Complex variables lecture 6: Taylor and Laurent series Hyo-Sung Ahn School of Mechatronics Gwangju Institute of Science and Technology (GIST) 1 Oryong-dong, Buk-gu, Gwangju, Korea Advanced Engineering Mathematics
Uniform convergence - For a function f defined on a set T, the sequence of functions {S n } converges to f at the point z 0 T, provided lim n S n (z 0 ) = f(z 0 ). Thus, for the particular point z 0, we know that for each ɛ > 0, there exists a positive integer N ɛ,z0 such that if n N ɛ,z0, then S n (z 0 ) f(z 0 ) < ɛ (1) - If S n (z) is the nth partial sum of the series k=0 c k (z α) k, the statement (1) becomes if n N ɛ,z0, then n 1 k=0 c k (z α) k f(z 0 ) < ɛ - Uniform convergence: N ɛ (depending only on ɛ) - Def. (Uniform convergence): The sequence {S n (z)} converges uniformly to f(z) on the set T if for every ɛ > 0, there exists a positive integer N ɛ such that if n N ɛ, then S n (z) f(z) < ɛ (2) - Ex. The sequence {S n (z)} = {e z + 1 n } converges uniformly to the function f(z) = e z on the entire complex plane.
Uniform convergence - Ex. The geometric series S n (z) = n 1 k=0 z k converges to f(z) = 1 1 z for z D 1 (0); but not uniformly. The negation of the statement (2): There exists ɛ > 0 such that, for all positive integers N, there are some n > N and some z 0 T such that S n (z 0 ) f(z 0 ) ɛ.
Uniform convergence - Theorem (Weierstrass M-test): Suppose that the infinite series k=0 u k (z) has the property that for each k, u k (z) M k for all z T. If k=0 M k converges, then k=0 u k (z) converges uniformly on T. Proof: Let S n (z) = n 1 k=0 u k (z) be the nth partial sum of the series. If n > m, then S n (z) S m (z) = u m (z) + u m+1 (z) + + u n 1 (z) n 1 k=m M k. Since k=0 M k converges, we can make the last expression as small as possible by choosing a large enough m. Thus, for ɛ > 0, there is a positive integer N ɛ such that if n, m > N ɛ, then S n (z) S m (z) < ɛ ( for all z T, {S n (z)} is a Cauchy sequence). Thus, the sequence must converge to a number (for example f(z)). That is, f(z) = lim n S n (z) = k=0 u k (z).
Uniform convergence (cont.) Now, we must show that the convergence is uniform. Let ɛ > 0 be given. Again, since k=0 M k converges, there exists N ɛ such that if n N ɛ, then k=n M k < ɛ. Thus, if n N ɛ and z T, then f(z) S n (z) = = < ɛ u k (z) k=0 u k (z) k=n k=n M k n 1 k=n u k (z) - Theorem: Suppose that the power series =0 c k (z α) k has radius of convergence ρ > 0. Then, for each r, 0 < r < ρ, the series converges uniformly on the closed disk D r (α).
