SOLUTION SET X FOR 18.075-Fall 2004
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SOLUTION SET X FOR 18.075–FALL 2004
5.11 Complete Fourier series
58. Expand each of the following functions in a Fourier series of period equal to the length
of the indicated interval of representation:
(a) f (x) = a + bx
0 (x < 0)
(b) f (x) =
1 (x > 0)
(c) f (x) = sin x
(d) f (x) = x
(0 < x < P ),
(−1 < x < 1),
(0 ≤ x ≤ π),
(1 < x < 2).
Solution. In this problem, we need to expand a function f (x) in a complete Fourier series
in (a, b); the periodic extension of f (x) has period equal to the length L = b − a of the
interval. With
f (x) = A0 +
∞
X
[An cos
n=1
nπx
nπx
],
+ Bn sin
L/2
L/2
the coefficients are given by the formulas
A0 =
1
L
An≥1 =
2
L
Bn≥1 =
2
L
Z
b
dx f (x),
a
Z
b
dx f (x) cos
2nπx
,
L
dx f (x) sin
2nπx
,
L
a
Z
a
b
which are direct generalizations of the formulas derived in class for the symmetric interval
(−l, l) where L = 2l.
SOLUTION SET X FOR 18.075–FALL 2004
(a) Let f (x) = A0 +
n=1
A0 =
An≥1
Bn≥1
2nπx
An cos 2nπx
in (0, P ); then,
P + Bn sin P
P∞
1
P
Z
P
(a + bx)dx
0
bP
, and, for n ≥ 1,
= a+
2
Z
2 P
2nπx
=
(a + bx) cos
dx
P 0
P
Z P
2nπx
1
=
(a + bx) d sin
nπ 0
P
!
P Z P
1
2nπx 2nπx
=
(a + bx) sin
−
b sin
dx
nπ
P 0
P
0
Z P 2nπx
bP
d cos
=
2n2 π 2 0
P
= 0,
Z
2 P
2nπx
=
(a + bx) sin
dx
P 0
P
Z P
1
2nπx
= −
(a + bx) d cos
P
nπ 0
!
P Z P
1
2nπx 2nπx
= −
(a + bx) cos
b cos
dx
−
P 0
P
nπ
0
= −
bP
.
nπ
It follows that
f (x) = a +
∞
bP X 1
2nπx
bP
sin
−
2
π
n
P
n=1
in (0, P ).
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SOLUTION SET X FOR 18.075–FALL 2004
(b) Let f (x) = A0 +
P∞
[An cos(nπx) + Bn sin(nπx)] in (−1, 1); then,
Z
1 1
A0 =
f (x) dx
2 −1
Z
1 1
=
dx
2 0
1
=
,
2
Z 1
An≥1 =
cos(nπx) dx = 0,
0
Z 1
Bn≥1 =
sin(nπx) dx
0
Z 1
1
d[cos(nπx)]
= −
nπ 0
1
=
[1 − cos(nπ)]
nπ
1
=
[1 − (−1)n ]
nπ
0
if n is even
=
2
nπ if n is odd
n=1
So,
1
2
f (x) = +
2 π
(c) Let f (x) = A0 +
P∞
[An cos(2nx) + Bn sin(2nx)] in [0, π]; then,
Z π
1
2
sin x dx = ,
π 0
π
Z
2 π
sin x cos(2nx) dx
π 0
Z
1 π
{sin[(2n + 1)x] − sin[(2n − 1)x]} dx
π 0
1
2
2
−
π 2n + 1 2n − 1
4
,
−
π(4n2 − 1)
Z
2 π
sin x sin(2nx) dx
π 0
Z π
1
{cos[(2n − 1)x] − cos[(2n + 1)x]} dx = 0.
π 0
n=1
A0 =
An≥1 =
=
=
=
Bn≥1 =
=
!
1
sin[(2n − 1)πx] .
2n − 1
n=1
∞
X
SOLUTION SET X FOR 18.075–FALL 2004
So,
2
4
f (x) = −
π π
(d) Let f (x) = A0 +
A0 =
P∞
∞
X
n=1
1
· cos(2nx)
2
4n − 1
n=1 [An cos(2nπx)
Z
2
x dx =
1
An≥1 = 2
Z
!
in [0, π].
+ Bn sin(2nπx)] in (1, 2); then,
3
,
2
2
x cos(2nπx) dx
1
Z 2
1
x d(sin(2nπx))
nπ 1
Z 2
1
2
sin(2nπx) dx = 0,
x sin(2nπx)|1 −
=
nπ
1
Z 2
= 2
x sin(2nπx) dx
1
Z 2
1
= −
x d(cos(2nπx))
nπ 1
Z 2
1
1
2
= −
x cos(2nπx)|1 −
cos(2nπx) dx = − .
nπ
nπ
1
=
Bn≥1
So
∞
3 1X1
f (x) = −
sin(2nπx) in (1, 2).
2 π
n
n=1
60. Obtain the Fourier series of period 2π which represents the solution of the problem
d2 y
+ Λy = h(x), y(−π) = y(π), y 0 (−π) = y 0 (π)
dx2
when

