FOURIER SERIES Presentation of given function

1. FOURIER SERIES
Periodic Functions
A function 𝑓(𝑥) is said to have a period T if for all 𝑥, 𝑓(𝑥 + 𝑇) = 𝑓(𝑥), where 𝑇 is a positive
constant. The least value of 𝑇 > 0 is called the period of 𝑓(𝑥). Example: sinx, cosx, tanx etc.
Example:
1. What is the period of 𝒔𝒊𝒏𝒙?
Solution:
𝑓(𝑥) = 𝑠𝑖𝑛𝑥 = sin(𝑥 + 2𝜋) = 𝑠𝑖𝑛(𝑥 + 4𝜋) =. . .therefore the function has periods
2𝜋, 4𝜋, 6𝜋, etc. However, 2𝜋 is the least value and therefore is the period of 𝑓(𝑥).
2. What is the period of 𝒕𝒂𝒏𝒙?
Solution:
𝑡𝑎𝑛𝑥 = 𝑡𝑎𝑛(𝜋 + 𝑥)
Hence 𝑡𝑎𝑛𝑥 is a periodic function with period 𝜋.
Piecewise Continuous Function:
A function 𝑓(𝑥) is said to be piecewise continuous in an interval if
(i)the interval can be divided into a finite number of subintervals in each of which 𝑓(𝑥) is
continuous.
(ii) the limits of 𝑓(𝑥) as 𝑥 approaches the end point of each subinterval are finite.
Dirichlet’s conditions for the existence of Fourier series of 𝒇(𝒙) in the interval
(𝟎, 𝟐𝝅):
A function 𝑓(𝑥) can be expanded as a Fourier series of 𝑓(𝑥) in the interval (0,2𝜋) if the
following conditions are satisfied.

2. (i) 𝑓(𝑥) is single valued and finite in (0, 2𝜋)
(ii) 𝑓(𝑥) is continuous or piecewise continues with finite number of finite discontinuities
in (0, 2𝜋)
(iii) 𝑓(𝑥) has a finite number of maxima or minima in (0, 2𝜋).
Problems in the interval (−𝝅, 𝝅)
1. Find the Fourier series for 𝒇(𝒙) = |𝐬𝐢𝐧 𝒙| in −π< 𝒙 < 𝝅.
Solution:
Given 𝑓(𝑥) = |sin 𝑥|
Since 𝑓(𝑥) is an even function 𝑏𝑛 = 0
𝑎0
∞
𝑓(𝑥) = + ∑ 𝑎𝑛 cos𝑛𝑥
2
𝑛=1
… … … . . (1)
2
𝑎0 =
π
∫ 𝑓(𝑥) 𝑑𝑥
π
0
2
π
= ∫ |sin 𝑥| 𝑑
𝑥
π
0
2
π
= ∫ sin 𝑥 𝑑
𝑥
π
0
2
π 0
= [− cos 𝑥]π
2
π
= − [cos π − cos 0]
2
π
= [−1 − 1] =
4
π

3. 2
𝑎𝑛 =
π
∫ 𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥
π
0
=
2
π
π
∫ |sin 𝑥| cos𝑛𝑥 𝑑
𝑥
0
=
2
π
π
∫ sin 𝑥 cos𝑛𝑥 𝑑
𝑥
0
=
2 1
π 2
π
∫ [sin(1 + 𝑛)𝑥 + sin(1 − 𝑛)𝑥]𝑑𝑥
0
=
1
π
π
∫ [sin(𝑛 + 1)𝑥 − sin(𝑛 − 1)𝑥]𝑑𝑥
0
= [– +
𝑛 + 1 𝑛 − 1
]
0
1 cos(𝑛 + 1)𝑥 cos(𝑛 − 1)𝑥 π
π
= [–
1 cos(𝑛 + 1)π
π 𝑛 + 1
+
cos(𝑛 − 1)π
𝑛 − 1
+
1 1
𝑛 + 1 𝑛 − 1
− ]
= [–
1 (−1)𝑛+1
π 𝑛 + 1
+
(−1)𝑛−1
𝑛 − 1
+ −
1 1
𝑛 + 1 𝑛 − 1
]
= [
1 (−1)𝑛 (−1)𝑛
π
1 1
𝑛 + 1 𝑛 − 1 𝑛 + 1 𝑛 − 1
− + − ]
=
1 (−1)𝑛 + 1 (−1)𝑛 + 1
π 𝑛 + 1 𝑛 − 1
[ − ]
=
π
[ −
(−1)𝑛 + 1 1 1
𝑛 + 1 𝑛 − 1
]
=
(−1)𝑛 + 1
π
[
𝑛 − 1 − 𝑛 − 1
𝑛2 − 1
]
=
(−1)𝑛 + 1
[
−2
2
π 𝑛 − 1
]
=
−2[(−1)𝑛 + 1]
(𝑛2 − 1)π
𝑖𝑓 𝑛 ≠ 1

