EI331 Signals and Systems Lecture 24 Bo Jiang John Hopcroft Center - PowerPoint PPT Presentation

EI331 Signals and Systems Lecture 24 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University May 21, 2019

Contents 1. Cauchy’s Integral Formula for Derivatives 2. Harmonic Functions 3. Power Series 1/24

Derivatives of Cauchy-type Integral Theorem. Assume (a) γ is a piecewise smooth simple (or Jordan) curve (b) f is continuous on γ Then the function defined by the following Cauchy-type integral f ( ζ ) 1 � F ( z ) � ζ − zd ζ, z / ∈ γ j 2 π γ is analytic on C \ γ . Moreover, it is infinitely differentiable and all its derivatives are analytic on C \ γ with F ( n ) ( z ) = n ! � f ( ζ ) ( ζ − z ) n + 1 d ζ, n = 1 , 2 , . . . j 2 π γ NB. The formula for F ( n ) can be obtained by differentiation under the integral sign. 2/24

Derivatives of Cauchy-type Integral Proof. If γ is a simple curve, then C \ γ is a domain. If γ is a Jordan curve, then C \ γ is the union of two domains by the Jordan Curve Theorem. (1). We first show F ( z ) is continuous on C \ γ . • For z 0 ∈ C \ γ , there exists an open disk B ( z 0 , δ ) ⊂ C \ γ , so | ζ − z 0 | ≥ δ for ζ ∈ γ . • Let z ∈ B ( z 0 , δ/ 2 ) . The triangle inequality yields | z − ζ | ≥ | z 0 − ζ | − | z − z 0 | ≥ δ/ 2 for ζ ∈ γ . • By the definition of F , F ( z ) − F ( z 0 ) = z − z 0 � f ( ζ ) ( ζ − z )( ζ − z 0 ) d ζ ( ⋆ ) j 2 π γ so | F ( z ) − F ( z 0 ) | ≤ | z − z 0 | � ( δ/ 2 ) δ | d ζ | = | z − z 0 | | f ( ζ ) | � | f ( ζ ) | d ζ 2 π πδ 2 γ γ 3/24

Derivatives of Cauchy-type Integral Proof (cont’d). (2). Next we show � f ( ζ ) 1 F ′ ( z ) = ( ζ − z ) 2 d ζ, ∀ z ∈ C \ γ. j 2 π γ • By ( ⋆ ) of the previous slide, F ( z ) − F ( z 0 ) 1 � f ( ζ ) = ( ζ − z )( ζ − z 0 ) d ζ = G ( z ) , z − z 0 j 2 π γ where g ( ζ ) � f ( ζ ) ζ − z 0 and � g ( ζ ) 1 G ( z ) � ζ − zd ζ j 2 π γ • g is continuous on γ . By (1), G is continuous on C \ γ , so F ( z ) − F ( z 0 ) � f ( ζ ) 1 F ′ ( z 0 ) = lim = G ( z 0 ) = ( ζ − z 0 ) 2 d ζ z − z 0 j 2 π z → z 0 γ 4/24

Derivatives of Cauchy-type Integral Proof (cont’d). (3). Finally we show the formula for higher order derivatives by induction. • Assume the formula holds for 1 ≤ k ≤ n , n ≥ 1 , F ( k ) ( z ) = k ! � f ( ζ ) ( ζ − z ) k + 1 d ζ. ( † ) j 2 π γ • For z 0 ∈ C \ γ , and g , G defined on the previous slide, F ( n ) ( z ) = n ! � ( ζ − z 0 ) g ( ζ ) ( ζ − z ) n + 1 d ζ j 2 π γ = n ! � ( ζ − z ) n d ζ + n ! g ( ζ ) � ( z − z 0 ) g ( ζ ) ( ζ − z ) n + 1 d ζ j 2 π j 2 π γ γ • Since g is continuous on γ , ( † ) holds with f replaced by g , F ( n ) ( z ) = nG ( n − 1 ) ( z ) + ( z − z 0 ) G ( n ) ( z ) 5/24

