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
f (t) dt,
L −L
�
1 L
π
f (t) cos(n t) dt,
an =
L −L
L
�
1 L
π
f (t) sin(n t) dt.
bn =
L −L
L
a0 =
�
(1)
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
⎨
�
1 L
π
π
L − L cos( n L t ) cos( m L t ) dt = ⎪ 0 n � = m
⎩
2 n=m=0
�
1 L
π
π
L − L cos( n L t ) sin( m L t ) dt = 0
�
�
1 n = m �= 0
1 L
π
π
L − L sin( n L t ) sin( m L t ) dt =
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:
iat
−iat
iat
−iat
Method 1: use cos at = e +2e , and sin at = e −2ie .
Method 2: use the trig identity cos(α) cos( β) = 21 (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 �
∞
a0
f (t) =
+
∑ an cos n t
+ bn sin n
t
(2)
2
n=1
L
L
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
�
� π �
1 L
f (t) cos 2 t dt.
L −L
L
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 2 πL t , to get
1
L
� π
�
� π �
� π
�
� π
�
� π
��
∞ �
cos
2
t
+
∑ an cos n t
cos
2
t
+ bn sin n
t
cos
2
t
dt
L
2
L
L
L
L
n =1
� L
a0
−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
� L
� π �
� π �
a2 cos 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.
a0
Why the denominator of 2 in
?
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.)
a0
Interpretation of the constant term .
2
We can also interpret the constant term a20 in the Fourier series of f (t) as the
�L
1
average of the function f (t) over one full period: a20 = 2L
− L f ( t ) dt.
2
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