Chapt_12

Prévia do material em texto

Part IV Fourier Series and Partial Differential Equations
1212 Orthogonal Functions andFourier Series
EXERCISES 12.1
Orthogonal Functions
1.
∫ 2
−2
xx2dx =
1
4
x4
∣∣∣∣ 2
−2
= 0
2.
∫ 1
−1
x3(x2 + 1)dx =
1
6
x6
∣∣∣∣ 1
−1
+
1
4
x4
∣∣∣∣ 1
−1
= 0
3.
∫ 2
0
ex(xe−x − e−x)dx =
∫ 2
0
(x− 1)dx =
(
1
2
x2 − x
) ∣∣∣∣2
0
= 0
4.
∫ π
0
cosx sin2 x dx =
1
3
sin3 x
∣∣∣∣π
0
= 0
5.
∫ π/2
−π/2
x cos 2x dx =
1
2
(
1
2
cos 2x + x sin 2x
) ∣∣∣∣π/2
−π/2
= 0
6.
∫ 5π/4
π/4
ex sinx dx =
(
1
2
ex sinx− 1
2
ex cosx
) ∣∣∣∣5π/4
π/4
= 0
7. For m �= n
∫ π/2
0
sin(2n + 1)x sin(2m + 1)x dx
=
1
2
∫ π/2
0
(
cos 2(n−m)x− cos 2(n + m + 1)x
)
dx
=
1
4(n−m) sin 2(n−m)x
∣∣∣∣π/2
0
− 1
4(n + m + 1)
sin 2(n + m + 1)x
∣∣∣∣π/2
0
= 0.
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12.1 Orthogonal Functions
For m = n ∫ π/2
0
sin2(2n + 1)x dx =
∫ π/2
0
(
1
2
− 1
2
cos 2(2n + 1)x
)
dx
=
1
2
x
∣∣∣∣π/2
0
− 1
4(2n + 1)
sin 2(2n + 1)x
∣∣∣∣π/2
0
=
π
4
so that
‖ sin(2n + 1)x‖ = 1
2
√
π .
8. For m �= n∫ π/2
0
cos(2n + 1)x cos(2m + 1)x dx
=
1
2
∫ π/2
0
(
cos 2(n−m)x + cos 2(n + m + 1)x
)
dx
=
1
4(n−m) sin 2(n−m)x
∣∣∣∣π/2
0
+
1
4(n + m + 1)
sin 2(n + m + 1)x
∣∣∣∣π/2
0
= 0.
For m = n ∫ π/2
0
cos2(2n + 1)x dx =
∫ π/2
0
(
1
2
+
1
2
cos 2(2n + 1)x
)
dx
=
1
2
x
∣∣∣∣π/2
0
+
1
4(2n + 1)
sin 2(2n + 1)x
∣∣∣∣π/2
0
=
π
4
so that
‖ cos(2n + 1)x‖ = 1
2
√
π .
9. For m �= n ∫ π
0
sinnx sinmxdx =
1
2
∫ π
0
(
cos(n−m)x− cos(n + m)x
)
dx
=
1
2(n−m) sin(n−m)x
∣∣∣∣π
0
− 1
2(n + m)
sin(n + m)x
∣∣∣∣π
0
= 0.
For m = n ∫ π
0
sin2 nx dx =
∫ π
0
(
1
2
− 1
2
cos 2nx
)
dx =
1
2
x
∣∣∣∣π
0
− 1
4n
sin 2nx
∣∣∣∣π
0
=
π
2
so that
‖ sinnx‖ =
√
π
2
.
10. For m �= n∫ p
0
sin
nπ
p
x sin
mπ
p
xdx =
1
2
∫ p
0
(
cos
(n−m)π
p
x− cos (n + m)π
p
x
)
dx
=
p
2(n−m)π sin
(n−m)π
p
x
∣∣∣∣p
0
− p
2(n + m)π
sin
(n + m)π
p
x
∣∣∣∣p
0
= 0.
For m = n ∫ p
0
sin2
nπ
p
x dx =
∫ p
0
(
1
2
− 1
2
cos
2nπ
p
x
)
dx =
1
2
x
∣∣∣∣p
0
− p
4nπ
sin
2nπ
p
x
∣∣∣∣p
0
=
p
2
635
12.1 Orthogonal Functions
so that ∥∥∥∥sin nπp x
∥∥∥∥ = √p2 .
11. For m �= n∫ p
0
cos
nπ
p
x cos
mπ
p
xdx =
1
2
∫ p
0
(
cos
(n−m)π
p
x + cos
(n + m)π
p
x
)
dx
=
p
2(n−m)π sin
(n−m)π
p
x
∣∣∣∣p
0
+
p
2(n + m)π
sin
(n + m)π
p
x
∣∣∣∣p
0
= 0.
For m = n ∫ p
0
cos2
nπ
p
x dx =
∫ p
0
(
1
2
+
1
2
cos
2nπ
p
x
)
dx =
1
2
x
∣∣∣∣p
0
+
p
4nπ
sin
2nπ
p
x
∣∣∣∣p
0
=
p
2
.
Also ∫ p
0
1 · cos nπ
p
x dx =
p
nπ
sin
nπ
p
x
∣∣∣∣p
0
= 0 and
∫ p
0
12dx = p
so that
‖1‖ = √p and
∥∥∥∥cos nπp x
∥∥∥∥ = √p2 .
12. For m �= n, we use Problems 11 and 10:∫ p
−p
cos
nπ
p
x cos
mπ
p
xdx = 2
∫ p
0
cos
nπ
p
x cos
mπ
p
xdx = 0
∫ p
−p
sin
nπ
p
x sin
mπ
p
xdx = 2
∫ p
0
sin
nπ
p
x sin
mπ
p
xdx = 0.
Also ∫ p
−p
sin
nπ
p
x cos
mπ
p
xdx =
1
2
∫ p
−p
(
sin
(n−m)π
p
x + sin
(n + m)π
p
x
)
dx = 0,
∫ p
−p
1 · cos nπ
p
x dx =
p
nπ
sin
nπ
p
x
∣∣∣∣p
−p
= 0,
∫ p
−p
1 · sin nπ
p
x dx = − p
nπ
cos
nπ
p
x
∣∣∣∣p
−p
= 0,
and ∫ p
−p
sin
nπ
p
x cos
nπ
p
x dx =
∫ p
−p
1
2
sin
2nπ
p
x dx = − p
4nπ
cos
2nπ
p
x
∣∣∣∣p
−p
= 0.
For m = n ∫ p
−p
cos2
nπ
p
x dx =
∫ p
−p
(
1
2
+
1
2
cos
2nπ
p
x
)
dx = p,
∫ p
−p
sin2
nπ
p
x dx =
∫ p
−p
(
1
2
− 1
2
cos
2nπ
p
x
)
dx = p,
and ∫ p
−p
12dx = 2p
so that
‖1‖ =
√
2p ,
∥∥∥∥cos nπp x
∥∥∥∥ = √p , and ∥∥∥∥sin nπp x
∥∥∥∥ = √p .
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12.1 Orthogonal Functions
13. Since ∫ ∞
−∞
e−x
2 · 1 · 2x dx = −e−x2
∣∣∣∣0
−∞
− e−x2
∣∣∣∣∞
0
= 0,
∫ ∞
−∞
e−x
2 · 1 · (4x2 − 2) dx = 2
∫ ∞
−∞
x
(
2xe−x
2
)
dx− 2
∫ ∞
−∞
e−x
2
dx
= 2
(
−xe−x2
∣∣∣∣∞
−∞
+
∫ ∞
−∞
e−x
2
dx
)
− 2
∫ ∞
−∞
e−x
2
dx
= 2
(
−xe−x2
∣∣∣∣0
−∞
−xe−x2
∣∣∣∣∞
0
)
= 0,
and ∫ ∞
−∞
e−x
2 · 2x · (4x2 − 2) dx = 4
∫ ∞
−∞
x2
(
2xe−x
2
)
dx− 4
∫ ∞
−∞
xe−x
2
dx
= 4
(
−x2e−x2
∣∣∣∣∞
−∞
+ 2
∫ ∞
−∞
xe−x
2
dx
)
− 4
∫ ∞
−∞
xe−x
2
dx
= 4
(
−x2e−x2
∣∣∣∣0
−∞
− x2e−x2
∣∣∣∣∞
0
)
+ 2
∫ ∞
−∞
2xe−x
2
dx = 0,
the functions are orthogonal.
14. Since ∫ ∞
0
e−x · 1(1− x) dx = (x− 1)e−x
∣∣∣∣∞
0
−
∫ ∞
0
e−xdx = 0,
∫ ∞
0
e−x · 1 ·
(
1
2
x2 − 2x + 1
)
dx =
(
2x− 1− 1
2
x2
)
e−x
∣∣∣∣∞
0
+
∫ ∞
0
e−x(x− 2) dx
= 1 + (2− x)e−x
∣∣∣∣∞
0
+
∫ ∞
0
e−xdx = 0,
and ∫ ∞
0
e−x · (1− x)
(
1
2
x2−2x + 1
)
dx
=
∫ ∞
0
e−x
(
−1
2
x3 +
5
2
x2 − 3x + 1
)
dx
= e−x
(
1
2
x3 − 5
2
x2 + 3x− 1
) ∣∣∣∣∞
0
+
∫ ∞
0
e−x
(
−3
2
x2 + 5x− 3
)
dx
= 1 + e−x
(
3
2
x2 − 5x + 3
) ∣∣∣∣∞
0
+
∫ ∞
0
e−x(5− 3x) dx
= 1− 3 + e−x(3x− 5)
∣∣∣∣∞
0
−3
∫ ∞
0
e−xdx = 0,
the functions are orthogonal.
15. By orthogonality
∫ b
a
φ0(x)φn(x)dx = 0 for n = 1, 2, 3, . . . ; that is,
∫ b
a
φn(x)dx = 0 for n = 1, 2, 3, . . . .
637
12.1 Orthogonal Functions
16. Using the facts that φ0 and φ1 are orthogonal to φn for n > 1, we have∫ b
a
(αx + β)φn(x) dx = α
∫ b
a
xφn(x) dx + β
∫ b
a
1 · φn(x) dx
= α
∫ b
a
φ1(x)φn(x) dx + β
∫ b
a
φ0(x)φn(x) dx
= α · 0 + β · 0 = 0
for n = 2, 3, 4, . . . .
17. Using the fact that φn and φm are orthogonal for n �= m we have
‖φm(x) + φn(x)‖2 =
∫ b
a
[φm(x) + φn(x)]2dx =
∫ b
a
[
φ2m(x) + 2φm(x)φn(x) + φ
2
n(x)
]
dx
=
∫ b
a
φ2m(x)dx + 2
∫ b
a
φm(x)φn(x)dx +
∫ b
a
φ2n(x) dx
= ‖φm(x)‖2 + ‖φn(x)‖2 .
18. Setting
0 =
∫ 2
−2
f3(x)f1(x) dx =
∫ 2
−2
(
x2 + c1x3 + c2x4
)
dx =
16
3
+
64
5
c2
and
0 =
∫ 2
−2
f3(x)f2(x) dx =
∫ 2
−2
(
x3 + c1x4 + c2x5
)
dx =
64
5
c1
we obtain c1 = 0 and c2 = −5/12.
19. Since sinnx is an odd function on [−π, π],
(1, sinnx) =
∫ π
−π
sinnx dx = 0
and f(x) = 1 is orthogonal to every member of {sinnx}. Thus {sinnx} is not complete.
20. (f1 + f2, f3) =
∫ b
a
[f1(x) + f2(x)]f3(x) dx =
∫ b
a
f1(x)f3(x) dx +
∫ b
a
f2(x)f3(x) dx = (f1, f3) + (f2, f3)
21. (a) The fundamental period is 2π/2π = 1.
(b) The fundamental period is 2π/(4/L) = 12πL.
(c) The fundamental period of sinx + sin 2x is 2π.
(d) The fundamental period of sin 2x + cos 4x is 2π/2 = π.
(e) The fundamental period of sin 3x+cos 4x is 2π since the smallest integer multiples of 2π/3 and 2π/4 = π/2
that are equal are 3 and 4, respectively.
(f) The fundamental period of f(x) is 2π/(nπ/p) = 2p/n.
22. (a) Following the pattern established by φ1(x) and φ2(x) we have
φ3(x) = f3(x)− (f3, φ0)(φ0, φ0) φ0(x)−
(f3, φ1)
(φ1, φ1)
φ1(x)− (f3, φ2)(φ2, φ2) φ2(x).
638
12.1 Orthogonal Functions
(b) To show mutual orthogonality we compute (φ0, φ1), (φ0, φ2), and (φ1, φ2) using properties (i), (ii), and (iii)
from this section in the text.
(φ0, φ1) =
(
φ0, f1 − (f1, φ0)(φ0, φ0) φ0
)
= (φ0, f1)− (f1, φ0)(φ0, φ0) (φ0, φ0) = (φ0, f1)− (f1, φ0) = 0
(φ0, φ2) =
(
φ0, f2 − (f2, φ0)(φ0, φ0) φ0 −
(f2, φ1)
(φ1, φ1)
φ1
)
= (φ0, f2)− (f2, φ0)(φ0, φ0) (φ0, φ0)−
(f2,φ1)
(φ1, φ1)
(φ0, φ1)
= (φ0, f2)− (f2, φ0)− 0 = 0
(φ1, φ2) =
(
φ1, f2 − (f2, φ0)(φ0, φ0) φ0 −
(f2, φ1)
(φ1, φ1)
φ1
)
= (φ1, f2)− (f2, φ0)(φ0, φ0) (φ1, φ0)−
(f2, φ1)
(φ1, φ1)
(φ1, φ1)
= (φ1, f2)− 0− (f2, φ1) = 0.
