UNIT II
FOURIER SERIES
A Fourier series of a periodic function consists of a sum of sine and cosine terms. Sines and cosines
are the most fundamental periodic functions.
The Fourier series is named after the French Mathematician and Physicist Jacques Fourier (1768 –
1830). Fourier series has its application in problems pertaining to Heat conduction, acoustics, etc.
The subject matter may be divided into the following sub topics.
FOURIER SERIES
Half-range series
Series with
arbitrary period
Harmonic Analysis
Complex series
FORMULA FOR FOURIER SERIES
Consider a real-valued function f(x) which obeys the following conditions called Dirichlet’s
conditions:
1. f(x) is defined in an interval (a,a+2l), and f(x+2l) = f(x) so that f(x) is a periodic function of
period 2l.
2. f(x) is continuous or has only a finite number of discontinuities in the interval (a,a+2l).
3. f(x) has no or only a finite number of maxima or minima in the interval (a,a+2l).
1 a2l
a0 
l  f (x)dx
Also, let


a
 n 
 n 
1 a  2l
1 a  2l
f
(x)
cos
f
(x)sin

n

1,2,3,.....
 xdx, 
xdx,
an  
(2) bn  
l a
l
 l 
 l 
a
n=1,2,3…
Then, the infinite series


 n 
 n 


  a n cos  x  bn sin  x
2 n 1
l 
l 

a0


(4)


is called the Fourier series of f(x) in the interval (a, a+2l). Also, the real numbers a0, a1, a2, ….an, and
b1, b2, ….bn are called the Fourier coefficients of f(x). The formulae (1), (2) and (3) are called Euler’s
formulae.

It can be proved that the
nsum of the series
n (4) is f(x) if f(x) is continuous at x. Thus we have

a





x

b
sin


 x …….
n
 l 
 l 
0

f(x) = 2  an cos

n 1

(5)

1
Suppose f(x) is discontinuous at x, then the sum of the series (4) would be
f (x )  f (x )Where


2
f(x+) and f(x-) are the values of f(x) immediately to the right and to the left of f(x) respectively.
Particular Cases:Case (i)
Suppose a=0. Then f(x) is defined over the interval (0,2l). Formulae (1), (2), (3) reduce to
12l
a0  f (x)dx
l0
2l
 n 
an  1 f (x) cos
xdx, n  1,2,. .... 
l
l


0
2l
 n 
bn  1  f (x)sin  xdx,
l
l


0
(6)

Then the right-hand side of (5) is the Fourier expansion of f(x) over the interval (0,2l).If we set l=,
then f(x) is defined over the interval (0, 2). Formulae (6) reduce to


1 2
a0 = 


 f (x)dx
0
2
an  1  f (x) cos nxdx
, n=1, 2… 
0
1 2
bn   f (x)sin nxdx
 
n=1, 2… 
0

Also, in this case, (5) becomes
a0
f(x) =
2

 
a
n 1
n
cos nx  bn sin nx
(8)
Case (ii) Suppose a=-l. Then f(x) is defined over the interval (-l,l). Formulae (1), (2) (3) reduce to



l
a0 
1
 f ( x)dx
l l
n =1, 2 … 
(9)

 n 
 n 
1 l
1l

xdx, n=1, 2 … 
an  l  f ( x) cos l  xdx bn   f (x) sin 
l
l




l
l

Then the right-hand side of (5) is the Fourier expansion of f(x) over the interval (-l, l).If we set l =,
then f(x) is defined over the interval (-,). Formulae (9) reduce to
1 
a0 = 

 f (x)dx


1

an    f (x) cos nxdx


,
n=1, 2… 
(10)
1


bn    f (x)sin nxdx
n=1, 2…



Putting l =  in (5), we get
f(x) =
a0
2

 
a
n 1
n
cos nx  bn sin nx
PARTIAL SUMS
The Fourier series gives the exact value of the function. It uses an infinite number of terms which is
impossible to calculate. However, we can find the sum through the partial sum S N defined as
follows:
S (x)  a
N
0
 n 
 n  
n N 

x  b  sin
x
  an cos


  Where N takes positive integral values.
n

 l 
 l  
n 1 

In particular, the partial sums for N=1, 2 are
 x 
 x 
S (x)  a 

a
cos

b
sin




1
1
1
0
 l 
 l 






 x 
 x 
 2x 
 2x 
S (x)  a 



a1 cos   b1 sin    a2 cos
  b sin 

2
0
l 
 l 
 l  2  l 


If we draw the graphs of partial sums and compare these with the graph of the original function f(x),
it may be verified that SN(x) approximates f(x) for some large N.
Some useful results:1. The following rule called Bernoulli’s generalized rule of integration by parts
is useful in
evaluating the Fourier coefficients.
 uvdx  uv
1
 u'v 2  u''v 3  .......
Here u, u … are the successive derivatives of u and
v1   vdx, v2   v1 dx,......

We illustrate the rule, through the following examples:

  cos nx 
  sin nx    cos nx 
x2 sin nxdx  x2 
 2x


 2 3 


2
n
 n


  2 x n 
2x
2x
2x






 e 
e
e
e
3
2
32x
 x e dx  x  2   3x  4   6x 8   6 16 





 












2. The following integrals are also useful:
eax
acosbx  bsin bx
a2  b2
eax
ax
sin
bxdx

e
asin bx  b cosbx

a2  b2
ax
 e cosbxdx 

3. If ‘n’ is integer, then
Sin n = 0, cosn = (-1)n , sin2n = 0, cos2n=1
HALF-RANGE FOURIER SERIES:
The Fourier expansion of the periodic function f(x) of period 2l may contain both sine and cosine
terms. Many a time it is required to obtain the Fourier expansion of f(x) in the interval (0,l) which is
regarded as half interval. The definition can be extended to the other half in such a manner that the
function becomes even or odd. This will result in cosine series or sine series only.
Sine series:
Suppose f(x) = (x) is given in the interval (0, l). Then we define f(x) = - (-x) in (-l, 0). Hence f(x)
becomes an odd function in (-l, l). The Fourier series then is
f (x)  b sin  nx 



n
 l 
n 1

where
bn 
2
(11)
 nx 
f (x)sin  dx
l


l
l
0

The series (11) is called half-range sine series over (0, l).
Putting l= in (11), we obtain the half-range sine series of f(x) over (0,) given by

f ( x) 
bn 
Cosine series:
b
n 1
2 
 
0
n
sin nx
f (x)sin nxdx
Let us define
 (x)
f (x)   (x) in (0, l) .....given in (-l,0) …..in order to make the


function even.


Then the Fourier series of f(x) is given by
a

f (x)  0  a cos  nx 



n
2 
 l 
n 1

(12)
where,
2l
a0 
l
 f (x)dx
0
l
nx
2 f (x) cos dx
an  
l
l


0

The series (12) is called half-range cosine series over (0,l)
Putting l =  in (12), we get
f (x) 
a0

2
 a

n
cos nx
n1
where
a0 
an 
2 
 
0
2 
 
0
f (x)dx
f (x) cos nxdx n  1,2,3,..
HARMONIC ANALYSIS
The Fourier series of a known function f(x) in a given interval may be found by finding the Fourier
coefficients. The method described cannot be employed when f(x) is not known explicitly, but
defined through the values of the function at some equidistant points. In such a case, the integrals in
Euler’s formulae cannot be evaluated. Harmonic analysis is the process of finding the Fourier
coefficients numerically.
To derive the relevant formulae for Fourier coefficients in Harmonic analysis, we employ the
following result:
The mean value of a continuous function f(x) over the interval (a,b) denoted by [f(x)] is defined as
 f (x) 
1
b
b  aa 
f (x)dx .
The Fourier coefficients defined through Euler’s formulae, (1), (2), (3) may be redefined as

 1 a  2l

a0  2  f (x)dx  2[ f (x)]
2l a


a  2l
 nx  
 nx 

an  2 1  f (x) cos
dx  2 f (x) cos 
l
l
2l



 

a


n

x

n

x

a  2l





1
dx  2 f (x)sin  
bn  2  f (x)sin 

l
l
2l




a
 

Using these in (5), we obtain the Fourier series of f(x). The term a 1cosx+b1sinx is called the first
harmonic or fundamental harmonic, the term a 2cos2x+b2sin2x is called the second harmonic and so
a12  b21
on. The amplitude of the first harmonic is
and so on.
and that of second harmonic is
a22  b22
PART –A
1. State Dirichlet’s conditions.(Nov’03) (May-13)( Apr-10)
(or)
State the Sufficient condition for the function f(x) expressed as Fourier series
(i)
f(x) is single valued periodic and well defined except possibly at a finite number of
points.s
(ii) f(x) has atmost a finite number of finite discontinuous and no infinite discontinuous
(iii) f(x) has atmost a finite number of maxima and minima.
2. If f(x) = x2 + x is expressed as a fourier series in the interval (-2,2) to which value this series
converges at x = 2.(May’03)
f(x) is discontinuous at x = 2 also x=2 is an end point .
Therefore f(2) =
𝑓(−2)+ 𝑓(2)
2
=
4−2+4+2
2
=4
3. Define Harmonic Analysis(may-05,Dec-08) (May-13)
The process of finding the Fourier series for a function given by numerical values is
as Harmonic analysis
4. State Parseval’s identity for the half range cosine expansion of f(x) in (0,1)(Dec’06)
𝑓(𝑥) =
𝑎0
2
∞
+ ∑ 𝑎 𝑐𝑜𝑠𝑛𝜋𝑥
𝑛
𝑛=1
known
1
|𝑓(𝑥)|2𝑑𝑥 =
∫0
5.
𝑎0
1
+2 ∑∞
2
𝑎2
2 𝑛=1 𝑛
State whether y = tanx can be expanded as a Fourirerseries.If so how? If not
why? (Dec 98,05,08)
tanx cannot be expanded as a Fourirer series. Since tanx not satisfies Dirichlet’s conditions.
6. If f(x) = sinhx is defined in –𝝅 < 𝑥 < 𝜋,write the value of fourirer constant a0 and an?
(Dec 08)
f(x) = sinhx
f(-x) = sinh(-x)=-sinhx
sinhx is an odd function => a0 =0,an =0
7. Define the root mean squae value of f(x)
(Dec 08, May-12)
𝑏
The root mean squae value of f(x) =y over the interval (a,b) is𝑦 =
√∫ {𝑓(𝑥)}2
𝑎
𝑏−𝑎
𝑑𝑥
8. Find the Fourirer series Constant bn for x sinx in (- ,  )(m-06)
bn =0 ,since functions is even function
9.
If f(x) is an odd function defined in (-l , l ) .What are the values of a0 and
an (Dec 05)
If f(x) is an odd function , a0 =0,an =0
10. State Parseval’s identity for the half range cosine expansion of f(x) in (0,2l) . (Dec’04)
∞
2
𝑎02 1
1 2𝑙
+ ∑ 𝑎𝑛
∫ (𝑓(𝑥))2 𝑑𝑥 =
4
2𝑙 0
2
𝑛=1
11.What are the constant term a0 and the coefficient of cosnx, an in the Fourirer series
Expansion of f(x)= x-x 3 in (- ,  )(Dec-04)
f(x)= x-x 3
f(-x)= - x-(-x) 3 = - f(x)
It is an odd function, => a0 =0,an =0
12. If f(x) is discontinuous at x = a,what does its Fourirer series represent at that point
( may-07,Dec-04) f(a) =
𝑓(𝑎−)+ 𝑓(𝑎+)
2

cos x,0  x  
13.If f (x)  

