340
College Mathematics
5.
FOURIER SERIES
5.1 Introduction
In various engineering problems it will be necessary to
express a function in a series of sines and cosines which are
periodic functions. Most of the single valued functions which are
used in applied mathematics can be expressed in the form.
1
a0 + a1 cos x + a 2 cos2x +KK
2
+ b1 sin x + b2 sin2x +KK
within a desired range of values of x. Such a series is called a
Fourier Series in the name of the French mathematician Jacques
Foureier (1768 - 1830)
5.2 Periodic Functions
Definition : If at equal intervals of the abscissa ‘x’ the value of
each ordinate f(x) repeats itself then f(x) is called a periodic
function. i.e., A function f(x) is said to be a periodic function if
there exists a real number α such that f(x + α ) = f(x) for all x.
The number α is called the period of f(x).
∴ we have f(x) = f(x + α ) = f(x + 2α ) = f(x + 3α )
= …………………..= f(x + n α ) = ………….
Ex : (i) sin x = sin (x + 2π ) = sin (x +4π ) = ……………
.......….= sin (x + 2n π ) = ……………
Hence sin x is a periodic function of the period 2 π .
(ii) cos x = cos(x + 2π ) = cos (x + 4π ) = ………
……….. = cos (x + 2n π ) = …………….
Hence cos x is a periodic function of the period 2 π .
We define the Fourier series in terms of these two periodic
functions.
Fourier Series
341
5.3 Fourier Series
Definition : A series of the form
∞
∞
a
nπ x
nπ x
f ( x) = 0 + ∑ a n cos(
) + ∑ bn sin(
)
2 n =1
l
l
n =1
is called a Fourier series of f(x) with period 2l in the interval
( c, c +2l ) where l is any positive real number and a 0, a n, bn are
given by the formulae called Euler’s Formulae :
1
a0 =
l
an =
1
l
c+ 2 l
∫
f (x ) dx ,
c
c+ 2 l
∫
f ( x )cos(
c
nπ x
)dx
l
c+ 2l
nπ x
)dx
l
c
These coefficients a 0, a n, bn are known as Fourier coefficients.
bn =
1
l
∫
f ( x)sin(
In particular if l = π , the Fourier series of f(x) with period 2π in
the interval (c, c+2 π ) is given by
∞
∞
a
f ( x) = 0 + ∑ an cos nx + ∑bn sin nx
2 n =1
n =1
and the Fourier coefficients are given by
c+ 2 π
1
a0 =
f ( x)dx,
π ∫c
1
an =
π
c+ 2π
∫
f ( x) cos nπ dx
c
c+ 2π
1
bn =
f ( x)sin nπ dx
π ∫c
We shall derive the Euler’s formulae’ for which the following
definite integrals are required.
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College Mathematics
c +2 l
∫
(i)
dx = 2l
c
c +2 l
(ii)
∫
cos
∫
cos
c
c +2 l
(iii)
c
mπ x
dx =
l
c+ 2 l
∫
sin
c
mπ x
dx = 0
l
mπ x
nπ x
sin
dx = 0 for all integers m and n
l
l
c +2 l
c+ 2 l
mπ x
nπ x
mπ x
nπ x
cos
dx = ∫ sin
sin
dx = 0
∫c
l
l
l
l
c
(for all integers m and n such that m ≠ n)
c +2 l
c +2 l
mπ x
2 mπ x
dx = l = ∫ sin2
dx
(v) ∫ cos
l
l
c
c
(iv)
cos
5.4 Derivation of Euler’s Formulae
∞
a
nπ x ∞
nπ x
We have f ( x) = 0 + ∑ a n cos
+ ∑ bn sin
2 n =1
l
l
n =1
. . .(1)
To find the coefficients a0 , a n and bn , we assume that the series
(1) can be integrated term by term from x = c to x = c + 2l
To find a0, integrate (1) w.r.t x from c to c + 2l.
∴
c +2 l
∫
f ( x) =
c
a0
2
c + 2l
c + 2l
∞
 nπ x 
dx
l 
∫ 1dx + ∑ a ∫ cos 
c
n=1
n
∞
c
c +2 l
n =1
c
+ ∑ bn
∫
nπ x
sin(
)dx
l
∞
∞
a0
( 2l ) + ∑ an (0) + ∑ bn (0)
2
n =1
n =1
= a0 (l ) (using the definite integrals (ii) above)
=
Fourier Series
c +2 l
1
∴ a0 = ∫ f ( x)dx
l c
343
..
. (a)
mπ x
where m is a
l
fixed positive integer and integrate w.r.t x from x= c to x= c+ 2l
c+ 2 l
mπ x
∴
∫c f ( x)cos l dx
To find an , multiply both sides of (1) by cos
a
= 0
2
c+ 2l
∫
c
∞
mπ x
cos
dx + ∑ an
l
n =1
∞
+ ∑ bn
n =1
c+ 2 l
∫ cos
c
c+ 2 l
∫
c
cos
mπ x
nπ x
cos
dx
l
l
mπ x
nπ x
sin
dx
l
l
c +2 l
∞
a0
mπ x
nπ x
(0) + ∑ an ∫ cos
cos
dx + ∑ bn (0)
2
l
l
n =1
n =1
c
[Using the definite integrals (ii) and (iii) above]
c+ 2 l
∞
mπ x
nπ x
= ∑ an ∫ cos
cos
dx ( m ≠ n)
l
l
n=1
c
∞
=
c+ 2 l
+ am
∫
c
cos2
mπ x
dx (m = n)
l
∞
= ∑ an (0) + am (l )
n =1
[Using the definite integrals (iv) and (v) above]
= a m (l )
c+ 2 l
1
mπ x
f ( x)cos
dx
∫
l c
l
Changing m to n we get
∴ am =
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College Mathematics
am =
1
l
c+ 2 l
∫
f ( x)cos
c
nπ x
dx
l
…(b)
mπ x
where m
l
is a fixed positive integer and integrate w.r.t x from x = c to
x= c+ 2l
c +2 l
mπ x
∴
∫c f ( x)sin l dx
To find bn, multiply both sides of (1) by sin
a
= 0
2
c+ 2l
∫
c
∞
mπ x
sin
dx + ∑ an
l
n =1
∞
+ ∑ bn
c +2 l
∫
c
c+ 2 l
n =1
∫
c
sin
sin
mπ x
nπ x
cos
dx
l
l
mπ x
nπ x
sin
dx
l
l
Fourier Series
c+ 2 l
1
nπ x
bn = ∫ f ( x)sin
dx
l c
l
(b)
Thus the Euler’s formulae (a), (b), (c) are proved.
Cor. 1 : In particular if l = π and c = 0, we get the Fourier series
∞
∞
a
f ( x) = 0 + ∑ an cos nx + ∑bn sin nx
2 n =1
n =1
where the Foureir coefficients are given by
2π
1
a0 = ∫ f ( x)dx ,
π 0
an =
c +2 l
∞
∞
a
mπ x
nπ x
= 0 (0) + ∑ an (0) + ∑ bn ∫ sin
sin
dx
2
l
l
n =1
n =1
c
[Using the definite integrals (ii) and (iii) above]
c+ 2l
∞
mπ x
nπ x
= ∑ an ∫ sin
sin
dx ( m ≠ n)
l
l
n =1
c
c+ 2l
+ bm
∫ sin
c
c +2 l
mπ x
nπ x
sin
dx ( m = n)
l
l
mπ x
dx
l
c
[Using the definite integrals (iv) above]
= bm (l ) [us ing the definite integral (v)]
= 0 + bm
c+ 2 l
∫
sin 2
1
mπ x
∴ bm = ∫ f ( x )sin
dx
l c
l
Changing m to n we get
345
bn =
1
π
1
π
2π
∫ f ( x)cos nπ dx
0
2π
∫ f ( x)sin nπ dx
0
Cor. 2: In the above formulae if l = π and c = −π , we get the
Fourier series
∞
∞
a
f ( x) = 0 + ∑ an cos nx + ∑bn sin nx
2 n =1
n =1
where the Fourier coefficients are given by
π
1
a 0 = ∫ f (x ) dx ,
π −π
π
an =
1
f ( x)cos nπ dx
π −∫π
π
1
bn = ∫ f ( x)sin nπ dx
π −π
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College Mathematics
5.5 Conditions for a Fourier series expansion
It should not be mistaken that every function can be
expanded as a Fourier series. In the above formulae we have only
shown that if f(x) is expressed as a Fourier series, then the Fourier
coefficients are given by Euler’s formula. It is very cumbersome
to discuss whether a function can be expressed as a Fourier series
and to discuss the convergence of this series. However the
following condition called Dirichlet’s condition cover all
problems.
∞
a
nπ x ∞
nπ x
f ( x) = 0 + ∑ a n cos
+ ∑ bn sin
2 n =1
l
l
n =1
provided
(i)
f(x) is bounded
(ii)
f(x) is periodic, single – valued and finite
(iii)
f(x) has a finite number of discontinuities in any one
period.
(iv)
f(x) has at the most a finite number of maxima and
minima.
These conditions are called Dirichlets conditions. In fact
expressing a function f(x) as a Fourier series depends on the
evaluation on the definite integrals
1
nπ x
1
nπ x
f ( x)cos
dx and ∫ f ( x)sin
dx
∫
l
l
π
l
within the limits c to c + 2l, 0 to 2π or - π to π according as
f(x) is defined for all x in (c, c + 2l) (0, 2π ) or (- π , π )
5.6 Interval with 0 as mid point
If c = -l then the interval (c, c + 2l) becomes (-l, l) and
further if c = - π , the interval becomes (- π , π ). These intervals
have 0 as the mid point. For functions defined in such intervals,
we consider the effect of changing x to –x and classify them as
even and odd functions.
Fourier Series
347
5.7 Even and odd functions
A function f(x) is said to be even if f(-x) = f(x) ∀x in the
given interval (c, c + 2l) and a function f(x) is said to be odd if
f(-x) = -f(x) ∀x in the given interval (c, c + 2l)
5.7.1 Tests for even and odd mature of a function
If f(x) is defined by one single expression, f(-x) = f(x)
implies f (x) is even and f(-x) = -f(x) implies f(x) is odd. If f(x) is
defined by two or more expressions on parts of the given interva l
with 0 as the mid point, f(-x) from the function as defined on one
side of 0 = f(x) from the corresponding function as defined on the
other side, implies f(x) is even.
f(-x) from the function as defined on one side of 0 = -f(x)
from the corresponding function as defined on the other side,
implies f(x) is odd.
Examples :
(1) f(x) = x2 + 1 in (-1, 1)
f(-x) = (-x) 2 + 1 = x2 + 1 = f(x)
∴ f(x) is even.
(2) f(x) = x3 in (-1, 1)
f(-x) = (-x3 ) =- x3 = - f(x)
∴ f(x) is odd.
 x + 1 in ( −π ,0)
(3) f ( x ) = 
 x − 1 in (0, π )
f (− x)in (0, π ) = − x − 1 = −( x + 1) = − f ( x ) i n (−π ,0)
∴ f(-x) = -f(x)
∴ f(x) is odd
5.7.2 Fourier coefficients when f(x) is even and odd
From definite integrals, we have
348
College Mathematics
a
a
−a
0
∫ φ ( x)dx = 2∫ φ ( x) dx
if φ ( x) is even.
a
∫ φ ( x) dx = 0 if φ ( x)
and
is odd.
−a
(a) If f(x) is even in (-l, l) i.e., iff f(-x) = f(x), then
nπ x
f(x) cos
is also even.
l
nπ (− x)
nπ x
∴ f(-x) cos
= f(x) cos
. Since cos(-θ )= cos θ
l
l
nπ x
and f(x) sin
is odd.
l
nπ (− x )
nπ x
= − f ( x )sin
Q f(x) sin
since sin(-θ ) = -sin θ
l
l
l
l
1
2
a0 = ∫ f ( x)dx = ∫ f ( x)dx (by above definite integral)
l −l
l 0
1
nπ x
2
nπ x
an = ∫ f ( x)cos
dx = ∫ f ( x)cos
dx
l −l
l
l 0
l
l
l
1
nπ x
f ( x)sin
dx = 0
∫
l −l
l
Fourier Series
nπ x
f(x) cos
is also odd in (-l, l)
l
nπ ( − x )
nπ x
= − f ( x )cos
Q f (− x )cos
l
l
nπ x
and f ( x)sin
is even in (-l, l)
l
nπ ( − x )
nπ x
f (− x )sin
= f ( x )sin
l
l
l
1
∴ a0 = ∫ f ( x)dx
l −l
an =
1
nπ x
f ( x )cos
dx = 0
∫
l −l
l
bn =
1
nπ x
2
nπ x
f ( x)sin
dx = ∫ f ( x)sin
dx
∫
l −l
l
l 0
l
349
l
l
l
If the interval is (-π, π ) then a0 = 0, an = 0
x
2
bn = ∫ f ( x )sin nxdx
π 0
l
bn =
2
2
nπ x
f (x )dx + ∫ f ( x)cos
dx
∫
l 0
l0
l
In this case if the interval is (-π , π ) we get
π
2
a 0 = ∫ f ( x)dx
π 0
l
l
∴ f ( x) =
π
an =
2
f ( x )cos nxdx
π ∫0
bn = 0
(b) If f(x) is odd in (-l, l) i.e., if f( -x) = -f(x) then
5.7.3 Intervals with 0 as an end point
Intervals like (0, 2l) and (0, 2π ) with 0 as end point have
special features.
0
0
and
= 0 if φ (2 a − x ) = −φ ( x)
f(2l - x) = f(x)
l
Then
a
∫ φ ( x)dx = 2∫ φ ( x) dx if φ (2a − x) = φ ( x)
We know that
If
2a
2
a 0 = ∫ f (x ) dx
l 0
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College Mathematics
Fourier Series
an =
2
nπ x
f ( x)cos
dx
∫
l 0
l
bn = 0
Similarly if l = π, i.e., if the interval is (0, 2π) we get
π
2
a0 = ∫ f ( x)dx
π 0
π
an =
2
f ( x)cos nx dx
π ∫0
bn = 0
If
f(2l – x) = -f(x) then
l
2
nπ x
dx
a0 = 0, an = 0, bn = ∫ f ( x)sin
l 0
l
Similarly If f(2π - x) = -f(x) then
π
2
a0 = 0, an = 0, bn = ∫ f ( x )sin nxdx
π 0
WORKED EXAMPLES
1) Find the Fourier coefficient a0 for f(x) = x sinx in (0, 2π )
(May 2003)
2π
1
a0 = ∫ xsin xdx
π 0
1
= [ x (− cos x ) + ∫ cos x.1dx ]
π
1
= [− x cos x + sin x]20 π = −2
π
2) Find the coefficient a 0 for f(x) = x-1 in (-π,π)
(A 1999)
a0 =
1
π
π
∫ ( x − 1)dx
−π
351
π
l

