9. Series representation for analytic functions

Transcription

1 9. Series representation for analytic functions 9.. Power series. Definition: A power series is the formal expression S(z) := c n (z a) n, a, c i, i =,,, fixed, z C. () The n.th partial sum S n (z) is the sum of the first n + terms. We say that the power series does converge for a fixed z C, if the sequence {S n (z)} converges, as n. If S n (z) diverges, then we say that the power series diverges at the point z. ℵ We will write S(z) < and S(z) =, respectively. We draw the reader s attention to the fact that the partial sums S n are polynomials of degree n. A convergent power series is the limes of polynomial sequences, e.g. S(z) := lim k k c n (z a) n, (2) so that all results from Chpr. 2 are applicable. Definition: The power series S(z) := c n (z a) n is absolutely convergent, if the series with nonnegative terms c n z a n < and uniformly convergent, if S n S uniformly in the metric of Chebyshev on compact sets. ℵ We recall some well known facts.

2 Theorem 9.. Suppose that S(z) := c n(z a) n is absolutely convergent at z. Then the power series converges in the regular sense at the same point. Proof: Fix an arbitrary ε. Taking into account the absolute convergence, we may write k+m k c n z a n c n z a n ε; this inequality is valid for every k great enough (say k k ) and for every m N. It turns out that S k+m (z ) S k (z ) ε, k k, m N. Our statement follows immediately from Cauchy s fundamental theorem Theorem 9.2. Let S(z) := c n(z a) n be convergent at z. Then c n (z a) n, n. Proof: The statement follows from the fact that S k (z ) S k (z ) = c k (z a) k, k. Theorem 9.3., Abel s theorem Suppose thats(z) := c n(z a) n converges at z. Then S is absolutely convergent at every z such that z a < z a. Proof: Take z with z a < z a. Fix ε >. By the previous theorem, for all n n. We have further, c n z a n = n c n (z a) n ε z a n c n z a z n a n + z a n c n z a z n a n n z a n c n z a z n a n + Cauchy s fundamental theorem: the infinite sequence {a n } converges iff for every ε > there exists a number k great enough such that a k+m a k < ε whenever k k and m N. 2

3 n z a n c n z a z n a n + Using common notations, we may write c n z a n n + ε z a n z a n. z a n z a. n To complete the proof, we need to remember that z a < z a. Corollary 9.4. : Suppose that S(z) := c n(z a) n diverges at z. Then it diverges at every point z with z a > z a. Naturally we come to the definition of a radius of convergence: Definition: Given S(z) := c n(z a) n, we set R := sup{ρ, S(z) < for z a < ρ}. The number R is called radius of convergence of the power series. ℵ Regarding Abel s theorem, we conclude that the power series converges in the disk D a (R) and diverges outside. Theorem 9.5., H Adamard s formula: R = lim sup n c n /n. Proof: If R = then we are done. We assume that R is positive. Fix ρ < R. We will show that the power series is uniformly convergent on D a (ρ). Select a positive number ε in such a way that ρ + ε < R. Viewing the definition of R, we get for every n > n (n large enough) n the estimation Consequently, c n z a n = n=n c n n=n c n z a n + (R ε) n. c n z a n + 3 n=n ( n=n c n z a n < ρ R ε )n,

4 or c n z a n ρ ( R ε )n. The right-hand side series is a convergent geometric progression, (recall the choice of ε. ) Therefore, S(z) is absolutely convergent, and thus, convergent in the regular sense. Remark: There is no statement about the behavior on the circle C a (R). Theorem 9.6. Suppose that the power series S(z) := c n(z a) n is of positive radius of convergence R. Then it converges uniformly on compact subsets of the disk D a (R) and absolutely at every point z D a (R). Proof: Fix ρ < R. By the previous theorem, c n ρ n <. Rearranging the difference S n+m S n yields n+m S n+m S n = Da(ρ) k=n+ c k (z a) k n+m k=n+ c k ρ k. Applying again Cauchy s fundamental theorem, for all n large enough we get S n+m S n, which means a uniform convergence on D Da(ρ) a(ρ) Taylor s theorem and consequences Theorem 9.7. (Taylor s theorem) Let f A(D a (ρ)), ρ >, a C. Then f can be represented as a Taylor series 2 c n (z a) n, which is convergent uniformly inside D a (ρ) and c n = 2πi dζ. (ζ a) n+ Proof: In view of the integral formula, f(z) = 2πi ζ z dζ, z D a(ρ). 2 it if a =, then we speak about MacLaurin series. 4

5 We remember that z a < ζ a. Consequently, 2πi (ζ a) (z a) dζ = 2πi ζ a Using known estimates, we obtain f(z) = (z a) n 2πi Remark: ( z a ζ a )n. ζ a) n+ dζ := c n (z a) n. (3) c n = f (n) (a). (4) n! Example: Write down the Taylor series of Logz(:= ln z + iargz,.) around z =. Solution; Since we get d j Logz dz j = ( ) j+ (j )!z j, j =, 2, Logz = + (z ) (z ) 2 /2! + 2!(z ) 3 /3! 3!(z ) 4 /4! + = = ( ) j+ (z ) j /j. j= The series converges uniformly on D (r) for every r <. Theorem 9.8. Suppose that f is analytic at the point z = a, f(z) = c n (z a) n, 3 Then f (z) = nc n (z a) n. n= Proof: Indeed, from Chpt. 8 we know that f (z) is also analytic at z = a. From THeorem 9.7, we get f (f (z)) (n) (a) (z) = (z a) n. n! 3 analytic in some domain D a (r), r >. 5

6 Recalling that (f (z)) (n) (a) = (f(z)) (n+) (a), we obtain the required statement. The proof of the following theorems is left to the reader. Theorem 9.9. Let the functions f(z) and g(z) be analytic at z =, f(z) = f n (z a) n, g(z) = g n (z a) n. Denote by R(f), R(g) the radii of convergence of both Taylor series. Then f ± g and fg are analytic at z = a and R(f ± g) min(r(f), R(g)), R(fg) min(r(f), R(g)). and (f ± g)(z) = (f n ± g n )(z a) n, f(z)g(z) = c n (z a) n with 9.3. The point of infinity. c n = n f k g n k. k= Definition: The function f(z), defined at infinity, is said to be analytic at z =, if the function g(ζ) := f( ζ ) is analytic at ζ =. ℵ The checking of the validity of the following theorem is left to the reader: Theorem 9.. Suppose that f is analytic at infinity. Then it is expandable into Taylor series c n f(z) = z. n The series converges at every point z, z > lim sup c n /n and is uniformly convergent in the exterior of every circle D (R) with R > lim sup c n /n. 6

7 Exercises:. Find the Taylor series of at z =. 2. Find the Taylor series of at z =. 3. Find the Taylor series of f(z) := z 2 cos 3z f(z) = z 2 f(z) = z(z 2) at z =. 7