CHAPTER 2
CHAPTER 2
FOURIER SERIES
PERIODIC FUNCTIONS
A function
)
(
x
f
is said to have a period T if for all x,
)
(
)
(
x
f
T
x
f
=
+
, where T is a positive constant. The least value of T>0 is
called the period of
)
(
x
f
.
EXAMPLES
We know that
)
(
x
f
= sin x = sin (x + 4
p
) = Therefore the function has period 2
p
, 4
p
, 6
p
, etc. However, 2 is the least value and therefore is the period
of f(x).
Similarly cos x is a periodic function with the period 2
p
and tan x has period
p
.
DIRICHLETS CONDITIONS
A function
)
(
x
f
defined in c
x
c+2l can be expanded as an infinite trigonometric series of the
form
2
o
a
+
+
,
sin
cos
l
x
n
b
l
x
n
a
n
n
p
p
provided
1.
)
(
x
f
is single- valued and finite in (c , c+2l)
2.
)
(
x
f
is continuous or piecewise continuous with finite number of
finite discontinuities in (c , c+2l).
3.
)
(
x
f
has no or finite number of maxima or minima in (c , c+2l).
EULERS FORMULAS
If a function
)
(
x
f
defined in (c , c+2l) can be expanded as the infinite
trigonometric series
2
o
a
+
l
x
n
b
l
x
n
a
n
n
n
n
p
p
sin
cos
1
1
=
=
+
then
0
,
cos
)
(
1
2
=
+
n
dx
l
x
n
x
f
l
a
l
c
c
n
p
1
,
sin
)
(
1
2
=
+
n
dx
l
x
n
x
f
l
b
l
c
c
n
p
[ Formulas given above for
n
a
and
n
b
are called Eulers formulas for Fourier coefficients]
DEFINITION OF FOURIER SERIES
The infinite trigonometric series
2
o
a
+
l
x
n
b
l
x
n
a
n
n
n
n
p
p
sin
cos
1
1
=
=
+
is called the Fourier series of
)
(
x
f
in the interval c
x
c+2l, provided the coefficients are given by the Eulers
formulas.
EVEN FUNCTION
If
)
(
x
f
=
)
(
x
f
in (-l , l) such that
)
(
x
-
f
=
)
(
x
f
, then
)
(
x
f
is said to be an even function of x in (-l , l).
If
-
=
)
,
0
(
)
(
)
0
,
(
)
(
)
(
2
1
l
in
x
l
in
x
x
f
f
f
Such that
)
(
1
x
-
f
=
)
(
2
x
f
or
)
(
2
x
-
f
=
)
(
1
x
f
, then
)
(
x
f
is said to be an even function of x in
(-l , l).
EXAMPLE
y = cos x , y =
2
x
are even functions.
ODD FUNCTION
If
)
(
x
f
=
)
(
x
f
in (-l , l) such that
)
(
x
-
f
= -
)
(
x
f
, then
)
(
x
f
is said to be an odd function of x in (-l , l).
If
-
=
)
,
0
(
)
(
)
0
,
(
)
(
)
(
2
1
l
in
x
l
in
x
x
f
f
f
Such that
)
(
1
x
-
f
= -
)
(
2
x
f
or
)
(
2
x
-
f
= -
)
(
1
x
f
, then
)
(
x
f
is said to be an odd function of x in
(-l , l).
EXAMPLE
y = sin x , y = x are odd functions.
FOURIER SERIES OF EVEN AND ODD FUNCTIONS
1. The Fourier series of an even function
)
(
x
f
in (-l , l) contains only cosine terms
(constant term included), i.e. the Fourier series of an even
function
)
(
x
f
in (-l , l) is given by
)
(
x
f
=
2
o
a
+
,
cos
l
x
n
a
n
p
where
.
cos
)
(
2
0
dx
l
x
n
x
f
l
a
l
n
=
p
2. The Fourier series of an odd function
)
(
x
f
in (-l , l) contains only sine terms, i.e. the Fourier series of
an odd function
)
(
x
f
in (-l , l) is given by
)
(
x
f
=
l
x
n
b
n
p
sin
,
where
.
