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Unformatted text preview: Chapter 10 Fourier Series 10.1 Periodic Functions and Orthogonality Relations
The differential equation ′′ + 2 = cos models a massspring system with natural frequency with a pure cosine
forcing function of frequency . If 2 ∕= 2 a particular solution is easily
found by undetermined coefficients (or by using Laplace transforms) to be = cos .
2 − 2 If the forcing function is a linear combination of simple cosine functions, so
that the differential equation is ′′ + 2 = ∑ cos =1 where 2 ∕= 2 for any , then, by linearity, a particular solution is obtained
as a sum ∑ () =
cos .
2 − 2 =1 This simple procedure can be extended to any function that can be represented as a sum of cosine (and sine) functions, even if that summation is not
a finite sum. It turns out that the functions that can be represented as sums
in this form are very general, and include most of the periodic functions that
are usually encountered in applications. 723 724 10 Fourier Series Periodic Functions
A function is said to be periodic with period > 0 if ( + ) = ()
for all in the domain of . This means that the graph of repeats in
successive intervals of length , as can be seen in the graph in Figure 10.1. 2 3 4 5 Fig. 10.1 An example of a periodic function with period . Notice how the graph repeats
on each interval of length . The functions sin and cos are periodic with period 2, while tan is
periodic with period since
tan( + ) = − sin sin( + )
=
= tan .
cos( + )
− cos The constant function () = is periodic with period where is any
positive number since ( + ) = = ().
Other examples of periodic functions are the square wave and triangular wave
whose graphs are shown in Figure 10.2. Both are periodic with period 2. 1 1 −1
0
1
2
3
(a) Square wave sw()
Fig. 10.2 4 −2 −1
0
1
2
3
(b) Triangular wave tw() 4 5 10.1 Periodic Functions and Orthogonality Relations 725 Since a periodic function of period repeats over any interval of length ,
it is possible to define a periodic function by giving the formula for on an
interval of length , and repeating this in subsequent intervals of length .
For example, the square wave sw() and triangular wave tw() from Figure
10.2 are described by
{
0 if −1 ≤ < 0
sw() =
;
sw( + 2) = sw().
1 if 0 ≤ < 1
{
− if −1 ≤ < 0
tw() =
;
tw( + 2) = tw(). if 0 ≤ < 1
There is not a unique period for a periodic function. Note that if is a
period of (), then 2 is also a period because ( + 2) = (( + ) + ) = ( + ) = ()
for all . In fact, a similar argument shows that is also a period for any
positive integer . Thus 2 is a period for sin and cos for all positive
integers .
If > 0 is a period of and there is no smaller period then we say is the
fundamental period of although we will usually just say the period. Not
all periodic functions have a smallest period. The constant function () = is an example of such a function since any positive is a period. The fundamental period of the sine and cosine functions is 2, while the fundamental
period of the square wave and triangular wave from Figure 10.2 is 2. Periodic functions under scaling
If () is periodic of period and is any positive number let () = ().
Then for all (
( (
) )) +
= +
= ( + ) = () = (). Thus If () is periodic with period , then () is periodic with period . It is also true that if is the fundamental period for the periodic function (), then the fundamental period of () = () is /. To see this it is only
necessary to verify that any period of () is at least as large as /, which
is already a period as observed above. But if is a period for (), then is
a period for (/) = () so that ≥ , or ≥ /. 726 10 Fourier Series Applying these observations to the functions sin and cos with fundamental period 2 gives the following facts.
Theorem 1. For any > 0 the functions cos and sin are periodic with
period 2/. In particular, if > 0 then the functions
cos sin and , = 1, 2, 3, . . . . are periodic with fundamental period = 2/. Note that since the fundamental period of the functions cos and sin is = 2/, it follows that 2 = is also a period for each of these
functions. Thus, a sum
∞
∑
( ) cos + sin =1 will be periodic of period 2.
