Part IV Fourier Series and Partial Differential Equations
Orthogonal Functions and
Fourier Series
12
EXERCISES 12.1
Orthogonal Functions
2
xx2 dx =
1.
−2
2
1 4 x =0
4
−2
1
x3 (x2 + 1)dx =
2.
−1
2
3.
1
1 6 1
x + x4
6
4
−1
ex (xe−x − e−x )dx =
0
−1
2
(x − 1)dx =
0
π
2
1 2
x − x = 0
2
0
π
1
3
cos x sin x dx = sin x = 0
3
0
2
4.
0
5.
1
=0
π/2
5π/4
1
x cos 2x dx =
2
−π/2
ex sin x dx =
6.
π/4
π/2
1
=0
cos 2x + x sin 2x 2
−π/2
5π/4
1 x
1
=0
e sin x − ex cos x 2
2
π/4
7. For m = n
π/2
sin(2n + 1)x sin(2m + 1)x dx
0
=
1
2
π/2 cos 2(n − m)x − cos 2(n + m + 1)x dx
0
π/2
π/2
1
1
=
−
sin 2(n − m)x sin 2(n + m + 1)x = 0.
4(n − m)
4(n + m + 1)
0
0
634
12.1 Orthogonal Functions
For m = n
π/2
π/2
2
sin (2n + 1)x dx =
0
0
1 1
− cos 2(2n + 1)x dx
2 2
π/2
π/2
1 1
π
= x
−
sin 2(2n + 1)x =
2 0
4(2n + 1)
4
0
so that
sin(2n + 1)x =
1√
π.
2
8. For m = n
π/2
cos(2n + 1)x cos(2m + 1)x dx
0
=
=
For m = n
1
2
π/2 cos 2(n − m)x + cos 2(n + m + 1)x dx
0
π/2
π/2
1
1
+
sin 2(n − m)x sin 2(n + m + 1)x = 0.
4(n − m)
4(n + m + 1)
0
0
π/2
2
cos (2n + 1)x dx =
0
0
=
π/2
1 1
+ cos 2(2n + 1)x dx
2 2
π/2
π/2
1 1
π
+
x
sin 2(2n + 1)x =
2 0
4(2n + 1)
4
0
so that
cos(2n + 1)x =
9. For m = n
π
1
2
sin nx sin mx dx =
0
π
cos(n − m)x − cos(n + m)x dx
0
π
π
1
1
=
sin(n − m)x −
sin(n + m)x = 0.
2(n − m)
2(n
+
m)
0
0
For m = n
π
π
sin2 nx dx =
0
0
π
π
1 1
1
π
1 − cos 2nx dx = x −
sin 2nx =
2 2
2 0
4n
2
0
so that
sin nx =
10. For m = n
0
For m = n
p
1√
π.
2
π
.
2
(n + m)π
(n − m)π
x − cos
x dx
cos
p
p
0
p
p
p
(n − m)π (n + m)π p
=
sin
x −
sin
x = 0.
2(n − m)π
p
2(n + m)π
p
0
0
nπ
mπ
1
sin
x sin
x dx =
p
p
2
p
sin2
0
nπ
x dx =
p
0
p
p
p
p
1 1
p
p
2nπ
1 2nπ − cos
x dx = x −
sin
x =
2 2
p
2 0
4nπ
p
2
0
635
12.1 Orthogonal Functions
so that
sin nπ x = p .
p 2
11. For m = n
0
For m = n
p
(n − m)π
(n + m)π
cos
x + cos
x dx
p
p
0
p
p
p
(n − m)π (n + m)π p
=
sin
x +
sin
x = 0.
2(n − m)π
p
2(n + m)π
p
0
0
nπ
mπ
1
cos
x cos
x dx =
p
p
2
p
nπ
cos
x dx =
p
p
2
0
Also
p
0
0
p
p
p
1 1
p
p
2nπ
1 2nπ + cos
x dx = x +
sin
x = .
2 2
p
2 0
4nπ
p
2
0
p
p
nπ nπ
x dx =
sin
x =0
1 · cos
p
nπ
p 0
so that
1 =
√
p
p
12 dx = p
and
0
cos nπ x = p .
p 2
and
12. For m = n, we use Problems 11 and 10:
p
p
mπ
mπ
nπ
nπ
x cos
x dx = 2
x cos
x dx = 0
cos
cos
p
p
p
p
−p
0
p
sin
−p
Also
nπ
mπ
x sin
x dx = 2
p
p
p
mπ
1
nπ
x cos
x dx =
sin
p
p
2
−p
p
−p
p
sin
0
nπ
mπ
x sin
x dx = 0.
p
p
(n + m)π
(n − m)π
x + sin
x dx = 0,
sin
p
p
p
nπ
p
nπ 1 · cos
= 0,
x dx =
sin
x
p
nπ
p −p
−p
p
p
−p
and
1 · sin
p
p
p
nπ nπ
x dx = −
cos
x = 0,
p
nπ
p
−p
nπ
nπ
sin
x cos
x dx =
p
p
−p
For m = n
p
cos2
−p
p
sin2
−p
and
p
1
2nπ
p
2nπ sin
x dx = −
cos
x = 0.
2
p
4nπ
p
−p
p
−p
nπ
x dx =
p
nπ
x dx =
p
p
1 1
2nπ
+ cos
x dx = p,
2 2
p
1 1
2nπ
− cos
x dx = p,
2 2
p
−p
p
−p
p
12 dx = 2p
−p
so that
1 =
2p ,
cos nπ x = √p ,
p 636
and
sin nπ x = √p .
p 12.1 Orthogonal Functions
13. Since
∞
−x2
e
−∞
∞
−x2
e
−∞
−x2
· 1 · 2x dx = −e
· 1 · (4x − 2) dx = 2
2
−x2
x 2xe
−∞
−e
0
dx − 2
∞
2 = 2 −xe−x +
−∞
and
∞
−x2
e
−∞
· 2x · (4x − 2) dx = 4
2
∞
2
x
−x2
2xe
−∞
2 −x2
= 4 −x e
= 4 −x e
e−x dx
2
dx − 4
−2
∞
e−x dx
2
−∞
= 0,
∞
xe−x dx
2
−∞
∞
−x2
xe
dx
−∞
2 −x2
−∞
2
0
∞
+2
0
e−x dx
−∞
∞
2 −xe−x −∞
2 −x2
∞
∞
−∞
−∞
0
2 = 2 −xe−x ∞
= 0,
−x2
−∞
∞
0
−x e
∞
0
−4
∞
+2
∞
xe−x dx
2
−∞
2xe−x dx = 0,
2
−∞
the functions are orthogonal.
14. Since
∞
−x
e
−x
· 1(1 − x) dx = (x − 1)e
0
∞
−x
e
·1·
0
∞ −
0
∞
e−x dx = 0,
0
0
∞ ∞
1 2
1 2 −x e−x (x − 2) dx
x − 2x + 1 dx = 2x − 1 − x e +
2
2
0
0
∞ ∞
−x = 1 + (2 − x)e +
e−x dx = 0,
0
0
and
∞
1
e−x · (1 − x) x2 −2x + 1 dx
2
∞
1 3 5 2
−x
=
− x + x − 3x + 1 dx
e
2
2
0
∞ ∞
1 3 5 2
3 2
−x
−x
=e
− x + 5x − 3 dx
e
x − x + 3x − 1 +
2
2
2
0
0
∞ ∞
3 2
= 1 + e−x
x − 5x + 3 +
e−x (5 − 3x) dx
2
0
0
∞
∞
−x
= 1 − 3 + e (3x − 5) −3
e−x dx = 0,
0
0
the functions are orthogonal.
15. By orthogonality
b
a
φ0 (x)φn (x)dx = 0 for n = 1, 2, 3, . . . ; that is,
637
b
a
φn (x)dx = 0 for n = 1, 2, 3, . . . .
12.1 Orthogonal Functions
16. Using the facts that φ0 and φ1 are orthogonal to φn for n > 1, we have
b
b
(αx + β)φn (x) dx = α
a
a
b
1 · φn (x) dx
xφn (x) dx + β
a
b
=α
b
φ1 (x)φn (x) dx + β
a
φ0 (x)φn (x) dx
a
=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
2
b
φm (x) + φn (x) =
2
[φm (x) + φn (x)] dx =
a
φ2m (x) + 2φm (x)φn (x) + φ2n (x) dx
a
b
b
φ2m (x)dx + 2
=
b
a
a
b
φ2n (x) dx
φm (x)φn (x)dx +
a
2
2
= φm (x) + φn (x) .
18. Setting
0=
−2
and
2
f3 (x)f1 (x) dx =
0=
−2
16 64
x2 + c1 x3 + c2 x4 dx =
+ c2
3
5
64
x3 + c1 x4 + c2 x5 dx =
c1
5
−2
2
2
f3 (x)f2 (x) dx =
2
−2
we obtain c1 = 0 and c2 = −5/12.
19. Since sin nx is an odd function on [−π, π],
π
(1, sin nx) =
sin nx dx = 0
−π
and f (x) = 1 is orthogonal to every member of {sin nx}. Thus {sin nx} is not complete.
b
b
b
20. (f1 + f2 , f3 ) =
[f1 (x) + f2 (x)]f3 (x) dx =
f1 (x)f3 (x) dx +
f2 (x)f3 (x) dx = (f1 , f3 ) + (f2 , f3 )
a
a
a
21. (a) The fundamental period is 2π/2π = 1.
(b) The fundamental period is 2π/(4/L) = 12 πL.
(c) The fundamental period of sin x + 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 )
(f3 , φ1 )
(f3 , φ2 )
φ0 (x) −
φ1 (x) −
φ2 (x).
(φ0 , φ0 )
(φ1 , φ1 )
(φ2 , φ2 )
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.
(f1 , φ0 )
(f1 , φ0 )
(φ0 , φ1 ) = φ0 , f1 −
φ0 = (φ0 , f1 ) −
(φ0 , φ0 ) = (φ0 , f1 ) − (f1 , φ0 ) = 0
(φ0 , φ0 )
(φ0 , φ0 )
(f2 , φ0 )
(f2 , φ1 )
(f2 , φ0 )
(f2 , φ1 )
(φ0 , φ2 ) = φ0 , f2 −
φ0 −
φ1 = (φ0 , f2 ) −
(φ0 , φ0 ) −
(φ0 , φ1 )
(φ0 , φ0 )
(φ1 , φ1 )
(φ0 , φ0 )
(φ1 , φ1 )
= (φ0 , f2 ) − (f2 , φ0 ) − 0 = 0
(f2 , φ0 )
(f2 , φ1 )
(f2 , φ0 )
(f2 , φ1 )
(φ1 , φ2 ) = φ1 , f2 −
φ0 −
φ1 = (φ1 , f2 ) −
(φ1 , φ0 ) −
(φ1 , φ1 )
(φ0 , φ0 )
(φ1 , φ1 )
(φ0 , φ0 )
(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
1
1
(f1 , φ0 ) = (x, 1) =
x dx = 0 and (φ0 , φ0 ) =
1 dx = 2,
−1
−1
so
φ1 (x) = f1 (x) −
Next
(f2 , φ0 ) = (x2 , 1) =
1
x2 dx =
−1
(f1 , φ0 )
0
φ0 (x) = x − (1) = 1.