1  x2
 − x
π2
 −π
2
 1 π 2

1 π
=  −π −  +π 
π 2
 π 2

= -2
for
− 2 < x <0
=
0
3) If 
for
0<x<2
1
find the Fourier coefficient an in the fourier series.
L
1
 nπ x 
a n = ∫ f ( x )cos 
dx
L −L
 L 
1
1
nπ x
0dx + ∫ 1.cos(
)dx
∫
2 −2
20
2
0
=
2
nπ x 
)
2  = 1 [sin( nπ ) − 0] = 0

nπ
nπ

2
0
2
=
1
2
sin(
4) Obtain the Fourier series for f(x) = x-1 in the interval (-π, π ).
(A 1999)
Solution :
x
x
1
1
a0 = ∫ f ( x)dx = ∫ ( x − 1) dx
π −x
π −x
x
x
1
1
xdx = ∫ dx
∫
π −x
π −x
1
π
1
= 0 − ( x )−π = [2π ] = −2
π
π
∴ a0 = − 2
=
352
π
1
an = ∫ f ( x)cos nxdx
π −π
College Mathematics
π
=
1
( x − 1)cos nxdx
π −∫π
π

1 π
x
cos
nx
−
cos nxdx 
∫
∫
π  −π
−π

π

1
= 0 − 2 ∫ cos nxdx 
π 
−π

=
π
π
1
bn = ∫ f ( x)sin nxdx
π −π
π
1
= ∫ ( x − 1)sin nxdx
π −π
π

1 π
 ∫ x sin nx − ∫ sin nxdx 
π  −π
−π

π

1
=  2 ∫ x sin nxdx − 0 
π  0

=
∴ bn = 0
π
a0 =
π
2   − cos nx 
 cos nx  
= x 
− ∫ −
 dx 
π  n

 n  0
π
353
∞
2 ∞
2(−1)n +1
= − +∑0 + ∑
sin nx
2 n =1
n
n =1
∞
( −1) n +1
∴ f ( x) = −1 + 2∑
sin nx is the required Fourier series
n
n =1
5) Expand f(x) = x2 as a Fourier series in the interval (-π , π ) and
1 1 1
π2
hence show that (i) 2 − 2 + 2 − + K =
1 2 3
12
1 1 1
π2
(ii) 2 + 2 + 2 + K =
(A 1999)
1 2 3
6
Solution :
f(x) = x2
∴ f (− x) = (− x ) 2 = x 2 = f ( x )
∴ f ( x ) is even in ( −π , π )
1  −2sin x 
= 
π  n  0
2  sin nx sin0
=− 
−
=0
π n
n 
2  x cos nx
sin nx 
= −
−+ 2 
π
n
n 0
Fourier Series
2  π cos nπ


=  −
+ 0  − ( −0 + 0) 
π 
n


n +1
2cos nπ
2
2( −1)
=−
= − ( −1) n =
n
n
n
∴ Fourier series is given by
∞
∞
a
f ( x) = 0 + ∑ an cos nx + ∑b n sin nx
2 n =1
n =1
π
1
1
f (x ) dx = ∫ x 2 dx
∫
π −π
π −π
π
π
2
2  x3 
2 π 3 2π 2
= ∫ x 2 dx =   =
=
π 0
π  3 0 π 3
3
1
an =
π
π
∫
−π
f ( x)cos nxdx
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College Mathematics
π
=
π
1
2
x 2 cos nxdx = ∫ x2 cos nxdx Q x2 cos nx is even
∫
π −π
π 0
π
2  sin nx
 − cos nx   − sin nx 
= x2
− 2x
 + 2

2
3
π
n
 n
  n
 0
2  sin nπ
cosnπ
− sin nπ  
=  π 2
+ 2π
−2
−0
2
π 
n
n
n 3  

2
(−1)n
0
+
2
π
− 0

2
π
n

n
4(−1)
i. e, an =
n2
∴ Foureir series is
∞
∞
a
f ( x) = 0 + ∑ a n cos nx + ∑b n sin nx
2 n =1
n =1
=
=
∴ f ( x) =
π
4( −1)
+∑
cos nx + 0
3 n =1 n 2
2
∞
n
π 2 ∞ 4( −1) n
+∑
cos nx is the required Fourier series.
3 n =1 n 2
1 1 1
π
− 2 + 2 − +K =
2
1 2 3
12
Put x = 0 in the above Fourier series
x2 ∞ 4( −1) n
∴ f (0) =
+∑
cos0
3 n =1 n 2
2
(i) To prove
∞
x2
(− 1)n
∴ 0 = + 4 ∑ 2 Q f (0) = 0 2 = 0
3
n=1 n
x2
 1 1 1 1

+ 4 − 2 + 2 − 2 + 2 − + K
3
1
2
3
4


2
x
1 1 1 1

i.e, 0 = − 4  2 − 2 + 2 − 2 − + K
3
1 2 3 4

i.e, 0 =
Fourier Series
2
1 1 1 1
 π
∴ 4  2 − 2 + 2 − 2 + − K =
1 2 3 4
 3
1 1 1 1
π2
−
+
−
+
K
=
12 22 32 42
12
1 1 1
π2
(ii) To prove that 2 + 2 + 2 + K =
1 2 3
6
Put x = π in the Fourier series of f(x)
π 2 ∞ 4(−1)n
f (π ) =
+∑
cos nπ
3 n =1 n 2
π 2 ∞ 4(− 1)n
i.e, π 2 =
+∑
(−1)n
3 n =1 n 2
∴
=
π2 ∞ 4
+ ∑ 2 (−1)2 n
3 n =1 n
∞
π2
1
+ 4∑ 2 Q(−1)2 n = 1
3
n =1 n
∞
1
π2
∴ 4∑ 2 = π 2 −
3
n =1 n
=
π2
1 1 1

i.e. , 4  2 + 2 + 2 + K = 2
3
1 2 3

2
1 1 1
π
∴ 2 + 2 + 2 + K=
1 2 3
6
6) Obtain the Fourier series for f(x) = ex in ( −π , π )
Solution :
π
π
1
1
a 0 = ∫ f (x ) dx = ∫ e x dx
π −π
π −π
π
1
= e x 
−π
π
1
= e π − e −π 
π
355
356
π
1
an = ∫ f ( x)cos nπ dx
π −π
ax
∫ e cos bxdx =
+∑
eax ( a cos bx + b sin bx )
a 2 + b2
π
 e x (cos nx + n sin nx ) 


12 + n 2

 −π
x
−x
1  e cos nπ − e cos nx)  (−1)n ( ex − e− x )
= 
=
π
12 + n2
π (1 + n2 )

(as sin nπ = 0 = sin(− nπ ) and cos nπ = (−1)n ) )
1
∴ an =
π
π
bn =
1
f ( x)sin nπ dx
π −∫π
π
=
1
ex sin nπ dx
π −∫π
We know that
ax
∫ e sin bxdx =
ea x ( a sin bx − b cos bx)
a 2 + b2
π
 e x (sin nx − n cos nx) 


12 + n 2

−π
π
−π
− n  e cos nπ − e cos( −nx)  − n  eπ (−1)n − e−π (− 1) n 
=

=


π 
12 + n2
π (1 + n2 )
 π 

n
π
−π
− n (−1) ( e − e )
=
π (1 + n 2 )
∴ Fourier series is
∞
∞
a
f ( x) = 0 + ∑ an cos nx + ∑b n sin nx
2 n =1
n =1
1
∴ bn =
π
− n (−1)n ( eπ − e−π )
sin nπ
π (1 + n 2 )
n =1
∞
eπ − e −π 
( −1) n 2cos nπ ∞ (−1)n 2sin nπ 
f
(
x
)
=
1
+
+∑
i.e.,
 ∑

2π  n =1
1 + n2
1 + n2
n =1

∞
π
1
= ∫ e xc os nπ dx
π −π
We know that
Fourier Series
∞
1 π
( −1) n (eπ − e−π )
=
( e − e −π ) + ∑
cos nπ
2π
π (1 + n 2 )
n =1
College Mathematics
∞

(−1)n 2cos nπ ∞ ( −1) n 2n sin nπ 
1
+
−∑
 ∑

1 + n2
1 + n2
n =1
 n =1

π
−π

e −e 
as  sinh π =

2 

7) Obtain the Fourier series for f(x) = x in ( −π , π ) and prove
1 1 1
π
that 1 − + − K =
3 5 7
4
Solution :
π
1
a 0 = ∫ f (x )dx
π −π
=
sinh π
2π
1
=
π
π
π
1  x2 
xdx
=
  =0
∫
π  2  −π
−π
π
1
an = ∫ x cos nxdx
π −π
π
1  sin nx
− cos nx 
= x
− 1(
)
π
n
n2
 −π
1  sin nπ cos nπ  
sin( −nπ ) cos nπ 
+
+
π
 −  −π


2
π 
n
n  
n
n 2 
1  ( −1) n ( −1) n 
= 0 + 2 − 2  = 0
π
n
n 
=
357
358
π
1
bn = ∫ f ( x)sin nxdx
π −π
College Mathematics
π
=
1
x sin nxdx
π −∫π
=
2
x sin nxdx Q x sin nx is even
π ∫0
=
2
π
π
π
  − cos nx 
− sin nx 
)
 − 1(
x 
n
n2

 
0
2  −π cos nπ   sin nπ 


 +  2  − (0 + 0) 

π 
n
  n 

n
n +1
− 2 π (− 1)
2(− 1)
=
=
π
n
n
Fourier series is
∞
∞
a
f ( x) = 0 + ∑ a n cos nx + ∑b n sin nx
2 n =1
n =1
=
(− 1)n +1 2sin nx
n
n =0
∞
= 0+ 0 + ∑
(− 1)n +1 2sin nx
is the Fourier series.
n
n=1
∞
∴ f ( x) = ∑
π
in the Fourier series
2
nπ
( −1) n +1 2sin
π  ∞
2
∴ f   =∑
n
 2  n =0
nπ
(−1)n +1 2sin
∞
2 since sin nπ = 0 if n is even.
= ∑
n
2
n =1,3,5
Put x =
Fourier Series
3π
5π
 π

 sin 2 sin 2 sin 2

π
∴ = 2
+
+
+ K
2
3
5
 1



π
1 1 1
∴ = 1− + − +K
4
3 5 7
1 1 1
π
ie.,1 − + − +K =
3 5 7
4
8) Find the Fourier series for e-x in the interval (-l, l)
Solution :
l
l
1
1 −x
a 0 = ∫ f (x ) dx = ∫ e dx
l −l
l −l
l
1
= e− x 
−l
l
1
= − e −l − e l 
l
−l
e − el 2sin hl
=
=
l
l
l
1
nπ x
a n = ∫ f ( x)cos
dx
l −l
l
1
nπ x
= ∫ e− x cos
dx
l −l
l
l
nπ x π n
nπ x 
 −x
e (− cos
+
sin
)

1
l
l
l − l
=
2
l
 nπ 
(− 1)2 + 

 l 
l
359
360
College Mathematics
nπ
nπ
 −l

e ( − cos nπ +
sin nπ ) − el ( − cos nπ −
sin nπ )

