approximate fourier series - school of mathematics and ... · • in this lecture, we will apply...
TRANSCRIPT
Lesson 3 Approximating Fourier series
1
2
• Last lecture, we saw that the trapezoidal rule was an effective method for calculatingintegrals of periodic functions
We used the Euler–McLaurin formula to prove that the error decayed fasterthan O
�1
n�
�for any �
• In this lecture, we will apply this to calculating Fourier coefficients and Fourier series
• This is practical function approximation
3
• Given an integrable function f defined on T = [��, �), the following exist forevery integer k:
f̂k =1
2�
� �
��f(�) � k� x
• We can formally define the Fourier series of f :
f(�) ���
k=��f̂k
k�
This sum will converge uniformly to f for periodic smooth functions
4
-3 -2 -1 0 1 2 3
• Define the m evenly spaced points on the periodic interval � = (�1, . . . , �m):
� :=
���,
�2
m� 1
��, . . . ,
�1 � 2
m
��
�=
• We can approximate the Fourier coefficients using the m point trapezoidal rule
f̂k =1
2�
� �
��f(�) � � 1
m
m�
j=1
f(�j)�k�j =: f̂m
k
5
-3 -2 -1 0 1 2 3
• Define the m evenly spaced points on the periodic interval � = (�1, . . . , �m):
� :=
���,
�2
m� 1
��, . . . ,
�1 � 2
m
��
�=
• We can approximate the Fourier coefficients using the m point trapezoidal rule
f̂k =1
2�
� �
��f(�) � � 1
m
m�
j=1
f(�j)�k�j =: f̂m
k
6
• Using these, and truncating the Fourier sum between integers � and � we obtainan approximate Fourier series
f(�) � f�,�,m(�) :=��
k=�
f̂mk
k�
• Big question: how to choose �, � and m?
• When we specify just � and � , we will choose m to be the same as the numberof coefficients
f�,�(�) := f�,�,���+1(�)
• When we specify just m, we will choose roughly equal number of negative andpositive coefficients:
fm(�) :=
�f 1�m
2 , m�12 ,m(�) m odd
f� m2 , m
2 �1,m(�) m even
7
• Using these, and truncating the Fourier sum between integers � and � we obtainan approximate Fourier series
f(�) � f�,�,m(�) :=��
k=�
f̂mk
k�
• Big question: how to choose �, � and m?
• When we specify just � and � , we will choose m to be the same as the numberof coefficients
f�,�(�) := f�,�,���+1(�)
• When we specify just m, we will choose roughly equal number of negative andpositive coefficients:
fm(�) :=
�f 1�m
2 , m�12 ,m(�) m odd
f� m2 , m
2 �1,m(�) m even
8
• Using these, and truncating the Fourier sum between integers � and � we obtainan approximate Fourier series
f(�) � f�,�,m(�) :=��
k=�
f̂mk
k�
• Big question: how to choose �, � and m?
• When we specify just � and � , we will choose m to be the same as the numberof coefficients
f�,�(�) := f�,�,���+1(�)
• When we specify just m, we will choose roughly equal number of negative andpositive coefficients:
fm(�) :=
�f 1�m
2 , m�12 ,m(�) m odd
f� m2 , m
2 �1,m(�) m even
Experimental results
9
10
�
-3 -2 -1 1 2 3q
5
10
m = 5
11
�
-3 -2 -1 1 2 3q
5
10
m = 5
-3 -2 -1 1 2 3q
0.5
1.0
1.5
2.0
2.5
3.0
m = 5
|� � .1|
12
-3 -2 -1 1 2 3q
-1.0
-0.5
0.5
1.0
m = 5
�
-3 -2 -1 1 2 3q
5
10
m = 5
20
��
-3 -2 -1 1 2 3q
0.5
1.0
1.5
2.0
2.5
3.0
m = 5
|� � .1| (� � .1)
-3 -2 -1 1 2 3q
-1.0
-0.5
0.5
1.0
1.5m = 5
-3 -2 -1 1 2 3q
-1.0
-0.5
0.5
1.0m = 5
5�
-3 -2 -1 1 2 3q
-0.5
0.5
1.0m = 5
2 �
13
�20
��
|� � .1| (� � .1)
5�
-3 -2 -1 1 2 3q
-0.4
-0.2
0.2
0.4
0.6
0.8
1.0m = 10
-3 -2 -1 1 2 3q
-1.0
-0.5
0.5
1.0m = 10
2 �
-3 -2 -1 1 2 3q
-1.0
-0.5
