EI331 Signals and Systems Lecture 25 Bo Jiang John Hopcroft Center for Computer Science Shanghai Jiao Tong University May 23, 2019
Contents 1. Power Series 2. Taylor Series 3. Laurent Series 1/26
Properties of Power Series ∞ c n ( z − z 0 ) n has radius of convergence R , then � If n = 0 • its sum f ( z ) is analytic on | z − z 0 | < R . • f can be differentiated term by term on | z − z 0 | < R , i.e. ∞ � f ′ ( z ) = nc n ( z − z 0 ) n − 1 n = 1 • f can be integrated term by term on | z − z 0 | < R , i.e., ∞ � � � ( z − z 0 ) n dz f ( z ) dz = c n for γ in | z − z 0 | < R γ γ n = 0 In particular, � z ∞ c n � n + 1 ( z − z 0 ) n + 1 f ( ζ ) d ζ = z 0 n = 0 2/26
Uniform Convergence ∞ � A series f n ( z ) converges to f ( z ) uniformly on Ω ⊂ C , if the n = 1 sequence of partial sums s k ( z ) converges to f ( z ) uniformly on Ω , k →∞ max lim z ∈ Ω | s k ( z ) − f ( z ) | = 0 . ∞ Theorem (Cauchy’s convergence test). The series � f n ( z ) n = 1 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 | f n + 1 ( z ) + f n + 2 ( z ) + · · · + f n + p ( z ) | < ǫ Theorem (Weierstrass M-test). If | f n ( z ) | ≤ M n on Ω for each n , ∞ ∞ � � and M n < + ∞ , then f n ( z ) converges uniformly. n = 1 n = 1 3/26
Properties of Uniformly Convergent Series ∞ Theorem. If f n ( z ) is continuous on Ω for each n , and � f n ( z ) n = 1 converges to f ( z ) uniformly on Ω , then f ( z ) is continuous on Ω . Theorem. If for each n , f n ( z ) is continuous on a piecewise ∞ smooth curve γ with finite length, and � f n ( z ) converges to f ( z ) n = 1 uniformly on γ , then � ∞ � � f ( z ) dz = f n ( z ) dz γ γ n = 1 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 f n ( z ) | ≤ ǫ/ L for all k ≥ N and z ∈ γ . Thus � � � � k k � � � � ds ≤ ǫ � � � � � � � f ( z ) dz − f n ( z ) dz � ≤ � f ( z ) − f n ( z ) ds = ǫ � � � � � � � � L γ γ γ γ � n = 1 n = 1 4/26
Properties of Uniformly Convergent Series Theorem. If for each n , f n ( z ) is analytic on a domain D , and ∞ � f n ( z ) converges to f ( z ) uniformly on D , then n = 1 ∞ � 1. f ( z ) = f n ( z ) is analytic on D . n = 1 ∞ f ( k ) 2. f ( k ) ( z ) = � n ( z ) for z ∈ D and k ∈ N . n = 1 Proof of 1. Fix an arbitrary z 0 ∈ D and an open disk B ( z 0 , r ) ⊂ D . For any piecewise smooth Jordan curve γ in B ( z 0 , r ) , term-by-term integration and Cauchy’s Theorem imply ∞ � � � f ( z ) dz = f n ( z ) dz = 0 γ γ n = 1 Uniform convergence implies f is continuous. By Morera’s Theorem, f ( z ) is analytic on B ( z 0 , r ) . 5/26
Properties of Uniformly Convergent Series Proof of 2. Fix an arbitrary z 0 ∈ D and a closed disk ¯ B ( z 0 , r ) ⊂ D . For z on the circle S ( z 0 , r ) , � � � � N N f ( z ) f n ( z ) 1 � � � � � � ( z − z 0 ) k + 1 − � = � f ( z ) − f n ( z ) � � � � ( z − z 0 ) k + 1 r k + 1 � � � � � � n = 0 n = 0 ∞ f n ( z ) f ( z ) � so ( z − z 0 ) k + 1 converges to ( z − z 0 ) k + 1 uniformly on S ( z 0 , r ) . n = 0 Term-by-term integration yields ∞ k ! f ( z ) k ! f n ( z ) � � � ( z − z 0 ) k + 1 dz = ( z − z 0 ) k + 1 dz j 2 π j 2 π | z − z 0 | = r | z − z 0 | = r n = 1 By Cauchy’s Integral Formula, ∞ � f ( k ) ( z ) = f ( k ) n ( z ) n = 1 6/26
Contents 1. Power Series 2. Taylor Series 3. Laurent Series 7/26
Taylor Series Suppose an analytic function f has a power series expansion on an open disk B ( z 0 , R ) , ∞ � c n ( z − z 0 ) n f ( z ) = n = 0 Term-by-term differentiation yields, ∞ n ! � f ( m ) ( z ) = ( n − m )! c n ( z − z 0 ) n − m n = m Setting z = z 0 , ⇒ c m = f ( m ) ( z 0 ) f ( m ) ( z 0 ) = m ! c m = m ! Thus the power series expansion of f at z 0 is unique if it exists. 8/26
