Ch. 15 Power Series, Taylor Series 서울대학교 조선해양공학과 서유택 2017.12 ※ 본 강의 자료는 이규열, 장범선, 노명일 교수님께서 만드신 자료를 바탕으로 일부 편집한 것입니다. Seoul National 1 Univ. 15.1 Sequences (수열), Series (급수), Convergence Tests (수렴판정) Sequences: Obtained by assigning to each positive integer n a number zn z . Term: zn z1, z 2, or z 1, z 2 , or briefly zn N . Real sequence (실수열): Sequence whose terms are real Convergence . Convergent sequence (수렴수열): Sequence that has a limit c limznn c or simply z c n . For every ε > 0, we can find N such that Convergent complex sequence |zn c | for all n N → all terms zn with n > N lie in the open disk of radius ε and center c. Divergent sequence (발산수열): Sequence that does not converge. Seoul National 2 Univ. 15.1 Sequences, Series, Convergence Tests Convergence . Convergent sequence: Sequence that has a limit c Ex. 1 Convergent and Divergent Sequences iin 11 Sequence i , , , , is convergent with limit 0. n 2 3 4 limznn c or simply z c n Sequence i n i , 1, i, 1, is divergent. n Sequence {zn} with zn = (1 + i ) is divergent. Seoul National 3 Univ. 15.1 Sequences, Series, Convergence Tests Theorem 1 Sequences of the Real and the Imaginary Parts . A sequence z1, z2, z3, … of complex numbers zn = xn + iyn converges to c = a + ib . if and only if the sequence of the real parts x1, x2, … converges to a . and the sequence of the imaginary parts y1, y2, … converges to b. Ex. 2 Sequences of the Real and the Imaginary Parts 14 Sequence {z } with zn x n iy n 12 2 i converges to c = 1+2i. n nn 1 4 xn 1 has the limit 1 = Re c and y n 2 has the limit 2 = Im c. n2 n Seoul National 4 Univ. 15.1 Sequences, Series, Convergence Tests Series (급수): zm z12 z m1 . Nth partial sum: snn z12 z z . Term of the series: zz12, , . Convergent series (수렴급수): Series whose sequence of partial sums converges limsnm s Then we write s= z z12 z n m1 . Sum or Value: s . Divergent series (발산급수): Series that is not convergent . Remainder: Rn z n1 z n 2 z n 3 Theorem 2 Real and the Imaginary Parts z A series m with zm = xm + iym converges and has the sum s = u + iv m1 if and only if x1 + x2 + … converges and has the sum u and y1 + y2 + … converges and has the sum v. Seoul National 5 Univ. 15.1 Sequences, Series, Convergence Tests Tests for Convergence and Divergence of Series Theorem 3 Divergence If a series z + z +… converges, then lim z m 0 . 1 2 m Hence if this does not hold, the series diverges. Proof) If a series z1 + z2 +… converges, with the sum s, zm s m s m11 lim z m s m s m s s 0 m . zm → 0 is necessary for convergence of series but not sufficient. 111 Ex) The harmonic series 1 , which satisfies this condition but 234 diverges. The practical difficulty in proving convergence is that, in most cases, the sum of a series is unknown. Cauchy overcame this by showing that a series converges if and only if its partial sums eventually get close to each other. Seoul National 6 Univ. 15.1 Sequences, Series, Convergence Tests Ex) The harmonic series diverges Proof) 1 1 1 1 1 1 1 1 1 S 1 2 3 4 5 6 7 8 9 16 1 1 1 1 1 1 1 1 2 4 4 8 8 8 8 111 1 diverge! 222 1111 1 1 Ex) The harmonic series 1 1 converges 2234 3 4 1 1 1 1 1 1 1 1 S 1 2 3 4 5 6 7 8 9 1 1 1 1 1 1 1 1 11 2 2 4 4 6 6 8 8 converge! Seoul National 7 Univ. 15.1 Sequences, Series, Convergence Tests Theorem 4 Cauchy’s Convergence Principle for Series A series z1 + z2 + … is convergent if and only if for every given ε > 0 (no matter how small) we can find an N (which depends on ε in general) such that zn12 z n z n p for every n N and p 1, 2, Absolute Convergence (절대 수렴) . Absolute convergent: Series1 of 1 the 1 absolute value of the terms 1 2 3 4 zm z12 z is convergent. m1 . Conditionally convergent (조건 수렴): z1+z2+… converges but |z1|+|z2|+… diverges. Ex. 3 The series converges but zz 12 diverges, then