sin
)
(
2
0
dx
l
x
n
x
f
l
b
l
n
=
p
PROBLEMS
1. Find the Fourier series of period 2l for the function
)
(
x
f
= x(2l x) in (0 , 2l). Deduce the sum of
)
(
x
f
=
L
-
+
-
2
2
2
3
1
2
1
1
1
Solution:
Let
)
(
x
f
=
2
o
a
+
l
x
n
b
l
x
n
a
n
n
n
n
p
p
sin
cos
1
1
=
=
+
in (0 , 2l) (1)
dx
l
x
n
x
l
x
l
a
l
n
-
=
2
0
cos
)
2
(
1
p
,
sin
)
2
(
cos
)
2
2
(
sin
)
2
(
1
2
0
3
3
3
2
2
2
2
l
l
n
l
x
n
l
n
l
x
n
x
l
l
n
l
x
n
x
lx
l
-
-
+
-
-
-
-
=
p
p
p
p
p
p
using Bernoullis formula.
=
[
]
2
2
2
2
2
4
2
2
cos
2
1
p
p
p
n
l
l
n
l
n
-
=
-
-
.
3
4
3
1
)
2
(
1
2
2
0
3
2
2
0
l
x
lx
l
dx
x
l
x
l
a
l
l
o
=
-
=
-
=
dx
l
x
n
x
l
x
l
b
l
n
-
=
2
0
sin
)
2
(
1
p
= 0
Using these values in (1), we have
x (2l - x) =
=
-
1
2
2
2
2
cos
1
4
3
2
n
l
x
n
n
l
l
p
p
in (0, 2l) ..(2)
The required series
-
+
-
2
2
2
3
1
2
1
1
1
can be obtained by putting x = l in the Fourier series in
(2).
x = l lies in (0 , 2l) and is a point of continuity of the
function
)
(
x
f
= x(2l x).
\
[
Sum the Fourier series in (2)
]
1
=
x
= f(l)
i.e.
p
p
n
n
l
l
n
cos
1
4
3
2
1
2
2
2
2
=
-
= l(2l - l)
i.e.. -
3
...
3
1
2
1
1
1
4
2
2
2
2
2
2
l
l
=
+
-
+
-
p
\
-
+
-
2
2
2
3
1
2
1
1
1
=
12
2
p
2. Find the Fourier series of period 2
p
for the function
)
(
x
f
= x cos x in 0 < x < 2
p
.
Solution:
Let
)
(
x
f
=
2
o
a
+
nx
b
nx
a
n
n
n
n
sin
cos
1
1
=
=
+
...(1)
dx
nx
x
x
a
n
=
p
p
2
0
cos
cos
1
[
]
,
)
1
(
)
1
cos(
1
)
1
sin(
.
)
1
(
)
1
cos(
1
)
1
sin(
.
2
1
)
1
cos(
)
1
cos(
2
1
2
0
2
2
0
2
2
0
-
-
+
-
-
+
+
+
+
+
+
=
-
+
+
=
p
p
p
p
p
n
x
n
n
x
n
x
n
x
n
n
x
n
x
dx
x
n
x
n
x
if n
1
=0, if n
1
o
a
= 0
+
=
=
p
p
p
p
2
0
2
0
2
)
2
cos
1
(
2
1
cos
1
dx
x
x
dx
x
x
a
n
.
4
2
cos
2
2
sin
2
2
1
2
0
2
p
p
p
=
+
+
=
x
x
x
x
dx
nx
x
x
b
n
=
p
p
2
0
sin
cos
1
[
]
,
)
1
(
)
1
sin(
1
)
1
cos(
.
)
1
(
)
1
sin(
1
)
1
cos(
.
