1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

68 Chapter 1 Fourier Series and Integrals

Let h(x)be an odd function defined in a symmetric interval−a<x<a. Then
∫a

−a

h(x)dx= 0. 

Suppose now thatgis an even function in the interval−a<x<a. Since the
sine function is odd and the productg(x)sin(nπx/a)is odd,

bn=

1

a

∫a

−a

g(x)sin

(nπx

a

)

dx= 0.

That is, all the sine coefficients are zero. Also, since the cosine is even, so is
g(x)cos(nπx/a),andthen

an=

1

a

∫a

−a

g(x)cos

(nπx

a

)

dx=

2

a

∫a

0

g(x)cos

(nπx

a

)

dx.

Thus the cosine coefficients can be computed from an integral over the interval
from 0 toa.
Parallel results hold for odd functions: the cosine coefficients are all zero
and the sine coefficients can be simplified. We summarize the results.

Theorem 2.If g(x)is even on the interval−a<x<a(g(−x)=g(x)),then

g(x)∼a 0 +

∑∞

n= 1

ancos

(nπx

a

)

, −a<x<a,

where

a 0 =^1

a

∫a

0

g(x)dx, an=^2

a

∫a

0

g(x)cos

(nπx

a

)

dx.

If h(x)is odd on the interval−a<x<a(h(−x)=−h(x)),then

h(x)∼

∑∞

n= 1

bnsin

(nπx

a

)

, −a<x<a,

where

bn=^2 a

∫a

0

h(x)sin

(

nπx

a

)

dx. 

Very frequently, a function given in an interval 0<x<amust be repre-
sented in the form of a Fourier series. There are infinitely many ways of do-
ing this, but two ways are especially simple and useful: extending the given
function to one defined on a symmetric interval−a<x<aby making the
extended function either odd or even.