(φ0 , φ0 )
2
2
, (f2 , φ1 ) = (x2 , x) =
3
1
−1
x3 dx = 0, and (φ1 , φ1 ) =
1
x2 dx =
−1
2
,
3
so
φ2 (x) = f2 (x) −
(f2 , φ0 )
(f2 , φ1 )
2/3
1
0
φ0 (x) −
φ1 (x) = x2 −
(1) − (x) = x2 − .
(φ0 , φ0 )
(φ1 , φ1 )
2
2
3
Finally,
(f3 , φ0 ) = (x3 , 1) =
and
1
x3 dx = 0,
(f3 , φ1 ) = (x3 , x) =
−1
x3 , x2 −
(f3 , φ2 ) =
1
3
1
=
−1
x5 −
1
x4 dx =
−1
2
,
5
1 3
x dx = 0,
3
so
φ3 (x) = f3 (x) −
(f3 , φ0 )
(f3 , φ1 )
(f3 , φ2 )
2/5
3
φ0 (x) −
φ1 (x) −
φ2 (x) = x3 − 0 −
(x) − 0 = x3 − x.
(φ0 , φ0 )
(φ1 , φ1 )
(φ2 , φ2 )
2/3
5
(b) Recall from Section 5.3 that the first four Legendre polynomials are P0 (x) = 1, P1 (x) = x, P2 (x) =
and P3 (x) = 52 x3 − 32 x. We then see that φ0 (x) = P0 (x), φ1 (x) = P1 (x), φ2 (x) = x2 −
2
3
2 5 3
3
2
3
3 P2 (x), and φ3 (x) = x − 5 x = 5 ( 2 x − 2 x) = 5 P3 (x).
24. (i): (f1 , f2 ) =
b
a
(ii): (kf1 , f2 ) =
f1 (x)f2 (x)dx =
b
a
b
a
(iv): (f1 + f2 , f3 ) =
=
b
[f (x)
a 1
b
a
b
a
x2 − 12 ,
= 23 ( 32 x − 12 ) =
f2 (x)f1 (x)dx = (f2 , f1 ).
kf1 (x)f2 (x)dx = k
(iii): If f1 (x) = 0 then (f1 , f1 ) =
1
3
3
2
2
b
a
f1 (x)f2 (x)dx = k(f1 , f2 ).
0 dx = 0; if f1 (x) = 0 then (f1 , f1 ) =
+ f2 (x)]f3 (x)dx =
f1 (x)f3 (x)dx +
b
a
b
[f (x)f3 (x)
a 1
b
[f (x)]2 dx
a 1
+ f2 (x)f3 (x)]dx
f2 (x)f3 (x)dx = (f1 , f3 ) + (f2 , f3 ).
639
> 0 since [f1 (x)]2 > 0.
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
1
an =
π
bn =
1
π
π
f (x) dx =
−π
π
1
π
π
1 dx = 1
0
nπ
1
f (x) cos
x dx =
π
π
−π
π
f (x) sin
−π
nπ
1
x dx =
π
π
π
cos nx dx = 0
0
π
1
1
(1 − cos nπ) =
[1 − (−1)n ]
nπ
nπ
sin nx dx =
0
∞
1
1 1 − (−1)n
f (x) = +
sin nx
2 π n=1
n
1 0
1 π
f (x) dx =
(−1) dx +
2 dx = 1
π −π
π 0
−π
1 π
1 0
1 π
an =
f (x) cos nx dx =
− cos nx dx +
2 cos nx dx = 0
π −π
π −π
π 0
1 π
1 0
1 π
3
bn =
[1 − (−1)n ]
f (x) sin nx dx =
− sin nx dx +
2 sin nx dx =
π −π
π −π
π 0
nπ
1
2. a0 =
π
f (x) =
3. a0 =
π
∞
1
3 1 − (−1)n
+
sin nx
2 π n=1
n
1
f (x) dx =
−1
1
bn =
x dx =
0
−1
1
1
cos nπx dx +
−1
0
f (x) sin nπx dx =
−1
3
2
0
f (x) cos nπx dx =
1
1 dx +
−1
an =
0
x cos nπx dx =
0
1
x sin nπx dx = −
sin nπx dx +
−1
0
3
1
(−1) − 1
f (x) = +
cos nπx −
sin nπx
4 n=1
n2 π 2
nπ
4. a0 =
n
1
f (x) dx =
−1
1
an =
1
1
2
x dx =
0
1
f (x) cos nπx dx =
bn =
∞ −1
1
x cos nπx dx =
0
1
f (x) sin nπx dx =
−1
x sin nπx dx =
0
1
[(−1)n − 1]
n2 π 2
1
[(−1)n − 1]
n2 π 2
(−1)n+1
nπ
640
1
nπ
12.2 Fourier Series
∞ 1 (−1)n − 1
(−1)n+1
f (x) = +
cos nπx +
sin nπx
4 n=1
n2 π 2
nπ
1 π
1 π 2
1
5. a0 =
f (x) dx =
x dx = π 2
π −π
π 0
3
π
π
π
1
1
1 x2
2 π
2(−1)n
2
an =
f (x) cos nx dx =
x cos nx dx =
x sin nx dx =
sin nx −
π −π
π 0
π π
n 0
n2
0
π
2
π
π
x
1
1
2
π
2
− cos nx +
bn =
x2 sin nx dx =
x cos nx dx = (−1)n+1 + 3 [(−1)n − 1]
π 0
π
n
n 0
n
n π
0
∞
2
n
n
2(−1)
π
π
2[(−1) − 1]
f (x) =
+
(−1)n+1 +
sin nx
cos nx +
2
6
n
n
n3 π
n=1
1 0 2
1 π 2
5
f (x) dx =
π dx +
π − x2 dx = π 2
π −π
π 0
3
−π
π
0
π
1
1
1
π 2 − x2 cos nx dx
an =
f (x) cos nx dx =
π 2 cos nx dx +
π −π
π −π
π 0
π
2
π
2
1 π −x
2
2
=
x sin nx dx = 2 (−1)n+1
sin nx +
π
n
n 0
n
0
π
0
1
1
1 π 2
2
bn =
f (x) sin nx dx =
π sin nx dx +
π − x2 sin nx dx
π −π
π −π
π 0
π
π
1 x2 − π 2
2 π
π
2
n
= [(−1) − 1] +
cos nx −
x cos nx dx = (−1)n + 3 [1 − (−1)n ]
n
π
n
n 0
n
n π
0
∞ 2
n
5π
2
π
2[1 − (−1) ]
f (x) =
+
(−1)n +
sin nx
(−1)n+1 cos nx +
2
6
n
n
n3 π
n=1
1 π
1 π
7. a0 =
f (x) dx =
(x + π) dx = 2π
π −π
π −π
1 π
1 π
an =
f (x) cos nx dx =
(x + π) cos nx dx = 0
π −π
π −π
π
1
2
bn =
f (x) sin nx dx = (−1)n+1
π −π
n
1
6. a0 =
π
π
∞
2
(−1)n+1 sin nx
n
n=1
1 π
1 π
8. a0 =
f (x) dx =
(3 − 2x) dx = 6
π −π
π −π
1 π
1 π
an =
f (x) cos nx dx =
(3 − 2x) cos nx dx = 0
π −π
π −π
1 π
4
bn =
(3 − 2x) sin nx dx = (−1)n
π −π
n
f (x) = π +
∞
(−1)n
sin nx
n
n=1
1 π
1 π
2
9. a0 =
f (x) dx =
sin x dx =
π −π
π 0
π
π
1 π
1 π
1
sin(1 + n)x + sin(1 − n)x dx
f (x) cos nx dx =
sin x cos nx dx =
an =
π −π
π 0
2π 0
f (x) = 3 + 4
641
12.2 Fourier Series
1 + (−1)n
for n = 2, 3, 4, . . .
π(1 − n2 )
π
1
a1 =
sin 2x dx = 0
2π 0
1 π
1 π
bn =
f (x) sin nx dx =
sin x sin nx dx
π −π
π 0
π
1
cos(1 − n)x − cos(1 + n)x dx = 0 for n = 2, 3, 4, . . .
