Numerical Complex Analysis · 2013. 7. 29. · Numerical complex analysis • A combination of...
Transcript of Numerical Complex Analysis · 2013. 7. 29. · Numerical complex analysis • A combination of...
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Numerical Complex Analysis
Dr Sheehan Olver
1Tuesday, 30 July 13
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• Introduction• What is “Numerical Complex Analysis”?• What are its applications?• Algebraic versus spectral convergence
• Project description• Lesson 1: Review of Fourier analysis• When and in what sense do Fourier series converge?• How fast?
Today’s lecture
2Tuesday, 30 July 13
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Numerical complex analysis
• A combination of harmonic analysis, numerical analysis, complex analysis, functional analysis and approximation theory in order to:• Use complex analysis to come up with “good” numerical schemes• Apply numerical analysis to problems which arise in complex analysis
3Tuesday, 30 July 13
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Root finding
-1.0 -0.5 0.5 1.0
-0.1
0.1
0.2
0.3
0.4
-1.0 -0.5 0.5 1.0
-0.1
0.1
0.2
0.3
0.4
4Tuesday, 30 July 13
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Root finding
-1.0 -0.5 0.5 1.0
-0.1
0.1
0.2
0.3
0.4
-1.0 -0.5 0.5 1.0
-0.1
0.1
0.2
0.3
0.4
4Tuesday, 30 July 13
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Signal smoothing
-3 -2 -1 1 2 3
-0.5
0.5
-3 -2 -1 1 2 3
-0.5
0.5
5Tuesday, 30 July 13
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Signal smoothing
-3 -2 -1 1 2 3
-0.5
0.5
-3 -2 -1 1 2 3
-0.5
0.5
5Tuesday, 30 July 13
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Laplace’s equation�u = 0, �u = f
6Tuesday, 30 July 13
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Integrable PDEsut + uxxx + 6uux = 0, u(0, x) = f(x)
7Tuesday, 30 July 13
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Spectral vs algebraic convergence
8Tuesday, 30 July 13
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Finite element methods Spectral methods
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
-1.0 -0.5 0.5 1.0
-1.0
-0.5
0.5
1.0
9Tuesday, 30 July 13
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n
1/np
p(x) =n�
k=0
ckxk
1/np p
��n � > 0
10Tuesday, 30 July 13
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n
1/np
p(x) =n�
k=0
ckxk
1/np p
��n � > 0
10Tuesday, 30 July 13
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Spectral methods for differential equationsu
00(x) + exu(x) = 0, u(�1) = 0, u(1) = 1
Classical finite difference Spectral
500 1000 1500 2000
5¥10-71¥10-65¥10-61¥10-55¥10-51¥10-4
Number of degrees of freedom
Error
50 100 150 200
10-1210-910-6
0.001
11Tuesday, 30 July 13
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Project
12Tuesday, 30 July 13
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• A major component of the courses evaluation will be a project
Project
12Tuesday, 30 July 13
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• A major component of the courses evaluation will be a project
• Timeline:
Project
12Tuesday, 30 July 13
-
• A major component of the courses evaluation will be a project
• Timeline:
• Short proposal (< 1 page) will be due before the spring vacation
Project
12Tuesday, 30 July 13
-
• A major component of the courses evaluation will be a project
• Timeline:
• Short proposal (< 1 page) will be due before the spring vacation
• Project will be due exam week
Project
12Tuesday, 30 July 13
-
• A major component of the courses evaluation will be a project
• Timeline:
• Short proposal (< 1 page) will be due before the spring vacation
• Project will be due exam week
• The goal of the project is to use ideas from the course to do something original
Project
12Tuesday, 30 July 13
-
• A major component of the courses evaluation will be a project
• Timeline:
• Short proposal (< 1 page) will be due before the spring vacation
• Project will be due exam week
• The goal of the project is to use ideas from the course to do something original
• Most projects will consist of a programming component and an approximately 10 page write-up describing the maths behind the programming
Project
12Tuesday, 30 July 13
-
• A major component of the courses evaluation will be a project
• Timeline:
• Short proposal (< 1 page) will be due before the spring vacation
• Project will be due exam week
• The goal of the project is to use ideas from the course to do something original
• Most projects will consist of a programming component and an approximately 10 page write-up describing the maths behind the programming
• A good example project would be to solve a problem from another course numerically using the methods we discuss
Project
12Tuesday, 30 July 13
-