1
l
l

= 
2
l
 nπ 
(−1) + 

 l 
1  ( −1)n (el − e−l ) 
= 
l
l  l 2 + n 2π 2 
1
l (−1)n 2sinh l
n
l
−l
(
−
1)
(
e
−
e
)
=
l 2 + n2π 2
l 2 + n 2π 2
l
1
nπ x
bn = ∫ f ( x)sin
dx
l −l
l
=
1
nπ x
= ∫ e −x sin
dx
l −l
l
l
l

nπ x n π
nπ x 
 e−l ( − sin
−
cos
)
1
l
l
l 
= 
2

l
 nπ 
2
(− 1) +  


 l 

− l
 −l

nπ
nπ
e ( − sin nπ −
cos nπ ) − el (sin nπ −
cos nπ ) 
1
l
l
= 

n 2π 2
l

1+ 2


l
 −l  −nπ 
n
l  π 
n 2
e 
 (− 1) + e  n l  (− 1)  l

l
1





= 
2
2 2
l
l +n π
2
n nπ
l
l ( −1)
(e − e− l )
1
(−1)n nπ 2sinh l
l
=
=
l
l 2 + n2π 2
l 2 + n 2π 2
∴
Fourier series is
Fourier Series
∞
a
nπ x ∞
nπ x
f ( x) = 0 + ∑ a n cos
+ ∑ bn sin
2 n =1
l
l
n =1
sinh l ∞ (− 1)n 2l sinh l
nπ x
+∑ 2
cos
2 2
l
l +n π
l
n =1
n
∞
( −1) 2 nπ sinh l
nπ x
+∑
sin
2
2 2
l +n π
l
n =1
n
2
∞
sinh l
( −1) 2l
nπ x
i.e., ∴ f ( x) =
[1 + ∑ 2 2 2 cos
l
l
n =1 l + n π
n
∞
( −1) 2nπ l
nπ x
+∑ 2
sin
2 2
l
n =1 l + n π
9) Expand f(x) = x sinx, 0 <x < 2π in a fourier series
∴ f ( x) =
1
a0 =
π
1
an =
π
2π
1
∫ xsinxdx = π [−x cos x + sin x]
2π
0
= −2
0
2π
∫ xsinx.cos nxdx
0
2π
1
x(sin(n+1)x -sin(n-1)xdx
2π ∫0
1
cos( n + 1) x cos(n − 1) x
=
[ x (−
+
)
2π
n +1
n −1
cos(n + 1) x cos(n − 1) x
−∫ ( −
+
)dx]
n +1
n −1
1
1
1
2
=
[2π (−
+
)] = 2
2π
n +1 n −1
n −1
=
bn =
=
1
x sin x.sin nxdx
π∫
1
2π
∫ x(cos(1− n )x − cos(1+ n)x )dx
361
362
College Mathematics
1
sin(1 − n ) x sin(1 + n ) x
sin(1 − n )x sin(1 + n) x 
=  x(
)−
−∫
−
dx

2π 
1−n
1+ n
1−n
1 + n 
2π

cos(1 − n) x cos(1 + n ) x 
0 + (1 − n) 2 − (1 + n )2  = 0

0
 2 
f ( x) = −1 + ∑  2
 cos nx
 n −1 
π
Note : When x = we derive that
2
1
1
1
π +2
−
+
K∞ =
1.3 3.5 5.7
4
10) Find the fourier series for the periodic function f ( x) = x in
(-l, l)
Given f ( x) = x which is even
1
=
2π
∴ The fourier series is
f ( x) =
Fourier Series
363
2
2
2
l
l 
= 0 + 2 2 cos(nπ ) − 2 2 
l
nπ
nπ 
2l
2l
= 2 2 (cos nπ − 1) = 2 2 (( −1) n − 1)
nπ
nπ
l
2l
 nπ x 
∴ f ( x ) = + ∑ 2 2 ((− 1) n − 1)cos 

2
nπ
 l 
for 0 ≤ x < π
 x
11) Expand f ( x ) = 
as a fourier
for π ≤ x < 2π
 2π − x
series.
π
2π
1
1
a 0 = ∫ xdx + ∫ (2π − x )
π 0
π π
a0 ∞
nπ x
+ ∑ an cos(
)
2
l
1
1 π2

π 2
=
π 1
π2
+ [(4π 2 − 2π 2) − (2π 2 − )]
2 π
2
l
l
2
2  x2 
a 0 = ∫ xdx =   = l
l 0
l  2 0
π π
+ =π
2 2
π
1
1
a n = ∫ x cos nxdx +
π 0
π
2
nπ x
f ( x)cos
dx
∫
l 0
l
2
nπ x
= ∫ x cos(
) dx
l 0
l
l
=
nπ x
nπ x


sin(
)
sin(
) 
2
l − sin
l dx
=  x.

∫
nπ
nπ
l

l
l


l
=
 1
x2 
+
(2
π
x
−


2 π
 π
=
l
an =
2π
=
2  lx
nπ x
l2
nπ x 
sin(
)
+
cos(
)

2 2
l  nπ
l
nπ
l 0
2π
∫ (2π − x)cos nxdx
π
1  x sin( nx)
sin(nx) 
−∫
dx 

π
n
n

1
sin nx
sin nx 
+ (2π − x)
+∫
.dx 
π
n
n

π
2π
1  cos nx 
1  cos nx 
= 0 + 2  +  0 −
π
n 0 π 
n2 π
=
1
1  1 ( −1) n 
n


(
−
1)
−
1
+
 π  − n2 + n2 
π n2 


364
College Mathematics
2
= 2 (− 1)n − 1
πn
π
2π
1
1
bn = ∫ x sin nxdx + ∫ (2π − x)sin nxdx
π 0
π π
π
=
1  − x cos nx
cos nx 
+∫
dx 

π
n
n
0
2π
1
 − cos nx  cos nx 
+ (2π − x) 
.dx 
+
π
n  ∫ n

π
π
2π
1 π cos nx 
1  π cos nx 
= 
+ 
=0

π  n 0 π  n π
∴ The fourier series is
π
2
f ( x ) = + ∑ 2 ((−1)n − 1)cos nx + 0
2
πn
π
2
= + ∑ 2 ((−1)n − 1)cos nx.
2
πn
12) Find a fourier series for the function
−π < x < 0
 −1

f ( x) =  0
x=0
 1
0 < x <π

π
1
a0 = ∫ f (x )dx
π −π
=
1
π
0
∫
−π
−1dx +
π
1
1
1dx = [−π + π ] = 0
π ∫0
π
π
an =
1
f ( x)cos( nx) dx
π −∫π
0
=
π
1
1
− cos nxdx + ∫ 1.cos nxdx
∫
π −π
π0
Fourier Series
1  sin nx sin nx 
= −
+
π
n
n 
1
= (0) = 0
π
π
1
bn = ∫ f ( x)sin( nx) dx
π −π
π
0
=
365
1
1
(−1)sin nxdx + ∫1.sin nxdx
∫
π −π
π0
0
π
1  cos nx 
 − cos nx  
= 
+


π  n  −π 
n
 0
1
=
[1 − cos nπ − cos nπ + 1]
nπ
2
=
(1 − ( −1) n )
πn
∴ bn is zero for n = 2, 4, 6, . . .
4
and bn =
for n = 1, 3, 5, . .
πn
∴ Required fourier series
4
f ( x) = 0 + ∑ 0.cos nx + ∑
.sin nx
πn
4
sin3x

= sin x +
+ ...∞ 
π
3

π
Note : when x =
2
4 1 1

f ( x ) = 1 = 1 − + − +...∞.
π 3 5

π
1 1
= 1− + − +...
4
3 5
13) Find the Fourier series for 1 − cos x in the interval
366
College Mathematics
-π < x < π
Let f ( x ) = 1 − cos x . It is an even function
a
∴ f ( x) = 0 + ∑ a nc os nx; bn = 0
2
π
2
2
x
a0 = ∫ 1− cos xdx = . 2 ∫ sin dx
π 0
π
2
π
x

cos

2 2
4 2
2
=
− 1  =
π 
π


2 0
π
an =
π
=
π
2
2 2
x
1 − cos x .cos nxdx =
sin .cos nxdx
∫
∫
π 0
π 0
2
π
=
2 2 1
x
x
{sin( + nx) + sin( − nx)}dx
∫
π 02
2
2
π
1
1 

cos(n + ) x cos( n − ) x 

2
2 +
2
=
−

1
1
π 

n+
n−

2
2 0


2
1
1 
=
−
(0 + 0) +
1
1
π 
n+
n− 

2
2
1
1

n − − n− 
2
2
2
=


1
π  n2 −


4

=−
4 2
π
 1 
 4n2 − 1 


Fourier Series
367
2 2 4 2
 1 
∴ f ( x) =
−
∑
 4n 2 − 1 c os nx
π
π


14) If a is not an integer show that for -π <x < π
2sin ax  sin x 2sin2 x 3sin3x

sin ax =
− 2
+ 2
− + K
2
2
2
2

π
2 −a
3 −a
1 − a

Since f(x) = sin ax is an odd function, a0 & an are equal to zero.
π
1
bn = ∫ sinax.sinn xdx
π −π
2 1
(cos(n -a ) x − cos(n + a) x) dx
π ∫0 2
π
1  sin(n − a) x sin( n + a) x 
−
π  n − a
n + a 0
1  cos nπ sin aπ cos nπ sin aπ 
= −
−

π
n−a
n−a
cos nπ sin aπ  2 n 
=−
 n2 − a 2 
π


− cos nπ sin aθ  2 n 
∴ sin ax = ∑
 n 2 − a 2  sin nx
π


2sin aπ
− n cos nπ .sin ax
=
∑ n2 − a 2
π
2cos nπ sin aπ  sin x 2sin2x 3sin3x

=
− 2
+ 2
− + K
2
2
2
2

π
2 −a
3 −a
1 − a

=
Exercise :
I A.
1. Define a Fourier series
2. Write the empherical formulae for the fourier coefficients.
3. Write the fourier series with period 2π in the interval
(c, c + 2π )
4. Derive the Euler’s formulae in the interval (c, c + 2l)
368
College Mathematics
5. Write the Fourier coefficients in the interval (-l, l ) when f(x) is
a) even and b) odd.
6. Mention dirichlets conditions.
7. Find the fourier coefficient a0 for the following functions : (i)
f(x) = x2 in -π < x < π
(ii)
f ( x) = x2 in − l <x <l
0< x<π
x
(iii)
f (x) = 
π < x < 2π
 2π − x
f ( x) = x
(iv)
-π <x < π
(v)
f ( x) = cos λ x in − π ≤ x ≤ π
(vi)
 −1

f ( x) =  0
1

−π < x < 0
x=0
0< x<π
 2x
−π < x ≤ 0
1 +
(vi) f ( x) =  π
1− 2x
0< x <π
 π
8. Find the fourier coefficients a n and bn for the above
problems.
(2 marks for each constants)
B . 1. Find the Fourier series for
a) f(x) = x2 in -π <x < π . Hence deduce
1 1
π2
1+ 2 + 2 + K ∞ =
3 5
8
in − π < x ≤ 0
−x
b) f ( x ) = 
in 0 < x < π
 x
1 1 1
π
Hence deduce 1 − + − K ∞ =
3 5 7
4
c) f ( x) = x in -π <x < π. Hence deduce
Fourier Series
1 1
π2
1+ 2 + 2 + K ∞ =
3 5
8
d) f ( x) = sin x
e) f ( x ) = 1 +
2x
, −π ≤ x ≤ 0
π
2x
, 0 ≤ x ≤π
π
f) f ( x ) = cos ax in − π ≤ x ≤ π , a is not an integer.
 π
in − π < x ≤ 0
 x + 2
g) f ( x) = 
π −x
in 0 ≤ x < π
 2
h) f ( x) = x(π − x)
0≤ x≤π
 x
in 0 < x < π

2
If f ( x) = 
in π < x < π
π − x
2
in 0 ≤ x ≤ 1
πx
j) f ( x ) = 
in 1 ≤ x ≤ 2
π (2 − x)
in (0, l )
x
k) f ( x ) = 
in (l , 2l )
π − 2 l
− x 2
−π < x < 0
l) f ( x) =  2
0 < x <π
x
3
m) f(x) = x in −π < x < π
−π < x < 0
1
n) f ( x ) = 
0< x<π
0
−π ≤ x < 0
1
o) f ( x ) = 
0< x ≤π
2
=1−
369
370
College Mathematics
−π < x < 0
−a
p) f ( x ) = 
0 < x <π
a
−π < x < 0
π + x
q) f ( x ) = 
0< x<π
π − x
x2
π2
1 1 1
r) f ( x) = , − π < x < π , Hence
= 1 + + + +K ∞
4
6
4 9 16
II. 1 . Show that the fourier series for f(x) = 1-x2 in (-1, 1) is
2 4
(−1)n
= − ∑
cos π x
3 π
n
( Hint f(x) is even )
2. Show that the fourier series of
 x
in − π < x ≤ π