=
2π 0
π
1
1
b1 =
(1 − cos 2x) dx =
2π 0
2
∞
1 + (−1)n
1
1
f (x) = + sin x +
cos nx
π 2
π(1 − n2 )
n=2
=
2 π/2
2
f (x) dx =
cos x dx =
π 0
π
−π/2
π/2
π/2
2
2
1 π/2 an =
cos(2n − 1)x + cos(2n + 1)x dx
f (x) cos 2nx dx =
cos x cos 2nx dx =
π −π/2
π 0
π 0
2
10. a0 =
π
π/2
2(−1)n+1
π(4n2 − 1)
2 π/2
2 π/2
1 π/2 bn =
sin(2n − 1)x + sin(2n + 1)x dx
f (x) sin 2nx dx =
cos x sin 2nx dx =
π −π/2
π 0
π 0
=
4n
π(4n2 − 1)
∞ 1 2(−1)n+1
4n
f (x) = +
cos
2nx
+
sin
2nx
π n=1 π(4n2 − 1)
π(4n2 − 1)
=
0
1
1
1
f (x) dx =
−2 dx +
1 dx = −
2
2
−2
−1
0
0
2
1
1
nπ
nπ
nπ
1
1
nπ
an =
f (x) cos
(−2) cos
cos
x dx =
x dx +
x dx = −
sin
2 −2
2
2
2
2
nπ
2
−1
0
0
2
1
1
nπ
nπ
nπ
1
3
nπ bn =
f (x) sin
(−2) sin
sin
x dx =
x dx +
x dx =
1 − cos
2 −2
2
2
2
2
nπ
2
−1
0
∞ 1
1
nπ
nπ
3
nπ
nπ
f (x) = − +
−
sin
cos
x+
1 − cos
sin
x
4 n=1
nπ
2
2
nπ
2
2
1
11. a0 =
2
2
1
2
1
3
f (x) dx =
x dx +
1 dx =
2
4
−2
0
1
1
2
2
nπ
1
nπ
nπ
nπ
1
2 an =
f (x) cos
x cos
cos
x dx =
x dx +
x dx = 2 2 cos
−1
2 −2
2
2
2
2
n π
2
0
1
1
2
2
1
nπ
nπ
nπ
1
bn =
f (x) sin
x sin
sin
x dx =
x dx +
x dx
2 −2
2
2
2
2
0
1
nπ nπ
2 +
(−1)n+1
= 2 2 sin
n π
2
2
∞ 3
2 nπ
nπ nπ
nπ
nπ
2 n+1
f (x) = +
cos
sin
sin
−
1
cos
x
+
+
(−1)
x
8 n=1 n2 π 2
2
2
n2 π 2
2
2
2
1
12. a0 =
2
2
642
12.2 Fourier Series
13. a0 =
1
5
5
f (x) dx =
−5
5
1
5
0
−5
(1 + x) dx
=
0
9
2
5
5
nπ
nπ
x dx +
x dx = 2 2 [(−1)n − 1]
(1 + x) cos
5
5
n π
−5
0
0
5
5
1
nπ
nπ
nπ
1
5
bn =
f (x) sin
sin
(1 + x) cos
x dx =
x dx +
x dx =
(−1)n+1
5 −5
5
5
5
5
nπ
−5
0
∞ 5
9 nπ
nπ
5
n
n+1
f (x) = +
[(−1) − 1] cos
sin
x+
(−1)
x
4 n=1 n2 π 2
5
nπ
5
0
2
1 2
1
14. a0 =
f (x) dx =
(2 + x) dx +
2 dx = 3
2 −2
2
−2
0
0
2
2
1
nπ
nπ
nπ
1
2
an =
f (x) cos
(2 + x) cos
2 cos
x dx =
x dx +
x dx = 2 2 [1 − (−1)n ]
2 −2
2
2
2
2
n π
−2
0
0
2
2
1
nπ
nπ
nπ
1
2
bn =
f (x) sin
(2 + x) sin
2 sin
x dx =
x dx +
x dx =
(−1)n+1
2 −2
2
2
2
2
nπ
−2
0
∞ 2
3 nπ
nπ
2
n
n+1
f (x) = +
[1 − (−1) ] cos
sin
x+
(−1)
x
2 n=1 n2 π 2
2
nπ
2
1 π
1 π x
1
15. a0 =
f (x) dx =
e dx = (eπ − e−π )
π −π
π −π
π
π
n π
1
(−1) (e − e−π )
an =
f (x) cos nx dx =
π −π
π(1 + n2 )
π
π
1
1
(−1)n n(e−π − eπ )
bn =
f (x) sin nx dx =
ex sin nx dx =
π −π
π −π
π(1 + n2 )
∞
eπ − e−π (−1)n (eπ − e−π )
(−1)n n(e−π − eπ )
f (x) =
+
cos nx +
sin nx
2π
π(1 + n2 )
π(1 + n2 )
n=1
1 π
1 π x
1
16. a0 =
f (x) dx =
(e − 1) dx = (eπ − π − 1)
π −π
π 0
π
1 π
1 π x
[eπ (−1)n − 1]
an =
f (x) cos nx dx =
(e − 1) cos nx dx =
π −π
π 0
π(1 + n2 )
π
π
π
1
1
1 ne (−1)n+1
n
(−1)n
1
bn =
f (x) sin nx dx =
(ex − 1) sin nx dx =
+
+
−
π −π
π 0
π
1 + n2
1 + n2
n
n
∞
π
π
n
n
(−1) − 1
e − π − 1 e (−1) − 1
n
π
n+1
f (x) =
+1 +
e (−1)
+
cos nx +
sin nx
2π
π(1 + n2 )
1 + n2
n
n=1
1
an =
5
1
nπ
x dx =
f (x) cos
5
5
−5
5
1 dx +
0
cos
17. The function in Problem 5 is discontinuous at x = π, so the corresponding Fourier series converges to π 2 /2 at
x = π. That is,
∞ π2
π 2 2(−1)n
π
2[(−1)n − 1]
n+1
=
+
(−1)
sin nπ
cos nπ +
+
2
6
n2
n
n3 π
n=1
=
and
∞
∞
π 2 2(−1)n
π2 2
π2
1
1
n
(−1)
=
=
+
+
·
·
·
+
+
+
2
1
+
6
n2
6
n2
6
22
32
n=1
n=1
π2
1
=
6
2
π2
π2
−
2
6
=1+
643
1
1
+ 2 + ···.
22
3
12.2 Fourier Series
At x = 0 the series converges to 0 and
∞
π 2 2(−1)n
π2
1
1
1
0=
=
+
+ 2 −1 + 2 − 2 + 2 − · · ·
6
n2
6
2
3
4
n=1
so
π2
1
1
1
= 1 − 2 + 2 − 2 + ···.
12
2
3
4
18. From Problem 17
π2
1
=
8
2
π2
π2
+
6
12
1
=
2
2
2
1
1
2 + 2 + 2 + ··· = 1 + 2 + 2 + ···.
3
5
3
5
19. The function in Problem 7 is continuous at x = π/2 so
∞
π
3π
2
nπ
1 1 1
=f
=π+
(−1)n+1 sin
= π + 2 1 − + − + ···
2
2
n
2
3 5 7
n=1
and
π
1 1 1
= 1 − + − + ···.
4
3 5 7
20. The function in Problem 9 is continuous at x = π/2 so
1=f
1=
and
π
1
1 1 + (−1)n
nπ
+ +
cos
π 2 n=2 π(1 − n2 )
2
1
1
2
2
2
+ +
−
+
− ···
π 2 3π 3 · 5π 5 · 7π
π =1+
or
2
∞
=
π 2
2
2
+ −
+
− ···
2
3 3·5 5·7
π
1
1
1
1
= +
−
+
− ···.
4
2 1·3 3·5 5·7
21. Writing
f (x) =
a0
π
nπ
π
nπ
+ a1 cos x + · · · + an cos
x + · · · + b1 sin x + · · · + bn sin
x + ···
2
p
p
p
p
we see that f 2 (x) consists exclusively of squared terms of the form
a20
,
4
a2n cos2
nπ
x,
p
b2n sin2
nπ
x
p
and cross-product terms, with m = n, of the form
nπ
nπ
mπ
nπ
a0 an cos
x,
a0 bn sin
x,
2am an cos
x cos
x,
p
p
p
p
mπ
mπ
nπ
nπ
2am bn cos
x sin
x,
2bm bn sin
x sin
x.
p
p
p
p
The integral of each cross-product term taken over the interval (−p, p) is zero by orthogonality. For the squared
terms we have
p
p
a20 p
nπ
nπ
a2 p
dx = 0 ,
cos2
b2n
sin2
a2n
x dx = a2n p,
x dx = b2n p.
4 −p
2
p
p
−p
−p
Thus
∞
1
1 2
RM S(f ) = a20 +
(a + b2n ) .
4
2 n=1 n
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 cos x = −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:
2 π
2
bn =
1 · sin nx dx =
[1 − (−1)n ] .
π 0
nπ
Thus
f (x) =
∞
2
[1 − (−1)n ] sin nx.
nπ
n=1
12. Since f (x) is an even function, we expand in a cosine series:
2
a0 =
1 dx = 1
1
an =
2
cos
1
Thus
2
nπ
nπ
x dx = −
sin
.
2
nπ
2
∞
f (x) =
1 −2
nπ
nπ
+
sin
cos
x.
2 n=1 nπ
2
2
13. Since f (x) is an even function, we expand in a cosine series:
2 π
a0 =
x dx = π
π 0
2 π
2
an =
x cos nx dx = 2 [(−1)n − 1].
π 0
n π
Thus
∞
f (x) =
π 2
+
[(−1)n − 1] cos nx.
2 n=1 n2 π
645
12.3 Fourier Cosine and Sine Series
14. Since f (x) is an odd function, we expand in a sine series:
2 π
2
bn =
x sin nx dx = (−1)n+1 .
π 0
n
Thus
f (x) =
∞
2
(−1)n+1 sin nx.
n
n=1
15. Since f (x) is an even function, we expand in a cosine series:
1
2
a0 = 2
x2 dx =
3
0
1
1
1
x2
2
2
an = 2
x cos nπx dx = 2
x sin nπx dx
sin nπx −
nπ
nπ 0
0
0
Thus
=
4
n2 π 2
(−1)n .
∞
1 4
f (x) = +
(−1)n cos nπx.
3 n=1 n2 π 2
16. Since f (x) is an odd function, we expand in a sine series:
1
1
1
x2
2
2
bn = 2
x sin nπx dx = 2 −
x cos nπx dx
cos nπx +
nπ
nπ 0
0
0
Thus
f (x) =
∞ 2(−1)n+1
nπ
n=1
=
2(−1)n+1
4
+ 3 3 [(−1)n − 1].
nπ
n π
4
n
+ 3 3 [(−1) − 1] sin nπx.
n π
17. Since f (x) is an even function, we expand in a cosine series:
2 π 2
4
a0 =
(π − x2 ) dx = π 2
π 0
3
π
π
2
2 π 2 − x2
2 π
4
an =
(π 2 − x2 ) cos nx dx =
x sin nx dx = 2 (−1)n+1 .
sin nx +
π 0
π
n
n
n
0
0
Thus
f (x) =
∞
2 2 4
(−1)n+1 cos nx dx.
π +
2
3
n
n=1
18. Since f (x) is an odd function, we expand in a sine series:
π
3
π
x
2 π 3
2
3 π 2
2π 2
12
bn =
− cos nx +
(−1)n+1 − 2
x sin nx dx =
x cos nx dx =
x sin nx dx
π 0
π
n
n 0
n
n π 0
0
π
x
2π 2
12
1 π
2π 2
12
n+1
(−1)
− cos nx +
(−1)n+1 + 3 (−1)n .
=
− 2
cos nx dx =
n
n π
n
n
n
n
0
0
Thus
f (x) =
∞ 2π 2
n=1
n
n+1
(−1)
12
n
+ 3 (−1) sin nx.
n
19. Since f (x) is an odd function, we expand in a sine series:
2 π
2(π + 1)
2
bn =
(x + 1) sin nx dx =
(−1)n+1 +
.
π 0
nπ
nπ
646
12.3 Fourier Cosine and Sine Series
Thus
f (x) =
∞ 2(π + 1)
nπ
n=1
(−1)n+1 +
2
nπ
sin nx.
20. Since f (x) is an odd function, we expand in a sine series:
1
1
bn = 2
(x − 1) sin nπx dx = 2
x sin nπx dx −
0
0
1
sin nπx dx
0
1
1
x
2
1
= 2 2 2 sin nπx −
cos nπx +
cos nπx = −
.
n π
nπ
nπ
nπ
0
Thus
∞
2
sin nπx.
nπ
n=1
f (x) = −
21. Since f (x) is an even function, we expand in a cosine series:
1
a0 =
x dx +
0
an =
0
2
1 dx =
1
1
nπ
x cos
x dx +
2
Thus
f (x) =
3
2
2
cos
1
nπ
nπ
4 x dx = 2 2 cos
−1 .
2
n π
2
∞
3 4 nπ
nπ
cos
+
−
1
cos
x.
2
2
4 n=1 n π
2
2
22. Since f (x) is an odd function, we expand in a sine series:
bn =
1
π
π
x sin
0
n
x dx +
2
Thus
f (x) =
2π
π sin
π
n
4
nπ
2
x dx = 2 sin
+ (−1)n+1 .
2
n π
2
n
∞ nπ
2
4
n
n+1
sin
+
(−1)
sin x.
2π
n
2
n
2
n=1
23. Since f (x) is an even function, we expand in a cosine series:
2 π
4
a0 =
sin x dx =
π 0
π
2 π
1 π
an =
sin(1 + n)x + sin(1 − n)x dx
sin x cos nx dx =
π 0
π 0
2
(1 + (−1)n ) for n = 2, 3, 4, . . .
π(1 − n2 )
1 π
a1 =
sin 2x dx = 0.
π 0
=
Thus
∞
f (x) =
2 2[1 + (−1)n ]
+
cos nx.
π n=2 π(1 − n2 )
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.]