0,  x  2 find the sum of the fourirer series of f(x) at x= π
Ans:
:
f() =
14. 𝑰𝒇 𝒙𝟐 =
𝝅𝟐
𝑓(−)+ 𝑓(+)
2
+ 𝟒 ∑∞
2
=
−1+ 50
49
2
2
==
(−𝟏)𝒏
𝟏
𝒏𝟐
𝜋2
𝟏𝟐
(
) 𝒄𝒐𝒔 𝒏𝒙 , & 𝐷𝑒𝑑𝑢𝑐𝑒 𝑡ℎ𝑎𝑡
𝒏=𝟏
𝟑
𝑐𝑜𝑠+ 50
=
Soln: 𝑃𝑢𝑡 𝑥 = 𝜋 𝑤𝑒 𝑔𝑒𝑡, 𝜋2 =
(−1)𝑛
+ 4 ∑∞
+
𝟏
+
𝟐𝟐
+
𝟑𝟐
) 𝑐𝑜𝑠 𝑛𝜋 = 4 ∑∞
𝟏
….=
𝟒𝟐
(−1)𝑛
𝝅𝟐
(𝑨𝒑𝒓 − 𝟏𝟎)
𝟔
) 𝑐𝑜𝑠 𝑛𝜋 = 𝜋2 −
𝑛=1 ( 𝑛2
𝑛=1 ( 𝑛2
3
𝟏
𝜋2
3
∞
2𝜋2 = 4 ∑ ( 1 ) = 𝟏 + 𝟏 + 𝟏 + 𝟏 … . = 𝝅𝟐
3
𝑛2
𝟏𝟐 𝟐𝟐 𝟑𝟐 𝟒𝟐
𝟔
(∴ 𝑐𝑜𝑠 𝑛𝜋 = (−1)𝑛)
𝑛=1
15 𝑶𝒃𝒕𝒂𝒊𝒏 𝒕𝒉𝒆 𝒇𝒊𝒓𝒔𝒕 𝒕𝒆𝒓𝒎 𝒐𝒇 𝒇𝒐𝒖𝒓𝒊𝒆𝒓 𝒔𝒆𝒓𝒊𝒆𝒔 𝒇(𝒙) = 𝒙𝟐 𝒊𝒏
1
𝜋
()
1
𝜋
Soln:𝑎0 = 2𝜋 ∫−𝜋 𝑓 𝑥 𝑑𝑥 = 2𝜋 ∫−𝜋 𝑥
2
𝑥3
2
𝑑𝑥 =
2𝜋
𝜋
1 𝜋3
𝜋2
𝜋
3
[ ] = []=
3 0
− 𝝅 < 𝑥 < 𝜋 (𝑫𝒆𝒄 − 𝟎𝟗)
3
16. 𝑭𝒊𝒏𝒅 𝑹𝑴𝑺 𝒐𝒇 𝒇(𝒙) = 𝒙𝟐 𝒊𝒏 (−𝒍, 𝒍) (𝑫𝒆𝒄 − 𝟏𝟎)
𝟒
𝟏𝒍
Soln:𝟐𝒍 ∫−𝒍 𝒙 𝒅𝒙 =
𝟒𝒍𝟒
+𝟏
𝟏𝟔𝒍𝟒
∞
∑
𝟐 𝒏=𝟏 𝒏𝟒𝝅𝟒
𝟑𝟔
⇒
𝟐 [ 𝒙𝟓 𝒍
]=
𝟐𝒍 𝟓 𝟎
𝒍𝟒
𝟗
+ 𝟖𝒍𝟒 ∑ ∞
𝝅𝟒
𝟏
𝒏=𝟏 𝒏𝟒
⇒ 𝝅𝟒 = 𝟏 +
𝟗𝟎
𝟏𝟒
𝒙, 𝟎 ≤ 𝒙 ≤ 𝟏
17.𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒔𝒖𝒎 𝒐𝒇 𝒇𝒐𝒖𝒓𝒊𝒆𝒓 𝒔𝒆𝒓𝒊𝒆𝒔 𝒇(𝒙) = {
𝒂𝒕 𝒙 = 𝟎
𝟐, 𝟏 ≤ 𝒙 ≤ 𝟐
𝟏
𝟐𝟒
+
𝟏
𝟑𝟒
….
(𝑫𝒆𝒄 − 𝟐𝟎𝟏𝟏)
Soln: 𝑥 = 0 𝑖𝑠 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑖𝑛 𝑡ℎ𝑒 𝑒𝑥𝑡𝑟𝑒𝑒𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙𝑠
𝑆𝑢𝑚 =
𝑓(0) + 𝑓(2) 0 + 2
=
=1
2
2
18.𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒄𝒐𝒔𝒊𝒏𝒆 𝒔𝒆𝒓𝒊𝒆𝒔 𝒇𝒐𝒓 𝒇(𝒙) = 𝒙 𝒔𝒊𝒏 𝒙 𝒊𝒏 (𝟎, 𝝅)&
(𝐷𝑒𝑐 − 2011)
𝟏
𝟏
𝟏
𝝅
𝑫𝒆𝒅𝒖𝒄𝒆 𝒕𝒉𝒂𝒕 𝟏 + 𝟐 (
−
+
−⋯….)=
𝟏. 𝟑 𝟑. 𝟓 𝟓. 𝟕
𝟐
(−1)
1
∞
𝑛
Soln: xsinx = 1 - cos 𝑥 − 2 ∑
) 𝑐𝑜𝑠 𝑛𝑥
(
𝑛=2 2
2
𝜋
𝑛 −1
put x= [ point of continuity ]
2
𝜋
2
𝜋
2
𝜋
𝜋
1
sin = 1 +
𝜋
2
2
(−1)𝑛
𝑛𝜋
𝑛=2 (𝑛 2 −1
(
𝑛=2
1
−1
1.3
=1+2(
1.3
2
) 𝑐𝑜𝑠
(𝑛−1)(𝑛+1)
= 1 − 2 ( (−1) +
1
) 𝑐𝑜𝑠
(−1)𝑛
∑∞
2
𝑛𝜋
) 𝑐𝑜𝑠 2
𝑛2 −1
2
= 1 − 2 ∑∞
=1−2
(−1)𝑛
∞
cos − 2 ∑𝑛=2 (
2
1
(0) + (1) + ⋯ )
2.4
−
1
3.5
+
𝑛𝜋
1
−
5.7
3.5
⋯)
19.𝑮𝒊𝒗𝒆 𝒕𝒉𝒆 𝒆𝒙𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒇𝒐𝒖𝒓𝒊𝒆𝒓 𝒔𝒆𝒓𝒊𝒆𝒔 𝒄𝒐𝒆𝒇𝒇𝒊𝒄𝒊𝒆𝒏𝒕 𝒃𝒏 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒇(𝒙)
𝒅𝒆𝒇𝒊𝒏𝒆𝒅 𝒊𝒏 (−𝟐, 𝟐)
(𝑨𝒑𝒓 − 𝟐𝟎𝟏𝟏)
1 2
𝑛𝜋𝑥
𝑏𝑛 = 2 ∫−2 𝑓(𝑥) sin
Soln:
2
𝑑𝑥
20. 𝑾𝒊𝒕𝒉𝒐𝒖𝒕 𝑭𝒊𝒏𝒅𝒊𝒏𝒈 𝒂𝟎, 𝒂𝒏, 𝒃𝒏 𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒇𝒐𝒖𝒓𝒊𝒆𝒓 𝒔𝒆𝒓𝒊𝒆𝒔 𝒇𝒐𝒓 𝒇(𝒙) = 𝒙𝟐 𝒊𝒏 (𝟎, 𝝅)
𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 (
𝒂𝟐
𝟎
𝟒
Soln: 1 ∫𝜋𝑥 𝑑𝑥
4 =
1 𝑥5
𝜋0
𝜋50
𝜋
5
𝜋4
[ ] = 1 [𝜋] =
𝜋5
=
5
𝟏
+
𝑎2
0
𝟐
∞
∑(𝒂𝟐 + 𝒃𝟐 )) (𝑨𝒑𝒓 − 𝟐𝟎𝟏𝟏)
𝒏
𝒏
𝒏=𝟏
1
+ ∑∞ (𝑎2 + 𝑏2)
4
2
𝑛=1
𝑛
𝑛
(∵ 𝐵𝑦 𝑃𝑎𝑟𝑠𝑣𝑒𝑙`𝑠 𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦)
21.𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒕𝒆𝒓𝒎 𝒊𝒏 𝒕𝒉𝒆 𝒆𝒙𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏 𝒐𝒇 𝒄𝒐𝒔𝟐𝒙 𝒂𝒔 𝒕𝒉𝒆 𝑭. 𝑺 𝒊𝒏 ( −𝝅, 𝝅)(𝑴𝒂𝒚 − 𝟏𝟐)
Soln: Given f(x) = 𝑐𝑜𝑠2𝑥 =
a0
W.K.T. f(x) =
To find a0 




2
2


 a
1+cos 2𝑥


n
n1
cos nx   bn sin nx
n1
1  cos2 x dx  2  cos2 x dx
 


0
2
1 
(1
cos
2x)/
2
dx


(1 cos 2x) dx

1


0
x  sin 2x / 2


0

1
0
(  0)  (0  0)

1
∴ 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑡𝑒𝑟𝑚 =
𝑎0
2
=
1
2
PART-B
1) Obtain the Fourier expansion of f(x) = x2 over the interval (-,). Deduce that
2
1
1
1
4
1
1
1
(i)   1 2  2  ......   (𝑖𝑖) 4 + 4 + 4 + 4 … … … … … . = 𝜋
1
90
2
3
4
2
3
6
(m/J 07, N/D 07, dec 08, jun 09)
May-2001,Nov-2005,May-13
Solution:
The function f(x) is even. Hence

1
a0



= 
f (x)dx

a0 
2 2
3
a 
1

=
2 
 f (x)dx

2  x3  
2
=  x dx   
0
  3 0
2 
0
or
n

2 
  f (x) cos nxdx =  
0
2 

= 0
f (x) cos nxdx,
since f(x)cosnx is even
x cos nxdx
2
Integrating by parts, we get


 sin nx 
 sin nx  2x
  cos nx 
a  2  x2 
  2

n

2
   n 
 n
  n3 
0








4(1)n
n2
Also,

1
bn   f (x) sin nxdx  0


since f(x)sinnx is odd.

Thus

3
n1 n
2
n
f (x)    4 (1) cos nx
 n2
 2 3 n11
2
 

4

1

 n2
1

2
2
6
 2  1  1  1  .....
Hence, 6
22
32
𝟏
(𝒊𝒊)
𝟏
𝟒+
𝟏
𝟐
𝟒+
𝟏
𝟏
𝟒+
𝟒… …… … … . =
𝟑
𝟒
𝝅𝟒
𝟗𝟎
Using Parsvel`s Identity
2
4(−1)𝑛
16
𝜋4
2
2
𝑎0 = , 𝑎 𝑛= (
, [𝑓(𝑥)]2 = 𝑥4
)
=
9
𝑛2
𝑛4
𝜋
∞
∞
∞ 1
1 𝑥5
1
1 𝜋5
𝜋4
16
𝜋4 1
𝜋4
4
4
+
∑
⇒
[
]
=
+
8
∑
⇒
[
]
=
+
8
∑
∫ 𝑥 𝑑𝑥 =
∫ 𝑥 𝑑𝑥 =
𝜋5
9 2
9
9
𝜋5
2𝜋 −𝜋
2𝜋 0
𝑛4
𝑛4
𝑛4
0
1
𝜋
𝜋
2
𝑛=1
𝑛=1
𝑛=1
∞
∞
1
4𝜋4
1
𝜋4
1
1
1
1
𝜋4 𝜋4
− =8 ∑ 4⇒
= 8 ∑ 4⇒
= 4 + 4 + 4 + 4 … … … … ….
5 9
𝑛
45
𝑛
90 1
2
3
4
𝑛=1
𝑛=1
2
𝐨𝐯𝐞𝐫 (−, ); 𝐃𝐞𝐝𝐮𝐜𝐞 𝐭𝐡𝐚𝐭
2) 𝐟(𝐱) = x

