Mathematical Physics Vol 1

Chapter 6. Trigonometric Fourier series. Fourier integral

314

coefficients

1 ℓ Z 1 ℓ Z 1 ℓ Z

ℓ

a 0 =

f ( x ) d x = 0 ,

− ℓ

k π x ℓ

ℓ

a k =

f ( x ) cos

d x = 0 ,

− ℓ

2 ℓ Z

k π x ℓ

k π x ℓ

ℓ

ℓ

b k =

f ( x ) sin

d x =

f ( x ) sin

d x .

(6.18)

0

− ℓ

In this special case, the Fourier series takes the form

∞ ∑ k = 1

k π ℓ

f ( x )=

b k sin

x .

(6.19)

In this case it is said that the function is expanded into a Fourier sine series , where coefficients b k are determined by relations (6.18).

Theorem21 An arbitrary function f ( x ) , defined on the interval [ − ℓ,ℓ ] , can be represented as the sum of an even and an odd function in the same interval.

Proof The function f ( x ) can be represented in the following form

1 2

1 2

f ( x )= f ( x )+

f ( − x ) −

f ( − x )=

1 2

1 2

1 2

1 2

f ( x )+

f ( x )+

f ( − x ) −

f ( − x )=

=

f ( x )+ f ( − x ) 2

f ( x ) − f ( − x ) 2 .

=

+

If we introduce the following substitutions

f 1 ( x )= f ( x ) − f ( − x ) 2 , it follows that f ( x )= f 1 ( x )+ f 2 ( x ) . Let us further prove that the functions f 1 and f 2 are even and odd, respectively. Given that f 1 ( − x )= 1 2 [ f ( − x )+ f ( − ( − x ))]= 1 2 [ f ( x )+ f ( − x )]= f 1 ( x ) , 1 2 [ f ( x )+ f ( − x )]= − f 2 ( x ) , we can conclude that f 1 is an even, and f 2 an odd function. The theorem is thus proved. f ( x )+ f ( − x ) 2 i f 2 ( x )= f 2 ( − x )= 1 2 [ f ( − x ) − f ( − ( − x ))]= −

Based on the previous theorem, we conclude that we can always apply the so-called Fourier sine and cosine transformation, but only to parts of a function.