Higher Engineering Mathematics

680 FOURIER SERIES

72.2 Half-range Fourier series for

functions defined over rangeL

(a) By making the substitution u=

πx

L

(see

Section 72.1), the rangex=0tox=Lcorre-

sponds to the rangeu=0tou=π. Hence a

function may be expanded in a series of either

cosine terms or sine terms only, i.e. ahalf-range

Fourier series.

(b) Ahalf-range cosine seriesin the range 0 toL

can be expanded as:

where

f(x)=a 0 +

∑∞

n= 1

ancos

(nπx

L

)

a 0 =

1

L

∫L

0

f(x)dx and

an=

2

L

∫L

0

f(x) cos

(nπx

L

)

dx

(c) Ahalf-range sine seriesin the range 0 toLcan

be expanded as:

f(x)=

∑∞

n= 1

bnsin

(nπx

L

)

where bn=

2

L

∫L

0

f(x) sin

(nπx

L

)

dx

Problem 4. Determine the half-range Fourier

cosine series for the functionf(x)=xin the

range 0≤x≤2. Sketch the function within and

outside of the given range.

A half-range Fourier cosine series indicates an even
function. Thus the graph off(x)=xin the range 0
to 2 is shown in Fig. 72.4 and is extended outside
of this range so as to be symmetrical about thef(x)
axis as shown by the broken lines.
From para. (b), for a half-range cosine series:

f(x)=a 0 +

∑∞

n= 1

ancos

(nπx

L

)

− 4 − 2 0 24 6x

2

f(x)

f(x) = x

Figure 72.4

a 0 =

1

L

∫L

0

f(x)dx=

1

2

∫ 2

0

xdx

=

1

2

[

x^2

2

] 2

0

= 1

an=

2

L

∫L

0

f(x) cos

(nπx

L

)

dx

=

2

2

∫ 2

0

xcos

(nπx

2

)

dx

=

⎡

⎢

⎣

xsin

(nπx

2

)

(nπ

2

) +

cos

(nπx

2

)

(nπ

2

) 2

⎤

⎥

⎦

2

0

=

⎡

⎢

⎣

⎛

⎜

⎝

2 sinnπ

(nπ

2

) +

cosnπ

(nπ

2

) 2

⎞

⎟

⎠

−

⎛

⎜

⎝^0 +

cos 0

(nπ

2

) 2

⎞

⎟

⎠

⎤

⎥

⎦

=

⎡

⎢

⎣

cosnπ

(nπ

2

) 2 −

1

(nπ

2

) 2

⎤

⎥

⎦

=

(

2

πn

) 2

(cosnπ−1)

Whennis even,an= 0

a 1 =

− 8

π^2

, a 3 =

− 8

π^232

, a 5 =

− 8

π^252

and so on.

Hence the half-range Fourier cosine series forf(x)

in the range 0 to 2 is given by:

f(x)= 1 −

8

π^2

[

cos

(πx

2

)

+

1

32

cos

(

3 πx

2

)

+

1

52

cos

(

5 πx

2

)

+ ···

]