EI331 Signals and Systems Lecture 25 Bo Jiang John Hopcroft Center - - PowerPoint PPT Presentation

EI331 Signals and Systems

Lecture 25 Bo Jiang

John Hopcroft Center for Computer Science Shanghai Jiao Tong University

May 23, 2019

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Contents

• 1. Power Series
• 2. Taylor Series
• 3. Laurent Series

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Properties of Power Series

If

∞

• n=0

cn(z − z0)n has radius of convergence R, then

• its sum f(z) is analytic on |z − z0| < R.
• f can be differentiated term by term on |z − z0| < R, i.e.

f ′(z) =

∞

• n=1

ncn(z − z0)n−1

• f can be integrated term by term on |z − z0| < R, i.e.,
• γ

f(z)dz =

∞

• n=0

cn

• γ

(z − z0)ndz for γ in |z − z0| < R In particular, z

z0

f(ζ)dζ =

∞

• n=0

cn n + 1(z − z0)n+1

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Uniform Convergence

A series

∞

• n=1

fn(z) converges to f(z) uniformly on Ω ⊂ C, if the sequence of partial sums sk(z) converges to f(z) uniformly on Ω, lim

k→∞ max z∈Ω |sk(z) − f(z)| = 0.

Theorem (Cauchy’s convergence test). The series

∞

• n=1

fn(z) converges uniformly on Ω iff given any ǫ > 0, there exists N ∈ N s.t. for any n > N and p ≥ 1, max

z∈Ω |fn+1(z) + fn+2(z) + · · · + fn+p(z)| < ǫ

Theorem (Weierstrass M-test). If |fn(z)| ≤ Mn on Ω for each n, and

∞

• n=1

Mn < +∞, then

∞

• n=1

fn(z) converges uniformly.

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Properties of Uniformly Convergent Series

• Theorem. If fn(z) is continuous on Ω for each n, and

∞

• n=1

fn(z) converges to f(z) uniformly on Ω, then f(z) is continuous on Ω.

• Theorem. If for each n, fn(z) is continuous on a piecewise

smooth curve γ with finite length, and

∞

• n=1

fn(z) converges to f(z) uniformly on γ, then

γ

f(z)dz =

∞

• n=1
• γ

fn(z)dz

• Proof. Let L be the length of γ. By uniform convergence, given

any ǫ > 0, there exists an N ∈ N s.t. |f(z) − k

n=1 fn(z)| ≤ ǫ/L for

all k ≥ N and z ∈ γ. Thus

• γ

f(z)dz −

k

• n=1
• γ

fn(z)dz

• ≤
• γ
• f(z) −

k

• n=1

fn(z)

• ds ≤ ǫ

L

• γ

ds = ǫ

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Properties of Uniformly Convergent Series

• Theorem. If for each n, fn(z) is analytic on a domain D, and

∞

• n=1

fn(z) converges to f(z) uniformly on D, then

• 1. f(z) =

∞

• n=1

fn(z) is analytic on D.

• 2. f (k)(z) =

∞

• n=1

f (k)

n (z) for z ∈ D and k ∈ N.

Proof of 1. Fix an arbitrary z0 ∈ D and an open disk B(z0, r) ⊂ D. For any piecewise smooth Jordan curve γ in B(z0, r), term-by-term integration and Cauchy’s Theorem imply

• γ

f(z)dz =

∞

• n=1
• γ

fn(z)dz = 0 Uniform convergence implies f is continuous. By Morera’s Theorem, f(z) is analytic on B(z0, r).

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Properties of Uniformly Convergent Series

Proof of 2. Fix an arbitrary z0 ∈ D and a closed disk ¯ B(z0, r) ⊂ D. For z on the circle S(z0, r),

• f(z)

(z − z0)k+1 −

N

• n=0

fn(z) (z − z0)k+1

• =

1 rk+1

• f(z) −

N

• n=0

fn(z)

• so

∞

• n=0

fn(z) (z−z0)k+1 converges to f(z) (z−z0)k+1 uniformly on S(z0, r).

Term-by-term integration yields k! j2π

• |z−z0|=r

f(z) (z − z0)k+1dz =

∞

• n=1

k! j2π

• |z−z0|=r

fn(z) (z − z0)k+1dz By Cauchy’s Integral Formula, f (k)(z) =

∞

• n=1

f (k)

n (z)

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Contents

• 1. Power Series
• 2. Taylor Series
• 3. Laurent Series

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Taylor Series

Suppose an analytic function f has a power series expansion

• n an open disk B(z0, R),

f(z) =

∞

• n=0

cn(z − z0)n Term-by-term differentiation yields, f (m)(z) =

∞

• n=m

n! (n − m)!cn(z − z0)n−m Setting z = z0, f (m)(z0) = m!cm = ⇒ cm = f (m)(z0) m! Thus the power series expansion of f at z0 is unique if it exists.

