Lesson 3 Approximating Fourier series 1 Last lecture, we saw that - PowerPoint PPT Presentation

Lesson 3 Approximating Fourier series 1

• Last lecture, we saw that the trapezoidal rule was an effective method for calculating integrals of periodic functions � We used the Euler–McLaurin formula to prove that the error decayed faster � 1 than O for any α � n α • In this lecture, we will apply this to calculating Fourier coefficients and Fourier series • This is practical function approximation 2

• Given an integrable function f defined on T = [ − π , π ) , the following exist for every integer k : � π f k = 1 ˆ f ( θ ) � − � k θ � x 2 π − π • We can formally define the Fourier series of f : ∞ ˆ � f k � � k θ f ( θ ) ∼ k = −∞ � This sum will converge uniformly to f for periodic smooth functions 3

� � • Define the m evenly spaced points on the periodic interval θ = ( θ 1 , . . . , θ m ) : � 2 � � � � � 1 − 2 θ := m − 1 = − π , π , . . . , π - 3 - 2 - 1 0 1 2 3 m • We can approximate the Fourier coefficients using the point trapezoidal rule 4

• Define the m evenly spaced points on the periodic interval θ = ( θ 1 , . . . , θ m ) : � 2 � � � � � 1 − 2 θ := m − 1 = − π , π , . . . , π - 3 - 2 - 1 0 1 2 3 m • We can approximate the Fourier coefficients using the m point trapezoidal rule � π m f k = 1 f ( θ ) � θ ≈ 1 f ( θ j ) � − k θ j =: ˆ ˆ � f m k 2 π m − π j =1 5

• Using these, and truncating the Fourier sum between integers α and β we obtain an approximate Fourier series β ˆ � f m k � � k θ f ( θ ) ≈ f α , β ,m ( θ ) := k = α • Big question: how to choose α , β and m ? • When we specify just and , we will choose to be the same as the number of coefficients • When we specify just , we will choose roughly equal number of negative and positive coefficients: odd even 6

• Using these, and truncating the Fourier sum between integers α and β we obtain an approximate Fourier series β ˆ � f m k � � k θ f ( θ ) ≈ f α , β ,m ( θ ) := k = α • Big question: how to choose α , β and m ? • When we specify just α and β , we will choose m to be the same as the number of coefficients f α , β ( θ ) := f α , β , β − α +1 ( θ ) • When we specify just , we will choose roughly equal number of negative and positive coefficients: odd even 7

• Using these, and truncating the Fourier sum between integers α and β we obtain an approximate Fourier series β ˆ � f m k � � k θ f ( θ ) ≈ f α , β ,m ( θ ) := k = α • Big question: how to choose α , β and m ? • When we specify just α and β , we will choose m to be the same as the number of coefficients f α , β ( θ ) := f α , β , β − α +1 ( θ ) • When we specify just m , we will choose roughly equal number of negative and positive coefficients: � m odd ,m ( θ ) f 1 − m , m − 1 f m ( θ ) := 2 2 m even 2 − 1 ,m ( θ ) f − m 2 , m 8

Experimental results 9

� θ m = 5 10 5 q - 3 - 2 - 1 1 2 3 10

� θ m = 5 10 5 q - 3 - 2 - 1 1 2 3 | θ − . 1 | m = 5 3.0 2.5 2.0 1.5 1.0 0.5 q - 3 - 2 - 1 1 2 3 11

��� 20 � θ π θ ��� 5 θ m = 5 m = 5 m = 5 1.0 1.0 0.5 10 0.5 q q - 3 - 2 - 1 1 2 3 - 3 - 2 - 1 1 2 3 5 - 0.5 - 0.5 q - 3 - 2 - 1 1 2 3 - 1.0 - 1.0 ��� 2 ��� θ | θ − . 1 | ��� ( θ − . 1) m = 5 m = 5 m = 5 1.0 1.5 3.0 2.5 1.0 0.5 2.0 0.5 1.5 q - 3 - 2 - 1 1 2 3 q 1.0 - 3 - 2 - 1 1 2 3 - 0.5 0.5 - 1.0 - 0.5 q - 3 - 2 - 1 1 2 3 12

��� 20 � θ π θ ��� 5 θ m = 10 m = 10 m = 10 1.0 1.0 15 0.5 0.5 10 q q - 3 - 2 - 1 1 2 3 - 3 - 2 - 1 1 2 3 5 - 0.5 - 0.5 q - 3 - 2 - 1 1 2 3 - 1.0 - 1.0 ��� 2 ��� θ | θ − . 1 | ��� ( θ − . 1) m = 10 m = 10 m = 10 1.0 3.0 1.0 0.8 2.5 0.6 0.5 2.0 0.4 1.5 q - 3 - 2 - 1 1 2 3 0.2 1.0 - 0.5 q - 3 - 2 - 1 1 2 3 0.5 - 0.2 - 1.0 q - 0.4 - 3 - 2 - 1 1 2 3 13