Notice that if is a positive integer, then cos and sin are periodic
with period 2/. Thus, each period of cos or sin contains periods of
cos and sin . This means that the functions cos and sin oscillate
more rapidly as increases, as can be seen in Figure 10.3 for = 3. 1 1 −
−1 (a) The graphs of cos and cos 3. −
−1 (b) The graphs of sin and sin 3. Fig. 10.3 Example 2. Find the fundamental period of each of the following periodic
functions.
1.
2.
3.
4. cos 2
sin 32 ( − )
1 + cos + cos 2
sin 2 + sin 3 ▶ Solution. 1. = 2/2 = .
2. sin 32 ( − ) = sin( 32 − 23 ) = sin 32 cos 32 − cos 32 sin 32 = cos 32 . Thus, = 2/(3/2) = 4/3. 10.1 Periodic Functions and Orthogonality Relations 727 3. The constant function 1 is periodic with any period , the fundamental
period of cos is 2 and all the periods are of the form 2 for a positive
integer , and the fundamental period of cos 2 is with being all
the possible periods. Thus, the smallest number that works as a period
for all the functions is 2 and this is also the smallest period for the sum.
Hence = 2.
4. The fundamental period of sin 2 is 2/2 = 1 so that all of the periods
have the form for a positive integer, and the fundamental period of
sin 3 is 2/3 = 2/3 so that all of the periods have the form 2/3 for a positive integer. Thus, the smallest number that works as a period
for both functions is 2 and thus = 2.
◀ Orthogonality Relations for Sine and Cosine
The family of linearly independent functions
{
} 2
3 2
3
1, cos , cos
, cos
, . . . , sin , sin
, sin
, . . . form what is called a mutually orthogonal set of functions on the interval
[−, ], analogous to a mutually perpendicular set of vectors. Two functions and defined on an interval ≤ ≤ are said to be orthogonal on the
interval [, ] if
∫ ()() = 0. A family of functions is mutually orthogonal on the interval [, ] if any
two distinct functions are orthogonal. The mutual orthogonality of the family
of cosine and sine functions on the interval [−, ] is a consequence of the
following identities.
Proposition 3 (Orthogonality Relations). Let and be positive integers, and let > 0. Then ∫ ∫ cos
− = ∫ − sin = 0 −
{
∫ ,
cos cos = 0
−
{
∫ ,
sin sin = 0
− sin = 0 (1)
(2) cos if = ,
if =
∕ . (3) if = ,
if =
∕ . (4) 728 10 Fourier Series Proof. For (1): =
sin =
(sin − sin(−)) = 0.
cos −
− ∫ For (3) with ∕= , use the identity
cos cos = 1
(cos( + ) + cos( − )),
2 to get
)
∫ ( 1
( + )
( − )
cos cos =
cos + cos −
− 2
(
) ( + ) ( − ) =
sin
+
sin = 0.
2( + ) 2( − ) − ∫ For (3) with = , use the identity cos2 = (1 + cos 2)/2 to get
∫
= ∫ ∫ ( )2 cos =
cos = −
(
)
(
) 2
1 2 1
= .
1 + cos =
+
sin 2 2
2 − cos
− − The proof of (4) is similar, making use of the identities sin2 = (1−cos 2)/2
in case = and
sin sin = 1
(cos( − ) − cos( + ))
2 in case ∕= . The proof of (2) is left as an exercise.
Even and Odd Functions
A function defined on a symmetric interval [−, ] is even if (−) = ()
for − ≤ ≤ , and is odd if (−) = − () for − ≤ ≤ .
Example 4. Determine whether each of the following functions is even,
odd, or neither.
1. () = 32 + cos 5
2. () = 3 − 2 sin 2
3. ℎ() = 2 + + 1
▶ Solution. 1. Since (−) = 3(−)2 + cos 5(−) = 32 + cos 5 = () for
all , it follows that is an even function. 10.1 Periodic Functions and Orthogonality Relations 729 2. Since (−) = 3(−) − (−)2 sin 3(−) = −3 + 2 sin 3 = −() for all ,
it follows that is an odd function.
