Section 5.2: Sequences of functions Section 5.3: Taylor series for analytic functions wi4243AP: Complex Analysis week 5, Monday K. P. Hart Faculty EEMCS TU Delft Delft, 29 september, 2014 K. P. Hart wi4243AP: Complex Analysis
Section 5.2: Sequences of functions Section 5.3: Taylor series for analytic functions Outline Section 5.2: Sequences of functions 1 Convergence Power series Differentiation and integration Section 5.3: Taylor series for analytic functions 2 Example K. P. Hart wi4243AP: Complex Analysis
Convergence Section 5.2: Sequences of functions Power series Section 5.3: Taylor series for analytic functions Differentiation and integration Convergence Definition A sequence { f n } of functions converges to a function f (on some domain) if for each individual z in the domain one has n →∞ f n ( z ) = f ( z ) lim Definition f n ( z ) → f ( z ) uniformly if for every ε > 0 there is an N ( ε ) such that for all n � N ( ε ) we have � < ε � � � f n ( z ) − f ( z ) for all z in the domain. K. P. Hart wi4243AP: Complex Analysis
Convergence Section 5.2: Sequences of functions Power series Section 5.3: Taylor series for analytic functions Differentiation and integration Standard example D = { z : | z | < 1 } and f n ( z ) = z n . f n ( z ) → 0 for each individual z f n ( z ) → 0 not uniformly on D f n ( z ) → 0 uniformly on D r = { z : | z | � r } if r < 1 K. P. Hart wi4243AP: Complex Analysis
Convergence Section 5.2: Sequences of functions Power series Section 5.3: Taylor series for analytic functions Differentiation and integration Uniform convergence: properties Theorem If f n → f uniformly on some curve C then � � lim f n ( z ) d z = f ( z ) d z n →∞ C C Proof. By the useful inequality � � � � � � � f n ( z ) d z − f ( z ) d z | f n ( z ) − f ( z ) | d z � M n · L � � � � � C C C where M n = sup {| f n ( z ) − f ( z ) | : z ∈ C } and L is the length of C . Uniform convergence: M n → 0. K. P. Hart wi4243AP: Complex Analysis
Convergence Section 5.2: Sequences of functions Power series Section 5.3: Taylor series for analytic functions Differentiation and integration Uniform convergence: properties Theorem n → g uniformly and f n → f (in just one point) then f ′ = g. If f ′ So n f ′ n = (lim f n ) ′ lim provided { f ′ n } is known to converge uniformly and { f n } converges somewhere. K. P. Hart wi4243AP: Complex Analysis
Convergence Section 5.2: Sequences of functions Power series Section 5.3: Taylor series for analytic functions Differentiation and integration Power series Special form: a fixed number z 0 and a sequence { a n } of numbers are given. Put f n ( z ) = a n ( z − z 0 ) n , we write ∞ � a n ( z − z 0 ) n n =0 for the resulting series. K. P. Hart wi4243AP: Complex Analysis
Convergence Section 5.2: Sequences of functions Power series Section 5.3: Taylor series for analytic functions Differentiation and integration Radius of convergence Theorem n a n ( z − z 0 ) n there is an R such that Given a power series � n a n ( z − z 0 ) n converges if | z − z 0 | < R � n a n ( z − z 0 ) n diverges if | z − z 0 | > R � In addition: if r < R then the series converges uniformly on { z : | z − z 0 | � r } . On the boundary — | z − z 0 | = R — anything can happen. R = 0, 0 < R < ∞ and R = ∞ are all possible. R is the radius of convergence of the series. K. P. Hart wi4243AP: Complex Analysis
Convergence Section 5.2: Sequences of functions Power series Section 5.3: Taylor series for analytic functions Differentiation and integration Differentiation Theorem n a n ( z − z 0 ) n be a power series, with radius R, and let Let � n na n ( z − z 0 ) n − 1 be its termwise derivative, with radius R ′ . � Then R = R ′ . Proof. n na n ( z − z 0 ) n − 1 converges absolutely then so does If � n a n ( z − z 0 ) n , by comparison. So R � R ′ . � K. P. Hart wi4243AP: Complex Analysis
Convergence Section 5.2: Sequences of functions Power series Section 5.3: Taylor series for analytic functions Differentiation and integration Differentiation Proof. n a n ( w − z 0 ) n converges then so does Conversely: if � n na n ( z − z 0 ) n − 1 whenever | z − z 0 | < | w − z 0 | . � Fix N such that | a n ( w − z 0 ) n | � 1 for n � N . For those n � z − z 0 � n � � 1 | na n ( z − z 0 ) n − 1 | = � a n ( w − z 0 ) n � � n � � z − z 0 w − z 0 � n n � z − z 0 � � � � � � | z − z 0 | w − z 0 � � n nz n has radius 1. Now use that � K. P. Hart wi4243AP: Complex Analysis