• A major component of the courses evaluation will be a project
• Timeline:
• Short proposal (< 1 page) will be due before the spring vacation
• Project will be due exam week
• The goal of the project is to use ideas from the course to do something original
• Most projects will consist of a programming component and an approximately 10 page write-up describing the maths behind the programming
• A good example project would be to solve a problem from another course numerically using the methods we discuss
• Check out examples from www.maths.ox.ac.uk/chebfun for inspiration on developing a project
Project
12Tuesday, 30 July 13
http://www.maths.ox.ac.uk/chebfunhttp://www.maths.ox.ac.uk/chebfun
-
• A major component of the courses evaluation will be a project
• Timeline:
• Short proposal (< 1 page) will be due before the spring vacation
• Project will be due exam week
• The goal of the project is to use ideas from the course to do something original
• Most projects will consist of a programming component and an approximately 10 page write-up describing the maths behind the programming
• A good example project would be to solve a problem from another course numerically using the methods we discuss
• Check out examples from www.maths.ox.ac.uk/chebfun for inspiration on developing a project
• Using chebfun (or other software packages) is permitted, but remember that it is expected that you understand what’s going on behind the scenes
Project
12Tuesday, 30 July 13
http://www.maths.ox.ac.uk/chebfunhttp://www.maths.ox.ac.uk/chebfun
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Lesson 1Review of Fourier Analysis
13Tuesday, 30 July 13
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f T = [��, �)k
f̂k =1
2�
� �
��f(�) � k� x
f
f(�) ���
k=��f̂k
k�
f
� � T
f(�) =n��
n�
k=�nf̂k
k�?
14Tuesday, 30 July 13
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f T = [��, �)k
f̂k =1
2�
� �
��f(�) � k� x
f
f(�) ���
k=��f̂k
k�
f
� � T
f(�) =n��
n�
k=�nf̂k
k�?
14Tuesday, 30 July 13
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f T = [��, �)k
f̂k =1
2�
� �
��f(�) � k� x
f
f(�) ���
k=��f̂k
k�
f
� � T
f(�) =n��
n�
k=�nf̂k
k�?
14Tuesday, 30 July 13
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-3 -2 -1 1 2 3
5
10
15
20
n = 5
-3 -2 -1 1 2 3
5
10
15
20
n = 5
e�
15Tuesday, 30 July 13
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-3 -2 -1 1 2 3
5
10
15
20
n = 10
-3 -2 -1 1 2 3
5
10
15
20
n = 10
e�
15Tuesday, 30 July 13
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-3 -2 -1 1 2 3
5
10
15
20
25n = 100
-3 -2 -1 1 2 3
5
10
15
20
25n = 100
e�
15Tuesday, 30 July 13
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-3 -2 -1 1 2 3
5
10
15
20
n = 1000
-3 -2 -1 1 2 3
5
10
15
20
n = 1000
e�
15Tuesday, 30 July 13
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-3 -2 -1 1 2 3
-40
-30
-20
-10
n = 5
-3 -2 -1 1 2 3
-40
-30
-20
-10
n = 5
(� � �)(� + �)e�
16Tuesday, 30 July 13
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-3 -2 -1 1 2 3
-40
-30
-20
-10
n = 10
-3 -2 -1 1 2 3
-40
-30
-20
-10
n = 10
(� � �)(� + �)e�
16Tuesday, 30 July 13
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-3 -2 -1 1 2 3
-40
-30
-20
-10
n = 100
-3 -2 -1 1 2 3
-40
-30
-20
-10
n = 100
(� � �)(� + �)e�
16Tuesday, 30 July 13
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-3 -2 -1 1 2 3
-40
-30
-20
-10
n = 1000
-3 -2 -1 1 2 3
-40
-30
-20
-10
n = 1000
(� � �)(� + �)e�
16Tuesday, 30 July 13
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-3 -2 -1 1 2 3
50
100
150
200
250
n = 5
-3 -2 -1 1 2 3
50
100
150
200
250
n = 5
(� � �)2(� + �)2e�
17Tuesday, 30 July 13
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-3 -2 -1 1 2 3
50
100
150
200
250
n = 10
-3 -2 -1 1 2 3
50
100
150
200
250
n = 10
(� � �)2(� + �)2e�
17Tuesday, 30 July 13
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-3 -2 -1 1 2 3
50
100
150
200
250
n = 100
-3 -2 -1 1 2 3
50
100
150
200
250
n = 100
(� � �)2(� + �)2e�
17Tuesday, 30 July 13
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-3 -2 -1 1 2 3
50
100
150
200
250
n = 1000
-3 -2 -1 1 2 3
50
100
150
200
250
n = 1000
(� � �)2(� + �)2e�
17Tuesday, 30 July 13
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ecos �
-3 -2 -1 1 2 3
0.5
1.0
1.5
2.0
2.5
n = 5
18Tuesday, 30 July 13
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10 20 30 40
10-13
10-10
10-7
10-4
0.1
Error at zero
ecos �
2000 4000 6000 8000 10000 1200010-12
10-10
10-8
10-6
10-4
0.01
1
10000 20000 30000 40000
10-6
10-4
0.01
1
(� � �)2(� + �)2e�
(� � �)(� + �)e�e�
10000 20000 30000 40000
10-8
10-7
10-6
19Tuesday, 30 July 13
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Coefficient spaces
20Tuesday, 30 July 13
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C�
Ff =
�
������������
...f̂�2f̂�1f̂0f̂1f̂2...
