Higher Engineering Mathematics, Sixth Edition

Even and odd functions and half-range Fourier series 627

22 ^02 

A

x

B

f(x)

f(x) 5 x

2 



Figure 68.4

(b) If ahalf-range cosine seriesis required for the

function f(x)=x in the range 0 toπthen an

evenperiodic function is required. In Figure 68.4,

f(x)=x is shown plotted fromx=0tox=π.

Since an even function is symmetrical about the

f(x)axis the lineABis constructed as shown.

If the triangular waveform produced is assumed

to be periodic of period 2πoutside of this range

then the waveform is as shown in Fig. 68.4. When

a half-range cosine series is required then the

Fourier coefficientsa 0 andanare calculated as

in Section 68.2(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 requiredfor the func-

tionf(x)=xin the range 0 toπthen an odd peri-

odic functionis required. In Figure 68.5,f(x)=x

is 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

period2πoutsideofthisrange, thenthe waveform

is as shown in Fig. 68.5. When a half-range sine

series is required then the Fourier coefficientbnis

22 

D

C

f(x)

f(x) 5 x

2  0 2  3 x

2 





Figure 68.5

calculated as in Section 68.2(b), i.e.

f(x)=

∑∞

n= 1

bnsinnx

where bn=

2

π

∫π

0

f(x)sinnxdx

Problem 6. Determine the half-range Fourier

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

the range 0≤x≤π.

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

f(x)=a 0 +

∑∞

n= 1

ancosnx

When f(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 +

cos0

n^2

)]

=

6

π

(

0 +

cosnπ

n^2

−

cos0

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

cos3x

+

1

52

cos5x+ ···

)