4. 2
𝑎1 =
π
∫ 𝑓(𝑥) cos 𝑥 𝑑𝑥
π
0
=
2
π
π
∫ |sin 𝑥| cos𝑥 𝑑
𝑥
0
=
2
π
π
∫ sin 𝑥 cos𝑥 𝑑
𝑥
0
=
2 1
π 2
π
∫ sin 2𝑥 𝑑𝑥
0
=
π 2
[ ]
0
1 − cos2𝑥 π
=
1 1 1
π 2 2
[− + ] = 0
π
𝑛
(𝑛2−1)π
−2[(−1) +1]
cos𝑛𝑥
Substitute in equation (1) we get 𝑓(𝑥) =
2
+ ∑∞
𝑛 =1
2. Find the Fourier series for the function 𝒇(𝒙) = {
𝒙 − 𝟏,
𝒙 + 𝟏,
− 𝛑 < 𝒙 < 0
𝟎 < 𝒙 < 𝝅
Solution:
Given 𝑓(𝑥) = {
𝑥 − 1, − π < 𝑥 < 0
𝑥 + 1, 0 < 𝑥 < 𝜋
𝑓(−𝑥) = {
−𝑥 − 1,
−𝑥 + 1,
0 < 𝑥 < 𝜋
− π < 𝑥 < 0
= {−(𝑥 + 1), 0 < 𝑥 < 𝜋
− π < 𝑥 < 0
−(𝑥 − 1),
= −𝑓(𝑥)
Therefore 𝑓(𝑥) is an odd function. Therefore 𝑎0 = 0, 𝑎𝑛 = 0

5. ∞
𝐻𝑒𝑛𝑐𝑒 𝑓(𝑥) = ∑ 𝑏𝑛 sin 𝑛𝑥
𝑛=1
… … … . . (1)
2
π
= ∫ 𝑓(𝑥) sin 𝑛𝑥 𝑑
𝑥
π
0
2
π
= ∫ (𝑥 + 1) sin 𝑛𝑥 𝑑
𝑥
π
0
= [(𝑥 + 1) (−
2 cos 𝑛𝑥
π 𝑛
) − (1) (−
sin 𝑛𝑥
𝑛2
)]
0
π
= [−(π + 1) (
2 cos𝑛π 1
π 𝑛 𝑛
) + ]
2
nπ
= [−(π + 1)(−1)𝑛 + 1]
2
nπ
= [1 − (1 + π)(−1)𝑛]
Substitute in (1) we get
2
nπ
𝑓(𝑥) = ∑ [1 − (1 + π)(−1)𝑛 ] sin 𝑛𝑥
∞
𝑛 =1
Problems in the interval (𝟎,𝟐𝒍)
1. Obtain the Fourier series for 𝒇(𝒙) of period 𝟐𝒍 and defined as follows
𝑓(𝑥) = { 𝑙 − 𝑥, 0 < 𝑥 ≤ 𝑙
0, 𝑙 ≤ 𝑥 < 2𝑙
Hence deduce that
1 1 1
3 5 7
(𝑖)1 − + − + ⋯ =
𝜋
4