Uniform convergence - Corollary: For each r, 0 < r < 1, the geometric series converges uniformly on the closed disk D r (0). - Theorem: Suppose that {S k } is a sequence of continuous functions defined on a set T containing the contour C. If {S k } converges uniformly to f on the set T, then f is continuous on T, and lim k C S k(z)dz = C lim k S k (z)dz = C f(z)dz Proof: Given z 0 T, we must prove lim z z0 f(z) = f(z 0 ). Let ɛ > 0 be given. Since {S k } converges uniformly to f on T, there exists a positive integer N ɛ such that for all z T, f(z) S k (z) < ɛ 3 whenever k N ɛ. And, as S Nɛ is continuous at z 0, there exists δ > 0 such that if z z 0 < δ, then S Nɛ (z) S Nɛ (z 0 ) < ɛ 3. Hence, if z z 0 < δ, we have f(z) f(z 0 ) = f(z) S Nɛ (z) + S Nɛ (z) S Nɛ (z 0 ) + S Nɛ (z 0 ) f(z 0 ) f(z) S Nɛ (z) + S Nɛ (z) S Nɛ (z 0 ) + S Nɛ (z 0 ) f(z 0 ) < ɛ 3 + ɛ 3 + ɛ 3 = ɛ
Uniform convergence (cont.) which complete the part (i). To show part (ii), let ɛ > 0 be given and let L be the length of the contour C. Because {S k } converges uniformly to f on T, there exists a positive integer N ɛ such that if k N ɛ, then S k (z) f(z) < ɛ for L all z T. Because C is contained in T, max z C S k (z) f(z) < ɛ if k N L ɛ, and we can use the ML inequality to get S k (z)dz f(z)dz = [S k (z) f(z)]dz C C C max z C S k(z) f(z) L < ( ɛ L L = ɛ
Taylor series representation - It was shown that functions defined by power series have derivatives of all orders, and analytic functions also have derivatives of all orders. - Def. (Taylor series): If f(z) is analytic at z = α, then the series f(α) + f (α)(z α) + f (2) (α) (z α) 2 + = 2! k=0 f (k) (α) (z α) k k! is called the Taylor series for f centered at α. When α = 0, it is called the Maclaurin series for f. - Theorem (Taylor s theorem): Suppose that f is analytic in a domain G and that D R (α) is any disk contained in G. Then the Taylor series for f converges to f(z) for all z in D R (α); that is, f(z) = k=0 f (k) (α) (z α) k, for all z D R (α) (3) k! Furthermore, for any r, 0 < r < R, the convergence is uniform on the closed subdisk D r (α) = {z : z α r}.
Taylor series representation - Corollary: Suppose that f is analytic in the domain G that contains the point α. Let z 0 be a nonremovable singular point of minimum distance to the point α. If z 0 α = R, then the Taylor series k=0 if z 1 α = S > R, the Taylor series k=0 to f(z 1 ). - Ex. Show that 1 (1 z) 2 f (k) (α) k! (z α) k converges to f(z) on all of D R (α) and = (n + 1)z n is valid for z D 1 (0). f (k) (α) k! (z α) k does not converge Solution: If 1 (1 z) 2, then we have f (n) (z) = (n+1)! (1 z) n+2 for z D 1 (0). Thus, f (n) (0) = (n + 1)!, and Taylor theorem gives f(z) = 1 (1 z) = 2 f (n) (0) z n = n! (n + 1)! z n = n! (n + 1)z n and since f is analytic in D 1 (0), this series expansion is valid for all z D 1 (0).
Taylor series representation - Ex. Show that, for z D 1 (0), 1 1 z = 2 Solution: For z D 1 (0), z 2n and 1 1 z = 1 1 + z = ( 1) n z 2n (4) 2 z n (5) If we let z 2 take the role of z in (5), we get the desired result. Letting z 2 take 1 the role of z in (5), we get the second result. Note that for, we have singular 1+z values at ±i. 2 - Theorem (Uniqueness of power series): Suppose that in some disk D r (α) we have f(z) = a n (z α) n = b n (z α) n Then, a n = b n, for n = 0, 1, 2,....
Taylor series representation - Theorem: Let f and g have the power series representations f(z) = a n (z α) n, for z D r1 (α); g(z) = b n (z α) n, for z D r1 (α) If r = min{r 1, r 2 } and β is any complex constant, then βf(z) = f(z) + g(z) = f(z)g(z) = βa n (z α) n, for z D r1 (α) (6) β(a n + b n )(z α) n, for z D r1 (α) (7) βc n (z α) n, for z D r1 (α) (8) where c n = n k=0 a k b n k. Note (8) is known as Cauchy product of series. - Ex. Use the Cauchy product of series to show 1 = (1 z) (n + 1)z n, for 2 z D 1 (0).