2
2
f ( x) = 
is
in π < x < 3π
π − x
2
2
4
(−1)n
sin(2 n + 1) x
∑
π
n
3. Show that the fourier series of
 π +x
−π < x < 0
 − 2
f ( x) = 
is
π − x
in 0 < x < π
 2
sin2 x sin3x
f ( x ) = sin x +
+
+K∞
2
3
( Hint : f(x) is odd )
4. Show that the fourier series of
f ( x) = cos x in ( −π ,π ) is
2 4 1
1
f ( x) = + ( cos2x − cos4x +…∞)
π π 3
15
5. If f(x ) = x + x 2 for −π < x < π , show that
f ( x) =
Fourier Series
1 1 1
π2
+
+
+
K
=
and
12 22 32
6
1 1 1
π2
+
+
+
K
=
12 32 52
8
6. If f(x) =x in ( −π ,π ) , show that
1
1
f ( x) = 2(sin x − sin2x + sin3x +K)
2
3
(Hint f(x) is odd)
371
Answers
2π 2
2l 2
2sinλπ
A. 7 (i)
(ii)
(iii) π (iv) π (v)
3
3
λπ
4
4l 2( − 1)n
8. (i) a n =
, bn = 0
(ii) a n = (−1)n 2 ; bn = 0
2 2
nπ
n
2
−4
(iii) a n = 2 ((−1)n − 1), bn = 0 (iv) a n =
, bn = 0
nπ
π (2 m − 1)2
( −1) n λ sin λπ
2
, bn = 0 (vi) a n = 0, bn =
[1 − ( −1) n ]
2
2
λ −n
nπ
4
n
(vii) a n = 2 2 [1 − ( −1) ], bn = 0
nπ
B. I
2 4  cos2 x cos4 x cos6x

d) f ( x ) = − 
+
+
+ K ∞
π π 3
15
35

8 
1
1

e) f ( x) = 2  cos x + cos3 x + cos5x + K ∞ 
π 
9
25

(v) a n =
1  1 ∞ 2a 
 −∑

π  a 1 n2 − a2 
4  cos x cos3 x cos3x

g) f ( x ) =  2 +
+
+K∞ 
2
2
π 1
3
5

f) f ( x ) =
372
College Mathematics
 cos x cos3x

h) f ( x ) = −π 2 − 8  2 +
+K 
2
3
 1

2
2
2
2
2  3π
4
π sin2 x  3π
4
π s i n 2x

+ 
− 2  sin x +
+
− 2  sin x +
+K 
π  1
1 
2
2
 1 1 

4  sin x sin3 x sin5 x

i)
(i) f ( x) =  2 −
+ 2 − + K
π 1
32
5

4 2  cos2 x cos6 x cos10 x

(ii) f ( x ) = −  2 +
+
+ K ∞
2
2
π π 1
3
5

π 4  cosπ x cos3π x cos3π x

j) f ( x ) = −  2 +
+
+K∞ 
2 π 1
32
52

n +1
2l
(− 1)
nπ x
k)
sin
∑
π
n
l
2
2
2

2  π
4
π
π2 4 
π
l)  − 2  sin x − sin2 x +  − 2  sin3x−
s i n 4x + K
π  1 1 
2
4
 3 3 

2
2
2
 π

6
π
6
π
6
m) 2  − 3  sin x − 
− 3  sin2x + 
− 3  sin3x K
 2 2 
 3 3 
 1 1 

1 2
sin(2n − 1) x
n) f ( x ) = − ∑
2 π
2n − 1
3 2 sin(2n − 1) x
o) f ( x) = + ∑
2 π
2n − 1
4a
sin(2 n − 1) x
p) f ( x ) =
∑ 2n − 1
π
3π 2
1
q) f ( x ) =
+ ∑ (1 − (−1)n ) 2 cos nx
8 π
n
5.8 Half – range cosine and sine series
Many times, it may be required to obtain a Fourier series
expansion of a function in the interval (0 , l) which is half the
period of the Fourier series. This is achieved by treating (0, l) as
half – range of (-l, l) and defining f(x) suitably in the other half
Fourier Series
373
i.e., in (-l, 0) so as to make the function even or odd according
as cosine series or sine series is required.
l
l
2
2
nπ x
∴ a0 = ∫ f ( x)dx, an = ∫ f ( x)cos
dx
l 0
l 0
l
∴ f ( x) =
a0 ∞
nπ x
for half – range cosine series and
+ ∑ an cos
2 n =1
l
2
nπ x
bn = ∫ f ( x)sin
dx and write the series as
l 0
l
l
∞
f ( x) = ∑ bn sin
n =1
nπ x
for half – range sine series.
l
Similarly, in (0, π)
π
π
2
2
a0 = ∫ f ( x)dx, an = ∫ f ( x)cos nx dx
π 0
π 0
and f ( x) =
a0 ∞
+ ∑ a cos nx
2 n=1 n
π
∞
2
bn = ∫ f ( x )sin nxdx f ( x) = ∑ bn sin nx
π 0
n=1
NOTE : (i) To solve a problem on Fourier series we have to find
a0, an and bn and substitute in
∞
a
nπ x ∞
nπ x
f ( x) = 0 + ∑ a n cos
+ ∑ bn sin
2 n =1
l
l
n =1
(ii) Finding of a 0, a n, bn , involves integration. In most of the
problems, f(x) consists of terms like x, x2 , x3, etc which after a
few differentiation will be zero.
The generalized formula for integration of the product of
two functions u and v called the Bernoulli’s rule may be used for
finding a n and bn.
1
2
3
4
∫ uvdx = uv −u′v +u′′v − u′′′v + −K
where dashes denote differentiation w.r.t x and suffixes 1,2,3,. . .
denote integration w.r.t. x
374
College Mathematics
 − cos nx 
 − sin nx 
 cos nx 
sin nxdx = x2 
− 2x
+ 2



2
3
n


 n

 n 
(iii) The following values of cosine and sine are useful
nπ
cos 0 = 1, cos n π = (-1)n = cos (-nπ), cos
= 0 if n is odd and
2
nπ
n
cos
= (−1) if n is even.
2
2
n −1
nπ
sin 0 = 0, sin n π = sin (nπ), sin
= (−1) 2 if n is odd and
2
nπ
sin
= 0 is even.
2
(iv) Integration work can be reduced to a great extent by using the
ideas of even and odd functions, whenever 0 is the mid point.
For eg.
∫x
2
(v) If f(x) is neither odd nor even, then f(x) may consist of some
terms which when taken individually may be odd or even and the
integr ation work can be reduced.
Worked Examples :
1) Find the half range sine series for f(x) =x in (0, 1)
(May 2003)
2
 nπ x 
 nπ x 
where bn = ∫ f ( x)sin 
f ( x) = ∑ bn sin 
dx

L0
 L 
 L 
L
1
bn =
2
x.sin nπ xdx
1 ∫0
 x cos nπ x 1
= 2 −
+
nπ
nπ


∫ cos nπ xdx 
1
 x cos nπ x sin nπ x 
= 2 −
+

nπ
(nπ )2 0

Fourier Series
375
n
 x cos nπ x  2( −1)
= 2 −
 = nπ
nπ


∴
Half = ramge Sine series is
∞
− 2(− 1) n
=∑
.sin( nπ ) x
nπ
1
2) Obtain the half -range Sine series for f(x) = x over the interval
(0, π)
(A 2003)
π
2
bn = ∫ x.sin nxdx
π 0
=
2
π
cos nx 1


 x( − n ) + n ∫ cos nxdx 
π
2  x cos nx sin nx 
= −
+ 2 
π
n
n 0
2  π cos nπ  2( −1) n
−
=
π 
n 
n
Half = ramge Sine series is
∴
− 2( −1) n
f ( x) = ∑
.sin nx.
n
3) Find the half – range Fourier sine series of f(x) = x2 in the
interval (0, 1)
(N 2000)
1
2
bn = ∫ x 2 sin nπ xdx
10
=
cos nπ x
1

= 2  x2 ( −
)+
cos( nπ x).2xdx
n
π
n
π∫

 cos nπ x 2  sin nπ x
sin nπ x 
= 2 −
+
−∫
.dx 
x
nπ
nπ 
nπ
nπ


 cos nπ x 2
= 2 −
+
nπ
nπ

1
cos nπ x  

 0 + n 2π 2  

0 
376
College Mathematics
 cos nπ x 2  cos nπ x
1 
= 2 −
+
 2 2 − 2 2 
nπ
nπ  n π
n π 

n
n
 (−1) 2(−1)
2 
∴ f ( x) = ∑ 2 
+ 3 3 − 3 3  sin(nπ x )
nπ
nπ 
 nπ
4) Find the half range cosine series for the function f(x) =x2 in
(0,π ) (A 2003)
It is required to find
∞
a
 nπ x 
f ( x) = 0 + ∑ an cos 
 where
2
 L 
1
L
L
2
2
 nπ x 
a0 = ∫ f ( x)dx; an = ∫ f ( x)cos 
 dx
L0
L0
 L 
π
2π
2  x3 
2π 2
a 0 = ∫ x 2 dx =   =
π 0
π  3 0
3
π
2
a0 = ∫ cos(nx)dx
π 0
sin(nx )
 2 sin( nx)

−∫
.2 xdx
 x
n
n

2  2   cos(nx) 
cos(nx)

= 0 −  x  −
.1dx 
+∫
π  n 
n 
n

=
2
π
2
=
π
π
 2  x cos(nx ) sin( nx)  
+

n−
n
n 2  0
 
 2  π ( −1) n  
−  −

n 
 n
4(−1) n
=
n2
2π 2
4(−1)n
∴ f ( x) =
+∑
cos( nx)
2(3)
n2
=
2
π
Fourier Series
π2
4( −1) n
=
+∑
cos(nx)
3
n2
377
5) Find the half range Sine series for f ( x) = π x − x2 in the
internal 0<x<π
π
2
bn = ∫ (π x − x 2 )sin xdx
π 0
π
2π
cos nx
sin nx
cos nx 
= ∫ (π x − x 2 )(−
) − (π − 2 x)( − 2 ) + ( −2)(−
)
π 0
n
n
n  0
4
= 3 (1− cos nπ )
πn
8
= 3
πn
8
∴ f ( x ) = ∑ 3 sin nx
πn
8
sin3x sin5x

= sin x + 3 + 3 + K
π
3
5

6) Find half – range sine series of
 x
0< x ≤π 2

f (x) = 
π < x<π
π − x
2
π
2
bn = ∫ f ( x)sin xdx
π 0
π
2
=
π
2
∫
0
π
2 2
x sin xdx + ∫ (π − x)sin xdx
ππ
2
π
2   cos nx   sin nx  2
= x −
 −  − 2 
π 
n  
n  0
378
College Mathematics
π
+
2
 cos nx 
 sin nx 
(π − x)  −
 − (−1)  − 2 

π
n 
n π


379
1
= 2∫ ( x − 1) 2 cos( nπ x)dx
2
 π cos( n π )
sin(n π ) π cos(n π ) sin n π 
2 +−
2 +
2 +
2
−
 2
n
n2
2
n
n2 


4
n
π


=
sin 

π n2
2


4
sin3 x sin5x

∴ f ( x) = sin x − 3 + 3 − + K ∞ 
π
3
5

7) Find the half – range sine series for f(x) = 2x-1 in the interval
(0, 1)
(A 2001)
1
2
bn = ∫ (2 x − 1)sin(nπ x)dx
10
2
=
π
Fourier Series
0
1

 sin nπ x 
 cos nπ x   − sin nπ x 
= 2 ( x − 1)2 
 − 2( x − 1)  − 2 2  + 2 

3 3
 nπ 
 nπ   nπ
 0

2cos(0) 
4
 − 2sin(nπ )
= 2
+− 2 2  = 2 2
2 2
nπ  nπ
 nπ
1
4
f ( x ) = + ∑ 2 2 cos nπ x
3
nπ
9) Expand f(x) = x as a cosine half – range series in 0 < x <2
Solution : The graph of f(x) = x is a straight line. Let us extend
the function f(x) in the interval (-2, 0) so that the new function is
symmetric al about they y – axis and hence it represents an even
function in (-2, 2)
1
− cos nπ x
− sin(nπ x) 

= 2 (2 x − 1)(
) − (2)(
n
π
n2π 2 0

1 
 − cos nπ
= 2
−
nπ 
 nπ
2
∴ f ( x) = ∑ −
(1 + cos nπ )sin(nxπ )
nπ
8. Find the half – range cosine series for the function of
f(x) = (x - 1)2 in the interval 0 < x < 1.
1
2
2( x − 1) 3 
2
2
a 0 = ∫ ( x − 1) 2 dx =
 = 0+ 3 = 3
10
3
0
1
2
nπ x
2
( x − 1).cos(
) dx
∫
10
l
1
an =
∴the Fourier coefficient bn= 0
∞
a
nπ x
∴ f ( x) = 0 + ∑ a n cos
2 n =1
2
2
a0 =
2
2
 x2 
1
1
4
xdx
=
2
xdx
=
  = =2
∫
∫
2 −2
2 −2
 2 0 2
an =
2
nπ x
x cos
dx
∫
2 −2
2
2

nπ x
nπ x 
 sin

− cos
2
2


= x
−1
2
nπ

n
π

 

 2  
2


 
2
0
2
nπ x
4
nπ x 
 2x
=  sin
+ 2 2 cos
2
nπ
2 0
 nπ
380
College Mathematics
4
4

 

=  0 + 2 2 cos nπ  −  0 + 2 2 cos0 
nπ
nπ

 

4
4
= 2 2 (cos nπ − 1) = 2 2 [(−1)n − 1]
nπ
nπ
a0 ∞
nπ x
∴ f ( x) = + ∑ an cos
2 n =1
2
∞
2
4
nπ x
= + ∑ 2 2 [(−1) n − 1]cos
2 n =1 n π
2
πx
3π x
5π x


− 2cos
−2cos
4  −2cos 2
2 +0+
2 + K
f ( x) = 1 + 2 
+0+

2
2
2
π 
1
3
5



πx
3π x
5π x


cos
8  cos 2 cos 2
2 + K
i.e., f ( x ) = 1− 2  2 +
+

2
2
π  1
3
5



Important Note : It must be clearly understood that we expand a
function in 0 < x < c as a series of sines and cosines merely
looking upon it as an odd or even function of period 2c. It hardly
matters whether the function is odd or even .
1
1
10) Expand f ( x) = − x if 0 < x <
4
2
3
1
= x − if < x < 1
4
2
in the Fourier series of sine terms
Solution : Let f(x) be an odd function in (-1, 1)
∴ a0 = 0 and an = 0
nπ x
and bn = 1 ∫ f ( x )sin
dx
1
−1
1
1
= 2∫ f ( x)sin nπ xdx
0
Fourier Series
1
1 2  1

3


= 2  ∫  − x  sin nπ xdx + ∫  x −  sin nπ xdx 
4

1 2
 0  4

 1
 − cos nπ
= 2  − x 
 nπ
 4
381
12

 − sin nπ x  
 − (−1) 