23. (a) First we identify f0(x) = 1, f1(x) = x, f2(x) = x2, and f3(x) = x3. Then, we use the formulas from
Problem 22. First, we have φ0(x) = f0(x) = 1. Then
(f1, φ0) = (x, 1) =
∫ 1
−1
x dx = 0 and (φ0, φ0) =
∫ 1
−1
1 dx = 2,
so
φ1(x) = f1(x)− (f1, φ0)(φ0, φ0) φ0(x) = x−
0
2
(1) = 1.
Next
(f2, φ0) = (x2, 1) =
∫ 1
−1
x2dx =
2
3
, (f2, φ1) = (x2, x) =
∫ 1
−1
x3dx = 0, and (φ1, φ1) =
∫ 1
−1
x2dx =
2
3
,
so
φ2(x) = f2(x)− (f2, φ0)(φ0, φ0) φ0(x)−
(f2, φ1)
(φ1, φ1)
φ1(x) = x2 − 2/32 (1)−
0
2
(x) = x2 − 1
3
.
Finally,
(f3, φ0) = (x3, 1) =
∫ 1
−1
x3dx = 0, (f3, φ1) = (x3, x) =
∫ 1
−1
x4dx =
2
5
,
and
(f3, φ2) =
(
x3, x2 − 1
3
)
=
∫ 1
−1
(
x5 − 1
3
x3
)
dx = 0,
so
φ3(x) = f3(x)− (f3, φ0)(φ0, φ0) φ0(x)−
(f3, φ1)
(φ1, φ1)
φ1(x)− (f3, φ2)(φ2, φ2) φ2(x) = x
3 − 0− 2/5
2/3
(x)− 0 = x3 − 3
5
x.
(b) Recall from Section 5.3 that the first four Legendre polynomials are P0(x) = 1, P1(x) = x, P2(x) = 32 x
2− 12 ,
and P3(x) = 52 x
3 − 32 x. We then see that φ0(x) = P0(x), φ1(x) = P1(x), φ2(x) = x2 − 13 = 23 ( 32 x2 − 12 ) =
2
3 P2(x), and φ3(x) = x
3 − 35 x = 25 ( 52 x3 − 32 x) = 25 P3(x).
24. (i): (f1, f2) =
∫ b
a
f1(x)f2(x)dx =
∫ b
a
f2(x)f1(x)dx = (f2, f1).
(ii): (kf1, f2) =
∫ b
a
kf1(x)f2(x)dx = k
∫ b
a
f1(x)f2(x)dx = k(f1, f2).
(iii): If f1(x) = 0 then (f1, f1) =
∫ b
a
0 dx = 0; if f1(x) �= 0 then (f1, f1) =
∫ b
a
[f1(x)]2dx > 0 since [f1(x)]2 > 0.
(iv): (f1 + f2, f3) =
∫ b
a
[f1(x) + f2(x)]f3(x)dx =
∫ b
a
[f1(x)f3(x) + f2(x)f3(x)]dx
=
∫ b
a
f1(x)f3(x)dx +
∫ b
a
f2(x)f3(x)dx = (f1, f3) + (f2, f3).
639
12.1 Orthogonal Functions
25. In R3 the set {i, j} is not complete since k is orthogonal to both i and j. The set {i, j,k} is complete. To see
this suppose that ai + bj + ck is orthogonal to i, j, and k. Then
0 = (ai + bj + ck, i) = a(i, i) + b(j, i) + c(k, i) = a(1) + b(0) + c(0) = a.
Similarly, b = 0 and c = 0. Thus, the only vector in R3 orthogonal to i, j, and k is 0, so {i, j,k} is complete.
EXERCISES 12.2
Fourier Series
1. a0 =
1
π
∫ π
−π
f(x) dx =
1
π
∫ π
0
1 dx = 1
an =
1
π
∫ π
−π
f(x) cos
nπ
π
x dx =
1
π
∫ π
0
cosnx dx = 0
bn =
1
π
∫ π
−π
f(x) sin
nπ
π
x dx =
1
π
∫ π
0
sinnx dx =
1
nπ
(1− cosnπ) = 1
nπ
[1− (−1)n]
f(x) =
1
2
+
1
π
∞∑
n=1
1− (−1)n
n
sinnx
2. a0 =
1
π
∫ π
−π
f(x) dx =
1
π
∫ 0
−π
(−1) dx + 1
π
∫ π
0
2 dx = 1
an =
1
π
∫ π
−π
f(x) cosnx dx =
1
π
∫ 0
−π
− cosnx dx + 1
π
∫ π
0
2 cosnx dx = 0
bn =
1
π
∫ π
−π
f(x) sinnx dx =
1
π
∫ 0
−π
− sinnx dx + 1
π
∫ π
0
2 sinnx dx =
3
nπ
[1− (−1)n]
f(x) =
1
2
+
3
π
∞∑
n=1
1− (−1)n
n
sinnx
3. a0 =
∫ 1
−1
f(x) dx =
∫ 0
−1
1 dx +
∫ 1
0
x dx =
3
2
an =
∫ 1
−1
f(x) cosnπx dx =
∫ 0
−1
cosnπx dx +
∫ 1
0
x cosnπx dx =
1
n2π2
[(−1)n − 1]
bn =
∫ 1
−1
f(x) sinnπx dx =
∫ 0
−1
sinnπx dx +
∫ 1
0
x sinnπx dx = − 1
nπ
f(x) =
3
4
+
∞∑
n=1
[
(−1)n − 1
n2π2
cosnπx− 1
nπ
sinnπx
]
4. a0 =
∫ 1
−1
f(x) dx =
∫ 1
0
x dx =
1
2
an =
∫ 1
−1
f(x) cosnπx dx =
∫ 1
0
x cosnπx dx =
1
n2π2
[(−1)n − 1]
bn =
∫ 1
−1
f(x) sinnπx dx =
∫ 1
0
x sinnπx dx =
(−1)n+1
nπ
640
12.2 Fourier Series
f(x) =
1
4
+
∞∑
n=1
[
(−1)n − 1
n2π2
cosnπx +
(−1)n+1
nπ
sinnπx
]
5. a0 =
1
π
∫ π
−π
f(x) dx =
1
π
∫ π
0
x2 dx =
1
3
π2
an =
1
π
∫ π
−π
f(x) cosnx dx =
1
π
∫ π
0
x2 cosnx dx =
1
π
(
x2
π
sinnx
∣∣∣∣π
0
− 2
n
∫ π
0
x sinnx dx
)
=
2(−1)n
n2
bn =
1
π
∫ π
0
x2 sinnx dx =
1
π
(
−x
2
n
cosnx
∣∣∣∣π
0
+
2
n
∫ π
0
x cosnx dx
)
=
π
n
(−1)n+1 + 2
n3π
[(−1)n − 1]
f(x) =
π2
6
+
∞∑
n=1
[
2(−1)n
n2
cosnx +
(
π
n
(−1)n+1 + 2[(−1)
n − 1]
n3π
)
sinnx
]
6. a0 =
1
π
∫ π
−π
f(x) dx =
1
π
∫ 0
−π
π2 dx +
1
π
∫ π
0
(
π2 − x2) dx = 5
3
π2
an =
1
π
∫ π
−π
f(x) cosnx dx =
1
π
∫ 0
−π
π2 cosnx dx +
1
π
∫ π
0
(
π2 − x2) cosnx dx
=
1
π
(
π2 − x2
n
sinnx
∣∣∣∣π
0
+
2
n
∫ π
0
x sinnx dx
)
=
2
n2
(−1)n+1
bn =
1
π
∫ π
−π
f(x) sinnx dx =
1
π
∫ 0
−π
π2 sinnx dx +
1
π
∫ π
0
(
π2 − x2) sinnx dx
=
π
n
[(−1)n − 1] + 1
π
(
x2 − π2
n
cosnx
∣∣∣∣π
0
− 2
n
∫ π
0
x cosnx dx
)
=
π
n
(−1)n + 2
n3π
[1− (−1)n]
f(x) =
5π2
6
+
∞∑
n=1
[
2
n2
(−1)n+1 cosnx +
(
π
n
(−1)n + 2[1− (−1)
n]
n3π
)
sinnx
]
7. a0 =
1
π
∫ π
−π
f(x) dx =
1
π
∫ π
−π
(x + π) dx = 2π
an =
1
π
∫ π
−π
f(x) cosnx dx =
1
π
∫ π
−π
(x + π) cosnx dx = 0
bn =
1
π
∫ π
−π
f(x) sinnx dx =
2
n
(−1)n+1
f(x) = π +
∞∑
n=1
2
n
(−1)n+1 sinnx
8. a0 =
1
π
∫ π
−π
f(x) dx =
1
π
∫ π
−π
(3− 2x) dx = 6
an =
1
π
∫ π
−π
f(x) cosnx dx =
1
π
∫ π
−π
(3− 2x) cosnx dx = 0
bn =
1
π
∫ π
−π
(3− 2x) sinnx dx = 4
n
(−1)n
f(x) = 3 + 4
∞∑
n=1
(−1)n
n
sinnx
9. a0 =
1
π
∫ π
−π
f(x) dx =
1
π
∫ π
0
sinx dx =
2
π
an =
1
π
∫ π
−π
f(x) cosnx dx =
1
π
∫ π
0
sinx cosnx dx =
1
2π
∫ π
0
(
sin(1 + n)x + sin(1− n)x
)
dx
641
12.2 Fourier Series
=
1 + (−1)n
π(1− n2) for n = 2, 3, 4, . . .
a1 =
1
2π
∫ π
0
sin 2x dx = 0
bn =
1
π
∫ π
−π
f(x) sinnx dx =
1
π
∫ π
0
sinx sinnx dx
=
1
2π
∫ π
0
(
cos(1− n)x− cos(1 + n)x
)
dx = 0 for n = 2, 3, 4, . . .
b1 =
1
2π
∫ π
0
(1− cos 2x) dx = 1
2
f(x) =
1
π
+
1
2
sinx +
∞∑
n=2
1 + (−1)n
π(1− n2) cosnx
10. a0 =
2
π
∫ π/2
−π/2
f(x) dx =
2
π
∫ π/2
0
cosx dx =
2
π
an =
2
π
∫ π/2
−π/2
f(x) cos 2nx dx =
2
π
∫ π/2
0
cosx cos 2nx dx =
1
π
∫ π/2
0
(
cos(2n− 1)x + cos(2n + 1)x
)
dx
=
2(−1)n+1
π(4n2 − 1)
bn =
2
π
∫ π/2
−π/2
f(x) sin 2nx dx =
2
π
∫ π/2
0
cosx sin 2nx dx =
1
π
∫ π/2
0
(
sin(2n− 1)x + sin(2n + 1)x
)
dx
=
4n
π(4n2 − 1)
f(x) =
1
π
+
∞∑
n=1
[
2(−1)n+1
π(4n2 − 1) cos 2nx +
4n
π(4n2 − 1) sin 2nx
]
11. a0 =
1
2
∫ 2
−2
f(x) dx =
1
2
(∫ 0
−1
−2 dx +
∫ 1
0
1 dx
)
= −1
2
an =
1
2
∫ 2
−2
f(x) cos
nπ
2
x dx =
1
2
(∫ 0
−1
(−2) cos nπ
2
x dx +
∫ 1
0
cos
nπ
2
x dx
)
= − 1
nπ
sin
nπ
2
bn =
1
2
∫ 2
−2
f(x) sin
nπ
2
x dx =
1
2
(∫ 0
−1
(−2) sin nπ
2
x dx +
∫ 1
0
sin
nπ
2
x dx
)
=
3
nπ
(
1− cos nπ
2
)
f(x) = −1
4
+
∞∑
n=1
[
− 1
nπ
sin
nπ
2
cos
nπ
2
x +
3
nπ
(
1− cos nπ
2
)
sin
nπ
2
x
]
12. a0 =
1
2
∫ 2
−2
f(x) dx =
1
2
(∫ 1
0
x dx +
∫ 2
1
1 dx
)
=
3
4
an =
1
2
∫ 2
−2
f(x) cos
nπ
2
x dx =
1
2
(∫ 1
0
x cos
nπ
2
x dx +
∫ 2
1
cos
nπ
2
x dx
)
=
2
n2π2
(
cos
nπ
2
− 1
)
bn =
1
2
∫ 2
−2
f(x) sin
nπ
2
x dx =
1
2
(∫ 1
0
x sin
nπ
2
x dx +
∫ 2
1
sin
nπ
2
x dx
)
=
2
n2π2
(
sin
nπ
2
+
nπ
2
(−1)n+1
)
f(x) =
3
8
+
∞∑
n=1
[
2
n2π2
(
cos
nπ
2
− 1
)
cos
nπ2
x +
2
n2π2
(
sin
nπ
2
+
nπ
2
(−1)n+1
)
sin
nπ
2
x
]
642
12.2 Fourier Series
13. a0 =
1
5
∫ 5
−5
f(x) dx =
1
5
(∫ 0
−5
1 dx +
∫ 5
0
(1 + x) dx
)
=
9
2
an =
1
5
∫ 5
−5
f(x) cos
nπ
5
x dx =
1
5
(∫ 0
−5
cos
nπ
5
x dx +
∫ 5
0
(1 + x) cos
nπ
5
x dx
)
=
5
n2π2
[(−1)n − 1]
bn =
1
5
∫ 5
−5
f(x) sin
nπ
5
x dx =
1
5
(∫ 0
−5
sin
nπ
5
x dx +
∫ 5
0
(1 + x) cos
nπ
5
x dx
)
=
5
nπ
(−1)n+1
f(x) =
9
4
+
∞∑
n=1
[
5
n2π2
[(−1)n − 1] cos nπ
5
x +
5
nπ
(−1)n+1 sin nπ
5
x
]
14. a0 =
1
2
∫ 2
−2
f(x) dx =
1
2
(∫ 0
−2
(2 + x) dx +
∫ 2
0
2 dx
)
= 3
an =
1
2
∫ 2
−2
f(x) cos
nπ
2
x dx =
1
2
(∫ 0
−2
(2 + x) cos
nπ
2
x dx +
∫ 2
0
2 cos
nπ
2
x dx
)
=
2
n2π2
[1− (−1)n]
bn =
1
2
∫ 2
−2
f(x) sin
nπ
2
x dx =
1
2
(∫ 0
−2
(2 + x) sin
nπ
2
x dx +
∫ 2
0
2 sin
nπ
2
x dx
)
=
2
nπ
(−1)n+1
f(x) =
3
2
+
∞∑
n=1
[
2
n2π2
[1− (−1)n] cos nπ
2
x +
2
nπ
(−1)n+1 sin nπ
2
x
]
15. a0 =
1
π
∫ π
−π
f(x) dx =
1
π
∫ π
−π
ex dx =
1
π
(eπ − e−π)
an =
1
π
∫ π
−π
f(x) cosnx dx =
(−1)n(eπ − e−π)
π(1 + n2)
bn =
1
π
∫ π
−π
f(x) sinnx dx =
1
π
∫ π
−π
ex sinnx dx =
(−1)nn(e−π − eπ)
π(1 + n2)
f(x) =
eπ − e−π
2π
+
∞∑
n=1
[
(−1)n(eπ − e−π)
π(1 + n2)
cosnx +
(−1)nn(e−π − eπ)
π(1 + n2)
sinnx
]
16. a0 =
1
π
∫ π
−π
f(x) dx =
1
π
∫ π
0
(ex − 1) dx = 1
π
(eπ − π − 1)
an =
1
π
∫ π
−π
f(x) cosnx dx =
1
π
∫ π
0
(ex − 1) cosnx dx = [e
π(−1)n − 1]
π(1 + n2)
bn =
1
π
∫ π
−π
f(x) sinnx dx =
1
π
∫ π
0
(ex − 1) sinnx dx = 1
π
(
neπ(−1)n+1
1 + n2
+
n
1 + n2
+
(−1)n
n
− 1
n
)
f(x) =
eπ − π − 1
2π
+
∞∑
n=1
[
eπ(−1)n − 1
π(1 + n2)
cosnx +
(
n
1 + n2
[
eπ(−1)n+1 + 1] + (−1)n − 1
n
)
sinnx
]
17. The function in Problem 5 is discontinuous at x = π, so the corresponding Fourier series converges to π2/2 at
x = π. That is,
π2
2
=
π2
6
+
∞∑
n=1
[
2(−1)n
n2
cosnπ +
(
π
n
(−1)n+1 + 2[(−1)
n − 1]
n3π
)
sinnπ
]
=
π2
6
+
∞∑
n=1
2(−1)n
n2
(−1)n = π
2
6
+
∞∑
n=1
2
n2
=
π2
6
+ 2
(
1 +
1
22
+
1
32
+ · · ·
)
and
π2
6
=
1
2
(
π2
2
− π
2
6
)
= 1 +
1
22
+
1
32
+ · · · .