8
∞
𝐃𝐞𝐝𝐮𝐜𝐞 𝒕𝒉𝒂𝒕 ∑
𝒏=𝟏
1 1
  .......
12 32
&
𝟏
,
𝝅𝟐
=
(𝟐𝒏 − 𝟏)𝟐
𝟖
(𝑴𝒂𝒚 − 𝟏𝟐)
𝐒𝐨𝐥𝐧: 𝑓(−𝑥) = |−𝑥| = |𝑥| = 𝑓(𝑥) ⇒ 𝑓(𝑥) 𝑖𝑠 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑠𝑜 𝑏𝑛 = 0
W. K. T 𝑓(𝑥) =
𝑎0
∞
2 𝜋
+ ∑ 𝑎 cos 𝑛𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝑎 = ∫ 𝑓(𝑥)𝑑𝑥, 𝑎 =
𝑛
0
𝑛
2
𝜋 0
𝑛=1
2 ∫ 𝜋𝑓(𝑥)𝑐𝑜𝑠𝑛𝑥𝑑𝑥,
𝜋 0
𝜋
1
2 𝑥2
𝜋
2 𝜋
𝜋2
2
𝑎0 = ∫ 𝑓(𝑥)𝑑𝑥, ⇒ ∫ 𝑥𝑑𝑥 = [ ] = [𝜋2 − 0] =
=𝜋
𝜋2
𝜋
𝜋
𝜋0
𝜋0
0
𝑎𝑛 =
2
sin 𝑛𝑥
− cos 𝑛𝑥 𝜋
2 𝜋
2 𝜋
)
−
(
∫ 𝑓(𝑥)𝑐𝑜𝑠𝑛𝑥𝑑𝑥 = ∫ 𝑥 𝑐𝑜𝑠𝑛𝑥𝑑𝑥 = [(𝑥
)]
𝑛
𝜋 0
𝜋 0
𝜋
𝑛2
0
𝑎𝑛 =
2
𝜋𝑛2
[(−1)𝑛 − 1] ⇒ 𝑎𝑛 = {
𝜋
∞
𝑓(𝑥) = + ∑
2
−4
𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
𝜋𝑛2
0, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
−4
𝑛=𝑜𝑑𝑑
𝜋𝑛2
cos 𝑛𝑥 .................. (1)
𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = 0, 𝑖𝑠 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 (1) 𝑏𝑒𝑐𝑜𝑚𝑒𝑠
𝜋
∞
4
∞
∞
𝑛=𝑜𝑑𝑑
𝑛=1
1
2
𝜋
4
1
1
⇒∑
=𝜋
0= − ∑
2
2
cos(0)
⇒
−
=
−
∑
(2𝑛 − 1)
𝑛
8
2 𝜋
𝑛2
2
𝜋
𝑛=𝑜𝑑𝑑
3) 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐇𝐚𝐥𝐟 𝐑𝐚𝐧𝐠𝐞 𝐂𝐨𝐬𝐢𝐧𝐞 𝐬𝐞𝐫𝐢𝐞𝐬 𝐟(𝐱) = (𝛑 − 𝐱𝟐) 𝒊𝒏 (𝟎, 𝝅) (𝑴𝒂𝒚 − 𝟏𝟐)
𝑆𝑜𝑙𝑛: 𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 ℎ𝑎𝑙𝑓 𝑟𝑎𝑛𝑔𝑒 𝑐𝑜𝑠𝑖𝑛𝑒 𝑠𝑒𝑟𝑖𝑒𝑠 𝑏𝑒 𝑓(𝑥) =
𝑎0
∞
+∑𝑎
2
𝑎0 =
2
𝜋
∫ 𝑓(𝑥) 𝑑𝑥 =
𝜋0
2
𝜋
∫ (π − x2) 𝑑𝑥 =
𝜋0
𝑎𝑛 =
2
𝜋
[𝜋𝑥 −
𝑥3
3
𝜋
]=
0
𝑛
cos 𝑛𝜋𝑥 ............. (1)
𝑛=1
2
𝜋3
2
[𝜋2 −
] = [𝜋(3 − 𝜋)],
𝜋
3
3
2 ∫ 𝜋𝑓(𝑥)𝑐𝑜𝑠𝑛𝑥 𝑑𝑥 = 2 ∫ 𝜋(π − x2)𝑐𝑜𝑠𝑛𝑥 𝑑𝑥
𝜋0
𝜋0
sin 𝑛𝑥
2 (π − x2) sin 𝑛𝑥
cos 𝑛𝑥
= [
− 2𝑥
]
+
2
𝜋
𝑛
𝑛3
𝑛2
𝜋
0
𝑎𝑛 =
2
𝜋
2
𝑓(𝑥)
[(0 −
2𝜋(−1)𝑛
[𝜋(3 − 𝜋)]
=3
+ 0) − (0 − 0 + 0)] =
𝑛2
∞
𝑛=1
𝑛2
𝑛2
[−2(−1)𝑛] =
[𝜋(3 − 𝜋)]
−4
+∑
2
2
(−1)𝑛
cos 𝑛𝜋 =
∞
−4
𝑛2
(−1)𝑛
−4 ∑
3
(−1)𝑛
𝑛2
𝑛=1
cos 𝑛𝜋𝑥
4) 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝒄𝒐𝒎𝒑𝒍𝒆𝒙 𝒇𝒐𝒓𝒎 𝒐𝒇 𝒇𝒐𝒖𝒓𝒊𝒆𝒓 𝒔𝒆𝒓𝒊𝒆𝒔 𝒇𝒐𝒓 𝒕𝒉𝒆 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏
𝒇(𝒙) = 𝒄𝒐𝒔 𝒂𝒙 𝒊𝒏 (– 𝝅, 𝝅), 𝑾𝒉𝒆𝒓𝒆 a𝒊𝒔 𝒏𝒐𝒕 𝒂𝒏 𝒊𝒏𝒕𝒆𝒈𝒆𝒓 (𝑴𝒂𝒚 − 𝟏𝟑)
𝐺𝑖𝑣𝑒𝑛 𝑓(𝑥) = cos 𝑎𝑥 𝑖𝑛 (−𝜋, 𝜋)
∞
𝐹𝑜𝑟𝑚𝑢𝑙𝑎: 𝑓(𝑥) = ∑ 𝑐𝑛 𝑒𝑖𝑛𝑥, 𝑤ℎ𝑒𝑟𝑒 𝑐𝑛 =
𝑛=−∞
1 𝜋
1 𝜋
∫ 𝑓(𝑥) 𝑒−𝑖𝑛𝑥𝑑𝑥 =
∫ cos 𝑎𝑥 𝑒−𝑖𝑛𝑥𝑑𝑥
2𝜋 −𝜋
2𝜋 −𝜋
𝑇𝑜 𝑓𝑖𝑛𝑑 𝑐𝑛 =
1
=
1 𝜋
∫ 𝑓(𝑥) 𝑒−𝑖𝑛𝑥𝑑𝑥
2𝜋 −𝜋
𝜋
𝑒−𝑖𝑛𝑥
[
2
2𝜋 (−𝑖𝑛) + 𝑎
{(−𝑖𝑛) cos 𝑎𝑥 + asin 𝑎𝑥) }]
2
–𝜋
𝜋
𝜋
1
𝑒−𝑖𝑛𝑥
1 [ 𝑒−𝑖𝑛𝑥
[
=
{(−𝑖𝑛)
}]
{(−𝑖𝑛) cos 𝑎𝑥 + asin 𝑎𝑥) }]
cos
𝑎𝑥
+
asin
𝑎𝑥)
2
2
2 + 𝑎2
2𝜋 𝑖 2 𝑛 + 𝑎
2𝜋
−𝑛
–𝜋
–𝜋
=
𝑖𝑛𝜋
𝑒−𝑖𝑛𝜋
{(−𝑖𝑛) cos 𝑎𝜋 + asin 𝑎𝜋) } − 𝑒
[ 2
2
2𝜋 𝑎 − 𝑛
𝑎2− 𝑛
1
2
{(−𝑖𝑛) cos(−𝑎𝜋) + asin(−𝑎𝜋) }]
(−1)𝑛
𝑎 sin 𝑎𝜋 (−1)𝑛
[
]
{2𝑎 sin 𝑎𝜋}] =
[
𝑐𝑛 =
2𝜋 𝑎2 − 𝑛 2
𝜋
𝑎2 − 𝑛2
1
∞
𝑓(𝑥) = ∑
𝑛=−∞
∞
𝑎 sin 𝑎𝜋 (−1)𝑛
𝑎 sin 𝑎𝜋 ∑ [ (−1)𝑛
𝑖𝑛𝑥
[
]𝑒 =
] 𝑒𝑖𝑛𝑥
𝜋
𝜋
𝑎2 − 𝑛2
𝑎2 − 𝑛2
𝑛=−∞
5) 𝑭𝒊𝒏𝒅 𝒕𝒉𝒆 𝑭𝒐𝒖𝒓𝒊𝒆𝒓 𝒄𝒐𝒔𝒊𝒏𝒆 𝒕𝒓𝒂𝒏𝒔𝒇𝒐𝒓𝒎 𝒐𝒇 (𝒙 − 𝟏)𝟐 𝒊𝒏 𝟎 < 𝑥 < 1 &
2  1  1 1
 2 .....
2
2
𝑫𝒆𝒅𝒖𝒄𝒆 𝒕𝒉𝒂𝒕 6
3
(𝑴𝒂𝒚 − 𝟏𝟑)
𝑆𝑜𝑙𝑛: 𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 ℎ𝑎𝑙𝑓 𝑟𝑎𝑛𝑔𝑒 𝑐𝑜𝑠𝑖𝑛𝑒 𝑠𝑒𝑟𝑖𝑒𝑠 𝑏𝑒 𝑓(𝑥) =
𝑎0
∞
+∑𝑎
𝑛
2
1
1
𝑎0 = 2 ∫ 𝑓(𝑥) 𝑑𝑥 = 2 ∫ (𝑥 − 1)2 𝑑𝑥 = 2 [
0
0
1
(𝑥 −
1
1)3
3
0
] =
cos 𝑛𝑥 ............. (1)
𝑛=1
2
2
[0 − (−1)] = ,
3
3
1
𝑎𝑛 = 2 ∫ 𝑓(𝑥)𝑐𝑜𝑠𝑛𝜋𝑥 𝑑𝑥 = 2 ∫ (𝑥 − 1)2𝑐𝑜𝑠𝑛𝜋𝑥 𝑑𝑥
0
0
(𝑥 − 1)2 sin 𝑛𝜋𝑥
=2[
𝜋𝑛
cos 𝑛𝜋𝑥
− 2(𝑥 − 1)
+2
𝜋2𝑛2
sin 𝑛𝜋𝑥
]
𝜋3𝑛3
1
0
2
𝑎𝑛 = 2 [(0 − 0 + 0) − (0—
𝑓(𝑥) =
2
∞
3
+∑
2
4
𝑛=1
2
4
+
0)]
=
2
[
]
=
𝜋2𝑛2
𝜋2𝑛2
𝜋2𝑛2
cos 𝑛𝜋𝑥 =
𝜋2𝑛2
𝑛=1
1
𝑓(𝑥) =
4 ∞ 1
+ ∑ cos 𝑛𝜋𝑥
𝑛2
3 𝜋2
1
+
3
∞
4
𝜋2
1
∑
𝑛=1
cos 𝑛𝜋𝑥
𝑛2
𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = 0 𝑖𝑠 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠,
∞
𝑓(0) = 1
∞
∞
2 𝜋2
1
1
4
1
1 = + 2 ∑ 2 cos(0) ⇒ 1 − = 2 ∑ 2 ⇒ ( ) = ∑ 2
3 𝜋
𝑛
3 𝜋
𝑛
3 4
𝑛
1
4
1
𝑛=1
𝑛=1
𝜋2
6
1
=
1
1
2+
22
1
+
32
+ ⋯ ….
6) 𝐎𝐛𝐭𝐚𝐢𝐧 𝐭𝐡𝐞 𝐡𝐚𝐥𝐟 − 𝐫𝐚𝐧𝐠𝐞 𝐜𝐨𝐬𝐢𝐧𝐞 𝐬𝐞𝐫𝐢𝐞𝐬 𝐟𝐨𝐫 𝐟(𝐱) =
𝑆𝑜𝑙𝑛: 𝑓(𝑥) =
𝑎0
∞
+ ∑ 𝑎 cos
𝑛
2
𝑛𝜋𝑥
𝑛=1
… … (1),
l