2 2

 nπ
 0

3  − cos nπ
+2  x − 
4  nπ

1

 − sin nπ x  
 − (−1)  n2π 2  


 1 2
nπ


 1

nπ sin 2
1
= 2
cos
− 2 2 +
cos0 + 0
2
nπ
4nπ
 4nπ



nπ

sin
 1
1
+2  −
cos nπ + 0 −
cos nπ − 2 22
4nπ
nπ
 4nπ

nπ
4sin
nπ
1
n
i.e., bn =
[1 − (−1) ] − 2 22 since cos
=0
2 nπ
nπ
2
1 4
∴ b1 = − 2 ; b2 = 0
π π
1
4
b3 =
+ 2 2 ; b4 = 0
3π 3 π
1
4
b5 =
− 2 2 ; b6 = 0 etc.
5π 5 π
∞
∴ f ( x) = ∑ bn sin nπ x
n =1
4 
1 4 
 1
+ 2 2  sin3π x
 − 2  sinπ x + 
π π 
 3π 3 π 
4 
 1
+
− 2 2  sin5π x + K
 5π 5 π 





382
College Mathematics
11) Find the sine and cosine series of the function f(x) = π - x in
0 < x < π.
(A 99)
Solution :
(i) Fourier sine series:
∞
2
bn = ∫ f ( x)sin nxdx
π 0
∞
=
2
(π − x)sin nxdx
π ∫0
x

 cos nx 
 − sin nx  
(π − x)  − n  − ( −1)  n  



 0

2
 − cos0   sin0  
= (0 − 0) − (π − 0) 
+

π
 n   n 
2π 2
=
=
π n n
∴ Fourier sine series is
∞
∞
2
f ( x) = ∑ bn sin nx = ∑ sin nx
n =1
n =1 n
1
1
1

= 2  sin x + sin2x + sin3x + K
2
3
1

(ii) Fourier cosine series:
π
π
2
2
a0 = ∫ f (x ) dx = ∫ (π − x) dx
π 0
π 0
2
=
π
π

x2 
π
x
−


2 0

2
π2  2 π2
= π 2 −  = .
=π
π
2 π 2
π
2
a 0 = ∫ (π − x)cos nxdx
π 0
=
2
π
Fourier Series
383
π
=
2
 sin nx 
 − cos nx 
(π − x) 
 − ( −1) 


2
π
 n 
 n
0
2  cos nπ  
cos0  
 0.
 −  0− 2 

π 
n  
n 
2  1 cos nπ 
=  2−
π n
n 2 
2
= 2 (1 − cos nπ )
nπ
2
= 2 [1 − (−1)n )]
nπ
a0 ∞
∴ f ( x) = + ∑ a n cos nx
2 n =1
π ∞ 2
= + ∑ 2 [1 − ( −1) n ]cos nx
2 n =1 n π
π 22
2
2

= +  2 cos x + 2 cos3 x + 2 cos5x + K
2 π 1
3
5

=
=
π 4 1
1
1

+  2 cos x + 2 cos3x + 2 cos5x +K
2 π 1
3
5

12) Find the Fourier series expansion with period 3 to represent
the function f(x) =2x – x2 in the range (0, 3)
Solution : We have c = 0 and 2l = 3
c +2 l
1
∴ a0 = ∫ f ( x)dx
l c
3
=
2
(2x − x 2 )dx
∫
30
=
2  2 x 2 x3 
2
x3 
−  = x2 − 

3 2
3 0 3 
3 0
3
3
384
College Mathematics
2
= [9 − 9] = 0
3
c +2 l
1
nπ x
an = ∫ f ( x)cos
dx
l c
l
2
2nπ x
(2 x − x 2 )cos
dx
∫
3c
3
Fourier Series

2nπ x

− cos

2
3
= (2 x − x 2 ) 
2
n
π
x
3



3

385
3
2nπ x
2nπ x 

sin
cos


3 + ( +2)
3 
+
(2
−
2
x
)

2
3
 2nπ x 
 2nπ x  




 

 3 
 3  0
3
=


2 nπ x
2nπ x 

 − cos
sin
2
3 − (2 − 2 x) 
3 
= (2 x − x 2 )

2
n
π
x
  2 nπ x 2 
3

 
 
3
  3  

3

2 nπ x 
 − sin

3


+ (+ 2)
  2 nπ x 2 
 
 
  3   0
2 9
36
27

= −
sin2nπ − 2 2 cos2nπ + 3 3 sin2nπ 
3  2 nπ
2n π
2n π

2
9

− 0 + 2 2 + 0 
3
2n π

2
9
3  2  12 
=  − 2 2 − 2 2  = − 2 2 
3 nπ
nπ  3 nπ 
8
=− 2 2
nπ
c+ 2 l
1
nπ x
bn = ∫ f ( x)sin
dx
l c
l
=
2
3
c+ 2 l
∫
c
(2x − x 2 )sin
2nπ x
dx
3
2  9cos2 nπ (− 4)9
2(27)

− 2 2 sin2 nπ + 3 3 cos2nπ 

3  2 nπ
2n π
8n π

2
27

= 0 + 0 + 2 × 3 3 cos0
3
8n π

2 9
27
27  3
= 
+ 3 3 − 3 3 =
3  2nπ 8n π 8 nπ  nπ
∞
a
nπ x ∞
nπ x
∴ f ( x) = 0 + ∑ a n cos
+ ∑ bn sin
2 n =1
l
l
n =1
∞
∞
−8
2 nπ x
3
2nπ x
= 0 + ∑ 2 2 cos
+∑
sin
3
3
n =1 n π
n =1 nπ
∞
∞
8
1
2 nπ x 3
3
2 nπ x
∴ f ( x ) = − 2 ∑ 2 cos
+ ∑ sin
π n =1 n
3
π n =1 n
3
=
π 2 ∞ cos nx
π − x 
13) If f ( x ) = 
,
show
that
f
(
x
)
=
+∑
in the

12 n =1 n 2
 2 
range of (0, 2 π)
Solution : It is an even function ∴b n = 0
2
1 2π
1 2π  π − x 
f
(
x
)
dx
=

 dx
π ∫0
π ∫0  2 
2
an =
2π
1 π − x 
=


4π  3( −1)  0
1
=−
[( −π )3 − π 3 ]
12π
386
−1
π2
=
( −2π 3 ) =
12π
6
2π
1
a n = ∫ f ( x )cos nxdx
π 0
College Mathematics
Fourier Series
π
2 eax + e− ax
= ∫
c os nxdx
π 0
2
=
1 2π  π − x 
= ∫
 cos nxdx
π 0 2 
2
1  π − x  sin nx

( π − x )  − cos nx    1   − sin nx 

  +  −  

2
2  n
   2   n 3  0
2π
2
=
+

 n
1 2π
π
=
= 2
2
π 2n
n
∴ The Fourier series is
∞
∞
a
∴ f ( x) = 0 + ∑ a n cos nx + ∑b n sin nπ x
2 n =1
n =1
π 
2
π 2 ∞ co s nx
+∑
+0
12 n =1 n 2
π 2 ∞ co s nx
∴ f ( x) =
+∑
12 n =1 n 2
14) Find the fourier Series expansion of cosh ax in(-π ,π )
Solution : f(x) = cosh ax
π
1
a0 = ∫ cosh axdx
π −π
=
π
2 π
2  sinh ax 
= ∫ cosh axdx = 
π −π
π  a 0
2
=
sinh aπ
πa
π
1
a0 = ∫ cosh ax cos nxdx
π −π
387
π

2 π ax
e
c
os
nxdx
+
e− ax cos nxdx 
∫
∫
2π  0
0

π
 ax  a cos nx + n sin nx  − ax  a sin nx − a cos nx 
+e 

e 
a2 + n2
a 2 + n2


 0
 
1  a cos nx + n sin nx  − ax  a sin nx − a cos nx 
= e ax 
+e 

π
a2 + n2
a2 + n2




0
 a cos nx + n sin nx 
 n sin0 − a cos0 
− e0 
 −e 

2
2
a
+
n
a2 + n2




aπ
− aπ
n
1 e a
e a(−1)
1
1 
=  2 2 ( −1) n − 2
− 2
+ 2

2
2
π a +n
a +n
a + n a + n2 
1 a ( −1) n aπ
=
(e − e− aπ )
2
2
π a +n
2 a (− 1)n
i.e., a n =
sinh aπ
π (a 2 + n 2 )
bn = 0
∴ The Fourier series for cosh ax is
∞
a
∴ cosh ax = 0 + ∑ a n cos nπ
2 n =1
1
=
π
=
∞
1
2a (−1) n
sinh π a + ∑
sinh aπ cos nπ
2
2
πa
n =1 π (a + n )
Exercise
1. Find the half – range Fourier cosine series for f(x) = x in
0< x≤π
388
College Mathematics
1
0< x < 1
 −x
2
2. Prove that f ( x ) =  4
1 < x < 1,
x− 3
4
2


 nπ  
4sin 

 1
 2   sin nπ x
the sine series is = = ∑ 
[1− ( −1)n ] −
n2π 2 
 2nπ




3. Find the half – range Fourier sine series for
f(x) = ex in the interval (0, 1)
4. Find the half – range cosine series for
a

0<x<
 x
2
f ( x) = 
a
a − x
< x < a,

2
5. Obtain a half – range cosine series for f(x) 2x – 1 for 0 < x < 1.
Hence show that
π2 1 1 1
= + + + K∞
8 12 32 52
6. Find a Fourier sine series for
1

0<x<
1
2
a) f ( x ) = 
1
0
< x<1

2
b) f ( x) = x(π − x) in 0 < x < π
7) Expand f(x) =1 –x 2 , -1 < x < 1 in a fourier series.
(N 2001)
−x
8) Obtain the Fourier series for f ( x ) = e
in (0,2π )
(N 2000)
2
9) Obtain the Fourier series for f(x) = x in (-π ,π )
(N 2001)
− ax
10) Obtain the Fourier series for f ( x ) = e
in ( −π , n )
Fourier Series
389
and hence deduce that cos ehx =
2 ( −1)
∑ n2 + 1
π
n
(A 2001)
11) Prove that in 0 < x < l
1 4l 
πx 1
3π x 1
5π x

x = − 2  cos + 2 cos
+ 2 cos
+ K
2 π 
l
3
l
5
l

and deduce that
Answers
1
π
1 π4
(i) ∑
=
(ii)
∑ n4 = 90
(2n − 1) 4 96
π
2
1) f ( x) = − ∑ 2 (− 1)n − 1 cos nx
2
nπ
n
3) f ( x ) = 2π ∑
1 − ( −1)n  sin π x
2 2 
1+ n π
a 8 1
2π x 1
6π x
1
10π x

4) f ( x ) = − 2  2 cos
+ 2 cos
+ 2 cos
+ K
4 π 2
a
6
a
10
a

2
nπ
6) a) f ( x) = ∑ (1 − cos )sin π x
nπ
2
8
sin(2 n − 1) x
b) f ( x ) = ∑
π
(2 n − 1)3
4
π2 ∞ 4
− ∑ 2 cos nπ cos nx
3
1 n
sin ax ∞ 2a sinh aπ
10) e− ax =
+∑
( −1) n cos nx
2
2
aπ
π
(
a
+
n
)
1
∞
2 a sinh aπ
+∑
( −1) n sin nx
2
2
π
(
a
+
n
)
1
9) x2 =
EXERCISE
A. Define Half range a) cosine
b) sine series
390
College Mathematics
1. Find the cosine and sine series for f(x) = x in 0 ≤ x ≤ π and
π2 1 1 1
hence show that
= + + +K
8 12 32 52
2. Obtain the Fourier series for the periodic function f(x) defined
for − π < x < 0
 1− x
by
f (x) = 
for
0< x<π
1 + x
π2 1 1 1
and hence show that
= + + +K
8 12 32 52
3. Obtain the Fourier series of f(x) defined by
π

in − π < x ≤ 0
 x + 2
f ( x) = 
π − x
in
0≤ x <π
 2
4. Prove that the Fourier series expansion of x(π - x) defined in
π 2  cos2 x cos4 x cos6x

the interval (0, π ) is
−  2 + 2 + 2 + K
6  1
2
3

5. Obtain the Fourier series for the function
 x2
for 0 < x < π
f ( x) =  2
for π ≤ x < π
− x

 x
6. f ( x) = 
π − x

in 0 < x <
π
2
π
< x <π
2
4  sin x sin3 x sin5 x

Show that (i) f ( x ) =  2 −
+
− + K
π 1
32
52

π 2  cos2 x cos6 x cos10x

(ii) f ( x ) = −  2 +
+
+ K
4 π 1
32
52

in
Fourier Series
391
πx
in 0 ≤ x ≤ 1

7. If f ( x ) = 
in
1≤ x ≤ 2
π (2 − x)
in the interval (0, 2) find the Fourier series of f(x)
x
in (0, l )

8. If f ( x ) = 
find the Fourier series
in
(l , 2l )
 x − 2l
in (-π, π )
9. Find the half-range cosine series for sinx in (0, π)
10. Find the half – range sine series for f(x) = 2x – 1 in (0, 1)
11. Find the half – range cosine series for f(x) = x2 on (0, π )
12. Find the half – range sine series for f(x) = x2 in (0, π)
13. Find the Fourier series for f(x) = 1 + x + x2 in (- π, π)
14. Express f(x) = 1 + x2 as a Fourier series in (0, π)
 x2
in (−π ,0)
15. Expand f ( x ) = 
as a Fourier series
in
(0, π )
0
in (- π , π )
∞
 (− 1)n π [(−1)n − 1]2 
π 2 ∞ 2(−1)n
+∑
cos nx + ∑ 
−
 sin nx
6 n =1 n 2
n
π n3
n =1 

for − π < x < 0
0
16. If f ( x ) = 
for
0< x <π
 sin x
1 sin x 2 ∞ cos2nx
Prove that +
and hence show that
− ∑ 2
π
2
π n =1 4 n − 1
1
1
1
1
−
+
− + K = (π − 2)
1.3 3.5 5.7
4

