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312Differential Equations 2301312: ISE Program, Chulalongkorn Univer- sity: First semester

Exercises for Fourier Series

Find Fourier series for For each of the given functions. Assume that the functions are periodically extended outside the original interval.

(1.) f(x) =

(−1, −1≤x <0 1, 0≤x <1 (2.) f(x) =

(0, −π≤x <0 1, 0≤x < π (3.) f(x) = sin2x, −π≤x≤π (4.) f(x) = 1−x2, −1≤x≤1 (5.) f(x) =

(0, −1≤x <0 x2, 0≤x <1

Example: For (1.), function f(x) is defined for −1 ≤ x < 1. We can extend the function periodically outside this interval with period T = 2 by setting f(x+ 2) =f(x).

The functionf(x) thus has the Fourier series, (L= 1), f(x) = a0

2 +

X

n=1

ancosnπx+bnsinnπx.

a0 = Z 1

1

f(x)dx= Z 0

1

(−1)dx+ Z 1

0

1dx=−1 + 1 = 0

an= Z 1

1

f(x) cosnπx dx= Z 0

1

−cosnπx dx+ Z 1

0

cosnπx dx

=

·

− 1

nπ sinnπx

¸0

1

+

· 1

nπ sinnπx

¸1 0

= 1

nπ (−sinnπ+ sinnπ) an= 0 (n= 1,2,3, . . .)

bn= Z 1

1

f(x) sinnπx dx= Z 0

1

−sinnπx dx+ Z 1

0

sinnπx dx

=

· 1

nπ cosnπx

¸0

1

+

·

− 1

nπ cosnπx

¸1 0

= 1

nπ (1−cosnπ−cosnπ+ 1) = 2

nπ (1−cosnπ)

∴ bn = 2 nπ

³

1−cosnπ 2

´

(n = 1,2,3, . . .) The Fourier series for f(x) is

X

n=1

2

nπ (1−cosnπ) sinnπx.

Fourier Cosine and Sine Series 1

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Expanding a function f defined on [0, L] in Fourier Series.

1. Fourier Cosine Series. We extend f to be an even periodic function with period 2Lusing

g(x) :=

(f(x), 0≤x≤L, f(−x), −L < x <0.

2. Fourier Sine Series. We extend f to be an odd periodic function with period 2Lusing

h(x) :=





f(x), 0< x < L,

0, x= 0, L

−f(−x), −L < x < 0.

3. Fourier Series (Normal Form). We extendf to be a periodic function with period L using f(x+L) =f(x).

Exercises Find the Fourier cosine and sine series forf: (1.) f(x) =

(1, 0< x <1, 0, 1< x <2 (2.) f(x) =x, 0≤x <1 (3.) f(x) =

(x, 0< x < π, 0, π < x <2π (4.) From the given Fourier series of a periodic function f(x)

f(x) = L

2 +−4L π2

X

n=1

cos(2n−L1)πx (2n−1)2 , f(x) =

(−x, −L≤x <0,

x, 0≤x < L; f(x+ 2L) =f(x).

Show that

π2

8 = 1 + 1 32 + 1

52 +· · ·=

X

n=1

1 (2n+ 1)2.

2

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