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Taylor Series

• Theorem. If f is analytic on an open disk B(z0, R), then f has the

following power series expansion for z ∈ B(z0, R), f(z) =

∞

• n=0

cn(z − z0)n (⋆) where cn = f (n)(z0) n! , n = 0, 1, 2, . . . The series in (⋆) is called the Taylor series of f at z0.

• NB. f is analytic at z0 iff (⋆) holds in B(z0, R) for some R.
• NB. If f is analytic at z0, then there is always a largest open disk

B(z0, Rmax) on which the Taylor series expansion (⋆) holds. The radius Rmax is the distance between z0 and the closest point at which f is not analytic1.

1A point at which f fails to be analytic is called a singular point of f.

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Taylor Series

z0 z R r ζ

• Proof. Let r ∈ (0, R). Fix z ∈ B(z0, r). For

ζ ∈ γ = S(z0, r),

• z−z0

ζ−z0

• = |z−z0|

r

< 1, so 1 ζ − z = 1 ζ − z0 · 1 1 − z−z0

ζ−z0

=

∞

• n=0

(z − z0)n (ζ − z0)n+1 The convergence is uniform for ζ ∈ γ. Since f(ζ) is bounded on γ, f(ζ) ζ − z =

∞

• n=0

(z − z0)nf(ζ) (ζ − z0)n+1 uniformly for ζ ∈ γ. Term-by-term integration yields

• γ

f(ζ)dζ ζ − z =

∞

• n=0
• γ

f(ζ)dζ (ζ − z0)n+1

• (z − z0)n

Cauchy’s Integral Formula then yields the desired result.

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Examples

• Example. f(z) = ez is analytic on C and f (n)(z) = ez for n ∈ N.

Thus ez =

∞

• n=0

e0 n!zn =

∞

• n=0

1 n!zn, z ∈ C Recall that this can be used an alternative definition of ez.

• Example. cos z and sin z are analytic on C. For z ∈ C,

cos z = 1 2(ejz + e−jz) = 1 2

∞

• n=0

1 n![(jz)n + (−jz)n] =

∞

• n=0

(−1)n (2n)! z2n and sin z = 1 2j(ejz−e−jz) = 1 2j

∞

• n=0

1 n![(jz)n−(−jz)n] =

∞

• n=0

(−1)n (2n + 1)!z2n+1

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Examples

Re Im

− 1

2

x

−1

Example.

1 (1+z)2 is analytic on C \ {−1}. Its

Taylor series at z = 0 converges on B(0, 1). 1 1 + z =

∞

• n=0

(−1)nzn, |z| < 1 Differentiation yields 1 (1 + z)2 = −

∞

• n=1

(−1)nnzn−1 =

∞

• n=0

(−1)n(n + 1)zn, |z| < 1

• Example. Find the Taylor series of

1 (1+z)2 at z = − 1 2.

1 1 + z = 2 1 + 2(z + 1

2) = ∞

• n=0

2(−2)n

• z + 1

2 n ,

• z + 1

2

• < 1

2 1 (1 + z)2 =

∞

• n=0

(−2)n+2(n + 1)

• z + 1

2 n ,

• z + 1

2

• < 1

2

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Examples

Re Im

x x

j −j

Example.

1 1+z2 is analytic on C \ {j, −j}. Its

Taylor series at z = 0 converges on B(0, 1). 1 1 + z2 =

∞

• n=0

(−1)nz2n, |z| < 1 Re Im

x x

j −j 1 R = √ 2

• Example. The Taylor series of

1 1+z2 at z = 1

converges on B(1, √ 2). To obtain the expansion, let ζ = z − 1. 1 1 + z2 = 1/(2j) ζ + 1 − j − 1/(2j) ζ + 1 + j = 1 2j

∞

• n=0

(−ζ)n (1 − j)n+1 − 1 2j

∞

• n=0

(−ζ)n (1 + j)n+1 =

∞

• n=0

sin 3π(n+1)