π/2
2
4
a0 =
cos x dx =
π/2 0
π
π/2
4(−1)n+1
2
nπ
an =
x dx =
cos x cos
π/2 0
π/2
π (4n2 − 1)
Thus
∞
f (x) =
2 4(−1)n+1
+
cos 2nx.
π n=1 π (4n2 − 1)
1/2
25. a0 = 2
1 dx = 1
0
1/2
2
nπ
sin
nπ
2
0
1/2
2
nπ bn = 2
1 · sin nπx dx =
1 − cos
nπ
2
0
∞
1 2
nπ
f (x) = +
sin
cos nπx
2 n=1 nπ
2
1 · cos nπx dx =
an = 2
∞
2 nπ 1 − cos
sin nπx
nπ
2
n=1
f (x) =
1
26. a0 = 2
1 dx = 1
1/2
1
nπ
2
sin
nπ
2
1
nπ
2
bn = 2
cos
+ (−1)n+1
1 · sin nπx dx =
nπ
2
1/2
∞
1 2
nπ
f (x) = +
−
sin
cos nπx
2 n=1
nπ
2
1 · cos nπx dx = −
an = 2
1/2
f (x) =
4
27. a0 =
π
an =
4
π
4
bn =
π
f (x) =
f (x) =
∞
2 nπ
cos
+ (−1)n+1 sin nπx
nπ
2
n=1
π/2
cos x dx =
0
4
π
π/2
cos x cos 2nx dx =
0
π/2
0
2
π
2
cos x sin 2nx dx =
π
∞
π/2
[cos(2n + 1)x + cos(2n − 1)x] dx =
0
π/2
[sin(2n + 1)x + sin(2n − 1)x] dx =
0
n
2
4(−1)
+
cos 2nx
π n=1 π(1 − 4n2 )
∞
8n
sin 2nx
2 − 1)
π(4n
n=1
648
4(−1)n
π(1 − 4n2 )
8n
π(4n2 − 1)
12.3 Fourier Cosine and Sine Series
2 π
4
sin x dx =
π 0
π
2 π
1 π
2[(−1)n + 1]
an =
sin x cos nx dx =
[sin(n + 1)x − sin(n − 1)x] dx =
for n = 2, 3, 4, . . .
π 0
π 0
π(1 − n2 )
2 π
1 π
bn =
sin x sin nx dx =
[cos(n − 1)x − cos(n + 1)x] dx = 0 for n = 2, 3, 4, . . .
π 0
π 0
1 π
a1 =
sin 2x dx = 0
π 0
2 π 2
b1 =
sin x dx = 1
π 0
f (x) = sin x
∞
2
2 (−1)n + 1
f (x) = +
cos nx
π π n=2 1 − n2
28. a0 =
29. a0 =
an =
bn =
2
π
π/2
π
(π − x) dx
x dx +
0
π/2
2
π
π/2
(π − x) cos nx dx
0
2
π
π
2
π
x cos nx dx +
=
=
π/2
π/2
π
(π − x) sin nx dx
x sin nx dx +
0
=
π/2
f (x) =
π
nπ
2 +
2 cos
+ (−1)n+1 − 1 cos nx
2
4 n=1 n π
2
f (x) =
∞
4
nπ
sin
sin nx
2
n π
2
n=1
30. a0 =
an =
1
π
1
π
1
bn =
π
nπ
2 n+1
2
cos
+
(−1)
−
1
n2 π
2
nπ
4
sin
2
n π
2
∞
2π
(x − π) dx =
π
2π
(x − π) cos
π
2π
π
2
n
nπ 4 x dx = 2
(−1)n − cos
2
n π
2
n
4
2
nπ
x dx = (−1)n+1 − 2 sin
2
n
n
π
2
π
∞
π 4 nπ n
f (x) = +
(−1)n − cos
cos x
2
4 n=1 n π
2
2
∞
2
4
nπ
n
f (x) =
(−1)n+1 − 2 sin
sin x
n
n π
2
2
n=1
(x − π) sin
1
31. a0 =
x dx +
0
2
1 dx =
1
3
2
nπ
4 nπ
x dx = 2 2 cos
−1
2
n π
2
0
1
2
nπ
nπ
nπ
4
2
bn =
x sin
1 · sin
x dx +
x dx = 2 2 sin
+
(−1)n+1
2
2
n π
2
nπ
0
1
∞
nπ
3 4 nπ
cos
f (x) = +
− 1 cos
x
2
2
4 n=1 n π
2
2
an =
1
x cos
649
12.3 Fourier Cosine and Sine Series
∞ 4
nπ
nπ
2
n+1
f (x) =
sin
sin
+
(−1)
x
2 π2
n
2
nπ
2
n=1
1
32. a0 =
0
1
1
nπ
x dx +
1 · cos
2
an =
0
1
2
(2 − x) dx =
1 dx +
3
2
2
(2 − x) cos
1
2
nπ
4 nπ
x dx = 2 2 cos
+ (−1)n+1
2
n π
2
nπ
nπ
nπ
2
4
(2 − x) sin
x dx +
x dx =
+ 2 2 sin
2
2
nπ
n
π
2
0
1
∞
3 4
nπ
nπ
f (x) = +
cos
+ (−1)n+1 cos
x
4 n=1 n2 π 2
2
2
∞ 4
nπ
2
nπ
f (x) =
+ 2 2 sin
sin
x
nπ n π
2
2
n=1
1 · sin
bn =
1
(x2 + x) dx =
33. a0 = 2
0
5
3
1
1
2(x2 + x)
2
2
(2x + 1) sin nπx dx = 2 2 [3(−1)n − 1]
sin nπx −
nπ
nπ
n
π
0
0
0
1
1
1
2(x2 + x)
2
bn = 2
(x2 + x) sin nπx dx = −
(2x + 1) cos nπx dx
cos nπx +
nπ
nπ
0
0
0
1
(x2 + x) cos nπx dx =
an = 2
4
4
(−1)n+1 + 3 3 [(−1)n − 1]
nπ
n π
∞
5 2
f (x) = +
[3(−1)n − 1] cos nπx
6 n=1 n2 π 2
∞ 4
4
f (x) =
(−1)n+1 + 3 3 [(−1)n − 1] sin nπx
nπ
n π
n=1
=
2
(2x − x2 ) dx =
34. a0 =
0
2
an =
(2x − x2 ) cos
nπ
8
x dx = 2 2 [(−1)n+1 − 1]
2
n π
(2x − x2 ) sin
nπ
16
x dx = 3 3 [1 − (−1)n ]
2
n π
0
2
bn =
4
3
0
∞
f (x) =
2 8
nπ
+
x
[(−1)n+1 − 1] cos
2
2
3 n=1 n π
2
f (x) =
∞
16
nπ
[1 − (−1)n ] sin
x
3 π3
n
2
n=1
35. a0 =
an =
bn =
1
π
1
π
1
π
2π
x2 dx =
0
8 2
π
3
2π
x2 cos nx dx =
0
2π
4
n2
x2 sin nx dx = −
0
4π
n
650
12.3 Fourier Cosine and Sine Series
∞ 4 2 4
4π
f (x) = π +
cos nx −
sin nx
3
n2
n
n=1
2 π
36. a0 =
x dx = π
π 0
2 π
an =
x cos 2nx dx = 0
π 0
π
2
1
bn =
x sin 2nx dx = −
π 0
n
∞
π 1
f (x) = −
sin 2nx
2 n=1 n
1
37. a0 = 2
(x + 1) dx = 3
0
1
an = 2
(x + 1) cos 2nπx dx = 0
0
1
(x + 1) sin 2nπx dx = −
bn = 2
0
1
nπ
∞
3 1
f (x) = −
sin 2nπx
2 n=1 nπ
38. a0 =
2
2
2
an =
2
bn =
2
2
2
(2 − x) dx = 2
0
2
(2 − x) cos nπx dx = 0
0
2
(2 − x) sin nπx dx =
0
f (x) = 1 +
2
nπ
∞
2
sin nπx
nπ
n=1
39. We have
2
π
bn =
π
5 sin nt dt =
0
so that
∞
10[1 − (−1)n ]
sin nt.
nπ
n=1
f (t) =
Substituting the assumption xp (t) =
xp + 10xp =
∞
n=1
∞
10
[1 − (−1)n ]
nπ
Bn sin nt into the differential equation then gives
Bn (10 − n2 ) sin nt =
n=1
∞
10[1 − (−1)n ]
sin nt
nπ
n=1
and so Bn = 10[1 − (−1) ]/nπ(10 − n2 ). Thus
n
xp (t) =
40. We have
bn =
2
π
∞
10 1 − (−1)n
sin nt.
π n=1 n(10 − n2 )
1
(1 − t) sin nπt dt =
0
651
2
nπ
12.3 Fourier Cosine and Sine Series
so that
f (t) =
Substituting the assumption xp (t) =
∞
n=1
xp + 10xp =
∞
2
sin nπt.
nπ
n=1
Bn sin nπt into the differential equation then gives
∞
Bn (10 − n2 π 2 ) sin nπt =
n=1
∞
2
sin nπt
nπ
n=1
and so Bn = 2/nπ(10 − n π ). Thus
2 2
xp (t) =
41. We have
2
π
a0 =
2
an =
π
∞
1
2
sin nπt.
π n=1 n(10 − n2 π 2 )
π
(2πt − t2 ) dt =
0
4 2
π
3
π
(2πt − t2 ) cos nt dt = −
0
so that
4
n2
∞
f (t) =
2π 2 4
−
cos nt.
3
n2
n=1
Substituting the assumption
∞
xp (t) =
A0 An cos nt
+
2
n=1
into the differential equation then gives
∞
∞
1 1 2
2π 2 4
An − n + 12 cos nt =
cos nt
xp + 12xp = 6A0 +
−
4
4
3
n2
n=1
n=1
and A0 = π 2 /9, An = 16/n2 (n2 − 48). Thus
xp (t) =
42. We have
a0 =
an =
2
1/2
2
1/2
∞
π2
1
+ 16
cos nt.
2 (n2 − 48)
18
n
n=1
1/2
t dt =
0
1
2
1/2
t cos 2nπt dt =
0
so that
1
[(−1)n − 1]
n2 π 2
∞
f (t) =
1 (−1)n − 1
cos 2nπt.
+
4 n=1 n2 π 2
Substituting the assumption
∞
xp (t) =
A0 An cos 2nπt
+
2
n=1
into the differential equation then gives
∞
∞
1 1 (−1)n − 1
An (12 − n2 π 2 ) cos 2nπt = +
cos 2nπt
xp + 12xp = 6A0 +
4
4 n=1 n2 π 2
n=1
652
12.3 Fourier Cosine and Sine Series
and A0 = 1/24, An = [(−1)n − 1]/n2 π 2 (12 − n2 π 2 ). Thus
xp (t) =
43. (a) The general solution is x(t) = c1 cos
∞
1
1 (−1)n − 1
+ 2
cos 2nπt.
48 π n=1 n2 (12 − n2 π 2 )
√
√
10t + c2 sin 10t + xp (t), where
xp (t) =
∞
10 1 − (−1)n
sin nt.
