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Part IV Fourier Series and Partial Differential Equations
Orthogonal Functions and
Fourier Series
EXERCISES 12.
Orthogonal Functions
− 2
xx
2 dx =
x
4
2
− 2
1
− 1
x
3 (x
2
x
6
1
− 1
x
4
1
− 1
2
0
e
x (xe
−x − e
−x )dx =
2
0
(x − 1)dx =
x
2 − x
2
0
π
0
cos x sin
2 x dx =
sin
3 x
π
0
π/ 2
−π/ 2
x cos 2x dx =
cos 2x + x sin 2x
π/ 2
−π/ 2
∫ (^5) π/ 4
π/ 4
e
x sin x dx =
e
x sin x −
e
x cos x
5 π/ 4
π/ 4
- For m = n
π/ 2
0
sin(2n + 1)x sin(2m + 1)x dx
π/ 2
0
cos 2(n − m)x − cos 2(n + m + 1)x
dx
4(n − m)
sin 2(n − m)x
π/ 2
0
4(n + m + 1)
sin 2(n + m + 1)x
π/ 2
0
12.1 Orthogonal Functions
For m = n ∫ π/ 2
0
sin
2 (2n + 1)x dx =
π/ 2
0
cos 2(2n + 1)x
dx
x
π/ 2
0
4(2n + 1)
sin 2(2n + 1)x
π/ 2
0
π
so that
‖ sin(2n + 1)x‖ =
π.
- For m = n
∫ (^) π/ 2
0
cos(2n + 1)x cos(2m + 1)x dx
π/ 2
0
cos 2(n − m)x + cos 2(n + m + 1)x
dx
4(n − m)
sin 2(n − m)x
π/ 2
0
4(n + m + 1)
sin 2(n + m + 1)x
π/ 2
0
For m = n ∫ (^) π/ 2
0
cos
2 (2n + 1)x dx =
∫ (^) π/ 2
0
cos 2(2n + 1)x
dx
x
π/ 2
0
4(2n + 1)
sin 2(2n + 1)x
π/ 2
0
π
so that
‖ cos(2n + 1)x‖ =
π.
- For m = n ∫ π
0
sin nx sin mx dx =
π
0
cos(n − m)x − cos(n + m)x
dx
2(n − m)
sin(n − m)x
π
0
2(n + m)
sin(n + m)x
π
0
For m = n ∫ π
0
sin
2 nx dx =
π
0
cos 2nx
dx =
x
π
0
4 n
sin 2nx
π
0
π
so that
‖ sin nx‖ =
π
- For m = n
∫ (^) p
0
sin
nπ
p
x sin
mπ
p
x dx =
∫ (^) 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
For m = n ∫ p
0
sin
2 nπ
p
x dx =
p
0
cos
2 nπ
p
x
dx =
x
p
0
p
4 nπ
sin
2 nπ
p
x
p
0
p
12.1 Orthogonal Functions
- Since ∫ (^) ∞
−∞
e
−x 2 · 1 · 2 x dx = −e
−x 2
0
−∞
− e
−x 2
∞
0
−∞
e
−x 2 · 1 · (4x
2 − 2) dx = 2
−∞
x
2 xe
−x 2
dx − 2
−∞
e
−x 2 dx
−xe
−x 2
∞
−∞
∞
−∞
e
−x 2 dx
∞
−∞
e
−x 2 dx
−xe
−x 2
0
−∞
−xe
−x 2
∞
0
and ∫ (^) ∞
−∞
e
−x 2 · 2 x · (4x
2 − 2) dx = 4
−∞
x
2
2 xe
−x 2
dx − 4
−∞
xe
−x 2 dx
−x
2 e
−x 2
∞
−∞
−∞
xe
−x 2 dx
−∞
xe
−x 2 dx
−x
2 e
−x 2
0
−∞
− x
2 e
−x 2
∞
0
−∞
2 xe
−x 2 dx = 0,
the functions are orthogonal.