3. Since ℎ(−) = (−)2 + (−) + 1 = 2 − + 1 = 2 + + 1 = ℎ() ⇐⇒
− = ⇐⇒ = 0, we conclude that ℎ is not even. Similarly, ℎ(−) =
−ℎ() ⇐⇒ 2 −+1 = −2 −−1 ⇐⇒ 2 +1 = (−2 +1) ⇐⇒ 2 +1 = 0,
which is not true for any . Thus ℎ is not odd, and hence it is neither
even or odd.
◀
The graph of an even function is symmetric with respect to the axis, while
the graph of an odd function is symmetric with respect to the origin, as
illustrated in Figure 10.4. () (−) ()
− − (−) = − () (a) The graph of an even function. (b) The graph of an odd function. Fig. 10.4 Here is a list of basic properties of even and odd functions that are useful
in applications to Fourier series. All of them follow easily from the definitions,
and the verifications will be left to the exercises.
Proposition 5. Suppose that and are functions defined on the interval
− ≤ ≤ .
1.
2.
3.
4. If
If
If
If both and are even then + and are even.
both and are odd, then + is odd and is even. is even and is odd, then is odd. is even, then
∫ ∫ () = 2 () .
0 − 5. If is odd, then ∫ () = 0. − Since the integral of computes the signed area under the graph of , the
integral equations can be seen from the graphs of even and odd functions in
Figure 10.4. 730 10 Fourier Series Exercises
1–9. Graph each of the following periodic functions. Graph at least 3 periods.
{
3
if 0 < < 3
1. () =
; ( + 6) = ().
−3 if −3 < < 0
⎧ ⎨−3 if −2 ≤ < −1
2. () = 0
if −1 ≤ ≤ 1 ; ⎩
3
if 1 < < 2
3. () = , 0 < ≤ 2; 4. () = , −1 < ≤ 1; ( + 4) = (). ( + 2) = (). ( + 2) = (). 5. () = sin , 0 < ≤ ; ( + ) = ().
{
0
if − ≤ < 0
6. () =
; ( + 2) = ().
sin if 0 < ≤ {
− if −1 ≤ < 0
7. () =
; ( + 2) = ().
1
if 0 ≤ < 1
8. () = 2 , −1 < ≤ 1;
9. () = 2 , 0 < ≤ 2; ( + 2) = (). ( + 2) = (). 10–17. Determine if the given function is periodic. If it is periodic find the
fundamental period.
10. 1
11. sin 2
12. 1 + cos 3
13. cos 2 + sin 3
14. + sin 2
15. sin2 16. cos + cos 17. sin + sin 2 + sin 3
18–26. Determine if the given function is even, odd, or neither. 10.2 Fourier Series 731 18. () = ∣∣
19. () = ∣∣
20. () = sin2 21. () = cos2 22. () = sin sin 3
23. () = 2 + sin 24. () = + ∣∣
25. () = ln ∣cos ∣
26. () = 5 + 2 sin 3
27. Verify the orthogonality property (2) from Proposition 3:
∫ − cos sin = 0 28. Use the properties of even and odd functions (Proposition 10.4) to evaluate the following integrals. (a)
(c) ∫ 1 −1
∫ sin − (e) ∫ (b) cos − sin (d)
(f) ∫ 1 −1
∫ ∫ − 4 cos 2 sin − 10.2 Fourier Series
We start by considering the possibility of representing a function as a sum
of a series of the form
)
∞ (
0 ∑ () =
+ cos
+ sin
(1)
2 =1 732 10 Fourier Series where the coefficients 0 , 1 , . . ., 1 , 2 , . . ., are to be determined. Since the
individual terms in the series (1) are periodic with periods 2, 2/2, 2/3,
. . ., the function () determined by the sum of the series, where it converges,
must be periodic with period 2. This means that only periodic functions of
period 2 can be represented by a series of the form (1). Our first problem is
to find the coefficients and in the series (1). The first term of the series
is written 0 /2, rather than simply as 0 , to make the formula to be derived
below the same for all , rather than a special case for 0 .