π π   π  π
for
0 ≤ x <  , f   =
 0
2  2   2 4
17. If f ( x ) = 
π
π
for
< x≤π
 2
2
prove that
π
cos3 x cos5x
f ( x ) = − cos x +
−
+ −K and hence show that
4
3
5
392
College Mathematics
1 1 1
π
1− + − + − K =
3 5 7
4
2
x

1 + π for − π ≤ x ≤ 0
18. f ( x ) = 
1 − 2 x for 0 ≤ x ≤ π
 π
8  cos x cos3 x cos5x

P rove that f ( x ) = 2  2 +
+
+ K
2
2
π  1
3
5

19. For f ( x) = x in ( −π ,π ) , prove that
π 4
cos3 x cos5x

−  cos x +
+
+ K
2 π
9
25

20. For f ( x) = x sin x in ( −π ,π ) find the Fourier series and hece
1
1
1
1
deduce that
−
+
− + K = (π − 2)
1.3 3.5 5.7
4
21. Prove that the Half – range fourier sine series for f(x) = π - x
∞
2
in (0, π) is ∑ sin nx
[2 Marks]
1 n
22. Prove that the Half range sine series for f(x) = ex in (0, 1) is
2π n
∑ 1+ n 2π 2 [1 −( −1) n e]sin nπ x
f ( x) =
ANSWERS
π 4
1
1

− cos x + 2 cos3 x + 2 cos5 x +K
2 π
3
5

1
1


(ii )2 sin x − sin2x + sin3x − + K
2
3


π +2 4 1
1
1

2.
−  2 cos x + 2 cos3 x + 2 cos5x + K
2
π 1
3
5

1. (i )
Fourier Series
4  cos x cos3 x cos5 x

3.
+
+
+ K

2
2
2
π 1
3
5

393
 cos x cos3x cos5x

5. −π 2 − 8  2 +
+
+K
2
2
3
5
 1

2
2
2
2
2  3π
4
π sin2 x  3π
4
π s i n 4x

6.
−
sin
x
+
+
−
sin
x
+
+K

3 

3 

π  1 1 
2
4
 3 3 

π 4  cosπ x cos3π x cos5π x

7.
−
+
+
+ K
2 π  12
32
52

n +1
∞
2l
(− 1)
nπ x
8.
sin
∑
π n=1 n
l
∞
2 4
cos2nx
9.
− ∑ 2
π π n =1 4n − 1
2  sin2π x sin2π x sin6π x

10. − 
+
+
+ K
π 1
2
3

2 ∞
n +1
π
( −1) 4cos nx
11.
∑
3 n =1
n2
∞
12.
 (− 1)n +1 2π
∑
+
[( −1) n − 1]4 
 sin nx
π n3

n2

∞
π 2 ∞ ( −1) n 4
(− 1)n +1 2
13. 1 +
cos
n
π
+
sin nx
∑
∑
3 n =1 n 2
n
n =1
∞
 [( −1) n − 1]4 2[( −1) n (1+ π 2 ) − 1] 
14. ∑ 
−
 sin x
n
πn
n =1 

1
2
2
2
cos2 x +
cos3 x −
cos4x −K
20. 1 − cos x −
2
1.3
2.4
3.5
n =1
5.9 Finite Sine and Cosine Transforms
Definitions: If f(x) is a sectionally continuous function over
some finite interval (0, l) of the variable x, then the finite Fourier
Sine and Cosine Transforms of f(x) over (0, l)are defined by
394
College Mathematics
l
Fs ( n) = ∫ Fs dx where n = 1,2,3, K
0
 nπ x 
Fs ( n) = ∫ f ( x)cos 
 dx where n = 1,2,K
 l 
0
In the interval (0, π ) we have
l
and
π
Fs ( n) = ∫ f ( x )cos nx dx where n = 1,2,3K
0
π
and Fc ( n) = ∫ f ( x)cos nx dx n = 1,2,K
0
Using Fourier Sine and Cosine half-range series, the inverse
transforms in the interval (0, l) are given by
2 ∞
 nπ x 
f ( x) = ∑ Fs (n ) sin 

l n =l
 l 
1
2 ∞
 nπ x 
and f ( x) = Fc (0) + ∑ Fc (n )cos 

l
l n =l
 l 
l
where Fc (0) = ∫ f ( x)d x
0
In the interval (0, π ), the above result becomes
2 ∞
F ( x) = ∑ Fs (n )sin nx
π n =l
1
2 ∞
F ( x) = Fc (0) + ∑ Fc (n )cos nx
π
π n =l
π
where Fc (0) = ∫ f ( x)d x
0
NOTE : If the interval is not given in the problems, then we have
to take the interval as (0, A).
WORKED EXAMPLES :
Fourier Series
395
(1) Find the finite Fourier sine and cosine transforms of f(x)
=1 in (0, π)
Solution : Given : f(x) =1, in (0, l) = (0, π)
→
(1)
 nπ x 
We know Fs ( n) = ∫ f (x ) sin 
 dx
 l 
0
l
π
= ∫ 1sin nxdx
[using (1)]
0
π
 Cosnx 
= −
n  0

1− cosnπ
=
n
1 − (−1)n
Fs ( n) =
n
l
 nπ x 
Also, Fc (0) = ∫ f (x )Cos 
 dx
 l 
0
π
= ∫ 1C os nxdx
[using (1)] → (2)
0
π
 Sinnx 
=

 n 0
F(n) = 0 if n = 1, 2, 3, . . .
π
If n= 0 then F c(0) = ∫ Cos0dx
[using (2)]
0
= [ x ]0 = π
(2) Find the finite Fourier sine and Cosine transforms of f(x) = x
in (0, l).
l
 nπ x 
Solution : We know Fs ( n) = ∫ F (x )sin 
 dx
 l 
0
π
396
College Mathematics
 nπ x 
= ∫ xsin 
 dx (using given data)
 l 
0
Using Bernoullie’s rule, we get
l
 
nπ x  
nπ x  
  − cos
  − sin

l  
l 


Fs ( n) = x
−1
2


 nπ 
 nπ 






 l 
 l 


l
l
nπ x
= ∫ x cos
dx
(using given data)
l
0
Using Bernoullie’s rule, we get
l
l

nπ x  

nπ x
 − cos l  
 sin

l −1
Fc ( n) =  x
2 2
n
π
nπ




l
l
0
=
l
nπ
l
nπ x 
l

 x sin l  + n2π 2

0
2
l
nπ x 

cos l 

0
397
l
If n = 0, Fc (0) = ∫ xdx
0
l
0
nπ x 
l2 
nπ x 

x
cos
+
− sin
2 2 


l 0 n π 
l 0

−l
l2
=
[ l cos nπ − 0] + 2 2 [sin nπ − sin0]
nπ
nπ
2
−l
=
[ −1]n
nπ
l 2 (−1)n +1
Fs ( n) =
where n = 1, 2, 3, . . .
nπ
l
nπ x
Now Fc ( n) = ∫ f ( x )cos
dx
l
0
=
−l
nπ
l
Fourier Series
l
l2
=
(0 − 0) + 2 2 (cos nπ − cos0)
nπ
nπ
2
l
Fc ( n ) = 2 2 [(− 1) n − 1] where n = 1, 2, 3, . . .
nπ
 x2 
= 
 2 0
l2
Fc (0) =
2
(3) For the func tion f(x) = x, find the finite Fourier sine and
Cosine transforms in (0, π )
Solution Given : f(x) = x, (0, l) =(0, π )
→
(1)
 nπ x 
We know Fs ( n) = ∫ f ( x)sin 
 dx
 l 
0
l
π
= ∫ x sin nxdx
[using (1)]
0
Using Bernoullie’s rule, we get
π
  − cos nx   − sin nx 
Fs ( n) =  x 
 − 1

2
n
  n
 0
 
1
π
= − [ x cos nx]0 (Q sin nπ = sin0 = 0)
n
1
= − [π cos nπ − 0 ]
n
n +1
(−1) π
Fs ( n) =
where n = 1, 2, 3, . . .
n
Also ,
398
College Mathematics
 nπ x 
Fc ( n) = ∫ f ( x)cos 
 dx
 l 
0
Fourier Series
399
2
l
l
= ∫ x cos nxdx
[usaing (1)]
0
2
π
  sin nx   − cos nx  
= x 
 − 1

2
 0
  n   n
1
π
Fc ( n ) = 2 [ cos nx ]0
n
1
Fc ( n) = 2 (cos nπ − cos0)
n
1
Fc ( n) = 2 {(−1)n − 1}
n
If n = 2, 4,6, . . . , Fc(n) = 0
−2
If n = 1, 3, 5, . . . ., Fc ( n) = 2
n

nπ x
nπ x
nπ x 
− sin
cos
 − cos

2 − (2 x)
2 + (2)
2 
Fs ( n) =  x2
2
3
nπ

n 2π
n 3π 


2

4
8
0
(using Bernoullie’s rule)
2
−2  2
16   nπ x  
 nπ x  
=
x cos 
  + 3 3 cos 
  (Q sin nπ = sin0 = 0)

nπ 
 2  0 n π   2  0
−2
16
=
(4cos nπ − 0) + 3 3 (cos nπ − cos0)
nπ
nπ
8
16
Fs ( n) =
(−1)n +1 + 3 3 [(−1)n − 1]
nπ
nπ
(5) Find the finite Fourier sine and Cosine transforms of
f(x) = π-x.
Solution : Since the range is not given we shall take the interval
as (0, π)
π
We know Fs ( n) = ∫ (π − x)sin nxdx(Q f ( x) = π − x and l = π )
π
If n = 0, Fc (0) = ∫ xdx
0
Using Bernoullie’s rule,
0
π
π
 x2 
= 
 2 0
π2
2
(4) Find the finite Fourier sine transform of f(x)= x2 in (0, 2)
Solution Given : f(x) = x2 , (0, l) = (0, 2)
→
(1)
l
 nπ x 
We know Fs ( n) = ∫ f (x ) sin 
 dx
 l 
0
Fc (0) =
 nπ x 
= ∫ x 2 sin 
 dx
 2 
0
Using Bernoullie’s rule,
2
[using (1)]

 − cos nx 
 − sin nx  
Fs ( n) = (π − x ) 
 − ( −1) 

2
n


 n
 0

−1
π
= [(π − x)cos nx ]0 (Q sin nπ = sin0 = 0)
n
−1
= [(0 − π )]
n
π
Fs ( n) =
n
π
Also Fc ( n ) = ∫ (π − x)cos nxdx
0
Using Bernoullie’s rule,
400
College Mathematics
π

 sin nx 
 − cos nx 
Fs ( n) = (π − x) 
 − (− 1) 

2
 n 
 n
 0

−1
π
= 2 [cos nx ]0 (Q sin nπ = sin0 = 0)
n
−1
= 2 [ cos nπ − cos0]
n
−1
Fc ( n ) = 2 (− 1) n − 1 where n ≠ 0.
n
π
When n = 0, Fc (0) = ∫ (π − x) dx
Fourier Series
π
−1
2
π
(2x − x 2 )cos nx  − 3 [cos nπ ]0
=
0
n
n
−1
−2
=
(2π − π 2 )cos nπ − 0  3 [cos nπ ]π0
n
n
−1
−
2
(2π − π 2 )(− 1) n  3 [( −1) n − 1]
=
n 
n
n +1
2
(−1) (2π − π ) −2
Fs ( n) =
[( −1) n − 1] where n = 1, 2, 3, . .
3
n
n
l
 nπ x 
Also Fc ( n) = ∫ f ( x)cos 
dx
 l 
0
401
0
π
= ∫ (2 x − x 2 )cos nxdx

x2 
= π x − 
2 0

0
π2
2
π2
Fc (0) =
2
(6) Find the finite Fourier Sine and Cosine transforms of
f(x) = 2x - x2
Solution :Since the range is not given, we shall take the interval
as (0, π)
Given : f ( x) = 2x − x 2 ,(0, l ) = (0,π )
→ (1)
=π2−
l
π
1

=  2 (2 − 2 x) cos nx 
n
0
=
=
2
2
n
2
n2
[(1 − x)cos nx]π0
[ (1 − π )cos nπ − cos0]
π
If n = 0, Fc (0) = ∫ (2x − x 2 )dx
π
= ∫ (2 x − x 2 )sin nxdx
π

 − cos nx 
 − sin nx 
2  sin nx 
Fc ( n) = (2 x − x ) 
 − (2 −2 x ) 
 + ( −2) 

2

 n 
 n 
 n 3  0
where n ≠ 0
 nπ x 
Fs ( n) = ∫ f ( x)sin 
dx
 l 
0
We know
Using Bernoullie’s rule,
[using (1)]
0
Using Bernoullie’s rule,
π