643
12.2 Fourier Series
At x = 0 the series converges to 0 and
0 =
π2
6
+
∞∑
n=1
2(−1)n
n2
=
π2
6
+ 2
(
−1 + 1
22
− 1
32
+
1
42
− · · ·
)
so
π2
12
= 1− 1
22
+
1
32
− 1
42
+ · · · .
18. From Problem 17
π2
8
=
1
2
(
π2
6
+
π2
12
)
=
1
2
(
2 +
2
32
+
2
52
+ · · ·
)
= 1 +
1
32
+
1
52
+ · · · .
19. The function in Problem 7 is continuous at x = π/2 so
3π
2
= f
(π
2
)
= π +
∞∑
n=1
2
n
(−1)n+1 sin nπ
2
= π + 2
(
1− 1
3
+
1
5
− 1
7
+ · · ·
)
and
π
4
= 1− 1
3
+
1
5
− 1
7
+ · · · .
20. The function in Problem 9 is continuous at x = π/2 so
1 = f
(π
2
)
=
1
π
+
1
2
+
∞∑
n=2
1 + (−1)n
π(1− n2) cos
nπ
2
1 =
1
π
+
1
2
+
2
3π
− 2
3 · 5π +
2
5 · 7π − · · ·
and
π = 1 +
π
2
+
2
3
− 2
3 · 5 +
2
5 · 7 − · · ·
or
π
4
=
1
2
+
1
1 · 3 −
1
3 · 5 +
1
5 · 7 − · · · .
21. Writing
f(x) =
a0
2
+ a1 cos
π
p
x + · · ·+ an cos nπ
p
x + · · ·+ b1 sin π
p
x + · · ·+ bn sin nπ
p
x + · · ·
we see that f2(x) consists exclusively of squared terms of the form
a20
4
, a2n cos
2 nπ
p
x, b2n sin
2 nπ
p
x
and cross-product terms, with m �= n, of the form
a0an cos
nπ
p
x, a0bn sin
nπ
p
x, 2aman cos
mπ
p
x cos
nπ
p
x,
2ambn cos
mπ
p
x sin
nπ
p
x, 2bmbn sin
mπ
p
x sin
nπ
p
x.
The integral of each cross-product term taken over the interval (−p, p) is zero by orthogonality. For the squared
terms we have
a20
4
∫ p
−p
dx =
a20p
2
, a2n
∫ p
−p
cos2
nπ
p
x dx = a2np, b
2
n
∫ p
−p
sin2
nπ
p
x dx = b2np.
Thus
RMS(f) =
√√√√1
4
a20 +
1
2
∞∑
n=1
(a2n + b2n) .
644
12.3 Fourier Cosine and Sine Series
EXERCISES 12.3
Fourier Cosine and Sine Series
1. Since f(−x) = sin(−3x) = − sin 3x = −f(x), f(x) is an odd function.
2. Since f(−x) = −x cos(−x) = −x cosx = −f(x), f(x) is an odd function.
3. Since f(−x) = (−x)2 − x = x2 − x, f(x) is neither even nor odd.
4. Since f(−x) = (−x)3 + 4x = −(x3 − 4x) = −f(x), f(x) is an odd function.
5. Since f(−x) = e|−x| = e|x| = f(x), f(x) is an even function.
6. Since f(−x) = e−x − ex = −f(x), f(x) is an odd function.
7. For 0 < x < 1, f(−x) = (−x)2 = x2 = −f(x), f(x) is an odd function.
8. For 0 ≤ x < 2, f(−x) = −x + 5 = f(x), f(x) is an even function.
9. Since f(x) is not defined for x < 0, it is neither even nor odd.
10. Since f(−x) = ∣∣(−x)5∣∣ = ∣∣x5∣∣ = f(x), f(x) is an even function.
11. Since f(x) is an odd function, we expand in a sine series:
bn =
2
π
∫ π
0
1 · sinnx dx = 2
nπ
[1− (−1)n] .
Thus
f(x) =
∞∑
n=1
2
nπ
[1− (−1)n] sinnx.
12. Since f(x) is an even function, we expand in a cosine series:
a0 =
∫ 2
1
1 dx = 1
an =
∫ 2
1
cos
nπ
2
x dx = − 2
nπ
sin
nπ
2
.
Thus
f(x) =
1
2
+
∞∑
n=1
−2
nπ
sin
nπ
2
cos
nπ
2
x.
13. Since f(x) is an even function, we expand in a cosine series:
a0 =
2
π
∫ π
0
x dx = π
an =
2
π
∫ π
0
x cosnx dx =
2
n2π
[(−1)n − 1].
Thus
f(x) =
π
2
+
∞∑
n=1
2
n2π
[(−1)n − 1] cosnx.
645
12.3 Fourier Cosine and Sine Series
14. Since f(x) is an odd function, we expand in a sine series:
bn =
2
π
∫ π
0
x sinnx dx =
2
n
(−1)n+1.
Thus
f(x) =
∞∑
n=1
2
n
(−1)n+1 sinnx.
15. Since f(x) is an even function, we expand in a cosine series:
a0 = 2
∫ 1
0
x2 dx =
2
3
an = 2
∫ 1
0
x2 cosnπx dx = 2
(
x2
nπ
sinnπx
∣∣∣∣1
0
− 2
nπ
∫ 1
0
x sinnπx dx
)
=
4
n2π2
(−1)n.
Thus
f(x) =
1
3
+
∞∑
n=1
4
n2π2
(−1)n cosnπx.
16. Since f(x) is an odd function, we expand in a sine series:
bn = 2
∫ 1
0
x2 sinnπx dx = 2
(
− x
2
nπ
cosnπx
∣∣∣∣1
0
+
2
nπ
∫ 1
0
x cosnπx dx
)
=
2(−1)n+1
nπ
+
4
n3π3
[(−1)n − 1].
Thus
f(x) =
∞∑
n=1
(
2(−1)n+1
nπ
+
4
n3π3
[(−1)n − 1]
)
sinnπx.
17. Since f(x) is an even function, we expand in a cosine series:
a0 =
2
π
∫ π
0
(π2 − x2) dx = 4
3
π2
an =
2
π
∫ π
0
(π2 − x2) cosnx dx = 2
π
(
π2 − x2
n
sinnx
∣∣∣∣π
0
+
2
n
∫ π
0
x sinnx dx
)
=
4
n2
(−1)n+1.
Thus
f(x) =
2
3
π2 +
∞∑
n=1
4
n2
(−1)n+1 cosnx dx.
18. Since f(x) is an odd function, we expand in a sine series:
bn =
2
π
∫ π
0
x3 sinnx dx =
2
π
(
−x
3
n
cosnx
∣∣∣∣π
0
+
3
n
∫ π
0
x2 cosnx dx
)
=
2π2
n
(−1)n+1 − 12
n2π
∫ π
0
x sinnx dx
=
2π2
n
(−1)n+1 − 12
n2π
(
−x
n
cosnx
∣∣∣∣π
0
+
1
n
∫ π
0
cosnx dx
)
=
2π2
n
(−1)n+1 + 12
n3
(−1)n.
Thus
f(x) =
∞∑
n=1
(
2π2
n
(−1)n+1 + 12
n3
(−1)n
)
sinnx.
19. Since f(x) is an odd function, we expand in a sine series:
bn =
2
π
∫ π
0
(x + 1) sinnx dx =
2(π + 1)
nπ
(−1)n+1 + 2
nπ
.
646
12.3 Fourier Cosine and Sine Series
Thus
f(x) =
∞∑
n=1
(
2(π+ 1)
nπ
(−1)n+1 + 2
nπ
)
sinnx.
20. Since f(x) is an odd function, we expand in a sine series:
bn = 2
∫ 1
0
(x− 1) sinnπx dx = 2
[∫ 1
0
x sinnπx dx−
∫ 1
0
sinnπx dx
]
= 2
[
1
n2π2
sinnπx− x
nπ
cosnπx +
1
nπ
cosnπx
]1
0
= − 2
nπ
.
Thus
f(x) = −
∞∑
n=1
2
nπ
sinnπx.
21. Since f(x) is an even function, we expand in a cosine series:
a0 =
∫ 1
0
x dx +
∫ 2
1
1 dx =
3
2
an =
∫ 1
0
x cos
nπ
2
x dx +
∫ 2
1
cos
nπ
2
x dx =
4
n2π2
(
cos
nπ
2
− 1
)
.
Thus
f(x) =
3
4
+
∞∑
n=1
4
n2π2
(
cos
nπ
2
− 1
)
cos
nπ
2
x.
22. Since f(x) is an odd function, we expand in a sine series:
bn =
1
π
∫ π
0
x sin
n
2
x dx +
∫ 2π
π
π sin
n
2
x dx =
4
n2π
sin
nπ
2
+
2
n
(−1)n+1.
Thus
f(x) =
∞∑
n=1
(
4
n2π
sin
nπ
2
+
2
n
(−1)n+1
)
sin
n
2
x.
23. Since f(x) is an even function, we expand in a cosine series:
a0 =
2
π
∫ π
0
sinx dx =
4
π
an =
2
π
∫ π
0
sinx cosnx dx =
1
π
∫ π
0
(
sin(1 + n)x + sin(1− n)x
)
dx
=
2
π(1− n2) (1 + (−1)
n) for n = 2, 3, 4, . . .
a1 =
1
π
∫ π
0
sin 2x dx = 0.
Thus
f(x) =
2
π
+
∞∑
n=2
2[1 + (−1)n]
π(1− n2) cosnx.
647
12.3 Fourier Cosine and Sine Series
24. Since f(x) is an even function, we expand in a cosine series. [See the solution of Problem 10 in Exercises 12.2
for the computation of the integrals.]
a0 =
2
π/2
∫ π/2
0
cosx dx =
4
π
an =
2
π/2
∫ π/2
0
cosx cos
nπ
π/2
x dx =
4(−1)n+1
π (4n2 − 1)
Thus
f(x) =
2
π
+
∞∑
n=1
4(−1)n+1
π (4n2 − 1) cos 2nx.
