Higher Engineering Mathematics

664 FOURIER SERIES

=

2

π

[

xsinnx

n

+

cosnx

n^2

]π

−π

=

2

π

[(

0 +

cosnπ

n^2

)

−

(

0 +

cosn(−π)

n^2

)]

= 0

bn=

1

π

∫π

−π

f(x) sinnxdx=

1

π

∫π

−π

2 xsinnxdx

=

2

π

[

−xcosnx

n

−

∫ (

−cosnx

n

)

dx

]π

−π

by parts

=

2

π

[

−xcosnx

n

+

sinnx

n^2

]π

−π

=

2

π

[(

−πcosnπ

n

+

sinnπ

n^2

)

−

(

−(−π) cosn(−π)

n

+

sinn(−π)

n^2

)]

=

2

π

[

−πcosnπ

n

−

πcos (−nπ)

n

]

=

− 4

n

cosnπ

When n is odd, bn=

4

n

. Thus b 1 =4, b 3 =

4

3

,

b 5 =

4

5

, and so on.

When n is even, bn=

− 4

n

. Thus b 2 =−

4

2

,

b 4 =−

4

4

,b 6 =−

4

6

, and so on.

Thus f(x)= 2 x=4 sinx−

4

2

sin 2x+

4

3

sin 3x

−

4

4

sin 4x+

4

5

sin 5x−

4

6

sin 6x+···

i.e. 2 x= 4

(

sinx−

1

2

sin 2x+

1

3

sin 3x−

1

4

sin 4x

+

1

5

sin 5x−

1

6

sin 6x+ ···

)

(1)

for values off(x) between−πandπ. For values

off(x) outside the range−πto+πthe sum of the

series is not equal tof(x).

Problem 2. In the Fourier series of Problem 1,

by lettingx=π/2, deduce a series forπ/4.

Whenx=π/2,f(x)=πfrom Fig. 70.2.

Thus, from the Fourier series of equation (1):

2

(π

2

)

= 4

(

sin

π

2

−

1

2

sin

2 π

2

+

1

3

sin

3 π

2

−

1

4

sin

4 π

2

+

1

5

sin

5 π

2

−

1

6

sin

6 π

2

+···

)

π= 4

(

1 − 0 −

1

3

− 0 +

1

5

− 0 −

1

7

−···

)

i.e.

π

4

= 1 −

1

3

+

1

5

−

1

7

+···

Problem 3. Obtain a Fourier series for the

function defined by:

f(x)=

{

x, when 0<x<π

0, whenπ<x< 2 π.

The defined function is shown in Fig. 70.3 between

0 and 2π. The function is constructed outside of this

range so that it is periodic of period 2π, as shown by

the broken line in Fig. 70.3.

− 2 π−π 0 π 2 π 3 π

π

f(x) f(x) = x

x

Figure 70.3

For a Fourier series:

f(x)=a 0 +

∑∞

n= 1

(ancosnx+bnsinnx)

It is more convenient in this case to take the limits

from0to2πinstead of from−πto+π. The value

of the Fourier coefficients are unaltered by this

change of limits. Hence

a 0 =

1

2 π

∫ 2 π

0

f(x)dx=

1

2 π

[∫π

0

xdx+

∫ 2 π

π

0dx

]

=

1

2 π

[

x^2

2

]π

0

=

1

2 π

(

π^2

2

)

=

π

4