The document discusses Fourier series and periodic functions. It provides: - Definitions of Fourier series and periodic functions. - Examples of periodic functions including trigonometric and other functions. - Euler's formulae for calculating the coefficients of a Fourier series. - Integration properties used to solve Fourier series problems. - Two examples of determining the Fourier series for given periodic functions and using it to deduce mathematical results.

2. Fazaia College of Education for Women
2
Presented to: Ma’am Javeria Hanif
Presented by: Hafsa Butt, Alishba Malik,Niha Iqbal
Discipline: BS.CS (III)
Subject: Multivariate Caculus

3. Fourier series of Periodic Function
3

4. 4
Overview
• What is Fourier series?
• What is periodic function?
• What are integration rules?
• How to solve Fourier series with given periodic function.
Through examples.
• Applications of Fourier series.

5. “
5
Periodic Functions
A function f: R →R is said to be periodic if there exists a positive
number T such that f(x+T) = f(x) for all x ϵ R. T is called the period
of f(x).
Examples:
i. Trigonometric functions sin x, cos x, cosec x are periodic
functions with primitive period 2π.
ii. Sin 2x, cos 2x are periodic functions with primitive period π.
iii. Tan x has primitive period π.

6. Fourier Series
Then, f(x)=
6
Let f(x) be a periodic function with period 2π. Then we can obtain the
Fourier series in any interval of length 2 π.
0
1 1
cos sin
2
n n
n n
a
a nx b nx
 
 
 
 
,
0
1
( )
a f x dx


 
 
1
( )cos
n
a f x nxdx


 
 
1
( )sin
n
b f x xdx


 
 
The formulae for the coefficients a0, an, bn are known as Euler’s formulae.

7. 7
Integration Properties
' '' '''
1 2 3 4 ......
uvdx uv uv u v u v
    

(suffixes(v) represent successive integrals and dashes(u’) represent
successive derivatives)
2 2
sin [ sin cos ]
ax
ax e
e bxdx a bx b bx
a b
 


2 2
cos [ cos sin ]
ax
ax e
e bxdx a bx b bx
a b
 



8. 8
Example# 01
Obtain a Fourier series to represent f(x)=x2 in (0, 2𝛑) and hence deduce that
2
2 2 2 2
1 1 1 1
........
1 2 3 4 12

    
2
2 0
1 1
2
2
0
0
2
3
0
2
2
2
2
0
0
2
2
Let ( )
cos sin
2
be the Fourier series of ( ) in (0,2 ).
1
then
1
3
8
3
4 1
= cos
2 3
1 sin cos
-2x
n
n n
n
f x x
a
x a nx bn nx
f x
a x dx
x
a
a x nxdx
nx nx
x
n n










 
 

  

 
  
 

 
 
   
 
  
 
   
 
 


2
3
0
cos
+2
(using generalised rule of integeration by parts)
nx
n


 
  
 

9. 9
2
3 3
1 4 2 2
4
n n n
n



 
 

  
 
 
 
 
 
2 2 2
2
2
0
2
2
2 3
0
sin 2 sin0 0
1 4 4
= =
cos2 n=(-1) 1
1
sin
1 cos sin cos
2 2
n
n
n
n n
b x nxdx
nx nx nx
x x
n n n




 


 
 
 
 
  
   

   
     
  
     
 
     
 


10. 10
0
2
2
2
1 1
2
2 2 2
Substituting the values of a , a , b in (1), we get,
4 4 4
cos sin
3
4 cos cos2 cos3 sin sin 2 sin3
= +4 ..... -4 ......
3 1 2 3 1 2 3
n n
n n
x nx nx
n n
x x x x x x
 


 
 
  
   
     
   
   
 
2
......(2)
is the Fourier expansion of f(x) = x in (0, 2 )
Take x = in (2), so that,


2
2
2 2 2 2
2
2 2 2 2
2
2 2 2 2
4 1 1 1 1
= +4 .....
3 1 2 3 4
1 1 1 1
= -4 .....
3 1 2 3 4
1 1 1 1
.....
12 1 2 3 4