π n=1 n(10 − n2 )
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 + xp (t) and x (0) = 0 implies
∞
√
10 1 − (−1)n
c2 10 +
cos 0 = 0.
π n=1 10 − n2
Thus
√
c2 = −
and
∞
10 1 − (−1)n
π n=1 10 − n2
∞
√
10 1 − (−1)n 1
1
x(t) =
sin nt − √ sin 10t .
π n=1 10 − n2 n
10
(b) The graph is plotted using eight nonzero terms in the series expansion of x(t).
x
4
2
20
40
60
80
t
-2
-4
√
√
44. (a) The general solution is x(t) = c1 cos 4 3t + c2 sin 4 3t + xp (t), where
xp (t) =
∞
1
π2
+ 16
cos nt.
2
2
18
n (n − 48)
n=1
The initial condition x(0) = 0 implies c1 + xp (0) = 1 or
c1 = 1 − xp (0) = 1 −
∞
1
π2
− 16
.
2
2
18
n (n − 48)
n=1
√
√
√
√
√
Now x (t) = −4 3c1 sin 4 3t + 4 3c2 cos 4 3t + xp (t), so x (0) = 0 implies 4 3c2 + xp (0) = 0. Since
xp (0) = 0, we have c2 = 0 and
x(t) =
1−
∞
π2
1
− 16
2 (n2 − 48)
18
n
n=1
∞
√
1
π2
cos 4 3t +
+ 16
cos nt
2 (n2 − 48)
18
n
n=1
∞
√
√ π2
1
π2
=
+ 1−
cos 4 3t + 16
cos nt − cos 4 3t .
2
2
18
18
n (n − 48)
n=1
653
12.3 Fourier Cosine and Sine Series
(b) The graph is plotted using five nonzero terms in the series expansion of x(t).
x
1.5
1
0.5
2
4
6
8
10
12
14
t
-0.5
-1
45. (a) We have
bn =
2
L
L
0
w0 x
nπ
2w0
sin
x dx =
(−1)n+1
L
L
nπ
so that
w(x) =
∞
(b) If we assume yp (x) =
n=1
∞
2w0
nπ
(−1)n+1 sin
x.
nπ
L
n=1
Bn sin(nπx/L) then
yp(4) =
∞
n4 π 4
nπ
Bn sin
x
4
L
L
n=1
(4)
and so the differential equation EIyp = w(x) gives
Bn =
Thus
yp (x) =
46. We have
bn =
so that
If we assume yp (x) =
2
L
2w0 (−1)n+1 L4
.
EIn5 π 5
∞
2w0 L4 (−1)n+1
nπ
sin
x.
5
5
EIπ n=1
n
L
2L/3
w0 sin
L/3
2w0
nπ
x dx =
L
nπ
cos
nπ
2nπ
− cos
3
3
∞
2w0
2nπ
nπ
nπ
w(x) =
cos
− cos
sin
x.
nπ
3
3
L
n=1
∞
n=1
Bn sin(nπx/L) then
yp(4) (x)
∞
n4 π 4
nπ
=
Bn sin
x
4
L
L
n=1
(4)
and so the differential equation EIyp (x) = w(x) gives
Bn = 2w0 L4
Thus
yp (x) =
2nπ
cos nπ
3 − cos 3
.
EIn5 π 5
∞
2nπ
2w0 L4 cos nπ
nπ
3 − cos 3
sin
x.
5
5
EIπ n=1
n
L
654
12.3 Fourier Cosine and Sine Series
47. The graph is obtained by summing the series from n = 1 to 20. It appears that
x,
0<x<π
f (x) =
−π, π < x < 2π.
S 20
3
2
1
2
4
6
8
10
12
14
x
-1
-2
-3
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.
S 20
1
0.5
2
4
6
8
10
x
49. The function in Problem 47 is not unique; it could also be defined as

 x,
f (x) = 1,

−π,
0<x<π
x=π
π < x < 2π.
The function in Problem 48 is not unique; it could also be defined as

0,
−2 < x < −1



 x + 1,
−1 < x < 0
f (x) =

−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
x
e ,
0<x<π
f (x) =
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
f (x) dx = −
−a
0
a
f (x) dx = −
−a
f (x) dx =
−a
a
f (x) dx.
0
0
f (x) dx =
0
a
0
a
f (−x) dx +
a
f (x) dx = 2
0
0
=−
a
f (u) du +
0
a
(g) If f is odd then
a
f (−u) du +
a
f (u) du +
f (x) dx
0
a
a
f (u) du +
f (x) dx = 0.
0
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
1
cn =
4
=
2
−inπx/2
f (x)e
−2
1
dx =
4
0
−inπx/2
(−1)e
dx +
−2
2
−inπx/2
e
dx
0
i
i
1 − (−1)n
−1 + einπ + e−inπ − 1 =
[−1 + (−1)n + (−1)n − 1] =
2nπ
2nπ
nπi
and
1
c0 =
4
2
f (x)dx = 0.
−2
Thus
f (x) =
∞
1 − (−1)n inπx/2
.
e
inπ
n=−∞
n=0
656
12.4 Complex Fourier Series
2. Identifying 2p = 2 or p = 1 we have
1 2
1 2 −inπx
1 −inπx 2
−inπx
cn =
f (x)e
dx =
e
dx = −
e
2 0
2 1
2inπ
1
1 −2inπ
i
1
=−
− e−inπ = −
e
[1 − (−1)n ] =
[1 − (−1)n ]
2inπ
2inπ
2nπ
and
1
c0 =
2
2
0
1
f (x) dx =
2
Thus
2
1
.
2
dx =
1
∞
1
i 1 − (−1)n inπx
f (x) = +
.
e
2 2π n=−∞
n
n=0
3. Identifying p = 1/2 we have
cn =
1/2
f (x)e−2inπx dx =
−1/2
=−
and
1/4
e−2inπx dx = −
0
1/4
1 −2inπx e
2inπ
0
1
1
i
e−inπ/2 − 1 = −
[(−i)n − 1] =
[(−i)n − 1]
2inπ
2inπ
2nπ
1/4
c0 =
1
.
4
dx =
0
Thus
∞
1
i (−i)n − 1 2inπx
f (x) = +
e
.
4 2π n=−∞
n
n=0
4. Identifying p = π we have
π
π
1
1
f (x)e−inx/π dx =
xe−inx/π dx
2π −π
2π 0
π
1 π
ix −inx/π π(1 + in) −in
π
=
+
=
e
− 2
e
2
2
2 n
n
2n
2n
0
cn =
and
c0 =
1
2π
Thus
f (x) =
π
x dx =
0
π
.
4
∞
π π 1
−in
(1
+
in)e
+
−
1
einx .
4
2 n=−∞ n2
n=0
5. Identifying 2p = 2π or p = π we have
2π
2π
1
1
cn =
f (x)e−inx dx =
xe−inx dx
2π 0
2π 0
2π
1
ix −inx 1 + 2inπ
1
1
i
=
+
=
e
− 2 =
2
2
2π n
n
2n π
2n π
n
0
657
12.4 Complex Fourier Series
and
c0 =
1
2π
2π
x dx = π.
0
Thus
f (x) = π +
∞
i inx
e .
n
n=−∞
n=0
6. Identifying p = 1 we have
0
1
1
x −inπx
−x −inπx
f (x)e
dx =
e e
dx +
e e
dx
2 −1
−1
0
0
1 1
1
1
(1−inπ)x −(1+inπ)x =
−
e
e
−
2
1 − inπ
1 + inπ
−1
0
1
cn =
2
=
1
−inπx
e − (−1)n
1 − e−1 (−1)n
2[e − (−1)n ]
+
=
.
e(1 − inπ)
1 + inπ
e(1 + n2 π 2 )
Thus
f (x) =
∞
2[e − (−1)n ] inπx
.
e
e(1 + n2 π 2 )
n=−∞
7. The fundamental period is T = 4, so ω = 2π/4 = π/2
cn
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.
0.6
0.4
0.2
5Π
2
3Π
2
Π
2
Π
2
3Π
2
5Π
2
frequency
n
-5
-4
-3
-2
-1
0
1
2
3
4
5
cn
0.1273
0.0000
0.2122
0.0000
0.6366
0.0000
0.6366
0.0000
0.2122
0.0000
0.1273
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,
cn
0.25
0.2
0.15
0.1
0.05
10 Π
6 Π
2 Π
2Π
6Π
10 Π
frequency
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.
n
-5
-4
-3
-2
-1
0
1
2
3
4
5
cn
0.0450
0.0000
0.0750
0.1592
0.2251
0.2500
0.2251
0.1592
0.0750
0.0000
0.0450
658
12.4 Complex Fourier Series
9. Identifying 2p = π or p = π/2, and using sin x =
f
(eix − e−ix )/2i, we have
1 π
1 π
cn =
f (x)e−2inx/π dx =
(sin x)e−2inx/π dx
π 0
π 0
1 π 1 ix
=
(e − e−ix )e−2inx/π dx
π 0 2i
π
1
=
e(1−2n/π)ix − e−(1+2n/π)ix dx
2πi 0
1
1
=
e(1−2n/π)ix
2πi i(1 − 2n/π)
π
1
+
e−(1+2n/π)ix
i(1 + 2n/π)
0
=
4
3
2
1
2 Π
Π
Π
x
2Π
cn
0.6
0.4
0.2
π(1 + e−2in )
.
π 2 − 4n2
-6
-4
-2
2
4
6
frequency
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.
n
-5
-4
-3
-2
-1
0
1
2
3
4
5
cn
0.0198
0.0759
0.2380
0.4265
0.5784
0.6366
0.5784
0.4265
0.2380
0.0759
0.0198
10. Identifying 2p = π or p = π/2, and using cos x =
(eix − e−ix )/2, we have
1 π
cn =
f (x)e−2inx/π dx
π 0
1 π/2
=
(cos x)e−2inx/π dx
π 0
1 π/2 1 ix
=
(e − e−ix )e−2inx/π dx
π 0
2
π/2 1
=
e(1−2n/π)ix − e−(1+2n/π)ix dx
2π 0
1
1
=
e(1−2n/π)ix
2π i(1 − 2n/π)
π/2
1
−(1+2n/π)ix
+
e
i(1 + 2n/π)
0
=
f
1
0.8
0.6
0.4
0.2
2 Π
Π
Π
2Π
x
cn
0.3
0.2
0.1
-5
-10
5
10
frequency
2ne−in + iπ
.
π 2 − 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.
n
-5
-4
-3
-2
-1
0
1
2
3
4
5
cn
0.1447
0.1954
0.2437
0.2833
0.3093
0.3183
0.3093
0.2833
0.2437
0.1954
0.1447
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
sinh π 1 − in
1 + in
2(−1)n sinh π
+
=
,
π
n2 + 1 n2 + 1
π(n2 + 1)
n = 0, 1, 2, . . .
and
bn = i(cn − c−n ) = i(−1)n
sinh π 1 − in
1 + in
2in
2(−1)n n sinh π
n sinh π
−
=
i(−1)
−
=
,
2
2
2
π
n +1 n +1
π
n +1
π(n2 + 1)
the Fourier series of f is
f (x) =
∞ sinh π 2 sinh π (−1)n
n(−1)n
+
cos
nx
+
sin
nx
.
