Mathematics Department

Dr. Mohammad Yasin

Worksheet (2) Solution

Fourier Series

Problem 1:
Knowing that 𝑓(𝑥) = 𝑓(𝑥 + 2𝜋) , sketch its graph in each of the following cases:

(a) 𝑓(𝑥) = |𝑥|, −𝜋 <𝑥 <𝜋 𝑥 if − 𝜋 < 𝑥 < 0

(c) 𝑓(𝑥) = {
(b) 𝑓(𝑥) = |sin 𝑥|, 0 < 𝑥 < 2𝜋 𝜋 − 𝑥 if 0 < 𝑥 < 𝜋

Solution:

(a) 𝑓(𝑥) = |𝑥|, −𝜋 <𝑥 <𝜋

(b) 𝑓(𝑥) = |sin 𝑥|, 0 < 𝑥 < 2𝜋

Page 1 of 11
 𝑥 if − 𝜋 < 𝑥 ≤ 0
(c) 𝑓(𝑥) = {
𝜋 − 𝑥 if 0 < 𝑥 ≤ 𝜋

Problem 2:
Find the Fourier series of the given function 𝑓(𝑥), which is assumed to have a period 𝟐𝝅.

(a) 𝑓(𝑥) = |𝑥|, −𝜋 <𝑥 <𝜋 (c)

𝑥 if − 𝜋 < 𝑥 < 0
(b) 𝑓(𝑥) = {
𝜋 − 𝑥 if 0 < 𝑥 < 𝜋

Solution:

(a) 𝑓(𝑥) = |𝑥|, −𝜋 <𝑥 <𝜋

It is clear that 𝑓(𝑥) is even. So, 𝑏𝑛 = 0.

𝐿=𝜋

∞
𝑎0
∴ 𝑓(𝑥) = + ∑ 𝑎𝑛 cos(𝑛𝑥)
2
𝑛=1

Page 2 of 11
 𝜋
1
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝜋
−𝜋

𝜋 𝜋 𝜋
2 2 2 𝑥2
= ∫ 𝑓(𝑥) 𝑑𝑥 = ∫ 𝑥 𝑑𝑥 = [ ] = 𝜋
𝜋 𝜋 𝜋 2 0
0 0

𝜋
1
𝑎𝑛 = ∫ 𝑓(𝑥) cos(𝑛𝑥) 𝑑𝑥
𝜋
−𝜋

𝜋 𝜋
2 2
= ∫ 𝑓(𝑥) cos(𝑛𝑥) 𝑑𝑥 = ∫ 𝑥 cos(𝑛𝑥) 𝑑𝑥
𝜋 𝜋
0 0

using integration by parts 𝑑 Τ𝑑𝑥 ∫ 𝑑𝑥

𝜋 𝑥 cos(𝑛𝑥)
2 𝑥 sin(𝑛𝑥) cos(𝑛𝑥) +
= [ + ] sin(𝑛𝑥)
𝜋 𝑛 𝑛2 0 1
− 𝑛
− cos(𝑛𝑥)
2 𝜋 sin(𝑛𝜋) cos(𝑛𝜋) 1 0
= [( + ) − (0 + )] 𝑛2
𝜋 𝑛 𝑛2 𝑛2

2 (−1)𝑛 1 Note that

= [ 2 − 2]
𝜋 𝑛 𝑛
𝐬𝐢𝐧(𝒏𝝅) = 𝟎
2 and 𝐜𝐨𝐬(𝒏𝝅) = (−𝟏)𝒏
= [(−1)𝑛 − 1]
𝑛2 𝜋

Hence, the Fourier series of 𝑓(𝑥) is:

∞
𝜋 2 [(−1)𝑛 − 1]
𝑓(𝑥) = + ∑ cos(𝑛𝑥)
2 𝜋 𝑛2
𝑛=1

Page 3 of 11
 𝑥 if − 𝜋 < 𝑥 ≤ 0
(b) 𝑓(𝑥) = {
𝜋 − 𝑥 if 0 < 𝑥 ≤ 𝜋

It is clear that 𝑓(𝑥) is neither odd nor even and 𝐿 = 𝜋 .