Derivatives of Cauchy-type Integral Proof (cont’d). • As in (1), let B ( z 0 , δ ) ⊂ C \ γ . G ( n ) is bounded on B ( z 0 , δ/ 2 ) | G ( n ) ( z ) | ≤ n ! � | ζ − z | n + 1 | d ζ | ≤ n ! | g ( ζ ) | � | g ( ζ ) | ( δ/ 2 ) n + 1 | d ζ | j 2 π j 2 π γ γ • G ( n − 1 ) ( z ) is differentiable and hence continuous z → z 0 F ( n ) ( z ) = lim z → z 0 [ nG ( n − 1 ) ( z ) + ( z − z 0 ) G ( n ) ( z )] = F ( n ) ( z 0 ) lim so F ( n ) is continuous. Similarly, G ( n ) is continuous • Let z → z 0 , F ( n ) ( z ) − F ( n ) ( z 0 ) = nG ( n − 1 ) ( z ) − G ( n − 1 ) ( z 0 ) + G ( n ) ( z ) z − z 0 z − z 0 → ( n + 1 ) G ( n ) ( z 0 ) = F ( n + 1 ) ( z 0 ) 6/24

Integral Formula for Derivatives of Analytic Functions Theorem. If f is analytic on a domain D , then its derivative f ′ is also analytic on D , D and z f ( n ) ( z ) = n ! f ( ζ ) � ( ζ − z ) n + 1 d ζ, γ j 2 π γ where γ is a positively oriented piecewise smooth Jordan curve encircling z whose interior lies entirely in D . Proof. By Cauchy’s Integral Formula � f ( z ) 1 f ( z ) = ζ − zdz j 2 π γ Since f is analytic on D , it is continuous on γ . By the previous theorem, f ′ is analytic and the derivatives of f are as given. 7/24

Example Let γ = { z : | z | = 2 } be positively oriented and r > 1 . y Let C 0 , C 1 be positively oriented circles centered at j and − j that lie in the interior of γ . By Cauchy’s γ C 0 Theorem, x C 1 e z dz e z dz e z dz � � � ( z 2 + 1 ) 2 = ( z 2 + 1 ) 2 + ( z 2 + 1 ) 2 γ C 0 C 1 e z � ′ ( z + j ) 2 dz e z dz e z = ( 1 − j ) e j � � � ( z 2 + 1 ) 2 = ( z − j ) 2 = j 2 π π ( z + j ) 2 dz 2 C 0 C 0 z = j e z � ′ ( z − j ) 2 dz e z dz e z = − ( 1 + j ) e − j � � � ( z 2 + 1 ) 2 = ( z + j ) 2 = j 2 π π ( z − j ) 2 dz 2 C 1 C 1 z = − j e z dz � ( z 2 + 1 ) 2 = j π Im [( 1 − j ) e j ] = j π (sin 1 − cos 1 ) γ 8/24

Morera’s Theorem � Theorem. If f is continuous on a domain D and γ f ( z ) dz = 0 for any piecewise smooth Jordan curve γ in D whose interior also lies in D , then f is analytic on D . Proof. Fix z 0 ∈ D and an open disk B ( z 0 , δ ) ⊂ D . It suffices to show f is analytic on B ( z 0 , δ ) . � 1. Because γ f ( z ) dz = 0 for any piecewise Jordan curves, the � z integral z 0 f ( z ) dz is independent of the path in B ( z 0 , δ ) that connects z 0 and z . Define � z F ( z ) = f ( z ) dz , z ∈ B ( z 0 , δ ) . z 0 2. Since f is continuous, the proof on slides 16-17 of Lecture 23 shows F ′ ( z ) = f ( z ) , so F is analytic on B ( z 0 , δ ) . 3. Thus f = F ′ is also analytic on B ( z 0 , δ ) . 9/24

Cauchy’s Inequality Theorem. If f is analytic on the open disk B ( z 0 , R ) , and | f ( z ) | ≤ M on B ( z 0 , R ) , then | f ( n ) ( z 0 ) | ≤ n ! M R n , n ∈ N . Proof. Let γ be the circle | z − z 0 | = r with r ∈ ( 0 , R ) . Then f ( n ) ( z 0 ) = n ! � f ( ζ ) ( ζ − z 0 ) n + 1 d ζ j 2 π γ Thus | f ( n ) ( z 0 ) | ≤ n ! � | ζ − z 0 | n + 1 | d ζ | ≤ n ! | f ( ζ ) | r n + 1 2 π r = n ! M M 2 π 2 π r γ Letting r → R , | f ( n ) ( z 0 ) | ≤ n ! M R n . 10/24

