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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.
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
MIT OpenCourseWare
http://ocw.mit.edu
18.03SC Differential Equations��
Fall 2011 ��
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

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Orthogonality relations pdf

  • 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
  • 3. MIT OpenCourseWare http://ocw.mit.edu 18.03SC Differential Equations�� Fall 2011 �� For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.