kx,0  x 

2

(𝑴𝒂𝒚 − 𝟏𝟑)
k(l  x), l
xl

2
𝑙
𝑎 = 2 ∫ 𝑓(𝑥) 𝑑𝑥
𝑜
𝑙 0
𝑊ℎ𝑒𝑟𝑒
𝑙
𝑛=1
𝑛𝜋𝑥
2 𝑙
𝑑𝑥
𝑎𝑛 = ∫ 𝑓(𝑥) cos
𝑙
𝑙 0
𝑙
𝑙
𝑙
𝑙
𝑙
2 2
2
𝑎0 = 2 [∫ 𝑘𝑥 𝑑𝑥 + ∫ 𝑘(𝑙 − 𝑥) 𝑑𝑥] = [∫ 𝑘𝑥 𝑑𝑥 + ∫ 𝑘𝑙 𝑑𝑥 − ∫ 𝑘𝑥 𝑑𝑥]
𝑙
𝑙
𝑙
𝑙 0
𝑙 0
2
2
𝑙
2
𝑙
2
𝑥2 2
2 𝑘 𝑙2
𝑙
𝑘 2 𝑙2
𝑥2
𝑎0 = [𝑘 ( ) + 𝑘𝑙(𝑥)𝑙 𝑙 − 𝑘 ( 2 ) ] = [ (
− 0) + 𝑘𝑙 (𝑙 − ) − (𝑙 − )]
𝑙
2
2
2
4
2
𝑙34
𝑙
0
2
2 𝑘𝑙2
𝑘𝑙2 3𝑘𝑙2
1 1 3
1
3
+
−
] = 2𝑘𝑙 [ + − ] = 𝑘𝑙 [ + 1 − ]
𝑎0 = [
4
𝑙 12
2
8
12
2 8
6
𝑎0 = 𝑘𝑙
2
𝑎𝑛 =
𝑙
𝑙
2
[4 + 24 − 18]
10
5𝑘𝑙
= 𝑘𝑙 ( ) =
24
24
12
[∫ 𝑘𝑥 cos
0
𝑛𝜋𝑥
𝑙
𝑙
𝑑𝑥 + ∫ 𝑘(𝑙 − 𝑥) cos
𝑙
2
𝑛𝜋𝑥
𝑙
𝑑𝑥]
𝑙
2
2
𝑎𝑛 = 2 [𝑘𝑥 ( 𝑙 ) (sin 𝑛𝜋𝑥 ) − 𝑘 ( 2𝑙 2) (− cos 𝑛𝜋𝑥)]
𝑙
𝑛 𝜋
𝑙
𝑛𝜋
𝑙
0
𝑙
𝑛𝜋𝑥
2
𝑙
𝑙2
𝑛𝜋𝑥
)
(sin
)
−
(−𝑘)
(
)]
2
+ [𝑘(𝑙
−
𝑥)
(
𝑙
2) (− cos
𝑛𝜋
𝑙
𝑛𝜋
𝑙
𝑙
2
2 𝑘 𝑙
𝑛𝜋
𝑛𝜋
𝑙2
𝑙2
𝑎𝑛 = [( ( )(sin ) + 𝑘 ( 2 2) (cos ) ) − (0 + 𝑘 ( 2 2))]
𝑙 2 𝑛𝜋
2
𝑛𝜋
2
𝑛𝜋
𝑘 𝑙
2
𝑛𝜋
𝑙2
𝑙2
𝑛𝜋
𝑛
+ [(0 − 𝑘 (𝑛2𝜋 2 )(−1) ) — ( ( ) (sin )) − 𝑘 ( 𝑛2𝜋 2 ) (cos ))
2
2
2 𝑛𝜋
𝑙
𝑎𝑛 =
2 𝑘 𝑙
𝑙2
𝑙2
𝑙2 )(−1)𝑛 − 𝑘 ( 𝑙 ) (sin 𝑛𝜋
) (cos 𝑛𝜋 ) − 𝑘 (
)−𝑘(
[ ( )(sin 𝑛𝜋 ) + 𝑘 (
)
𝑛2𝜋2
𝑛2𝜋2
𝑙 2 𝑛𝜋
2
𝑛2 𝜋2
2
2 𝑛𝜋
2
𝑛𝜋
𝑙2
+ 𝑘 ( 2 2)(cos )]
𝑛 𝜋
2
2
𝑙2
𝑙2
𝑙2
𝑛𝜋
𝑎𝑛 = [2𝑘 ( 2 2 ) (cos ) + −𝑘 ( 2 2 ) − 𝑘 ( 2 2 ) (−1)𝑛]
𝑛 𝜋
𝑛 𝜋
𝑛 𝜋
2
𝑙
𝑎𝑛 =
2
𝑙2
𝑛𝜋
2𝑘𝑙
𝑛𝜋
+ [(−1)𝑛 + 1])] = 2 2 (2 cos
+ [(−1)𝑛 + 1])
[(𝑘 (𝑛2𝜋 2 )) (2 cos
2
2
𝑛 𝜋
𝑙
𝑓(𝑥) =
∞
∞
𝑛=1
𝑛=1
𝑛𝜋
𝑛𝜋𝑥
+ [(−1)𝑛 + 1]) cos
+ ∑ 𝑎 cos 𝑛𝜋𝑥 = 5𝑘𝑙 + ∑ 2𝑘𝑙 (2 cos
𝑛
2
𝑙
24
𝑛2 𝜋2
2
𝑙
𝑎0
5𝑘𝑙
𝑓(𝑥) =
+
∞
2𝑘𝑙
𝜋2
24
1
∑
𝑛=1
(2 cos
𝑛𝜋
𝑛2
+ [(−1)𝑛 + 1]) cos
2
𝑙
7) 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐅𝐨𝐮𝐫𝐢𝐞𝐫 𝐬𝐞𝐫𝐢𝐞𝐬 𝐟(𝐱) = (𝛑 − 𝐱)𝟐 𝒊𝒏 (𝟎, 𝟐𝝅)
𝑆𝑜𝑙𝑛: 𝑓(𝑥) =
𝑎0
𝑛
𝑛
1
𝑎𝑜 =
1
𝜋
𝑤ℎ𝑒𝑟𝑒 𝑎 =
𝑜
𝑛=1
𝑎𝑛 =
𝑎𝑜 =
(𝑴𝒂𝒚 − 𝟏𝟐)
∞
+ ∑(𝑎 cos 𝑛𝑥 + 𝑏 𝑠𝑖𝑛𝑛𝑥) ,
2
1
∫
𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥, 𝑏𝑛 =
0
2𝜋
𝜋0
[π2x +
𝜋
2𝜋
∫
𝑓(𝑥) 𝑑𝑥 =
x3
3
− 2π (
x2
2
1
∫
2
𝜋
𝜋0
2𝜋
)]
0
𝑛𝜋𝑥
1
∫
2𝜋
𝑓(𝑥) 𝑑𝑥
𝜋0
1 2𝜋
∫ 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥
𝜋 0
(π − x)2 𝑑𝑥 =
1
∫
𝜋0
2𝜋
(π2 + x 2 − 2πx) 𝑑𝑥
1
π3
1 6π3 + π3 − 12π3
= [2π3 +
− π(4π2)] = [
]
𝜋
3
𝜋
3
1
𝑎𝑜 =
𝜋
5π3
[−
3
]=−
5π2
3
𝑎𝑛 =
1 ∫ 2𝜋 𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥 = 1 ∫ 2𝜋 (π − x)2 cos 𝑛𝑥 𝑑𝑥
𝜋0
𝜋0
1
𝑎𝑛 =
𝜋
[(π − x)2
1
𝑎𝑛 =
𝜋
[(0 +
𝑏𝑛 =
1
𝑏𝑛 =
sin 𝑛𝑥
𝑛
sin 𝑛𝑥 2𝜋
+ 2(x − π) cos 𝑛𝑥 − 2
]
𝑛2
𝑛3 0
1 4𝜋
4
2𝜋
2𝜋
+
0)
−
(0
−
+
0)]
⇒
𝑎
=
[
]
=
𝑛
𝑛2
𝑛2
𝜋 𝑛2
𝑛2
1 ∫ 2𝜋 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥 = 1 ∫ 2𝜋 (π − x)2 sin 𝑛𝑥 𝑑𝑥
𝜋0
𝜋0
[(π − x)2
−cos 𝑛𝑥
𝜋
𝑛
+ 2(x − π) sin 𝑛𝑥 + 2 cos 𝑛𝑥]
𝑛2
𝑛3 0
2𝜋
2
1
𝜋2
2
𝜋2
𝑏𝑛 = [(− + 0 + 3) − (− + 0 + 3)] = 0
𝜋
𝑛
𝑛
𝑛
𝑛
𝑓(𝑥) =
𝑎0
∞
+ ∑(𝑎 cos 𝑛𝑥 + 𝑏 𝑠𝑖𝑛𝑛𝑥) = −
𝑛
2
5π2
𝑛
6
𝑛=1
∞
−5π2
𝑓(𝑥) =
8) 𝐅𝐢𝐧𝐝 𝐭𝐡𝐞 𝐅𝐨𝐮𝐫𝐢𝐞𝐫 𝐬𝐞𝐫𝐢𝐞𝐬 𝐟(𝐱) = {
𝑬𝒗𝒍𝒖𝒂𝒕𝒆 𝒕𝒉𝒂𝒕
0,
𝟏
𝟏. 𝟑
𝑛=1
4
cos 𝑛𝑥 + (0 )sin 𝑛𝑥)
𝑛2
1
+4∑
6
∞
+ ∑(
𝑛=1
n2
cos 𝑛𝑥
𝟎, −𝝅 ≤ 𝒙 ≤ 𝟎
&
𝒔𝒊𝒏 𝒙, 𝟎 ≤ 𝒙 ≤ 𝝅
+
𝟏
𝟑. 𝟓
+
−𝜋 ≤ 𝑥 ≤ 0
𝟏
+ ⋯ . =?
𝟓. 𝟕
(𝑫𝒆𝒄 − 𝟐𝟎𝟏𝟏)
𝜑1(𝑥), 𝜋 ≤ 𝑥 ≤ 0
𝑠𝑖𝑛 𝑥,
0≤𝑥≤ 𝜋
𝜑2(𝑥), 0 ≤ 𝑥 ≤ 𝜋
𝜑 (𝑥)
𝑊ℎ𝑒𝑟𝑒 𝜑 (𝑥) = 0 𝜑 (𝑥) = sin 𝑥 , 𝐻𝑒𝑟𝑒 𝜑 (−𝑥) = 0 ≠ { 2
1
2
1
−𝜑2(𝑥)
𝑆𝑜𝑙𝑛: f(x) = {
,
𝑙𝑒𝑡 𝑓(𝑥) = {
∴ 𝑓(𝑥)𝑖𝑠 𝑛𝑒𝑖𝑡ℎ𝑒𝑟 𝑒𝑣𝑒𝑛 𝑛𝑜𝑟 𝑜𝑑𝑑 𝑖𝑛 [−𝜋, 𝜋]
𝑓(𝑥) =
𝑎0
2
∞
+ ∑(𝑎 cos 𝑛𝑥 + 𝑏 𝑠𝑖𝑛𝑛𝑥) ,
𝑛
𝑛
𝑛=1
1 𝜋
𝑤ℎ𝑒𝑟𝑒 𝑎 = ∫ 𝑓(𝑥) 𝑑𝑥
𝑜
𝜋 −𝜋
1 𝜋
1 𝜋
∫ 𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥, 𝑏𝑛 = ∫ 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥
𝜋 −𝜋
𝜋 −𝜋
1
1
𝜋
𝜋
𝑎𝑜 = 1 ∫ 𝑓(𝑥) 𝑑𝑥 = 1 ∫ sin x 𝑑𝑥 = [− cos 𝑥]𝜋 = − [cos 𝜋 − cos 0] = − 1 (−2) = 2
0
𝜋 −𝜋
𝜋
𝜋
𝜋
𝜋
𝜋0
𝑎𝑛 =
𝑎𝑛 =
1 𝜋
1 𝜋
1 𝜋
∫ 𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥 = ∫ sin x cos 𝑛𝑥 𝑑𝑥 =
∫ (sin(n + 1)x − sin(n − 1) x) 𝑑𝑥
𝜋 −𝜋
𝜋 0
2𝜋 0
cos(𝑛 + 1)𝑥
1
𝑎𝑛 =
2𝜋
[−
+
𝑛+1
𝑎𝑛 =
1
cos(𝑛 − 1)𝑥
𝑛−1
[(−1)𝑛 {(
2𝜋
𝜋
] =
0
1
𝑛+1
𝑎𝑛 =
1
[−
2𝜋
−
2
𝑛2 − 1
1
2𝜋
[(
cos 𝑛𝜋
𝑛+1
−
cos 𝑛𝜋
1
1
)+(
−
)]
𝑛−1
𝑛+1 𝑛−1
1
1
𝑛−1
)} + (
𝑛+1
−
1
𝑛−1
)]
[1 + (−1)𝑛]] , 𝑛 ≠ 1
0, 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
−2
𝑎𝑛 = {
, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
𝜋(𝑛2 − 1)
𝑎1 =
𝑏𝑛 =
1
cos 2𝑥 𝜋
1
1 𝜋
1 𝜋
]
=
−
∫ sin 𝑥 𝑐𝑜𝑠 𝑥 𝑑𝑥 =
∫ sin 2𝑥 𝑑𝑥 =
[−
2
4𝜋 (1 − 1) = 0
𝜋 0
2𝜋 0
2𝜋
0
1 𝜋
1 𝜋
1 𝜋
∫ 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥 = ∫ sin x sin 𝑛𝑥 𝑑𝑥 =
∫ (cos(n − 1)x − cos(n + 1) x) 𝑑𝑥
𝜋 −𝜋
𝜋 0
2𝜋 0
1 sin(𝑛 − 1)𝑥
𝑏𝑛 =
[
2𝜋
𝑛−1
−
sin(𝑛 + 1)𝑥
𝑛−1
𝜋
] =
0
1
2𝜋
[(0 − 0) + (0 − 0)]
1
𝑏𝑛 =
𝑏1 =
1
2𝜋
[0] = 0, 𝑛 ≠ 1
𝜋
1 𝜋 (1 − cos 2 𝑥)
1
𝑠𝑖𝑛2𝑥 𝜋
∫ sin2 𝑥 𝑑𝑥 =
𝑑𝑥 =
∫
[𝑥 −
]
𝜋0
𝜋 0
2
2𝜋
2 0
𝜋
1
∫ sin 𝑥 𝑠𝑖𝑛 𝑥 𝑑𝑥 =
𝜋0
1
𝑏1 =
𝑓(𝑥) =
𝑎0
2
1
2𝜋
[(𝜋 − 0) − (0 − 0)] =
∞
+ ((𝑎 cos 𝑥 + 𝑏
1
1
1
𝑓(𝑥) = + (((0) cos 𝑥 + sin 𝑥) +
𝜋
2
1
∞
sin 𝑥) + ∑
(𝑎 cos 𝑛𝑥) + ∑ 𝑏 𝑠𝑖𝑛𝑛𝑥)
𝑛
𝑛=𝑒𝑣𝑒𝑛
1
1
2
∞
2
𝑓(𝑥) = + sin 𝑥 −
∑
𝜋 2
𝜋
𝑛=𝑒𝑣𝑒𝑛
∞
∑
𝑛=𝑒𝑣𝑒𝑛
𝑛
𝑛=2
−2
𝜋(𝑛2 − 1)
cos 𝑛𝑥 + (0)𝑠𝑖𝑛𝑛𝑥)
∞
cos 𝑛𝑥
cos 2𝑛𝑥
1 1
2
⇒ 𝑓(𝑥) = + sin 𝑥 − ∑
2
(𝑛 − 1)
((2𝑛)2 − 1)
𝜋 2
𝜋
𝑛=1
𝑃𝑢𝑡 𝑛 = 2𝑛
Put x=0 is a point of continuous,
𝑓(0) = sin 0 = 0,
⇒ 0=
1
1
∞
2
cos 2𝑛(0)
+ sin(0) −
∑
𝜋 2
𝜋
𝑛=𝑒𝑣𝑒𝑛
∞
∞
𝑛=𝑒𝑣𝑒𝑛
𝑛=𝑒𝑣𝑒𝑛
−1 −2
1
1
⇒ = ∑
=
∑
(𝑛
+
1)(𝑛
−
1)
2
𝜋
𝜋
(𝑛2 − 1)
1
(𝑛 + 1)(𝑛 − 1)
−𝟏 + 𝒙, − < 𝑥 < 0
𝟏 + 𝐱, 𝟎 < 𝑥 < .
9) Find the Fourier series of 𝒇(𝒙) = {

