1000 Solved Problems in Modern Physics

40 1 Mathematical Physics

1.18 The given function is of the square form. Asf(x) is defined in the interval
(−π,π), the Fourier expansion is given by

f(x)=

1

2

a 0 +

∑∞

n= 1

(ancosnx+bnsinnx)(1)

wherean=(1/π)

∫π

−π

f(x) cosnxdx (2)

a 0 =(1/π)

∫π

−π

f(x)dx (3)

bn=

(

1

π

)∫π

−π

f(x)sinnxdx (4)

By (3)

a 0 =(1/π)

(∫ 0

−π

0dx+

∫π

0

πdx

)

=π (5)

By (2)

an=(1/π)

∫π

0

cosnxdx= 0 ,n≥1(6)

By (4)

bn=(1/π)

∫π

0

πsinnx dx=

(

1

n

)

(1−cosnπ)(7)

Using (5), (6) and (7) in (1)

f(x)=

π

2

+ 2

(

sin(x)+

(

1

3

)

sin 3x+

(

1

5

)

sin 5x+···

)

The graph off(x) is shown in Fig. 1.8. It consists of thex-axis from−πto 0

and of the line AB from 0toπ. A simple discontinuity occurs atx =0at

which point the series reduces toπ/2.

Now,

π/ 2 = 1 /2[f(0−)+f(0+)]

Fig. 1.8Fourier expansion of
a square wave