the series z1 + z2 + … is called conditionally convergent. Seoul National 8 Univ. 15.1 Sequences, Series, Convergence Tests Theorem 5 Comparison Test (비교 판정법) If a series z1 + z2 + … is given and we can find a convergent series b1 + b2 + … with nonnegative real terms such that |z1| < b1, |z2| < b2,…, then the given series converges, even absolutely. Proof) by Cauchy’s principle, bn12 b n b n p for every n N and p 1, 2, . From this and |z1| < b1, |z2| < b2, …, |zn11 | | z n p | b n b n p . |z1| + |z2| + … converges, so that z1 + z2 + … is absolutely convergent. Seoul National 9 Univ. 15.1 Sequences, Series, Convergence Tests Theorem 6 Geometric Series (기하 급수) m 2 The geometric series q 1 q q . m0 1 converges with the sum if q 1 and diverges if q 1 . 1 q 2 n Proof) sn 1 q q q 21n qsn q q q n1 sn qs n (1 q ) s n 1 q 11 qqnn11 qn1 s since qn1, 0 n 1q 1 q 1 q 1 q 1 s n 1 q Seoul National 10 Univ. 15.1 Sequences, Series, Convergence Tests Theorem 7 Ratio Test (비 판정법) If a series z1 + z2 + … with zn n 0 1, 2, has the property that for every n greater than some N, z n1 q 1 n N zn (where q < 1 is fixed), this series converges absolutely. If for every n > N z , the series diverges. n1 1 nN zn z Proof) i) n1 1 znn1 z z 1 z 2 diverges zn 2 ii) zn1 zq nfor nNz N 2 zq N 1 z N 3 zqzq N 2 N 1 p1 zN p z N 1 q 1 z z z z(1 q q2 ) z NNNNN1 2 3 1 1 1 q Absolutely convergence follows from Theorem 5 Comparison Test Seoul National 11 Univ. 15.1 Sequences, Series, Convergence Tests Theorem 8 Ratio Test zn0 1, 2, z If a series z + z + … with n is such that lim n 1 L , then: 1 2 n zn a. If L < 1, the series converges absolutely. b. If L > 1, the series diverges. c. If L = 1, the series may converge or diverge, so that the test fails and permits no conclusion. z n1 1 nN Theorem 7 Ratio Test Proof) (a) kn z n1 / z n z ,n let L 1 b 1 z 1 n1 q 1 n N kbn 1 saykn q 1 b 1 z 2 n for n N ⇒ the series converges Theorem 7 Ratio Test z1 + z2 + … converges (b) kn z n1 / z n , let L 1 c 1 1 kcn 1 saykcn 1 2 1 ⇒ the series diverge Theorem 7 Ratio Test z1 + z2 +… diverges Seoul National 12 Univ. 15.1 Sequences, Series, Convergence Tests Theorem 8 Ratio Test zn0 1, 2, z If a series z + z + … with n is such that lim n 1 L , then: 1 2 n zn a. If L < 1, the series converges absolutely. b. If L > 1, the series diverges. c. If L = 1, the series may converge or diverge, so that the test fails and permits no conclusion. Proof) (c) harmonic series zznn11n 111 ,as n 1 , 1, L 1 diverge! znn n1 z234 2 1 1 1 zznn11n another series 1 2 ,as n , 1, L 1 4 9 16 znn( n 1) z 1 1 1n dx 1 s 1 1 2 n 49 n221 x n converge! Seoul National 13 Univ. 15.1 Sequences, Series, Convergence Tests Theorem 8 Ratio Test zn0 1, 2, z If a series z + z + … with n is such that lim n 1 L , then: 1 2 n zn a. If L < 1, the series converges absolutely. b. If L > 1, the series diverges. c. If L = 1, the series may converge or diverge, so that the test fails and permits no conclusion. Ex. 4 Ratio Test Is the following series convergent or divergent? n 100 75i 1 2 1 100 75ii 100 75 n0 n! 2! Sol) The series is convergent, since 100 75in1 z n1! 100 75i 125 n1 0 L z100 75in n11 n n n! Seoul National 14 Univ. 15.1 Sequences, Series, Convergence Tests Theorem 9 Root Test (근 판정법) If a series z1 + z2 + … is such that for every n greater than some N n zn q 1 n N (where q < 1 is fixed), this series converges absolutely. n If for infinitely many n zn 1, the series diverges. n n Proof) (a) zn q1 zn q 1 for all n N 1 z z z (1 q q2 ) 1 2 3 1 q Absolutely convergence follows from Theorem 5 Comparison Test (b) zn 1 z1 + z2 + … diverges. This diverges from Theorem 3 Divergence. Seoul National 15 Univ. 15.1 Sequences, Series, Convergence Tests Caution! harmonic series n n zn 1/ n 1 but diverge! because q < 1 is not fixed.