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

Lesson 3 Approximating Fourier series

1

2

• 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 than O 1

nα

• for any α
• 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 = [−π, π), the following exist for

every integer k: ˆ fk = 1 2π π

−π

f(θ)−kθ x

• We can formally define the Fourier series of f:

f(θ) ∼

∞

• k=−∞

ˆ fkkθ This sum will converge uniformly to f for periodic smooth functions

4

• 3
• 2
• 1

1 2 3

• Define the m evenly spaced points on the periodic interval θ = (θ1, . . . , θm):

θ :=

• −π,

2 m − 1

• π, . . . ,
• 1 − 2

m

• π
• =
• We can approximate the Fourier coefficients using the

point trapezoidal rule

5

• 3
• 2
• 1

1 2 3

• Define the m evenly spaced points on the periodic interval θ = (θ1, . . . , θm):

θ :=

• −π,

2 m − 1

• π, . . . ,
• 1 − 2

m

• π
• =
• We can approximate the Fourier coefficients using the m point trapezoidal rule

ˆ fk = 1 2π π

−π

f(θ) θ ≈ 1 m

m

• j=1

f(θj)−kθj =: ˆ f m

k

6

• Using these, and truncating the Fourier sum between integers α and β we obtain

an approximate Fourier series f(θ) ≈ fα,β,m(θ) :=

β

• k=α

ˆ f m

k kθ

• Big question: how to choose α, β and m?
• When we specify just

and , we will choose to be the same as the number

• f coefficients
• When we specify just

, we will choose roughly equal number of negative and positive coefficients:

• dd

even

7

• Using these, and truncating the Fourier sum between integers α and β we obtain

an approximate Fourier series f(θ) ≈ fα,β,m(θ) :=

β

• k=α

ˆ f m

k kθ

• 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
• f coefficients

fα,β(θ) := fα,β,β−α+1(θ)

• When we specify just

, we will choose roughly equal number of negative and positive coefficients:

• dd

even

8

• Using these, and truncating the Fourier sum between integers α and β we obtain

an approximate Fourier series f(θ) ≈ fα,β,m(θ) :=

β

• k=α

ˆ f m

k kθ

• 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
• f coefficients

fα,β(θ) := fα,β,β−α+1(θ)

• When we specify just m, we will choose roughly equal number of negative and

positive coefficients: fm(θ) :=

• f 1−m

2

, m−1

2

,m(θ)

m odd f− m

2 , m 2 −1,m(θ)

m even

Experimental results

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10

θ

• 3
• 2
• 1

1 2 3 q 5 10

m = 5

11

θ

• 3
• 2
• 1

1 2 3 q 5 10

m = 5

• 3
• 2
• 1

1 2 3 q 0.5 1.0 1.5 2.0 2.5 3.0

m = 5

|θ − .1|

12

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 5

θ

• 3
• 2
• 1

1 2 3 q 5 10

m = 5

20 π θ

• 3
• 2
• 1

1 2 3 q 0.5 1.0 1.5 2.0 2.5 3.0

m = 5

|θ − .1| (θ − .1)

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0 1.5

m = 5

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 5

5θ

• 3
• 2
• 1

1 2 3 q

• 0.5

0.5 1.0

m = 5

2 θ

13

θ 20 π θ |θ − .1| (θ − .1) 5θ

• 3
• 2
• 1

1 2 3 q

• 0.4
• 0.2

0.2 0.4 0.6 0.8 1.0

m = 10

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 10

2 θ

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 10

• 3
• 2
• 1

1 2 3 q 0.5 1.0 1.5 2.0 2.5 3.0

m = 10

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 10

• 3
• 2
• 1

1 2 3 q 5 10 15

m = 10

14

θ 20 π θ |θ − .1| (θ − .1) 5θ

• 3
• 2
• 1

1 2 3 q

• 0.4
• 0.2

0.2 0.4 0.6 0.8 1.0

m = 100

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 100

2 θ

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 100

• 3
• 2
• 1

1 2 3 q 0.5 1.0 1.5 2.0 2.5 3.0

m = 100

• 3
• 2
• 1

1 2 3 q

• 1.0
• 0.5

0.5 1.0

m = 100

• 3
• 2
• 1

1 2 3 q 5 10 15 20 25

m = 100

15

• 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 fk

• This will follow naturally from orthogonality properties of θ

Orthogonality of complex exponentials

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17

The L2 inner product and norm

• We define the 2 inner product (on T) by

f, g = 1 2π π

−π

¯ f(θ)g(θ) θ

• Associated with this inner product is the 2 norm:

f =

• 1

2π π

−π

|f(θ)|2 θ

• 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

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• 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:

, 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

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• 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