4

2

n+1 2

(z − 1)n, |z − 1| < √ 2

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Examples

Re Im

x

−1

• Example. Let log(1 + z) be the principal

branch of Log(1 + z) with value 0 at z = 0. It is analytic on C \ (−∞, −1]. Its Taylor series converges on B(0, 1). [log(1 + z)]′ = 1 1 + z =

∞

• n=0

(−1)nzn, |z| < 1 For any path connecting 0 and z, term-by-term integration yields log(1 + z) =

∞

• n=0

(−1)n z ζndζ =

∞

• n=1

(−1)n−1 n zn, |z| < 1

• Example. Replacing z by −z in the above expansion,

log(1 − z) = −

∞

• n=1

zn n , |z| < 1

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Example

Re Im

x

−1

The principal branch of (1 + z)α with value 1 at z = 0 is (1 + z)α = eα log(1+z). It is analytic

• n C \ (−∞, −1]. Its Taylor series converges
• n B(0, 1).

By the chain rule, [(1 + z)α](n) = α(α − 1) · · · (α − n + 1)(1 + z)α−n If α = m ∈ N, then [(1 + z)α](n) ≡ 0 for n > m. Setting z = 0, [(1 + z)α](n)

• z=0 = α(α − 1) · · · (α − n + 1)

Thus (1 + z)α =

∞

• n=0

α(α − 1) · · · (α − n + 1) n! zn

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Contents

• 1. Power Series
• 2. Taylor Series
• 3. Laurent Series

17/26

Laurent Series

A Laurent series is a series of the form

∞

• n=−∞

cn(z − z0)n (†) where cn, z0 ∈ C are constants. If cn = 0 for n ≤ −1, a Laurent series reduces to a power series. The Laurent series (†) converges at z iff the following two series both converge at z

∞

• n=0

cn(z − z0)n,

∞

• n=1

c−n(z − z0)−n The series (†) diverges if either of the above series diverges.

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Laurent Series

The power series

∞

• n=0

cn(z − z0)n has a radius of convergence R1 ∈ [0, ∞], i.e. it converges for |z − z0| < R1. Changing variables by ζ =

1 z−z0, the series ∞

• n=1

c−n(z − z0)−n reduces to a power series

∞

• n=1

c−nζn with radius of convergence R ∈ [0, ∞], so

∞

• n=1

c−n(z − z0)−n converges for |z − z0| > R2 = 1

R

The Laurent series (†) converges on the annulus2 (or ring) R2 < |z − z0| < R1, which is nonempty only if R2 < R1.

2As in the case of power series, the convergence on the boundary

• f the annulus has to be considered case by case.

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Example

Consider the Laurent series

−1

• n=−∞

zn an +

∞

• n=0

zn bn

• ∞
• n=0

zn bn converges on |z| < |b| and diverges on |z| ≥ |b|

• −1
• n=−∞

zn an converges on |z| > |a| and diverges on |z| ≤ |a|

a b a b

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Properties of Laurent Series

• Theorem. If the Laurent series

∞

• n=−∞

cn(z − z0)n converges on a domain D, then D is an annulus, i.e. D = {z : R2 < |z − z0| < R1}, where 0 ≤ R2 < R1 ≤ ∞. Moreover,

• 1. the series converges absolutely on D, and uniformly on

r2 ≤ |z − z0| ≤ r1 for any R2 < r2 ≤ r1 < R1

• 2. its sum f(z) is analytic on D
• 3. f can be differentiated term by term on D, i.e.

f ′(z) =

∞

• n=−∞

ncn(z − z0)n−1

• 4. f can be integrated term by term on D, i.e.,
• γ

f(z)dz =

∞

• n=−∞

cn

• γ

(z − z0)ndz for γ in D

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Laurent Series Expansion

• Theorem. If f is analytic on D = {z : R2 < |z − z0| < R1}, where

0 ≤ R2 < R1 ≤ ∞, then f has the following unique Laurent series expansion at z0, f(z) =

∞

• n=−∞

cn(z − z0)n, z ∈ D where cn = 1 j2π

• S(z0,r)

f(ζ) (ζ − z0)n+1dζ, R2 < r < R2, n ∈ Z

• NB. cn = f (n)(z0)

n!

in general. In fact, f (n)(z0) may not even exist.

• NB. The annulus on which the Laurent series expansion holds

can be expanded by increasing R1 and decreasing R2 until the circles |z − z0| = R1 and |z − z0| = R2 hit singular points of f.

• NB. The expansion is unique on a given annulus, but may be

different on different annuli.

