ch15

Advanced Engineering Mathematics

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CHAPTER 15 Power Series, Taylor Series Power series and, in particular, Taylor series, play a much more fundamental role in complex analysis than they do in calculus. The student may do well to review what has been presented about power series in calculus but should become aware that many new ideas appear in complex, mainly owing to the use of complex integration. SECTION 15.1. Sequences, Series, Convergence Tests, page 664 Purpose. The beginnings on sequences and series in complex is similar to that in calculus (differences between real and complex appear only later). Hence this section can almost be regarded as a review from calculus plus a presentation of convergence tests for later use. Main Content, Important Concepts Sequences, series, convergence, divergence Comparison test (Theorem 5) Ratio test (Theorem 8) Root test (Theorem 10) SOLUTIONS TO PROBLEM SET 15.1, page 672 2. Bounded, divergent, 8 limit points (the values of Ï 8 1 w ) 4. Unbounded, divergent 6. Bounded, convergent to 0 (the terms of the Maclaurin series of e 3 1 4 i ) 8. Divergent. All terms have absolute value 1. 10. Convergent to 0 12. Let < 1 and < 2 be two limits, d 5 u < 1 2 < 2 u and e 5 d /3. Then there is an N ( ) such that u z n 2 < 1 u , , u z n 2 < 2 u , for all n . N . This is impossible because the disks u z 2 < 1 u , and u z 2 < 2 u , are disjoint. 14. The sequences are bounded, u z n u , K , u z n * u , K . Since they converge, for an . 0 there is an N such that u z n 2 < u , /(3 K ), u z n * 2 < * u , /(3 u < u ) ( < Þ 0; the case < 5 0 is rather trivial), hence u z n z n * 2 < * u 5 u ( z n 2 < ) z n * 1 ( z n * 2 < *) < u % u z n 2 < uu z n * u 1 u z n * 2 < * < u , /3 1 /3 , ( n . N ). 16. Convergent. Sum e 10 2 15 i 18. Convergent because , and o ` n 5 1 converges. 20. Divergent because 1/ln n . 1/ n and the harmonic series diverges. 22. By the ratio test it converges because after simplification jj 5 * . Ï 2 w ± 27 ( n 1 1) 3 u 1 1 i u ±±± (3 n 1 3)(3 n 1 2)(3 n 1 1) z n 1 1 ± z n 1 ± n 2 1 ± n 2 1 ± u n 2 2 2 i u 260 im15.qxd 9/21/05 12:56 PM Page 260
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24. Convergent by the ratio test because jj * 30. Team Project. (a) By the generalized triangle inequality (6*), Sec. 13.2, we have u z n 1 1 1 ••• 1 z n 1 p u % u z n 1 1 u 1 u z n 1 2 u 1 1 u z n 1 p u . Since u z 1 u 1 u z 2 u 1 converges by assumption, the sum on the right becomes less than any given e . 0 for every n greater than a sufficiently large N and p 5 1, 2, , by Cauchy’s convergence principle. Hence the same is true for the left side, which proves convergence of z 1 1 z 2 1 by the same theorem. (c) The form of the estimate of R n suggests we use the fact that the ratio test is a comparison test based on the geometric series. This gives R n 5 z n 1 1 1 z n 1 2 1 5 z n 1 1 ( 1 111 ) , % q , 5 % q 2 , etc., u R n u % u z n 1 1 u (1 1 q 1 q 2 1 ) 5 . (d) For this series we obtain the test ratio j j 5 ! § 5 ! § , ; from this with q 5 1/2 we have u R n u %5 5 , 0.05. Hence n 5 5 (by computation), and s 51 i 5 0.96875 1 0.688542 i .
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ch15 - im15.qxd 9/21/05 12:56 PM Page 260 CHAPTER 15 Power...

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