6. 1 1 1 𝜋2
(𝑖𝑖) + + + ⋯ =
12 32 52 8
Solution:
Given 𝑓(𝑥) = { 𝑙 − 𝑥, 0 < 𝑥 ≤ 𝑙
0, 𝑙 ≤ 𝑥 < 2𝑙
The Fourier series is
𝑓(𝑥) =
𝑎0
+ ∑∞ (𝑎 cos
𝑛𝜋𝑥
+ 𝑏 sin
𝑛𝜋𝑥
)
2 𝑛=1 𝑛 𝑙 𝑛 𝑙
2𝑙
1
𝑎0 =
𝑙
∫ 𝑓(𝑥)𝑑𝑥
0
𝑙 𝑙
1 1 (𝑙 − 𝑥)2
= ∫(𝑙 − 𝑥)𝑑𝑥 = [ ]
𝑙 𝑙 −2
0 0
1 𝑙2 𝑙
= [ ] =
𝑙 2 2
2𝑙
1 𝑛𝜋𝑥
𝑎𝑛 =
𝑙
∫ 𝑓(𝑥) cos
𝑙
𝑑𝑥
0
=
1
∫
𝑙
(𝑙 − 𝑥) cos
𝑛𝜋𝑥
𝑑𝑥
𝑙 0 𝑙
1 𝑛𝜋𝑥 𝑙 𝑛𝜋𝑥 𝑙2 𝑙
= [(𝑙 − 𝑥) (sin ) − (−1) (− cos ) ]
𝑙 𝑙 𝑛𝜋 𝑙 𝑛2𝜋2
0
1 𝑙2
= [− (cos𝑛𝜋 − 1)]
𝑙 𝑛2𝜋2
𝑙 0, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
= − [(−1)𝑛 − 1] = { 2𝑙
𝑛2𝜋2 , 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
𝑛2𝜋2

7. 1 𝑛𝜋𝑥
𝑙
𝑑𝑥
2𝑙
𝑏𝑛 =
𝑙
∫ 𝑓(𝑥) sin
0
1
𝑙
= ∫ (𝑙 − 𝑥) sin
𝑙
0
𝑛𝜋𝑥
𝑙
𝑑𝑥
1
= [(𝑙 − 𝑥) (− 𝑐𝑜𝑠
𝑙
𝑛𝜋𝑥 𝑙
) − (−1) (− sin
𝑙 𝑛𝜋
𝑛𝜋𝑥
𝑙
𝑙2
) ]
𝑛2𝜋2
0
𝑙
𝑙
1 𝑙
= [𝑙 cos0 ] =
𝑙
𝑛𝜋 𝑛𝜋
𝑛𝜋𝑥
𝑙
∞
𝑙 2𝑙
𝑓(𝑥) = + ∑ [ cos
4 𝑛2𝜋2
𝑛=1,3,5…
𝑙 𝑛𝜋𝑥
𝑙
∞
] + ∑ sin
𝑛𝜋
𝑛=1
Deduction (i)
Put 𝑥 =
𝑙
2
𝑓(𝑥) =
𝑙
2
𝑙 𝑙 𝑙
= + ∑ sin
𝑙 𝑛𝜋
𝑙
2 4 𝜋 𝑛
𝑛=1
∞
𝑛𝜋
2
[𝑠𝑖𝑛𝑐𝑒 cos = 0 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑]
4 𝜋
𝑙 𝑙 𝜋 1
2 2 3
1 3𝜋
2
= [sin + sin 𝜋 + sin + ⋯]
𝑙 𝑙
4 𝜋
1 1
3 5
⇒ = [1 − + + ⋯ ]
1 1
3 5
1 − + + ⋯ =
𝜋
4
Deduction (ii)
Put 𝑥 = 𝑙 𝑓(𝑥) = 0
𝑙 2𝑙
0 = +
4 𝜋2
∞
𝑙
∑ cos 𝑛𝜋
𝑛2
𝑛=1,3,5…
[𝑠𝑖𝑛𝑐𝑒 sin 𝑛𝜋 = 0 ]

8. 4
− = [
𝑙 2𝑙 cos𝜋
+
cos3𝜋
𝜋2 12 32
+
cos 5𝜋
52
+ ⋯ ]
⇒
−𝜋2
8
1 1 1
12 32 52
= − − − − ⋯
1 1 1 𝜋2
12 +
32 +
52 + ⋯ =
8
Problems in the interval (−𝒍, 𝒍)
1. Expand 𝒇(𝒙) = 𝒆−𝒙 as a Fourier series in the interval (– 𝒍, 𝒍)
Solution:
Given 𝑓(𝑥) = 𝑒−𝑥
𝑓(𝑥) is neither even nor odd
The Fourier series is
𝑎0 𝑛𝜋𝑥
𝑙
+ 𝑏𝑛 sin
𝑛𝜋𝑥
𝑙
)
∞
𝑓(𝑥) = + ∑ (𝑎𝑛 cos
2
𝑛=1
… … … . (1)
1
𝑙
𝑎0 =
𝑙
∫ 𝑓(𝑥)𝑑𝑥
−𝑙
1
𝑙
1
𝑙 −𝑙
= ∫ 𝑒−𝑥𝑑𝑥 = [−𝑒−𝑥]𝑙
𝑙
−𝑙
1
𝑙
1
𝑙
2
𝑙
= [−𝑒−𝑙 + 𝑒𝑙 ] = [𝑒𝑙 − 𝑒−𝑙 ] = sinh 𝑙
1 𝑛𝜋𝑥
𝑙
𝑑𝑥
𝑙
𝑎𝑛 =
𝑙
∫ 𝑓(𝑥) cos
−𝑙