 − sin nx 
 cos nx 
2  − cos nx 
Fs ( n) = (2 x − x ) 
 − (2 − 2 x) 
 + (−2)  3  
2

 n 
 n 
 n 0
0
π

x3 
= x2 − 
3 0

3
π
Fc (0) = π 2 −
3
402
College Mathematics
7) Show that the finite Fourier sine transform of f(x) =x(π -x) is
4
if n is odd and 0 if n is even.
n3
Solution : Since the range is not given, we shall take the interval
as (0, π)
Given : f(x) =x( π-x), (0, l) = (0, π )
→
(1)
 nπ x 
We know Fs ( n) = ∫ f ( x)sin 
dx
 l 
0
l
[using (1)]
0
π
= ∫ (π x − x2 )sin nxdx
0
Using Bernoullie’s rule,
π
π

 − sin nx 
 cos nx 
2  − cos nx 
 − (π − 2x ) 
 + (−2)  3 
2
 n 
 n

 n  0

π
−1
2
=
(π x − x 2 )cos nx  − 3 [cos nx ]π0 (Q sin nx = sin0 = 0)
0
n
n
−1
2
= [0 − 0] − 3 [cos nπ − cos0]
n
n
2
= − 3 [(−1)n − 1]
n
2
Fs ( n) = 3 [1 − (−1)n ]
n
2
If n is odd, Fs ( n) = 3 [1 − ( − 1)]
n
4
Fs ( n ) = 3
n
2
If n is even, Fs ( n) = 3 [1 − 1] = 0
n
Fs (n) = (π x − x ) 
403
2
 x
Given : f ( x) = 1 −  ,(0, l ) = (0,π )
 π
l
 nπ x 
We know Fs ( n) = ∫ f ( x)cos 
dx
 l 
0
l
= ∫ x(π − x )sin nxdx
Fourier Series
4
Thus, Fs (n ) = 3 if n is odd and 0 if n is even.
n
(8) Show that the finite Fourier Cosine transform of
 2
, if n = 1,2,3, K
2
 x   π n 2
f ( x) = 1 −  is 
 π  π
,
if n = 0.
 3
Solution : We shall take the interval as (0, π)
→ (1)
2
x

= ∫  1 −  cos nxdx
π
0
Using Bernoullie’s rule,
 x 
Fc ( n ) =  1 − 
 π 

2
( )
sin nx
n
[using (1)]
 x  −1 
− 2  1 − 

 π  π 
π
( )
cos nx
n
2
 −1 
+2 2 
π 
(

−2  x 
=
1 −  cos nx  (Q sin nπ = sin0 = 0)
2 
π n  π 
0
−2
= 2 [0 − cos0]
πn
2
Fc ( n ) = 2 for n ≠ 0
πn
π
2
 x
If n = 0, Fc (0) = ∫  1 −  dx
π
0
− sin nx
n
3
)
π


0
404
College Mathematics
π
  x 
 1 − π  

= 
−
3


 π 
0
−π
=
3
π 2
π
 x 3 
1 −  
 π   0
When n = 0, Fc (0) =
 0,

( n −1) 2
F c (n ) = 
2
,
 (−1)
n
[Using (2) & (3)]
=
∫
f ( x)cos nxdx
π 2
π 2
π
0
π 2
∫ 1.cos nxdx +
∫ ( −1)cos nxdx
→
(1)
(using given data)
π
π
 sin nx  2  sin nx 
=
−
 n 0
 n π 2
π
2
→
for
for
n = 0,2,4,6, K
n = 1,3,5, K
(10) Find the finite Fourier Cosine transform of the function
1, for 0 < x ≤ π

2
f ( x) = 
 0, for π 2 < x < π
Solution : Given : (0, l) = (0, π)
We know
l
 nπ x 
Fc ( n) = ∫ f ( x)cos 
dx
 l 
0
π
= ∫ f ( x)cos nxdx
0
(2)
[using (1)]
Thus,
π
0
∫ (−1) dx
π
π
 
=  − 0  − π − 
2
2
 
Fc(0) = 0
0
f ( x)cos nx +
→
π 2
π
π
π 2
405
π
= [ x]0 2 − [ x ]π
= ∫ f ( x)cos nxdx
∫
∫
1.dx +
0
−π
=
[0 − 1]
3
π
Fc (0) =
3
(9) Find the Fourier Cosine transform of f(x) defined by
π

 1, 0 < x < 2
f ( x) = 
 −1, π < x < π

2
Solution : Given (0, l) = (0, π)
l
 nπ x 
We know Fc ( n ) = ∫ f ( x)cos 
dx
 l 
0
=
Fourier Series
1
nπ 
π   1 
= sin n   − 0  − 0 − sin
n
2 
2  n
2
nπ
Fc ( n ) = sin
for n ≠ 0
n
2
(3)
406
College Mathematics
π 2
=
∫
π
f ( x)cos nxdx
π 2
π 2
=
∫
f ( x)cos nx +
0
∫ 1.cos nxdx +
0
π
∫ 0cos nxdx
(using given data)
Fourier Series
407
(11) Find the finite Fourier Cosine and Sine transforms of the
function f(x) = eax in (0, l).
l
 nπ x 
Solution : We know Fc ( n) = ∫ f ( x)cos 
dx
 l 
0
π 2
 nπ x 
Fc ( n) = ∫ e ax cos 
dx
 l 
0
l
π 2
=
∫ cos nxdx
→
(1)
l
0
Using ∫ eax cos bxdx =
π
 sin nx  2
=

 n 0
1  nπ

= sin
− sin0
n
2

1  nπ 
= sin
n
2 
0, if n iseven


(n − 1) 1
F c (n ) = 
2
, if nisodd
 (−1)
n
π
If n = 0 Fc (0) =
0
l 2a
= 2 2
l a + n 2π 2
nπ x 
nπ l
 ax
e cos l  + l 2 a 2 + n 2π 2

0
l
l
 ax
 nπ x  
e sin  l  

 0

l 2a
nπ l
eal cosnπ − 1 + 2 2
eal sin nπ − 0 
2 2
2 2 
2 2 
l a +n π
l a +n π
l 2a
Fc ( n ) = 2 2
[(− 1)n e al − 1] where n = 1,2,3, . . .
l a + n 2π 2
when n = 0, we get
=
∫ cos0 dx
[using (1)]
2
∫ 1.dx
l
Fc ( n) = ∫ eaxdx
0
π
Fc (0) =
2
Thus,

 π 2,

Fc ( n) = 0,
1
(n −1)
 ( −1) 2 ,
n
eax (a cos bx + b sin bx )
we get
a2 + b2
l
0
=
[using (1)]
0
l
 eax 
= 
 a 0
for
n=0
n = 2,4,6, K
for
(1)
 ax 
 nπ x  nπ
 nπ x   
sin 
 e  a cos 
+

 l  l
 l 


Fc ( n ) =

n 2π 2

2
a
+

2

l

0
2
π
→
n = 1,3,5, K
Fc (0) =
eal − 1
a
 nπ x 
Also, Fs ( n) = ∫ f ( x)sin 
dx
 l 
0
l
408
College Mathematics
 nπ x 
Fs ( n) = ∫ eax s in 
dx
 l 
0
l
l
Using
ax
∫ e sin bxdx =
0
eax (a sin bx − b cos bx)
, we get
a 2 + b2
l
 
 nπ x  n π
 nπ x  
cos 
  a sin 
−
 

 l  l
 l  
Fc ( n ) = eax 

n 2π 2
 
2

a
+
2
 
 0
l
 
l 2a
= 2 2
l a + n 2π 2
l
 ax
nπ l
 nπ x 
 e sin  l  − l 2 a 2 + n 2π 2

 0

l
 ax
 nπ x  
 e cos  l  

 0

nπ l
eal cos nπ − 1
l a + n 2π 2 
nπ l
Fs ( n) = 2 2
[1 − ( − 1)n e al ] where n = 1,2,3, . . .
l a + n 2π 2
=
2
2
(12) Find f(x) in (0, π) given that the finite Fourier Cosine
cos(2 nπ 3)
transform is Fc ( n) =
(2 n + 1)2
Solution : In the interval l = π , we know
1
2 ∞
 nπ x 
f ( x) = f c (0) + ∑ f c (n )cos 

l
l n =1
 l 
Here l = π
1
2 ∞
→
(1)
f ( x) = f (0) + ∑ fc ( n)cos nx
π
π n =1
cos(2 nπ 3)
Given : f c (n ) =
(2 n + 1)2
∴ fc(0) = 1
Using these in (1), we get
Fourier Series
1 2 ∞ cos(2nπ 3)
f ( x) = + ∑
cos nx
π π n =1 (2 n + 1)2
409
(13) Find f(x) in (0, π) given that the finite Fourier sine transform
1 − cos nπ
is f s (n ) =
n 2π 2
2 ∞
 nπ x 
Solution : We know f s (n) = ∑ f s ( n )sin 
 in (0, l)
l n =1
 l 
Here l = π
2 ∞
f ( x) = ∑ Fs (n)sin nx
π n =1
2 ∞ (1 − cos nπ )
(Using given data)
f ( x) = ∑
sin nx
π n =1
n 2π 2
Q1 − cos nπ = 1− ( − 1) n = 0
if n is even and 2 if n is odd.
Using this, we get
2 ∞
2
f ( x) =
sin nx
∑
2 2
π n =13,5,K n π
f (x) =
4
π3
 sin x sin3 x sin5x

 (1) 2 + (3) 2 + (5) 2 +K


(14) Find f(x) in 0 < x < 4 Given that Fc(0)=16,
3
f c (n ) = 2 2 ( −1) n − 1 where n = 1, 2, 3, . . .
nπ
1
2 ∞
 nπ x 
Solution : We know f ( x) = f c (0) + ∑ f c (n )cos 
 in
l
l n =1
 l 
0<x< l
Given : l = 4
1
1 ∞ 3
 nπ x 
∴ f ( x ) = (16) + ∑ 2 2 [(−1)n − 1]cos 

4
2 n =1 n π
 4 
(using given data)
410
Fourier Series
1
 1 − C os nπ
S n  x(π − x)  =
n3
2

College Mathematics
−2
 nπ x 
c os 

2
 4 
n =1,3,5,K n
∞
f ( x) = 4 +
3
2π 2
f ( x) = 4 −
3 1
πx 1
3π x 1
5π x

cos
+ 2 cos
+ 2 cos
+K
2  2
π 1
4 3
4
5
4

∑
[NOTE : Sn [f(x)] denote the finite Fourier sine transform of f(x)
and Sn -1 is its inverse.
Similarly C n [f(x)] denote the finite Fourier Cosine transform of
f(x) and Cn -1 is its inverse.]
1 − cos nπ
(15) Show that Sn−1 
n3

 1
 = 2 x(π − x)
1 − cos nπ
1

Solution : We shall prove that
= S n  x(π − x)
3
n
2


Here l = π
π
0
=
π
π
π
1
+ ∫ [ co s{kπ − ( k + n) x}]dx
20
π
π
2
π
 −(π x − x )cos nx 2cos nx 
−
3

2 
n
n
0
1 −2cos nπ
2
= 
+ 3
3

2
n
n 
1
2
1
[co s( kπ − kx + nx) + cos(kπ − kx − nx)]dx
2 ∫0
π
 − cos nx 
 − sin nx 
 cos nx 
= (π − x ) 
 − (π − 2 x) 
 + ( −2)  3 
2
2
 n 
 n 
 n  0
=
π
1
= ∫ [co s{kπ − ( k − n) x}]dx
20
1
x(π − x )sin nxdx
2 ∫0
Using Bernoullie’s rule,
 1 x (π − x) 
Sn
 2

1
1 − cos nπ  1
∴ Sn−1 
 = 2 x(π − x)
n3


 k sin kπ 
(16) Show that Cn−1  2
= cos k(π − x) where k ≠ n.
2
 k − n 
k sin k π
Solution : To prove that Cn [cos k (π − x )] = 2
k − n2
Here l = π.
We know
[cos k (π − x)] = ∫ cos k (π − x)cos nxdx
1
1

Sn  x(π − x) = ∫ x(π − x )sin nxdx
2
 02
=
411
π
1  sin[kπ − ( k − n ) x]  1  sin[ kπ − ( k + n ) x] 
= 
 + 