Laurent series representation - Suppose that f(z) is not analytic in D R (α) but is analytic in the punctured disk D R(α) = {0 < z α < R}. For example, the function f(z) = 1 z 3 e z is not analytic when z = 0 but is analytic for z > 0. Clearly the function does not have a Maclaurin series representation. If we use the Maclaurin series for g(z) = e z, however, and formally divide each term in that series by z 3, we obtain the representation f(z) = 1 z 3ez = 1 z 3 + 1 z 2 + 1 2!z + 1 3! + which is valid for all z such that z > 0 ( generalize Taylor series to an annulus). - Def. (Laurent series): Let c n be a complex number for n = 0, ±1, ±2, ±3,.... The doubly infinite series n= c n (z α) n defined by n= c n (z α) n = is called a Laurent series. c n (z α) n + c 0 + n=1 c n (z α) n (9) n=1
Laurent series representation - Def. (Annulus): Given 0 r < R, the annulus A(α, r, R) = {z : r < z α < R}, and the closed annulus A(α, r, R) = {z : r z α R}. - Theorem: Suppose that the Laurent series n= c n (z α) n converges on the annulus A(α, r, R). Then, the series converges uniformly on any closed subannulus A(α, s, t), where r < s < t < R. - Theorem (Laurent s theorem): Suppose that 0 r < R and that f is analytic in the annulus A = A(α, r, R). If ρ is any number such that r < ρ < R, then for all z 0 A, f has the Laurent series representation f(z 0 ) = n= c n (z 0 α) n = c n (z 0 α) n + + n=1 c n (z 0 α) n (10) where for n = 0, 1, 2,..., the coefficients c n and c n are given by c n = 1 f(z) 2πi (z α) n+1dz and c n = 1 f(z) 2πi (z α) n+1dz (11) C + ρ (α) C + ρ (α) Moreover, the convergence in (10) is uniform on any closed subannulus A(α, s, t), where r < s < t < R.
Laurent series representation - Ex. Find three different Laurent series representations for the function f(z) = 3 2+z z 2 involving powers of z. Solution: The function f has singularities at z = 1, 2, and is analytic in the disk D : z < 1, in the annulus A : 1 < z < 2, and in the region R : z > 2. Three Laurent series for f in D, A, and R. From f(z) = From the above fractions, 3 2 + z z = 1 2 1 + z + 1 2 1 1 z 2 1 1 + z = ( 1) n z n (valid for z < 1)) (12) 1 1 + z = n=1 1 1/2 1 z/2 = ( 1) n+1 z n (valid for z > 1)) (13) z n (valid for z < 2)) (14) 2n+1
Laurent series representation 1 1/2 1 z/2 = n=1 2 n+1 z n (valid for z > 1)) (15) Thus, (12) and (14) are both valid in D, f(z) = [( 1) n + 1 ]z n for z < 1. 2 n+1 In the annulus A, (13) and (14) are valid, hence f(z) = ( 1) n+1 n=1 + z n z n 2 for 1 < z < 2. Finally, in the region R, (13) and (15) are valid; so f(z) n+1 = ( 1) n+1 2 n 1 n=1 for z > 2. z n
Singularities, Zeros, and Poles - The point α is called a singular point, or singularity, of the complex function f if f is not analytic at the point α, but every neighborhood D R (α) of α contains at least one point at which f is analytic. - Ex. The function f(z) = 1 is not analytic at α = 1 but is analytic for all 1 z other values of z. Thus, the point α = 1 is a singular point of f. - Ex. Consider the function g(z) = Logz. g is analytic for all z except at the origin and at the points on the negative real axis. Thus, the origin and each point on the negative real axis are singularities of g. - The point α is called an isolated singularity of a complex function f if f is not analytic at α but there exists a real number R > 0 such that f is analytic everywhere in the punctured disk DR(α). (Functions with isolated singularities have a Laurent series because the punctured disk DR(α) is the same as the annulus A(α, 0, R). - Def. (Classification of singularities) Let f have an isolated singularity at α with Laurent series f(z) = c n (z α) n (valid for all z A(α, 0, R) n=