9. 𝑙
−𝑙
𝑙 𝑙
1 𝑛𝜋𝑥
= ∫ 𝑒−𝑥 cos 𝑑𝑥
= I
𝑙
1 𝖥
𝑒−𝑥
I1 +
[ 𝑙2
𝑛2𝜋2 [− cos
𝑛𝜋𝑥
𝑙
+ sin
𝑙 𝑙
𝑛𝜋 𝑛𝜋𝑥 1
]I
I
]−𝑙
𝑙
=
𝑙2
𝑙(𝑙2 + 𝑛2𝜋2)
[−𝑒−𝑙 cos𝑛𝜋 + 𝑒𝑙 cos𝑛𝜋]
=
𝑙(−1)𝑛
𝑙2 + 𝑛2𝜋2
[𝑒𝑙 − 𝑒−𝑙 ]
2𝑙(−1)𝑛
= sinh 𝑙
𝑙2 + 𝑛2𝜋2
1 𝑛𝜋𝑥
𝑙
𝑑𝑥
𝑙
𝑏𝑛 =
𝑙
∫ 𝑓(𝑥) sin
−𝑙
𝑙
−𝑙
𝑙 𝑙
1 𝑛𝜋𝑥
= ∫ 𝑒−𝑥 sin 𝑑𝑥
= I
𝑙
1 𝖥
𝑒−𝑥
I1 +
[ 𝑙2
𝑛2𝜋2 [− sin
𝑛𝜋𝑥
𝑙
− cos
𝑙 𝑙
𝑛𝜋 𝑛𝜋𝑥 1
]I
I
]−𝑙
𝑙
=
𝑙2
𝑙(𝑙2 + 𝑛2𝜋2)
𝑛𝜋
𝑙
𝑛𝜋
𝑙
[−𝑒−𝑙 cos𝑛𝜋 + 𝑒𝑙 cos 𝑛𝜋]
𝑙(−1)𝑛 𝑛𝜋
= 𝑙 [𝑒𝑙 − 𝑒−𝑙 ]
𝑙2 + 𝑛2𝜋2
2(−1)𝑛𝑛𝜋
= sinh 𝑙
𝑙2 + 𝑛2𝜋2
Substituting in (1), we get

10. 𝑓(𝑥) =
sinh 𝑙
𝑙
2𝑙(−1)𝑛
𝑙2 + 𝑛2𝜋2
sinh 𝑙 cos
𝑙
+ sinh 𝑙 sin
𝑛𝜋𝑥 2(−1)𝑛𝑛𝜋 𝑛𝜋𝑥
𝑙2 + 𝑛2𝜋2 𝑙
]
∞
+ ∑ [
𝑛 =1
=
sinh 𝑙
𝑙
𝑙(−1)𝑛
𝑙2 + 𝑛2𝜋2
cos
∞
+ 2 sinh 𝑙 ∑ [
𝑛=1
𝑛𝜋𝑥
𝑙
+ sin
𝑛𝜋(−1)𝑛 𝑛𝜋𝑥
𝑙2 + 𝑛2𝜋2 𝑙
]
Half Range Expansions:
In many Engineering problems it is required to expand a function 𝑓(𝑥) in the range (0, 𝜋)
In a Fourier series of period 2𝜋 or in the range (0, 𝑙) in a Fourier series of period 2𝑙. If it is
required to expand 𝑓(𝑥) in the interval (0, 𝑙), then it is immaterial what the function may be
outside the range 0 < 𝑥 < 𝑙.
If we extend the function 𝑓(𝑥) by reflecting it in the Y – axis so that 𝑓(−𝑥) = 𝑓(𝑥), then the
extended function is even for which 𝑏𝑛 = 0. The Fourier expansion of 𝑓(𝑥) will contain only
cosine terms.
If we extend the function 𝑓(𝑥) by reflecting it in the origin so that 𝑓(−𝑥) = −𝑓(𝑥), then the
extended function is odd for which 𝑎0 = 𝑎𝑛 = 0. The Fourier expansion of 𝑓(𝑥) will contain
only sine terms.
Here a function 𝑓(𝑥) defined over the interval 0 < 𝑥 < 𝑙 is capable of two distinct half range
series.
(i) Sine Series
(ii) Cosine Series
Problems under Half Range Sine series and Cosine series

