2Fourier series

IB Methods

2.1 Fourier series
So. As mentioned in the previous chapter, we want to find a set of “basis
functions” for periodic functions. We could go with the simplest case of periodic
functions we know of the exponential function
e
inθ
. These functions have a
period of 2
π
, and are rather easy to work with. We all know how to integrate
and differentiate the exponential function.
More importantly, this set of basis functions is orthogonal. We have
(e
imθ
, e
inθ
) =
Z
π
π
e
imθ
e
inθ
dθ =
Z
π
π
e
i(nm)θ
dθ =
(
2π n = m
0 n 6= m
= 2πδ
nm
We can normalize these to get a set of orthonormal functions
n
1
2π
e
inθ
: n Z
o
.
Fourier’s idea was to use this as a basis for any periodic function. Fourier
claimed that any f : S
1
C can be expanded in this basis:
f(θ) =
X
nZ
ˆ
f
n
e
inθ
,
where
ˆ
f
n
=
1
2π
(e
inθ
, f) =
1
2π
Z
π
π
e
inθ
f(θ) dθ.
These really should be defined as
f
(
θ
) =
P
ˆ
f
n
e
inθ
2π
with
ˆ
f
n
=
e
inθ
2π
, f
, but for
convenience reasons, we move all the constant factors to the
ˆ
f
n
coefficients.
We can consider the special case where
f
:
S
1
R
is a real function. We
might want to make our expansion look a bit more “real”. We get
(
ˆ
f
n
)
=
1
2π
Z
π
π
e
inθ
f(θ) dθ
=
1
2π
Z
π
π
e
inθ
f(θ) dθ =
ˆ
f
n
.
So we can replace our Fourier series by
f(θ) =
ˆ
f
0
+
X
n=1
ˆ
f
n
e
inθ
+
ˆ
f
n
e
inθ
=
ˆ
f
0
+
X
n=1
ˆ
f
n
e
inθ
+
ˆ
f
n
e
inθ
.
Setting
ˆ
f
n
=
a
n
+ ib
n
2
, we can write this as
f(θ) =
ˆ
f
0
+
X
n=1
(a
n
cos + b
n
sin )
=
a
0
2
+
X
n=1
(a
n
cos + b
n
sin ).
Here the coefficients are
a
n
=
1
π
Z
π
π
cos f(θ) dθ, b
n
=
1
π
Z
π
π
sin f(θ) dθ.
This is an alternative formulation of the Fourier series in terms of sin and cos.
So when given a real function, which expansion should we use? It depends. If
our function is odd (or even), it would be useful to pick the sine/cosine expansion,
since the cosine (or sine) terms will simply disappear. On the other hand, if we
want to stick our function into a differential equation, exponential functions are