 
   
 
 

 
     
 
 

     

11. 11
Example# 02
Express the function F(x) = -π, - π<x<0
= x, 0<x<π,
as a Fourier series. Hence deduce that
1
12 +
1
32 +
1
52 + ⋯ =
𝜋2
8
Let, f(x) =
𝑎0
2
+ σ𝑛=1
∞
𝑎𝑛 cos nx + σ𝑛=1
∞
𝑏𝑛
sin nx
Then,
𝒂𝟎=
1
𝜋
‫׬‬
−𝜋
𝜋
𝑓 𝑥 𝑑𝑥
=
1
𝜋
‫׬‬
−𝜋
0
− 𝜋𝑑𝑥 + ‫׬‬
0
𝜋
𝑥𝑑𝑥
=
1
𝜋
[−𝜋𝑥]−𝜋
0
+
1
𝜋
𝑥2
2 0
𝜋
=
1
𝜋
(0 –𝜋2
) +
1
𝜋
𝜋2
2
=
−𝜋
2

12. 12
𝒂𝒏 = න
−𝜋
𝜋
𝑓 𝑥 cos nxdx =
1
𝜋
න
−𝜋
0
−𝜋cos𝑛𝑥𝑑𝑥 + න
0
𝜋
𝑥𝑐𝑜𝑠𝑛𝑥𝑑𝑥
=
1
𝜋
ቚ
−𝜋𝑠𝑖𝑛𝑛𝑥
𝑛
0
−𝜋
+ 𝑥
𝑠𝑖𝑛𝑛𝑥
𝑛
+ 1
𝑐𝑜𝑠𝑛𝑥
𝑛2
𝜋
0 =
1
𝜋
(−1)𝑛−1
𝑛2
𝒃𝒏=
1
𝜋
‫׬‬
−𝜋
𝜋
𝑓 𝑥 𝑠𝑖𝑛𝑛𝑥𝑑𝑥
By chain rule of integration
=
1
𝜋
න
−𝜋
0
−𝜋sin𝑛𝑥𝑑𝑥 + න
0
𝜋
𝑥𝑠𝑖𝑛𝑛𝑥𝑑𝑥 =
1
𝜋
ቚ
𝜋𝑐𝑜𝑠𝑛𝑥
𝑛
0
−𝜋
+ 𝑥
−𝑐𝑜𝑠𝑛𝑥
𝑛
− 1
−𝑠𝑖𝑛𝑛𝑥
𝑛2
𝜋
0
=
1
𝜋
𝜋
𝑛
(1 − (−1))𝑛
+
1
𝜋
𝜋
(−1)𝑛+1
𝑛
=
1
𝜋
𝜋
𝑛
+
(−1)𝑛
𝑛
(−𝜋 − 𝜋)
=
1
𝑛
1 − (−1))𝑛
2

13. 13
f(x) =
−𝜋
4
+
1
𝜋
σ𝑛=1
∞ (−1)𝑛−1
𝑛2 𝑐𝑜𝑠𝑛𝑥 + σ𝑛=1
∞ 1−(−1))𝑛2
𝑛
sinnx
=
−𝜋
4
−
2
𝜋
𝑐𝑜𝑠𝑥
12 +
𝑐𝑜𝑠3𝑥
32 +
𝑐𝑜𝑠5𝑥
52 + ⋯ +
3𝑠𝑖𝑛𝑥
1
−
𝑠𝑖𝑛2𝑥
2
+
3𝑠𝑖𝑛3𝑥
3
− ⋯ …(2)
As the Fourier expansion of f(x) , F(x) is no defined when x=0
F(0) =
1
2
𝑙𝑡 𝑓 𝑥 + 𝑙𝑡 𝑓(𝑥)
𝑥 → 0+
𝑥 → 0−
=
1
2
0 − 𝜋 =
−𝜋
2
Substituting x=0
−𝜋
2
=
−𝜋
4
−
2
𝜋
1
12
+
1
32
+
1
52
−𝜋
4
=
−2
𝜋
1
12
+
1
32
+
1
52
=
𝜋2
8
𝑎𝑛𝑠. .