- Since ∫ ∞
0
e
−x · 1(1 − x) dx = (x − 1)e
−x
∞
0
∞
0
e
−x dx = 0,
∞
0
e
−x · 1 ·
x
2 − 2 x + 1
dx =
2 x − 1 −
x
2
e
−x
∞
0
∞
0
e
−x (x − 2) dx
= 1 + (2 − x)e
−x
∞
0
∞
0
e
−x dx = 0,
and ∫ (^) ∞
0
e
−x · (1 − x)
x
2 − 2 x + 1
dx
∞
0
e
−x
x
3
x
2 − 3 x + 1
dx
= e
−x
x
3 −
x
2
∞
0
0
e
−x
x
2
dx
= 1 + e
−x
x
2 − 5 x + 3
∞
0
0
e
−x (5 − 3 x) dx
= 1 − 3 + e
−x (3x − 5)
∞
0
0
e
−x dx = 0,
the functions are orthogonal.
- By orthogonality
∫ (^) b
a φ 0 (x)φn(x)dx = 0 for n = 1, 2, 3,... ; that is,
∫ (^) b
a φn(x)dx = 0 for n = 1, 2, 3,....
12.1 Orthogonal Functions
- Using the facts that φ 0 and φ 1 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
φ 1 (x)φn(x) dx + β
b
a
φ 0 (x)φn(x) dx
= α · 0 + β · 0 = 0
for n = 2, 3, 4,....
- 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)]
2 dx =
∫ (^) b
a
[
φ
2 m (x) + 2φm(x)φn(x) + φ
2 n (x)
]
dx
∫ (^) b
a
φ
2 m (x)dx + 2
∫ (^) b
a
φm(x)φn(x)dx +
∫ (^) b
a
φ
2 n (x) dx
= ‖φm(x)‖
2
2 .
- Setting
− 2
f 3 (x)f 1 (x) dx =
− 2
x
2
3
4
dx =
c 2
and
− 2
f 3 (x)f 2 (x) dx =
− 2
x
3
4
5
dx =
c 1
we obtain c 1 = 0 and c 2 = − 5 /12.
- Since sin nx is an odd function on [−π, π],
(1, sin nx) =
π
−π
sin nx dx = 0
and f (x) = 1 is orthogonal to every member of {sin nx}. Thus {sin nx} is not complete.
- (f 1 + f 2 , f 3 ) =
∫ (^) b
a
[f 1 (x) + f 2 (x)]f 3 (x) dx =
∫ (^) b
a
f 1 (x)f 3 (x) dx +
∫ (^) b
a
f 2 (x)f 3 (x) dx = (f 1 , f 3 ) + (f 2 , f 3 )
- (a) The fundamental period is 2π/ 2 π = 1.
(b) The fundamental period is 2π/(4/L) =
1 2 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.
- (a) Following the pattern established by φ 1 (x) and φ 2 (x) we have
φ 3 (x) = f 3 (x) −
(f 3 , φ 0 )
(φ 0 , φ 0 )
φ 0 (x) −
(f 3 , φ 1 )
(φ 1 , φ 1 )
φ 1 (x) −
(f 3 , φ 2 )
(φ 2 , φ 2 )
φ 2 (x).
12.1 Orthogonal Functions
- In R 3 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 R
3 orthogonal to i, j, and k is 0 , so {i, j, k} is complete.
EXERCISES 12.