�
������������
f Ff
21Tuesday, 30 July 13
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C�
Ff =
�
������������
...f̂�2f̂�1f̂0f̂1f̂2...
�
������������
f Ff
Underline indicates zero entry
21Tuesday, 30 July 13
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�p
�f��1 =��
k=��|fk| ,
�f��2 =
������
k=��|fk|2,
�f��� =��
-
�p
�f��1 =��
k=��|fk| ,
�f��2 =
������
k=��|fk|2,
�f��� =��
-
�p
�f��1 =��
k=��|fk| ,
�f��2 =
������
k=��|fk|2,
�f��� =��
-
�p
�f��1 =��
k=��|fk| ,
�f��2 =
������
k=��|fk|2,
�f��� =��
-
�p
�f��1 =��
k=��|fk| ,
�f��2 =
������
k=��|fk|2,
�f��� =��
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Continuous spaces
23Tuesday, 30 July 13
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p
�f� 1 =� �
��|f(�)| �,
�f� 2 =
�� �
��|f(�)|2 �,
�f� � =����
-
p
�f� 1 =� �
��|f(�)| �,
�f� 2 =
�� �
��|f(�)|2 �,
�f� � =����
-
fn(�) =n�
k=�nfk
k�
f � �1
f(�) =n��
fn(�)
f
�f � fn� � � 0.
Ff = f
25Tuesday, 30 July 13
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f � �1 � �n�1�
k=��+
��
k=n+1
�|fk| � 0
n � �
�����f(�) �n�
k=�nfk
k�
����� �� �n�1�
k=��+
��
k=n+1
�|fk| � 0
26Tuesday, 30 July 13
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f � �1 � �n�1�
k=��+
��
k=n+1
�|fk| � 0
n � �
�����f(�) �n�
k=�nfk
k�
����� �� �n�1�
k=��+
��
k=n+1
�|fk| � 0
26Tuesday, 30 July 13
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Ff = f
�n < j < n
1
2�
� �
��fn(�)
� j� � =n�
k=�nfk
1
2�
� �
��
(k�j)� �
= fj
����fj �1
2�
� �
��f(�) � j� �
���� =����
1
2�
� �
��[fn(�) � f(�)] � j� �
����
� 12�
� �
��|fn(�) � f(�)| � � 0
27Tuesday, 30 July 13
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Ff = f
�n < j < n
1
2�
� �
��fn(�)
� j� � =n�
k=�nfk
1
2�
� �
��
(k�j)� �
= fj
����fj �1
2�
� �
��f(�) � j� �
���� =����
1
2�
� �
��[fn(�) � f(�)] � j� �
����
� 12�
� �
��|fn(�) � f(�)| � � 0
27Tuesday, 30 July 13
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Ff = f
�n < j < n
1
2�
� �
��fn(�)
� j� � =n�
k=�nfk
1
2�
� �
��
(k�j)� �
= fj
����fj �1
2�
� �
��f(�) � j� �
���� =����
1
2�
� �
��[fn(�) � f(�)] � j� �
����
� 12�
� �
��|fn(�) � f(�)| � � 0
27Tuesday, 30 July 13
-
Ff = f
�n < j < n
1
2�
� �
��fn(�)
� j� � =n�
k=�nfk
1
2�
� �
��
(k�j)� �
= fj
����fj �1
2�
� �
��f(�) � j� �
���� =����
1
2�
� �
��[fn(�) � f(�)] � j� �
����
� 12�
� �
��|fn(�) � f(�)| � � 0
27Tuesday, 30 July 13
-
Ff = f
�n < j < n
1
2�
� �
��fn(�)
� j� � =n�
k=�nfk
1
2�
� �
��
(k�j)� �
= fj
����fj �1
2�
� �
��f(�) � j� �
���� =����
1
2�
� �
��[fn(�) � f(�)] � j� �
����
� 12�
� �
��|fn(�) � f(�)| � � 0
27Tuesday, 30 July 13
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Smoothness implies convergence
28Tuesday, 30 July 13
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T C[T]
k T Ck[T]
f � Ck[T] f � C[T] f f � � Ck�1[T]
f � C2[T] Ff � �1
f(�) =��
k=��f̂k
k�
�=
n��
n�
k=�nf̂k
k�
�.