14. Exercise # 9b
14
•Obtain the Fourier series expansions for the following function in the
Intervals given against them.
( ) ( ) in (0, 2 )
x
a f x e 

0
1 1
Let cos sin ...(i)
2
x
n n
n n
a
e a nx b nx
 
 
  
 
2
2
0 0
0
1 1
[ ]
x
x
a e
e dx


 
 

2
2
0
1
[ 1]
1
e
e
a




 

 

15. 15
 
2
0
2
2
0
2 0
2
2
2
2
2
1
cos
1
cos sin here a=1 and b=n
1
1
(cos2 sin2 ) (cos0 sin0 )
( 1)
sin2 sin0 0
1
[ (1 0) 1(1 0)]
cos2 ( 1) 1
( 1)
1
[
( 1)
x
n
x
n
e nxdx
a
e
nx n nx
n
e n n n e n n n
n
n
e
n
n
n






 






 
 
 

 
 
   
 

 
 
     
  
  



2
1]
e 

2
2
1
( 1)
n
e
a
n



 


16. 16
 
   
   
   
2
0
2
2
0
2 0
2
2
2
2
2
2
2
2
2
1
sin
1
sin cos here a=1 and b=n
1
1
[ sin(2 ) cos(2 ) sin0 cos0 ]
( 1)
1
[ 0 (1) 1 0 (1) ]
( 1)
1
[ 1 ]
( 1)
1
[ ]
( 1)
( 1
x
n
x
n
e nxdx
b
e
nx n nx
n
e n n n e n
n
e n n
n
e n n
n
n ne
n
n ne
b
n









 






 
 
 

 
   

   

   

 


 


)

17. 17
0
a , a , b
n n
Putting values of in equation (i)
2 2 2
2 2
1 1
2 2 2
2 2
1 1
2
2 2
1 1
1 1
cos sin
2 ( 1) ( 1)
1 1 ( 1)
= cos sin
2 ( 1) ( 1)
1 1 cos sin
2 ( 1) ( 1)
x
n n
n n
n n
e e n ne
e nx nx
n n
e e n e
nx nx
n n
e nx nx
n
n n
  
  

  
  

 
 
 
 
 
 
  
  
 
   
 
 
 

  
 
 
 
 
 
 

18. (b) f(x) = 𝒆𝒙
…..in (-π, π)
18
f(x) =
𝑎0
2
+ σ𝑛=1
∞
𝑎𝑛 cos nx + σ𝑛=1
∞
𝑏𝑛
sin nx …(i)
𝑎0 =
1
𝜋
‫׬‬
−𝜋
𝜋
𝑓 𝑥 𝑑𝑥
=
1
𝜋
𝑒𝑥
−𝜋
𝜋
=
1
𝜋
𝑒𝜋 − 𝑒−𝜋
= 2
𝑒𝜋−𝑒−𝜋
2𝜋
=
2
𝜋
𝑠𝑖𝑛ℎπ

19. 𝑎𝑛 = න
−𝜋
𝜋
𝑓 𝑥 cos nxdx = න
−𝜋
𝜋
𝑒𝑥
cos nxdx
=
1
𝜋
[
𝑒𝑥
𝑛2+1
𝑐𝑜𝑠𝑛𝑥 + 𝑛𝑠𝑖𝑛𝑛𝑥 ]−𝜋
𝜋
=
1
𝜋(𝑛2+1)
𝑒𝜋 𝑐𝑜𝑠𝑛𝜋 + 𝑛𝑠𝑖𝑛𝑛𝜋 − 𝑒−𝜋 cos 𝑛 −𝜋 + 𝑛𝑠𝑖𝑛𝑛(−𝜋)
=
1
𝜋(𝑛2+1)
𝑒𝜋 𝑐𝑜𝑠𝑛𝜋 + 𝑛𝑠𝑖𝑛𝑛𝜋) − 𝑒−𝜋(cos 𝑛 −𝜋 + 𝑛𝑠𝑖𝑛𝑛(−𝜋)
=
1
𝜋(𝑛2+1)
𝑒𝜋(−1)𝑛+0 − 𝑒−𝜋( −1 𝑛 + 0)
=
1
𝜋(𝑛2+1)
𝑒𝜋
(−1)𝑛
− 𝑒−𝜋
−1 𝑛
=
(𝑒𝜋− 𝑒−𝜋)(−1)𝑛
𝜋(𝑛2+1)
=
2(−1)𝑛
𝜋(𝑛2+1)
𝑠𝑖𝑛ℎ𝜋