25. a0 = 2
∫ 1/2
0
1 dx = 1
an = 2
∫ 1/2
0
1 · cosnπx dx = 2
nπ
sin
nπ
2
bn = 2
∫ 1/2
0
1 · sinnπx dx = 2
nπ
(
1− cos nπ
2
)
f(x) =
1
2
+
∞∑
n=1
2
nπ
sin
nπ
2
cosnπx
f(x) =
∞∑
n=1
2
nπ
(
1− cos nπ
2
)
sinnπx
26. a0 = 2
∫ 1
1/2
1 dx = 1
an = 2
∫ 1
1/2
1 · cosnπx dx = − 2
nπ
sin
nπ
2
bn = 2
∫ 1
1/2
1 · sinnπx dx = 2
nπ
(
cos
nπ
2
+ (−1)n+1
)
f(x) =
1
2
+
∞∑
n=1
(
− 2
nπ
sin
nπ
2
)
cosnπx
f(x) =
∞∑
n=1
2
nπ
(
cos
nπ
2
+ (−1)n+1
)
sinnπx
27. a0 =
4
π
∫ π/2
0
cosx dx =
4
π
an =
4
π
∫ π/2
0
cosx cos 2nx dx =
2
π
∫ π/2
0
[cos(2n + 1)x + cos(2n− 1)x] dx = 4(−1)
n
π(1− 4n2)
bn =
4
π
∫ π/2
0
cosx sin 2nx dx =
2
π
∫ π/2
0
[sin(2n + 1)x + sin(2n− 1)x] dx = 8n
π(4n2 − 1)
f(x) =
2
π
+
∞∑
n=1
4(−1)n
π(1− 4n2) cos 2nx
f(x) =
∞∑
n=1
8n
π(4n2 − 1) sin 2nx
648
12.3 Fourier Cosine and Sine Series
28. a0 =
2
π
∫ π
0
sinx dx =
4
π
an =
2
π
∫ π
0
sinx cosnx dx =
1
π
∫ π
0
[sin(n + 1)x− sin(n− 1)x] dx = 2[(−1)
n + 1]
π(1− n2) for n = 2, 3, 4, . . .
bn =
2
π
∫ π
0
sinx sinnx dx =
1
π
∫ π
0
[cos(n− 1)x− cos(n + 1)x] dx = 0 for n = 2, 3, 4, . . .
a1 =
1
π
∫ π
0
sin 2x dx = 0
b1 =
2
π
∫ π
0
sin2 x dx = 1
f(x) = sinx
f(x) =
2
π
+
2
π
∞∑
n=2
(−1)n + 1
1− n2 cosnx
29. a0 =
2
π
(∫ π/2
0
x dx +
∫ π
π/2
(π − x) dx
)
=
π
2
an =
2
π
(∫ π/2
0
x cosnx dx +
∫ π
π/2
(π − x) cosnx dx
)
=
2
n2π
(
2 cos
nπ
2
+ (−1)n+1 − 1
)
bn =
2
π
(∫ π/2
0
x sinnx dx +
∫ π
π/2
(π − x) sinnx dx
)
=
4
n2π
sin
nπ
2
f(x) =
π
4
+
∞∑
n=1
2
n2π
(
2 cos
nπ
2
+ (−1)n+1 − 1
)
cosnx
f(x) =
∞∑
n=1
4
n2π
sin
nπ
2
sinnx
30. a0 =
1
π
∫ 2π
π
(x− π) dx = π
2
an =
1
π
∫ 2π
π
(x− π) cos n
2
x dx =
4
n2π
(
(−1)n − cos nπ
2
)
bn =
1
π
∫ 2π
π
(x− π) sin n
2
x dx =
2
n
(−1)n+1 − 4
n2π
sin
nπ
2
f(x) =
π
4
+
∞∑
n=1
4
n2π
(
(−1)n − cos nπ
2
)
cos
n
2
x
f(x) =
∞∑
n=1
(
2
n
(−1)n+1 − 4
n2π
sin
nπ
2
)
sin
n
2
x
31. a0 =
∫ 1
0
x dx +
∫ 2
1
1 dx =
3
2
an =
∫ 1
0
x cos
nπ
2
x dx =
4
n2π2
(
cos
nπ
2
− 1
)
bn =
∫ 1
0
x sin
nπ
2
x dx +
∫ 2
1
1 · sin nπ
2
x dx =
4
n2π2
sin
nπ
2
+
2
nπ
(−1)n+1
f(x) =
3
4
+
∞∑
n=1
4
n2π2
(
cos
nπ
2
− 1
)
cos
nπ
2
x
649
12.3 Fourier Cosine and Sine Series
f(x) =
∞∑
n=1
(
4
n2π2
sin
nπ
2
+
2
nπ
(−1)n+1
)
sin
nπ
2
x
32. a0 =
∫ 1
0
1 dx +
∫ 2
1
(2− x) dx = 3
2
an =
∫ 1
0
1 · cos nπ
2
x dx +
∫ 2
1
(2− x) cos nπ
2
x dx =
4
n2π2
(
cos
nπ
2
+ (−1)n+1
)
bn =
∫ 1
0
1 · sin nπ
2
x dx +
∫ 2
1
(2− x) sin nπ
2
x dx =
2
nπ
+
4
n2π2
sin
nπ
2
f(x) =
3
4
+
∞∑
n=1
4
n2π2
(
cos
nπ
2
+ (−1)n+1
)
cos
nπ
2
x
f(x) =
∞∑
n=1
(
2
nπ
+
4
n2π2
sin
nπ
2
)
sin
nπ
2
x
33. a0 = 2
∫ 1
0
(x2 + x) dx =
5
3
an = 2
∫ 1
0
(x2 + x) cosnπx dx =
2(x2 + x)
nπ
sinnπx
∣∣∣∣1
0
− 2
nπ
∫ 1
0
(2x + 1) sinnπx dx =
2
n2π2
[3(−1)n − 1]
bn = 2
∫ 1
0
(x2 + x) sinnπx dx = −2(x
2 + x)
nπ
cosnπx
∣∣∣∣1
0
+
2
nπ
∫ 1
0
(2x + 1) cosnπx dx
=
4
nπ
(−1)n+1 + 4
n3π3
[(−1)n − 1]
f(x) =
5
6
+
∞∑
n=1
2
n2π2
[3(−1)n − 1] cosnπx
f(x) =
∞∑
n=1
(
4
nπ
(−1)n+1 + 4
n3π3
[(−1)n − 1]
)
sinnπx
34. a0 =
∫ 2
0
(2x− x2) dx = 4
3
an =
∫ 2
0
(2x− x2) cos nπ
2
x dx =
8
n2π2
[(−1)n+1 − 1]
bn =
∫ 2
0
(2x− x2) sin nπ
2
x dx =
16
n3π3
[1− (−1)n]
f(x) =
2
3
+
∞∑
n=1
8
n2π2
[(−1)n+1 − 1] cos nπ
2
x
f(x) =
∞∑
n=1
16
n3π3
[1− (−1)n] sin nπ
2
x
35. a0 =
1
π
∫ 2π
0
x2 dx =
8
3
π2
an =
1
π
∫ 2π
0
x2 cosnx dx =
4
n2
bn =
1
π
∫ 2π
0
x2 sinnx dx = −4π
n
650
12.3 Fourier Cosine and Sine Series
f(x) =
4
3
π2 +
∞∑
n=1
(
4
n2
cosnx− 4π
n
sinnx
)
36. a0 =
2
π
∫ π
0
x dx = π
an =
2
π
∫ π
0
x cos 2nx dx = 0
bn =
2
π
∫ π
0
x sin 2nx dx = − 1
n
f(x) =
π
2
−
∞∑
n=1
1
n
sin 2nx
37. a0 = 2
∫ 1
0
(x + 1) dx = 3
an = 2
∫ 1
0
(x + 1) cos 2nπx dx = 0
bn = 2
∫ 1
0
(x + 1) sin 2nπx dx = − 1
nπ
f(x) =
3
2
−
∞∑
n=1
1
nπ
sin 2nπx
38. a0 =
2
2
∫ 2
0
(2− x) dx = 2
an =
2
2
∫ 2
0
(2− x) cosnπx dx = 0
bn =
2
2
∫ 2
0
(2− x) sinnπx dx = 2
nπ
f(x) = 1 +
∞∑
n=1
2
nπ
sinnπx
39. We have
bn =
2
π
∫ π
0
5 sinnt dt =
10
nπ
[1− (−1)n]
so that
f(t) =
∞∑
n=1
10[1− (−1)n]
nπ
sinnt.
Substituting the assumption xp(t) =
∑∞
n=1 Bn sinnt into the differential equation then gives
x′′p + 10xp =
∞∑
n=1
Bn(10− n2) sinnt =
∞∑
n=1
10[1− (−1)n]
nπ
sinnt
and so Bn = 10[1− (−1)n]/nπ(10− n2). Thus
xp(t) =
10
π
∞∑
n=1
1− (−1)n
n(10− n2) sinnt.
40. We have
bn =
2
π
∫ 1
0
(1− t) sinnπt dt = 2
nπ
651
12.3 Fourier Cosine and Sine Series
so that
f(t) =
∞∑
n=1
2
nπ
sinnπt.
Substituting the assumption xp(t) =
∑∞
n=1 Bn sinnπt into the differential equation then gives
x′′p + 10xp =
∞∑
n=1
Bn(10− n2π2) sinnπt =
∞∑
n=1
2
nπ
sinnπt
and so Bn = 2/nπ(10− n2π2). Thus
xp(t) =
2
π
∞∑
n=1
1
n(10− n2π2) sinnπt.
41. We have
a0 =
2
π
∫ π
0
(2πt− t2) dt = 4
3
π2
an =
2
π
∫ π
0
(2πt− t2) cosnt dt = − 4
n2
so that
f(t) =
2π2
3
−
∞∑
n=1
4
n2
cosnt.
Substituting the assumptionxp(t) =
A0
2
+
∞∑
n=1
An cosnt
into the differential equation then gives
1
4
x′′p + 12xp = 6A0 +
∞∑
n=1
An
(
−1
4
n2 + 12
)
cosnt =
2π2
3
−
∞∑
n=1
4
n2
cosnt
and A0 = π2/9, An = 16/n2(n2 − 48). Thus
xp(t) =
π2
18
+ 16
∞∑
n=1
1
n2(n2 − 48) cosnt.
42. We have
a0 =
2
1/2
∫ 1/2
0
t dt =
1
2
an =
2
1/2
∫ 1/2
0
t cos 2nπt dt =
1
n2π2
[(−1)n − 1]
so that
f(t) =
1
4
+
∞∑
n=1
(−1)n − 1
n2π2
cos 2nπt.
Substituting the assumption
xp(t) =
A0
2
+
∞∑
n=1
An cos 2nπt
into the differential equation then gives
1
4
x′′p + 12xp = 6A0 +
∞∑
n=1
An(12− n2π2) cos 2nπt = 14 +
∞∑
n=1
(−1)n − 1
n2π2
cos 2nπt
652
20 40 60 80
t
-4
-2
2
4
x
12.3 Fourier Cosine and Sine Series
and A0 = 1/24, An = [(−1)n − 1]/n2π2(12− n2π2). Thus
xp(t) =
1
48
+
1
π2
∞∑
n=1
(−1)n − 1
n2(12− n2π2) cos 2nπt.
43. (a) The general solution is x(t) = c1 cos
√
10t + c2 sin
√
10t + xp(t), where
xp(t) =
10
π
∞∑
n=1
1− (−1)n
n(10− n2) sinnt.
The initial condition x(0) = 0 implies c1 + xp(0) = 0. Since xp(0) = 0, we have c1 = 0 and x(t) =
c2 sin
√
10t + xp(t). Then x′(t) = c2
√
10 cos
√
10t + x′p(t) and x
′(0) = 0 implies
c2
√
10 +
10
π
∞∑
n=1
1− (−1)n
10− n2 cos 0 = 0.
Thus
c2 = −
√
10
π
∞∑
n=1
1− (−1)n
10− n2
and
x(t) =
10
π
∞∑
n=1
1− (−1)n
10− n2
[
1
n
sinnt− 1√
10
sin
√
10t
]
.
(b) The graph is plotted using eight nonzero terms in the series expansion of x(t).
44. (a) The general solution is x(t) = c1 cos 4
√
3t + c2 sin 4
√
3t + xp(t), where
xp(t) =
π2
18
+ 16
∞∑
n=1
1
n2(n2 − 48) cosnt.
The initial condition x(0) = 0 implies c1 + xp(0) = 1 or
c1 = 1− xp(0) = 1− π
2
18
− 16
∞∑
n=1
1
n2(n2 − 48) .
Now x′(t) = −4√3c1 sin 4
√
3t + 4
√
3c2 cos 4
√
3t + x′p(t), so x
′(0) = 0 implies 4
√
3c2 + x′p(0) = 0. Since
x′p(0) = 0, we have c2 = 0 and
x(t) =
(
1− π
2
18
− 16
∞∑
n=1
1
n2(n2 − 48)
)
cos 4
√
3t +
π2
18
+ 16
∞∑
n=1
1
n2(n2 − 48) cosnt
=
π2
18
+
(
1− π
2
18
)
cos 4
√
3t + 16
∞∑
n=1
1
n2(n2 − 48)
[
cosnt− cos 4
√
3t
]
.
653
2 4 6 8 10 12 14
t
-1
-0.5
0.5
1
1.5
x
12.3 Fourier Cosine and Sine Series
(b) The graph is plotted using five nonzero terms in the series expansion of x(t).
45. (a) We have
bn =
2
L
∫ L
0
w0x
L
sin
nπ
L
xdx =
2w0
nπ
(−1)n+1
so that
w(x) =
∞∑
n=1
2w0
nπ
(−1)n+1 sin nπ
L
x.
(b) If we assume yp(x) =
∑∞
n=1 Bn sin(nπx/L) then
y(4)p =
∞∑
n=1
n4π4
L4
Bn sin
nπ
L
x
and so the differential equation EIy(4)p = w(x) gives
Bn =
2w0(−1)n+1L4
EIn5π5
.
Thus
yp(x) =
2w0L4
EIπ5
∞∑
n=1
(−1)n+1
n5
sin
nπ
L
x.
