The Chemistry Maths Book, Second Edition

15.4 Fourier series 431

The following example demonstrates the Fourier series for an arbitrary period, and it

also prepares the ground for the completion of the solution of the wave equation for

the vibrating string discussed in Section 14.7.

EXAMPLE 15.7The Fourier series of the function

(15.51)

Figure 15.9 shows that the function is an oddfunction

of x, withf(−x) 1 = 1 −f(x). It follows from the discussion of

even and odd functions in Section 5.3 that an integral

is zero unless g(x)is also an odd function of x, or

if it contains an odd component. Of the expansion

functions in (15.48), the sine functions are odd but the

cosine functions are even. All the integrals (15.49) fora

n

are therefore zero, and the

Fourier series reduces to the Fourier sine series for the expansion of an odd function:

(15.52)

with coefficients given by (15.50). It also follows from the discussion of Section 5.3,

equation (5.25), that the integral (15.50) over the whole interval−l 1 ≤ 1 x 1 ≤ 1 lis equal to

twice the integral over the half-interval 01 ≤ 1 x 1 ≤ 1 l:

(15.53)

For the function (15.51), we then have

=+−







4

2

0

2

A

l

x

nx

l

dx l x

nx

l

dx

l

ZZ

/

sin ( ) sin

ππ















b

l

fx

nx

l

dx f x

nx

l

dx

n

l

=+



2

0

2

ZZ

/

()sin ()sin

ππ



















b

l

fx

nx

l

dx

n

l

=

+

2

0

Z ()sin

π

fx b

nx

l

n

()= sin

=

∑

1

∞

π

Z

−

+

l

fxgxdx()()

fx

A

l

lx lx

Ax

l

() x

()

=

−+ −≤≤−

−≤

21

2

21

2

for

for ≤≤+

−+≤≤+











1

2

21

2

A

l

() forlx x l

+A

−A

+l

−l

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0

x

f(x)

Figure 15.9