, we can construct a projection of a vector into

• by

If

• then

is equal to its projection: In other words

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

• then

is equal to its projection: In other words

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
• kθ, kθ

= 1 2π π

−π

θ = 1 and for k = j

• kθ, jθ

= 1 2π π

−π

(j−k)θ θ = (j−k)π −(j−k)π 2π(j k) = 0

• In other words, the complex exponentials are orthonormal!
• Thus we can think of the Fourier series as an infinite projection

f(θ)

∞

• k=−∞
• kθ, f
• kθ

Since this sum is infinite, we cannot appeal to the simple argument of equality from the last slide

Discrete orthogonality of complex exponentials

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24

• We have shown that the complex exponentials are orthogonal with respect to the

inner product f, g = 1 2π

−

¯ f(θ)g(θ) θ

• A remarkable fact we now show is that they are also orthogonal with respect to

the following discrete semi-inner product: f, gm = 1 m

m

• j=1

¯ f(θj)g(θj) = f(θ)g(θ) m where θ = (θ1, . . . , θm) are again evenly spaced points: θ =

• π,

2 m 1

• π, . . . ,
• 1 2

m

• π
• =
• 3
• 2
• 1

1 2 3

25

Evenly spaced points on the unit circle

eiθ

• 1.0
• 0.5

0.5 1.0

• 1.0
• 0.5

0.5 1.0

• 3
• 2
• 1

1 2 3

θ = (θ1, . . . , θm) z = (z1, . . . , zm)

26

Some identities (shown for even m):

z

m

• j=1

eiθj =

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

ei2θj =

m

• j=1

z2

j = 0

• 1.0
• 0.5

0.5 1.0

• 0.5

0.5

28

:

m

• j=1

kθj = (−)km for k = . . . , −2m, −m, 0, m, 2m . . .

m

• j=1

kθj = 0 for all other integer k

• Note that

zj = −2π(j−1)/m = −ωj−1 for ω = 11/m = 2π/m

• Therefore,
• If

is a multiple of , then we have

• Note that

zj = −2π(j−1)/m = −ωj−1 for ω = 11/m = 2π/m

• Therefore,

m

• j=1

zk

j = (−)k m−1

• j=0

ωkj

• If

is a multiple of , then we have

• Note that

zj = −2π(j−1)/m = −ωj−1 for ω = 11/m = 2π/m

• Therefore,

m

• j=1

zk

j = (−)k m−1

• j=0

ωkj

• If k = αm is a multiple of m, then we have

m−1

• j=0

ωkj =

m−1

• j=0

(ωm)αj =

m−1

• j=0

1αj = m

• Note that

zj = −2π(j−1)/m = −ωj−1 for ω = 11/m = 2π/m

• Therefore,

m

• j=1

zk

j = (−)k m−1

• j=0

ωkj

• If k = αm is a multiple of m, then we have

m−1

• j=0

ωkj =

m−1

• j=0

(ωm)αj =

m−1

• j=0

1αj = m

• For k not a multiple of m, recall the geometric sum formula:

m−1

• j=0

zj = zm − 1 z − 1

• Thus
• Because

is not a multiple of , the denominator is nonzero

• Because

, the numerator is zero

• Note that

zj = −2π(j−1)/m = −ωj−1 for ω = 11/m = 2π/m

• Therefore,

m

• j=1

zk

j = (−)k m−1

• j=0

ωkj

• If k = αm is a multiple of m, then we have

m−1

• j=0

ωkj =

m−1

• j=0

(ωm)αj =

m−1

• j=0

1αj = m

• For k not a multiple of m, recall the geometric sum formula:

m−1

• j=0

zj = zm − 1 z − 1

• Thus

m−1

• j=0

ωkj =

m−1

• j=0

(ωk)j = ωkm − 1 ωk − 1

• Because

is not a multiple of , the denominator is nonzero

• Because

, the numerator is zero

• Note that

zj = −2π(j−1)/m = −ωj−1 for ω = 11/m = 2π/m

• Therefore,

m

• j=1

zk

j = (−)k m−1

• j=0

ωkj

• If k = αm is a multiple of m, then we have

m−1

• j=0

ωkj =

m−1

• j=0

(ωm)αj =

m−1

• j=0

1αj = m

• For k not a multiple of m, recall the geometric sum formula:

m−1

• j=0

zj = zm − 1 z − 1

• Thus

m−1

• j=0

ωkj =

m−1

• j=0

(ωk)j = ωkm − 1 ωk − 1

• Because k is not a multiple of m, the denominator is nonzero
• Because ωkm = (ωm)k = 1k, the numerator is zero

• kθ, jθ

m = (−1)j−k

• j − k = . . . , −2m, −m, 0, m, 2m, . . .
• kθ, jθ

m = 0

• kθ, jθ

m = 1

m

m

• j=1

(j−k)θj