Fourier Series
- a 0 =
π
π
−π
f (x) dx =
π
π
0
1 dx = 1
an =
π
π
−π
f (x) cos
nπ
π
x dx =
π
π
0
cos nx dx = 0
bn =
π
π
−π
f (x) sin
nπ
π
x dx =
π
π
0
sin nx dx =
nπ
(1 − cos nπ) =
nπ
[1 − (−1)
n ]
f (x) =
π
∞ ∑
n=
n
n
sin nx
- a 0 =
π
∫ (^) π
−π
f (x) dx =
π
−π
(−1) dx +
π
∫ (^) π
0
2 dx = 1
an =
π
π
−π
f (x) cos nx dx =
π
0
−π
− cos nx dx +
π
π
0
2 cos nx dx = 0
bn =
π
∫ (^) π
−π
f (x) sin nx dx =
π
−π
− sin nx dx +
π
∫ (^) π
0
2 sin nx dx =
nπ
[1 − (−1)
n ]
f (x) =
π
∞ ∑
n=
n
n
sin nx
- a 0 =
− 1
f (x) dx =
− 1
1 dx +
0
x dx =
an =
1
− 1
f (x) cos nπx dx =
0
− 1
cos nπx dx +
1
0
x cos nπx dx =
n 2 π 2
[(−1)
n − 1]
bn =
− 1
f (x) sin nπx dx =
− 1
sin nπx dx +
0
x sin nπx dx = −
nπ
f (x) =
∞ ∑
n=
[
n − 1
n 2 π 2
cos nπx −
nπ
sin nπx
]
- a 0 =
1
− 1
f (x) dx =
1
0
x dx =
an =
− 1
f (x) cos nπx dx =
0
x cos nπx dx =
n 2 π 2
[(−1)
n − 1]
bn =
1
− 1
f (x) sin nπx dx =
1
0
x sin nπx dx =
n+
nπ
12.2 Fourier Series
f (x) =
∞ ∑
n=
[
n − 1
n 2 π 2
cos nπx +
n+
nπ
sin nπx
]
- a 0 =
π
π
−π
f (x) dx =
π
π
0
x
2 dx =
π
2
an =
π
∫ (^) π
−π
f (x) cos nx dx =
π
∫ (^) π
0
x
2 cos nx dx =
π
x
2
π
sin nx
π
0
n
∫ (^) π
0
x sin nx dx
n
n 2
bn =
π
∫ (^) π
0
x
2 sin nx dx =
π
x
2
n
cos nx
π
0
n
∫ (^) π
0
x cos nx dx
π
n
n+
n 3 π
[(−1)
n − 1]
f (x) =
π
2
∞ ∑
n=
[
n
n 2
cos nx +
π
n
n+
2[(−1)
n − 1]
n 3 π
sin nx
]
- a 0 =
π
∫ (^) π
−π
f (x) dx =
π
−π
π
2 dx +
π
∫ (^) π
0
π
2 − x
2
dx =
π
2
an =
π
π
−π
f (x) cos nx dx =
π
0
−π
π
2 cos nx dx +
π
π
0
π
2 − x
2
cos nx dx
π
π
2 − x
2
n
sin nx
π
0
n
∫ (^) π
0
x sin nx dx
n^2
n+
bn =
π
∫ (^) π
−π
f (x) sin nx dx =
π
−π
π
2 sin nx dx +
π
∫ (^) π
0
π
2 − x
2
sin nx dx
π
n
[(−1)
n − 1] +
π
x 2 − π 2
n
cos nx
π
0
n
π
0
x cos nx dx
π
n
n
n 3 π
[1 − (−1)
n ]
f (x) =
5 π
2
∞ ∑
n=
[
n 2
n+ cos nx +
π
n
n
2[1 − (−1)
n ]
n 3 π
sin nx
]
- a 0 =
π
∫ (^) π
−π
f (x) dx =
π
∫ (^) π
−π
(x + π) dx = 2π
an =
π
∫ (^) π
−π
f (x) cos nx dx =
π
∫ (^) π
−π
(x + π) cos nx dx = 0
bn =
π
∫ (^) π
−π
f (x) sin nx dx =
n
n+
f (x) = π +
∞ ∑
n=
n
n+ sin nx
- a 0 =
π
∫ (^) π
−π
f (x) dx =
π
∫ (^) π
−π
(3 − 2 x) dx = 6
an =
π
∫ (^) π
−π
f (x) cos nx dx =
π
∫ (^) π
−π
(3 − 2 x) cos nx dx = 0
bn =
π
∫ (^) π
−π
(3 − 2 x) sin nx dx =
n
n
f (x) = 3 + 4
∞ ∑
n=
n
n
sin nx
- a 0 =
π
π
−π
f (x) dx =
π
π
0
sin x dx =
π
an =
π
π
−π
f (x) cos nx dx =
π
π
0
sin x cos nx dx =
2 π
π
0
sin(1 + n)x + sin(1 − n)x
dx
12.2 Fourier Series
- a 0 =
5
− 5
f (x) dx =
0
− 5
1 dx +
5
0
(1 + x) dx
an =