𝟏
𝟏
Hence deduce that 𝟏 − 𝟑 + 𝟓 − ⋯ … = 𝟒 (May’07,05,2011)
1 x,in    x  0
Sol: Given f(-x) = 
1  x,in0   x 
f(-x) = 

1  x,in  x  0
1  x,in    x  0
f(-x) = 



1  x,in0  x   ie.,
1 x,in0  x  
=>f(-x)
1 𝜋= - f(x) => f(x) is an odd function . Therefore
1 𝜋
𝑎= ∫
𝑓(𝑥)𝑑𝑥 = 0&𝑎
= ∫ 𝑓(𝑥)𝑐𝑜𝑠𝑛𝑥𝑑𝑥 = 0
0
𝑛
𝜋 −𝜋
𝜋 −𝜋
Let the required series be 𝑓(𝑥) = ∑∞𝑛=1 𝑏𝑛𝑠𝑖𝑛𝑛𝑥 where
2 𝜋
𝑏𝑛 = ∫ 𝑓(𝑥)𝑠𝑖𝑛𝑛𝑥𝑑𝑥
𝜋 0
=
=
=
2
𝜋
[−(𝜋 + 1)
(−1)𝑛
𝑛
𝜋
𝜋
𝜋
0
−𝑐𝑜𝑠𝑛𝑥
−𝑠𝑖𝑛𝑛𝑥 𝜋
{(𝑥 + 1) [
] − (1) [ (𝑛)2 ]}
𝜋
𝑛
0
1
+ ] {cosn𝜋 = (−1)n1 , cos 0 = 1, sin n𝜋 = 0 , , sin 0 = 0 }
𝑛
[1−(𝜋+1)(−1) n]
2
∫ (𝑥 + 1)𝑠𝑖𝑛𝑛𝑥𝑑𝑥
2
𝑏𝑛 =
Therefore 𝑓(𝑥) = ∑∞
𝜋
2
𝑛=2
2
𝑛𝜋
[1 − (𝜋 + 1)(−1)n]
𝑠𝑖𝑛𝑛𝑥 -------- (1)
𝑛
𝜋
Put 𝑥 = 2 {𝑥 = 2 𝑖𝑠 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦}
∴ [𝑓(𝑥)𝑎𝑡 (𝑥 =
∴ (1) =>
𝜋
2 ∞
1 + = ∑𝑛=1
2
(
(
2+𝜋
2
2+𝜋
2
[1−(𝜋+1)(−1)n]
𝜋
2
[1−(𝜋+1)(−1)n]
𝜋
𝑛
∞
)= ∑𝑛=1
𝑛
𝑠𝑖𝑛𝑛
𝜋
2
) = 2 [2+𝜋 − 2+𝜋 + 2+𝜋 − 2+𝜋 + ⋯ ]
𝜋 1
3
5
7
𝑠𝑖𝑛𝑛
𝜋
2
𝜋
𝜋
)] = 1 +
2
2
(
2+𝜋
2
)=
1
1
1
3
5
7
2(2+𝜋) 1
𝜋
[ − + − +⋯]
1
𝜋
1
1
4
1
7
1
1
( ) = [ − 3+ 5− + ⋯ ]
𝟏
𝟏
𝟏
10) 𝑭𝒊𝒏𝒅 𝑭. 𝑺 𝒐𝒇 𝒇(𝒙) = 𝒙(𝝅 − 𝒙) 𝒊𝒏 (𝟎, 𝟐𝝅)& 𝐷𝑒𝑑𝑢𝑐𝑒 + + … . =
𝟏𝟐
𝑆𝑜𝑙𝑛: 𝑓(𝑥) =
𝑎0
∞
+ ∑(𝑎 cos 𝑛𝑥 + 𝑏 𝑠𝑖𝑛𝑛𝑥) ,
𝑛
2
𝑛
1
𝑜
𝑎𝑜 =
1
∫
2𝜋
∫
𝜋
2𝜋
𝜋0
𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥, 𝑏𝑛 =
0
𝑓(𝑥) 𝑑𝑥 =
2
𝜋
𝑥3
𝑎𝑜 = [𝜋 ( ) − ]
𝜋
2
3
1
𝑥2
𝑎𝑛 =
𝑎𝑛 =
𝜋
𝜋
𝜋0
2
1 2𝜋
∫ 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥
𝜋 0
𝑥(π − x) 𝑑𝑥 =
1
∫
2𝜋
𝜋0
(𝑥π − x 2 ) 𝑑𝑥
1 4𝜋3
8𝜋3
1 6𝜋3 − 8𝜋3
2𝜋3
2𝜋2
[
−
]= [
]=−
=−
𝜋 2
3
𝜋
3
3𝜋
3
1 −4𝜋
−4
−3𝜋
𝜋
+
0)
−
(0
+
+
0)]
⇒
𝑎
=
[
]
=
𝑛
𝑛2
𝑛2
𝜋 𝑛2
𝑛2
1 ∫ 2𝜋 𝑓(𝑥) sin 𝑛𝑥 𝑑𝑥 = 1 ∫ 2𝜋 (xπ − x 2) sin 𝑛𝑥 𝑑𝑥
𝜋0
𝜋0
1
2𝜋
[(xπ − x2) −cos 𝑛𝑥 + (π − 2x) sin 𝑛𝑥 − 2 cos 𝑛𝑥]
𝑛2
𝑛3 0
𝜋
𝑛
2π2
[{(
n
2
2
)+0−
𝑛
3} − (0 + 0 −
𝑏𝑛 =
𝑎0
𝑓(𝑥) 𝑑𝑥
𝜋0
2𝜋
[(xπ − x2) sin 𝑛𝑥 + (π − 2x) cos 𝑛𝑥 + 2 sin 𝑛𝑥]
𝑛2
𝑛3 0
𝜋
𝑛
[(0 +
1
𝑏𝑛 =
2
𝜋
2𝜋
1
𝑏𝑛 =
𝑏𝑛 =
0
∫
∫
(𝑨𝒑𝒓 − 𝟏𝟏)
𝟖
1 ∫ 2𝜋 𝑓(𝑥) cos 𝑛𝑥 𝑑𝑥 = 1 ∫ 2𝜋 (xπ − x 2 ) cos 𝑛𝑥 𝑑𝑥
𝜋0
𝜋0
1
𝑎𝑛 =
=
1
𝟑𝟐
1
𝑤ℎ𝑒𝑟𝑒 𝑎 =
𝑛=1
𝑎𝑛 =
𝑓(𝑥) =
𝟐𝟐
𝝅𝟐
∞
𝑛=1
𝑛
𝜋
[(
2
2
2π2
) − 3 + 3]
n
𝑛
𝑛
1 2π2
2𝜋
[
]=
𝜋 n
𝑛
+ ∑(𝑎 cos 𝑛𝑥 + 𝑏 𝑠𝑖𝑛𝑛𝑥) = −
𝑛
𝑛
1
3)] =
𝜋2
3
∞
+ ∑(
𝑛=1
−4
cos 𝑛𝑥 +
𝑛2
𝑃𝑢𝑡 𝑥 = 0 𝑖𝑠 𝑎 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑑𝑖𝑠𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑠𝑜 𝑓(𝑥) =
2𝜋
sin 𝑛𝑥)
𝑛
𝑓(0) + 𝑓(2𝜋)
2
𝑓(𝑥) = 𝑥(𝜋 − 𝑥),
𝑓(0) = 0, 𝑓(2𝜋) = 2𝜋(𝜋 − 2𝜋) = 2𝜋(−𝜋) = −2𝜋2
𝑓(𝑥) =
−𝜋2
𝑓(𝑥) =
−𝜋2
=
3
𝑓(0) + 𝑓(2𝜋) 0 + (−2𝜋2)
=
= −𝜋2
2
2
∞
+ ∑(
−4
𝑛=1
𝑛2
cos 𝑛(0) +
∞
−2𝜋2
3
2𝜋
= −4 ∑
𝑛=1
1
𝑛
𝑛
𝜋2
2⇒
6
=
sin 𝑛(0)) ⇒ −𝜋2 +
∞
𝜋2
3
= −4 ∑
𝑛=1
1
𝑛2
1
1
1
+
+
+ ⋯ ..
12 22 32
11) Obtain the Fourier series of f(x) = 1-x2 over the interval (-1, 1).(NOV 05, Dec-10)
The given function is even, as f(-x) = f(x). Also period of f(x) is 1-(-1) =2
Solution:
1
1
1
3
Here a0 = 1  f (x)dx = 2 f (x)dx = 2 (1  x2 )dx  2x  x   4
1 1
3 0 3
0
0

1

an 

11
 f (x) cos(nx)dx
1 1
1
 2 f (x) cos(nx)dx
as f(x) cos(nx) is even
0
1
= 2(1  x ) cos(nx)dx
2
0


 sin nx 


 2
Integrating parts, a  21 x 

n
 n

bn 
11
1 1


1
  sin nx  4(1)n 1
  cos nx 
  (2)
 =
  (2x)

2
3
 (n ) 
 (n )




0
n2 2



f (x)sin( nx)dx =0, since f(x) sin (nx) is odd.
The Fourier series of f(x) is
f(x) =
2
3

4  (1)n1

2 n1 n 2
cos(nx)