π
π
n2 + 1
n2 + 1
n=1
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) =
∞
2[1 − (−1)n
i=1
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 cos αx + c2 sin αx.
Now
y (x) = −c1 α sin αx + c2 α cos αx
and y (0) = 0 implies c2 = 0, so
y(1) + y (1) = c1 (cos α − α sin α) = 0
or
cot α = α.
The eigenvalues are λn = αn2 where α1 , α2 , α3 , . . . are the consecutive positive solutions of cot α = α. The
corresponding eigenfunctions are cos αn x for n = 1, 2, 3, . . . . Using a CAS we find that the first four eigenvalues 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 = c1 x + 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 cos αx + c2 sin αx
660
12.5
Sturm-Liouville Problem
and
y (x) = −c1 α sin αx + c2 α cos αx.
The boundary conditions imply
c1 + c2 α = 0
c1 cos α + c2 sin α = 0
which gives
−c2 α cos α + c2 sin α = 0
or
tan α = α.
The eigenvalues are λn = αn2 where α1 , α2 , α3 , . . . are the consecutive positive solutions of tan α = α. The
corresponding eigenfunctions are α cos αx − sin αx (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 = c1 x + 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 cosh αx + c2 sinh αx and y = c1 α sinh αx + c2 α cosh αx. The condition
y (0) = 0 implies c2 = 0 and so y = c1 cosh αx. 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 cos αx + c2 sin αx and y = −c1 α sin αx + c2 α cos αx. The condition y (0) = 0
implies c2 = 0 and so y = c1 cos αx. Now the condition y (L) = 0 implies −c1 α sin αL = 0. For c1 = 0 this
condition will hold when αL = nπ or λ = α2 = n2 π 2 /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 cosh αx + c2 sinh αx
y = c1 α sinh αx + c2 α cosh αx.
Using the fact that cosh x is an even function and sinh x is odd we have
y(−L) = c1 cosh(−αL) + c2 sinh(−αL)
= c1 cosh αL − c2 sinh αL
and
y (−L) = c1 α sinh(−αL) + c2 α cosh(−αL)
= −c1 α sinh αL + c2 α cosh αL.
The boundary conditions imply
c1 cosh αL − c2 sinh αL = c1 cosh αL + c2 sinh αL
or
2c2 sinh αL = 0
and
−c1 α sinh αL + c2 α cosh αL = c1 α sinh αL + c2 α cosh αL
or
2c1 α sinh αL = 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 = c1 x + c2
y = c1 .
From y(−L) = y(L) we obtain
−c1 L + c2 = c1 L + c2 .
Then c1 = 0 and y = 1 is an eigenfunction corresponding to the eigenvalue λ = 0.
For λ = α2 > 0 we have
y = c1 cos αx + c2 sin αx
y = −c1 α sin αx + c2 α cos αx.
The first boundary condition implies
c1 cos αL − c2 sin αL = c1 cos αL + c2 sin αL
or
2c2 sin αL = 0.
Thus, if c1 = 0 and c2 = 0,
αL = nπ
or λ = α2 =
n2 π 2
, n = 1, 2, 3, . . . .
L2
The corresponding eigenfunctions are sin(nπx/L), for n = 1, 2, 3, . . . . Similarly, the second boundary condition
implies
2c1 α sin αL = 0.
If c1 = 0 and c2 = 0,
αL = nπ
or λ = α2 =
n2 π 2
, n = 1, 2, 3, . . . ,
L2
and the corresponding eigenfunctions are cos(nπx/L), for n = 1, 2, 3, . . . .
5. The eigenfunctions are cos αn x where cot αn = αn . Thus
cos αn x =
2
=
=
=
=
1 1
cos αn x dx =
(1 + cos 2αn x) dx
2 0
0
1
1
1
1
1
sin 2αn x =
sin 2αn
x+
1+
2
2αn
2
2αn
0
1
1
1+
(2 sin αn cos αn )
2
2αn
1
1
sin αn cot αn sin αn
1+
2
αn
1
1
1
(sin αn ) αn (sin αn ) =
1+
1 + sin2 αn .
2
αn
2
1
2
662
12.5
Sturm-Liouville Problem
6. The eigenfunctions are sin αn x where tan αn = −αn . Thus
1
1 1
2
2
sin αn x =
sin αn x dx =
(1 − cos 2αn x) dx
2 0
0
1
1
1
1
1
=
sin 2αn x =
sin 2αn
x−
1−
2
2αn
2
2αn
0
1
1
=
(2 sin αn cos αn )
1−
2
2αn
1
1
=
1−
tan αn cos αn cos αn
2
αn
1
1 1
2
=
1−
1 + cos2 αn .
−αn cos αn =
2
αn
2
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 ln x) + c2 sin(α ln x). The condition y(1) = 0 implies c1 = 0, so that
y = c2 sin(α ln x). The condition y(5) = 0 implies α ln 5 = nπ, n = 1, 2, 3, . . . . Thus, the eigenvalues are
n2 π 2 /(ln 5)2 for n = 1, 2, 3, . . . , with corresponding eigenfunctions sin[(nπ/ ln 5) ln x].
(b) The self-adjoint form is
d
λ
[xy ] + y = 0.
dx
x
(c) An orthogonality relation is
5
1
mπ
nπ
1
sin
ln x sin
ln x dx = 0,
x
ln 5
ln 5
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 + c2 e−x . The boundary conditions imply c1 + c2 = 0 and c1 + c2 e−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 =
1
4
, the general solution of the differential equation is c1 e−x/2 + c2 xe−x/2 .
The boundary conditions imply c1 = 0 and c1 + 2c2 = 0, so c1 = c2 = 0 and y = 0. Similarly, if 0 < λ <
the general solution is
1
y = c1 e 2 (−1+
√
1−4λ )x
1
√
+ c2 e 2 (−1−
1−4λ )x
1
4
,
.
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 = c1 e−x/2 cos 4λ − 1 x + c2 e−x/2 sin 4λ − 1 x.
√
The condition y(0) = 0 implies c1 = 0 so y = c2 e−x/2 sin 4λ − 1 x. From
√
y(2) = c2 e−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 n2 π 2 /42 + 1/4 for n = 1, 2, 3, . . . , with corresponding eigenfunctions e−x/2 sin(nπx/2).
(b) The self-adjoint form is
d x [e y ] + λex y = 0.
dx
663
12.5 Sturm-Liouville Problem
(c) An orthogonality relation is
2 2
nπ
mπ −x/2
nπ mπ
x
−x/2
x e
x dx =
x cos
x dx = 0.
e e
sin
cos
sin
2
2
2
2
0
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−x y + (1 − x)e−x y + ne−x y = 0
and the self-adjoint form is
d
xe−x y + ne−x y = 0.
dx
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
∞
e−x Lm (x)Ln (x) dx = 0, m = n.
0
10. To obtain the self-adjoint form we note that an integrating factor is e
−2x dx
= e−x . Thus, the differential
2
equation is
e−x y − 2xe−x y + 2ne−x y = 0
2
and the self-adjoint form is
2
2
2
d −x2 y + 2ne−x y = 0.
e
dx
Identifying the weight function p(x) = e−x and noting that since r(x) = e−x , limx→−∞ r(x) = limx→∞ r(x) =
0, we have the orthogonality relation
∞
2
e−x Hm (x)Hn (x) dx = 0, m = n.
2
2
−∞
11. (a) The differential equation is
(1 + x2 )y + 2xy +
λ
y = 0.
1 + x2
Letting x = tan θ we have θ = tan−1 x and
dy
dy
dy dθ
1
=
=
2
dx
dθ dx
1 + x dθ
2
2
d y
dy
dy
d
1
1
2x
d y dθ
=
=
−
2
2
2
2
2
2
dx
dx 1 + x dθ
1+x
dθ dx
(1 + x ) dθ
=
1
d2 y
dy
2x
.
−
2
2
(1 + x ) dθ2
(1 + x2 )2 dθ
The differential equation can then be written in terms of y(θ) as
1
λ
d2 y
dy
dy
1
2x
(1 + x2 )
+
2x
+
−
y
2
2
2
2
2
2
(1 + x ) dθ
(1 + x ) dθ
1 + x dθ
1 + x2
=
or
1
d2 y
λ
+
y=0
2
1 + x dθ2
1 + x2
d2 y
+ λy = 0.
dθ2
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 cos αθ + c2 sin αθ.
The condition y(0) = 0 implies c1 = 0 so y = c2 sin αθ. Now the condition y(π/4) = 0 implies c2 sin απ/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
sin(4m tan−1 x) sin(4n tan−1 x) dx = 0,
x2 + 1
m = n.
12. (a) Letting λ = α2 the differential equation becomes x2 y + xy + (α2 x2 − 1)y = 0. This is the parametric
Bessel equation with ν = 1. The general solution is
y = c1 J1 (αx) + c2 Y1 (αx).
Since Y is unbounded at 0 we must have c2 = 0, so that y = c1 J1 (αx). The condition J1 (3α) = 0 defines
the eigenvalues λn = αn2 for n = 1, 2, 3, . . . . The corresponding eigenfunctions are J1 (αn x).
(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 = α12 = 1.6313, λ2 = α22 = 5.4687,
λ3 = α32 = 11.4999, and λ4 = α42 = 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
cos xm x cos xn x dx = 0
0
where xm = xn are positive solutions of cot x = x.
(b) Referring to Problem 1 we use a CAS to compute
1
(cos 0.8603x)(cos 3.4256x) dx = −1.8771 × 10−6 ≈ 0.
0
15. (a) An orthogonality relation is
1
(xm cos xm x − sin xm x)(xn cos xn x − sin xn x) dx = 0
0
where xm = xn are positive solutions of tan x = x.
(b) Referring to Problem 2 we use a CAS to compute
1
(4.4934 cos 4.4934x − sin 4.4934x)(7.7253 cos 7.7253x − sin 7.7253x) dx = −2.5650 × 10−4 ≈ 0.
0
665
and Legendre
Series
12.6
12.5 Bessel
Sturm-Liouville
Problem
EXERCISES 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 J0 (2α) = −J1 (2α). Thus, J0 (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
2
2
ci =
xJ0 (αi x) dx
4J12 (2αi ) 0
t = αi x
2αi
1
1
=
tJ0 (t) dt
·
2J12 (2αi ) αi2 0
=
1
2
2αi J12 (2αi )
2αi
0
d
[tJ1 (t)] dt
dt
=
2αi
1
(t)
tJ
1
2
2
2αi J1 (2αi )
0
=
1
.
αi J1 (2αi )
Thus
f (x) =
∞
i=1
dt = αi dx
[From (5) in the text]
1
J0 (αi x).
αi J1 (2αi )
4. The boundary condition indicates that we use (19) and (20) in the text. With b = 2 we obtain
2
2 2
2 x2 c1 =
x dx =
= 1,
4 0
4 2 0
2
2
ci =
xJ0 (αi x) dx
4J02 (2αi ) 0
2αi
1
1
=
tJ0 (t) dt
·
2J02 (2αi ) αi2 0
=
1
2
2αi J02 (2αi )
2αi
0
d
[tJ1 (t)] dt
dt
=
2αi
1
(t)
tJ
1
2
2
2αi J0 (2αi )
0
=
J1 (2αi )
.
