The document explains orthogonality relations, which are integral formulas for sines and cosines that show that integrating the product of two distinct trigonometric functions over a period equals zero. These relations allow the Fourier coefficient formulas to work by showing that when replacing a function in a Fourier series with the series itself, only the term with the same frequency remains after integrating. This proves the formulas for computing the coefficients from integrals of the function multiplied by sines and cosines.

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

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18.03SC Differential Equations��
Fall 2011 ��
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