20. 𝑏𝑛 =
1
𝜋
න
−𝜋
𝜋
𝑓 𝑥 𝑠𝑖𝑛𝑛𝑥𝑑𝑥
=
1
𝜋
[
𝑒𝑥
𝑛2+1
𝑠𝑖𝑛𝑛𝑥 − 𝑛𝑐𝑜𝑠𝑛𝑥 ]−𝜋
𝜋
=
1
𝜋(𝑛2+1)
𝑒𝜋( 𝑠𝑖𝑛𝑛𝜋 − 𝑛𝑐𝑜𝑠𝑛𝜋) − 𝑒−𝜋(sin 𝑛 −𝜋 − 𝑛𝑐𝑜𝑠𝑛 −𝜋 )
=
1
𝜋(𝑛2+1)
𝑒𝜋
(0 − 𝑛 −1 𝑛
) − 𝑒−𝜋
(0 − 𝑛 −1 𝑛
)
=
1
𝜋(𝑛2+1)
−𝑛𝑒𝜋(−1)𝑛+ 𝑛𝑒−𝜋 −1 𝑛
=
−2(−1)𝑛
𝜋(𝑛2+1)
(𝑒𝜋− 𝑒−𝜋)
2
=
−2(−1)𝑛
𝜋(𝑛2+1)
sinhπ
20

21. 21
Putting values of 𝑎0 ,𝑎𝑛 , 𝑏𝑛 in equation(i)
𝑎0
2
+ σ𝑛=1
∞
𝑎𝑛 cos nx + σ𝑛=1
∞
𝑏𝑛
sin nx
=
2 𝑠𝑖𝑛ℎπ
2𝜋
+ σ𝑛=1
∞ 2(−1)𝑛
𝜋(𝑛2+1)
𝑠𝑖𝑛ℎ𝜋 cosnx + σ𝑛=1
∞ −2(−1)𝑛
𝜋(𝑛2+1)
sinhπ sinnx
=
𝑠𝑖𝑛ℎπ
𝜋
+ 2 σ𝑛=1
∞ (−1)𝑛
𝜋(𝑛2+1)
𝑠𝑖𝑛ℎ𝜋 cosnx - 2σ𝑛=1
∞ (−1)𝑛
𝜋(𝑛2+1)
sinhπ sinnx
=
𝑠𝑖𝑛ℎπ
𝜋
+ 1 + 2 ෍
𝑛=1
∞
(
(−1)𝑛
𝑐𝑜𝑠𝑛𝑥
(𝑛2 + 1)
−
−1 𝑛
𝑠𝑖𝑛𝑛𝑥
𝑛2 + 1
)

22. (c) f(x ) =
𝒙 − 𝝅 ≤ 𝒙 ≤ 𝟎
𝟎 𝟎 ≤ 𝒙 ≤ 𝝅
22
𝑎0
2
+ σ𝑛=1
∞
𝑎𝑛 cos nxdx + σ𝑛=1
∞
𝑏𝑛
sin nxdx …(i)
𝑎0 =
1
𝜋
‫׬‬
−𝜋
𝜋
𝑓 𝑥 𝑑𝑥
=
1
𝜋
‫׬‬
−𝜋
0
𝑥 𝑑𝑥 + ‫׬‬
0
𝜋
0 𝑑𝑥
=
1
𝜋
𝑥2
2 −𝜋
0
+ 0
=
1
𝜋
0
2
−
(−𝜋)2
2
𝑎0=
−𝜋
2

