Higher Engineering Mathematics

FOURIER SERIES FOR PERIODIC FUNCTIONS OF PERIOD 2π 659

L

Hencea 1 ,a 2 ,a 3 ,...are all zero (since sin 0=
sin (−nπ)=sinnπ=0), and therefore no cosine
terms will appear in the Fourier series.

From Section 69.3(i):

bn=

1

π

∫π

−π

f(x) sinnxdx

=

1

π

{∫ 0

−π

−ksinnxdx+

∫π

0

ksinnxdx

}

=

1

π

{[

kcosnx

n

] 0

−π

+

[

−kcosnx

n

]π

0

}

Whennis odd:

bn=

k

π

{[(

1

n

)

−

(

−

1

n

)]

+

[

−

(

−

1

n

)

−

(

−

1

n

)]}

=

k

π

{

2

n

+

2

n

}

=

4 k

nπ

Henceb 1 =

4 k

π

,b 3 =

4 k

3 π

,b 5 =

4 k

5 π

, and so on.

Whennis even:

bn=

k

π

{[

1

n

−

1

n

]

+

[

−

1

n

−

(

−

1

n

)]}

= 0

Hence, from equation (1), the Fourier series for the
function shown in Fig. 69.3 is given by:

f(x)=a 0 +

∑∞

n= 1

(ancosnx+bnsinnx)

= 0 +

∑∞

n= 1

(0+bnsinnx)

i.e.f(x)=

4 k

π

sinx+

4 k

3 π

sin 3x+

4 k

5 π

sin 5x+···

i.e.f(x)=

4 k

π

(

sinx+

1

3

sin 3x

+

1

5

sin 5x+ ···

)

Problem 2. For the Fourier series of Prob-

lem 1 letk=π. Show by plotting the first three

partial sums of this Fourier series that as the

series is added together term by term the result

approximates more and more closely to the

function it represents.

Ifk=πin the Fourier series of Problem 1 then:

f(x)=4(sinx+^13 sin 3x+^15 sin 5x+···)

4 sinxis termed the first partial sum of the Fourier

series of f(x), (4 sinx+^43 sin 3x) is termed the

second partial sum of the Fourier series, and

(4 sinx+^43 sin 3x+^45 sin 5x) is termed the third

partial sum, and so on.

Let P 1 =4 sinx,

P 2 =

(

4 sinx+^43 sin 3x

)

and P 3 =

(

4 sinx+^43 sin 3x+^45 sin 5x

)

.

Graphs ofP 1 ,P 2 andP 3 , obtained by drawing up

tables of values, and adding waveforms, are shown

in Figs. 69.4(a) to (c) and they show that the series is

convergent, i.e. continually approximating towards

a definite limit as more and more partial sums are

taken, and in the limit will have the sumf(x)=π.

Even with just three partial sums, the waveform is

starting to approach the rectangular wave the Fourier

series is representing.

Problem 3. If in the Fourier series of Prob-

lem 1, k=1, deduce a series for

π

4

at the

pointx=

π

2

.

Ifk=1 in the Fourier series of Problem 1:

f(x)=

4

π

(

sinx+

1

3

sin 3x+

1

5

sin 5x+···

)

Whenx=

π

2

,f(x)=1,

sinx=sin

π

2

=1,

sin 3x=sin

3 π

2

=−1,

sin 5x=sin

5 π

2

=1, and so on.

Hence 1=

4

π

[

1 +

1

3

(−1)+

1

5

(1)+

1

7

(−1)+···

]

i.e.

π

4

= 1 −

1

3

+

1

5

−

1

7

+···