12) 𝐎𝐛𝐭𝐚𝐢𝐧 𝐭𝐡𝐞 𝐡𝐚𝐥𝐟 − 𝐫𝐚𝐧𝐠𝐞 𝐬𝐢𝐧𝐞 𝐬𝐞𝐫𝐢𝐞𝐬 𝐟𝐨𝐫 𝐟(𝐱) =
Soln: 𝑓(𝑥) = ∑∞
𝑏 sin
𝑛=1 𝑛
𝑛𝜋𝑥
𝑙
… … (1), 𝑊ℎ𝑒𝑟𝑒 𝑏
2
𝑛
𝑙
l
x,0  x 

2

(𝑨𝒑𝒓`𝟏𝟏, 𝒎𝒂𝒚`𝟎𝟓)
l
(l  x),
xl

2
= ∫ 𝑓(𝑥) sin
𝑙 0
𝑛𝜋𝑥
𝑙
𝑑𝑥
𝑏𝑛 =
2
𝑙
𝑙
2
[∫ 𝑥 sin
0
𝑙
𝑛𝜋𝑥
𝑛𝜋𝑥
𝑑𝑥 + ∫ (𝑙 − 𝑥) sin
𝑑𝑥]
𝑙
𝑙
𝑙
2
𝑙
2
2
𝑏𝑛 = 2 [𝑥 ( 𝑙 )(− cos 𝑛𝜋𝑥 ) − ( 2𝑙 ) (− sin 𝑛𝜋𝑥)]
𝑙
2
𝑙
𝑛𝜋
𝑙
𝑛𝜋
0
𝑙
𝑛𝜋𝑥
2
𝑙
𝑛𝜋𝑥
𝑙2
+ [(𝑙 − 𝑥) (
)]
) + ( 2 2) (− sin
) (− cos
𝑙
𝑛𝜋
𝑙
𝑙
𝑛𝜋
𝑙
2
2
𝑙2
𝑛𝜋
2
𝑛𝜋
𝑙2
𝑏𝑛 = [(−0 + ( 2 2) (sin
)) − (−0 + 0)] + [(−0 + 0) − (−0 − ( 2 2 ) (sin ))]
𝑙
𝑛 𝜋
2
2
𝑙
𝑛 𝜋
2
𝑙2
𝑛𝜋
4𝑙
𝑛𝜋
𝑏𝑛 = [2 ( 2 2) (sin )] = 2 2 sin
𝑙
𝑛 𝜋
2
𝑛 𝜋
2
∞
𝑛𝜋𝑥
𝑓(𝑥) = ∑ 𝑏𝑛 sin
𝑙
𝑛=1
∞
=∑
𝑛𝜋
4𝑙
𝑛 2𝜋
2 sin
𝑛=1
2
sin
𝑛𝜋𝑥
𝑙
4𝑙
=
𝜋
∞
2∑(
𝑛=1
𝑛𝜋
𝑛𝜋𝑥
) sin
sin
𝑛2
2
𝑙
1
13) Obtain the Fourier expansion of
x,0  x  
f (x)  
2  x,  x  2 May-2004, Dec-10
Deduce that
2
8
1 
1 1
  ......
32 52
Solution:
The graph of f(x) is shown below.
Here OA represents the line f(x)=x, AB represents the line f(x)=(2-x) and AC represents the line x=.
Note that the graph is symmetrical about the line AC, which in turn is parallel to y-axis. Hence the
function f(x) is an even function.
Here,
a0 =
1

2 
f (x)dx =  f (x)dx
 

0

2 
 
0
an 
xdx  
Since f(x) cosnx is even.

2 
f (x) cos nxdx   f (x) cos nxdx
1
  

0
2 
 x cos nxdx
=
0

2   sin nx    cos nx 
x
n   1 n 2 

0

 

=
2
 2 (1) n  1
n 



1 
Also, bn  



 f (x)sin nxdx  0 ,
since f(x)sinnx is odd


 1
(1)n 1 cos nx
Thus the Fourier series of f(x) is f (x)    2 

2  n 1 n2


 2  1
For x=, we get f ( )    
 2 (1) n  1 cos n
 n 1 n
2


or
Thus,  
2

8



n 1
1
2


2   2cos(2n 1)


 
n1
(2n 1)2
 2  1  1 1  ......
 2
2
2
(2n 1) 8
3
5
14) 𝐎𝐛𝐭𝐚𝐢𝐧 𝐭𝐡𝐞 𝐡𝐚𝐥𝐟 − 𝐫𝐚𝐧𝐠𝐞 𝐜𝐨𝐬𝐢𝐧𝐞 𝐬𝐞𝐫𝐢𝐞𝐬 𝐟𝐨𝐫 𝐟(𝐱) = 𝒙 𝒊𝒏 (𝟎, 𝝅)
Soln: 𝐿𝑒𝑡 𝑡ℎ𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 ℎ𝑎𝑙𝑓 𝑟𝑎𝑛𝑔𝑒 𝑐𝑜𝑠𝑖𝑛𝑒 𝑠𝑒𝑟𝑖𝑒𝑠 𝑏𝑒 𝑓(𝑥) =
𝑎0
+ ∑∞
2
𝑎 cos 𝑛𝑥 ............. (1)
𝑛=1 𝑛
𝜋
2
2
𝑎0 = 2 ∫ 𝜋𝑓(𝑥) 𝑑𝑥 = 2 ∫ 𝜋𝑥 𝑑𝑥 = 2 [𝑥 ] = 2 [𝜋 ] = 𝜋,
𝜋
𝜋
𝜋2
𝜋20
0
0
𝑎𝑛 =
𝑎𝑛 =
𝜋
2 𝑥 sin 𝑛𝑥 + cos 𝑛𝑥
2 𝜋
2 𝜋
]
∫ 𝑓(𝑥)𝑐𝑜𝑠𝑛𝑥 𝑑𝑥 = ∫ 𝑥𝑐𝑜𝑠𝑛𝑥 𝑑𝑥 = [
𝑛
𝑛2
𝜋 0
𝜋 0
𝜋
0
0, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
𝑛
1
2
[(−1)𝑛 − 1] ⇒ 𝑎𝑛 = {
4
[(0 + (−1) ) − (0 + )] =
−
𝜋
𝑛2
𝜋𝑛2
𝑛2
, 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑
𝜋𝑛 2
2
∞
∞
𝜋
4
1
𝜋 4
𝑓(𝑥) = + ∑ −
cos
𝑛𝑥
=
−
∑
cos 𝑛𝑥
2
𝜋𝑛2
2 𝜋
𝑛2
𝑛=𝑜𝑑𝑑
𝑛=𝑜𝑑𝑑
15) Expand f(x) = x (-x) as half-range sine series over the interval (0,)
(JUN 07,09,Ap-10)

We have,

bn 

2
 0

f (x) sin nxdx 
2
 0
2
(x  x ) sin nxdx

2 
Integrating by parts, we get
bn 

2

  cos nx 

x  x 
 

The sine series of f(x) is f ( x) 
4
n



1
 1  (1) n sin nx

 n 1 n3
16)
Obtain the Fourier expansion of f(x) =e-ax in the interval (-,).
Dec-2008,Apr-2010
a0 
1

e
 
dx 
 ax
1

 e ax  


   a 
ea  e a  2sinh a

a
a
 
1 e ax cos nxdx
an  

 


1  e ax 
 a cos nx  n sin nx 
 a2  n2

n
2a (1) sinh a 


2
2
 

 a n



1  ax

1  eax 
bn =
e sin nxdx =  2 2  a sin nx  n cos nx 
 a  n
 

an 


 cos nx  
4
  sin nx 
n


    2x  n2   (2) 3   n3 1  (1)



 n  0




17)
2n (1)n sinh a 
=
  a2  n2 

sinh a 2a sinh a  (1)n
2
n(1)n
f(x) =
cos nx  sinh a  2
sin nx

a 

2
2
2

n1 a  n
n1 a  n
Find the fourier series of periodicity 3 for f(x) = 2x-x2 in 0<x<3. Apr-10,May’08
3
Sol : Here 2l = 3 =>𝑙 = Let the Fourierseries be
f (x) 
a0
2
2

 
a
n
cos
2nx
3
n1
2


b
n
sin
n1
3
2nx
3
2nx
23
Where 𝑎0 = 3 ∫0 𝑓(𝑥)𝑑𝑥 , 𝑎𝑛 = 3 ∫0 𝑓(𝑥)𝑐𝑜𝑠
2
2nx
3
𝑑𝑥 and 𝑏𝑛 = 3 ∫0 𝑓(𝑥)𝑠𝑖𝑛
3
𝑑𝑥
3
2 3
𝑎0 = ∫ (2𝑥 − 𝑥2)𝑑𝑥
30
2
𝑥3 3
3
3 0
= [{𝑥2 −
} ]
1
= [9 − 9] =>𝑎 = 0
0
𝜋
2
2
3
𝑎 = ∫ (2𝑥 − 𝑥2)𝑐𝑜𝑠
𝑛
3 0
2nx
2nx
3
𝑎𝑛 = ∫ 𝑓(𝑥)𝑐𝑜𝑠
3 0
3
𝑑𝑥
𝑑𝑥 By Bernoullis Theorem ∫ 𝑢𝑑𝑣 = 𝑢𝑣 − 𝑢′𝑣 + 𝑢′′𝑣 − ⋯
1
3
2nx
𝑠𝑖𝑛 3
− (2 − 2𝑥)
2
= (2𝑥 − 𝑥2)
3
2nx
−𝑐𝑜𝑠
3
2𝑛𝜋
[
(
]
+ (−2)
2𝑛𝜋 2
3
{
−𝑠𝑖𝑛
[
3
)
]
2
2nx
3
2𝑛𝜋 3
(3)
2 −9
−9
)−(
= [(
)] {cos 2n𝜋 = 1 , sin2n𝜋 = 0, cos 0 = 1 . sin0 = 0}
3 𝑛2𝜋 2
2𝑛2𝜋 2
−9
2
1
) (1 + )
)
(
2
𝑛2𝜋2 3
−9
2 3
𝑎𝑛 = ( 2 2)( ) ( )
3 2
𝑛𝜋
−9
𝑎𝑛 = ( 2 2 )
𝑛 𝜋
=(
2
𝑏𝑛 =
3
3
∫ 𝑓(𝑥)𝑠𝑖𝑛
0
2nx
3
𝑑𝑥
3
}0
2
3
𝑏 = ∫ (2𝑥 − 𝑥2)𝑠𝑖𝑛
𝑛
2nx
3 0
𝑑𝑥 By Bernoullis Theorem ∫ 𝑢𝑑𝑣 = 𝑢𝑣 − 𝑢′𝑣 + 𝑢′′𝑣 − ⋯
1
3
2nx
−𝑐𝑜𝑠 3
− (2 − 2𝑥)
2
= (2𝑥 − 𝑥2)
3
2nx
−𝑠𝑖𝑛
3
(
3
{
[
]
𝑐𝑜𝑠
+ (−2)
2𝑛𝜋 2
2𝑛𝜋
[
3
2
)
2nx
3
2𝑛𝜋 3
(3)
]
3
}0
2
9
27
)−(
= [(
) (1 − 1)] {cos 2n𝜋 = 1 , sin2n𝜋 = 0, cos 0 = 1 . sin0 = 0}
3 2𝑛𝜋
4𝑛3𝜋 3
𝑏 =
𝑛
3
9
Therefore 𝑓(𝑥) = − ∑∞
𝜋2
𝑛𝜋
1
2𝑛𝜋𝑥
3
1
2𝑛𝜋𝑥
𝑐𝑜𝑠
+ ∑∞𝑛=1 𝑠𝑖𝑛
𝑛=1 𝑛2
3
𝜋
𝑛
3
18) Obtain the Complex form of Fourier series of the function f(x) = e-x in the interval
0<x<2𝝅(A/M 10) (Dec-09)
Ans : The complex form of Fourier series of the function is given by
𝑖𝑛𝑥
f(x) = ∑𝑛=∞
𝑛=−∞ 𝐶𝑛 𝑒
1
where𝐶 = ∫
𝑛
2𝜋
1
𝑓(𝑥)𝑒−𝑖𝑛𝑥𝑑𝑥 = ∫
2𝜋 0
2𝜋 −𝑥 −𝑖𝑛𝑥
𝑒 𝑒 𝑑𝑥
2𝜋 0
1
2𝜋
−(1+𝑖𝑛)𝑥
= 2𝜋 ∫0 𝑒
𝑑𝑥 =
1
2𝜋
{[−
−2𝜋(𝑒−𝑖2𝑛𝜋)+1 2𝜋
1
= {− 𝑒
2𝜋
} =
1+𝑖𝑛
0
𝑒−(1+𝑖𝑛)𝑥
1+𝑖𝑛
2𝜋
]}
0
1 1−𝑒−2𝜋(1)
2
𝜋
(
1+𝑖𝑛
)
Where 𝑒−𝑖2𝑛𝜋 = 𝑐𝑜𝑠2𝑛𝜋 − 𝑖 𝑠𝑖𝑛2𝑛𝜋 = 1
𝐶𝑛 = (
(1 − 𝑒−2𝜋)(1 − 𝑖𝑛)
)
2𝜋(1 + 𝑛2 )
The required complex form of the series is ( )
𝑓𝑥=(
19) Obtain the Fourier expansion of
  ,  x  0
f(x) = 
x,0  x  
Deduce that
2
8
 1
1
2
3
a0 