11. 1. Expand 𝒇(𝒙) = 𝒙 as a cosine series in 𝟎 < 𝒙 < 𝒍 and deduce the value of
14 34 54
(𝑖) 1
+
1
+
1
+ ⋯ =
𝜋 4
96 14 24 34
(𝑖𝑖) 1
+
1
+
1
+ ⋯ =
𝜋 4
90
Solution:
Given 𝑓(𝑥) = 𝑥
𝑎0
2
The cosine series is 𝑓(𝑥) = + ∑ 𝑎 cos𝑛𝜋𝑥
𝑙
∞
𝑛=1 𝑛 … … . . (1)
2
𝑙
𝑎0 =
𝑙
∫ 𝑓(𝑥)𝑑𝑥
0
2
𝑙 𝑙 2
= ∫ 𝑥 𝑑𝑥 = [ ]
0
2 𝑥2 𝑙
𝑙
0
1
𝑙
= [𝑙2] = 𝑙
2 𝑛𝜋𝑥
𝑙
𝑙
𝑎𝑛 =
𝑙
∫ 𝑓(𝑥) cos
0
2
𝑙
𝑑𝑥 = ∫ 𝑥 cos
𝑙
0
𝑛𝜋𝑥
𝑙
𝑑𝑥
𝑙
= [𝑥 (sin
2 𝑛𝜋𝑥 𝑙
𝑙 𝑛𝜋
) − (− cos
𝑛𝜋𝑥
𝑙
)
𝑙2
𝑛2𝜋2
𝑙
]
0
2
𝑙
= [
𝑙2
𝑛2𝜋2
(−1)𝑛 −
𝑙2
𝑛2𝜋2
]
=
2𝑙
𝑛2𝜋2
[(−1)𝑛 − 1]
= 0 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
=
−4𝑙
𝑛2𝜋2
𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
Substituting in equation (1) we get

12. 𝑛𝜋𝑥
𝑙
𝑙 4𝑙 1
𝑓(𝑥) = − ∑ cos
2 𝜋2 𝑛2
𝑛=𝑜𝑑𝑑
… … . (2)
Deduction (i)
By Parseval’s identity
2
𝑙
𝑙
0
𝑎2
2 𝑛
∫[𝑓(𝑥)]2 𝑑𝑥 = 0
+ ∑ 𝑎2
∞
𝑛 =1
2
𝑙
⇒ ∫ 𝑥2 𝑑𝑥 =
𝑙
0
𝑙2
2
−4𝑙
𝑛2𝜋2
)
2
+ ∑ (
𝑛=𝑜𝑑𝑑
0
2 𝑥3 𝑙
𝑙2
𝑙 3 2
⇒ [ ] = +
16 𝑙2
𝑛 𝜋
4 4
∑
𝑛=𝑜𝑑𝑑
2 𝑙3 𝑙2
𝑙 3 2
⇒ [ ] = +
16 𝑙2
𝜋4
1
𝑛4
∑
𝑛=𝑜𝑑𝑑
⇒
2𝑙2
3
𝑙2
= +
2
16 𝑙2
𝜋4
1
𝑛4
∑
𝑛=𝑜𝑑𝑑
⇒
16 𝑙2
𝜋4 𝑛4
∑
𝑛=𝑜𝑑𝑑
1 2𝑙2 𝑙2
3 2
2 1
3 2
= − = 𝑙2 [ − ] =
𝑙2
6
1
𝑛 =𝑜𝑑
𝑑
𝑙2 𝜋4
6 16 𝑙
∑ = [ ] =
𝑛4 2
𝜋4
96
1 1 1 𝜋4
(𝑖. 𝑒. ) + + + ⋯ =
14 34 54 96
Deduction (ii)
1 1 1
𝐿𝑒𝑡 𝑆 = + + + ⋯
14 24 34
1 1 1 1 1 1
= [+ + + ⋯ ] + [ + + + ⋯ ]
14 34 54 24 44 64

