NOTES
THEOREM 4.2 Suppose that lim
34. The energy density of electromagnetic radiation at wave- wave-length λ from a blackbody at temperature T degrees
7.9 FOURIER SERIES
Over the last several sections, we have seen how we can represent a function by a power series (i.e., a Taylor series). This is an extraordinary development, in particular because we can use the partial sums of such a series expansion (i.e., the Taylor polynomials) to compute approximate values of the function for values of x close to the point c about which you expanded. For the first time, this gave us the means of calculating approximate values of transcendental functions like ex, ln x and sin x, which we otherwise could not compute.
Although this is a good reason for studying series expansions of functions, it is not the only reason we need them.
Observe that many phenomena we encounter in the world around us are periodic in nature. That is, they repeat themselves over and over again. For instance, light, sound,
radio waves and x-rays (to mention only a few) are all periodic. For such phenomena, Taylor polynomial approximations have some obvious shortcomings. Look back at any of the graphs you (or we) constructed of the Taylor polynomials of a periodic function and you’ll notice that as x gets farther away from c (the point about which you expanded), the difference between the function and a given Taylor polynomial grows. Such behavior, as we illustrate in Figure 7.44a for the case of f (x)= sin x expanded about x = π2, is typical of the convergence of Taylor series.
Because the Taylor polynomials provide an accurate approximation only in the vicinity of c, we say that they are accurate locally. In general, no matter how large you make n, the approximation is still only valid locally. In many situations, notably in communications, we need to find an approximation to a given periodic function that is valid globally (i.e., for all x). Consider the Taylor polynomial graphed in Figure 7.44b and convince yourself that Taylor polynomials will not satisfy this need. For this reason, we construct a different type of series expansion for periodic functions, one where each of the terms in the expansion is periodic.
1 1
q q p w
y sin x y P4(x) y
x
y
1 1
q w r t
y sin x y P8(x)
q
w
r
x
FIGURE 7.44a
y= sin x and y = P4(x).
FIGURE 7.44b
y= sin x and y = P8(x).
Recall that we say that a function f is periodic of period T> 0 if f (x + T ) = f (x), for all x in the domain of f . Can you think of any periodic functions? Surely, sin x and cos x come to mind. These are both periodic of period 2π. Further, sin (2x), cos (2x), sin (3x), cos (3x) and so on are all periodic of period 2π. In fact, the simplest periodic functions you can think of (aside from constant functions) are the functions
sin (kx) and cos (kx), for k = 1, 2, 3,. . . .
Note that each of these is periodic of period 2π, as follows. For any integer k, let f (x) = sin (kx). We then have
f (x+ 2π) = sin[k(x + 2π)] = sin (kx + 2kπ) = sin (kx) = f (x).
Likewise, you can show that cos (kx) has period 2π.
So, if you wanted to expand a periodic function in a series, the simplest periodic functions to use in the terms of the series are just these functions. Consequently, we consider a series of the form
a0
2 +∞
k=1
[akcos (kx)+ bksin (kx)],
Fourier series
called a Fourier series. Notice that if the series converges, it will converge to a periodic function whose period is 2π, since every term in the series has period 2π. The coefficients of the series, a0, a1, a2,. . . and b1, b2,. . . are called the Fourier coefficients. You may have noticed the unusual way in which we wrote the leading term of the series
a0
2
. We did this in order to simplify the formulas for computing these coefficients, as we’ll see later.
There are a number of important questions we must address.
r What functions can be expanded in a Fourier series?
r How do we compute the Fourier coefficients?
r Does the Fourier series converge? If so, to what function does the series converge?
We begin our investigation much as we did with power series. Suppose that a given Fourier series converges on the interval [−π, π]. It then represents a function f on that interval,
f (x)= a0 2 +∞
k=1
[akcos (kx)+ bksin (kx)], (9.1)
where f must be periodic outside of [−π, π]. Although some of the details of the proof are beyond the level of this course, we want to give you some idea of how the Fourier coeffi-cients are computed. If we integrate both sides of equation (9.1) with respect to x on the interval [−π, π], we get
π
−π f (x) d x =
π
−π
a0 2 d x+
π
−π
∞ k=1
[akcos (kx)+ bksin (kx)] d x
=
π
−π
a0
2 d x+∞
k=1
ak
π
−πcos (kx) d x+ bk
π
−πsin (kx) d x
, (9.2)
assuming we can interchange the order of integration and summation. In general, the order may not be interchanged (this is beyond the level of this course), but for many Fourier series, doing so is permissible. Observe that for every k = 1, 2, 3,. . . , we have
π
−πcos (kx) d x = 1
ksin (kx)
π
−π =1
k[sin (kπ) − sin (−kπ)] = 0
HISTORICAL NOTES
Jean Baptiste Joseph Fourier (1768–1830) French mathematician who invented Fourier series. Fourier was heavily involved in French politics, becoming a member of the Revolutionary Committee, serving as scientific advisor to Napoleon and establishing educational facilities in Egypt.