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Laurent Series Expansion

z0 R1 R2 z γ1 ζ γ2 ζ

• Proof. Fix z ∈ D. Choose two circles γ1

and γ2 centered at z0 with radii r1 and r2 satisfying R2 < r2 < |z − z0| < r1 < R1. By a generalization of Cauchy’s Integral Formula to multiply connected domain proved by cutting the domain into simply connected domains (e.g. by the red lines) f(z) = 1 j2π

• γ1

f(ζ) ζ − zdζ − 1 j2π

• γ2

f(ζ) ζ − zdζ For ζ ∈ γ1,

• z−z0

ζ−z0

• = |z−z0|

r1

< 1, so

1 ζ−z = ∞

• n=0

(z−z0)n (ζ−z0)n+1 uniformly for

ζ ∈ γ1. Term-by-term integration and Cauchy’s Theorem yield

• γ1

f(ζ) ζ − zdζ =

∞

• n=0
• γ1

(z − z0)nf(ζ) (ζ − z0)n+1 dζ =

∞

• n=0
• S(z0,r)

(z − z0)nf(ζ) (ζ − z0)n+1 dζ

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Laurent Series Expansion

Proof (cont’d). Similarly, for ζ ∈ γ2,

• ζ−z0

z−z0

• =

r2 |z−z0| < 1, so 1 ζ−z = − ∞

• n=1

(ζ−z0)n−1 (z−z0)n

uniformly for ζ ∈ γ2. Term-by-term integration and Cauchy’s Theorem yield

• γ2

f(ζ) ζ − zdζ = −

∞

• n=0
• γ2

(ζ − z0)n−1f(ζ) (z − z0)n dζ = −

∞

• n=0
• S(z0,r)

(ζ − z0)n−1f(ζ) (z − z0)n dζ = −

−1

• n=−∞
• S(z0,r)

(z − z0)nf(ζ) (ζ − z0)n+1 dζ Adding the two integrals along γ1 and γ2 together yields the desired expansion.

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Laurent Series Expansion

Proof (cont’d). For uniqueness, suppose f has the following Laurent series expansion, f(z) =

∞

• n=−∞

an(z − z0)n The convergence is uniform for z ∈ S(z0, r). Dividing both sides by (z − z0)k+1, the resulting series given below still converges uniformly on S(z0, r), f(z) (z − z0)k+1 =

∞

• n=−∞

an(z − z0)n−k−1 Term-by-term integration yields

• S(z0,r)

f(z)dz (z − z0)k+1 =

∞

• n=−∞

an

• S(z0,r)

(z − z0)n−k−1dz = j2πak Thus an = cn.

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Example

Find the Laurent series of f(z) = ez

z2 on 0 < |z| < ∞.

Method 1. Let γ be the unit circle S(0, 1). f(z) =

∞

• n=−∞

cnzn, where cn = 1 j2π

• γ

f(z) zn+1dz = 1 j2π

• γ

ez zn+3dz

• For n ≤ −3, ez/(zn+3) is analytic on C. By Cauchy’s

Theorem, cn = 0 for n ≤ −3.

• For n > −3, cn =

1 (n+2)!(ez)(n+2)

• z=0 =

1 (n+2)! by Cauchy’s

Integral Formula. Thus ez z2 =

∞

• n=−2

zn (n + 2)! Method 2. ez z2 = 1 z2

∞

• n=0

zn n! =

∞

• n=0

zn−2 n! =

∞

• n=−2

zn (n + 2)!

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Example

Re Im

x x

1 2

Find the Laurent series of f(z) =

1 (z−1)(z−2)

• n each of the following domains.
• 1. |z| < 1
• 2. 1 < |z| < 2
• 3. 2 < |z| < ∞
• Solution. f(z) =

1 1 − z − 1 2 − z

• 1. The Laurent series reduces to the Taylor series on |z| < 1

f(z) =

1 1−z − 1 2 · 1 1− z

2 =

∞

• n=0
• zn − 1

2

z

2

n =

∞

• n=0

(1 −

1 2n+1)zn

• 2. f(z) = − 1

z · 1 1− 1

z − 1

2 1 1− z

2 = −

∞

• n=0

1 zn+1 − ∞

• n=0

1 2n+1zn

• 3. f(z) = − 1

z · 1 1− 1

z + 1

z 1 1− 2

z = −

∞

• n=0

1 zn+1 + ∞

• n=0

2n zn+1 = ∞

• n=2

2n−1−1 zn