Numerical Complex Analysis · 2013. 7. 29. · Numerical complex analysis • A combination of harmonic analysis, numerical analysis, complex analysis, functional analysis and approximation

Numerical Complex Analysis

Dr Sheehan Olver

1Tuesday, 30 July 13

• 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


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


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


Signal smoothing

-3 -2 -1 1 2 3

-0.5

0.5

-3 -2 -1 1 2 3

-0.5

0.5


Laplace’s equation�u = 0, �u = f


Integrable PDEsut + uxxx + 6uux = 0, u(0, x) = f(x)


Spectral vs algebraic convergence


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


n

1/np

p(x) =n�

k=0

ckxk

1/np p

��n � > 0


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


Project


• A major component of the courses evaluation will be a project

Project



• Timeline:

Project



• Timeline:

• Short proposal (< 1 page) will be due before the spring vacation

Project



• Timeline:


• Project will be due exam week

Project



• Timeline:



• The goal of the project is to use ideas from the course to do something original

Project



• Timeline:




• Most projects will consist of a programming component and an approximately 10 page write-up describing the maths behind the programming

Project



• Timeline:





• A good example project would be to solve a problem from another course numerically using the methods we discuss

Project



• Timeline:






• Check out examples from www.maths.ox.ac.uk/chebfun for inspiration on developing a project

Project


http://www.maths.ox.ac.uk/chebfunhttp://www.maths.ox.ac.uk/chebfun


• Timeline:






• 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


http://www.maths.ox.ac.uk/chebfunhttp://www.maths.ox.ac.uk/chebfun

Lesson 1Review of Fourier Analysis


f T = [��, �)k

f̂k =1

2�

� �

��f(�) � k� x

f

f(�) ��

k=��f̂k

k�

f

� � T

f(�) =n��

n�

k=�nf̂k

k�?


-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�


-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�


-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�


-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�


-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�


-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�


-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�


-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�


-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�


-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�


-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�


-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�


ecos �

-3 -2 -1 1 2 3

0.5

1.0

1.5

2.0

2.5

n = 5


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


Coefficient spaces


C�

Ff =

�

��

...f̂�2f̂�1f̂0f̂1f̂2...

�

��

f Ff


C�

Ff =

�

��

...f̂�2f̂�1f̂0f̂1f̂2...

�

��

f Ff

Underline indicates zero entry


�p

�f��1 =��

k=��|fk| ,

�f��2 =

��

k=��|fk|2,

�f�� =��

Continuous spaces


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


f � �1 � �n�1�

k=��+

��

k=n+1

�|fk| � 0

n � �

��f(�) �n�

k=�nfk

k�

�� n�1�

k=��+

��

k=n+1

�|fk| � 0


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


Smoothness implies convergence


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�

�.


��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


• 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






• A second question is how fast does the approximation

converge

f(x)−n∑

k=−nf̂k

kx =

(−1−n∑

k=−∞+

∞∑

k=n+1

)f̂k

kx


�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�


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�


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


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


Numerical Complex Analysis · 2013. 7. 29. · Numerical complex analysis • A combination of harmonic analysis, numerical analysis, complex analysis, functional analysis and approximation

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