13. =
𝜋4
96
1 1 1 1
24 14 24 34
+ [ + + + ⋯ ] 𝑏𝑦 (𝑖)
(𝑖.𝑒. ) 𝑆 =
𝜋4
96
1
24
+ 𝑆 =
𝜋4
1
96 16
+ 𝑆
1
16
𝑆 − 𝑆 =
𝜋4
96
1
16
𝑆 (1 − ) =
𝜋4
96
15
16
𝑆 ( ) =
𝜋4
96
𝜋4
16
96 15
𝑆 = ( ) =
𝜋4
90
1 1 1
14 24 34
(𝑖. 𝑒. ) + + + ⋯ =
𝜋4
90
𝟏
= 𝝅 𝟐
2. Obtain the Sine series for 𝒇(𝒙) = 𝒙 in 𝟎 < 𝒙 < 𝝅 and hence deduce that ∑∞
𝒏=𝟏 𝒏𝟐 𝟔
Solution:
Given 𝑓(𝑥) = 𝑥
The Sine series is 𝑓(𝑥) = ∑∞ 𝑏𝑛 sin 𝑛𝑥
𝑛 =1 … … … . . (1)
2
𝜋
𝑏𝑛 =
𝜋
∫ 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥
0
2
𝜋
= ∫ 𝑥 sin 𝑛𝑥 𝑑𝑥
𝜋
0
2
𝜋
= [𝑥 (
− cos 𝑛𝑥
𝑛
) − (
− sin 𝑛𝑥
𝑛2
)]
0
𝜋

14. 2 −1 𝑛
= [−𝜋 ( ) ] =
𝜋 𝑛 𝑛
−2(−1)𝑛
Substitute in equation (1) we get
−2(−1)𝑛
∞
𝑓(𝑥) = ∑
𝑛 =1
𝑛
sin 𝑛𝑥
Deduction:
By Parseval’s identity
2
𝜋
𝜋
0
𝑛
∫[𝑓(𝑥)]2 𝑑𝑥 = ∑ 𝑏2
∞
𝑛 =1
2
𝜋
∫ 𝑥2 𝑑𝑥 = ∑ (
𝜋
0
−2(−1)𝑛
𝑛
)
2
∞
𝑛 =1
2 𝑥3
𝜋 3
( ) = ∑
0
𝜋
4
𝑛2
∞
𝑛 =1
𝜋3
2 1
∞
⇒ [ ] = 4 ∑
𝜋 3 𝑛2
𝑛=1
⇒
2𝜋2
3
1
𝑛2
∞
= 4 ∑
𝑛 =1
𝑛2
∞
𝑛 =1
(𝑖.𝑒. ) ∑ =
1 2𝜋2 1
3 4
( ) =
𝜋2
6
Complex or Exponential Form of Fourier series:
1. Find the complex form of the Fourier series of 𝒇(𝒙) = 𝒆−𝒙 in −𝟏 ≤ 𝒙 ≤ 𝟏
Solution:

15. The complex form of the Fourier series in (−1,1) is given by
𝑐 𝑒𝑖𝑛𝜋𝑥
𝑓(𝑥) = ∑∞
𝑛=−∞ 𝑛 … …. . (1)
Where 𝑐𝑛
=
1
∫
1
𝑓(𝑥) 𝑒−𝑖𝑛𝜋𝑥 𝑑𝑥
2 −1
1 1
1 1
= ∫ 𝑒−𝑥 𝑒−𝑖𝑛𝜋𝑥 𝑑𝑥 = ∫ 𝑒−(1+𝑖𝑛𝜋)𝑥
𝑑𝑥
2 2
−1 −1
1
= [
𝑒−(1+𝑖𝑛𝜋)𝑥
2 −(1 + 𝑖𝑛𝜋)]
−1
1
=
𝑒1+𝑖𝑛𝜋 − 𝑒−(1+𝑖𝑛𝜋)
2(1 + 𝑖𝑛𝜋)
=
𝑒(cos𝑛𝜋 + 𝑖 sin 𝑛𝜋) − 𝑒−1(cos𝑛𝜋 − 𝑖 sin 𝑛𝜋)
2(1 + 𝑖𝑛𝜋)
=
𝑒(−1)𝑛 − 𝑒−1(−1)𝑛
2(1 + 𝑖𝑛𝜋)
𝑐𝑛 =
(𝑒 − 𝑒−1)(−1)𝑛
2
1 − 𝑖𝑛𝜋
( )
1 + 𝑛2𝜋2
=
(−1)𝑛(1 − 𝑖𝑛𝜋)
1 + 𝑛2𝜋2
sinh 1
Hence (1) becomes
𝑒−𝑥 =
(−1)𝑛(1 − 𝑖𝑛𝜋)
1 + 𝑛2𝜋2
sinh 1 𝑒𝑖𝑛𝜋𝑥
∞
∑
𝑛=−∞