��� 20 � θ π θ ��� 5 θ m = 100 m = 100 m = 100 25 1.0 1.0 20 0.5 0.5 15 q q - 3 - 2 - 1 1 2 3 - 3 - 2 - 1 1 2 3 10 - 0.5 5 - 0.5 q - 3 - 2 - 1 1 2 3 - 1.0 - 1.0 ��� 2 ��� θ | θ − . 1 | ��� ( θ − . 1) m = 100 m = 100 m = 100 1.0 3.0 1.0 0.8 2.5 0.6 0.5 2.0 0.4 q 1.5 - 3 - 2 - 1 1 2 3 0.2 1.0 - 0.5 q - 3 - 2 - 1 1 2 3 0.5 - 0.2 - 1.0 q - 0.4 - 3 - 2 - 1 1 2 3 14

• Observed convergence properties: � Fast convergence for periodic functions, just like trapezoidal rule � Slow convergence for non-periodic functions away from singularities � No convergence in neighbourhood of jump singularities (including ± π ) • We also observed interpolation at the quadrature points θ • To understand this, we need to related the approximate Fourier coefficients ˆ f m k to the true Fourier coefficients ˆ f k • This will follow naturally from orthogonality properties of � � θ 15

Orthogonality of complex exponentials 16

The L 2 inner product and norm • We define the � 2 inner product (on T ) by � π � f, g � = 1 ¯ f ( θ ) g ( θ ) � θ 2 π − π • Associated with this inner product is the � 2 norm: � � π 1 | f ( θ ) | 2 � θ � f � = 2 π − π • � 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 17

���� ���� • A set of nonzero vectors v 1 , . . . , v n in a vector space V are called orthogonal if � v i , v j � = 0 whenever i � = k. • They are called orthonormal if they are orthogonal and all vectors are of unit norm: , or equivalently, • For orthonormal vectors , we can construct a projection of a vector into by � If then is equal to its projection: � In other words 18

���� ���� • A set of nonzero vectors v 1 , . . . , v n in a vector space V are called orthogonal if � v i , v j � = 0 whenever i � = k. • They are called orthonormal if they are orthogonal and all vectors are of unit norm: 1 = � v i � , or equivalently, � v i , v i � = 1 . • For orthonormal vectors , we can construct a projection of a vector into by � If then is equal to its projection: � In other words 19

���� • A set of nonzero vectors v 1 , . . . , v n in a vector space V are called orthogonal if � v i , v j � = 0 whenever i � = k. • They are called orthonormal if they are orthogonal and all vectors are of unit norm: 1 = � v i � , or equivalently, � v i , v i � = 1 . • For orthonormal vectors v k , we can construct a projection of a vector f � V into ���� { v 1 , . . . , v n } by n � P f := � v k , f � v k k =1 � If then is equal to its projection: � In other words 20

• A set of nonzero vectors v 1 , . . . , v n in a vector space V are called orthogonal if � v i , v j � = 0 whenever i � = k. • They are called orthonormal if they are orthogonal and all vectors are of unit norm: 1 = � v i � , or equivalently, � v i , v i � = 1 . • For orthonormal vectors v k , we can construct a projection of a vector f � V into ���� { v 1 , . . . , v n } by n � P f := � v k , f � v k k =1 � If f � ���� { v 1 , . . . , v n } then f is equal to its projection: f = P f � In other words P 2 f = P f 21

• We have � π = 1 � � k θ , � � k θ � � � θ = 1 2 π − π and for k � = j � π � � ( j − k ) θ � θ = � � ( j − k ) π � � − � ( j − k ) π = 1 � � k θ , � � j θ � � = 0 2 π 2 π � ( j � k ) − π • In other words, the complex exponentials are orthonormal! • Thus we can think of the Fourier series as an infinite projection ∞ � � � k θ , f � � k θ � � f ( θ ) � k = −∞ � Since this sum is infinite, we cannot appeal to the simple argument of equality from the last slide 22

Discrete orthogonality of complex exponentials 23

• We have shown that the complex exponentials are orthogonal with respect to the inner product � � � f, g � = 1 ¯ f ( θ ) g ( θ ) � θ 2 π − � • A remarkable fact we now show is that they are also orthogonal with respect to the following discrete semi-inner product : m � f, g � m = 1 f ( θ j ) g ( θ j ) = f ( θ ) � g ( θ ) ¯ � m m j =1 where θ = ( θ 1 , . . . , θ m ) are again evenly spaced points: � 2 � � � � � 1 � 2 θ = � π , m � 1 = π , . . . , π - 3 - 2 - 1 0 1 2 3 m 24

Evenly spaced points on the unit circle θ = ( θ 1 , . . . , θ m ) z = ( z 1 , . . . , z m ) e i θ 1.0 0.5 - 3 - 2 - 1 0 1 2 3 - 1.0 - 0.5 0.5 1.0 - 0.5 - 1.0 25

Some identities (shown for even m ): z 1.0 0.5 m m e i θ j = � � z j = 0 - 1.0 - 0.5 0.5 1.0 j =1 j =1 - 0.5 - 1.0 26

Some identities (shown for even m ): z 0.5 m m e i2 θ j = � � z 2 j = 0 - 1.0 - 0.5 0.5 1.0 j =1 j =1 - 0.5 27

������� : m � � k θ j = ( − ) k m � for k = . . . , − 2 m, − m, 0 , m, 2 m . . . j =1 m � � k θ j = 0 � for all other integer k j =1 28

• Note that z j = − � 2 π � ( j − 1)/ m = − ω j − 1 for ω = 1 1/ m = � 2 π � / m • Therefore, • If is a multiple of , then we have