46. We have
bn =
2
L
∫ 2L/3
L/3
w0 sin
nπ
L
xdx =
2w0
nπ
(
cos
nπ
3
− cos 2nπ
3
)
so that
w(x) =
∞∑
n=1
2w0
nπ
(
cos
nπ
3
− cos 2nπ
3
)
sin
nπ
L
x.
If we assume yp(x) =
∑∞
n=1 Bn sin(nπx/L) then
y(4)p (x) =
∞∑
n=1
n4π4
L4
Bn sin
nπ
L
x
and so the differential equation EIy(4)p (x) = w(x) gives
Bn = 2w0L4
cos nπ3 − cos 2nπ3
EIn5π5
.
Thus
yp(x) =
2w0L4
EIπ5
∞∑
n=1
cos nπ3 − cos 2nπ3
n5
sin
nπ
L
x.
654
2 4 6 8 10 12 14
x
-3
-2
-1
1
2
3
S20
2 4 6 8 10
x
0.5
1
S20
12.3 Fourier Cosine and Sine Series
47. The graph is obtained by summing the series from n = 1 to 20. It appears that
f(x) =
{
x, 0 < x < π
−π, π < x < 2π.
48. The graph is obtained by summing the series from n = 1 to 10. It appears that
f(x) =
{
1− x, 0 < x < 1
0, 1 < x < 2.
49. The function in Problem 47 is not unique; it could also be defined as
f(x) =

x, 0 < x < π
1, x = π
−π, π < x < 2π.
The function in Problem 48 is not unique; it could also be defined as
f(x) =

0, −2 < x < −1
x + 1, −1 < x < 0
−x + 1, 0 < x < 1
0, 1 < x < 2.
50. The cosine series converges to an even extension of the function on the interval (−π, 0). Since the even extension
of f(x) is f(−x), in this case f(−x) = e−x on (−π, 0).
51. No, it is not a full Fourier series. A full Fourier series of f(x) = ex, 0 < x < π, would converge to the π-periodic
extension of f . The cosine and sine series converge to a 2π-periodic extension (even and odd, respectively). The
average of the two series converges to a 2π-periodic extension of
f(x) =
{
ex, 0 < x < π
0, −π < x < 0.
52. (a) If f and g are even and h(x) = f(x)g(x) then
h(−x) = f(−x)g(−x) = f(x)g(x) = h(x)
and h is even.
655
12.3 Fourier Cosine and Sine Series
(c) If f is even and g is odd and h(x) = f(x)g(x) then
h(−x) = f(−x)g(−x) = f(x)[−g(x)] = −h(x)
and h is odd.
(d) Let h(x) = f(x)± g(x) where f and g are even. Then
h(−x) = f(−x)± g(−x) = f(x)± g(x) = h(x),
and so h is an even function.
(f) If f is even then∫ a
−a
f(x) dx = −
∫ 0
a
f(−u) du +
∫ a
0
f(x) dx =
∫ a
0
f(u) du +
∫ a
0
f(x) dx = 2
∫ a
0
f(x) dx.
(g) If f is odd then∫ a
−a
f(x) dx = −
∫ 0
−a
f(−x) dx +
∫ a
0
f(x) dx =
∫ 0
a
f(u) du +
∫ a
0
f(x) dx
= −
∫ a
0
f(u) du +
∫ a
0
f(x) dx = 0.
EXERCISES 12.4
Complex Fourier Series
In this section we make use of the following identities due to Euler’s formula:
einπ = e−inπ = (−1)n, e−2inπ = 1, e−inπ/2 = (−i)n.
1. Identifying p = 2 we have
cn =
1
4
∫ 2
−2
f(x)e−inπx/2dx =
1
4
[∫ 0
−2
(−1)e−inπx/2dx +
∫ 2
0
e−inπx/2dx
]
=
i
2nπ
[−1 + einπ + e−inπ − 1] = i
2nπ
[−1 + (−1)n + (−1)n − 1] = 1− (−1)
n
nπi
and
c0 =
1
4
∫ 2
−2
f(x)dx = 0.
Thus
f(x) =
∞∑
n=−∞
n�=0
1− (−1)n
inπ
einπx/2.
656
12.4 Complex Fourier Series
2. Identifying 2p = 2 or p = 1 we have
cn =
1
2
∫ 2
0
f(x)e−inπxdx =
1
2
∫ 2
1
e−inπxdx = − 1
2inπ
e−inπx
∣∣∣2
1
= − 1
2inπ
(
e−2inπ − e−inπ) = − 1
2inπ
[1− (−1)n] = i
2nπ
[1− (−1)n]
and
c0 =
1
2
∫ 2
0
f(x) dx =
1
2
∫ 2
1
dx =
1
2
.
Thus
f(x) =
1
2
+
i
2π
∞∑
n=−∞
n�=0
1− (−1)n
n
einπx.
3. Identifying p = 1/2 we have
cn =
∫ 1/2
−1/2
f(x)e−2inπxdx =
∫ 1/4
0
e−2inπxdx = − 1
2inπ
e−2inπx
∣∣∣∣1/4
0
= − 1
2inπ
[
e−inπ/2 − 1] = − 1
2inπ
[(−i)n − 1] = i
2nπ
[(−i)n − 1]
and
c0 =
∫ 1/4
0
dx =
1
4
.
Thus
f(x) =
1
4
+
i
2π
∞∑
n=−∞
n�=0
(−i)n − 1
n
e2inπx.
4. Identifying p = π we have
cn =
1
2π
∫ π
−π
f(x)e−inx/πdx =
1
2π
∫ π
0
xe−inx/πdx
=
1
2
(
π
n2
+
ix
n
)
e−inx/π
∣∣∣∣π
0
=
π(1 + in)
2n2
e−in − π
2n2
and
c0 =
1
2π
∫ π
0
x dx =
π
4
.
Thus
f(x) =
π
4
+
π
2
∞∑
n=−∞
n�=0
1
n2
[
(1 + in)e−in − 1
]
einx.
5. Identifying 2p = 2π or p = π we have
cn =
1
2π
∫ 2π
0
f(x)e−inxdx =
1
2π
∫ 2π
0
xe−inxdx
=
1
2π
(
1
n2
+
ix
n
)
e−inx
∣∣∣∣2π
0
=
1 + 2inπ
2n2π
− 1
2n2π
=
i
n
657
�
5 Π
��������
2
�
3 Π
��������
2
�
Π
����
2
Π
����
2
3 Π
��������
2
5 Π
��������
2
frequency
0.2
0.4
0.6
cn
n -5 -4 -3 -2 -1 0 1 2 3 4 5
c n 0.1273 0.0000 0.2122 0.0000 0.6366 0.0000 0.6366 0.0000 0.2122 0.0000 0.1273
�10 Π �6 Π �2 Π 2 Π 6 Π 10 Π
frequency
0.05
0.1
0.15
0.2
0.25
cn
n -5 -4 -3 -2 -1 0 1 2 3 4 5
c n 0.0450 0.0000 0.0750 0.1592 0.2251 0.2500 0.2251 0.1592 0.0750 0.0000 0.0450
12.4 Complex Fourier Series
and
c0 =
1
2π
∫2π
0
x dx = π.
Thus
f(x) = π +
∞∑
n=−∞
n�=0
i
n
einx.
6. Identifying p = 1 we have
cn =
1
2
∫ 1
−1
f(x)e−inπxdx =
1
2
[∫ 0
−1
exe−inπxdx +
∫ 1
0
e−xe−inπxdx
]
=
1
2
[
− 1
1− inπ e
(1−inπ)x
∣∣∣0
−1
− 1
1 + inπ
e−(1+inπ)x
∣∣∣1
0
]
=
e− (−1)n
e(1− inπ) +
1− e−1(−1)n
1 + inπ
=
2[e− (−1)n]
e(1 + n2π2)
.
Thus
f(x) =
∞∑
n=−∞
2[e− (−1)n]
e(1 + n2π2)
einπx.
7. The fundamental period is T = 4, so ω = 2π/4 = π/2
and the values of nω are 0, ±π/2, ±π, ±3π/2, . . . . From
Problem 1, c0 = 0 and |cn| = (1 − (−1)n)/nπ. The table
shows some values of n with corresponding values of |cn|.
The graph is a portion of the frequency spectrum.
8. The fundamental period is T = 1, so ω = 2π and the
values of nω are 0, ±2π, ±4π, ±6π, . . . . From Problem 3,
c0 = 14 and |cn| = |(−i)n − 1|/2nπ, or c1 = c−1 =
√
2/2π,
c2 = c−2 = 1/2π, c3 = c−3 =
√
2/6π, c4 = c−4 = 0,
c5 = c−5 =
√
2/10π, c6 = c−6 = 1/6π, c7 = c−7 =
√
2/14π,
c8 = c−8 = 0, . . . . The table shows some values of n with
corresponding values of |cn|. The graph is a portion of the
frequency spectrum.
658
-6 -4 -2 2 4 6
frequency
0.2
0.4
0.6
cn
�2 Π �Π Π 2 Π
x
1
2
3
4
f
n -5 -4 -3 -2 -1 0 1 2 3 4 5
c n 0.0198 0.0759 0.2380 0.4265 0.5784 0.6366 0.5784 0.4265 0.2380 0.0759 0.0198
-10 -5 5 10
frequency
0.1
0.2
0.3
cn
�2 Π �Π Π 2 Π
x
0.2
0.4
0.6
0.8
1
f
n -5 -4 -3 -2 -1 0 1 2 3 4 5
c n 0.1447 0.1954 0.2437 0.2833 0.3093 0.3183 0.3093 0.2833 0.2437 0.1954 0.1447
12.4 Complex Fourier Series
9. Identifying 2p = π or p = π/2, and using sinx =
(eix − e−ix)/2i, we have
cn =
1
π
∫ π
0
f(x)e−2inx/πdx =
1
π
∫ π
0
(sinx)e−2inx/πdx
=
1
π
∫ π
0
1
2i
(eix − e−ix)e−2inx/πdx
=
1
2πi
∫ π
0
(
e(1−2n/π)ix − e−(1+2n/π)ix
)
dx
=
1
2πi
[
1
i(1− 2n/π)e
(1−2n/π)ix
+
1
i(1 + 2n/π)
e−(1+2n/π)ix
]π
0
=
π(1 + e−2in)
π2 − 4n2 .
The fundamental period is T = π, so ω = 2π/π = 2 and the
values of nω are 0, ±2, ±4, ±6, . . . . Values of |cn| for n = 0,
±1, ±2, ±3, ±4, and ±5 are shown in the table. The bottom
graph is a portion of the frequency spectrum.
10. Identifying 2p = π or p = π/2, and using cosx =
(eix − e−ix)/2, we have
cn =
1
π
∫ π
0
f(x)e−2inx/πdx
=
1
π
∫ π/2
0
(cosx)e−2inx/πdx
=
1
π
∫ π/2
0
1
2
(eix − e−ix)e−2inx/πdx
=
1
2π
∫ π/2
0
(
e(1−2n/π)ix − e−(1+2n/π)ix
)
dx
=
1
2π
[
1
i(1− 2n/π)e
(1−2n/π)ix
+
1
i(1 + 2n/π)
e−(1+2n/π)ix
]π/2
0
=
2ne−in + iπ
π2 − 4n2 .
The fundamental period is T = π, so ω = 2π/π = 2 and the values of nω are 0, ±2, ±4, ±6, . . . . Values of
|cn| for n = 0, ±1, ±2, ±3, ±4, and ±5 are shown in the table. The bottom graph is a portion of the frequency
spectrum.
659
12.4 Complex Fourier Series
11. (a) Adding cn = 12 (an− ibn) and c−n = 12 (an + ibn) we get cn +c−n = an. Subtracting, we get cn−c−n = −ibn.
Multiplying both sides by i we obtain i(cn − c−n) = bn.
(b) From
an = cn + c−n = (−1)n sinhπ
π
[
1− in
n2 + 1
+
1 + in
n2 + 1
]
=
2(−1)n sinhπ
π(n2 + 1)
, n = 0, 1, 2, . . .
and
bn = i(cn − c−n) = i(−1)n sinhπ
π
[
1− in
n2 + 1
− 1 + in
n2 + 1
]
= i(−1)n sinhπ
π
[
− 2in
n2 + 1
]
=
2(−1)nn sinhπ
π(n2 + 1)
,
the Fourier series of f is
f(x) =
sinhπ
π
+
2 sinhπ
π
∞∑
n=1
[
(−1)n
n2 + 1
cosnx +
n(−1)n
n2 + 1
sinnx
]
.
12. From Problem 11 and the fact that f is odd, cn + c−n = an = 0, so c−n = −cn. Then bn = i(cn − c−n) = 2icn.
From Problem 1, bn = 2i[1− (−1)n]/nπi = 2[1− (−1)n]/nπ, and the Fourier sine series of f is
f(x) =
∞∑
i=1
2[1− (−1)n
nπ
sin
nπx
2
.
EXERCISES 12.5
Sturm-Liouville Problem
1. For λ ≤ 0 the only solution of the boundary-value problem is y = 0. For λ = α2 > 0 we have
y = c1 cosαx + c2 sinαx.
Now
y′(x) = −c1α sinαx + c2α cosαx
and y′(0) = 0 implies c2 = 0, so
y(1) + y′(1) = c1(cosα− α sinα) = 0 or cotα = α.
The eigenvalues are λn = α2n where α1, α2, α3, . . . are the consecutive positive solutions of cotα = α. The
corresponding eigenfunctions are cosαnx for n = 1, 2, 3, . . . . Using a CAS we find that the first four eigen-
values are approximately 0.7402, 11.7349, 41.4388, and 90.8082 with corresponding approximate eigenfunctions
cos 0.8603x, cos 3.4256x, cos 6.4373x, and cos 9.5293x.