23. 23
𝑎𝑛 = න
−𝜋
𝜋
𝑓 𝑥 cos nxdx
=
1
𝜋
‫׬‬−𝜋
0
𝑥 . cos nxdx +
1
𝜋
‫׬‬0
𝜋
0 cos nxdx
=
1
𝜋
𝑥.
−𝑠𝑖𝑛𝑥
𝑛
− (1)
𝑐𝑜𝑠𝑛𝑥
𝑛2
=
1
𝜋
[0 −
−𝜋 sin −𝜋
𝑛
] +
𝑐𝑜𝑠𝑛𝑥
𝑛2
−𝜋
0
=
1
𝜋
0 +
1
𝑛2 −
𝑐𝑜𝑠𝑛𝜋
𝑛2
=
1
𝑛2𝜋
1 − (−1)𝑛 =
1
𝑛2𝜋
1 − 1
= 0

24. 24
For odd:
=
1
𝜋
1
𝑛2 −
𝑐𝑜𝑠𝑛𝜋
𝑛2
=
1
𝑛2𝜋
1 − (−1)
=
1
𝑛2𝜋
1 + 1)
=
2
𝑛2𝜋
2n is an even number; when it is 2n-1, it becomes odd
𝒂𝒏 =
2
(2𝑛−1)2𝜋

25. 25
𝑏𝑛 =
1
𝜋
න
−𝜋
𝜋
𝑓 𝑥 𝑠𝑖𝑛𝑛𝑥 𝑑𝑥
=
1
𝜋
‫׬‬−𝜋
0
𝑥. 𝑠𝑖𝑛𝑛𝑥 𝑑𝑥 + ‫׬‬0
𝜋
0. 𝑠𝑖𝑛𝑛𝑥 𝑑𝑥
=
1
𝜋
𝑥.
−𝑐𝑜𝑠𝑛𝑥
𝑛
+
𝑠𝑖𝑛𝑛𝑥
𝑛2
−𝜋
0
=
1
𝜋
0 −
−𝜋 −cos(−𝜋)
𝑛
+
𝑠𝑖𝑛𝑛(−𝜋)
𝑛2
=
1
𝜋
−𝜋 −cos(𝜋)
𝑛
+
𝑠𝑖𝑛𝑛(−𝜋)
𝑛2
=
1
𝜋
−𝜋 𝑐𝑜𝑠𝑛𝜋
𝑛
⸫ cosnπ = (−1)𝑛

26. 26
When,
cosnπ is even = 1
cosnπ is odd = -1
For odd;
=
𝑐𝑜𝑠𝑛𝜋
𝑛
=
−(−1)𝑛
𝑛
=
1
𝑛
For even:
= −
1
𝑛
=
−1
𝑛
By substituting even and odd equations
𝑏𝑛=
(−1)𝑛+1
𝑛

27. 27
Now,
Putting values of 𝑎0 ,𝑎𝑛 , 𝑏𝑛 in equation …1
f(x) =
−𝜋
2
2
+ σ𝑛=1
∞ 2
(2𝑛−1)2𝜋
𝑐𝑜𝑠𝑛𝑥 + σ𝑛=1
∞ (−1)𝑛+1
𝑛
𝑠𝑖𝑛 𝑛𝑥
=
−𝜋
4
+
2
𝜋
σ𝑛=1
∞ cos(2𝑛−1)𝑥
(2𝑛−1)2 + σ𝑛=1
∞ (−1)𝑛+1
𝑛
𝑠𝑖𝑛 𝑛𝑥

28. Applications
28
• A Fourier Series has many applications in mathematical analysis
as it is defined as the sum of multiple sines and cosines.
• Thus, it can be easily differentiated and integrated, which
usually analyses the functions such as saw waves which are
periodic signals in experimentation.
• It also provides an analytical approach to solve the
discontinuity problem. In calculus, this helps in solving complex
differential equations.

29. Thanks!
Any questions?
29