∞
𝑎0
∴ 𝑓(𝑥) = + ∑ [𝑎𝑛 cos(𝑛𝑥) + 𝑏𝑛 sin(𝑛𝑥)]
2
𝑛=1

𝜋
1
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝜋
−𝜋

0 𝜋 0 𝜋
1 1 𝑥2 𝑥2
= [ ∫(𝑥)𝑑𝑥 + ∫(𝜋 − 𝑥) 𝑑𝑥 ] = [( )| + (𝜋𝑥 − )| ] = 0
𝜋 𝜋 2 −𝜋 2 0
−𝜋 0

𝜋
𝑑 Τ𝑑𝑥 ∫ 𝑑𝑥
1
𝑎𝑛 = ∫ 𝑓(𝑥) cos(𝑛𝑥) 𝑑𝑥
𝜋 𝑥 cos(𝑛𝑥)
+
−𝜋
sin(𝑛𝑥)
1
0 𝜋 − 𝑛
1 − cos(𝑛𝑥)
= [ ∫ 𝑥 cos(𝑛𝑥) 𝑑𝑥 + ∫(𝜋 − 𝑥) cos(𝑛𝑥) 𝑑𝑥] 0
𝜋 𝑛2
−𝜋 0

0 𝜋
1 𝑥 sin(𝑛𝑥) cos(𝑛𝑥) (𝜋 − 𝑥) sin(𝑛𝑥) cos(𝑛𝑥)
= ([ + ] + [ − ] )
𝜋 𝑛 𝑛2 −𝜋
𝑛 𝑛 2
0

1 2 2 cos(𝑛𝜋) 2 𝑑 Τ𝑑𝑥 ∫ 𝑑𝑥
= [ 2− 2
] = 2
[1 − (−1)𝑛 ]
𝜋 𝑛 𝑛 𝑛 𝜋 𝜋−𝑥 cos(𝑛𝑥)
+

sin(𝑛𝑥)
−1
− 𝑛
− cos(𝑛𝑥)
0
𝑛2

Page 4 of 11
 𝜋
𝑑 Τ𝑑𝑥 ∫ 𝑑𝑥
1
𝑏𝑛 = ∫ 𝑓(𝑥) sin(𝑛𝑥) 𝑑𝑥
𝜋 𝑥 sin(𝑛𝑥)
+
−𝜋
− cos(𝑛𝑥)
1
0 𝜋 − 𝑛
1 − sin(𝑛𝑥)
= [ ∫ 𝑥 sin(𝑛𝑥) 𝑑𝑥 + ∫(𝜋 − 𝑥) sin(𝑛𝑥) 𝑑𝑥] 0
𝜋 𝑛2
−𝜋 0

0 𝜋
1 −𝑥 cos(𝑛𝑥) sin(𝑛𝑥) −(𝜋 − 𝑥) cos(𝑛𝑥) sin(𝑛𝑥)
= ([ + ] + [ − ] )
𝜋 𝑛 𝑛2 −𝜋
𝑛 𝑛 2
0

1 −𝜋 cos(𝑛𝜋) 𝜋 1 𝑑 Τ𝑑𝑥 ∫ 𝑑𝑥
= [ + ] = [1 − cos(𝑛𝜋)]
𝜋 𝑛 𝑛 𝑛 𝜋−𝑥 + sin(𝑛𝑥)

− cos(𝑛𝑥)
1 −1
= [1 − (−1)𝑛 ] − 𝑛
𝑛 − sin(𝑛𝑥)
0
𝑛2
Hence, the Fourier series of 𝑓(𝑥) is

∞
2 𝑛]
1
𝑓(𝑥) = ∑ { 2
[1 − (−1) cos(𝑛𝑥) + [1 − (−1)𝑛 ] sin(𝑛𝑥) }
𝑛 𝜋 𝑛
𝑛=1

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(c) The given sketch of 𝑓(𝑥) can be described over its period as follows:

0 if − 𝜋 < 𝑥 < −𝜋Τ2

𝑓(𝑥) = 𝑥 if −𝜋Τ2 < 𝑥 < 𝜋Τ2

{0 if 𝜋 Τ2 < 𝑥 < 𝜋

It is clear that 𝑓(𝑥) is odd. So, 𝑎0 = 𝑎𝑛 = 0 and 𝐿 = 𝜋 .