Singularities, Zeros, and Poles Then, we distinguish the following types of singularities at α. If c n = 0, for n = 1, 2, 3,..., then f has a removable singularity at α. If k is a positive integer such that c k 0, but c n = 0 for n < k, then f has a pole of order k at α. If c n 0 for infinitely many negative integers n, then f has an essential singularity at α - Ex. f(z) = sin z. It is undefined at z = 0 and has an isolated singularity at z z = 0 because the Laurent series for f is f(z) = sin z = 1 z3 (z + z5 z7 + ) = z z 3! 5! 7! 1 z2 + z4 z6 + (valid for z > 0). We can remove this singularity if we 3! 5! 7! define f(0) = 1, for then f will be analytic at 0 in accordance with (Theorem 4.17)
Singularities, Zeros, and Poles cos z 1 - Ex. g(z) =, which has an isolated singularity at the point 0 because the z 2 Laurent series for g is g(z) = / 1 z2 ( z2 + z4 z6 + ) = 2! 4! 6! 1 + z2 z4 + (valid 2 4! 6! for z > 0). If we define g(0) = 1, then g will be analytic for all z. 2 - If f has a pole of order k at α, the Laurent series for f is f(z) = n= k where c k 0. For example, f(z) = sin z z 3 c n (z α) n (valid for all z A(α, 0, R) = 1 z 2 1 3! + z2 5! z4 7! + has a pole of order 2 at 0. If infinitely many negative powers of (z α) occur in the Laurent series, then f has an essential singularity at α. For example, f(z) = z 2 sin 1 z = z 1 3! z 1 + 1 5! z 3 1 7! z 5 + has an essential singularity at the origin.
Singularities, Zeros, and Poles - Def. (zero of order k). A function f analytic in D R (α) has a zero of order k at the point α iff f (n) (α) = 0, for n = 0, 1,..., k 1, but f (k) (α) 0 A zero of order 1 is sometimes called a simple zero. - Theorem. A function f analytic in D R (α) has a zero of order k at the point α iff its Taylor series given by f(z) = c n (z α) n has c 0 = c 1 = = c k 1 = 0, but c k 0 - Ex. The function f(z) = z sin z 2 = z 3 z7 + z11 z15 + has a zero of order 3! 5! 7! 3 at z = 0. - Theorem: Suppose that the function f is analytic in D R (α). Then f has a zero of order k at the point α iff f can be expressed in the form where g is analytic at the point α and g(α) 0. f(z) = (z α) k g(z) (16)
Singularities, Zeros, and Poles - Theorem: A function f analytic in the punctured disk D R(α) has a pole of order k at the point α iff f can be expressed in the form f(z) = where h is analytic at the point α and h(α) 0. h(z) (z α) k (17)
Applications of Taylor and Laurent series - Theorem: Suppose that f is analytic in a domain D containing α and that f(α) = 0. If f is not identically zero in D, then there exists a punctured disk D R(α) in which f has no zeros. - Corollary: Suppose that f is analytic in the domain D and that α D. If there exists a sequence of points {z n } in D such that z n α, and f(z n ) = 0, then f(z) = 0 for all z D. - Corollary: Suppose that f and g are analytic in the domain D and that α D. If there exists a sequence of points {z n } in D such that z n α, and f(z n ) = g(z n ) for all n, then f(z) = g(z) for all z D. - Corollary (L Hopital s rule): Suppose that f and g are analytic at α. If f(α) = 0 and g(α) = 0, but g (α) 0, then f(z) lim z α g(z) = f (α) g (α)
Applications of Taylor and Laurent series - Theorem (Division of power series): Suppose that f and g are analytic at α with the power series representations f(z) = a n (z α) n and g(z) = b n (z α) n, z D R (α) If g(α) 0, then the quotient f g has the power series representation f(z) g(z) = c n (z α) n where the coefficients satisfy a n = b 0 c n + + b n 1 c 1 + b n c 0. - Theorem (Riemann): Suppose that f is analytic in D r(α). If f is bounded in D r(α), then either f is analytic at α or f has a removable singularities at α. - Theorem: Suppose that f is analytic in D r(α). If f has a pole or order k at α iff lim z α f(z) =. - Theorem: The function f has an essential singularity at α iff lim z α f(z) = does not exist.
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