632 Higher Engineering Mathematics

Thus, from para. (c),

f(x)=a 0 +

∑∞

n= 1

ancos

(

2 πnx

L

)

a 0 =

1

L

∫ L

2

−L

2

f(x)dx=

1

4

∫ 2

− 2

f(x)dx

=

1

4

{∫− 1

− 2

0dx+

∫ 1

− 1

5dx+

∫ 2

1

0dx

}

=

1

4

[5x]^1 − 1 =

1

4

[( 5 )−(− 5 )]=

10

4

=

5

2

an=

2

L

∫ L 2

−L

2

f(x)cos

(

2 πnx

L

)

dx

=

2

4

∫ 2

− 2

f(x)cos

(

2 πnx

4

)

dx

=

1

2

{∫− 1

− 2

0cos

(πnx

2

)

dx

+

∫ 1

− 1

5cos

(πnx

2

)

dx

+

∫ 2

1

0cos

(πnx

2

)

dx

}

=

5

2

⎡

⎢

⎣

sin

πnx

2

πn

2

⎤

⎥

⎦

1

− 1

=

5

πn

[

sin

(πn

2

)

−sin

(

−πn

2

)]

Whennis even,an= 0

Whennis odd,

a 1 =

5

π

( 1 −(− 1 ))=

10

π

a 3 =

5

3 π

(− 1 − 1 )=

− 10

3 π

a 5 =

5

5 π

( 1 −(− 1 ))=

10

5 π

and so on.

Hence the Fourier series for the function f(x) is

given by:

f(x)=

5

2

+

10

π

[

cos

(πx

2

)

−

1

3

cos

(

3 πx

2

)

+

1

5

cos

(

5 πx

2

)

−

1

7

cos

(

7 πx

2

)

+ ···

]

Problem 3. Determine the Fourier series for the

functionf(t)=tin the ranget=0tot=3.

The functionf(t)=tin the interval 0 to 3 is shown in

Fig. 69.3. Although the function is not periodic it may

be constructed outside of this range so that it is periodic

of period 3, as shown by the broken lines in Fig. 69.3.

Period L 53

f (t )

f (t ) 5 t

(^23306) t
Figure 69.3
From para.(c), the Fourier series is given by:
f(t)=a 0 +
∑∞
n= 1
[
ancos
(
2 πnt
L
)
+bnsin
(
2 πnt
L
)]
a 0 =
1
L
∫ L 2
−L
2
f(t)dx=
1
L
∫ L
0
f(t)dx

1
3
∫ 3
0
tdt=
1
3
[
t^2
2
] 3
0

3
2
an=
2
L
∫ L 2
−L
2
f(t)cos
(
2 πnt
L
)
dt

2
L
∫L
0
tcos
(
2 πnt
L
)
dt

2
3
∫ 3
0
tcos
(
2 πnt
3
)
dt