2
1
)
1
sin(
)
1
sin(
2
1
2
0
2
2
0
2
2
0
-
-
+
-
-
-
+
+
+
+
+
+
-
=
-
+
+
=
p
p
p
p
p
n
x
n
n
x
n
x
n
x
n
n
x
n
x
dx
x
n
x
n
x
if n
1
=
1
2
1
1
1
1
1
1
1
1
2
-
-
=
-
+
+
-
=
-
-
+
-
n
n
n
n
n
n
, if n
1
p
p
p
2
1
sin
cos
1
2
0
1
=
=
dx
x
x
x
b
p
2
0
2
sin
xdx
x
=
2
1
4
2
sin
2
2
cos
2
1
2
0
-
=
+
-
p
p
x
x
x
Using these values in (1), we get
f(x) =
=
-
-
-
,...
3
,
2
2
sin
1
2
sin
2
1
cos
n
nx
n
n
x
x
p
3. Find the Fourier series expansion of
)
(
x
f
= sin ax in (-l , l).
Solution:
Since
)
(
x
f
is defined in a range of length 2l, we can expand
)
(
x
f
in Fourier series of period 2l.
Also
)
(
x
f
-
= sin[a(-x)] = -sin ax = -
)
(
x
f
\
)
(
x
f
is an odd function of x in (-l , l).
Hence Fourier series of
)
(
x
f
will not contain cosine terms.
Let
)
(
x
f
=
l
x
n
b
n
n
p
sin
1
=
.(1)
l
l
a
l
n
x
a
l
n
a
l
n
x
a
l
n
l
dx
x
a
l
n
a
l
n
l
0
0
sin
sin
1
cos
cos
1
+
+
-
-
-
=
+
-
-
=
p
p
p
p
p
p
{
}
{
}
2
2
2
2
1
1
sin
2
)
1
(
1
1
sin
)
1
(
sin
)
1
(
1
sin
)
1
(
1
sin
1
sin
1
l
a
n
al
n
al
n
al
n
al
al
al
n
al
al
n
l
a
l
n
la
n
l
a
l
n
la
n
n
n
n
n
-
-
=
+
+
-
-
=
-
+
-
-
-
-
=
+
+
-
-
-
=
+
+
p
p
p
p
p
p
p
p
p
p
Using these values in (1), we get
=
+
-
-
=
1
2
2
2
2
1
sin
)
1
(
sin
2
sin
n
n
l
x
n
l
a
n
n
al
ax
p
p
p
4. Find the Fourier series expansion of
)
(
x
f
=
)
,
(
p
p
-
-
in
e
x
. Hence obtain a series for cosec
p
Solution:
Though the range
)
,
(
p
p
-
is symmetric about the origin,
x
e
-
is neither an even function nor an odd function.
\
Let
)
(
x
f
=
2
o
a
+
nx
b
nx
a
n
n
n
n
sin
cos
1
1
=
=
+
....(1)
in
)
,
(
p
p
-
[
]
p
2
is
range
the
of
length
the
Q
(
)
(
)
{
}
p
p
p
p
p
p
p
p
p
p
p
sinh
)
1
(
)
1
(
2
)
1
(
)
1
(
1
1
sin
cos
1
1
cos
1
2
2
2
+
-
=
-
-
-
+
-
=
+
-
+
=
=
-
-
-
-
-
n
e
e
n
nx
n
nx
n
e
dx
nx
e
a
n
n
n
x
x
n
p
p
sinh
2
=
o
a
(
)
p
p
p
p
p
p
-
-
-
-
-
-
+
=
=
nx
n
nx
n
e
nxdx
e
b
x
x
n
cos
sin
1
1
sin
1
2
(
)
{
}
p
p
p
p
p
sinh
)
1
(
)
1
(
2
)
1
(
)
1
(
1
2
2
+
-
=
-
-
-
+
-
=
-
n
n
e
e
n
n
n
n
n
Using these values in (1), we get
x
e
-
=
=
=
+
-
+
+
-
+
1
1
2
2
sin
1
)
1
(
sinh
2
cos
1
)
1
(
sinh
2
sinh
n
n
n
n
nx
n
n
nx
n
p
p
p
p
p
p
in
)
,
(
p
p
-
[
]
),
0
(
)
(
0
f
x
f
of
series
Fourier
the
of
Sum
x
=
=
[Since x=0 is a point of continuity of f(x)]
i.e.,
1
1
)
1
(
2
1
sinh
0
1
2
=
=
+
-
+
-
=
e
n
n
n
p
p
i.e.,
=
+
-
+
-
+
=
2
2
1
)
1
(
2
2
1
2
1
cos
n
n
n
ech
p
p
i.e.,
=
+
-
=
2
2
1
)
1
(
2
cos
n
n
n
ech
p
p
HALF-RANGE FOURIER SERIES AND PARSEVALS THEOREM
(i) The half range cosine series in (0 , l) is
)
(
x
f
=
2
o
a
+
=
1
cos
n
n
l
x
n
a
p
where
.