1
52
 ......
(Nov-09)


10

  dx   xdx  
 
2
0

1−𝑒−2𝜋
1−𝑖𝑛
2𝜋
1+𝑛2
(
) ∑𝑛=∞
𝑛=−∞
) 𝑒𝑖𝑛𝑥


an 




10


cos
nxdx

x
cos
nxdx



 
0

1
n 
2
(1)

1
n


10
bn      sin nxdx   x sin nxdx
 
0

1

1  2(1)n
n

Fourier series is

(1)

 n
f(x) =    1
4


1
n
2
n1



1 cos nx  
 1 2(1)n sin nx
n
n1

Note that the point x=0 is a point of discontinuity of f(x). Here f(x+) =0, f(x-) =- at x=0. Hence
1
1

[ f (x  )  f (x  )]  0    

2
2
2
The Fourier expansion of f(x) at x=0 becomes
    1  1 [(1)n  1]

 
2 2 4  n 1 n 2

1
n
or
  2 [(1)  1]
4
n 1 n
Simplifying we get,
2
8
1
1 1
  ......
32 52
20) Find the complex form of the Fourirer Series of f(x)= e –x in -1< x <1.(Nov 2009)
Solution:
The complex form of the Fourirer Series of f(x)= e –x in -1< x <1, is given by
f(x) =∑∞𝑛=−∞ 𝐶𝑛 𝑒𝑖𝑛𝑥𝜋
C =
n
1 1
1
𝑒−𝑥 𝑒−𝑖𝑥𝜋𝑛 𝑑𝑥
1 −(1+𝑖𝑛𝜋)𝑥
𝑒
1
1 𝑒−(1+𝑖𝑛𝜋)𝑥
=2 [
𝑑𝑥
=2∫−1
∫
2 −1
−(1+𝑖𝑛𝜋) ]at (-1 to 1)
=−2(1+𝑖𝑛𝜋) [𝑒
−1
(1+𝑖𝑛𝜋)𝑥
−(1+𝑖𝑛𝜋)
−𝑒
= 2(1+𝑖𝑛𝜋) [𝑒−1(𝑐𝑜𝑠𝑛𝜋 − 𝑖 sin 𝑛𝜋) − 𝑒1(𝑐𝑜𝑠𝑛𝜋 + 𝑖 sin 𝑛𝜋)]
cos 𝑛𝜋
=−2 (𝑖+𝑛𝜋) [𝑒
−1
− 𝑒1 ] =
cos 𝑛𝜋
2 (𝑖+𝑛𝜋)
2 sin h 1
]
=
(−1)𝑛 sin h1
1+𝑖𝑛𝜋
Sub in f(x)
f(x)= ∑∞
−∞
(−1)𝑛 sin h1 𝑖𝑛𝑥𝜋
𝑒
1+𝑖𝑛𝜋
= sinh 1 ∑∞
−∞
(−1)𝑛 (1−inπ) 𝑖𝑛𝑥𝜋
𝑒
1+𝑛2 𝜋2
21) Expand f(x)= x-x2 in –L <X<L and using this series find the root mean square values of f(x)
in the interval. (Nov-09)
The fourirer series of f(x) in ( -L , L) is given by f (x) 
a0

 a

2 n1

n
cos nx / L   bn sin nx / L
n1
………(1)
1 𝐿
𝑤ℎ𝑒𝑟𝑒 𝑎0 = ∫ 𝑓(𝑥)𝑑𝑥
𝐿 −𝐿
1
𝐿
= ∫ (𝑥 − 𝑥2)𝑑𝑥
𝐿 −𝐿
1 𝑥2 𝑥 3 𝐿
= (2 − )
3 −𝐿
𝐿
1
3
2
3
2
= (𝐿 − 𝐿 − 𝐿 − 𝐿 )
2
3
2
3
𝐿
=−
2𝐿2
𝑎0 = −
3
2𝐿3
3𝐿
… … … … … … … … … … … … … … … . (2)
𝑛𝜋𝑥
1 𝐿
)𝑑𝑥
𝑊ℎ𝑒𝑟𝑒 𝑎𝑛 = ∫ 𝑓(𝑥)cos(
𝐿 −𝐿
𝐿
𝑎𝑛 =
𝑎 = 1 [(𝑥 − 𝑥 2 ) (
𝑛
𝐿
𝑛𝜋𝑥
sin ( 𝐿 )
𝑛𝜋
𝐿
1
=
𝐿
[
1
𝐿
𝐿
∫ (𝑥 − 𝑥2)cos(
−𝐿
) − (1 − 2𝑥) (
(1 − 2𝐿)𝑐𝑜𝑠𝑛𝜋
𝑛2𝜋 2
𝐿2
𝑛𝜋𝑥
𝐿
−cos (
)𝑑𝑥
𝑛𝜋𝑥
𝐿 )
−sin (
) + (−2) (
𝑛 2𝜋 2
𝑛3𝜋 3
𝐿2
𝐿3
(1 + 2𝐿)𝑐𝑜𝑠𝑛𝜋
−
𝑛𝜋𝑥
𝐿 )
𝑛2𝜋 2
𝐿2
]
𝐿
)]
−𝐿
1 𝐿2
=
𝑎𝑛 =
4𝐿2
𝑛2𝜋2
𝐿 𝑛2𝜋2
[−4𝐿𝑐𝑜𝑠𝑛𝜋]
(−1)𝑛+1 … … … … … … … … … … … … … (3)
𝑛𝜋𝑥
1 𝐿
)𝑑𝑥
𝑊ℎ𝑒𝑟𝑒 𝑏𝑛 = ∫ 𝑓(𝑥)sin(
𝐿 −𝐿
𝐿
𝑏𝑛 =
𝑛𝜋𝑥
𝐿 )
−cos (
𝑏 = 1 [(𝑥 − 𝑥 2 ) (
𝑛
𝑛𝜋
𝐿
1
𝐿
𝐿
∫ (𝑥 − 𝑥2)sin(
𝑛𝜋𝑥
) − (1 − 2𝑥) (
𝑛𝜋𝑥
𝐿 )
−sin (
𝑛
1
𝑐𝑜𝑠𝑛𝜋
[(𝐿 − 𝐿2) (−
𝐿
)−
2𝑐𝑜𝑠𝑛𝜋
𝑛𝜋
𝐿
𝑛3 𝜋 3
𝐿2
𝐿3
− (−𝐿 − 𝐿2) (−
𝑐𝑜𝑠𝑛𝜋
𝑛𝜋
𝐿
𝑛3𝜋 3
𝐿3
1
𝑛𝜋𝑥
𝐿 )
cos (
) + (−2) (
𝑛2𝜋 2
𝐿
𝑏=
)𝑑𝑥
𝐿
−𝐿
)+
2𝑐𝑜𝑠𝑛𝜋
𝑛3 𝜋 3
𝐿3
(−2𝐿𝑐𝑜𝑠𝑛𝜋)
=𝐿
𝑛𝜋
𝐿
𝑏𝑛 =
2𝐿
(−1)𝑛+1 … … … … … … … … … . . (4)
𝑛𝜋
Sub (2)(3) and (4) in (1) we get final answer
∞
∞
𝑛𝜋𝑥
𝑛𝜋𝑥
𝐿2
4𝐿2 (−1)𝑛+1
2𝐿
)
(−1)𝑛+1 sin (
𝑓(𝑥) = − + ∑ 2 2
) +∑
cos (
3
𝐿
𝐿
𝑛𝜋
𝑛𝜋
𝑛=1
𝑛=1
22) Obtain the Fourier expansion of
1
f(x) =
  x in - < x < 
2
We have,
1
a0 


11

  f (x)dx   2 (  x)dx



x2  
1 
=
x    
2 
2 
1 

an   f (x) cos nxdx 

 
11
 2
(  x) cos nxdx
𝐿
)]
−𝐿
]



Here we use integration by parts, so that

  cos nx  
1 
sin nx
 (1)
an 
  x

2 
n
 n2  
1
 0  0
2
1 1
bn  

 2(  x)sin nxdx


 sin nx  
1 
  x   cos nx  (1)


2
2 

n
 n
n
(1)

n

Using the values of a0, an and bn in the Fourier expansion
f (x) 
a0
2



 a
n 1

n cos nx   bn sin nx
n 1
We get,

f (x) 

2
(1)n
 
n1
sin nx
n
This is the required Fourier expansion of the given function.
23) Obtain the Fourier expansion of
  ,  x  0
f(x) = 
x,0  x  

Deduce that
2
8
1 
1 1
  ......
32 52
Here,
a0 


10



dx

xdx



  2
 
0



an 
1




10


cos
nxdx

x
cos
nxdx



 
0

n 
2
(1)
n

1


10
bn      sin nxdx   x sin nxdx
 
0

1

1  2(1)n
n

Fourier series is

(1)

 n
f(x) =   1
4

1
2
n1
n



1 cos nx  
 1 2(1)n sin nx
n
n1

Note that the point x=0 is a point of discontinuity of f(x). Here f(x +) =0, f(x-) =- at x=0. Hence
1
[ f (x  )  f (x  )] 
2
1
0      
2
2
The Fourier expansion of f(x) at x=0 becomes
    1  1 [(1)n  1]

 
2 2 4  n 1 n 2

1
n
or

[(1)  1]
2
4 
n 1 n
Simplifying we get,
2
8
1 
1 1
  ......
32 52
24) Obtain the Fourier expansion of
3
 4x
1 3 in  2  x  0
2
1
1

 1 2  2  ......
Deduce
that
f(x) =  4x
8
3
5
1 in0  x  3
2
 3

The period of f(x) is 3    3   3


2  2 
Also
f(-x) = f(x). Hence f(x) is even
1
a 
3 /2

2
f (x)dx 
3/2
f (x)dx
3 / 2 3 / 2
3 / 2 0
4x 
3/2 
4  1  dx  0
3
3

0 
 nx 
an  1 3/ 2
3 / 2  f (x) cos
dx
3 / 2

3 / 2
2nx 
3/2
2 f (x) cos
dx
 3 / 2 
3


0
3/ 2

  2nx  
 2nx  
  cos
 sin 
 
 
32  
4
4x   3     4 

 1     2n      

   3   2n 

3
3  



  3  
  3  0
0




4
=
n
2



1  (1) 
n
2
Also,

b
f (x)sin dx 2 0
1
n
3 3


n x
 3 
 2 
3

2


Thus
4 1

f(x) =

1  (  
n
2
2
n 1 n


 2nx 
1) cos

 3 

Putting x=0, we get


1  (1)n
1


 2 n 1 n 2
8
1 1
1=
......
1  

2

 2  3 52
f(0) =
or
4


Thus,
2  1  1 1
 2  ......
2
8
3
25) Obtain the cosine series of
5



x,0  x  

2
f (x)  

  x, 
 x  

2




over(0, )
Here

 

xdx

(


x)dx



 2
 0


2



2


2
an    x cos nxdx  (  x) cos nxdx
 0



2


a0 
2 

2



Performing integration by parts and simplifying, we get

2

 n 
 2 cos 
1  (1)
 2 
n2 
a
n


n
8

, n  2,6,10,.....
n2
Thus, the Fourier cosine series is
f(x) =   2  cos 2x  cos 6x  cos10x  