The coefficients and can be found from the orthogonality relations
of the family of functions cos(/) and sin(/) on the interval [−, ]
given in Proposition 3 of Sect. 10.1. To compute the coefficient for = 1, 2,
3, . . ., multiply both sides of the series (1) by cos(/), with a positive
integer and then integrate from − to . For the moment we will assume
that the integrals exist and that it is justified to integrate term by term. Then
using (1), (2), and (3) from Sect. 10.1, we get
∫ () cos − 0 = 2 ∫ cos −

{z
}
=0 + ∞ [
∑ =1 ∫ ∫ cos +
sin cos = . −
{z
}

{z
} cos
−  = ⎧ ⎨0 ⎩ Thus, 1
=
=0 if ∕= if = ∫ () cos − , = 1, 2, 3, . . . , = 1, 2, 3, . . . or, replacing the index by ,
1 = ∫ − () cos (2) To compute 0 , integrate both sides of (1) from − to to get
∫ − () = 0
2 ∫ ∫ ∫ ∞ [
∑ + + cos
sin −
−
−
 {z } =1

{z
}

{z
} =0 =2 =0 = 0 .
Thus, 1
0 = ∫ − () . (3) 10.2 Fourier Series 733 Hence, 0 is two times the average value of the function () over the interval
− ≤ ≤ . Observe that the value of 0 is obtained from (2) by setting = 0. Of course, if the constant 0 in (1) were not divided by 2, we would
need a separate formula for 0 . It is for this reason that the constant term
in (1) is labeled 0 /2. Thus, for all ≥ 0, the coefficients are given by a
single formula = 1 ∫ , () cos − (4) = 0, 1, 2, . . . To compute for = 1, 2, 3, . . ., multiply both sides of the series (1)
by sin(/), with a positive integer and then integrate from − to .
Then using (1), (2), and (4) from Sect. 10.1, we get
∫ 0 () sin = 2
− ∫ sin −

{z
}
=0 ∫ ∫ ∞ [
∑ + cos
sin sin + sin = . =1
 −
{z
}
 − ⎧
{z
}
=0 = ⎨0 ⎩ if ∕= if = Thus, replacing the index by , we find that
1 = ∫ () sin − , (5) = 1, 2, 3, . . . We have arrived at what are known as the Euler Formulas for a function () that is the sum of a trigonometric series as in (1): 0 = = = 1 1 1 ∫ −
∫ −
∫ − (6) () () cos , = 1, 2, 3, . . . (7) () sin , = 1, 2, 3, . . . (8) The numbers and are known as the Fourier coefficients of the function . Note that while we started with a periodic function of period 2, the
formulas for and only use the values of () on the interval [−, ]. 734 10 Fourier Series We can then reverse the process, and start with any function () defined on
the symmetric interval [−, ] and use the Euler formulas to determine a
trigonometric series. We will write () ∼ )
∞ (
0 ∑ + cos
+ sin
,
2 =1 (9) where the , are defined by (6), (7), and (8), to indicate that the right
hand side of (9) is the Fourier series of the function () defined on [−, ].
Note that the symbol ∼ indicates that the trigonometric series on the right
of (9) is associated with the function (); it does not imply that the series
converges to () for any value of . In fact, there are functions whose Fourier
series do not converge to the function. Of course, we will be interested in the
conditions under which the Fourier series of () converges to (), in which
case ∼ can be replaced by =; but for now we associate a specific series using
Equations (6), (7), and (8) with () and call it the Fourier series. The mild
conditions under which the Fourier series of () converges to () will be
considered in the next section.
Remark 1. If an integrable function () is periodic with period , then the
integral of () over any interval of length is the same; that is
∫ ∫ + () = (10) () 0 for any choice of . To see this, first observe that for any and , if we use
the change of variables = − , then
∫ () = ∫ + ∫ + ( − ) = ∫ + () = + Letting = and = 0 gives
∫ 0
∫ () = + () .
+ () + so that
∫ + () = = ∫
∫ 0 () + ∫ + () 0 () +
+ ∫ 0 + () = ∫ () , 0 which is (10). This formula means that when computing the Fourier coefficients, the integrals can be computed over any convenient interval of length
2. For example, 10.2 Fourier Series 735 = 1 ∫ () cos − 1 = ∫ 2 () cos 0 . We now consider some examples of the calculation of Fourier series.
