FOURIER SERIES

4. Integrals
Formula for integration by parts: ¸ b u dv dx = [uv]b ? ¸ b du v dx

a dx

a a dx




[g (x)]

g (x)












sinh x
cosh x











sin a

ln . a .







2 2
2 a 2
a2

? 2 2
a a2

5. Useful trig results
When calculating the Fourier coefficients an and bn , for which n = 1, 2, 3, . . . , the following trig. results are useful. Each of these results, which are also true for n = 0, ?1, ?2, ?3, . . . , can be deduced from the graph of sin x or that of cos x




? sin n? = 0





????????????


?


?? ?


sin(x)




x
? ?? ??


??





? cos n? = (?1)n





????????????


?


?? ?

cos(x)




x
? ?? ??


??



sin(x)
?

cos(x)
?




????????????


?? ?

x
? ?? ??


????????????


?? ?

x
? ?? ??


?? ??


?
? sin n ? = ?


0 , n even
1 , n = 1, 5, 9, ...

?
? cos n ? = ?


0 , n odd
1 , n = 0, 4, 8, ...

2 ? ?1 , n = 3, 7, 11, ...

2 ? ?1 , n = 2, 6, 10, ...



Areas cancel when when integrating over whole periods


???sin(x)
+ + + x

? ¸ sin nx dx = 0
2?
? ¸ cos nx dx = 0
2?

??? ???

?? ?

??

? ?? ??


? For a waveform f (x) with period L = 2?

f (x) = a0 + . [a
2 n
n=1


cos nkx + bn


sin nkx]

The corresponding Fourier coefficients are



STEP ONE


a0 =

2 ¸
f (x) dx L
L
2 ¸

STEP TWO an =


f (x) cos nkx dx L
L


STEP THREE bn =

2 ¸
f (x) sin nkx dx L

L
and integrations are over a single interval in x of L

k = 2? ?

n?x k

2L = L and nkx = L
?

f (x) = a0 + . .a
2 n
n=1

cos n?x + b
L n

sin n?x .
L

The corresponding Fourier coefficients are



STEP ONE


a0 =

1 ¸
f (x) dx L
2L
1 ¸





n?x

STEP TWO an =


f (x) cos dx L L
2L


STEP THREE bn =

1 ¸
f (x) sin
L
2L

n?x
dx L

and integrations are over a single interval in x of 2L

? For a waveform f (t) with period T = 2?

f (t) = a0 + . [a
2 n
n=1


cos n?t + bn


sin n?t]

The corresponding Fourier coefficients are



STEP ONE


a0 =

2 ¸
f (t) dt T
T
2 ¸

STEP TWO an =


f (t) cos n?t dt T
T


STEP THREE bn =

2 ¸
f (t) sin n?t dt T

T
and integrations are over a single interval in t of T

Section 7: Tips on using solutions 24
7. Tips on using solutions


? When looking at the THEORY, ANSWERS, INTEGRALS, TRIG or NOTATION pages, use the Back button (at the bottom of the page) to return to the exercises


? Use the solutions intelligently. For example, they can help you get started on an exercise, or they can allow you to check whether your intermediate results are correct


? Try to make less use of the full solutions as you work your way through the Tutorial

Full worked solutions
Exercise 1.
. 1, ?? < x < 0

f (x) =

0, 0 < x < ?, and has period 2?



a) Sketch a graph of f (x) in the interval ? 2? < x < 2?


f(x)





? ?
x


b) Fourier series representation of f (x)


STEP ONE

1 ¸ ?

1 ¸ 0

1 ¸ ?

a0 = ?

f (x)dx =
??


? ??

f (x)dx +
?

f (x)dx
0

1 ¸ 0
=


1 · dx +

1 ¸ ?


0 · dx

? ?? ? 0
1 ¸ 0
= dx
? ??
= 1 [ ]0
? ??
1
= ? (0 ? (??))
1
= ? · (?)
i.e. a0 = 1 .


STEP TWO
1 ¸ ?



