1. Orthogonality Relations
We now explain the basic reason why the remarkable Fourier coefficent
formulas work. We begin by repeating them from the last note:
1
� L
a0 = f (t) dt,
L −L
1
� L π
an = f (t) cos(n t) dt, (1)
L
�
L L
−
1 L π
bn = f (t) sin(n t) dt.
L −L L
�
� � � �
The key fact is the following collection of integral formulas for sines and
cosines, which go by the name of orthogonality relations:
⎧
⎪ 1 n = m = 0
� L
⎨ �
L
1
−L
cos(nπ
L t) cos(mπ
L t) dt =
⎪
0 n �= m
⎩
2 n = m = 0
� L
1
−L
cos(nπ
L t) sin(mπ
L t) dt = 0
L
1
� L
sin(nπ
L t) sin(mπ
L t) dt =
1 n = m �= 0
L −L
0 n �= m
Proof of the orthogonality relations: This is just a straightforward calcu
lation using the periodicity of sine and cosine and either (or both) of these
two methods:
Method 1: use cos at = eiat+
2
e−iat
, and sin at = eiat−
2i
e−iat
.
Method 2: use the trig identity cos(α) cos(β) = 1
2 (cos(α + β) +cos(α − β)),
and the similar trig identies for cos(α) sin(β) and sin(α) sin(β).
Using the orthogonality relations to prove the Fourier coefficient formula
Suppose we know that a periodic function f (t) has a Fourier series expan
sion
π π
∞
∑
f (t) =
a0
2
+
+ bn sin (2)
t
t
an cos n n
L
L
n=1
How can we find the values of the coefficients? Let’s choose one coefficient,
say a2, and compute it; you will easily how to generalize this to any other
coefficient. The claim is that the right-hand side of the Fourier coefficient
formula (1), namely the integral
1
� L � π �
L −L
f (t) cos 2
L
t dt.
2. Orthogonality Relations OCW 18.03SC
is in fact the coefficent a2 in the series (2). We can�
replace f (t) in this integral
by the series in (2) and multiply through by cos 2π
L t
1 L a0 π ∞
�
, to get
� �
t
�
cos
2
+
∑
� � π � π π
a
� π
n cos n t
cos
2
t
�
+ bn sin
�
n
t
cos
2
t
dt
L L 2
L
n=1
L
L
L
�
−
�
L
��
Now the orthogonality relations tell us that almost every term in this sum
will integrate to 0. In fact, the only non-zero term is the n = 2 cosine term
1
� L π π
a2 cos
L
�
2 t
�
cos
�
2 t
�
dt
−L L L
and the orthogonality relations for the case n = m = 2 show this integral is
equal to a2 as claimed.
a
Why the denominator of 2 in
0
?
2
Answer: it is in fact just a convention, but the one which allows us to have
the same Fourier coefficent formula for an when n = 0 and n ≥ 1. (Notice
that in the n = m case for cosine, there is a factor of 2 only for n = m = 0.)
a
Interpretation of the constant term
0
.
2
We can also interpret the constant term a0
2 in the Fourier series of f (t) as the
average of the function f (t) over one full period: a0
� L
= 1
2 2L −L
f (t) dt.
2