Example 2. Compute the Fourier series of the odd square wave function
of period 2 and amplitude 1 given by
{
−1 − ≤ < 0, () =
; ( + 2) = ().
1
0 ≤ < ,
See Figure 10.5 for the graph of (). 1 −3 −2 − 2 3 −1 Fig. 10.5 The odd square wave of period 2 ▶ Solution. Use the Euler formulas for (Equations (6) and (7)) to conclude
∫
1 = () cos = 0 − for all ≥ 0. This is because the function () cos is the product of an
odd and even function, and hence is odd, which implies by Proposition 5
(Part 5) of Section 10.1 that the integral is 0. It remains to compute the
coefficients from (8). = 1 ∫ () sin
−
0 1 = ∫ 0 () sin − ∫ 1 + 1 +
(+1) sin −
0
{[0
[ } 1
=
cos −
cos − 0 = 1 ∫ (−1) sin 1
[(1 − cos(−)) + (1 − cos())] 2
2
=
(1 − cos ) =
(1 − (−1) ). = Therefore, ∫ 0 () sin 736 10 Fourier Series = { if is even,
if is odd, 0
4 and the Fourier series is
(
)
4 1
3
1
5
1
7 () ∼
sin + sin + sin + sin
+ ⋅⋅⋅ . 3 5 7 (11)
◀ Example 3. Compute the Fourier series of the even square wave function
of period 2 and amplitude 1 given by
⎧ ⎨−1 − ≤ < −/2, () = 1 ( + 2) = ().
−/2 ≤ < /2, ; ⎩
−1 /2 ≤ < ,
See Figure 10.6 for the graph of (). 1 − 5
2 −
2
−1 − 3
2 2 3
2 5
2 Fig. 10.6 The even square wave of period 2 ▶ Solution. Use the Euler formulas for (Equation (8)) to conclude
1 = ∫ () sin − = 0, for all ≥ 1. As in the previous example, this is because the function () sin is the product of an even and odd function, and hence is odd,
which implies by Proposition 5 (Part 5) of Section 10.1 that the integral is
0. It remains to compute the coefficients from (6) and (7).
For = 0, 0 is twice the average of () over the period [−, ], which
is easily seen to be 0 from the graph of (). For ≥ 1,
1 = = 1 ∫ () cos −
−/2 ∫ − () cos 1 + ∫ /2 () cos
−/2 1 + ∫ /2 () cos 10.2 Fourier Series ∫ 737 ∫ ∫ 1 /2 1 +
(+1) cos +
(−1) cos −
−/2
/2
{ [
}−/2 [/2
[ 1
−
sin +
sin −
sin = − −/2 /2
( )
( )
( )
( )]
1 [
=
− sin −
+ sin(−) + sin
− sin −
− sin + sin 2
2
2
2
4 =
sin
. 2
= 1 −/2 (−1) cos Therefore, = ⎧ ⎨0 if is even,
if = 4 + 1
if = 4 + 3 4 ⎩ 4
− and the Fourier series is
(
) 1
3
1
5
1
7
4 () ∼
cos − cos + cos − cos + ⋅⋅⋅ 3 5 7 ∞ 4 ∑ (−1)
(2 + 1)
=
cos
. 2 + 1 =0 ◀
Example 4. Compute the Fourier series of the even triangular wave function of period 2 given by
{
− − ≤ < 0, () =
; ( + 2) = (). 0 ≤ < ,
See Figure 10.7 for the graph of (). −3 −2 − 0 2 3 Fig. 10.7 The even triangular wave of period 2. ▶ Solution. The period is 2 = 2 so = . Again, since the function () is even, the coefficients = 0. It remains to compute the coefficients from the Euler formulas (6) and (7).
