Mathematical Physics Vol 1

6.3 Examples

327

π Z

π Z 0

2 n 2 π Z

n π

1 π

2 π

a n =

| x | cos nx d x =

x cos nx d x =

y cos y d y =

0

− π

=( y = u , d u = d y , cos y d y = d v , v = sin y )=

2 n 2 π 

sin y d y 

 y sin y   π 0 − cos y   n π 0 =

n π Z 0

 =

=

2 n 2 π

2 n 2 π

(( − 1 ) n − 1 )=

=

=   −

0 for even values of n ,

4 n 2 π

for odd values of n .

1 π Z

π

b n = | x | sin nx d x = 0 . Thus, the Fourier series of the function f ( x )= | x | is f ( x )= π 2 − 4 π ∞ ∑ n = 0 1 ( 2 n + 1 ) 2 − π

cos ( 2 n + 1 ) x .

Problem 220

Expand the following function into a Fourier series f ( x )= ax , − π < x ≤ 0 , bx , 0 < x < π .

Solution

∞ ∑ k = 0

∞ ∑ k = 1

( − 1 ) k + 1 sin kx k .

π ( b − a ) 4

2 ( a − b ) π

cos ( 2 k + 1 ) x ( 2 k + 1 ) 2

f ( x )=

+( a + b )

+

Problem 221 Expand the function f ( x )= cos ax into a Fourier series in the interval ( − π , π ) , ( a̸ = ± 1 2 , ± 3 2 ,... ) .

Result

π "

a cos nx n 2 − a 2 #

∞ ∑ n = 1

2sin π a

1 2 a

( − 1 ) n + 1

f ( x )=

+

.