1 ¸ 0



1 ¸ ?

an = ?

f (x) cos nx dx =
?? ?

f (x) cos nx dx +
?? ?

f (x) cos nx dx
0

1 ¸ 0
=


1 · cos nx dx +

1 ¸ ?
0 · cos nx dx

? ?? ? 0

1 ¸ 0
=


cos nx dx

? ??

1 . sin nx .0
=

= 1 [sin nx]0

? n ?? n? ?
1
= n? (sin 0 ? sin(?n?))
1
= (0 + sin n?) n?
1

i.e. an =

(0 + 0) = 0.
n?


STEP THREE

bn =


=



1 ¸ ?
? ??
1 ¸ 0




f (x) sin nx dx


f (x) sin nx dx +






1 ¸ ?







f (x) sin nx dx

? ?? ? 0

1 ¸ 0
=


1 · sin nx dx +

1 ¸ ?


0 · sin nx dx

? ?? ? 0

1 ¸ 0

1 . ? cos nx .0

i.e. bn =


? ??
1

sin nx dx =
?

0


??
1

= ? n? [cos nx]?? = ? n? (cos 0 ? cos(?n?))

1

1 n

= ? n? (1 ? cos n?) = ? n? (1 ? (?1)
. 0 , n even

) , see Trig
. 1 , n even

i.e. bn =


? n? , n odd

, since (?1)n =


?1 , n odd


We now have that
?

f (x) = a0 + .[a
2 n
n=1

cos nx + bn

sin nx]

with the three steps giving


a0 = 1, an = 0 , and bn =



. 0 , n even
? n? , n odd

It may be helpful to construct a table of values of bn
n 1 2 3 4 5

bn ? 2

0 ? 2 . 1 . 0 ? 2 . 1 .

? ? 3 ? 5
Substituting our results now gives the required series
1 2 . 1 1 .

f (x) = 2 ? ?

sin x + 3 sin 3x + 5 sin 5x + . . .


c) Pick an appropriate value of x, to show that

? 1 1 1
4 = 1 ? 3 + 5 ? 7 + . . .
Comparing this series with
1 2 . 1 1 .

f (x) = 2 ? ?

sin x + 3 sin 3x + 5 sin 5x + . . . ,

we need to introduce a minus sign in front of the constants 1 , 1 ,. . .
3 7
So we need sin x = 1, sin 3x = ?1, sin 5x = 1, sin 7x = ?1, etc
The first condition of sin x = 1 suggests trying x = ? .

This choice gives sin ?

+ 1 sin 3 ?

+ 1 sin 5 ?

+ 1 sin 7 ?

2 3 2 5 2 7 2
i.e. 1 ? 1 1 ? 1
Looking at the graph of f (x), we also have that f ( ? ) = 0.


Picking x = ? thus gives
0 = 1 ? 2 . sin ?




+ 1 sin 3?




+ 1 sin 5?

2 ? 2 3

2 5
7 sin

2
.
2 + . . .


i.e. 0 = 1 ? 2 . 1 ? 1 + 1

2 ? 3
?

5
.
7 + . . .


A little manipulation then gives a series representation of ?

2 . 1 1
? 1 ? 3 + 5
1 1

1 . 1
? 7 + . . . = 2
1 ?

1 ? 3 + 5 ? 7 + . . . = 4 .



Return to Exercise 1


Exercise 2.
. 0, ?? < x < 0

f (x) =

x, 0 < x < ?, and has period 2?



a) Sketch a graph of f (x) in the interval ? 3? < x < 3?


f(x)
?




??? ??? ??

? ?? ??
x


b) Fourier series representation of f (x)

STEP ONE


1 ¸ ?
a0 =
??


f (x)dx =

1 ¸ 0
? ??
1 ¸ 0

1 ¸ ?
f (x)dx +
? 0
1 ¸ ?


f (x)dx

=
? ??

0 · dx + ?

x dx
0

1 . x2 .?
=
? 2 0
1 . ?2 .
= ? 0
? 2
?
i.e. a0 = 2 .

an = ?

f (x) cos nx dx =
?? ?

f (x) cos nx dx +
?? ?

f (x) cos nx dx
0

1 ¸ 0
=


0 · cos nx dx +

1 ¸ ?
x cos nx dx

? ?? ? 0

1 ¸ ?
i.e. an =
0


x cos nx dx =

1 ..
x ?

sin nx .?
n 0

¸ ? sin nx .
dx
0 n

(using integration by parts)

1 ..

sin n?