For = 0, using the fact that () is even, 738 10 Fourier Series ∫ 2 () 0
− ∫
2 2 2 = =
= . 0 2 0 0 = 1 ∫ () = For ≥ 1, using the fact that () is even, and taking advantage of the
integration by parts formula
∫ cos = sin + cos + ,
∫
∫
1 2 () cos = () cos − 0
(
)
∫ 2 cos let = so = and =
= 0 ∫ 2
2 =
cos = 2 [ sin + cos ]=0
= 0 2
2
= 2 [cos − 1] = 2 [(−1) − 1] = Therefore, = { 0
− 42 if is even,
if is odd and the Fourier series is 4 () ∼ −
2 ( cos cos 3 cos 5 cos 7
+
+
+
+ ⋅⋅⋅
12
32
52
72
∞ 4 ∑ cos(2 + 1)
= −
.
2 (2 + 1)2 ) =0 ◀
Example 5. Compute the Fourier series of the sawtooth wave function of
period 2 given by () = for − ≤ < ; ( + 2) = (). See Figure 10.8 for the graph of ().
▶ Solution. As in Example 2, the function () is odd, so the cosine terms are all 0. Now compute the coefficients from (8). Using the integration
by parts formula 10.2 Fourier Series 739 −3 −2 − 2 3 − Fig. 10.8 The sawtooth wave of period 2 ∫ = 1 ∫ () sin
− sin = sin − cos + , (
)
∫
2 = sin let = so = and = 0 ∫ ∫ 2
2
= sin = 2 2 sin 0 0
2
=
= 2 2 [sin − cos ]=0 2
2
= − 2 2 ( cos ) = − (−1) . Therefore, the Fourier series is
(
)
2 1
2
1
3
1
4 () ∼
sin − sin + sin − sin
+ ⋅⋅⋅ 2 3 4 ∞
+1 2 ∑ (−1)
sin
.
= =1 ◀
All of the examples so far have been of functions that are either even or
odd. If a function () is even, the resulting Fourier series will only have
cosine terms, as in the case of Examples 3 and 4, while if () is odd, the
resulting Fourier series will only have sine terms, as in Examples 2 and 5.
Here are some examples where both sine an cosine terms appear.
Example 6. Compute the Fourier series of the function of period 4 given
by 740 10 Fourier Series () = { 0 −2 ≤ < 0,
; 0 ≤ < 2, ( + 4) = (). See Figure 10.9 for the graph of (). 2 −6 −4 2 −2 4 6 Fig. 10.9 A half sawtooth wave of period 4 ▶ Solution. This function is neither even nor odd, so we expect both sine
and cosine terms to be present. The period is 4 = 2 so = 2. Because () = 0 on the interval (−2, 0), each of the integrals in the Euler formulas,
which should be an integral from = −2 to = 2, can be replaced with an
integral from = 0 to = 2. Thus, the Euler formulas give 0 = =
=
=
= [ ]2
1 2
= 1;
2 2 0
0
(
)
∫
1 2 2
2 cos let = so =
and =
2 0
2
2 ∫ ∫ 1
2
2
2
cos = 2 2 cos 2 0 0
2
=
[cos + sin ]=0
2 2
2
2
(cos − 1) = 2 2 ((−1) − 1).
2
2 1
2 ∫ 2 = Therefore, ⎧ ⎨1 = 0 ⎩
− 242 if = 0,
if is even and ≥ 2,
if is odd. Now compute : = 1
2 ∫ 0 2 sin 2 (
) 2
2
let = so =
and =
2 10.2 Fourier Series = 1
2 ∫ 0 741 2
2
2
sin = 2 2 ∫ sin 0 2
[sin − cos ]=
=0
2 2
2
2
= 2 2 (− cos ) = − (−1) . = Thus, 2(−1)+1 Therefore, the Fourier series is for all ≥ 1. = () ∼ ∞
∞
1
4 ∑
1
(2 + 1)
2 ∑ (−1)+1 − 2
cos +
sin
.
2 (2 + 1)2
2 =1 2
=0 ◀
Example 7. Compute the Fourier series of the square pulse wave function
of period 2 given by
{
1 0 ≤ < ℎ, () =
; ( + 2) = ().
0 ℎ ≤ < 2,
See Figure 10.10 for the graph of ().
1 −4 0 −2 2 ℎ 4 Fig. 10.10 A squ...
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 Fourier Series, D461 D451D461