− 5
f (x) cos
nπ
x dx =
0
− 5
cos
nπ
x dx +
0
(1 + x) cos
nπ
x dx
n 2 π 2
[(−1)
n − 1]
bn =
5
− 5
f (x) sin
nπ
x dx =
0
− 5
sin
nπ
x dx +
5
0
(1 + x) cos
nπ
x dx
nπ
n+
f (x) =
∞ ∑
n=
[
n 2 π 2
[(−1)
n − 1] cos
nπ
x +
nπ
n+ sin
nπ
x
]
- a 0 =
2
− 2
f (x) dx =
0
− 2
(2 + x) dx +
2
0
2 dx
an =
− 2
f (x) cos
nπ
x dx =
0
− 2
(2 + x) cos
nπ
x dx +
0
2 cos
nπ
x dx
n 2 π 2
[1 − (−1)
n ]
bn =
2
− 2
f (x) sin
nπ
x dx =
0
− 2
(2 + x) sin
nπ
x dx +
2
0
2 sin
nπ
x dx
nπ
n+
f (x) =
∞ ∑
n=
[
n 2 π 2
[1 − (−1)
n ] cos
nπ
x +
nπ
n+ sin
nπ
x
]
- a 0 =
π
∫ (^) π
−π
f (x) dx =
π
∫ (^) π
−π
e
x dx =
π
(e
π − e
−π )
an =
π
∫ (^) π
−π
f (x) cos nx dx =
n (e π − e −π )
π(1 + n 2 )
bn =
π
π
−π
f (x) sin nx dx =
π
π
−π
e
x sin nx dx =
n n(e
−π − e
π )
π(1 + n 2 )
f (x) =
e π − e −π
2 π
∞ ∑
n=
[
n (e π − e −π )
π(1 + n 2 )
cos nx +
n n(e −π − e π )
π(1 + n 2 )
sin nx
]
- a 0 =
π
∫ (^) π
−π
f (x) dx =
π
∫ (^) π
0
(e
x − 1) dx =
π
(e
π − π − 1)
an =
π
∫ (^) π
−π
f (x) cos nx dx =
π
∫ (^) π
0
(e
x − 1) cos nx dx =
[e
π (−1)
n − 1]
π(1 + n 2 )
bn =
π
∫ (^) π
−π
f (x) sin nx dx =
π
∫ (^) π
0
(e
x − 1) sin nx dx =
π
ne π (−1) n+
1 + n 2
n
1 + n 2
n
n
n
f (x) =
e
π − π − 1
2 π
∞ ∑
n=
[
e
π (−1)
n − 1
π(1 + n 2 )
cos nx +
n
1 + n 2
[
e
π (−1)
n+
]
n − 1
n
sin nx
]
- 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,
π 2
π 2
∞ ∑
n=
[
n
n 2
cos nπ +
π
n
n+
2[(−1)
n − 1]
n 3 π
sin nπ
]
π 2
∞ ∑
n=
n
n 2
n
π 2
∞ ∑
n=
n 2
π 2
2
2
and
π 2
π 2
π 2
2
2
12.2 Fourier Series
At x = 0 the series converges to 0 and
π 2
∞ ∑
n=
n
n 2
π 2
2
2
2
so
π 2
2
2
2
- From Problem 17
π 2
π 2
π 2
2
2
2
2
- The function in Problem 7 is continuous at x = π/2 so
3 π
= f
π
= π +
∞ ∑
n=
n
n+ sin
nπ
= π + 2
and π
- The function in Problem 9 is continuous at x = π/2 so
1 = f
π
π
∞ ∑
n=
n
π(1 − n 2 )
cos
nπ
π
3 π
3 · 5 π
5 · 7 π
and
π = 1 +
π
or π
- Writing
f (x) =
a 0
π
p
x + · · · + an cos
nπ
p
x + · · · + b 1 sin
π
p
x + · · · + bn sin
nπ
p
x + · · ·
we see that f
2 (x) consists exclusively of squared terms of the form
a
2 0
4
, a
2 n cos
2 nπ
p
x, b
2 n sin
2 nπ
p
x
and cross-product terms, with m = n, of the form
a 0 an cos
nπ
p
x, a 0 bn sin
nπ
p
x, 2 aman cos
mπ
p
x cos
nπ
p
x,
2 ambn cos
mπ
p
x sin
nπ
p
x, 2 bmbn 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
a
2 0
4
p
−p
dx =
a
2 0 p
2
, a
2 n
p
−p
cos
2 nπ
p
x dx = a
2 np,^ b
2 n
p
−p
sin
2 nπ
p
x dx = b
2 np.
Thus
RM S(f ) =
a
2 0 +
∞ ∑
n=
(a 2 n +^ b
2 n)^.