2. For λ < 0 the only solution of the boundary-value problem is y = 0. For λ = 0 we have y = c1x + c2. Now
y′ = c1 and the boundary conditions both imply c1 + c2 = 0. Thus, λ = 0 is an eigenvalue with corresponding
eigenfunction y0 = x− 1.
For λ = α2 > 0 we have
y = c1 cosαx + c2 sinαx
660
12.5 Sturm-Liouville Problem
and
y′(x) = −c1α sinαx + c2α cosαx.
The boundary conditions imply
c1 + c2α = 0
c1 cosα + c2 sinα = 0
which gives
−c2α cosα + c2 sinα = 0 or tanα = α.
The eigenvalues are λn = α2n where α1, α2, α3, . . . are the consecutive positive solutions of tanα = α. The
corresponding eigenfunctions are α cosαx − sinαx (obtained by taking c2 = −1 in the first equation of the
system.) Using a CAS we find that the first four positive eigenvalues are 20.1907, 59.6795, 118.9000, and
197.858 with corresponding eigenfunctions 4.4934 cos 4.4934x − sin 4.4934x, 7.7253 cos 7.7253x − sin 7.7253x,
10.9041 cos 10.9041x− sin 10.9041x, and 14.0662 cos 14.0662x− sin 14.0662x.
3. For λ = 0 the solution of y′′ = 0 is y = c1x + c2. The condition y′(0) = 0 implies c1 = 0, so λ = 0 is an
eigenvalue with corresponding eigenfunction 1.
For λ = −α2 < 0 we have y = c1 coshαx + c2 sinhαx and y′ = c1α sinhαx + c2α coshαx. The condition
y′(0) = 0 implies c2 = 0 and so y = c1 coshαx. Now the condition y′(L) = 0 implies c1 = 0. Thus y = 0 and
there are no negative eigenvalues.
For λ = α2 > 0 we have y = c1 cosαx + c2 sinαx and y′ = −c1α sinαx + c2α cosαx. The condition y′(0) = 0
implies c2 = 0 and so y = c1 cosαx. Now the condition y′(L) = 0 implies −c1α sinαL = 0. For c1 �= 0 this
condition will hold when αL = nπ or λ = α2 = n2π2/L2, where n = 1, 2, 3, . . . . These are the positive
eigenvalues with corresponding eigenfunctions cos(nπx/L), n = 1, 2, 3, . . . .
4. For λ = −α2 < 0 we have
y = c1 coshαx + c2 sinhαx
y′ = c1α sinhαx + c2α coshαx.
Using the fact that coshx is an even function and sinhx is odd we have
y(−L) = c1 cosh(−αL) + c2 sinh(−αL)
= c1 coshαL− c2 sinhαL
and
y′(−L) = c1α sinh(−αL) + c2α cosh(−αL)
= −c1α sinhαL + c2α coshαL.
The boundary conditions imply
c1 coshαL− c2 sinhαL = c1 coshαL + c2 sinhαL
or
2c2 sinhαL = 0
and
−c1α sinhαL + c2α coshαL = c1α sinhαL + c2α coshαL
or
2c1α sinhαL = 0.
661
12.5 Sturm-Liouville Problem
Since αL �= 0, c1 = c2 = 0 and the only solution of the boundary-value problem in this case is y = 0.
For λ = 0 we have
y = c1x + c2
y′ = c1.
From y(−L) = y(L) we obtain
−c1L + c2 = c1L + c2.
Then c1 = 0 and y = 1 is an eigenfunction corresponding to the eigenvalue λ = 0.
For λ = α2 > 0 we have
y = c1 cosαx + c2 sinαx
y′ = −c1α sinαx + c2α cosαx.
The first boundary condition implies
c1 cosαL− c2 sinαL = c1 cosαL + c2 sinαL
or
2c2 sinαL = 0.
Thus, if c1 = 0 and c2 �= 0,
αL = nπ or λ = α2 =
n2π2
L2
, n = 1, 2, 3, . . . .
The corresponding eigenfunctions are sin(nπx/L), for n = 1, 2, 3, . . . . Similarly, the second boundary condition
implies
2c1α sinαL = 0.
If c1 �= 0 and c2 = 0,
αL = nπ or λ = α2 =
n2π2
L2
, n = 1, 2, 3, . . . ,
and the corresponding eigenfunctions are cos(nπx/L), for n = 1, 2, 3, . . . .
5. The eigenfunctions are cosαnx where cotαn = αn. Thus
‖ cosαnx‖2 =
∫ 1
0
cos2 αnx dx =
1
2
∫ 1
0
(1 + cos 2αnx) dx
=
1
2
(
x +
1
2αn
sin 2αnx
) ∣∣∣∣1
0
=
1
2
(
1+
1
2αn
sin 2αn
)
=
1
2
[
1 +
1
2αn
(2 sinαn cosαn)
]
=
1
2
[
1 +
1
αn
sinαn cotαn sinαn
]
=
1
2
[
1 +
1
αn
(sinαn)αn (sinαn)
]
=
1
2
(
1 + sin2 αn
)
.
662
12.5 Sturm-Liouville Problem
6. The eigenfunctions are sinαnx where tanαn = −αn. Thus
‖ sinαnx‖2 =
∫ 1
0
sin2 αnx dx =
1
2
∫ 1
0
(1− cos 2αnx) dx
=
1
2
(
x− 1
2αn
sin 2αnx
) ∣∣∣∣1
0
=
1
2
(
1− 1
2αn
sin 2αn
)
=
1
2
[
1− 1
2αn
(2 sinαn cosαn)
]
=
1
2
[
1− 1
αn
tanαn cosαn cosαn
]
=
1
2
[
1− 1
αn
(−αn cos2 αn)] = 12 (1 + cos2 αn) .
7. (a) If λ ≤ 0 the initial conditions imply y = 0. For λ = α2 > 0 the general solution of the Cauchy-Euler
differential equation is y = c1 cos(α lnx) + c2 sin(α lnx). The condition y(1) = 0 implies c1 = 0, so that
y = c2 sin(α lnx). The condition y(5) = 0 implies α ln 5 = nπ, n = 1, 2, 3, . . . . Thus, the eigenvalues are
n2π2/(ln 5)2 for n = 1, 2, 3, . . . , with corresponding eigenfunctions sin[(nπ/ ln 5) lnx].
(b) The self-adjoint form is
d
dx
[xy′] +
λ
x
y = 0.
(c) An orthogonality relation is∫ 5
1
1
x
sin
(mπ
ln 5
lnx
)
sin
( nπ
ln 5
lnx
)
dx = 0, m �= n.
8. (a) The roots of the auxiliary equation m2+m+λ = 0 are 12 (−1±
√
1− 4λ ). When λ = 0 the general solution of
the differential equation is c1 +c2e−x. The boundary conditions imply c1 +c2 = 0 and c1 +c2e−2 = 0. Since
the determinant of the coefficients is not 0, the only solution of this homogeneous system is c1 = c2 = 0,
in which case y = 0. When λ = 14 , the general solution of the differential equation is c1e
−x/2 + c2xe−x/2.
The boundary conditions imply c1 = 0 and c1 + 2c2 = 0, so c1 = c2 = 0 and y = 0. Similarly, if 0 < λ < 14 ,
the general solution is
y = c1e
1
2 (−1+
√
1−4λ )x + c2e
1
2 (−1−
√
1−4λ )x.
In this case the boundary conditions again imply c1 = c2 = 0, and so y = 0. Now, for λ > 14 , the general
solution of the differential equation is
y = c1e−x/2 cos
√
4λ− 1x + c2e−x/2 sin
√
4λ− 1x.
The condition y(0) = 0 implies c1 = 0 so y = c2e−x/2 sin
√
4λ− 1x. From
y(2) = c2e−1 sin 2
√
4λ− 1 = 0
we see that the eigenvalues are determined by 2
√
4λ− 1 = nπ for n = 1, 2, 3, . . . . Thus, the eigenvalues
are n2π2/42 + 1/4 for n = 1, 2, 3, . . . , with corresponding eigenfunctions e−x/2 sin(nπx/2).
(b) The self-adjoint form is
d
dx
[exy′] + λexy = 0.
663
12.5 Sturm-Liouville Problem
(c) An orthogonality relation is∫ 2
0
ex
(
e−x/2 sin
mπ
2
x
) (
e−x/2 cos
nπ
2
x
)
dx =
∫ 2
0
sin
mπ
2
x cos
nπ
2
x dx = 0.
9. To obtain the self-adjoint form we note that an integrating factor is (1/x)e
∫
(1−x)dx/x = e−x. Thus, the
differential equation is
xe−xy′′ + (1− x)e−xy′ + ne−xy = 0
and the self-adjoint form is
d
dx
[
xe−xy′
]
+ ne−xy = 0.
Identifying the weight function p(x) = e−x and noting that since r(x) = xe−x, r(0) = 0 and limx→∞ r(x) = 0,
we have the orthogonality relation ∫ ∞
0
e−xLm(x)Ln(x) dx = 0, m �= n.
10. To obtain the self-adjoint form we note that an integrating factor is e
∫
−2x dx = e−x
2
. Thus, the differential
equation is
e−x
2
y′′ − 2xe−x2y′ + 2ne−x2y = 0
and the self-adjoint form is
d
dx
[
e−x
2
y′
]
+ 2ne−x
2
y = 0.
Identifying the weight function p(x) = e−x
2
and noting that since r(x) = e−x
2
, limx→−∞ r(x) = limx→∞ r(x) =
0, we have the orthogonality relation∫ ∞
−∞
e−x
2
Hm(x)Hn(x) dx = 0, m �= n.
11. (a) The differential equation is
(1 + x2)y′′ + 2xy′ +
λ
1 + x2
y = 0.
Letting x = tan θ we have θ = tan−1 x and
dy
dx
=
dy
dθ
dθ
dx
=
1
1 + x2
dy
dθ
d2y
dx2
=
d
dx
[
1
1 + x2
dy
dθ
]
=
1
1 + x2
(
d2y
dθ2
dθ
dx
)
− 2x
(1 + x2)2
dy
dθ
=
1
(1 + x2)2
d2y
dθ2
− 2x
(1 + x2)2
dy
dθ
.
The differential equation can then be written in terms of y(θ) as
(1 + x2)
[
1
(1 + x2)2
d2y
dθ2
− 2x
(1 + x2)2
dy
dθ
]
+ 2x
[
1
1 + x2
dy
dθ
]
+
λ
1 + x2
y
=
1
1 + x2
d2y
dθ2
+
λ
1 + x2
y = 0
or
d2y
dθ2
+ λy = 0.
664
12.5 Sturm-Liouville Problem
The boundary conditions become y(0) = y(π/4) = 0. For λ ≤ 0 the only solution of the boundary-value
problem is y = 0. For λ = α2 > 0 the general solution of the differential equation is y = c1 cosαθ+c2 sinαθ.
The condition y(0) = 0 implies c1 = 0 so y = c2 sinαθ. Now the condition y(π/4) = 0 implies c2 sinαπ/4 =
0. For c2 �= 0 this condition will hold when απ/4 = nπ or λ = α2 = 16n2, where n = 1, 2, 3, . . . . These are
the eigenvalues with corresponding eigenfunctions sin 4nθ = sin(4n tan−1 x), for n = 1, 2, 3, . . . .
(b) An orthogonality relation is
∫ 1
0
1
x2 + 1
sin(4m tan−1 x) sin(4n tan−1 x) dx = 0, m �= n.
12. (a) Letting λ = α2 the differential equation becomes x2y′′ + xy′ + (α2x2 − 1)y = 0. This is the parametric
Bessel equation with ν = 1. The general solution is
y = c1J1(αx) + c2Y1(αx).
Since Y is unbounded at 0 we must have c2 = 0, so that y = c1J1(αx). The condition J1(3α) = 0 defines
the eigenvalues λn = α2n for n = 1, 2, 3, . . . . The corresponding eigenfunctions are J1(αnx).
(b) Using a CAS or Table 5.1 in the text to solve J1(3α) = 0 we find 3α1 = 3.8317, 3α2 = 7.0156, 3α3 =
10.1735, and 3α4 = 13.3237. The corresponding eigenvalues are λ1 = α21 = 1.6313, λ2 = α
2
2 = 5.4687,
λ3 = α23 = 11.4999, and λ4 = α
2
4 = 19.7245.
13. When λ = 0 the differential equation is r(x)y′′ + r′(x)y′ = 0. By inspection we see that y = 1 is a solution of
the boundary-value problem. Thus, λ = 0 is an eigenvalue.
14. (a) An orthogonality relation is ∫ 1
0
cosxmx cosxnx dx = 0
where xm �= xn are positive solutions of cotx = x.
(b) Referring to Problem 1 we use a CAS to compute
∫ 1
0
(cos 0.8603x)(cos 3.4256x) dx = −1.8771× 10−6 ≈ 0.
15. (a) An orthogonality relation is
∫ 1
0
(xm cosxmx− sinxmx)(xn cosxnx− sinxnx) dx = 0
where xm �= xn are positive solutions of tanx = x.
(b) Referring to Problem 2 we use a CAS to compute
∫ 1
0
(4.4934 cos 4.4934x− sin 4.4934x)(7.7253 cos 7.7253x− sin 7.7253x) dx = −2.5650× 10−4 ≈ 0.
665
12.5 Sturm-Liouville Problem
EXERCISES 12.6
Bessel and Legendre Series
12.6 Bessel and Legendre Series
1. Identifying b = 3, we have α1 = 1.2772, α2 = 2.3385, α3 = 3.3912, and α4 = 4.4412.
2. By (6) in the text J ′0(2α) = −J1(2α). Thus, J ′0(2α) = 0 is equivalent to J1(2α). Then α1 = 1.9159, α2 = 3.5078,
α3 = 5.0867, and α4 = 6.6618.