Liouville’s Theorem A functions that is analytic on C is called an entire function. Theorem. If f is entire and bounded, then it is constant. Proof. 1. Since f is bounded, there exists an M s.t. | f ( z ) | ≤ M , ∀ z ∈ C . 2. For any z 0 ∈ C and R > 0 , Cauchy’s inequality on B ( z 0 , R ) yields | f ′ ( z 0 ) | ≤ M R . 3. Letting R → ∞ , f ′ ( z 0 ) = 0 4. Since z 0 is arbitrary, f ′ ( z ) ≡ 0 on C , so f is constant. 11/24

Fundamental Theorem of Algebra Theorem. A polynomial P of degree n ≥ 1 , i.e. P ( z ) = a 0 + a 1 z + · · · + a n z n , a n � = 0 has exactly n roots in C . Proof Sketch. First show P has at least one root. 1. If P does not have a root, then Q ( z ) = 1 / P ( z ) is entire 2. As | z | → ∞ , Q ( z ) → 0 , so ∃ R > 0 s.t. | Q ( z ) | ≤ 1 for | z | > R 3. Being continuous, Q is bounded on | z | ≤ R and hence on C 4. By Liouville’s Theorem, Q is constant. So P is also constant, contradiction. Let z 0 be a root of P . Factor P ( z ) = ( z − z 0 ) P 1 ( z ) , where P 1 has degree n − 1 . We can repeat this process until P 1 has degree 0. 12/24

Contents 1. Cauchy’s Integral Formula for Derivatives 2. Harmonic Functions 3. Power Series 13/24

Harmonic Functions A real function φ ( x , y ) of two real variables is called a harmonic function on a domain D if it satisfies the Laplace equation on D , ∂φ 2 ∂ x 2 + ∂φ 2 ∂ y 2 = 0 Theorem. If f ( z ) = u ( x , y ) + jv ( x , y ) is analytic on a domain D , then u and v are harmonic functions on D . Proof. Since f is analytic, the Cauchy-Riemann equations hold u x = v y , u y = − v x Since f ′ is analytic, u and v are continuously differentiable, so u xx = v yx = v xy = − u yy So u xx + u yy = 0 . Similarly, v xx + v yy = 0 . 14/24

Harmonic Conjugate If f = u + jv is analytic, v is called a harmonic conjugate of u . Theorem. A harmonic function on a simply connected domain has a harmonic conjugate. Example. u ( x , y ) = y 3 − 3 x 2 y is harmonic on C . For its conjugate, 1. By the Cauchy-Riemann equations v x = − u y = − 3 y 2 + 3 x 2 , v y = u x = − 6 xy 2. Integrate w.r.t. y , � � ( − 6 xy ) dy = − 3 xy 2 + g ( x ) v ( x , y ) = v y dy = 3. Differentiate w.r.t. x , � v x = − 3 y 2 + g ′ ( x ) = − 3 y 2 + 3 x 2 = 3 x 2 dx = x 3 + c ⇒ g ( x ) = 4. v ( x , y ) = − 3 xy 2 + x 3 + c 15/24

Mean-value Property Theorem. If f ( z ) is analytic on an open disk B ( z 0 , R ) , then � 2 π f ( z 0 ) = 1 f ( z 0 + re jt ) dt , 0 < r < R . 2 π 0 i.e. the mean value of an analytic function on a circle is equal to the value at the center. Proof. Use Cauchy’s Integral Formula and the parameterization z ( t ) = re jt , t ∈ [ 0 , 2 π ] . Theorem. If u ( x , y ) is harmonic on an open disk B ( z 0 , R ) , then � 2 π u ( z 0 ) = 1 u ( z 0 + re jt ) dt 2 π 0 Proof. Use the previous theorem and the fact that u is the real part of an analytic function. 16/24

Contents 1. Cauchy’s Integral Formula for Derivatives 2. Harmonic Functions 3. Power Series 17/24