12.3 Fourier Cosine and Sine Series
- Since f (x) is an odd function, we expand in a sine series:
bn =
π
∫ (^) π
0
x sin nx dx =
n
n+ .
Thus
f (x) =
∞ ∑
n=
n
n+ sin nx.
- Since f (x) is an even function, we expand in a cosine series:
a 0 = 2
0
x
2 dx =
an = 2
0
x
2 cos nπx dx = 2
x
2
nπ
sin nπx
1
0
nπ
0
x sin nπx dx
n^2 π^2
n .
Thus
f (x) =
∞ ∑
n=
n 2 π 2
n cos nπx.
- Since f (x) is an odd function, we expand in a sine series:
bn = 2
1
0
x
2 sin nπx dx = 2
x
2
nπ
cos nπx
1
0
nπ
1
0
x cos nπx dx
n+
nπ
n 3 π 3
[(−1)
n − 1].
Thus
f (x) =
∞ ∑
n=
n+
nπ
n 3 π 3
[(−1)
n − 1]
sin nπx.
- Since f (x) is an even function, we expand in a cosine series:
a 0 =
π
π
0
(π
2 − x
2 ) dx =
π
2
an =
π
π
0
(π
2 − x
2 ) cos nx dx =
π
π
2 − x
2
n
sin nx
π
0
n
π
0
x sin nx dx
n 2
n+ .
Thus
f (x) =
π
2
∞ ∑
n=
n 2
n+ cos nx dx.
- Since f (x) is an odd function, we expand in a sine series:
bn =
π
∫ (^) π
0
x
3 sin nx dx =
π
x 3
n
cos nx
π
0
n
∫ (^) π
0
x
2 cos nx dx
2 π 2
n
n+ −
n 2 π
∫ (^) π
0
x sin nx dx
2 π 2
n
n+ −
n 2 π
x
n
cos nx
π
0
n
π
0
cos nx dx
2 π 2
n
n+
n 3
n .
Thus
f (x) =
∞ ∑
n=
2 π 2
n
n+
n 3
n
sin nx.
- Since f (x) is an odd function, we expand in a sine series:
bn =
π
∫ (^) π
0
(x + 1) sin nx dx =
2(π + 1)
nπ
n+
nπ
12.3 Fourier Cosine and Sine Series
Thus
f (x) =
∞ ∑
n=
2(π + 1)
nπ
n+
nπ
sin nx.
- Since f (x) is an odd function, we expand in a sine series:
bn = 2
1
0
(x − 1) sin nπx dx = 2
[∫
1
0
x sin nπx dx −
1
0
sin nπx dx
]
[
n^2 π^2
sin nπx −
x
nπ
cos nπx +
nπ
cos nπx
] 1
0
nπ
Thus
f (x) = −
∞ ∑
n=
nπ
sin nπx.
- Since f (x) is an even function, we expand in a cosine series:
a 0 =
1
0
x dx +
2
1
1 dx =
an =
0
x cos
nπ
x dx +
1
cos
nπ
x dx =
n 2 π 2
cos
nπ
Thus
f (x) =
∞ ∑
n=
n 2 π 2
cos
nπ
cos
nπ
x.
- Since f (x) is an odd function, we expand in a sine series:
bn =
π
π
0
x sin
n
x dx +
2 π
π
π sin
n
x dx =
n^2 π
sin
nπ
n
n+ .
Thus
f (x) =
∞ ∑
n=
n 2 π
sin
nπ
n
n+
sin
n
x.
- Since f (x) is an even function, we expand in a cosine series:
a 0 =
π
∫ (^) π
0
sin x dx =
π
an =
π
∫ (^) π
0
sin x cos nx dx =
π
∫ (^) π
0
sin(1 + n)x + sin(1 − n)x
dx
π(1 − n 2 )
n ) for n = 2, 3 , 4 ,...
a 1 =
π
∫ (^) π
0
sin 2x dx = 0.
Thus
f (x) =
π
∞ ∑
n=
2[1 + (−1)
n ]
π(1 − n 2 )
cos nx.