16. 2. Find the complex form of the Fourier series of 𝒇(𝒙) = 𝐜𝐨𝐬 𝒂𝒙 in (– 𝝅, 𝝅) where ‘a’ is
neither zero nor an integer.
Solution:
Here 2𝑐 = 2𝜋 or 𝑐 = 𝜋
Let the complex form of the Fourier series be
𝑐𝑛𝑒𝑖𝑛𝑥
∞
𝑓(𝑥) = ∑
𝑛=−∞
… … . . (1)
𝑤ℎ𝑒𝑟𝑒 𝑐𝑛 =
1
2𝜋
𝜋
∫ 𝑓(𝑥) 𝑒−𝑖𝑛𝑥 𝑑𝑥
−𝜋
𝑐𝑛 =
1
2𝜋
𝜋
∫ cos 𝑎𝑥 𝑒−𝑖𝑛𝑥 𝑑𝑥
−𝜋
1 𝑒−𝑖𝑛𝑥
= [ (−𝑖𝑛 cos𝑎𝑥 + 𝑎 sin 𝑎𝑥)]
2𝜋 𝑎2 − 𝑛2
−𝜋
𝜋
=
1
2𝜋(𝑎2 − 𝑛2)
[𝑒−𝑖𝑛𝜋(−𝑖𝑛 cos𝑎𝜋 + 𝑎 sin 𝑎𝜋) − 𝑒𝑖𝑛𝜋(−𝑖𝑛 cos𝑎𝜋 + 𝑎 sin 𝑎𝜋)]
=
1
2𝜋(𝑎2 − 𝑛2)
[𝑖𝑛 cos 𝑎𝜋 ( 𝑒𝑖𝑛𝜋 − 𝑒−𝑖𝑛𝜋) + 𝑎 sin 𝑎𝜋( 𝑒𝑖𝑛𝜋 + 𝑒−𝑖𝑛𝜋) ]
=
1
2𝜋(𝑎2 − 𝑛2)
[𝑖𝑛 cos 𝑎𝜋 (2𝑖 sin 𝑛𝜋) + 𝑎 sin 𝑎𝜋(2cos𝑛𝜋) ]
𝑐𝑛 =
1
2𝜋(𝑎2 − 𝑛2)
(−1)𝑛 2 a sin 𝑎𝜋

17. (−1)𝑛
Hence (1) becomes cos𝑎𝑥 =
𝑎 sin 𝑎𝜋
∑∞
𝜋 𝑛=−∞ (𝑎2−𝑛2)
𝑒𝑖𝑛𝑥
Harmonic Analysis:
The process of finding the Fourier series for a function given by numerical values is known as
harmonic analysis.
In harmonic analysis the Fourier coefficients 𝑎0, 𝑎𝑛 and 𝑏𝑛 of the function 𝑦 = 𝑓(𝑥) in
(0,2𝜋) are given by
𝑎0 = 2[𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 𝑖𝑛(0,2𝜋) ]
𝑎𝑛 = 2[𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 cos 𝑛𝑥 𝑖𝑛(0,2𝜋)]
𝑏𝑛 = 2[𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 sin 𝑛𝑥 𝑖𝑛(0, 2𝜋)]
Problems under Harmonic Analysis:
1. The following table gives the variations of a periodic function over a period T
𝑥 0 𝑇⁄6 𝑇⁄3 𝑇⁄2 2𝑇⁄3 5𝑇⁄6 𝑇
𝑓(𝑥) 1.98 1.3 1.05 1.3 –
0.88
–
0.25
1.98
Find 𝒇(𝒙) upto first harmonic.
Solution: 𝑃𝑢𝑡 𝜃 =
2𝜋𝑥
𝑇
When 𝑥 takes the values of 0,
𝑇
,
𝑇
,
𝑇
,
2𝑇
,
6 3 2 3 6
5𝑇
, 𝑇
then θ takes values 0,
𝜋
,
2𝜋
, 𝜋,
4𝜋
,
5𝜋
, 2𝜋
3 3 3 3

