Higher Engineering Mathematics

EVEN AND ODD FUNCTIONS AND HALF-RANGE FOURIER SERIES 673

L

as in Section 71.2(a), i.e.

f(x)=a 0 +

∑∞

n= 1

ancosnx

where a 0 =

1

π

∫π

0

f(x)dx

and an=

2

π

∫π

0

f(x) cosnxdx

(c) If ahalf-range sine seriesis required for the

functionf(x)=xin the range 0 toπthen an

odd periodic function is required. In Figure 71.5,

f(x)=xis shown plotted fromx=0tox=π.

Since an odd function is symmetrical about the

origin the lineCDis constructed as shown. If

the sawtooth waveform produced is assumed to

be periodic of period 2πoutside of this range,

then the waveform is as shown in Fig. 71.5.

When a half-range sine series is required then

the Fourier coefficientbn is calculated as in

Section 71.2(b), i.e.

f(x)=

∑∞

n= 1

bnsinnx

where bn=

2

π

∫π

0

f(x) sinnxdx

f(x)

f(x) = x

π

−π

− 2 π −π 0 2 π 3 πx

C

π

D

Figure 71.5

Problem 6. Determine the half-range Fourier

cosine series to represent the functionf(x)= 3 x

in the range 0≤x≤π.

From para. (b), for a half-range cosine series:

f(x)=a 0 +

∑∞

n= 1

ancosnx

Whenf(x)= 3 x,

a 0 =

1

π

∫π

0

f(x)dx=

1

π

∫π

0

3 xdx

=

3

π

[

x^2

2

]π

0

=

3 π

2

an=

2

π

∫π

0

f(x) cosnxdx

=

2

π

∫π

0

3 xcosnxdx

=

6

π

[

xsinnx

n

+

cosnx

n^2

]π

0

by parts

=

6

π

[(

πsinnπ

n

+

cosnπ

n^2

)

−

(

0 +

cos 0

n^2

)]

=

6

π

(

0 +

cosnπ

n^2

−

cos 0

n^2

)

=

6

πn^2

(cosnπ−1)

Whennis even,an= 0

Whennis odd,an=

6

πn^2

(− 1 −1)=

− 12

πn^2

Hencea 1 =

− 12

π

,a 3 =

− 12

π 32

,a 5 =

− 12

π 52

, and so on.

Hence the half-range Fourier cosine series is

given by:

f(x)= 3 x=

3 π

2

−

12

π

(

cosx+

1

32

cos 3x

+

1

52

cos 5x+···

)

Problem 7. Find the half-range Fourier sine

series to represent the functionf(x)= 3 xin the

range 0≤x≤π.

From para. (c), for a half-range sine series:

f(x)=

∑∞

n= 1

bnsinnx

Whenf(x)= 3 x,

bn=

2

π

∫π

0

f(x) sinnxdx=

2

π

∫π

0

3 xsinnxdx

=

6

π

[

−xcosnx

n

+

sinnx

n^2

]π

0

by parts