29Tuesday, 30 July 13
-
T C[T]
k T Ck[T]
f � Ck[T] f � C[T] f f � � Ck�1[T]
f � C2[T] Ff � �1
f(�) =��
k=��f̂k
k�
�=
n��
n�
k=�nf̂k
k�
�.
29Tuesday, 30 July 13
-
T C[T]
k T Ck[T]
f � Ck[T] f � C[T] f f � � Ck�1[T]
f � C2[T] Ff � �1
f(�) =��
k=��f̂k
k�
�=
n��
n�
k=�nf̂k
k�
�.
29Tuesday, 30 July 13
-
T C[T]
k T Ck[T]
f � Ck[T] f � C[T] f f � � Ck�1[T]
f � C2[T] Ff � �1
f(�) =��
k=��f̂k
k�
�=
n��
n�
k=�nf̂k
k�
�.
29Tuesday, 30 July 13
-
��k=��
���f̂k���
f
f
� �
��f(�) � k� � = � 1
k
� �
��f(�) [ � k�]
= �f(�)� k� � f(��) k�
k+
1
k
� �
��f �(�) � k� �
=1
k
� �
��f �(�) � k� � =
1
k2
� �
��f �(�) [ � k�]
= � 1k2
� �
��f ��(�) � k� �
f ��
����� �
��f ��(�) � k� �
���� �� �
��|f ��(�)| � < C
��
k=��
���f̂k��� �
���f̂0��� +
C
�
��
k=1
k�2
30Tuesday, 30 July 13
-
��k=��
���f̂k���
f
f
� �
��f(�) � k� � = � 1
k
� �
��f(�) [ � k�]
= �f(�)� k� � f(��) k�
k+
1
k
� �
��f �(�) � k� �
=1
k
� �
��f �(�) � k� � =
1
k2
� �
��f �(�) [ � k�]
= � 1k2
� �
��f ��(�) � k� �
f ��
����� �
��f ��(�) � k� �
���� �� �
��|f ��(�)| � < C
��
k=��
���f̂k��� �
���f̂0��� +
C
�
��
k=1
k�2
30Tuesday, 30 July 13
-
��k=��
���f̂k���
f
f
� �
��f(�) � k� � = � 1
k
� �
��f(�) [ � k�]
= �f(�)� k� � f(��) k�
k+
1
k
� �
��f �(�) � k� �
=1
k
� �
��f �(�) � k� � =
1
k2
� �
��f �(�) [ � k�]
= � 1k2
� �
��f ��(�) � k� �
f ��
����� �
��f ��(�) � k� �
���� �� �
��|f ��(�)| � < C
��
k=��
���f̂k��� �
���f̂0��� +
C
�
��
k=1
k�2
30Tuesday, 30 July 13
-
��k=��
���f̂k���
f
f
� �
��f(�) � k� � = � 1
k
� �
��f(�) [ � k�]
= �f(�)� k� � f(��) k�
k+
1
k
� �
��f �(�) � k� �
=1
k
� �
��f �(�) � k� � =
1
k2
� �
��f �(�) [ � k�]
= � 1k2
� �
��f ��(�) � k� �
f ��
����� �
��f ��(�) � k� �
���� �� �
��|f ��(�)| � < C
��
k=��
���f̂k��� �
���f̂0��� +
C
�
��
k=1
k�2
30Tuesday, 30 July 13
-
��k=��
���f̂k���
f
f
� �
��f(�) � k� � = � 1
k
� �
��f(�) [ � k�]
= �f(�)� k� � f(��) k�
k+
1
k
� �
��f �(�) � k� �
=1
k
� �
��f �(�) � k� � =
1
k2
� �
��f �(�) [ � k�]
= � 1k2
� �
��f ��(�) � k� �
f ��
����� �
��f ��(�) � k� �
���� �� �
��|f ��(�)| � < C
��
k=��
���f̂k��� �
���f̂0��� +
C
�
��
k=1
k�2
30Tuesday, 30 July 13
-
��k=��
���f̂k���
f
f
� �
��f(�) � k� � = � 1
k
� �
��f(�) [ � k�]
= �f(�)� k� � f(��) k�
k+
1
k
� �
��f �(�) � k� �
=1
k
� �
��f �(�) � k� � =
1
k2
� �
��f �(�) [ � k�]
= � 1k2
� �
��f ��(�) � k� �
f ��
����� �
��f ��(�) � k� �
���� �� �
��|f ��(�)| � < C
��
k=��
���f̂k��� �
���f̂0��� +
C
�
��
k=1
k�2
30Tuesday, 30 July 13
-
31Tuesday, 30 July 13
-