4   12
32
52





26) Obtain the half-range cosine series of f(x) = c-x in 0<x<c
Here
2c
a0  (c  x)dx  c
c0
an 
 nx 
(c

x)
cos
 dx
c
c


0
2
c

Integrating by parts and simplifying we get,
an 
1  (1) 
n
2c
n
2


2
The cosine series is given by


c
f(x) =

2

2c 


2
1
n

n 1
2
1  (1) cos nx 
n


 c 
27) Find the first two harmonics of the Fourier series of f(x) given the following table:
x
0
f(x)
1.0

3
2
3

4
3
5
3
2
1.4
1.9
1.7
1.5
1.2
1.0
Note that the values of y = f(x) are spread over the interval 0 x  2 and f (0) = f (2) = Hence the
function is periodic and so we omit the last value f (2) = 0. We prepare the following table to
compute the first two harmonics. (dec 04, 06, 08,Apr-11)
x0
y = f(x) cosx cos2x
sinx
sin2x
ycosx ycos2x
ysinx
ysin2x
0
1.0
1
1
0
0
1
1
0
0
60
1.4
0.5
-0.5
0.866
0.866
0.7
-0.7
1.2124
1.2124
We have


120
1.9
-0.5
-0.5
0.866
-0.866
-0.95
-0.95
1.6454
-1.6454
180
1.7
-1
1
0
0
-1.7
1.7
0
0
240
1.5
-0.5
-0.5
-0.866
0.866
-0.75
-0.75
1.299
1.299
300
1.2
0.5
-0.5
-0.866 -0.866
0.6
-0.6
-1.0392 -1.0392
-1.1
-0.3
3.1176
Total
-0.1732

as
the
length of
interval= 2l
= 2 or
l=
Putting,
n=1,2, we
get
a  2[ y cos x] 
2 y cos x

2(1.1)
 0.367
1
6
6
2 y cos 2x 2(0.3)
a  2[ y cos 2x] 

 0.1
2
6
6
b  [ y sin x] 
2 y sin x
1
b  [ y sin 2x] 
2
 1.0392
6
2 y sin 2x



2 f (x) cos
n



n
bn  2 f (x)sin 


a 
 0.0577
6
The first two harmonics are a1cosx+b1sinx and a2cos2x+b2sin2x. That is (-0.367cosx +
1.0392 sinx) and (-0.1cos2x – 0.0577sin2x)
28) Obtain the constant term and the coefficients of the first sine and cosine terms in the Fourier
expansion of y as given in the following table (Jun-12,Dec-10)
x
0
1
2
3
4
5
y
9
18
24
28
26
20
Soln:
𝐻𝑒𝑟𝑒 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑖𝑠 2𝑙 = 6,
𝑦=
⇒𝑙 =3
∴ 𝑇ℎ𝑒 𝑓𝑜𝑢𝑟𝑖𝑒𝑟 𝑠𝑒𝑟𝑖𝑒𝑠 𝑐𝑎𝑛 𝑏𝑒 𝑟𝑒𝑝𝑒𝑟𝑠𝑒𝑛𝑡𝑒𝑑 𝑏𝑦
𝑎0
𝜋𝑥
𝜋𝑥
2𝜋𝑥
2𝜋𝑥
)) + ⋯.
+ (𝑎1 cos ( ) + 𝑏1 sin ( )) + (𝑎2 cos (
) + 𝑏2 sin ( 𝑙
2
𝑙 𝑎0
𝑙
𝑙
𝐻𝑒𝑟𝑒 𝑦 =
𝜋𝑥
𝜋𝑥 )) + ⋯ … … . (1)
+ (𝑎1 cos ( ) + 𝑏1 sin (
2
3
3
𝜋𝑥
𝜋𝑥
𝜋𝑥
𝜋𝑥
y
cos ( )
sin ( )
𝑦 cos ( )
𝑦 sin ( )
3
3
3
3
9
1
0
9
0
18
0.5
0.866
9
15.588
24
-0.5
0.866
-12
20.785
28
-1
0
-28
0
26
-0.5
-0.866
-13
-22.517
20
0.5
-0.866
10
-17.321
125
-25
-3.465
x
0
1
2
3
4
5
∑
2
𝑎0 =
𝑛
1
(∑ 𝑦) =
3
2
(125) = 41.67,
𝑎1 =
(∑ 𝑦 cos (
𝜋𝑥
1
)) = (−25) = −8.33
3
3
𝑛
𝜋𝑥
1
𝑏1 = (∑ 𝑦 sin ( )) = (−3.465) = −1.16
𝑛
3
3
𝜋𝑥
41.67
𝜋𝑥
𝑦=
+ ((−8.33) cos ( ) + (−1.16) sin ( ))
3
2
3
𝜋𝑥
𝜋𝑥
𝑦 = 20.84 − 8.33 cos ( ) − 1.16 sin ( )
3
3
2
29) The following table gives the variations of a periodic current A over a period T :
t(secs)
0
T/6
T/3
T/2
2T/3
5T/6
T
A (amp)
1.98
1.30
1.05
1.30
-0.88
-0.25
1.98
Show that there is a constant part of 0.75amp. in the current A and obtain the amplitude of the
first harmonic.
(A/M 03, Nov09)
Note that the values of A at t=0 and t=T are the same. Hence A(t) is a periodic function of period
T. Let us denote  2 
 t . We have

 T 


a0  2[A]
a  
 2  
Acos
2
 t  2[ Acos ]
1

 T 

b  
 2  
2 Asin  t  2[ Asin  ]
1
 T  


(1)
We prepare the following table:
t

2 t
T
A
cos
sin
Acos
Asin
0
0
1.98
1
0
1.98
0
T/6
600
1.30
0.5
0.866
0.65
1.1258
T/3
1200
1.05
-0.5
0.866
-0.525
0.9093
T/2
1800
1.30
-1
0
-1.30
0
2T/3
2400
-0.88
-0.5
-0.866
0.44
0.7621
5T/6
3000
-0.25
0.5
-0.866
-0.125
0.2165
1.12
3.0137
Total
4.5
Using the values of the table in (1), we get
a
2 A

4.5
 1.5
0
6
3
2 Acos  1.12  0.3733
a1  
6
3
3.0137
2 Asin  
 1.0046
b1  
6
3
The Fourier expansion upto the first harmonic is
 2t 
 2t 
 a cos
 b sin



  1  T 
1
2
T 2t 
 2t 
 0.75  0.3733cos
 1.0046sin





 T 
 T 
A



a0
The expression shows that A has a constant part 0.75 in it. Also the amplitude of the first harmonic
is
a12  b21 = 1.0717.
30) The displacement y of a part of a mechanism is tabulated with corresponding angular
movement x0 of the crank. Express y as a Fourier series upto the third harmonic. (Dec-11)
x0
0
30
60
90
120
150
180
210
240
270
300
330
f(x)
1.80
1.10
0.30
0.16
0.50
1.30
2.16
1.25
1.30
1.52
1.76
2.00
Soln:
x0
0
30
60
90
120
150
180
210
240
270
300
330
∑
y=f(x)
1.8
1.10
0.30
0.16
0.50
1.30
2.16
1.25
1.30
1.52
1.76
2.00
15.15
Cosx
1
0.86
0.5
0
-0.5
-0.86
-1
-0.86
-0.5
0
0.5
0.86
Cos2x
1
0.5
-0.5
-1
-0.5
0.5
1
0.5
-0.5
-1
-0.5
0.5
Cos3x
1
0
-1
0
1
0
-1
0
1
0
-1
0
Sinx
0
0.5
0.86
1
0.86
0.5
0
-0.5
-0.86
-1
-0.86
-0.5
Sin2x
0
0.86
0.86
0
-0.86
-0.86
0
0.86
0.86
0
-0.86
-0.86
Sin3x
0
1
0
-1
0
1
0
-1
0
1
0
-1
yCosx
1.8
0.94
0.15
0
-0.25
-1.11
-2.16
-1.07
-0.65
0
0.88
1.72
yCos2x
1.8
0.55
-0.15
-0.16
-0.25
0.65
2.16
0.625
-0.65
-1.52
-0.88
1
yCos3x
1.8
0
-0.3
0
0.5
0
-2.16
0
1.3
0
-1.76
0
-0.062
ySinx
0
0.55
0.25
0.16
0.43
0.65
0
-0.625
-1.118
-1.52
-1.514
-1
-2.73
ySin2x
0
0.946
0.25
0
-0.43
-1.11
0
1.075
1.118
0
-1.514
-1.72
-1.39
ySin3x
0
1.1
0
-0.16
0
1.3
0
-1.25
0
1.52
0
-2
0.51
𝒇(𝒙) =
𝒂𝟎
+ 𝒂 𝒄𝒐𝒔 𝒙 + 𝒂 𝒄𝒐𝒔𝟐𝒙 + 𝒂
𝟏
𝟐
𝑎 =
0
𝟐
2
𝑛
(∑ 𝑦) =
2
𝟑
𝒄𝒐𝒔 𝟑𝒙 + 𝒃𝟏𝒔𝒊𝒏 𝒙 + 𝒃𝟐 𝒔𝒊𝒏 𝟐𝒙 + 𝒃𝟑 𝒔𝒊𝒏 𝟑𝒙 + ⋯
(15.15) = 2.525,
12
𝑎0
2
=
2.525
2
= 1.2625
∑
0.243
3.175
2
𝑎1 =
𝑎3 =
𝑛
2
2
(∑ 𝑦𝑐𝑜𝑠 𝑥) = 0.0405, 𝑎2 =
𝑛
(∑ 𝑦𝑐𝑜𝑠 2𝑥) = 0.5292
2
(∑ 𝑦𝑠𝑖𝑛𝑥) = −0.4548
𝑛
2
𝑏2 = (∑ 𝑦𝑠𝑖𝑛 2𝑥) = −0.2308, 𝑏3 = (∑ 𝑦𝑠𝑖𝑛3𝑥) = 0.085
𝑛
𝑛
𝑛
2
(∑ 𝑦𝑐𝑜𝑠 3𝑥) = −0.103, 𝑏1 =
𝑓(𝑥) = 1.2625 + (0.0405 ) cos 𝑥 + (0.5292 )𝑐𝑜𝑠2𝑥 − (0.103 )cos 3𝑥
−(0.4548)𝑠𝑖𝑛 𝑥 − (0.2308)𝑠𝑖𝑛 2𝑥 + (0.085 )𝑠𝑖𝑛 3𝑥 + ⋯
31) Express y as a Fourier series upto the third harmonic given the following values:
x
0
1
2
3
4
5
y
4
8
15
7
6
2
The values of y at x=0,1,2,3,4,5 are given and hence the interval of x should be 0  x < 6. The
length of the interval = 6-0 = 6, so that 2l = 6 or l = 3.
The Fourier series upto the third harmonic is
y
a0 
x
x  
2x
2x  
3x
3x 
 a cos  b sin
 a cos
 b sin
 a cos
 b sin

2

1
l
 2
l  
1
l
2
 3
l  

l
3

l 
or
a0 
x
x  
2x
2x  
3x
3x 
 a cos  b sin
 a cos
 b sin
 a cos
 b sin



1
2
1
 3
2

3
3  
3
2 
3
3  
3
3 
x

Put  
, then

3

a0
y   a cos  b sin    a cos 2  b sin 2   a cos3  b sin 3 
(1)
1
1
2
2
3
3
2
y

We prepare the following table using the given values :
x
=
x
3
y
ycos
ycos2
ycos3
ysin
ysin2
ysin3
0
0
04
4
4
4
0
0
0
1
600
08
4
-4
-8
6.928
6.928
0
2
1200
15
-7.5
-7.5
15
12.99
-12.99
0
3
1800
07
-7
7
-7
0
0
0
4
2400
06
-3
-3
6
-5.196
5.196
0
5
3000
02
1
-1
-2
-1.732
-1.732
0
42
-8.5
-4.5
8
12.99
-2.598
0
Total
a  2[ f (x)]  2[ y] 
0
2 y
6
1
 (42)  14
3
2
a  2[ y cos ]  (8.5)  2.833
1
6
2
b  2[ y sin  ]  (12.99)  4.33
1
6
2
a  2[ y cos 2 ]  (4.5)  1.5
2
6
2
b  2[ y sin 2 ]  (2.598)  0.866
2
6
2
a  2[ y cos3 ]  (8)  2.667
3
6
b3  2[ y sin 3 ]  0
Using these in (1), we get
 x 
 x 
 2x 
 2x 
y  7 

2,833cos   
(4.33)sin   1.5cos 
  0.866sin 
  2.667 cos x
 3 
 3 
 3 
 3 




This is the required Fourier series upto the third harmonic.