Fourier held numerous offices, including secretary of the Cairo Institute and Prefect of Grenoble.
Fourier introduced his trigonometric series as an essential technique for developing his highly original and
revolutionary theory of heat.
and
π
−πsin (kx) d x= −1
kcos (kx)
π
−π = −1
k[cos (kπ) − cos (−kπ)] = 0.
This reduces equation (9.2) to simply
π
−π f (x) d x=
π
−π
a0
2 d x= a0π.
Solving this for a0, we have
a0= 1 π
π
−πf (x) d x. (9.3)
If we multiply both sides of equation (9.1) by cos (nx) (where n is an integer, n≥ 1), and then integrate with respect to x on the interval [−π, π], observe that we get
π
−π f (x) cos (nx) d x
=
π
−π
a0
2 cos (nx) d x +
π
−π
∞ k=1
[akcos (kx) cos (nx)+ bksin (kx) cos (nx)] d x
= a0
2
π
−πcos (nx) d x +∞
k=1
ak
π
−πcos (kx) cos (nx) d x+ bk
π
−πsin (kx) cos (nx) d x
, (9.4)
again assuming we can interchange the order of integration and summation. Next, recall
that π
−πcos (nx) d x = 0, for all n = 1, 2,. . . . It’s an easy exercise to show that
π
−πsin (kx) cos (nx) d x = 0, for all n = 1, 2, . . . and for all k = 1, 2,. . . and that
π
−πcos (kx) cos (nx) d x=
0, if n = k π, if n = k.
Notice that this says that every term in the series in equation (9.4) except one (the term corresponding to k= n) is zero and equation (9.4) reduces to simply
π
−π f (x) cos (nx) d x= anπ.
This gives us (after substituting k for n) ak = 1
π
π
−π f (x) cos (kx) d x, for k = 1, 2, 3,. . . . (9.5)
Fourier coefficients
Likewise, multiplying both sides of equation (9.1) by sin (nx) and integrating from−π to π gives us
bk= 1 π
π
−π f (x) sin (kx) d x, for k = 1, 2, 3,. . . . (9.6) Equations (9.3), (9.5) and (9.6) are called the Euler-Fourier formulas. Notice that equation (9.3) is the same as (9.5) with k= 0. (This was the reason we chose the leading term of the series to bea0
2 , instead of simply a0.)
Let’s summarize what we’ve done so far. We observed that if a Fourier series converges on some interval, then it converges to a function f where the Fourier coefficients satisfy the Euler-Fourier formulas (9.3), (9.5) and (9.6).
Just as we did with power series, given any integrable function f , we can compute the coefficients in (9.3), (9.5) and (9.6) and write down a Fourier series. But, will the series converge and if it does, to what function will it converge? We’ll answer these questions
shortly. For the moment, let’s try to compute the terms of a Fourier series and see what we can observe.
EXAMPLE 9.1 Finding a Fourier Series Expansion
Find the Fourier series corresponding to the square-wave function f (x)=
0, if −π < x ≤ 0 1, if 0 < x ≤ π ,
where f is assumed to be periodic outside of the interval [−π, π] (see the graph in Figure 7.45).
y
x 1
0.5
2p p p 2p
FIGURE 7.45
Square-wave function.
Solution Notice that even though a0 satisfies the same formula as ak, for k≥ 1, we must always compute a0separately from the rest of the ak’s. From equation (9.3), we get
a0= 1 π
π
−π f (x) d x= 1 π
0
−π0 d x+ 1 π
π
0
1 d x= 0 +π π = 1.
From (9.5), we also have that for k≥ 1, ak= 1
π
π
−π f (x) cos (kx) d x = 1 π
0
−π(0) cos (kx) d x+ 1 π
π
0
1 cos (kx) d x
= 1
πksin (kx)
π
0
= 1
πk[sin (kπ) − sin (0)] = 0.
Finally, from (9.6), we have bk= 1
π
π
−π f (x) sin (kx) d x= 1 π
0
−π(0) sin (kx) d x+ 1 π
π
0
(1) sin (kx) d x
= − 1
πkcos (kx)
π
0
= − 1
πk[cos (kπ) − cos (0)] = − 1
πk[(−1)k− 1]
=
0, if k is even 2
πk, if k is odd .
Notice that we can write the even- and odd-indexed coefficients separately as b2k= 0, for k= 1, 2,. . . and b2k−1 = 2
(2k− 1)π, for k= 1, 2,. . . . We then have the Fourier series a0
2 +∞
k=1
[akcos (kx)+ bksin (kx)]= 1 2 +∞
k=1
bksin (kx)=1 2+∞
k=1
b2k−1sin (2k− 1)x
= 1 2 +∞
k=1
2
(2k− 1)π sin[(2k− 1)x]
= 1 2 + 2
πsin x+ 2
3π sin (3x)+ 2
5π sin (5x)+ · · · . Notice that none of our existing convergence tests are appropriate for Fourier series.
Since we can’t test this, we consider the graphs of the first few partial sums of the series defined by
Fn(x)= 1 2 +
n k=1
2
(2k− 1)π sin[(2k− 1)x].
