2.5 5

−x if − 4 ≤ x<0 9. Use f(x)= ,L=4. 0 if 0 ≤ x ≤ 4

1 a = L f(x) dx = 1 0 −xdx=1 0 2L −L 8 −4

1 L nπx 1 0 nπx 4 an = f(x)cos dx = −x cos dx = (cos (nπ) − 1) = L −L L 4 −4 4 (nπ)2 0 if n is even −8/(nπ)2 if n is odd 1 nπx nπx 4 4/nπ if n is even b = L f(x)sin dx = 1 0 −x sin dx = cos (nπ)= n −L 4 −4 L L 4 nπ −4/nπ if n is odd

Fourier Series: ∞ 4 π 8 π 4 π 1+ − sin (2k − 1)x − cos (2k − 1) x + sin (2k)x 2 2 k=1 (2k − 1)π 4 (2k − 1) π 4 (2k)π 4

y 4

0 -5 -2.5 0 2.5 5

11. Use f (x)={sin(3πt) if − 1 ≤ t ≤ 1 ,L=1. Note: sin(3πt) 2 f(x)=2 This can be done instantly if one observes that the period of is 3 , and the period of which 2 f(x) sin(3πt) t is an integer multiple of 3 . Therefore is the same as for all , and its Fourier series is therefore sin(3πt). 1 a = L f(x) dx = 1 1 sin(3πx) dx =0 We can get this result using the standard coefﬁcient formulas: 0 2L −L 2 −1 1 nπx a = 1 f(x)cos dx = 1 sin(3πx)cos(nπx) dx n L −1 L −1 =0 [applying change of variables to a formula in the section]

1 nπx b = L f(x)sin dx = 1 sin(3πx)sin(nπx) dx n L −L L −1  sin nπ 6 n =3 0 n =3 2 if if = π (−9+n ) [using integral table and addition formula =  1 if n =3 1 if n =3 Thomson Brooks-Cole copyright 2008 14 ■ FOURIER SERIES

Fourier Series: sin(3πx)

y 1

0.5

0 -5 -2.5 0 2.5 5

x -0.5

-1

y 3

 3 −5 ≤ x<−4 2  if  −1 if −4 ≤ x<0 13.  1  3 if 0 ≤ x<4  −1 4 ≤ x<5 if 0 -5 -2.5 0 2.5 5

-1

15. y 1

0.5

0 -1.25 0 1.25 2.5 3.75

-0.5

-1

17. (a) We ﬁnd the Fourier series for f(x)={x2 if −1 ≤ x ≤ 1, L =1 1 a = L f(x) dx = 1 1 x2 dx = 1 0 2L −L 2 −1 3  4  if n even  (nπ)2 1 1 nπx 1 2 4 an = f(x)cos dx = x cos(nπx) dx = cos nπ = L −1 L −1 (nπ)2  4 − if n odd (nπ)2 1 nπx b = L f (x)sin dx = 1 x2 sin(nπx) dx =0 x2 sin(nπx) n L −L L −1 because is odd.

∞ 2 1 n 4 So we have x = 3 + (−1) 2 cos(nπx) for −1 ≤ x ≤ 1. n=1 (nπ)

2 19. 0 ≤ x<π 1= 1 + ∞ sin((2k − 1)x) Example 1 says that, for , 2 k=1 (2k − 1)π . x = π Let 2 to obtain ∞ 2 π 1=1 + sin (2k − 1) 2 k=1 (2k − 1)π 2 π ∞ 1 = sin((2k − 1)) 4 k=1 (2k − 1) π =1− 1 + 1 − 1 + ··· 4 3 5 7 Thomson Brooks-Cole copyright 2008