. 1 .

cos nx .? .

i.e. an = ? ?
1 .

n ? 0

? n ? n 0
1 .

= ? ( 0 ? 0) +
1

[cos nx]? n2
1

=
?n2

{cos n? ? cos 0} =


?n2

{(?1)n ? 1}


i.e. an =

. 0 , n even
? ?n2 , n odd


, see Trig.


STEP THREE
1 ¸ ?



1 ¸ 0



1 ¸ ?

bn = ?

f (x) sin nx dx =
?? ?

f (x) sin nx dx +
?? ?

f (x) sin nx dx
0

1 ¸ 0
=


0 · sin nx dx +

1 ¸ ?


x sin nx dx

? ?? ? 0

1 ¸ ?

1 .. .

cos nx ..?

¸ ? .

cos nx . .

i.e. bn = ?

x sin nx dx = x
0 ? n

0 ? 0 ? n

(using integration by parts)

1 . 1
= ?

[x cos nx]? +

1 ¸ ? .
cos nx dx

? n
1 . 1

n 0
1 . sin nx .? .

= ? ? n

(? cos n? ? 0) + n n

1 n
= ? n (?1)
1 n
= ? n (?1)

1
+ (0 0), see Trig
?n2

. 1
i.e. bn = n
+ 1

, n even

n , n odd


We now have



f (x) = a0 + . [a
2 n
n=1




cos nx + bn




sin nx]


where a = ? , a =
0 2 n

. 0 , n even

2

. 1
, b = n
1


, n even

? ?n2 , n odd

Constructing a table of values gives

n , n odd






? ? 32

? 52


This table of coefficients gives


1 . ? .

. 2 .

f (x) = 2 2

+ ? ?
. 2

cos x + 0 · cos 2x
1 .

+ ? ? · 32

cos 3x + 0 · cos 4x

. 2 1 .

+ ? ? · 52
1

cos 5x + ...

1

+ sin x ? 2 sin 2x + 3 sin 3x ? ...


i.e. f (x) = ?
4

2 .
? ? cos x +
. 1


1
32 cos 3x +
1

1 .
52 cos 5x + ...
.

+ sin x ? 2 sin 2x + 3 sin 3x ? ...

and we have found the required series!


c) Pick an appropriate value of x, to show that

(i) ? = 1 ? 1 + 1 ? 1 + ...
4 3 5 7

Comparing this series with
2 . 1 1 .

f (x) = ?
4

? ? cos x +
. 1

32 cos 3x +
1

52 cos 5x + ...
.

+ sin x ? 2 sin 2x + 3 sin 3x ? ... ,
the required series of constants does not involve terms like 1 , 1 , 1 , ....
32 52 72
So we need to pick a value of x that sets the cos nx terms to zero.
The Trig section shows that cos n ? = 0 when n is odd, and note also that cos nx terms in the Fourier series all have odd n

i.e. cos x = cos 3x = cos 5x = ... = 0 when x = ? ,
i.e. cos ? = cos 3 ? = cos 5 ? = ... = 0
2 2 2


Setting x = ? in the series for f (x) gives

f . ? . = ?


cos ? + 1


3? 1
cos +

5? .
cos + ...

2 4 ? ?
.

2 32
1 2?

2 52
1 3?

2
1 4? 1 5? .

+ sin ?
2

sin
2 2

+ sin
3 2

? 4 sin 2

+ sin
5 2

? ...

? 2
=
4 ? ? [0 + 0 + 0 + ...]
? ?
1 1 1 1
+ ?1 ? 2 sin ? + 3 · (?1) ? 4 sin 2? + 5 · (1) ? ...?

s=¸¸0x

s =¸¸0 x


The graph of f (x) shows that f . ? . = ? , so that






i.e.

2 2


+

3 + ...



+ ...