0.5
1.0
m = 10
-3 -2 -1 1 2 3q
0.5
1.0
1.5
2.0
2.5
3.0
m = 10
-3 -2 -1 1 2 3q
-1.0
-0.5
0.5
1.0m = 10
-3 -2 -1 1 2 3q
5
10
15
m = 10
14
�20
��
|� � .1| (� � .1)
5�
-3 -2 -1 1 2 3q
-0.4
-0.2
0.2
0.4
0.6
0.8
1.0m = 100
-3 -2 -1 1 2 3q
-1.0
-0.5
0.5
1.0m = 100
2 �
-3 -2 -1 1 2 3q
-1.0
-0.5
0.5
1.0
m = 100
-3 -2 -1 1 2 3q
0.5
1.0
1.5
2.0
2.5
3.0
m = 100
-3 -2 -1 1 2 3q
-1.0
-0.5
0.5
1.0m = 100
-3 -2 -1 1 2 3q
5
10
15
20
25m = 100
15
• Observed convergence properties:
Fast convergence for periodic functions, just like trapezoidal ruleSlow convergence for non-periodic functions away from singularitiesNo convergence in neighbourhood of jump singularities (including ±�)
• We also observed interpolation at the quadrature points �
• To understand this, we need to related the approximate Fourier coefficients f̂mk
to the true Fourier coefficients f̂k
• This will follow naturally from orthogonality properties of �
Orthogonality of complex exponentials
16
17
The L2 inner product and norm• We define the 2 inner product (on T) by
�f, g� =1
2�
� �
��f̄(�)g(�) �
• Associated with this inner product is the 2 norm:
�f� =
�1
2�
� �
��|f(�)|2 �
• 2 space is all integrable functions f such that �f� < �
Exercise: verify 2 is a vector space and �f, g� is an inner product on 2
18
• A set of nonzero vectors v1, . . . , vn in a vector space V are called orthogonal if
�vi, vj� = 0 whenever i �= k.
• They are called orthonormal if they are orthogonal and all vectors are of unit norm:
1 = �vi� , or equivalently, �vi, vi� = 1.
• For orthonormal vectors vk , we can construct a projection of a vector f � V into{v1, . . . , vn} by
Pf :=n�
k=1
�vk, f� vk
If f � {v1, . . . , vn} then f is equal to its projection:
f = Pf
In other words P2f = Pf
19
• A set of nonzero vectors v1, . . . , vn in a vector space V are called orthogonal if
�vi, vj� = 0 whenever i �= k.
• They are called orthonormal if they are orthogonal and all vectors are of unit norm:
1 = �vi� , or equivalently, �vi, vi� = 1.
• For orthonormal vectors vk , we can construct a projection of a vector f � V into{v1, . . . , vn} by
Pf :=n�
k=1
�vk, f� vk
If f � {v1, . . . , vn} then f is equal to its projection:
f = Pf
In other words P2f = Pf
20
• A set of nonzero vectors v1, . . . , vn in a vector space V are called orthogonal if
�vi, vj� = 0 whenever i �= k.
• They are called orthonormal if they are orthogonal and all vectors are of unit norm:
1 = �vi� , or equivalently, �vi, vi� = 1.
• For orthonormal vectors vk , we can construct a projection of a vector f � V into{v1, . . . , vn} by
Pf :=n�
k=1
�vk, f� vk
If f � {v1, . . . , vn} then f is equal to its projection:
f = Pf
In other words P2f = Pf
21
• A set of nonzero vectors v1, . . . , vn in a vector space V are called orthogonal if
�vi, vj� = 0 whenever i �= k.
• They are called orthonormal if they are orthogonal and all vectors are of unit norm:
1 = �vi� , or equivalently, �vi, vi� = 1.
• For orthonormal vectors vk , we can construct a projection of a vector f � V into{v1, . . . , vn} by
Pf :=n�
k=1
�vk, f� vk
If f � {v1, . . . , vn} then f is equal to its projection:
f = Pf
In other words P2f = Pf
22
• We have�
k�, k��
=1
2�
� �
��� = 1
and for k �= j
�k�, j�
�=
1
2�
� �
��
(j�k)� � =(j�k)� � � (j�k)�
2� (j � k)= 0
• In other words, the complex exponentials are orthonormal!