∴ 𝑓(𝑥) = ∑ 𝑏𝑛 sin(𝑛𝑥)
𝑛=1
𝜋
1
𝑏𝑛 = ∫ 𝑓(𝑥) sin(𝑛𝑥) 𝑑𝑥
𝜋
−𝜋

𝜋 𝜋/2 𝜋
2 2
= ∫ 𝑓(𝑥) sin(𝑛𝑥) 𝑑𝑥 = [ ∫ 𝑥 sin(𝑛𝑥) 𝑑𝑥 + ∫ 0 𝑑𝑥 ]
𝜋 𝜋
0 0 𝜋/2

𝜋
2 −𝑥 cos(𝑛𝑥) sin(𝑛𝑥) 2
𝑑 Τ𝑑𝑥 ∫ 𝑑𝑥
= [ + ]
𝜋 𝑛 𝑛2 0 𝑥 sin(𝑛𝑥)
+

𝑛𝜋 𝑛𝜋 − cos(𝑛𝑥)
1
2 −𝜋 cos ( 2 ) sin ( 2 ) − 𝑛
= [ + ]
𝜋 2𝑛 𝑛2 0
− sin(𝑛𝑥)
𝑛2

Hence, the Fourier series of 𝑓(𝑥) is

∞ 𝑛𝜋 𝑛𝜋
− cos ( ) 2 sin ( )
𝑓(𝑥) = ∑ [ 2 + 2 ] sin(𝑛𝑥)
𝑛 2
𝑛 𝜋
𝑛=1

Page 6 of 11
Problem 3:
Find the Fourier series of the given periodic functions, defined over its fundamental interval.

Solution:

(a) The given sketch of 𝑓(𝑥) can be described over its fundamental interval as follows:

−1 if − 2 < 𝑥 < 0
𝑓(𝑥) = {
1 if 0 < 𝑥 < 2

2𝐿 = 4 → 𝐿=2

It is clear that 𝑓(𝑥) is odd. So, 𝑎0 = 𝑎𝑛 = 0.

∞
𝑛𝜋𝑥
𝑓(𝑥) = ∑ 𝑏𝑛 sin ( )
2
𝑛=1
𝐿
1 𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sin ( ) 𝑑𝑥
𝐿 𝐿
−𝐿

𝐿 2
2 𝑛𝜋𝑥 𝑛𝜋𝑥 −2 𝑛𝜋𝑥 2 2
= ∫ 𝑓(𝑥) sin ( ) 𝑑𝑥 = ∫ sin ( ) 𝑑𝑥 = [cos ( )] = [1 − (−1)𝑛 ]
𝐿 𝐿 2 𝑛𝜋 2 0 𝑛𝜋
0 0

The Fourier series of 𝑓(𝑥) is

∞
2 𝑛𝜋𝑥
𝑓(𝑥) = ∑ [1 − (−1)𝑛 ] sin ( )
𝑛𝜋 2
𝑛=1

Page 7 of 11
(b) The given sketch of 𝑓(𝑥) can be described over its fundamental interval as follows:

0 if − 1Τ2 < 𝑥 < 0

𝑓(𝑥) = {
𝑥 if 0 < 𝑥 < 1Τ2

2𝐿 = 1 → 𝐿 = 1Τ2

It is clear that 𝑓(𝑥) is neither odd nor even. So,

∞
𝑎0
∴ 𝑓(𝑥) = + ∑ [𝑎𝑛 cos(2𝑛𝜋𝑥) + 𝑏𝑛 sin(2𝑛𝜋𝑥)]
2
𝑛=1

𝐿
1
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝐿
−𝐿

1
0 2 1
𝑥2 2 1
= 2 ∫ 0 𝑑𝑥 + ∫ 𝑥 𝑑𝑥 = 2 ( )| =
2 0 4
−1 0
[ 2 ]

𝐿
1 𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos ( ) 𝑑𝑥
𝐿 𝐿 𝑑 Τ𝑑𝑥 ∫ 𝑑𝑥
−𝐿
𝑥 cos(2𝑛𝜋𝑥)
+
1
0 2 sin(2𝑛𝜋𝑥)
1 1
= ∫ 0 cos(2𝑛𝜋𝑥) 𝑑𝑥 + ∫ 𝑥 cos(2𝑛𝜋𝑥) 𝑑𝑥 − 2𝑛𝜋
0.5 − cos(2𝑛𝜋𝑥)
−1 0 0
[ 2 ] 4𝑛2 𝜋 2

1
𝑥 sin(2𝑛𝜋𝑥) cos(2𝑛𝜋𝑥) 1 2
= 2[ + 2 2
] = 2 2
[(−1)𝑛 − 1]
2𝑛𝜋 4𝑛 𝜋 0
2𝑛 𝜋

Page 8 of 11
 𝐿
1 𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sin ( ) 𝑑𝑥
𝐿 𝐿
−𝐿 𝑑Τ𝑑𝑥 ∫ 𝑑𝑥

