November 29, 2010
22-1
22. Periodic Functions and Fourier Series
1
Periodic Functions
A real-valued function f (x) of a real variable is called
periodic of period T > 0 if f (x + T ) = f (x) for all
x ∈ R.
For instance the functions sin(x), cos(x) are periodic
of period 2π. It is also periodic of period 2nπ, for any
positive integer n. So, there may be infinitely many periods. If needed we may specify the least period as the
number T > 0 such that f (x + T ) = f (x) for all x, but
f (x + s) 6= f (x) for 0 < s < T .
For later convenience, let us consider piecewise C 1
functions f (x) which are periodic of period 2L > 0 where
L is a positive real number. Denote this class of functions
by P erL(R).
nπx
Note that for each integer n, the functions cos( nπx
L ), sin( L )
are in examples of such functions. Also, note that if
f (x), g(x) ∈ P erL(R), and α, β are constants, then
αf + βg is also in P erL(R).
In particular, any finite sum
k
a0
mπx
mπx !
X
+
am cos(
) + bm sin(
)
2 m=1
L
L
is in P erL(R). Here the numbers a0, am, bm are constants.
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2
22-2
Fourier Series
The next result shows that in many cases the infinite sum
∞
a0
mπx
mπx !
X
f (x) =
+
am cos(
) + bm sin(
) (1)
2 m=1
L
L
determines a well-defined function f (x) which again is
in P erL(R).
An infinite sum as in formula (1) is called a Fourier
series (after the French engineer Fourier who first considered properties of these series).
Fourier Convergence Theorem. Let f (x) be a
piecewise C 1 function in P erL(R). Then, there are
constants a0, am, bm (uniquely defined by f ) such that
at each point of continuity of f (x) the expression on
the right side of (1) converges to f (x). At the points
y of discontinuity of f (x), the series converges to
1
(f (y−) + f (y+)).
2
The values f (y−), f (y+) denote the left and right limits of f as x → y, respectively.
That is,
f (y−) = limx→y,x<y f (x), f (y+) = limx→y,x>y f (x).
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22-3
It turns out that the constants a0, am, bm above are
determined by the formulas
1 ZL
a0 =
f (x)dx
(2)
L −L
mπx
1 ZL
f
(x)
cos(
am =
)dx, and
(3)
L −L
L
1 ZL
mπx
bm =
)dx.
(4)
f
(x)
sin(
L −L
L
We will justify this a bit later, but for now, let us
use these formulas to compute some Fourier series. The
constants a−, am, bm are called the Fourier coefficients
of f .
Example
1. Let f (x) be defined by


 −x,
−2 ≤ x < 0
f (x) = 

x,
0≤x<2
f (x + 4) = f (x)
for all x.
Determine the Fourier coefficients of f .
Note that the graph of this function f (x) looks like a
“triangular wave.”
Here L = 2, and we compute
1Z2
1Z0
a0 =
(−x)dx + 0 xdx
2 −2
2
= 1 + 1 = 2,
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22-4
and, for m > 0,
1Z0
mπx
mπx
1Z2
am =
(−x)
cos(
x
cos(
)dx
+
)dx
2 −2
2
2 0
2
mπx
mπx
1Z2
1Z0
(−x)
sin(
x
sin(
)dx
+
)dx.
bm =
2 −2
2
2 0
2
To compute these integrals, we note that, integration
by parts gives the formulas
Z sin(ax)
x
sin(ax) −
dx
a
a
x
cos(ax)
= sin(ax) +
a
a2
Z
sin(ax)
x
x sin(ax)dx = − cos(ax) +
a
a2
Let us compute am.
First note that the integrand
mπx
)
h(x) = f (x)cos(
2
Z
x cos(ax)dx =
(5)
satisfies h(−x) = h(x). Hence, the two integrals are
equal and
Z 2
Z 2
mπx
mπx
am = 2(1/2) 0 x cos(
)dx = 0 x cos(
)dx
2
2
Using the formula for (5) above, we get
November 29, 2010
22-5
mπx
)dx
0
2
2


2

mπx
mπx
2
2
 x sin(
 cos(
)+
)
= 
mπ
2
mπ
2
0


4
= 0 +  2 2  (cos(mπ) − 1)
mπ

8

 −
, m odd
(mπ)2
= 

0, m even
am =
Z
2
x cos(
A similar calculation shows that bm = 0 for all m. We
will see later that this last fact follows from the fact that
f (−x) = f (x) for all x.
Example 2. Let f (x) be defined by







0,
f (x) = 
1,




 0,
−3 < x < −1,
−1 < x < 1,
1<x<3
f (x + 6) = f (x)
for all x.
Find the Fourier series for f .
Here L = 3, and we compute
1Z3
f (x)dx
3 −3
1Z1
=
f (x)dx
3 −1
a0 =
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22-6
=
2
3
For m > 0, we have
an
bn =
=
=
=
3
1Z1
nπx
=
cos(
)dx
−1
3
3
1

nπx
1
)
=  sin(
nπ
3 −1
nπ
2
sin( )
=
nπ
3
1Z1
nπx
sin(
)dx
3 −1
3
1
1
nπx
−  cos(
)
nπ
3 −1

