Mathematical Physics Vol 1

Chapter 6. Trigonometric Fourier series. Fourier integral

324

6.3 Examples

Problem 217 Expand the function f ( x ) shown in the figure into a Fourier series

Figure 6.7: Figure with Example 217.

in the interval ( − π , π ) . Its analytic form is f ( x )= −

k , for − π < x < 0 k for 0 < x < π

f ( x + 2 π )= f ( x ) .

(6.44)

Solution From (6.11) it follows that a 0 = 0, and from (6.12) we obtain

π Z

1 π

a n =

f ( x ) cos nx d x =

− π

1 π  

k cos nx d x 

0 Z

( − k ) cos nx d x + Z

π

 =

(6.45)

0

− π 1 π " −

n   π 0 #

n   0 − π

sin nx

sin nx

k

+ k

= 0 ,

=

because sin nx = 0 at − π , 0 and π , for each n = 1 , 2 ,..., . 7 In the same way, from (6.12) we obtain

1 π Z

k sin nx d x =

1 π Z 1 π "

( − k ) sin nx d x + Z

π

π

0

b n =

f ( x ) sin nx d x =

− π

− π

0

n   π 0 # .

n   0 − π

cos nx

cos nx

k

− k

(6.46)

=

Given that cos ( − α )= cos α and cos0 = 1, we finally obtain b n = k n π [ cos0 − cos n π − cos n π + cos0 ]= 2 k n π

( 1 − cos n π ) .

cos n π = −

1 for odd values of n , 1 for even values of n .