• Thus we can think of the Fourier series as an infinite projection
f(�) ���
k=��
�k�, f
�k�
Since this sum is infinite, we cannot appeal to the simple argument of equalityfrom the last slide
Discrete orthogonality of complex exponentials
23
24
• We have shown that the complex exponentials are orthogonal with respect to theinner product
�f, g� =1
2�
� �
��f̄(�)g(�) �
• A remarkable fact we now show is that they are also orthogonal with respect tothe following discrete semi-inner product:
�f, g�m =1
m
m�
j=1
f̄(�j)g(�j) =f(�)�g(�)
m
where � = (�1, . . . , �m) are again evenly spaced points:
� =
���,
�2
m� 1
��, . . . ,
�1 � 2
m
��
�= -3 -2 -1 0 1 2 3
25
Evenly spaced points on the unit circle
ei✓
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
-3 -2 -1 0 1 2 3
� = (�1, . . . , �m) z = (z1, . . . , zm)
26
Some identities (shown for even m):
z
m�
j=1
ei�j =m�
j=1
zj = 0-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
27
Some identities (shown for even m):
z
m�
j=1
ei2�j =m�
j=1
z2j = 0
-1.0 -0.5 0.5 1.0
-0.5
0.5
28
:
m�
j=1
k�j = (�)km for k = . . . , �2m, �m, 0, m, 2m. . .
m�
j=1
k�j = 0 for all other integer k
• Note thatzj = � 2� (j�1)/m = ��j�1
for � = 11/m = 2� /m
• Therefore,m�
j=1
zkj = (�)k
m�1�
j=0
�kj
• If k = �m is a multiple of m, then we have
m�1�
j=0
�kj =m�1�
j=0
(�m)�j =m�1�
j=0
1�j = m
• Note thatzj = � 2� (j�1)/m = ��j�1
for � = 11/m = 2� /m
• Therefore,m�
j=1
zkj = (�)k
m�1�
j=0
�kj
• If k = �m is a multiple of m, then we have
m�1�
j=0
�kj =m�1�
j=0
(�m)�j =m�1�
j=0
1�j = m
• Note thatzj = � 2� (j�1)/m = ��j�1
for � = 11/m = 2� /m
• Therefore,m�
j=1
zkj = (�)k
m�1�
j=0
�kj
• If k = �m is a multiple of m, then we have
m�1�
j=0
�kj =m�1�
j=0
(�m)�j =m�1�
j=0
1�j = m
• Note thatzj = � 2� (j�1)/m = ��j�1
for � = 11/m = 2� /m
• Therefore,m�
j=1
zkj = (�)k
m�1�
j=0
�kj
• If k = �m is a multiple of m, then we have
m�1�
j=0
�kj =m�1�
j=0
(�m)�j =m�1�
j=0
1�j = m
• For k not a multiple of m, recall the geometric sum formula:
m�1�
j=0
zj =zm � 1
z � 1
• Thusm�1�
j=0
�kj =m�1�
j=0
(�k)j =�km � 1
�k � 1
• Because k is not a multiple of m, the denominator is nonzero
• Because �km = (�m)k = 1k , the numerator is zero
• Note thatzj = � 2� (j�1)/m = ��j�1
for � = 11/m = 2� /m
• Therefore,m�
j=1
zkj = (�)k
m�1�
j=0
�kj
• If k = �m is a multiple of m, then we have
m�1�
j=0
�kj =m�1�
j=0
(�m)�j =m�1�
j=0
1�j = m
• For k not a multiple of m, recall the geometric sum formula:
m�1�
j=0
zj =zm � 1
z � 1
• Thusm�1�
j=0
�kj =m�1�
j=0
(�k)j =�km � 1
�k � 1
• Because k is not a multiple of m, the denominator is nonzero
• Because �km = (�m)k = 1k , the numerator is zero
• Note thatzj = � 2� (j�1)/m = ��j�1
for � = 11/m = 2� /m
• Therefore,m�
j=1
zkj = (�)k
m�1�
j=0
�kj
• If k = �m is a multiple of m, then we have
m�1�
j=0
�kj =m�1�
j=0
(�m)�j =m�1�
j=0
1�j = m
• For k not a multiple of m, recall the geometric sum formula:
m�1�
j=0
zj =zm � 1
z � 1
• Thusm�1�
j=0
�kj =m�1�
j=0
(�k)j =�km � 1
�k � 1
• Because k is not a multiple of m, the denominator is nonzero
• Because �km = (�m)k = 1k , the numerator is zero
��
k�, j��
m= (�1)j�k JSV j � k = . . . , �2m, �m, 0, m, 2m, . . .
�k�, j�
�m
= 0 SXLIV[MWI
�
� *SPPS[W�JVSQ�TVIGIHMRK�XLISVIQ�
�k�, j�
�m
=1
m
m�
j=1
(j�k)�j