αi J02 (2αi )
t = αi x
dt = αi dx
[From (5) in the text]
Now since J0 (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 = 2 and h = 1 we obtain
2
2αi2
ci =
xJ0 (αi x) dx
(4αi2 + 1)J02 (2αi ) 0
t = αi x
dt = αi dx
2αi
2
1
2αi
· 2
=
tJ0 (t) dt
2
2
(4αi + 1)J0 (2αi ) αi 0
=
2
2
(4αi + 1)J02 (2αi )
2αi
0
d
[tJ1 (t)] dt
dt
=
2αi
2
(t)
tJ
1
2
2
(4αi + 1)J0 (2αi )
0
=
4αi J1 (2αi )
.
(4αi2 + 1)J02 (2αi )
Thus
f (x) = 4
∞
[From (5) in the text]
αi J1 (2αi )
J0 (αi x).
+ 1)J02 (2αi )
(4αi2
i=1
6. Writing the boundary condition in the form
2J0 (2α) + 2αJ0 (2α) = 0
we identify b = 2 and h = 2. Using (17) and (18) in the text we obtain
2
2αi2
ci =
xJ0 (αi x) dx
(4αi2 + 4)J02 (2αi ) 0
2αi
αi2
1
=
tJ0 (t) dt
·
2(αi2 + 1)J02 (2αi ) αi2 0
=
1
2(αi2 + 1)J02 (2αi )
2αi
0
d
[tJ1 (t)] dt
dt
t = αi x
dt = αi dx
[From (5) in the text]
2αi
1
=
tJ1 (t)
2(αi2 + 1)J02 (2αi )
0
=
αi J1 (2αi )
.
(αi2 + 1)J02 (2αi )
Thus
f (x) =
∞
i=1
αi J1 (2αi )
J0 (αi x).
(αi2 + 1)J02 (2αi )
7. The boundary condition indicates that we use (17) and (18) in the text. With n = 1, b = 4, and h = 3 we
obtain
4
2αi2
ci =
xJ1 (αi x)5x dx
(16αi2 − 1 + 9)J12 (4αi ) 0
t = αi x
dt = αi dx
4α
i
1
5αi2
· 3
=
t2 J1 (t) dt
2
2
4(2αi + 1)J1 (4αi ) αi 0
=
5
4αi (2αi2 + 1)J12 (4αi )
667
0
4αi
d 2
[t J2 (t)] dt
dt
[From (5) in the text]
12.6 Bessel and Legendre Series
4αi
5
2
=
t J2 (t)
4αi (2αi2 + 1)J12 (4αi )
0
=
20αi J2 (4αi )
.
(2αi2 + 1)J12 (4αi )
Thus
f (x) = 20
∞
αi J2 (4αi )
J1 (αi x).
+ 1)J12 (4αi )
(2αi2
i=1
8. The boundary condition indicates that we use (15) and (16) in the text. With n = 2 and b = 1 we obtain
2
c1 = 2
J3 (αi )
xJ2 (αi x)x2 dx
0
2
=
1
αi4
·
J32 (αi )
2
=
1
t = αi x
αi
2
αi4 J32 (αi )
dt = αi dx
t3 J2 (t) dt
0
αi
d 3
[t J3 (t)] dt
dt
0
αi
3
t J3 (t)
αi4 J32 (αi )
=
[From (5) in the text]
0
2
.
αi J3 (αi )
=
Thus
f (x) = 2
∞
i=1
1
J2 (αi x).
αi J3 (αi )
9. The boundary condition indicates that we use (19) and (20) in the text. With b = 3 we obtain
2
c1 =
9
ci =
3
3
2 x4 9
xx dx =
= ,
9 4 0 2
2
0
2
2
9J0 (3αi )
3
xJ0 (αi x)x2 dx
0
t = αi x
2
1
=
· 4
2
9J0 (3αi ) αi
=
2
4
9αi J02 (3αi )
3αi
t3 J0 (t) dt
0
3αi
0
t2
d
[tJ1 (t)] dt
dt
u = t2
du = 2t dt
2
=
9αi4 J02 (3αi )
dt = αi dx
3αi
−2
t J1 (t) 3
0
668
0
3αi
t2 J1 (t) dt .
d
dv = dt
[tJ1 (t)] dt
v = tJ1 (t)
12.6 Bessel and Legendre Series
With n = 0 in equation (6) in the text we have J0 (x) = −J1 (x), so the boundary condition J0 (3αi ) = 0 implies
J1 (3αi ) = 0. Then
3αi
3αi
d 2
2
2
2
ci =
−2
t J2 (t) dt =
−2t J2 (t)
4 J 2 (3α )
9αi4 J02 (3αi )
dt
9α
i
0
i 0
0
=
2
9αi4 J02 (3αi )
−4J2 (3αi )
−18αi2 J2 (3αi ) = 2 2
.
αi J0 (3αi )
Thus
f (x) =
∞
9
J2 (3αi )
−4
J0 (αi x).
2
α2 J 2 (3αi )
i=1 i 0
10. The boundary condition indicates that we use (15) and (16) in the text. With b = 1 it follows that
1
2
ci = 2
x 1 − x2 J0 (αi x) dx
J1 (αi ) 0
2
= 2
J1 (αi )
1
xJ0 (αi x) dx −
0
1
3
x J0 (αi x) dx
0
αi
αi
2
1
1
3
tJ
(t)
dt
−
t
J
(t)
dt
0
0
J12 (αi ) αi2 0
αi4 0
αi
αi
2
d
1
1
2 d
= 2
t
[tJ1 (t)] dt − 4
[tJ1 (t)] dt
J1 (αi ) αi2 0 dt
αi 0
dt
t = αi x
dt = αi dx
=
u = t2
du = 2t dt
αi
−2
t2 J1 (t) dt
αi
αi
1
2
1
3
t
tJ
(t)
−
J
(t)
1
1
2
2
4
J1 (αi ) αi
αi
0
0
0
αi
2
d 2
J1 (αi ) J1 (αi )
2
= 2
t J2 (t) dt
−
+ 4
J1 (αi )
αi
αi
αi 0 dt
αi 2
4J2 (αi )
2 2
= 2
= 2 2
t J2 (t)
.
4
J1 (αi ) αi
αi J1 (αi )
0
=
Thus
f (x) = 4
11. (a)
∞
J2 (αi )
2 J 2 (α ) J0 (αi x).
α
i
i=1 i 1
y
4
2
5
10
15
20
25
30
-2
-4
669
x
d
dv = dt
[tJ1 (t)] dt
v = tJ1 (t)
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 =
20αi J2 (4αi )
.
(2αi2 + 1)J12 (4αi )
Using a CAS we find c1 = 26.7896, c2 = −12.4624, c3 = 7.1404, c4 = −4.68705, and c5 = 3.35619.
(b)
S1
20
S2
20
S3
20
S4
20
S5
20
15
15
15
15
15
10
10
10
10
10
5
5
5
5
5
1 2 3 4 5x
(c)
1 2 3 4 5x
S10
20
1 2 3 4 5x
1 2 3 4 5x
1 2 3 4 5x
S10
20
15
15
10
10
5
5
10
1
2
3
4x
20
30
40
50
x
-5
-10
13. Since f is expanded as a series of Bessel functions, J1 (αi x) 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).
y
2
1.5
1
0.5
-2
670
-1
1
2
x
12.6 Bessel and Legendre Series
(b) In Section 5.3 we saw that J2 (x) = 2J2 (x)/x − J3 (x). Since J2 is even and J3 is
odd we see that
y
20
J2 (−x) = 2J2 (−x)/(−x) − J3 (−x)
15
= −2J2 (x)/x + J3 (x) = −J2 (x),
so that J2 is an odd function. Now, if f (x) = 3J2 (x) + 2xJ2 (x), we see that
10
f (−x) = 3J2 (−x) − 2xJ2 (−x)
= 3J2 (x) + 2xJ2 (x) = f (x),
5
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).
-4 -2
15. We compute
c0 =
c1 =
c2 =
c3 =
c4 =
c5 =
c6 =
1
2
3
2
5
2
7
2
9
2
1
xP0 (x) dx =
0
1
xP1 (x) dx =
0
1
xP2 (x) dx =
0
1
xP3 (x) dx =
0
11
2
13
2
1
xP4 (x) dx =
0
1
2
xP6 (x) dx =
0
f (x) =
1
0
13
2
-1
1
1
(5x4 − 3x2 )dx = 0
2
1
1
3
(35x5 − 30x3 + 3x)dx = −
8
32
0
11
2
1
2
1
5
(3x3 − x)dx =
2
16
0
0.5
1
0
9
2
1
1
4
x2 dx =
7
2
0
Thus
0
5
2
xP5 (x) dx =
S5
1
1
x dx =
3
2
1
4x
2
1
1
(63x6 − 70x4 + 15x2 )dx = 0
8
1
1
13
(231x7 − 315x5 + 105x3 − 5x)dx =
.
16
256
0
0
-0.5
1
1
5
3
13
P0 (x) + P1 (x) + P2 (x) − P4 (x) +
P6 (x) + · · · .
4
2
16
32
256
The figure above is the graph of S5 (x) = 14 P0 (x) + 12 P1 (x) +
671
5
16 P2 (x)
−
3
32 P4 (x)
+
13
256 P6 (x).
0.5
1
x
12.6 Bessel and Legendre Series
16. We compute
c0 =
1
2
3
c1 =
2
5
c2 =
2
=
1
−1
1
2
ex P0 (x) dx =
S5
1
ex dx =
−1
1
3
e P1 (x) dx =
2
−1
2
1
x
1
(e − e−1 )
2
3
−1
x
xe dx = 3e
−1
1
5
ex P2 (x) dx =
2
−1
1
1
1
(3x2 ex − ex )dx
2
−1
-1
-0.5
5
(e − 7e−1 )
2
c3 =
7
2
c4 =
9
2
1
−1
ex P3 (x) dx =
7
2
ex P4 (x) dx =
9
2
1
−1
Thus
f (x) =
1
−1
1
−1
1
7
(5x3 ex − 3xex )dx = (−5e + 37e−1 )
2
2
1
9
(35x4 ex − 30x2 ex + 3ex )dx = (36e − 266e−1 ).
8
2
1
5
(e − e−1 )P0 (x) + 3e−1 P1 (x) + (e − 7e−1 )P2 (x)
2
2
7
9
+ (−5e + 37e−1 )P3 (x) + (36e − 266e−1 )P4 (x) + · · · .
2
2
The figure above is the graph of S5 (x).
17. Using cos2 θ = 12 (cos 2θ + 1) we have
1
3
1
(3 cos2 θ − 1) = cos2 θ −
2
2
2
3
1
3
1
1
= (cos 2θ + 1) − = cos 2θ + = (3 cos 2θ + 1).
4
2
4
4
4
P2 (cos θ) =
18. From Problem 17 we have
1
(3 cos 2θ + 1)
4
P2 (cos θ) =
or
cos 2θ =
1
4
P2 (cos θ) − .
