dydu,,,,,,,,,,,uytm

1. 488 CHAP. 11 Fourier Analysis
n ∫ p
p
Fig. 269. Partial sums S1, S2, S3, S20 in Example 5
Solution. We have f = f1 + f2, where f1 = x and f2 = p. The Fourier coefficients of f2 are zero, except for
the first one (the constant term), which is p. Hence, by Theorem 1, the Fourier coefficients an, bn are those of
f1, except for a0, which is p. Since f1 is odd, an = 0 for n = 1, 2, Á , and
b =
2
p f1 (x) sin nx dx =
2
∫ x sin nx dx.
Integrating by parts, we obtain
2
0 0
p
—x cos nx 1 2
bn = p c n
2 +
n
0 0
cos nx dx d = — n cos np.
Hence b1 = 2, b2 = —2, b3 = 2, b4 = —2, Á , and the Fourier series of f(x) is
2 3 4
f(x) = p + 2 asin x —
1
sin 2x +
1
sin 3x — + Áb. (Fig. 269) ■
2 3
3. Half-Range Expansions
Half-range expansions are Fourier series. The idea is simple and useful. Figure 270
explains it. We want to represent f (x) in Fig. 270.0 by a Fourier series, where f (x)
may be the shape of a distorted violin string or the temperature in a metal bar of length
L, for example. (Corresponding problems will be discussed in Chap. 12.) Now comes
the idea.
We could extend f (x) as a function of period L and develop the extended function into
a Fourier series. But this series would, in general, contain both cosine and sine terms. We
can do better and get simpler series. Indeed, for our given f we can calculate Fourier
coefficients from (6*) or from (6**). And we have a choice and can take what seems
more practical. If we use (6*), we get (5*). This is the even periodic extension f1 of f
in Fig. 270a. If we choose (6**) instead, we get (5**), the odd periodic extension f2 of
f in Fig. 270b.
Both extensions have period 2L. This motivates the name half-range expansions: f is
given (and of physical interest) only on half the range, that is, on half the interval of
periodicity of length 2L.
Let us illustrate these ideas with an example that we shall also need in Chap. 12.
y S20
5
S3
S2
S1
y
– 0  x
∫
p p

2. SEC. 11.2 Arbitrary Period. Even and Odd Functions. Half-Range Expansions 489
0 L
L
L L
∫
f2(x)
–L L x
f(x)
L x
(0) The given function f(x)
–L L x
(a) f (x) continued as an even periodic function of period 2L
(b) f(x) continued as an odd periodic function of period 2L
Fig. 270. Even and odd extensions of period 2L
E X A M P L E 6 “Triangle” and Its Half-Range Expansions
k Find the two half-range expansions of the function (Fig. 271)
2k L
0 L/2 L x x if 0 < x <
L 2
Fig. 271. The given
function in Example 6
f(x) = e
2k
(L — x) if
L
< x < L.
L 2
Solution. (a) Even periodic extension. From (6*) we obtain
1 2k L>2 2k L k
a0 =
L
c
L ∫ x dx + ∫
L>2
(L — x) dx d =
2
,
2 2k L>2
np 2k L np
an =
L
c
L ∫ x cos x dx + ∫
L>2
(L — x) cos
L
x dx d .
We consider an. For the first integral we obtain by integration by parts
∫
L>2
x cos
np
x dx = Lx
sin
np
L>2
x 2 — L
∫
L>2
sin
np
x dx
L np
0
L2
L 0 np 0
L
np L2 np
= sin
2np
Similarly, for the second integral we obtain
2
+
n2
p2
acos
2
— 1b .
∫
L
(L — x) cos
np
x dx =
L
(L — x) sin
np
x 2
L
L L np
+ sin x dx
L>2 L np
L
L L>2
L np
np L>2
L
L2 np
= a0 —
np
aL —
2
b sin
2
b —
n2
p2
acos np — cos
2
b.
f1(x)