Taylor Series Theorem. If f is analytic on an open disk B ( z 0 , R ) , then f has the following power series expansion for z ∈ B ( z 0 , R ) , ∞ � c n ( z − z 0 ) n f ( z ) = ( ⋆ ) n = 0 where c n = f ( n ) ( z 0 ) , n = 0 , 1 , 2 , . . . n ! The series in ( ⋆ ) is called the Taylor series of f at z 0 . NB. f is analytic at z 0 iff ( ⋆ ) holds in B ( z 0 , R ) for some R . NB. If f is analytic at z 0 , then there is always a largest open disk B ( z 0 , R max ) on which the Taylor series expansion ( ⋆ ) holds. The radius R max is the distance between z 0 and the closest point at which f is not analytic 1 . 1 A point at which f fails to be analytic is called a singular point of f . 9/26
Taylor Series Proof. Let r ∈ ( 0 , R ) . Fix z ∈ B ( z 0 , r ) . For � � � = | z − z 0 | � z − z 0 ζ ∈ γ = S ( z 0 , r ) , < 1 , so R � � ζ − z 0 r r ∞ ( z − z 0 ) n 1 1 1 � ζ − z = · = z z 0 1 − z − z 0 ( ζ − z 0 ) n + 1 ζ − z 0 ζ ζ − z 0 n = 0 The convergence is uniform for ζ ∈ γ . Since f ( ζ ) is bounded on γ , ∞ f ( ζ ) ( z − z 0 ) n f ( ζ ) � ζ − z = uniformly for ζ ∈ γ. ( ζ − z 0 ) n + 1 n = 0 Term-by-term integration yields ∞ f ( ζ ) d ζ �� f ( ζ ) d ζ � � � ( z − z 0 ) n ζ − z = ( ζ − z 0 ) n + 1 γ γ n = 0 Cauchy’s Integral Formula then yields the desired result. 10/26
Examples Example. f ( z ) = e z is analytic on C and f ( n ) ( z ) = e z for n ∈ N . Thus ∞ ∞ e 0 1 e z = n ! z n = � � n ! z n , z ∈ C n = 0 n = 0 Recall that this can be used an alternative definition of e z . Example. cos z and sin z are analytic on C . For z ∈ C , ∞ ∞ ( − 1 ) n cos z = 1 2 ( e jz + e − jz ) = 1 1 n ![( jz ) n + ( − jz ) n ] = � � ( 2 n )! z 2 n 2 n = 0 n = 0 and ∞ ∞ ( − 1 ) n sin z = 1 2 j ( e jz − e − jz ) = 1 1 � � n ![( jz ) n − ( − jz ) n ] = ( 2 n + 1 )! z 2 n + 1 2 j n = 0 n = 0 11/26
Examples 1 Im Example. ( 1 + z ) 2 is analytic on C \ {− 1 } . Its Taylor series at z = 0 converges on B ( 0 , 1 ) . ∞ 1 x � ( − 1 ) n z n , 1 + z = | z | < 1 − 1 − 1 Re 2 n = 0 Differentiation yields ∞ ∞ 1 ( − 1 ) n nz n − 1 = � � ( − 1 ) n ( n + 1 ) z n , ( 1 + z ) 2 = − | z | < 1 n = 1 n = 0 ( 1 + z ) 2 at z = − 1 1 Example. Find the Taylor series of 2 . ∞ � n � � � 1 2 z + 1 � z + 1 � < 1 � � � 2 ( − 2 ) n 1 + z = 2 ) = , � � 1 + 2 ( z + 1 2 2 2 n = 0 ∞ � n � � � 1 z + 1 � z + 1 � < 1 � ( − 2 ) n + 2 ( n + 1 ) � � ( 1 + z ) 2 = , � � 2 2 2 n = 0 12/26
Examples 1 Im Example. 1 + z 2 is analytic on C \ { j , − j } . Its j x Taylor series at z = 0 converges on B ( 0 , 1 ) . ∞ 1 � ( − 1 ) n z 2 n , 1 + z 2 = | z | < 1 Re n = 0 x 1 1 + z 2 at z = 1 Example. The Taylor series of − j √ converges on B ( 1 , 2 ) . To obtain the expansion, let ζ = z − 1 . Im j x 1 / ( 2 j ) 1 / ( 2 j ) 1 1 + z 2 = ζ + 1 − j − √ ζ + 1 + j R = 2 ∞ ∞ ( − ζ ) n ( − ζ ) n = 1 ( 1 − j ) n + 1 − 1 � � 1 Re ( 1 + j ) n + 1 2 j 2 j x n = 0 n = 0 − j ∞ sin 3 π ( n + 1 ) √ � ( z − 1 ) n , 4 = | z − 1 | < 2 n + 1 2 2 n = 0 13/26
Examples Example. Let log( 1 + z ) be the principal Im branch of Log( 1 + z ) with value 0 at z = 0 . It is analytic on C \ ( −∞ , − 1 ] . Its Taylor series x − 1 converges on B ( 0 , 1 ) . Re ∞ 1 [log( 1 + z )] ′ = � ( − 1 ) n z n , 1 + z = | z | < 1 n = 0 For any path connecting 0 and z , term-by-term integration yields � z ∞ ∞ ( − 1 ) n − 1 � � ( − 1 ) n ζ n d ζ = z n , log( 1 + z ) = | z | < 1 n 0 n = 0 n = 1 Example. Replacing z by − z in the above expansion, ∞ z n � log( 1 − z ) = − n , | z | < 1 n = 1 14/26
Example The principal branch of ( 1 + z ) α with value 1 Im at z = 0 is ( 1 + z ) α = e α log( 1 + z ) . It is analytic on C \ ( −∞ , − 1 ] . Its Taylor series converges x − 1 on B ( 0 , 1 ) . Re 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 ∞ α ( α − 1 ) · · · ( α − n + 1 ) ( 1 + z ) α = � z n n ! n = 0 15/26
Contents 1. Power Series 2. Taylor Series 3. Laurent Series 16/26
Laurent Series A Laurent series is a series of the form ∞ � c n ( z − z 0 ) n ( † ) n = −∞ where c n , z 0 ∈ C are constants. If c n = 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 ∞ ∞ � � c n ( z − z 0 ) n , c − n ( z − z 0 ) − n n = 0 n = 1 The series ( † ) diverges if either of the above series diverges. 17/26
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