18. The given data becomes
𝜃 0 𝜋
3
2𝜋
3
𝜋 4𝜋
3
5𝜋
3
2𝜋
𝑦 1.98 1.3 1.05 1.3 –
0.88
–
0.25
1.98
𝑎0
2 1 1
The Fourier series is 𝑓(𝑥) = + 𝑎 cos 𝜃 + 𝑏 sin 𝜃
[ ]
4.6
𝑎0 = 2 𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 = 2 [ 6
] = 1.5
[ ]
1.12
𝑎1 = 2 𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 cos𝜃 = 2 [ 6
] = 0.373
[ ]
𝑏1 = 2 𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 sin 𝜃 = 2 [
3.013
6
] = 1.004
𝜽 𝒚 𝒚 𝐜𝐨𝐬 𝜽 𝒚 𝐬𝐢𝐧 𝜽
0 1.98 1.98 0
𝜋
3
1.3 0.65 1.1258
2𝜋
3
1.05 –0.525 0.9093
𝜋 1.3 –1.3 0
4𝜋
3
– 0.88 0.44 0.762
5𝜋
3
– 0.25 –
0.125
0.2165
Total 4.6 1.12 3.013

19. 𝑓(𝑥) = 0.75 + 0.373 cos𝜃 + 1.004 sin 𝜃
= 0.75 + 0.373 cos
2𝜋𝑥
+ 1.004 sin
2𝜋𝑥
𝑇 𝑇
2. Obtain the first three harmonic for the data
𝑥 0 1 2 3 4 5
𝑦 4 8 15 7 6 2
Solution:
Given that the length of the interval is 6.
i.e. 2𝑙 = 6, 𝑙 = 3
The Fourier series upto third harmonic is
2 3 3 3
1 2 3 1 3 3
𝑓(𝑥) =
𝑎0
+ 𝑎 cos
𝜋𝑥
+ 𝑎 cos
2𝜋𝑥
+ 𝑎 cos
3𝜋𝑥
+ 𝑏 sin
𝜋𝑥
+ 𝑏 sin
2𝜋𝑥
+ 𝑏
2 3 sin 3𝜋𝑥
3
𝑥 𝑦 𝜋
𝑦 cos
3
2𝜋
𝑦 cos
3
3𝜋
𝑦 cos
3
𝜋
𝑦 sin
3
2𝜋
𝑦 sin
3
3𝜋
𝑦 sin
3
0 4 4 4 4 0 0 0
1 8 4 –4 –8 6.93 6.93 0
2 15 –7.5 –7.5 15 12.99 –12.99 0
3 7 –7 7 –7 0 0 0
4 6 –3 –3 6 –3.46 5.19 0
5 2 1 –1 –2 –4.33 –1.73 0
Total 42 –8.5 –4.5 8 12.13 –2.6 0
[ ]
42
𝑎0 = 2 𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 = 2 [ 6
] = 14

20. 1 3
𝑎 = 2 [𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 cos
𝜋𝑥
] = 2 [−
8.5] = −2.83
6
2 3
𝑎 = 2 [𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 cos
2𝜋𝑥
] = 2 [−
4.5
= −1.5
]
6
3
8
3 6
𝑎 = 2 [𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 cos
3𝜋𝑥
] = 2 [ ] = 2.67
1
12.13
𝑏 = 2 [𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 sin
𝜋𝑥
] = 2 [
3 6
] = 4.04
2 3
𝑏 = 2 [𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 sin
2𝜋𝑥
] = 2 [−
2.6] = −0.87
6
3 3
𝑏 = 2 [𝑚𝑒𝑎𝑛 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 sin
3𝜋𝑥
] = 0
𝑓(𝑥) = 7 − 2.83 cos − 1.5 cos
𝜋𝑥 2𝜋𝑥
3 3
+ 2.67 cos
3𝜋𝑥
3
𝜋𝑥
3
+ 4.04 sin − 0.87 sin
2𝜋𝑥
3