1 
nπ
−nπ 
−
cos( ) − cos(
)
nπ
3
3
0
Justification of the Fourier coefficient formulas
We need the following basic facts about the integrals of
certain products of sines and cosines.
November 29, 2010
22-7







0,
mπx
nπx
cos(
)cos(
)dx = 
L,
−L


L
L


 L,
L
Z
Z
L
cos(
−L
nπx
mπx
) sin(
)dx = 0 for all m, n;
L
L







0,
mπx
nπx
sin(
) sin(
)dx = 
L,
−L


L
L


 0,
Z
L
m 6= n,
m = n 6= 0,
m=n=0
m 6= n,
m = n 6= 0,
m=n=0
(6)
(7)
(8)
We justify formula (8), leaving the other similar calculations to the reader. First recall some formulas related
to the sine and cosine functions.
The sum and difference formulas are:
cos(α + β) = cos(α)(cos(β) − sin(α) sin(β)
cos(α − β) = cos(α)(cos(β) + sin(α) sin(β).
Applying the first formula with α = β gives
cos(2α) = cos(α)2 − sin(α)2
(9)
(10)
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22-8
This implies that
1 + cos(2α) = cos(α)2 + sin(α)2 + cos(α)2 − sin(α)2
= 2 cos(α)2
or the so-called cosine half-angle formula
1
cos(α)2 = (1 + cos(2α)).
2
Similarly, the sine half-angle formula is
1
sin(α)2 = (1 − cos(2α)).
2
Formulas (9) and (10) imply that
cos(α − β) − cos(α + β) = 2 sin(α) sin(β).
Using α = nx, β = mx then gives
cos((n − m)x) − cos((n + m)x) = 2 sin(nx) sin(mx).
(11)
This implies, for m 6= n and both positive,
mπx
nπx
1ZL
(m − n)πx
)
sin(
)dx
=
cos(
)dx
sin(
−L
L
L
2 −L
L
1ZL
(m + n)πx
− −L cos(
)dx
2
L
Z
L
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22-9

(m−n)πx
)
 sin(
L

L

2π
= 0.
=
m−n
−

(m+n)πx L
sin( L ) 
m+n
If m = n = 0, then
Z L
Z L
mπx
mπx
sin(
) sin(
)dx = −L 0 dx = 0,
−L
L
L
while if m = n 6= 0, we have
mπx !2
sin(
) dx
−L
L


1ZL 
2mπx 
=
1 − cos(
) dx
2 −L
L


2mπx L
1
sin(
)
= x − 2mπL 
2
L
−L
= L.
mπx
mπx
)
sin(
)dx =
sin(
−L
L
L
Z
L
Z
L
Now, suppose that
a0
mπx
mπx
X
+
am cos(
) + bm sin(
). (12)
2 m≥1
L
L
Since the integrals of cosine and sine functions over
intervals of lengths equal to their periods vanish, we have
f (x) =
Z
L
f (x)dx =
−L
Z
L
(
−L
a0
mπx
mπx
X
+
am cos(
) + bm sin(
))dx
2 m≥1
L
L

−L
November 29, 2010
22-10
a0
mπx
X Z L
=
a
cos(
(2L) +
)dx
m
2
L
m≥1 −L
mπx
X Z L
b
sin(
)dx
+
m
−L
L
m≥1
= a0 L
Analogously, using the orthogonality relations above,
we have that, for n ≥ 1,
nπx a0
mπx
X
)( +
am cos(
))dx
L
2 m≥1
L
Z L
nπx X
mπx
+ −L cos(
)(
bm sin(
))dx
L m≥1
L
= am L
nπx
)dx =
f
(x)
cos(
−L
L
Z
L
Z
L
cos(
−L
which gives (3). Formula (4) is justified in a similar
way.
4
Even and Odd functions
A function f (x) is called even if f (−x) = f (x) for all
x. Analogously, a function f (x) is called odd if f (−x) =
−f (x) for all x. For example, cos(x) is even, and sin(x)
is odd.
Also, one sees easily that linear combinations of even
(odd) functions are again even (odd).
The following facts are useful.
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22-11
1. The product of two odd functions is even.
2. The product of two even functions is even.
3. The product of and even function and an odd function is odd.
The following comments are useful.
4. The Fourier series of a periodic odd function only
involves sine terms: i.e.; am = 0 for all m ≥ 1. This
is because
mπx
1 ZL
f
(x)
cos(
)dx
am =
L −L
L
and the function f (x) cos( mπx
L ) is odd.
5. Similarly, the Fourier series of a periodic even function only involves cosine terms: i.e.; bm = 0 for all
m ≥ 1.
6. Any function f (x) defined on [0, L] (with L > 0 has
an even extension feven(x) to [−L, L] and also has
an odd extension fodd(x) to [−L, L].
The definitions are



f (x)
f (−x)
0≤x≤L
−L ≤ x ≤ 0
f (x)
−f (−x)
0≤x≤L
−L ≤ x ≤ 0
feven(x) = 




fodd(x) = 

November 29, 2010
22-12
Hence, any piecewise continuous function f (x) on [0, L]
can be represented both as Fourier cosine series and a
Fourier sine series.