• We found a class of functions where Fourier series converges
31Tuesday, 30 July 13
-
• We found a class of functions where Fourier series converges
• The usual approach is to try to find the weakest class such that this property holds true
31Tuesday, 30 July 13
-
• We found a class of functions where Fourier series converges
• The usual approach is to try to find the weakest class such that this property holds true
• I.e., will Fourier series converge for periodic continuously differentiable functions? (Yes!)
31Tuesday, 30 July 13
-
• We found a class of functions where Fourier series converges
• The usual approach is to try to find the weakest class such that this property holds true
• I.e., will Fourier series converge for periodic continuously differentiable functions? (Yes!)
• What about just continuous functions? (Not always...)
31Tuesday, 30 July 13
-
• We found a class of functions where Fourier series converges
• The usual approach is to try to find the weakest class such that this property holds true
• I.e., will Fourier series converge for periodic continuously differentiable functions? (Yes!)
• What about just continuous functions? (Not always...)
• A second question is how fast does the approximation
converge
31Tuesday, 30 July 13
-
• We found a class of functions where Fourier series converges
• The usual approach is to try to find the weakest class such that this property holds true
• I.e., will Fourier series converge for periodic continuously differentiable functions? (Yes!)
• What about just continuous functions? (Not always...)
• A second question is how fast does the approximation
converge
f(x)−n∑
k=−nf̂k
kx =
(−1−n∑
k=−∞+
∞∑
k=n+1
)f̂k
kx
31Tuesday, 30 July 13
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�p�
�f��p� =
����������������
�
�����������
. . .3�
2�
12�
3�
. . .
�
�����������
f
�����������������p
=
� ��
k=��(|k| + 1)p |fk|p
�1/p
f � �p�32Tuesday, 30 July 13
-
fn(�) =n�
k=�nfk
k�
f � �1� �
�f � fn� � � (n + 2)�� �f��1� .
�f � fn� � =
�����
� �n�1�
k=��+
��
k=n+1
�fk
k�
������
�� �n�1�
k=��+
��
k=n+1
�|fk|
� (n + 2)��� �n�1�
k=��+
��
k=n+1
�(|k| + 1)� |fk|
� (n + 2)�� �f��1�
33Tuesday, 30 July 13
-
fn(�) =n�
k=�nfk
k�
f � �1� �
�f � fn� � � (n + 2)�� �f��1� .
�f � fn� � =
�����
� �n�1�
k=��+
��
k=n+1
�fk
k�
������
�� �n�1�
k=��+
��
k=n+1
�|fk|
� (n + 2)��� �n�1�
k=��+
��
k=n+1
�(|k| + 1)� |fk|
� (n + 2)�� �f��1�
33Tuesday, 30 July 13
-
fn(�) =n�
k=�nfk
k�
f � �1� �
�f � fn� � � (n + 2)�� �f��1� .