)
(
2
0
dx
x
f
l
a
l
o
=
.
cos
)
(
2
0
dx
l
x
n
x
f
l
a
l
n
=
p
(ii) The half range sine series in (0 , l) is
)
(
x
f
=
=
1
sin
n
n
l
x
n
b
p
,
where
.
sin
)
(
2
0
dx
l
x
n
x
f
l
b
l
n
=
p
(iii) The half range cosine series in (0 ,
p
) is given by
)
(
x
f
=
2
o
a
+
=
1
cos
n
n
nx
a
where
.
)
(
2
0
dx
x
f
a
o
=
p
p
.
cos
)
(
2
0
dx
nx
x
f
a
n
=
p
p
(iv) The half range sine series in (0 ,
p
) is given by
)
(
x
f
=
=
1
sin
n
n
nx
b
,
where
.
sin
)
(
2
0
dx
nx
x
f
b
n
=
p
p
ROOT-MEAN SQUARE VALUE OF A FUNCTION
Definition
If a function y =
)
(
x
f
is defined in (c , c+2l), then
dx
y
l
l
c
c
+
2
2
2
1
is called the root mean-square(R.M.S.) value of y in (c , c+2l)
and is denoted by
.
y
Thus
.
2
1
2
2
2
+
=
l
c
c
dx
y
l
y
PARSEVALS THEOREM
If y =
)
(
x
f
can be expanded as a Fourier series of the form
2
o
a
+
l
x
n
b
l
x
n
a
n
n
n
n
p
p
sin
cos
1
1
=
=
+
in (c , c+2l), then the root-mean square value
y
of y =
)
(
x
f
in (c , c+2l) is given by
=
=
+
+
=
1
2
1
2
2
2
2
1
2
1
4
1
n
n
n
n
o
b
a
a
y
PROOF
)
(
x
f
=
2
o
a
+
l
x
n
b
l
x
n
a
n
n
n
n
p
p
sin
cos
1
1
=
=
+
in (c , c+2l) .....(1)
\
By Eulers formulas for the Fourier coefficients,
0
,
cos
)
(
1
2
=
+
n
dx
l
x
n
x
f
l
a
l
c
c
n
p
..(2)
1
,
sin
)
(
1
2
=
+
n
dx
l
x
n
x
f
l
b
l
c
c
n
p
......(3)
Now, by definition,
x
d
y
l
y
l
c
c
+
=
2
2
2
2
1
=
[
]
+
l
c
c
dx
x
f
l
2
2
)
(
2
1
=
,
sin
cos
2
)
(
2
1
1
1
2
dx
l
x
n
b
l
x
n
a
a
x
f
l
n
n
n
n
o
l
c
c
+
+
=
=
+
p
p
using (1)
=
+
+
+
=
+
l
c
c
n
n
l
c
c
o
dx
l
x
n
x
f
l
a
dx
x
f
l
a
2
1
2
cos
)
(
1
2
)
(
1
4
p
EMBED Equation.3
+
=
l
c
c
n
n
dx
l
x
n
x
f
l
b
2
1
sin
)
(
1
2
p
=
n
n
n
o
o
a
a
a
a
.
2
.
4
1
=
+
EMBED Equation.3
=
+
1
.
2
n
n
n
b
b
, by using (2) and (3)
=
=
+
1
2
2
2
4
n
n
o
a
a
EMBED Equation.3
=
+
1
2
.