Convergence Section 5.2: Sequences of functions Power series Section 5.3: Taylor series for analytic functions Differentiation and integration Differentiation/Integration Theorem n a n ( z − z 0 ) n for | z − z 0 | < R then If R > 0 and f ( z ) = � n na n ( z − z 0 ) n − 1 . f ′ ( z ) = � n +1 ( z − z 0 ) n +1 is a primitive function of f a n � n K. P. Hart wi4243AP: Complex Analysis
Convergence Section 5.2: Sequences of functions Power series Section 5.3: Taylor series for analytic functions Differentiation and integration Differentiation/Integration n =0 z n = 1 We have � ∞ 1 − z for | z | < 1. n =1 nz n − 1 = 1 So � ∞ (1 − z ) 2 . n =0 z n = 1 1 Also, � ∞ 1 − z is the derivative of � ∞ n z n . n =1 n z n = − Log(1 − z ) + c for some c ; 1 So, � ∞ n =1 put in z = 0: we get c = 0, and so ∞ 1 nz n = − Log(1 − z ) � ( | z | < 1) n =1 K. P. Hart wi4243AP: Complex Analysis
Convergence Section 5.2: Sequences of functions Power series Section 5.3: Taylor series for analytic functions Differentiation and integration Derivatives If f ( z ) = � ∞ n =0 a n ( z − z 0 ) n , with R > 0, then f ( z 0 ) = a 0 f ′ ( z 0 ) = 1 · a 1 f ′′ ( z 0 ) = 2 · 1 · a 2 f ( k ) ( z 0 ) = k ! a k and we get ∞ f ( n ) ( z 0 ) � ( z − z 0 ) n f ( z ) = n ! n =0 K. P. Hart wi4243AP: Complex Analysis
Section 5.2: Sequences of functions Example Section 5.3: Taylor series for analytic functions Main result Theorem Let f : D → C be analytic, let z 0 ∈ D and let R be the distance from z 0 to the complement of D (if D = C then R = ∞ ). Then on the disc { z : | z − z 0 | < R } we have ∞ � a n ( z − z 0 ) n f ( z ) = n =0 where ( ζ − z 0 ) n +1 d ζ = f ( n ) ( z 0 ) 1 f ( ζ ) � a n = 2 π i n ! C and C is any simple closed contour around z 0 lying inside D. K. P. Hart wi4243AP: Complex Analysis
Section 5.2: Sequences of functions Example Section 5.3: Taylor series for analytic functions Why? z r z 0 R 1 R Take z inside the circle { w : | w − z 0 | = R } and take R 1 such that | z − z 0 | = r < R 1 < R . Work on the circle C 1 of radius R 1 around z 0 . K. P. Hart wi4243AP: Complex Analysis
Section 5.2: Sequences of functions Example Section 5.3: Taylor series for analytic functions Why? f ( ζ ) 1 � Apply Cauchy’s formula: f ( z ) = ζ − z d ζ . 2 π i C 1 Transform 1 / ( ζ − z ): ∞ � n 1 1 1 1 � z − z 0 � ζ − z = = 1 − z − z 0 ζ − z 0 ζ − z 0 ζ − z 0 ζ − z 0 n =0 The modulus, r / R 1 , of the quotient is less than 1 on C 1 , so this series converges uniformly on C 1 . We may interchange sum and integral. K. P. Hart wi4243AP: Complex Analysis
Section 5.2: Sequences of functions Example Section 5.3: Taylor series for analytic functions Why? 1 � f ( ζ ) f ( z ) = ζ − z d ζ 2 π i C 1 ∞ � n 1 f ( ζ ) � z − z 0 � � = d ζ 2 π i ζ − z 0 ζ − z 0 C 1 n =0 ∞ 1 � f ( ζ ) ( ζ − z 0 ) n +1 ( z − z 0 ) n d ζ � = 2 π i C 1 n =0 ∞ 1 � f ( ζ ) � ( ζ − z 0 ) n +1 d ζ × ( z − z 0 ) n = 2 π i C 1 n =0 Done, by Cauchy’s general formula. K. P. Hart wi4243AP: Complex Analysis
Section 5.2: Sequences of functions Example Section 5.3: Taylor series for analytic functions What is the radius? The radius of convergence of the series is the largest R such that f is analytic on { z : | z − z 0 | < R } , possibly R = ∞ . For example: the Taylor series of arctan z centered at 0 has radius 1, because i and − i are branch points: arctan z is analytic on { z : | z | < 1 } but on no larger disc centered at 0. K. P. Hart wi4243AP: Complex Analysis
Section 5.2: Sequences of functions Example Section 5.3: Taylor series for analytic functions arctan z Remember: arctan z = 1 � 1 + iz � 2 i log 1 − iz Also ∞ ( − 1) n +1 � z n Log(1 + z ) = n n =1 We stick in iz and − iz and subtract the results. K. P. Hart wi4243AP: Complex Analysis
Section 5.2: Sequences of functions Example Section 5.3: Taylor series for analytic functions arctan z ∞ ( − 1) n +1 � ( iz ) n Log(1 + iz ) = n n =1 ∞ ( − 1) n +1 � ( − iz ) n Log(1 − iz ) = n n =1 the even-numbered terms drop out; if n is odd, say n = 2 k + 1, the n th terms give ( − 1) 2 k +2 2 k + 1 ( iz ) 2 k +1 − ( − 1) 2 k +2 2 k + 1 ( − iz ) 2 k +1 = 2 i 2 k +1 2 k + 1 z 2 k +1 K. P. Hart wi4243AP: Complex Analysis
Recommend
More recommend