3. The boundary condition indicates that we use (15) and (16) in the text. With b = 2 we obtain
ci =
2
4J21 (2αi)
∫ 2
0
xJ0(αix) dx
t = αix dt = αi dx
=
1
2J21 (2αi)
· 1
α2i
∫ 2αi
0
tJ0(t) dt
=
1
2α2i J
2
1 (2αi)
∫ 2αi
0
d
dt
[tJ1(t)] dt [From (5) in the text]
=
1
2α2i J
2
1 (2αi)
tJ1(t)
∣∣∣∣2αi
0
=
1
αiJ1(2αi)
.
Thus
f(x) =
∞∑
i=1
1
αiJ1(2αi)
J0(αix).
4. The boundary condition indicates that we use (19) and (20) in the text. With b = 2 we obtain
c1 =
2
4
∫ 2
0
x dx =
2
4
x2
2
∣∣∣∣2
0
= 1,
ci =
2
4J20 (2αi)
∫ 2
0
xJ0(αix) dx
t = αix dt = αi dx
=
1
2J20 (2αi)
· 1
α2i
∫ 2αi
0
tJ0(t) dt
=
1
2α2i J
2
0 (2αi)
∫ 2αi
0
d
dt
[tJ1(t)] dt [From (5) in the text]
=
1
2α2i J
2
0 (2αi)
tJ1(t)
∣∣∣∣2αi
0
=
J1(2αi)
αiJ20 (2αi)
.
Now since J ′0(2αi) = 0 is equivalent to J1(2αi) = 0 we conclude ci = 0 for i = 2, 3, 4, . . . . Thus the expansion
of f on 0 < x < 2 consists of a series with one nontrivial term:
f(x) = c1 = 1.
666
12.6 Bessel and Legendre Series
5. The boundary condition indicates that we use (17) and (18) in the text. With b = 2and h = 1 we obtain
ci =
2α2i
(4α2i + 1)J
2
0 (2αi)
∫ 2
0
xJ0(αix) dx
t = αix dt = αi dx
=
2α2i
(4α2i + 1)J
2
0 (2αi)
· 1
α2i
∫ 2αi
0
tJ0(t) dt
=
2
(4α2i + 1)J
2
0 (2αi)
∫ 2αi
0
d
dt
[tJ1(t)] dt [From (5) in the text]
=
2
(4α2i + 1)J
2
0 (2αi)
tJ1(t)
∣∣∣∣2αi
0
=
4αiJ1(2αi)
(4α2i + 1)J
2
0 (2αi)
.
Thus
f(x) = 4
∞∑
i=1
αiJ1(2αi)
(4α2i + 1)J
2
0 (2αi)
J0(αix).
6. Writing the boundary condition in the form
2J0(2α) + 2αJ ′0(2α) = 0
we identify b = 2 and h = 2. Using (17) and (18) in the text we obtain
ci =
2α2i
(4α2i + 4)J
2
0 (2αi)
∫ 2
0
xJ0(αix) dx
t = αix dt = αi dx
=
α2i
2(α2i + 1)J
2
0 (2αi)
· 1
α2i
∫ 2αi
0
tJ0(t) dt
=
1
2(α2i + 1)J
2
0 (2αi)
∫ 2αi
0
d
dt
[tJ1(t)] dt [From (5) in the text]
=
1
2(α2i + 1)J
2
0 (2αi)
tJ1(t)
∣∣∣∣2αi
0
=
αiJ1(2αi)
(α2i + 1)J
2
0 (2αi)
.
Thus
f(x) =
∞∑
i=1
αiJ1(2αi)
(α2i + 1)J
2
0 (2αi)
J0(αix).
7. The boundary condition indicates that we use (17) and (18) in the text. With n = 1, b = 4, and h = 3 we
obtain
ci =
2α2i
(16α2i − 1 + 9)J21 (4αi)
∫ 4
0
xJ1(αix)5x dx
t = αix dt = αi dx
=
5α2i
4(2α2i + 1)J
2
1 (4αi)
· 1
α3i
∫ 4αi
0
t2J1(t) dt
=
5
4αi(2α2i + 1)J
2
1 (4αi)
∫ 4αi
0
d
dt
[t2J2(t)] dt [From (5) in the text]
667
12.6 Bessel and Legendre Series
=
5
4αi(2α2i + 1)J
2
1 (4αi)
t2J2(t)
∣∣∣∣4αi
0
=
20αiJ2(4αi)
(2α2i + 1)J
2
1 (4αi)
.
Thus
f(x) = 20
∞∑
i=1
αiJ2(4αi)
(2α2i + 1)J
2
1 (4αi)
J1(αix).
8. The boundary condition indicates that we use (15) and (16) in the text. With n = 2 and b = 1 we obtain
c1 =
2
J23 (αi)
∫ 1
0
xJ2(αix)x2 dx
t = αix dt = αi dx
=
2
J23 (αi)
· 1
α4i
∫ αi
0
t3J2(t) dt
=
2
α4i J
2
3 (αi)
∫ αi
0
d
dt
[t3J3(t)] dt [From (5) in the text]
=
2
α4i J
2
3 (αi)
t3J3(t)
∣∣∣∣αi
0
=
2
αiJ3(αi)
.
Thus
f(x) = 2
∞∑
i=1
1
αiJ3(αi)
J2(αix).
9. The boundary condition indicates that we use (19) and (20) in the text. With b = 3 we obtain
c1 =
2
9
∫ 3
0
xx2 dx =
2
9
x4
4
∣∣∣∣3
0
=
9
2
,
ci =
2
9J20 (3αi)
∫ 3
0
xJ0(αix)x2 dx
t = αix dt = αi dx
=
2
9J20 (3αi)
· 1
α4i
∫ 3αi
0
t3J0(t) dt
=
2
9α4i J
2
0 (3αi)
∫ 3αi
0
t2
d
dt
[tJ1(t)] dt
u = t2 dv = ddt [tJ1(t)] dt
du = 2t dt v = tJ1(t)
=
2
9α4i J
2
0 (3αi)
(
t3J1(t)
∣∣∣∣3αi
0
−2
∫ 3αi
0
t2J1(t) dt
)
.
668
5 10 15 20 25 30
x
-4
-2
2
4
y
12.6 Bessel and Legendre Series
With n = 0 in equation (6) in the text we have J ′0(x) = −J1(x), so the boundary condition J ′0(3αi) = 0 implies
J1(3αi) = 0. Then
ci =
2
9α4i J
2
0 (3αi)
(
−2
∫ 3αi
0
d
dt
[
t2J2(t)
]
dt
)
=
2
9α4i J
2
0 (3αi)
(
−2t2J2(t)
∣∣∣∣3αi
0
)
=
2
9α4i J
2
0 (3αi)
[−18α2i J2(3αi)] = −4J2(3αi)α2i J20 (3αi) .
Thus
f(x) =
9
2
− 4
∞∑
i=1
J2(3αi)
α2i J
2
0 (3αi)
J0(αix).
10. The boundary condition indicates that we use (15) and (16) in the text. With b = 1 it follows that
ci =
2
J21 (αi)
∫ 1
0
x
(
1− x2) J0(αix) dx
=
2
J21 (αi)
[∫ 1
0
xJ0(αix) dx−
∫ 1
0
x3J0(αix) dx
]
t = αix dt = αi dx
=
2
J21 (αi)
[
1
α2i
∫ αi
0
tJ0(t) dt− 1
α4i
∫ αi
0
t3J0(t) dt
]
=
2
J21 (αi)
[
1
α2i
∫ αi
0
d
dt
[tJ1(t)] dt− 1
α4i
∫ αi
0
t2
d
dt
[tJ1(t)] dt
]
u = t2 dv = ddt [tJ1(t)] dt
du = 2t dt v = tJ1(t)
=
2
J21 (αi)
[
1
α2i
tJ1(t)
∣∣∣∣αi
0
− 1
α4i
(
t3J1(t)
∣∣∣∣αi
0
− 2
∫ αi
0
t2J1(t) dt
)]
=
2
J21 (αi)
[
J1(αi)
αi
− J1(αi)
αi
+
2
α4i
∫ αi
0
d
dt
[
t2J2(t)
]
dt
]
=
2
J21 (αi)
[
2
α4i
t2J2(t)
∣∣∣∣αi
0
]
=
4J2(αi)
α2i J
2
1 (αi)
.
Thus
f(x) = 4
∞∑
i=1
J2(αi)
α2i J
2
1 (αi)
J0(αix).
11. (a)
669
1 2 3 4 5 x
5
10
15
20
S1
1 2 3 4 5 x
5
10
15
20
S2
1 2 3 4 5 x
5
10
15
20
S3
1 2 3 4 5 x
5
10
15
20
S4
1 2 3 4 5 x
5
10
15
20
S5
1 2 3 4 x
5
10
15
20
S10
10 20 30 40 50
x
-10
-5
5
10
15
20
S10
-2 -1 1 2 x
0.5
1
1.5
2
y
12.6 Bessel and Legendre Series
(b) Using FindRoot in Mathematica we find the roots x1 = 2.9496, x2 = 5.8411, x3 = 8.8727, x4 = 11.9561,
and x5 = 15.0624.
(c) Dividing the roots in part (b) by 4 we find the eigenvalues α1 = 0.7374, α2 = 1.4603,
α3 = 2.2182, α4 = 2.9890, and α5 = 3.7656.
(d) The next five eigenvalues are α6 = 4.5451, α7 = 5.3263, α8 = 6.1085, α9 = 6.8915, and α10 = 7.6749.
12. (a) From Problem 7, the coefficients of the Fourier-Bessel series are
ci =
20αiJ2(4αi)
(2α2i + 1)J
2
1 (4αi)
.
Using a CAS we find c1 = 26.7896, c2 = −12.4624, c3 = 7.1404, c4 = −4.68705, and c5 = 3.35619.
(b)
(c)
13. Since f is expanded as a series of Bessel functions, J1(αix) and J1 is an odd function, the series should represent
an odd function.
14. (a) Since J0 is an even function, a series expansion of a function
defined on (0, 2) would converge to the even extension of the
function on (−2, 0).
670
-4 -2 2 4 x
5
10
15
20
y
-1 -0.5 0.5 1
x
0.5
1
S5
12.6 Bessel and Legendre Series
(b) In Section 5.3 we saw that J ′2(x) = 2J2(x)/x − J3(x). Since J2 is even and J3 is
odd we see that
J ′2(−x) = 2J2(−x)/(−x)− J3(−x)
= −2J2(x)/x + J3(x) = −J ′2(x),
so that J ′2 is an odd function. Now, if f(x) = 3J2(x) + 2xJ
′
2(x), we see that
f(−x) = 3J2(−x)− 2xJ ′2(−x)
= 3J2(x) + 2xJ ′2(x) = f(x),
so that f is an even function. Thus, a series expansion of a function defined on
(0, 4) would converge to the even extension of the function on (−4, 0).
15. We compute
c0 =
1
2
∫ 1
0
xP0(x) dx =
1
2
∫ 1
0
x dx =
1
4
c1 =
3
2
∫ 1
0
xP1(x) dx =
3
2
∫ 1
0
x2 dx =
1
2
c2 =
5
2
∫ 1
0
xP2(x) dx =
5
2
∫ 1
0
1
2
(3x3 − x)dx = 5
16
c3 =
7
2
∫ 1
0
xP3(x) dx =
7
2
∫ 1
0
1
2
(5x4 − 3x2)dx = 0
c4 =
9
2
∫ 1
0
xP4(x) dx =
9
2
∫ 1
0
1
8
(35x5 − 30x3 + 3x)dx = − 3
32
c5 =
11
2
∫ 1
0
xP5(x) dx =
11
2
∫ 1
0
1
8
(63x6 − 70x4 + 15x2)dx = 0
c6 =
13
2
∫ 1
0
xP6(x) dx =
13
2
∫ 1
0
1
16
(231x7 − 315x5 + 105x3 − 5x)dx = 13
256
.
Thus
f(x) =
1
4
P0(x) +
1
2
P1(x) +
5
16
P2(x)− 332P4(x) +
13
256
P6(x) + · · · .
The figure above is the graph of S5(x) = 14P0(x) +
1
2P1(x) +
5
16P2(x)− 332P4(x) + 13256P6(x).
671
-1 -0.5 0.5 1
x
1
2
3
S5
12.6 Bessel and Legendre Series
16. We compute
c0 =
1
2
∫ 1
−1
exP0(x) dx =
1
2
∫ 1
−1
exdx =
1
2
(e− e−1)
c1 =
3
2
∫ 1
−1
exP1(x) dx =
3
2
∫ 1
−1
xex dx = 3e−1
c2 =
5
2
∫ 1
−1
exP2(x) dx =
5
2
∫ 1
−1
1
2
(3x2ex − ex)dx
=
5
2
(e− 7e−1)
c3 =
7
2
∫ 1
−1
exP3(x) dx =
7
2
∫ 1
−1
1
2
(5x3ex − 3xex)dx = 7
2
(−5e + 37e−1)
c4 =
9
2
∫ 1
−1
exP4(x) dx =
9
2
∫ 1
−1
1
8
(35x4ex − 30x2ex + 3ex)dx = 9
2
(36e− 266e−1).
Thus
f(x) =
1
2
(e− e−1)P0(x) + 3e−1P1(x) + 52(e− 7e
−1)P2(x)
+
7
2
(−5e + 37e−1)P3(x) + 92(36e− 266e
−1)P4(x) + · · · .
The figure above is the graph of S5(x).
17. Using cos2 θ = 12 (cos 2θ + 1) we have
P2(cos θ) =
1
2
(3 cos2 θ − 1) = 3
2
cos2 θ − 1
2
=
3
4
(cos 2θ + 1)− 1
2
=
3
4
cos 2θ +
1
4
=
1
4
(3 cos 2θ + 1).