2
n
n
b
EXAMPLES
1. Find the half-range (i) cosine series and (ii) sine series
for
)
(
x
f
=
2
x
in (0 ,
p
)
Solution:
(i) To get the half-range cosine series for
)
(
x
f
in (0 ,
p
), we should give an even extension for
)
(
x
f
in (
p
-
, 0).
i.e. put
)
(
x
f
=
(
)
2
x
-
=
2
x
in (
p
-
, 0)
Now
)
(
x
f
is even in (
p
-
,
p
).
\
)
(
x
f
=
2
o
a
+
=
1
cos
n
n
nx
a
.(1)
.
cos
)
(
2
0
dx
nx
x
f
a
n
=
p
p
dx
nx
x
=
p
p
0
2
cos
2
0
,
)
1
(
4
)
1
(
.
4
sin
2
cos
2
sin
2
2
2
0
3
2
2
-
=
-
=
-
+
-
-
=
n
n
n
n
nx
n
nx
x
n
nx
x
n
n
p
p
p
p
=
=
=
p
p
p
p
p
0
2
2
0
3
2
2
)
(
2
dx
x
dx
x
f
a
o
\
The Fourier half-range cosine series of
2
x
is given by
nx
n
x
n
n
cos
)
1
(
4
3
1
2
2
2
=
-
+
=
p
in (0 ,
p
).
(ii) To get the half-range sine series of
)
(
x
f
in (0 ,
p
), we should give an odd extension for
)
(
x
f
in (-
p
, 0).
i.e. Put
)
(
x
f
= -
(
)
2
x
-
in (-
p
, 0)
= -
2
x
in (-
p
, 0)
Now
)
(
x
f
is odd in (-
p
,
p
).
\
)
(
x
f
=
=
1
sin
n
n
nx
b
.(2)
=
=
p
p
p
p
0
2
0
sin
2
sin
)
(
2
nxdx
x
dx
nx
x
f
b
n
{
}
-
-
=
-
-
+
-
=
+
-
-
-
=
+
even
is
n
if
odd
is
n
if
n
n
n
n
n
nx
n
nx
x
n
nx
x
n
n
n
,
,
4
2
1
)
1
(
2
)
1
(
2
cos
2
sin
2
cos
2
2
3
2
3
1
2
0
3
2
2
p
p
p
p
p
p
p
Using this value in(2), we get the half-range sine series of
2
x
in (0 ,
p
).
2. Find the half-range sine series of
)
(
x
f
= sin ax in (0 , l).
Solution:
We give an odd extension for
)
(
x
f
in (-l , 0).
i.e. we put
)
(
x
f
= -sin[a(-x)] = sin ax in (-l , 0)
\
)
(
x
f
is odd in (-l , l)
Let
)
(
x
f
=
=
1
sin
n
n
l
x
n
b
p
dx
l
x
n
ax
l
b
l
n
=
0
sin
.
sin
2
p
l
l
a
l
n
x
a
l
n
a
l
n
x
a
l
n
l
dx
x
a
l
n
x
a
l
n
l
0
0
sin
sin
1
cos
cos
1
+
+
-
-
-
=
+
-
-
=
p
p
p
p
p
p
(
)
(
)
al
n
al
n
al
n
al
n
n
+
+
-
-
-
-
=
+
p
p
p
p
sin
1
sin
)
1
(
1
1
2
2
2
2
1
1
1
2
.
sin
)
1
(
sin
)
1
(
1
sin
)
1
(
1
l
a
n
n
al
al
al
n
al
al
n
n
n
n
-
-
=
-
+
+
-
-
=
+
+
+
p
p
p
p
Using this values in (1), we get the half-range sine series
as
=
+
-
-
=
1
2
2
2
2
1
sin
.
)
1
(
sin
2
sin
n
n
l
x
n
l
a
n
n
al
ax
p
p
p
3. Find the half-range cosine series of
)
(
x
f
= a in (0 , l). Deduce the sum of
+
+
+
L
2
2
2
5
1
3
1
1
1
.
Solution:
Giving an odd extension for
)
(
x
f
in (-l , 0),
)
(
x
f
is made an odd function in (-l , l).