𝑥 sin(2𝑛𝜋𝑥)
1 +
0 2
− cos(2𝑛𝜋𝑥)
1 1
= ∫ 0 sin(2𝑛𝜋𝑥) 𝑑𝑥 + ∫ 𝑥 sin(2𝑛𝜋𝑥) 𝑑𝑥 − 2𝑛𝜋
0.5
−1 0 − sin(2𝑛𝜋𝑥)
[ 2 ] 0
4𝑛2 𝜋 2
1
−𝑥 cos(2𝑛𝜋𝑥) sin(2𝑛𝜋𝑥) 2 − cos(𝑛𝜋) (−1)𝑛+1
= 2[ + ] = =
2𝑛𝜋 4𝑛2 𝜋 2 0 2𝑛𝜋 2𝑛𝜋

Hence, the Fourier series of 𝑓(𝑥) is

∞
1 [(−1)𝑛 − 1] (−1)𝑛+1
𝑓(𝑥) = + ∑ { cos (2𝑛𝜋𝑥) + sin(2𝑛𝜋𝑥) }
8 2𝑛2 𝜋 2 2𝑛𝜋
𝑛=1

(c) The given sketch of 𝑓(𝑥) can be described over its fundamental interval as follows:

1+𝑥 if − 1 < 𝑥 < 0

𝑓(𝑥) = {
1−𝑥 if 0 < 𝑥 < 1

2𝐿 = 2 → 𝐿=1

It is clear that 𝑓(𝑥) is even. So, 𝑏𝑛 = 0.

∞
𝑎0
∴ 𝑓(𝑥) = + ∑ 𝑎𝑛 cos(𝑛𝜋𝑥)
2
𝑛=1

Page 9 of 11
 𝐿
1
𝑎0 = ∫ 𝑓(𝑥) 𝑑𝑥
𝐿
−𝐿

𝐿 1 1
2 𝑥2
= ∫ 𝑓(𝑥) 𝑑𝑥 = 2 ∫(1 − 𝑥) 𝑑𝑥 = 2 (𝑥 − )| = 1
𝐿 2 0
0 0

𝐿
1 𝑛𝜋𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos ( ) 𝑑𝑥
𝐿 𝐿
−𝐿

𝐿 1
2 𝑛𝜋𝑥 𝑑 Τ𝑑𝑥 ∫ 𝑑𝑥
= ∫ 𝑓(𝑥) cos ( ) 𝑑𝑥 = 2 ∫(1 − 𝑥) cos(𝑛𝜋𝑥) 𝑑𝑥
𝐿 𝐿
0 0 1−𝑥 +
cos(𝑛𝜋𝑥)

sin(𝑛𝜋𝑥)
1 −1
(1 − 𝑥) sin(𝑛𝜋𝑥) cos(𝑛𝜋𝑥) − 𝑛𝜋
= 2[ − ]
𝑛𝜋 𝑛2 𝜋 2 − cos(𝑛𝜋𝑥)
0 0
𝑛2 𝜋 2
− cos(𝑛𝜋) 1 2
= 2[ 2 2
+ 2 2
] = 2 2
[1 − (−1)𝑛 ]
𝑛 𝜋 𝑛 𝜋 𝑛 𝜋

Hence, the Fourier series of 𝑓(𝑥) is

∞
1 2 [1 − (−1)𝑛 ]
𝑓(𝑥) = + 2 ∑ cos (𝑛𝜋𝑥)
2 𝜋 𝑛2
𝑛=1

Page 10 of 11
Problem 4:

The Fourier series of the functions 𝑓(𝑥) and ℎ(𝑥) are as follows
∞
2 𝑛𝜋𝑥
𝑓(𝑥) = ∑ (1 − (−1)𝑛 ) sin ( ) and
𝑛𝜋 2
𝑛=1

∞
2 𝑛)
1
ℎ(𝑥) = ∑ [ 2
(1 − (−1) cos(𝑛𝑥) + (1 − (−1)𝑛 ) sin(𝑛𝑥)].
𝑛 𝜋 𝑛
𝑛=1

Complete the following:

1. The period of 𝑓(𝑥) equals 4 (𝑳 = 𝟐).

2. The function 𝑓(𝑥) is symmetric about origin (odd function).

3. ∫ ℎ(𝑥) 𝑑𝑥 = 𝜋 𝑎0 = 0
−𝜋

𝜋
2
4. ∫ ℎ(𝑥) cos(𝑛𝑥) 𝑑𝑥 = 𝜋 𝑎𝑛 = 2
(1 − (−1)𝑛 )
𝑛
−𝜋

Page 11 of 11