In Figures 7.46a–d, we graph a number of these partial sums.
y
x 1
0.5
p p 2p
2p
y
x 1
0.5
p p 2p
2p FIGURE 7.46a
y= F4(x) and y= f (x).
FIGURE 7.46b
y= F8(x) and y= f (x).
y
x 1
0.5
p p 2p
2p
y
x 1
0.5
p p 2p
2p FIGURE 7.46c
y= F20(x) and y= f (x).
FIGURE 7.46d
y= F50(x) and y= f (x).
Notice that as n gets larger and larger, the graph of Fn(x) appears to be approaching the graph of the square-wave function f (x) shown in red and seen in Figure 7.45. Based on this, we might conjecture that the Fourier series converges to the function f (x). As it turns out, this is not quite correct. We’ll soon see that the series converges to f (x), everywhere, except at points of discontinuity.
Next, we give an example of constructing a Fourier series for another common waveform.
EXAMPLE 9.2 A Fourier Series Expansion for the Triangular Wave Function
Find the Fourier series expansion of f (x)= |x|, for −π ≤ x ≤ π, where f is assumed to be periodic outside of the interval [−π, π], of period 2π.
Solution In this case, f is the triangular wave function graphed in Figure 7.47. From the Euler-Fourier formulas, we have
a0= 1 π
π
−π|x| dx = 1 π
0
−π−x dx + 1 π
π
0
x d x
= −1 π
x2 2
0
−π+ 1 π
x2 2
π
0
= π 2 +π
2 = π.
Similarly, for each k≥ 1, we get ak = 1
π
π
−π|x| cos (kx) dx = 1 π
0
−π(−x) cos (kx) dx + 1 π
π
0
x cos (kx) d x.
y 3
2
p 2p
p
2p
3p
4p 3p 4p x
FIGURE 7.47 Triangular wave.
Both integrals require the same integration by parts. We let u = x dv = cos (kx) dx
du= dx v = 1
ksin (kx) so that
ak= −1 π
0
−πx cos (kx) d x+ 1 π
π
0
x cos (kx) d x
= −1 π
x
ksin (kx)
0
−π+ 1 πk
0
−πsin (kx) d x+ 1 π
x
ksin (kx)
π
0
− 1 πk
π
0
sin (kx) d x
= −1 π
0+π
k sin (−πk)
− 1
πk2cos (kx)
0
−π+1 π
π
k sin (πk) − 0
+ 1
πk2cos (kx)
π
0
= 0 − 1
πk2[cos 0− cos (−kπ)] + 0 + 1
πk2[cos (kπ) − cos 0] Since sinπk = 0 and sin (−πk) = 0.
= 2
πk2[cos (kπ) − 1] =
0, if k is even
−4
πk2, if k is odd. Since cos (kπ) = 1, when k is even and cos (kπ) = −1, when k is odd.
Writing the even- and odd-indexed coefficients separately, we have a2k= 0, for k = 1, 2,. . . and a2k−1 = −4
π(2k − 1)2, for k= 1, 2,. . . . We leave it as an exercise to show that bk= 0, for all k.
This gives us the Fourier series a0
2 +∞
k=1
[akcos (kx)+ bksin (kx)]= π 2 +∞
k=1
akcos (kx)=π 2 +∞
k=1
a2k−1cos (2k− 1)x
= π 2 −∞
k=1
4
π(2k − 1)2 cos (2k− 1)x
= π 2 −4
π cos x− 4
9π cos (3x)− 4
25πcos (5x)− · · · .
You can show that this series converges absolutely for all x, by using the Comparison Test, since
|ak| =
4
π(2k − 1)2cos (2k− 1)x
≤ 4
π(2k − 1)2
and the series∞
k=1
4
π(2k − 1)2converges. (Hint: Compare this last series to the convergent p-series∞
k=1
1
k2, using the Limit Comparison Test.) To get an idea of the function to which the series is converging, we plot several of the partial sums of the series,
Fn(x)= π 2 −
n k=1
4
π(2k − 1)2 cos (2k− 1)x.
See if you can conjecture the sum of the series by looking at Figures 7.48a and b. Notice how quickly the partial sums of the series appear to converge to the triangular wave function f (shown in red; also see Figure 7.47). As it turns out, the graph of the partial sums will not change appreciably if you plot Fn(x) for much larger values of n. (Try this!) We’ll see later how to be sure that the Fourier series converges to f (x) for all x. There’s something further to note here: the accuracy of the approximation is fairly uniform. That is, the difference between a given partial sum and f is roughly the same for each x. Take care to distinguish this behavior from that of Taylor polynomial approximations, where the farther you get away from the point about which you’ve expanded, the worse the approximation tends to get.
y 3
2
p 2p
p
2p
3p
4p 3p 4p
x
y 3
2
p 2p
p
2p
3p
4p 3p 4p
x
FIGURE 7.48a
y= F1(x) and y= f (x).
FIGURE 7.48b
y= F4(x) and y= f (x).