�f � fn� � =
�����
� �n�1�
k=��+
��
k=n+1
�fk
k�
������
�� �n�1�
k=��+
��
k=n+1
�|fk|
� (n + 2)��� �n�1�
k=��+
��
k=n+1
�(|k| + 1)� |fk|
� (n + 2)�� �f��1�
33Tuesday, 30 July 13
-
fn(�) =n�
k=�nfk
k�
f � �1� �
�f � fn� � � (n + 2)�� �f��1� .
�f � fn� � =
�����
� �n�1�
k=��+
��
k=n+1
�fk
k�
������
�� �n�1�
k=��+
��
k=n+1
�|fk|
� (n + 2)��� �n�1�
k=��+
��
k=n+1
�(|k| + 1)� |fk|
� (n + 2)�� �f��1�
33Tuesday, 30 July 13
-
fn(�) =n�
k=�nfk
k�
f � �1� �
�f � fn� � � (n + 2)�� �f��1� .
�f � fn� � =
�����
� �n�1�
k=��+
��
k=n+1
�fk
k�
������
�� �n�1�
k=��+
��
k=n+1
�|fk|
� (n + 2)��� �n�1�
k=��+
��
k=n+1
�(|k| + 1)� |fk|
� (n + 2)�� �f��1�33Tuesday, 30 July 13
-
fn(�) =n�
k=�nfk
k�
f � �1� �
�f � fn� � � (n + 2)�� �f��1� .
�f � fn� � =
�����
� �n�1�
k=��+
��
k=n+1
�fk
k�
������
�� �n�1�
k=��+
��
k=n+1
�|fk|
� (n + 2)��� �n�1�
k=��+
��
k=n+1
�(|k| + 1)� |fk|
� (n + 2)�� �f��1�
33Tuesday, 30 July 13
-
f � C3[T]� �
��f(�) � k� � = � 1
k2
� �
��f ��(�) � k� �
=1
k3
� �
��f ��(�) [ � k�]
=f ��(�) � k� � f ��(��) k�
k3� 1
k3
� �
��f ���(�) � k� �
=1
k3
� �
��f ���(�) � k� � = O
�k�3
�
f � C3[T] Ff � �11f � Cd+2[T] Ff � �1d
34Tuesday, 30 July 13
-
f � C3[T]� �
��f(�) � k� � = � 1
k2
� �
��f ��(�) � k� �
=1
k3
� �
��f ��(�) [ � k�]
=f ��(�) � k� � f ��(��) k�
k3� 1
k3
� �
��f ���(�) � k� �
=1
k3
� �
��f ���(�) � k� � = O
�k�3
�
f � C3[T] Ff � �11f � Cd+2[T] Ff � �1d
34Tuesday, 30 July 13
-
f � C3[T]� �
��f(�) � k� � = � 1
k2
� �
��f ��(�) � k� �
=1
k3
� �
��f ��(�) [ � k�]
=f ��(�) � k� � f ��(��) k�
k3� 1
k3
� �
��f ���(�) � k� �
=1
k3
� �
��f ���(�) � k� � = O
�k�3
�
f � C3[T] Ff � �11f � Cd+2[T] Ff � �1d
34Tuesday, 30 July 13
-
f � C3[T]� �
��f(�) � k� � = � 1
k2
� �
��f ��(�) � k� �
=1
k3
� �
��f ��(�) [ � k�]
=f ��(�) � k� � f ��(��) k�
k3� 1
k3
� �
��f ���(�) � k� �
=1
k3
� �
��f ���(�) � k� � = O
�k�3
�
f � C3[T] Ff � �11f � Cd+2[T] Ff � �1d
34Tuesday, 30 July 13
-
f � C3[T]� �
��f(�) � k� � = � 1
k2
� �
��f ��(�) � k� �
=1
k3
� �
��f ��(�) [ � k�]
=f ��(�) � k� � f ��(��) k�
k3� 1
k3
� �
��f ���(�) � k� �
=1
k3
� �
��f ���(�) � k� � = O
�k�3
�
f � C3[T] Ff � �11f � Cd+2[T] Ff � �1d
34Tuesday, 30 July 13
-
Summary
• We established conditions for which Fourier series converge pointwise• We proved that the smoother (i.e., more differentiable) the function, the faster the
convergence rate• We accomplished this using:• Weierstrass M-test• Integration by parts
35Tuesday, 30 July 13