AB2.12: Fourier Series: Even and Odd Functions. Half-RangeExpansions

The Fourier series of an even function of a period p = 2L is the Fourier cosine series

f(x) = a0 +∞∑n=1

an cosnπ

Lx (1)

The coefficients are

a0 =1L

∫ L

0f(x)dx,

an =2L

∫ L

0f(x) cos

nπ

Lxdx, n = 1, 2, . . . .

The Fourier series of an odd function of a period p = 2L is the Fourier sine series

f(x) =∞∑n=1

bn sinnπ

Lx. (2)

The coefficients are

bn =2L

∫ L

0f(x) sin

nπ

Lxdx, n = 1, 2, . . . .

The case of period 2π

Assume that f(x) is a periodic function of period 2π that can be represented in the formof trigonometric series. Then the Fourier series of an even function is

f(x) = a0 +∞∑n=1

an cosnx

The coefficients are

a0 =1π

∫ π

0f(x)dx,

an =2π

∫ π

0f(x) cosnxdx, n = 1, 2, . . . .

The Fourier series of an odd function is

f(x) =∞∑n=1

bn sinnx

The coefficients are

bn =2π

∫ π

0f(x) sinnxdx, n = 1, 2, . . . .

THEOREM 10.4.2

The Fourier coefficients of a sum f1 + f2 are the sums of the corresponding Fourier coefficientsof f1 and f2.

EXAMPLE 1 Rectangular pulse.

Let

f(x) ={−k if −π < x < 0,k if 0 < x < π

and

f ∗(x) = f(x) + k ={

0 if −π < x < 0,2k if 0 < x < π,

Find the Fourier coefficients.Solution. We have

f(x) =∞∑n=1

bn sinnx =4kπ

∞∑l=1

sin(2l − 1)x2l − 1

=

4kπ

(sinx+

sin 3x3

+sin 5x

5+ . . .

).

Thus,

f∗(x) = f(x) + k = k +4kπ

∞∑l=1

sin(2l − 1)x2l − 1

=

k +4kπ

(sinx+

sin 3x3

+sin 5x

5+ . . .

).

EXAMPLE 2 Sawtooth wave.

Letf(x) = x+ π if −π < x < π, f(x) = f(x+ 2π).

Find the Fourier coefficients.Solution. We have

f(x) = f1 + f2, f1 = x, f2 = π,

bn =2π

∫ π

0f1(x) sinnxdx, n = 1, 2, . . . .

Thusbn =

2π

∫ π

0x sinnxdx =

2n2π

∫ nπ

0v sin vdv =

2n2π

∫ nπ

0vd(− cos v) =

2n2π

[−v cos v|v=nπ

v=0 +∫ nπ

0cos vdv

]

=2n2π

(−nπ cosnπ) = 2(−1)n+1

n.

f1(x) =∞∑n=1

bn sinnx = 2∞∑n=1

(−1)n+1

nsinnx =

2 sinx− sin 2x+23

sin 3x+ . . . ,

and

f(x) = f1(x) + f2(x) = π + 2∞∑n=1

(−1)n+1

nsinnx = π + 2 sinx− sin 2x+

23

sin 3x+ . . . .

Half-range expansions

Given function f(x), 0 ≤ x ≤ L

Even periodic extension of f(x) with a period 2L.

Odd periodic extension of f(x) with a period 2L.

EXAMPLE 3

Let

f(x) ={

2kLx if 0 < x < L/2,

2kL

(L− x) if L/2 < x < L,

Find the Fourier coefficients.

Given function f(x), 0 ≤ x ≤ L

Even periodic extension.

Odd periodic extension.

Solution. 1. Even periodic extension

a0 =1L

[2kL

∫ L/2

0xdx+

2kL

∫ L

L/2(L− x)dx

]=k

2.

an =2L

[2kL

∫ L/2

0x cos

nπ

Lxdx+

2kL

∫ L

L/2(L− x) cos

nπ

Lxdx

]=

4kL2

Lx

πnsin

nπ

Lx∣∣∣∣L/20− 4kL2

L

πn

∫ L/2

0sin

nπ

Lxdx+

4kL2

L(L− x)πn

sinnπ

Lx

∣∣∣∣∣L

L/2

+4kL2

L

πn

∫ L

L/2sin

nπ

Lxdx =

4kL2

L2

2πnsin

nπ

2+

4kL2

L2

n2π2

(cos

nπ

2− 1

)+

−4kL2

[L

πn

(L− L

2

)sin

nπ

2+

L2

n2π2

(cosnπ − cos

nπ

2

)]=

4kn2π2

(2 cos

nπ

2− cosnπ − 1

).

Thusa2 = − 16k

22π2 , a6 = − 16k62π2 , a10 = − 16k

102π2 , . . . .

and

f(x) = a0 +∞∑n=1

an cosnπ

Lx =

k

2− 16k

π2

(14

cos2πLx+

136

cos6πLx+ . . .

).

2. Odd periodic extension

bn =2L

∫ L

0f(x) sin

nπ

Lxdx =

8kn2π2 sin

nπ

2, n = 1, 2, . . . .

and

f(x) =∞∑n=1

bn sinnπ

Lx =

8kπ2

(11

sinπ

Lx− 1

9sin

3πLx+ . . .

).

PROBLEM 10.4.11

Find the Fourier series of

f(x) ={k if −π/2 < x < π/2,0 if π/2 < x < 3π/2.

Solution. f(x) is an even periodic function of period 2π that can be represented in theform of trigonometric series. The Fourier coefficients are

a0 =1π

∫ π

0f(x)dx =

k

π

∫ π/2

0dx =

k

2.

an =2π

∫ π

0f(x) cosnxdx =

2kπ

∫ π/2

0cosnxdx =

2knπ

sinnπ

2=

2kπ

(−1)l+1

2l − 1.

f(x) = a0 +∞∑n=1

an cosnx =k

2+

2kπ

∞∑l=1

(−1)l+1

2l − 1cos(2l − 1)x =

=k

2+

2kπ

(cosx− 1

3cos 3x+

15

cos 5x− . . .).

PROBLEM 10.4.13

Find the Fourier series of

f(x) ={x if −π/2 < x < π/2,π − x if π/2 < x < 3π/2.

Solution. f(x) is an odd periodic function of period 2π that can be represented in theform of trigonometric series. The Fourier coefficients are

bn =2π

∫ π/2

0x sinnxdx+

2π

∫ π

π/2(π − x) sinnxdx =

− 2nπ

x cosnx|π/20 +2nπ

∫ π/2

0cosnxdx−

− 2nπ

(π − x) cosnx|ππ/2 −2nπ

∫ π

π/2cosnxdx =

− 1n

cosnπ

2+

2n2π

sinnπ

2+

1n

cosnπ

2+

2n2π

sinnπ

2= (3)

2n2π

sinnπ

2+

2n2π

sinnπ

2=

4n2π

sinnπ

2=

4(2l − 1)2π

(−1)l+1 =4π

(−1)l+1

(2l − 1)2 .

Thus

f(x) =∞∑n=1

an sinnx =4π

∞∑l=1

(−1)l+1

(2l − 1)2 sin(2l − 1)x =

=4π

(sinx− 1

9sin 3x+

125

sin 5x− . . .).