3
3
Then, using P0 (cos θ) = 1,
F (θ) = 1 − cos 2θ = 1 −
4
1
P2 (cos θ) −
3
3
4 4
4
4
− P2 (cos θ) = P0 (cos θ) − P2 (cos θ).
3 3
3
3
=
19. If f is an even function on (−1, 1) then
−1
and
1
f (x)P2n (x) dx = 2
−1
f (x)P2n (x) dx
0
1
f (x)P2n+1 (x) dx = 0.
672
1
0.5
1
x
12.6 Bessel and Legendre Series
Thus
c2n
2(2n) + 1
=
2
1
4n + 1
f (x)P2n (x) dx =
2
−1
2
1
f (x)P2n (x) dx
0
1
= (4n + 1)
f (x)P2n (x) dx,
0
c2n+1 = 0, and
f (x) =
∞
c2n P2n (x).
n=0
20. If f is an odd function on (−1, 1) then
1
−1
and
1
−1
f (x)P2n+1 (x) dx = 2
1
f (x)P2n+1 (x) dx.
0
Thus
c2n+1 =
f (x)P2n (x) dx = 0
2(2n + 1) + 1
2
1
−1
f (x)P2n+1 (x) dx =
4n + 3
2
2
1
f (x)P2n+1 (x) dx
0
= (4n + 3)
1
f (x)P2n+1 (x) dx,
0
c2n = 0, and
f (x) =
∞
c2n+1 P2n+1 (x).
n=0
21. From (26) in Problem 19 in the text we find
1
1
1
c0 =
xP0 (x) dx =
x dx = ,
2
0
0
1
c2 = 5
1
0
1
c4 = 9
1
0
c6 = 13
1
xP6 (x) dx = 13
0
0.5
-1
1
3
(35x5 − 30x3 + 3x)dx = − ,
8
16
xP4 (x) dx = 9
0
and
1
1
5
(3x3 − x)dx = ,
2
8
xP2 (x) dx = 5
0
S4
0
1
1
13
(231x7 − 315x5 + 105x3 − 5x)dx =
.
16
128
Hence, from (25) in the text,
f (x) =
1
5
3
13
P0 (x) + P2 (x) − P4 (x) +
P6 + · · · .
2
8
16
128
On the interval −1 < x < 1 this series represents the function f (x) = |x|.
673
-0.5
0.5
1 x
12.6 Bessel and Legendre Series
22. From (28) in Problem 20 in the text we find
1
1
3
c1 = 3
P1 (x) dx = 3
x dx = ,
2
0
0
1
1
1 3
7
c3 = 7
P3 (x) dx = 7
5x − 3x dx = − ,
8
0
0 2
1
1
1
11
c5 = 11
P5 (x) dx = 11
63x5 − 70x3 + 15x dx =
8
16
0
0
and
c7 = 15
1
P7 (x) dx = 15
0
0
S4
1
0.5
-1
-0.5
0.5
1
x
-0.5
-1
1
1 75
429x7 − 693x5 + 315x3 − 35x dx = −
.
16
128
Hence, from (27) in the text,
3
7
11
75
P1 (x) − P3 (x) + P5 (x) −
P7 (x) + · · · .
2
8
16
128
On the interval −1 < x < 1 this series represents the odd function
−1, −1 < x < 0
f (x) =
1, 0 < x < 1.
f (x) =
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 (3x2 − 1), and
P3 (x) = 12 (5x3 − 3x). Setting
1
3
3
x2 = c0 P0 (x) + c1 P1 (x) + c2 P2 (x) = c0 + c1 x + c2 (3x2 − 1) = c0 − c2 + c1 x + c2 x2 ,
2
2
2
we obtain the system
1
c2 = 0
2
c1 = 0
3
c2 = 1.
2
The solution is c0 = 13 , c1 = 0, c2 = 23 . Thus, x2 = 13 P0 (x) + 23 P2 (x). Setting
1
1
x3 = c0 P0 (x) + c1 P1 (x) + c2 P2 (x) + c3 P3 (x) = c0 + c1 x + c2 (3x2 − 1) + c3 (5x3 − 3x)
2
2
3
1
3
5
= c0 − c2 + c1 − c3 x + c2 x2 + c3 x3 ,
2
2
2
2
c0 −
we obtain the system
1
c2 = 0
2
3
c1 − c3 = 0
2
3
c2 = 0
2
5
c3 = 1.
2
The solution is c0 = 0, c1 = 35 , c2 = 0, c3 = 25 . Thus x3 = 35 P1 (x) +
c0 −
674
2
5
P3 (x).
CHAPTER 12 REVIEW EXERCISES
CHAPTER 12 REVIEW EXERCISES
1. True, since
π
(x2
−π
− 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
[r(x)y ] + λp(x)y = 0,
dx
on the interval [a, b], has eigenvalue λ = 0.
y (a) = 0,
y (b) = 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) = 12 , and at x = 1 it converges to 12 (−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 cos αx + c2 sin αx and y (0) = 0 implies c2 = 0.
9. True, since
1
0
P2m (x)P2n (x) dx =
1
−1
1
2
P2m (x)P2n (x) dx = 0 when m = n.
10. Since Pn (x) is orthogonal to P0 (x) = 1 for n > 0,
1
Pn (x) dx =
−1
1
−1
P0 (x)Pn (x) dx = 0.
11. We know from a half-angle formula in trigonometry that cos2 x =
1
2
+
1
2
cos 2x, which is a cosine series.
12. (a) For m = n
L
n−m
(2n + 1)π
(2m + 1)π
1 L
n+m+1
cos
sin
x sin
x dx =
πx − cos
πx dx = 0.
2L
2L
2 0
L
L
0
(b) From
L
(2n + 1)π
x dx =
sin
2L
L
2
0
we see that
0
(2n + 1)π
L
1 1
− cos
x dx =
2 2
L
2
(2n
+
1)π
L
sin
x
= 2 .
2L
13. Since
a0 =
(−2x) dx = 1,
−1
an =
0
0
(−2x) cos nπx dx =
−1
and
675
2
[(−1)n − 1],
n2 π 2
CHAPTER 12 REVIEW EXERCISES
bn =
0
(−2x) sin nπx dx =
−1
4
(−1)n
nπ
for n = 1, 2, 3, . . . we have
∞
f (x) =
1 +
2 n=1
14. Since
2
4
n
n
[(−1)
−
1]
cos
nπx
+
sin
nπx
.
(−1)
n2 π 2
nπ
1
2
(2x2 − 1) dx = − ,
3
−1
1
an =
(2x2 − 1) cos nπx dx =
a0 =
−1
and
bn =
for n = 1, 2, 3, . . . we have
8
(−1)n ,
n2 π 2
1
−1
(2x2 − 1) sin nπx dx = 0
∞
1 8
f (x) = − +
(−1)n cos nπx.
3 n=1 n2 π 2
15. Since
a0 =
and
an =
2
1
2
1
1
ex dx = 2(e − 1)
0
1
ex cos nπx dx =
0
2
[e(−1)n − 1],
1 + n2 π 2
for n = 1, 2, 3, . . . , we have the cosine series
f (x) = e − 1 + 2
Since
bn =
2
1
∞
e(−1)n − 1
cos nπx.
1 + n2 π 2
n=1
1
ex sin nπx dx =
0
2nπ
[1 − e(−1)n ],
1 + n2 π 2
for n = 1, 2, 3, . . . , we have the sine series
f (x) = 2π
∞
n[1 − e(−1)n ]
sin nπx.
1 + n2 π 2
n=1
676
CHAPTER 12 REVIEW EXERCISES
16.
-3
-2
f
3
f
3
2
2
1
1
-1
1
2
3
x
-3
-2
-2
-1
-2
-2
-3
-3
f
3
2
2
1
1
f (x) =
1
e−x ,
x
e ,
2
2
x
3
f (x) = 2x2 − 1, −1 < x < 1
f
3
-1
1
-1
f (x) = |x| − x, −1 < x < 1
-3
-1
3
x
-3
-2
-1
1
-1
-1
-2
-2
-3
-3
−1 < x < 0
f (x) =
0<x<1
−e−x ,
ex ,
2
3
x
−1 < x < 0
0<x<1
17. For λ = α2 > 0 a general solution of the given differential equation is
y = c1 cos(3α ln x) + c2 sin(3α ln x)
and
3c2 α
3c1 α
sin(3α ln x) +
cos(3α ln x).
x
x
Since ln 1 = 0, the boundary condition y (1) = 0 implies c2 = 0. Therefore
y = −
y = c1 cos(3α ln x).
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)2 π 2 /36 with corresponding eigenfunctions
cos[(2n − 1)π(ln x)/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
1
e
1
cos
x
2n − 1
2m − 1
π ln x cos
π ln x dx = 0, m = n.
2
2
19. Since the coefficient of y in the differential equation is n2 , the weight function is the integrating factor
1
e
a(x)
(b/a)dx
1
=
e
1 − x2
−
x
1−x2
√
1
1 − x2
1
1
ln(1−x2 )
2
=
e
=
=√
2
2
1−x
1−x
1 − x2
dx
on the interval [−1, 1]. The orthogonality relation is
1
−1
√
1
Tm (x)Tn (x) dx = 0,
1 − x2
m = n.
20. Expanding in a full Fourier series we have
1
a0 =
2
2
x dx +
0
2 dx
=3
2
nπx
(−1)n − 1
dx = 2
2
n2 π 2
0
2
2
4
−1
1
nπx
nπx
dx +
dx = 4
bn =
x sin
2 sin
2
2
2
nπ
0
2
an =
1
2
4
so
f (x) =
2
x cos
nπx
dx +
2
4
2 cos
∞ 3
2
nπx
(−1)n − 1
nπx
+2
−
sin
.
cos
2
n2 π 2
2
nπ
2
n=1
21. The boundary condition indicates that we use (15) and (16) of Section 12.6 in the text. With b = 4 we obtain
2
ci =
16J12 (4αi )
1
=
8J12 (4αi )
=
=
=
4
xJ0 (αi x)f (x) dx
0
2
xJ0 (αi x) dx
0
t = αi x
1
1
· 2
2
8J1 (4αi ) αi
1
8J12 (4αi )
0
2αi
2αi
tJ0 (t) dt
0
d
[tJ1 (t)] dt
dt
2αi
1
J1 (2αi )
tJ
.
(t)
=
1
2
2
8J1 (4αi )
4α
i J1 (4αi )
0
Thus
∞
f (x) =
dt = αi dx
1 J1 (2αi )
J0 (αi x).
4 i=1 αi J12 (4αi )
678
[From (5) in 12.6 in the text]
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 x4 P1 (x) and x4 P3 (x) are odd functions, we see
immediately that c1 = c3 = 0. Now
1 1 4
1 1 4
1
c0 =
x P0 (x) dx =
x dx =
2 −1
2 −1
5
c2 =
5
2
c4 =
9
2
1
−1
x4 P2 (x) dx =
5
2
x4 P4 (x) dx =
9
2
1
−1
Thus
f (x) =
1
−1
1
−1
1
4
(3x6 − x4 )dx =
2
7
1
8
(35x8 − 30x6 + 3x4 )dx =
.
8
35
1
4
8
P0 (x) + P2 (x) + P4 (x).
5
7
35
679