1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

1.2 Arbitrary Period and Half-Range Expansions 65

case, the coefficients are

a 0 =^1

2 a

∫a

−a

f(x)dx, an=^1

a

∫a

−a

f(x)cos

(nπx

a

)

dx,

bn=^1

a

∫a

−a

f(x)sin

(nπx

a

)

dx.

(1)

Example.
Find the Fourier series off(x)=|sin(πx)|.
Solution:This function is periodic with period 1, soa=p/ 2 = 1 /2. To do the
integrals for the Fourier coefficients, we need to get rid of the absolute value
signs:

f(x)=

{sin(πx), 0 <x<1,

−sin(πx), − 1 <x<0.

Then, it is easy to calculate

a 0 =^11

∫ 1 / 2

− 1 / 2

∣∣

sin(πx)

∣∣

dx=

∫ 0

− 1 / 2

−sin(πx)dx+

∫ 1 / 2

0

sin(πx)dx

=cos(ππx)

∣∣

∣∣

0

− 1 / 2

−cos(ππx)

∣∣

∣∣

1 / 2

0

=π^1 −−π^1 =π^2.

(Recall that cos(±π/ 2 )=0.) The other coefficients are found similarly:

an=^21

∫ 1 / 2

− 1 / 2

∣∣

sin(πx)

∣∣

cos( 2 nπx)dx

= 2

[∫ 0

− 1 / 2

−sin(πx)cos( 2 nπx)dx+

∫ 1 / 2

0

sin(πx)cos( 2 nπx)dx

]

=−π^4 · 4 n (^21) − 1.
Andbnis found to be 0 for alln. Consequently, the Fourier series of the function
is
∣∣
sin(πx)

∣∣

∼π^2 −π^4

∑∞

n= 1

1

4 n^2 − 1 cos(^2 nπx).
We will see later that|sin(πx)|is equal to its series.

It is often necessary to use a Fourier series to represent a function that has
been defined only in a finite interval. We can justify such a representation by
making the given function part of a periodic function. If the given functionfis
defined on the interval−a<x<a, we may construct ̄f,theperiodic extension