 0 (−π < x < 0),
1 (0 < x < π2 ),
h(x) =

0 ( π2 < x < π),
assuming that Λ 6= p2 (p = 0, 1, 2, . . .).
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SOLUTION SET X FOR 18.075–FALL 2004
Solution. Let h(x) = a0 +
P∞
n=1 [an
cos(nx) + bn sin(nx)] in [−π, π]; then,
Z
π
2
1
1
dx = ,
2π 0
4
Z π
1 2
1
nπ
sin
,
cos(nx) dx =
π 0
nπ
2
Z π
nπ 1 2
1 .
sin(nx) dx =
1 − cos
π 0
nπ
2
a0 =
an≥1 =
bn≥1 =
So,
∞
i
nπ 1 1 X 1 h nπ
sin(nx) .
+
sin
cos(nx) + 1 − cos
4 π
n
2
2
h(x) =
n=1
P
Let y = A0 + ∞
n=1 (An cos nx + Bn sin nx) be the solution of
y(π), y 0 (−π) = y 0 (π). We determine An and Bn as follows:
h(x) =
d2 y
dx2
+ Λy = h(x), y(−π) =
∞
i
1
1X1h
nπ
nπ +
sin
cos(nx) + 1 − cos
sin(nx)
4 π n=1
n
2
2
d2 y
+ Λy
dx2
∞
X
= ΛA0 +
(A − n2 )[An cos(nx) + Bn sin(nx)],
=
n=1
by differentiating the Fourier series term by term. This relation gives A 0 =
n ≥ 1,
nπ
2
sin
An≥1 =
nπ(Λ−n2 ) ,
Bn≥1 =
1−cos nπ
2
nπ(Λ−n2 )
. Thus, the solution is
∞
nπ
1
1 X sin nπ
2 cos(nx) + 1 − cos 2 sin(nx)
in [−π, π].
y=
+
4Λ π n=1
n(Λ − n2 )
61. a) If the representation
f (x) = A0 +
∞
X
n=1
An cos
nπx
l
(0 ≤ x ≤ l)
is valid, show formally that
2
l
Z
l
0
(f (x))2 dx = 2A20 +
∞
X
n=1
A2n .
1
4Λ ,
and, for
SOLUTION SET X FOR 18.075–FALL 2004
Solution: Clearly,
Z l
0
But
!2
nπx
(f (x)) dx =
An cos
dx
l
0
n=0
Z l
∞
X
nπx
mπx
cos
dx .
=
An Am
cos
l
l
0
m,n=0
Z
2
Z
It follows that
l
0
∞
X
l

 0,
mπx
nπx
l,
cos
cos
dx =

l
l
l/2,
Z
Hence,
l
(f (x)) dx = l ·
0
2
l
2
Z
l
A20
if m =
6 n
if m = n = 0
if m = n ≥ 1.
∞
lX 2
A .
+
2 n=1 n
∞
X
(f (x))2 dx = 2A20 +
0
n=1
b) If the representation
f (x) =
∞
X
Bn sin
n=1
is valid, show formally that
2
l
l
Z
nπx
l
(f (x))2 dx =
0
But
l
0
Z
l
l
Bn2
∞
X
mπx
nπx
sin
sin
dx =
l
l
Z
l
2
l
Z
0
Hence,
∞
X
!2
nπx
dx
(f (x)) dx =
Bn sin
l
0
n=1
Z l
∞
X
mπx
nπx
sin
=
Bm Bn
sin
dx.
l
l
0
m,n=1
Z
2
0
This implies that
(0 < x < l)
n=1
Solution:
Z
A2n .
l
2
0
m=n
m 6= n
∞
lX 2
B .
(f (x)) dx =
2 n=1 n
2
l
0
(f (x))2 dx =
∞
X
n=1
Bn2 .
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SOLUTION SET X FOR 18.075–FALL 2004
c) If the representation
f (x) = A0 +
∞ X
An cos
n=1
nπx
nπx + Bn sin
l
l
(−l < x < l)
is valid, show formally that
Z
∞
X
1 l
2
2
(f (x)) dx = 2A0 +
(A2n + Bn2 ).
l −l
n=1
Solution:
Z
l
Z
(f (x))2 dx =
∞ X
l
A0 +
−l
−l
n=1
∞
X
=
m,n=0
∞
X
+
l
Am An
Z
Bm Bn
Z
m,n=1
∞ X
∞
X
+ 2
nπx
nπx + Bn sin
An cos
l
l
Z
mπx
nπx
cos
dx
l
l
sin
mπx
nπx
sin
dx
l
l
−l
l
−l
Am Bn
m=0 n=1
By virtue of
cos
Z
l
cos
nπx
mπx
sin
dx
l
l
cos
−l
−l
and
Z
l
−l
nπx 2
dx =
l
cos
Z
l
1
l
Z
2
Z
l
−l
(f (x)) dx =
−l
Hence,
dx
Z l
mπx
nπx
mπx
nπx
cos
dx =
sin
sin
dx = 0 for m =
6 n,
l
l
l
l
−l
Z l
mπx
nπx
cos
sin
dx = 0, for all m, n,
l
l
−l
Z l
dx = 2l,
l
−l
we find that
!2
sin
2lA20
nπx 2
dx = l,
l
+l
∞
X
(A2n + Bn2 ).
n=1
l
−l
(f (x))2 dx = 2A20 +
∞
X
n=1
(A2n + Bn2 ).
n ≥ 1,