2
− (k − n )
− (k + n)
0 2 
0
−1
−1
=
[sin nπ − sin kπ ] +
[sin(− nπ ) − sin k π ]
2(k − n )
2(k + n )
1  sin kπ sin kπ 
= 
+
2 k−n
k + n 
k sin kπ
Cn [cos k (π − x )] = 2
k − n2
 k sin kπ 
Cn−1  2
= cos k(π − x)
2
 k − n 
412
College Mathematics
5.12 Finite Sine and Cosine Transforms of Derivatives.
In the interval (0, l) , we prove the following results.
− nπ
(1) Fs  f ( r ) ( x )  =
Fc  f (r −1) ( x ) 
l
nπ
(2) Fs  f ( r ) ( x)  = ( −1) n f (r −1) (l ) − f ( r −1) (0) +
Fs  f (r −1) ( x) 
l
Proof : Fourier finite sine transform is given by
l
 nπ x 
Fs  f ( r ) ( x)  = ∫ f (r ) ( x)sin 
 dx
 l 
0
Using integration by parts,
 ( r −1) ( x )sin  nπ x  − nπ
(r )


 f ( x )  =  f

 l  0 l
l
Fs
=  f ( r −1 ) (l) s i n nπ − f ( r −1) (0)sin0 −
nπ
l
l
∫f
(r −1)
0
l
∫f
( r −1)
Also , Fc  f
l
( x) = ∫ f
( r)
0
 nπ x  dx

 l 
( x )cos 
0
nπ
Fs  f (r ) ( x) = −
Fc  f (r −1) ( x) 
l
(r )
 nπ x  dx

 l 
( x )cos 
→
(1)
Fourier Series
Fc  f 1 ( x)  = [( −1) n f (l ) − f (0)] +
nπ
Fs [ f ( x )]
l
→
(4)
Using r = 2 in (1) and (2), we get
nπ
Fs [ f ′′( x)] = −
Fc [ f ′( x) ]
l
− nπ
n 2π 2
Fs [ f ′( x)] =
[( −1) n f (l ) − f (0)] − 2 Fs [ f ( x )]
l
l
→ (5)
[Using (4)]
Also, Fc [ f ′′( x )] = [( −1) n f ′(l ) − f ′(0)] +
nπ
Fs [ f ′( x )]
l
n 2π 2
Fc [ f ( x )] → (6)[using (3)]
l2
In the interval (0, π), the above results becomes
Fs [ f ′( x)] = −nFc [ f ( x)]
→ (7)
Fc [ f ′′′( x)] = ( −1) n f ′(l ) − f ′(0) −
Fc [ f ′( x) ] = [(−1)n F (π ) − f (0)] + nFs[ f ( x )]
Fs [ f ′′( x)] = −n[(− 1)n f (π ) − f (0)] − n 2 Fs [ f ( x )]
 nπ x 
( x)cos 
 dx
 l 
413
Fs [ f ′′′( x)] = [( −1) f ′(π ) − f ′(0)] − n Fc [ f ( x)]
n
2
→ (8)
→ (9)
→ (10)
Using integration by parts,

 nπ x  + nπ
 f ( r ) ( x )  =  f ( r −1) ( x) cos 


 l 0 l
l
Fc
( r −1)
=  f ( r −1) ()cos
l
nπ − f
(0)cos0 +
nπ
l
l
∫f
( r −1 )
 nπ x  dx

 l 
( x )sin 
0
Fs  f (r −1) ( x) 
nπ
Fs  f (r −1) ( x )  → (2)
l
[NOTE : Using the above results (1) and (2), we obtain the
following results in the interval (0, l)]
Using r = 1 in (1) and (2), we get
nπ
Fs  F 1 ( x )  = −
Fc [ f ( x ) ]
→
(3)
l
Fc  f (r ) ( x ) = ( −1) n f ( r −1) (l ) − f ( r −1) (0) +
WORKED EXAMPLES
(17) By employing the finite Fourier Cosine transform, solve the
equation Y ′ + 3Y = e− x , Y ′(0) = Y ′(π ) = 0 .
Solution : Given : Y ′ + 3Y = e − x
Using finite Fourier Cosine transform, we get,
Fc[Y ′′] +3 Fc[ Y ] = Fc [e −x ]
→ (1)
In the interval, (0, l), we have
n 2π 2
Fc [ f ′′( x)] = (−1) n f ′′(l ) − f ′(0) − 2 Fc [ f ( x )]
l
Here (0, l) = (0, π ) and Y = f(x)
∴ Fc [ y′′] = ( −1) n y′ (π ) − y ′(0) − n2 Fc ( y )
414
Given : Y ′(0) = Y ′(π ) = 0
College Mathematics
∴ Fc [ y′′] = −n 2 Fc [ y ]
→ (2)
π
Also, Fc[ e− x ] = ∫e − x cos nxdx
0
π
 e− x ( −1cos nx + n sin nx) 
=

12 + n 2

0
π
−1
= 2
e − x cos nx 
0
1 + n2 
−1
e−π cos nπ − 1
= 2
2 
1 +n
−1
(−1)n e −π − 1 for n ≠ 0
1 + n2 
Using (2) and (3) in (1), we get
−1
− n 2 Fc[ y ]+ 3Fc[ y ] = 2
 (−1)n e −π − 1
1 + n2 
1
( −1) n e−π − 1
(n 2 − 3) Fc [ y] = 2
2 
1 +n
+1[(−1) n e−π − 1]
Fc[ y ] =
(1 + n 2 )( n 2 − 3)
This is denoted by fc(n) for n ≠ 0
+ (−1)n e −π −1
∴ f c (0) = 2
(n + 1)( n 2 − 3)
Put n = 0 in (4)
e −π − 1 1 − e −π
∴ f%c (0) =
=
−3
3
Using inverse Fourier Cosine transform,
1
2 ∞
y = f% (0) + ∑ f% (n )cos nx
π
π n =1
Fc[ e − x ] =
2
→ (3)
Fourier Series
415
n −π
∞
1
2
(− 1) e − 1
y=
(1 − e−π ) + ∑ 2
co s nx
3π
π n =1 ( n + 1)(n 2 − 3)
(18) Employing the finite Fourier sine transform, solve the
differential equation
in 0 ≤ x ≤ l , given
2y′′ + y = x 2
y(0) = y(l) = 0.
Solution : 2 y ′′ + y = x 2
Using finite Fourier sine transform, we get
2 Fs [ y ′′] + Fs [ y ] = Fs [ x 2 ]
→
(1)
In (0, l), we get
− nπ
n 2π 2
Fs [ f ′′( x)] =
[( −1) n f (l ) − f (0)] − 2 Fs [ f ( x )]
l
l
2 2
−nπ
nπ
Fs [ y′′] =
[( −1) n y(l ) − y(0)] − 2 Fs [Y ] [Q Y = f ( x)] Using
l
l
Y(0) = Y ( l)= 0, we get
n 2π 2
Fs [ y′′] = 2 Fs [ y]
→
(2)
l
l
 nπ x 
2
Also, Fs  x  = ∫ x 2 sin 
 dx
 l 
0
Using Bernoullies rule,
l
→ (4)
→ (5)
 
 nπ x  
nπ x 
nπ x  


− sin
cos
  − cos 





 l 
l 
l 

Fs  x2  =  x 2 
− 2x 
+
2
nπ

n 2π 2
n 3π 3 


l
l2
l3

0
l
l
nπ x 
2 l3 
nπ x 
 2
x
cos
+
cos
3 3


l 0 n π 
l  0

l 2
2l 3
l cos nπ − 0  + 3 3 [cos nπ − 1]
=−
nπ
nπ
n +1 3
3
( −1) l
2l
Fs [ x 2 ] =
+ 3 3 (−1)n − 1
nπ
nπ
l
=−
nπ
→
(3)
416
College Mathematics
Using (2) and (3) in (1), we get,
 −n2π 2

(−1)n +1 l 3 2l 3
−2  2 Fs [ y ] + Fs [ y ] =
+ 3 3 (−1)n − 1
nπ
nπ
 l

 l 2 − 2n2π 2 
( −1) n +1 l 3
2l 3
(−1)n − 1
F
[
y
]
=
+

 s
2
3 3 
l
nπ
nπ


3
n
 ( −1) n +1 l 3 2 l (−1) − 1  

l2


Fs [ y ] = 
+
 2
3 3
2 2 
nπ
 nπ
  l − 2n π 
Using inverse finite Fourier sine transform, we get
∞
nπ x
y = ∑ Fs ( y)sin
l
n=1
∞
 (− 1)n +1 2 (−1)n − 1 
2l 4
nπ x
y=∑ 2

+  3 3   sin
2 2
nπ
nπ
l

n =1 l − 2 n π 

(19) Using the finite Fourier Sine transform, solve the differential
equation y′′ + ky = x 3 in 0 < x < π given that y(0) = y( π ) = 0 and
k is a non-integral constant.
Solution : Given : y ′′ + ky = x 3
Using finite Fourier sine transform,
Fs [ y ′′] + kFs [ y ] = Fs [ x3 ]
→ (1)
In (0, π )
Fs [ y′′] = − n[( −1) n y (π ) − y (0)] − n2 Fs [ y ]
Using y(0) = y( π ) = 0, we get
Fs [ y′′] = − n2 Fs [ y ]
→ (2)
π
Also, Fs [ x3 ] = ∫ x3 sin nxdx
0
Using Bernoullie’s rule,
π
− cos nx 

 sin nx 
 cos nx 
2  − sin nx 
Fs [ x 3 ] =  x 3 
 − 3x 
 + 6 x  3  − 6  4 
2
 n 
 n 
  n 
 n  0
Fourier Series
417
π
 − x cos nx 6x cos nx 
Fs [ x 3 ] = 
+
3

n
n

0
3
 −π cos nπ 6π cos nπ 
=
+
3

n
n


2
6 π 
= π cos nπ  3 − 
n 
n
 6 π2 
Fs [ x 3 ] = ( −1) n π  3 − 
→ (3)
n 
n
Using (2) and (3) in (1) we get
(6 − n 2π 2 )
− n 2 Fs [ y]+ kFs [ y] = ( −1) n π
n3
(− 1)n π (6 − n 2π 2 )
∴ Fs [ y ] =
→ (4)
n3 ( k − n2 )
Using inverse finite Fourier sine transform, we get
2 ∞
y = ∑ Fs ( y)sin nx
π n =1
3
∞
y = 2∑
n =1
(−1)(6 − π 2 n 2 )
sin nx
( k − n 2 )n 3
418
College Mathematics
EXERCISES
1. Find the finite Fourier Sine transforms of the following
(a) x in (0, 1)
(b) 2-x in (0, 2)
(c) ax – x2 in (0, a)
(d) cos x
(e) e-x
2. Find the finite Fourier Cosine transforms of the following
(a) x2 in (0, 1)
(b) x (3 - x) in (0, 3)
x
(c) 1 − in (0, a)
a
3. Find the Fourier Cosine transform of the function
π − x in 0 < x < π

2
f ( x) = 
 x in π 2 < x < π
4. Find the Fourier Cosine transform of the function
 1 in 0 < x < 1
f (x) = 
 0 in 1 < x < 2
x
( −1) n +1
5. Show that the Finite Fourier Sine transform of
is
π
n
6. Show that the finite Fourier Sine transform of f ( x ) = e a x in
n
(0, π ) is 2
[1 + (−1)n +1 eax ]
a + n2
7. Find the finite Fourier Sine transform of
(i) sin ax and (ii) cos ax.
8. Find the finite Fourier Cosine transform of sin ax.
9. Find f(x) in (0,π ) given.
1 − cos nπ
(a) Fs ( n ) =
n3
Fourier Series
π
n
1 − cos nπ
π2
(c) Fc ( n ) =
,
n
=
1,2,3,.....,
F
(0)
=
c
n3
2
π
π
(d) Fc ( n ) = 2 , n = 1,2,3,....., Fc (0) =
2n
3
10. If k is a constant and 0 < x < l, then prove that
419
(b) Fs ( n) =

 cosh k (l − k )
kl 2
Cn−1  2 2 2 2  =
sinh al
k l + n π 
11. Solve the following differential equations.
(a)
y ′′ − 2 y = e −2 x , 0 ≤ x ≤ π , given y ′(0) = y ′ (π ) = 0
using Fourier finite Cosine transform.
(b) y′′ − y = xsin x in 0 ≤ x ≤ π given y(0) = y (π ) = 0
using Fourier finite Sine transform.
(c) y ′′ − y = e x in 0 ≤ x ≤ π , given y(0) = y(π ) = 0 using
Fourier finite Sine transform.
(d) 2 y ′ + y = sin 2 x in 0 ≤ x ≤ π , given y (0) = y (π ) = 0
using Fourier finite Sine transform.
x
(e) y′′ + y = sin , 0 < x < π give n y ′(0) = y ′(π ) = 0 using
2
Fourier finite Cosine transform.
ANSWERS
n +1
1.
( −1)
4
(b)
nπ
nπ
{1 − ( − 1) n }2 a3
(c)
(d) 0 for n = 1 and
n3π 3
n
{1 − (−1)n +1}n
for n = 2, 3, 4, …. (e)
[1− ( −1)n e −π ]
2
n −1
1 + n2
(a)
420
College Mathematics
2(− 1)
2(−1)
(b)
2 2
nπ
n2π 2
[( −1) n − 1]a
( c)
n 2π 2
1 + (−1)n 2
nπ
− 2 cos
2
n
n
2
2
nπ
sin
nπ
2
 0 if n ≠ a , a is an int eger and n = 1,2,3,...

(i) Fs ( n) =  π
in n = a , nisapositive int eger.
 2
n[1 + (−1)n cos aπ ]
(ii) Fc ( n) =
n 2 − a2
if
n ≠ a, niseven
0

Fc ( n) =  2a
n ≠ a, nisodd
 a 2 − n 2 if
∞
2 ∞  1 − cos nπ 
1
(a) ∑ 
(b)
sin
n
π
2
sin nx
∑

3
π n=1 
n

n =1 n
π 2 ∞  1 − cos nπ 
1 ∞ 1
(c)
+ ∑
sin
n
π
(d)
+ ∑ sin nx

2 π n=1 
n2
3 n =1 n 2

1 − e−2 x 4 ∞ ( −1) n e−2 x − 1
y
=
+ ∑ 2
cos nx
(a)
4π
π n =1 ( n + 2)( n 2 + 4)
n
2.
3.
4.
7.
8.
9.
11.
n
(a)
(b) y =
∞
−π 2
4n
sin x + ∑
sin nx
2
2
8
n = 2,4,6,.. (n − 1)
2 ∞ [(−1)n eπ − 1]
(c) y = ∑
sin nx
π n =1 (1 + n2 )2
Fourier Series
(d) y = 2
(e) y =
421
∞
1 
 1
 + 2
 sin nx
n −4
n =1,3,5,...  2 n
∑
2 4 ∞
1
− ∑
cos nx
2
π π n =1 4 n − 1 n 2 + 1
(
)(
)