Example. Solution - UCL

Continuing from the last lecture.... Example. (ii) Let f be a periodic function of period 2 such that

f (x) = 2 - x2 for - x < Solution: So f is periodic with period 2 and its graph is:

We first check if f is even or odd. f (-x) = 2 - (-x)2 = 2 - x2 = f (x), so f (x) is even.

Since f is even,

bn = 0 2

an = 0 f (x) cos(nx)dx

Using the formulas for the Fourier coefficients we have

2

an = 0 f (x) cos(nx)dx

2 =

(2 - x2) cos(nx)dx

0

=

2 (

(2 - x2) sin nx n

0

-

sin nx

-2x

dx)

0

n

=

2 ([(2

- 2) sin n

- (2

sin 0 - 0) ] +

sin nx

2x

dx)

n

n

0

n

22

=

x sin(nx)dx

n 0

22 =(

cos nx -x

-

cos nx

(-

)dx)

n

n0

0

n

=

2

2 n

1 ([-

n

cos n]

+

[

sin nx n2

]0

)

4 = - cos n

n2

=

-4/n2 if n is even 4/n2 if n is odd

It remains to calculate a0.

2 a0 =

(2 - x2)dx

0

=

2 [2x

-

x3 3

]0

42 =

3

The Fourier Series of f is therefore

1 f (x) = 2 a0 + a1 cos x + a2 cos 2x + a3 cos 3x + ...

b1 sin x + b2 sin 2x + b3 sin 3x + ...

22

1

1

1

1

= + 4(cos x - cos 2x + cos 3x - cos 4x + cos 5x + ....)

3

4

9

16

25

10.6 Functions with period 2L

If a function has period other than 2, we can find its Fourier series by making a change of variable. Suppose f (x) has period 2L, that is f (x + 2L) = f (x) for all x. If we let

x

Lt

t = and g(t) = f (x) = f ( )

L

then, g has period 2 and x = ?L corresponds to t = ?. We know the Fourier series of g(t):

1

2 a0 + (an cos nt + bn sin nt).

n=1

where

1

a0

=

f (t)dt

-

1

an

=

f (t) cos(nt)dt

-

1

bn

=

f (t) sin(nt)dt

-

If

we

now

use

the

Substitution

Rule

with

x

=

Lt

,

then

t

=

x L

,

dt

=

(

L

)dx,

we have the following:

Definition. If f (x) is a piecewise continuous function on [-L, L], its Fourier

Series is

1

nx

nx

2 a0 + an cos( L ) + bn sin( L ),

n=1

where

1L

a0

=

L

f (x)dx

-L

1L

nx

an

=

L

f (x) cos( )dx

-L

L

1L

nx

bn

=

L

f (x) sin( )dx

-L

L

Example. Find the Fourier series of the triangular wave function defined by f (x) = |x| - a for -a x a and f (x + 2) = f (x) for all x.

Solution: So f is periodic with period 2a and its graph is:

We first check if f is even or odd.

f (-x) = | - x| - a = |x| - a = f (x), so f (x) is even.

Therefore,

bn = 0

2a

nx

an = a 0 f (x) cos( a )dx

Using the formulas for the Fourier coefficients we have

2a

nx

an = a

(|x| - a) cos(

0

a

)dx

2a

nx

= (x - a) cos( )dx

a0

a

=

2 (

a

(x

-

a)

sin(

nx a

n

)

a

a-

0

a

a

sin(

nx a

n

)

dx)

0

a

2 a2 =(

nx cos( )

a)

a n22

a0

2a = (cos n - 1)

n22

0 if n is even

=

4a

-

if n is odd

n22

It remains to determine a0.

2a a0 = a 0 |x| - adx

2a

=

x - adx

a0

=

2 x2 [

a2

-

ax]a0

= -a

The Fourier Series of f is therefore

f (x) = a0 + a1 cos x + a2 cos 2x + a3 cos 3x + ...

a 4a x 1 3x 1 5x

= - - (cos( ) + cos( ) + cos( ) + ...)

2 2

a9

a 25

a

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