12.3 Fourier Cosine and Sine Series
- a 0 =
π
∫ (^) π
0
sin x dx =
π
an =
π
π
0
sin x cos nx dx =
π
π
0
[sin(n + 1)x − sin(n − 1)x] dx =
2[(−1)
n
π(1 − n 2 )
for n = 2, 3 , 4 ,...
bn =
π
∫ (^) π
0
sin x sin nx dx =
π
∫ (^) π
0
[cos(n − 1)x − cos(n + 1)x] dx = 0 for n = 2, 3 , 4 ,...
a 1 =
π
∫ (^) π
0
sin 2x dx = 0
b 1 =
π
∫ (^) π
0
sin
2 x dx = 1
f (x) = sin x
f (x) =
π
π
∞ ∑
n=
n
1 − n 2
cos nx
- a 0 =
π
∫ (^) π/ 2
0
x dx +
∫ (^) π
π/ 2
(π − x) dx
π
an =
π
∫ (^) π/ 2
0
x cos nx dx +
∫ (^) π
π/ 2
(π − x) cos nx dx
n 2 π
2 cos
nπ
n+ − 1
bn =
π
∫ (^) π/ 2
0
x sin nx dx +
∫ (^) π
π/ 2
(π − x) sin nx dx
n 2 π
sin
nπ
f (x) =
π
∞ ∑
n=
n 2 π
2 cos
nπ
n+ − 1
cos nx
f (x) =
∞ ∑
n=
n 2 π
sin
nπ
sin nx
- a 0 =
π
∫ (^2) π
π
(x − π) dx =
π
an =
π
∫ (^2) π
π
(x − π) cos
n
x dx =
n 2 π
n − cos
nπ
bn =
π
2 π
π
(x − π) sin
n
x dx =
n
n+ −
n 2 π
sin
nπ
f (x) =
π
∞ ∑
n=
n^2 π
n − cos
nπ
cos
n
x
f (x) =
∞ ∑
n=
n
n+ −
n 2 π
sin
nπ
sin
n
x
- a 0 =
1
0
x dx +
2
1
1 dx =
an =
1
0
x cos
nπ
x dx =
n 2 π 2
cos
nπ
bn =
0
x sin
nπ
x dx +
1
1 · sin
nπ
x dx =
n 2 π 2
sin
nπ
nπ
n+
f (x) =
∞ ∑
n=
n 2 π 2
cos
nπ
cos
nπ
x
12.3 Fourier Cosine and Sine Series
f (x) =
∞ ∑
n=
n 2 π 2
sin
nπ
nπ
n+
sin
nπ
x
- a 0 =
0
1 dx +
1
(2 − x) dx =
an =
1
0
1 · cos
nπ
x dx +
2
1
(2 − x) cos
nπ
x dx =
n 2 π 2
cos
nπ
n+
bn =
0
1 · sin
nπ
x dx +
1
(2 − x) sin
nπ
x dx =
nπ
n 2 π 2
sin
nπ
f (x) =
∞ ∑
n=
n 2 π 2
cos
nπ
n+
cos
nπ
x
f (x) =
∞ ∑
n=
nπ
n^2 π^2
sin
nπ
sin
nπ
x
- a 0 = 2
0
(x
2
an = 2
1
0
(x
2
2(x
2
nπ
sin nπx
1
0
nπ
1
0
(2x + 1) sin nπx dx =
n 2 π 2
[3(−1)
n − 1]
bn = 2
1
0
(x
2
2(x 2
nπ
cos nπx
1
0
nπ
1
0
(2x + 1) cos nπx dx
nπ
n+
n 3 π 3
[(−1)
n − 1]
f (x) =
∞ ∑
n=
n 2 π 2
[3(−1)
n − 1] cos nπx
f (x) =
∞ ∑
n=
nπ
n+
n 3 π 3
[(−1)
n − 1]
sin nπx
- a 0 =
2
0
(2x − x
2 ) dx =
an =
0
(2x − x
2 ) cos
nπ
x dx =
n 2 π 2
[(−1)
n+ − 1]
bn =
2
0
(2x − x
2 ) sin
nπ
x dx =
n^3 π^3
[1 − (−1)
n ]
f (x) =
∞ ∑
n=
n 2 π 2
[(−1)
n+ − 1] cos
nπ
x
f (x) =
∞ ∑
n=
n 3 π 3
[1 − (−1)
n ] sin
nπ
x
- a 0 =
π
2 π
0
x
2 dx =
π
2
an =
π
∫ (^2) π
0
x
2 cos nx dx =
n 2
bn =
π
∫ (^2) π
0
x
2 sin nx dx = −
4 π
n
12.3 Fourier Cosine and Sine Series
so that
f (t) =
∞ ∑
n=
nπ
sin nπt.
