Mathematical Physics Vol 1

6.3 Examples

325

Thus the Fourier coefficients for the given function have the following values

4 k π

4 k 3 π

4 k 5 π

b 1 =

b 2 = 0 , b 3 =

b 4 = 0 , b 5 =

,

,

,...,

that is

b n = (

4 k n π

, for odd values of n , 0 , for even values of n . As coefficients a n are equal to zero, the corresponding Fourier series is 4 k π sin x + 1 3 sin3 x + ... .

(6.47)

The partial sums are

4 k π

1 3

4 k π

S 1 =

sin x , S 2 =

( sin x +

sin3 x ) etc.

Graphs of partial sums are given in Figure 6.8 and they show how the series converges and how its sum is equal to the given function f ( x ) . At points x = 0 and x = π , which are the points of discontinuity of f ( x ) , all partial sums have a value of zero, which is the arithmetic mean of the values − k and k .

Figure 6.8:

Problem 218 Expand the function f ( x )= x into a Fourier series in the interval ( − π , π ) .

Solution

∞ ∑ n = 1 ( a n cos nx + b n sin nx ) x ∈ ( − π , π ) .

a 0 2

f ( x )=

+

x 2   π − π

1 π Z

π

1 π

1 2

1 2 π

( π 2 − π 2 )= 0 .

a 0 =

xdx =

=

− π

π Z

1 π

a n =

x cos nx d x .

− π

7 Note that this result can be obtained directly, as the observed function is odd.