\
Let f(x) =
l
x
n
b
n
p
sin
..(1)
dx
l
x
n
a
l
b
l
n
=
0
sin
2
p
(
)
{
}
n
l
n
a
l
n
l
x
n
l
a
1
1
2
cos
2
0
-
-
=
-
=
p
p
p
=
even
is
n
if
odd
is
n
if
n
a
,
0
,
4
p
Using this value in (1), we get
a =
)
,
0
(
sin
1
4
5
,
3
,
1
l
in
l
x
n
n
a
n
=
L
p
p
Since the series whose sum is required contains constant
multiples of squares of
n
b
, we apply Parsevals theorem.
[
]
=
l
n
dx
x
f
l
b
0
2
2
)
(
1
2
1
(
)
(
)
(
)
.
8
1
2
1
1
2
1
8
.
.
1
2
1
16
.
2
1
.
.
2
1
2
2
1
2
2
2
2
5
,
3
,
1
2
2
2
p
p
p
=
-
\
=
-
=
-
=
=
=
n
n
n
n
a
n
a
e
i
a
n
a
e
i
L
4. Expand
)
(
x
f
=
x
-
2
x
as a Fourier series in -1 < x < 1 and using this series
find the r.m.s. value of
)
(
x
f
in the interval.
Solution:
The Fourier series of
)
(
x
f
in (-1 , -1) is given by
)
(
x
f
=
2
o
a
+
x
n
b
x
n
a
n
n
n
n
p
p
sin
cos
1
1
=
=
+
.(1)
(
)
)
2
......(
..........
..........
3
2
3
1
2
1
3
1
2
1
3
2
)
(
1
1
1
1
3
2
1
1
1
1
2
-
=
+
-
-
=
-
=
-
=
=
-
-
-
o
o
a
x
x
dx
x
x
dx
x
f
a
(
)
(
)
(
)
1
1
3
2
2
1
1
1
1
2
sin
)
2
(
cos
2
1
sin
cos
cos
)
(
1
1
-
-
-
-
-
+
-
-
-
-
=
-
=
=
n
x
n
n
x
n
x
n
x
n
x
x
dx
x
n
x
x
dx
x
n
x
f
a
n
p
p
p
p
p
2
2
cos
3
cos
n
n
n
n
p
p
-
-
=
2
cos
4
n
n
a
n
p
-
=
.(3)
(
)
(
)
(
)
)
4
......(
..........
..........
)
1
(
2
cos
2
cos
2
cos
2
cos
)
2
(
sin
2
1
cos
sin
sin
)
(
1
1
1
3
3
3
3
1
1
3
3
2
2
2
1
1
1
1
2
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
n
b
n
n
n
n
n
n
n
x
n
n
x
n
x
n
x
n
x
x
dx
x
n
x
x
dx
x
n
x
f
b
n
n
n
+
-
-
-
-
=
+
-
-
=
-
+
-
-
-
-
-
=
-
=
=
Substituting (2), (3), (4) in (1) we get
)
(
x
f
=
x
n
n
x
n
n
n
n
n
n
p
p
p
sin
)
1
(
2
cos
)
1
(
4
3
1
1
1
1
2
1
=
+
=
+
-
+
-
+
-
We know that r.m.s. value of f(x) in (-l , l) is
=
=
+
+
=
1
2
1
2
2
2
2
1
2
1
4
1
n
n
n
n
o
b
a
a
y
.(5)
From (2) we get
9
4
3
2
2
=
-
=
o
o
a
a
...(6)
From (3) we get
4
2
2
1
16
)
1
(
4
n
a
n
a
n
n
n
=
-
=
+
..(7)
From (4) we get
2
2
2
1
4
)
1
(
2
p
p
n
b
n
b
n
n
n
=
-
=
+
..(8)
Substituting (6), (7) and (8) in (5) we get
=
+
+
=
1
2
2
4
2
4
16
2
1
9
1
n
n
n
y
p
5. Find the Fourier series for
)
(
x
f
=
2
x
in
.
p
p
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