18. From Problem 17 we have
P2(cos θ) =
1
4
(3 cos 2θ + 1) or cos 2θ =
4
3
P2(cos θ)− 13 .
Then, using P0(cos θ) = 1,
F (θ) = 1− cos 2θ= 1−
[
4
3
P2(cos θ)− 13
]
=
4
3
− 4
3
P2(cos θ) =
4
3
P0(cos θ)− 43P2(cos θ).
19. If f is an even function on (−1, 1) then∫ 1
−1
f(x)P2n(x) dx = 2
∫ 1
0
f(x)P2n(x) dx
and ∫ 1
−1
f(x)P2n+1(x) dx = 0.
672
-1 -0.5 0.5 1 x
0.5
1
S4
12.6 Bessel and Legendre Series
Thus
c2n =
2(2n) + 1
2
∫ 1
−1
f(x)P2n(x) dx =
4n + 1
2
(
2
∫ 1
0
f(x)P2n(x) dx
)
= (4n + 1)
∫ 1
0
f(x)P2n(x) dx,
c2n+1 = 0, and
f(x) =
∞∑
n=0
c2nP2n(x).
20. If f is an odd function on (−1, 1) then∫ 1
−1
f(x)P2n(x) dx = 0
and ∫ 1
−1
f(x)P2n+1(x) dx = 2
∫ 1
0
f(x)P2n+1(x) dx.
Thus
c2n+1 =
2(2n + 1) + 1
2
∫ 1
−1
f(x)P2n+1(x) dx =
4n + 3
2
(
2
∫ 1
0
f(x)P2n+1(x) dx
)
= (4n + 3)
∫ 1
0
f(x)P2n+1(x) dx,
c2n = 0, and
f(x) =
∞∑
n=0
c2n+1P2n+1(x).
21. From (26) in Problem 19 in the text we find
c0 =
∫ 1
0
xP0(x) dx =
∫ 1
0
x dx =
1
2
,
c2 = 5
∫ 1
0
xP2(x) dx = 5
∫ 1
0
1
2
(3x3 − x)dx = 5
8
,
c4 = 9
∫ 1
0
xP4(x) dx = 9
∫ 1
0
1
8
(35x5 − 30x3 + 3x)dx = − 3
16
,
and
c6 = 13
∫ 1
0
xP6(x) dx = 13
∫ 1
0
1
16
(231x7 − 315x5 + 105x3 − 5x)dx = 13
128
.
Hence, from (25) in the text,
f(x) =
1
2
P0(x) +
5
8
P2(x)− 316P4(x) +
13
128
P6 + · · · .
On the interval −1 < x < 1 this series represents the function f(x) = |x|.
673
-1 -0.5 0.5 1 x
-1
-0.5
0.5
1
S4
12.6 Bessel and Legendre Series
22. From (28) in Problem 20 in the text we find
c1 = 3
∫ 1
0
P1(x) dx = 3
∫ 1
0
x dx =
3
2
,
c3 = 7
∫ 1
0
P3(x) dx = 7
∫ 1
0
1
2
(
5x3 − 3x) dx = −7
8
,
c5 = 11
∫ 1
0
P5(x) dx = 11
∫ 1
0
1
8
(
63x5 − 70x3 + 15x) dx = 11
16
and
c7 = 15
∫ 1
0
P7(x) dx = 15
∫ 1
0
1
16
(
429x7 − 693x5 + 315x3 − 35x) dx = − 75
128
.
Hence, from (27) in the text,
f(x) =
3
2
P1(x)− 78P3(x) +
11
16
P5(x)− 75128P7(x) + · · · .
On the interval −1 < x < 1 this series represents the odd function
f(x) =
{−1, −1 < x < 0
1, 0 < x < 1.
23. Since there is a Legendre polynomial of any specified degree, every polynomial can be represented as a finite
linear combination of Legendre polynomials.
24. We want to express both x2 and x3 as linear combinations of P0(x) = 1, P1(x) = x, P2(x) = 12 (3x
2 − 1), and
P3(x) = 12 (5x
3 − 3x). Setting
x2 = c0P0(x) + c1P1(x) + c2P2(x) = c0 + c1x + c2
[
1
2
(3x2 − 1)
]
=
(
c0 − 32c2
)
+ c1x +
3
2
c2x
2,
we obtain the system
c0 − 12 c2 = 0
c1 = 0
3
2
c2 = 1.
The solution is c0 = 13 , c1 = 0, c2 =
2
3 . Thus, x
2 = 13 P0(x) +
2
3 P2(x). Setting
x3 = c0P0(x) + c1P1(x) + c2P2(x) + c3P3(x) = c0 + c1x + c2
[
1
2
(3x2 − 1)
]
+ c3
[
1
2
(5x3 − 3x)
]
=
(
c0 − 12 c2
)
+
(
c1 − 32 c3
)
x +
3
2
c2x
2 +
5
2
c3x
3,
we obtain the system
c0 − 12 c2 = 0
c1 − 32 c3 = 0
3
2
c2 = 0
5
2
c3 = 1.
The solution is c0 = 0, c1 = 35 , c2 = 0, c3 =
2
5 . Thus x
3 = 35 P1(x) +
2
5 P3(x).
674
CHAPTER 12 REVIEW EXERCISES
CHAPTER 12 REVIEW EXERCISES
1. True, since
∫ π
−π(x
2 − 1)x5 dx = 0
2. Even, since if f and g are odd then h(−x) = f(−x)g(−x) = −f(x)[−g(x)] = f(x)g(x) = h(x)
3. cosine, since f is even
4. True
5. False; the Sturm-Liouville problem,
d
dx
[r(x)y′] + λp(x)y = 0, y′(a) = 0, y′(b) = 0,
on the interval [a, b], has eigenvalue λ = 0.
6. Periodically extending the function we see that at x = −1 the function converges to 12 (−1 + 0) = − 12 ; at x = 0
it converges to 12 (0 + 1) =
1
2 , and at x = 1 it converges to
1
2 (−1 + 0) = − 12 .
7. The Fourier series will converge to 1, the cosine series to 1, and the sine series to 0 at x = 0. Respectively, this
is because the rule (x2 + 1) defining f(x) determines a continuous function on (−3, 3), the even extension of f
to (−3, 0) is continuous at 0, and the odd extension of f to (−3, 0) approaches −1 as x approaches 0 from the
left.
8. cos 5x, since the general solution is y = c1 cosαx + c2 sinαx and y′(0) = 0 implies c2 = 0.
9. True, since
∫ 1
0
P2m(x)P2n(x) dx = 12
∫ 1
−1 P2m(x)P2n(x) dx = 0 when m �= n.
10. Since Pn(x) is orthogonal to P0(x) = 1 for n > 0,∫ 1
−1
Pn(x) dx =
∫ 1
−1
P0(x)Pn(x) dx = 0.
11. We know from a half-angle formula in trigonometry that cos2 x = 12 +
1
2 cos 2x, which is a cosine series.
12. (a) For m �= n∫ L
0
sin
(2n + 1)π
2L
x sin
(2m + 1)π
2L
xdx =
1
2
∫ L
0
(
cos
n−m
L
πx− cos n + m + 1
L
πx
)
dx = 0.
(b) From ∫ L
0
sin2
(2n + 1)π
2L
xdx =
∫ L
0
(
1
2
− 1
2
cos
(2n + 1)π
L
x
)
dx =
L
2
we see that ∥∥∥∥sin (2n + 1)π2L x
∥∥∥∥ =
√
L
2
.
13. Since
a0 =
∫ 0
−1
(−2x) dx = 1,
an =
∫ 0
−1
(−2x) cosnπx dx = 2
n2π2
[(−1)n − 1],
and
675
CHAPTER 12 REVIEW EXERCISES
bn =
∫ 0
−1
(−2x) sinnπx dx = 4
nπ
(−1)n
for n = 1, 2, 3, . . . we have
f(x) =
1
2
+
∞∑
n=1
(
2
n2π2
[(−1)n − 1] cosnπx + 4
nπ
(−1)n sinnπx
)
.
14. Since
a0 =
∫ 1
−1
(2x2 − 1) dx = −2
3
,
an =
∫ 1
−1
(2x2 − 1) cosnπx dx = 8
n2π2
(−1)n,
and
bn =
∫ 1
−1
(2x2 − 1) sinnπx dx = 0
for n = 1, 2, 3, . . . we have
f(x) = −1
3
+
∞∑
n=1
8
n2π2
(−1)n cosnπx.
15. Since
a0 =
2
1
∫ 1
0
exdx = 2(e− 1)
and
an =
2
1
∫ 1
0
ex cosnπx dx =
2
1 + n2π2
[e(−1)n − 1],
for n = 1, 2, 3, . . . , we have the cosine series
f(x) = e− 1 + 2
∞∑
n=1
e(−1)n − 1
1 + n2π2
cosnπx.
Since
bn =
2
1
∫ 1
0
ex sinnπx dx =
2nπ
1 + n2π2
[1− e(−1)n],
for n = 1, 2, 3, . . . , we have the sine series
f(x) = 2π
∞∑
n=1
n[1− e(−1)n]
1 + n2π2
sinnπx.
676
-3 -2 -1 1 2 3
x
-3
-2
-1
1
2
3
f
-3 -2 -1 1 2 3
x
-3
-2
-1
1
2
3
f
-3 -2 -1 1 2 3
x
-3
-2
-1
1
2
3
f
-3 -2 -1 1 2 3
x
-3
-2
-1
1
2
3
f
CHAPTER 12 REVIEW EXERCISES
16.
f(x) = |x| − x, −1 < x < 1 f(x) = 2x2 − 1, −1 < x < 1
f(x) =
{
e−x, −1 < x < 0
ex, 0 < x < 1
f(x) =
{−e−x, −1 < x < 0
ex, 0 < x < 1
17. For λ = α2 > 0 a general solution of the given differential equation is
y = c1 cos(3α lnx) + c2 sin(3α lnx)
and
y′ = −3c1α
x
sin(3α lnx) +
3c2α
x
cos(3α lnx).
Since ln 1 = 0, the boundary condition y′(1) = 0 implies c2 = 0. Therefore
y = c1 cos(3α lnx).
Using ln e = 1 we find that y(e) = 0 implies c1 cos 3α = 0 or 3α = (2n − 1)π/2, for
n = 1, 2, 3, . . . . The eigenvalues are λ = α2 = (2n− 1)2π2/36 with corresponding eigenfunctions
cos[(2n− 1)π(lnx)/2] for n = 1, 2, 3, . . . .
677
CHAPTER 12 REVIEW EXERCISES
18. To obtain the self-adjoint form of the differential equation in Problem 17 we note that an integrating factor is
(1/x2)e
∫
dx/x = 1/x. Thus the weight function is 1/x and an orthogonality relation is∫ e
1
1
x
cos
(
2n− 1
2
π lnx
)
cos
(
2m− 1
2
π lnx
)
dx = 0, m �= n.
19. Since the coefficient of y in the differential equation is n2, the weight function is the integrating factor
1
a(x)
e
∫
(b/a)dx =
1
1− x2 e
∫
− x
1−x2 dx =
1
1− x2 e
1
2 ln(1−x2) =
√
1− x2
1− x2 =
1√
1− x2
on the interval [−1, 1]. The orthogonality relation is∫ 1
−1
1√
1− x2 Tm(x)Tn(x) dx = 0, m �= n.
20. Expanding in a full Fourier series we have
a0 =
1
2
(∫ 2
0
x dx +
∫ 4
2
2 dx
)
= 3
an =
1
2
(∫ 2
0
x cos
nπx
2
dx +
∫ 4
2
2 cos
nπx
2
dx
)
= 2
(−1)n − 1
n2π2
bn =
1
2
(∫ 2
0
x sin
nπx
2
dx +
∫ 4
2
2 sin
nπx
2
dx
)
= 4
−1
nπ
so
f(x) =
3
2
+ 2
∞∑
n=1
((−1)n − 1
n2π2
cos
nπx
2
− 2
nπ
sin
nπx
2
)
.
21. The boundary condition indicates that we use (15) and (16) of Section 12.6 in the text. With b = 4 we obtain
ci =
2
16J21 (4αi)
∫ 4
0
xJ0(αix)f(x) dx
=
1
8J21 (4αi)
∫ 2
0
xJ0(αix) dx
t = αix dt = αi dx
=
1
8J21 (4αi)
· 1
α2i
∫ 2αi
0
tJ0(t) dt
=
1
8J21 (4αi)
∫ 2αi
0
d
dt
[tJ1(t)] dt [From (5) in 12.6 in the text]
=
1
8J21 (4αi)
tJ1(t)
∣∣∣∣2αi
0
=
J1(2αi)
4αiJ21 (4αi)
.
Thus
f(x) =
1
4
∞∑
i=1
J1(2αi)
αiJ21 (4αi)
J0(αix).
678
CHAPTER 12 REVIEW EXERCISES
22. Since f(x) = x4 is a polynomial in x, an expansion of f in Legendre polynomials in x must terminate with
the term having the same degree as f . Using the fact that x4P1(x) and x4P3(x) are odd functions, we see
immediately that c1 = c3 = 0. Now
c0 =
1
2
∫ 1
−1
x4P0(x) dx =
1
2
∫ 1
−1
x4dx =
1
5
c2 =
5
2
∫ 1
−1
x4P2(x) dx =
5
2
∫ 1
−1
1
2
(3x6 − x4)dx = 4
7
c4 =
9
2
∫ 1
−1
x4P4(x) dx =
9
2
∫ 1
−1
1
8
(35x8 − 30x6 + 3x4)dx = 8
35
.
Thus
f(x) =
1
5
P0(x) +
4
7
P2(x) +
8
35
P4(x).
679