Substituting the assumption xp(t) =
n= Bn sin nπt into the differential equation then gives
x
′′ p + 10xp =
∞ ∑
n=
Bn(10 − n
2 π
2 ) sin nπt =
∞ ∑
n=
nπ
sin nπt
and so Bn = 2/nπ(10 − n
2 π
2 ). Thus
xp(t) =
π
∞ ∑
n=
n(10 − n 2 π 2 )
sin nπt.
- We have
a 0 =
π
π
0
(2πt − t
2 ) dt =
π
2
an =
π
π
0
(2πt − t
2 ) cos nt dt = −
n 2
so that
f (t) =
2 π
2
∞ ∑
n=
n 2
cos nt.
Substituting the assumption
xp(t) =
A 0
∞ ∑
n=
An cos nt
into the differential equation then gives
x
′′ p
∞ ∑
n=
An
n
2
cos nt =
2 π
2
∞ ∑
n=
n 2
cos nt
and A 0 = π
2 /9, An = 16/n
2 (n
2 − 48). Thus
xp(t) =
π
2
∞ ∑
n=
n 2 (n 2 − 48)
cos nt.
- We have
a 0 =
0
t dt =
an =
1 / 2
0
t cos 2nπt dt =
n 2 π 2
[(−1)
n − 1]
so that
f (t) =
∞ ∑
n=
n − 1
n 2 π 2
cos 2nπt.
Substituting the assumption
xp(t) =
A 0
∞ ∑
n=
An cos 2nπt
into the differential equation then gives
x
′′ p + 12xp^ = 6A^0 +
∞ ∑
n=
An(12 − n
2 π
2 ) cos 2nπt =
∞ ∑
n=
n − 1
n 2 π 2
cos 2nπt
t
x
12.3 Fourier Cosine and Sine Series
and A 0 = 1/24, An = [(−1) n − 1]/n 2 π 2 (12 − n 2 π 2 ). Thus
xp(t) =
π 2
∞ ∑
n=
n − 1
n 2 (12 − n 2 π 2 )
cos 2nπt.
- (a) The general solution is x(t) = c 1 cos
10 t + c 2 sin
10 t + xp(t), where
xp(t) =
π
∞ ∑
n=
n
n(10 − n 2 )
sin nt.
The initial condition x(0) = 0 implies c 1 + xp(0) = 0. Since xp(0) = 0, we have c 1 = 0 and x(t) =
c 2 sin
10 t + xp(t). Then x ′ (t) = c 2
10 cos
10 t + x ′ p (t) and x ′ (0) = 0 implies
c 2
π
∞ ∑
n=
n
10 − n^2
cos 0 = 0.
Thus
c 2 = −
π
∞ ∑
n=
n
10 − n 2
and
x(t) =
π
∞ ∑
n=
n
10 − n 2
[
n
sin nt −
sin
10 t
]
(b) The graph is plotted using eight nonzero terms in the series expansion of x(t).
- (a) The general solution is x(t) = c 1 cos 4
3 t + c 2 sin 4
3 t + xp(t), where
xp(t) =
π
2
∞ ∑
n=
n^2 (n^2 − 48)
cos nt.
The initial condition x(0) = 0 implies c 1 + xp(0) = 1 or
c 1 = 1 − xp(0) = 1 −
π 2
∞ ∑
n=
n 2 (n 2 − 48)
Now x
′ (t) = − 4
3 c 1 sin 4
3 t + 4
3 c 2 cos 4
3 t + x
′ p(t), so^ x
′ (0) = 0 implies 4
3 c 2 + x
′ p(0) = 0.^ Since
x
′ p(0) = 0, we have^ c^2 = 0 and
x(t) =
π 2
∞ ∑
n=
n 2 (n 2 − 48)
cos 4
3 t +
π 2
∞ ∑
n=
n 2 (n 2 − 48)
cos nt
π 2
π 2
cos 4
3 t + 16
∞ ∑
n=
n 2 (n 2 − 48)
[
cos nt − cos 4
3 t
]