3. 490 CHAP. 11 Fourier Analysis
a b
–L 0 L x
We insert these two results into the formula for an. The sine terms cancel and so does a factor L2. This gives
4k np
a = 2 cos — cos np — 1 .
Thus,
n
n2p2 2
a2 = —16k>(22p2), a6 = —16k>(62p2), a10 = —16k>(102p2), Á
and an = 0 if n G 2, 6, 10, 14, Á . Hence the first half-range expansion of f(x) is (Fig. 272a)
f(x) =
k
—
16k
a
1
cos
2p
x +
1
cos
6p
x + Áb .
2 p2 22 L 62 L
This Fourier cosine series represents the even periodic extension of the given function f(x), of period 2L.
(b) Odd periodic extension. Similarly, from (6**) we obtain
(5)
8k
bn =
n2
p2
sin
np
.
2
Hence the other half-range expansion of f(x) is (Fig. 272b)
f(x) =
8k
a
1
sin
p
x —
1
sin
3p
x +
1
sin
5p
x — + Á b.
p2 12 L 32 L 52 L
The series represents the odd periodic extension of f(x), of period 2L.
Basic applications of these results will be shown in Secs. 12.3 and 12.5. ■
–L 0 L x
(a) Even extension
(b) Odd extension
Fig. 272. Periodic extensions of f(x) in Example 6
EVEN AND ODD FUNCTIONS
Are the following functions even or odd or neither even nor
odd?
1. ex, e—ƒ x ƒ, x3 cos nx, x2 tan px, sinh x — cosh x
2. sin2 x, sin (x2), ln x, x>(x2 + 1), x cot x
3. Sums and products of even functions
4. Sums and products of odd functions
5. Absolute values of odd functions
6. Product of an odd times an even function
7. Find all functions that are both even and odd.
FOURIER SERIES FOR PERIOD p = 2L
Is the given function even or odd or neither even nor
odd? Find its Fourier series. Show details of your
work.
8.
–1 0 1
P R O B L E M SET 11.2
1
8–17
1–7

4. SEC. 11.2 Arbitrary Period. Even and Odd Functions. Half-Range Expansions 491
4
4
9.
10.
(b) Apply the program to Probs. 8–11, graphing the first
few partial sums of each of the four series on common
axes. Choose the first five or more partial sums until
they approximate the given function reasonably well.
Compare and comment.
22. Obtain the Fourier series in Prob. 8 from that in
Prob. 17.
11. f(x) = x2 (—1 < x < 1), p = 2
HALF-RANGE EXPANSIONS
Find (a) the Fourier cosine series, (b) the Fourier sine series.
Sketch f (x) and its two periodic extensions. Show the
details.
23.
12. f(x) = 1 — x2>4 (—2 < x < 2), p = 4
13. 1
2
24.
–1 1
2 2
14. f(x) = cos px (—1 < x < 1), p = 1
2 2
15.
16.
17.
f(x) = x ƒ x ƒ (—1 < x < 1), p = 2
26. 
–
2
–
18. Rectifier. Find the Fourier series of the function
obtained by passing the voltage v(t) = V0 cos 100pt
through a half-wave rectifier that clips the negative
half-waves.
19. Trigonometric Identities. Show that the familiar
identities cos3 x = 3 cos x + 1 cos 3x and sin3 x = 3
27. 
–
2
28.

– 
2

– 
2
4 4 4
sin x — 1 sin 3x can be interpreted as Fourier series
expansions. Develop cos4 x.
20. Numeric Values. Using Prob. 11, show that 1 + 1 +
1
+ 1
+ Á = 1
p2
.
9 16 6
21. CAS PROJECT. Fourier Series of 2L-Periodic
Functions. (a) Write a program for obtaining partial
sums of a Fourier series (5).
29. f(x) = sin x (0 < x < p)
30. Obtain the solution to Prob. 26 from that of
Prob. 27.
2

–
– 
– 
–
2
4
–4 4
–4
1
–2 2
–1
L
L
25. 

1
2 4
1
4
23–29
1
1 1