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# SSM-Ch15 - Chap 15 Power Series Taylor Series These are...

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Chap . 15 Power Series , Taylor Series These are analogs of power series and Taylor series (which are power series, too) known to you from calculus. Perhaps you take a look at your calculus book to refresh your memory before you begin to study this chapter. These series are important by themselves, in fact much more important than real power series in calculus. Moreover, this chapter is also needed for the understanding of the next chapter on Laurent series (which have no useful analog in calculus) and a complex integration method based on these series. Sec . 15 . 1 Sequences , Series , Convergence Tests This is similar to sequences and series in calculus. Most important from a practical point of view is the ratio test in Theorem 7. Problem Set 15 . 1 . Page 672 1 . Sequence . The terms z n u003d ue0a2 u2212 1 ue0a3 n u002b i /2 n , n u003d 0,1, u22ef , are i u002b i , u2212 1 u002b 1 2 i , 1 u002b 1 4 i , u22ef . For the absolute value you obtain | z n | 2 u003d 1 u002b 1 2 2 n u2264 1 u002b 1 u003d 2, which shows boundedness. Since Im z u003d 1/2 n u2192 0 as n u2192 u221e and Re z u003d u00b1 1, you see that ue0a6 z n ue0a7 has the two limit points 1 and u2212 1. Conclude from this that ue0a6 z n ue0a7 diverges. 9 . Sequence . Calculate | z n | u003d |0.9 u002b 0.1 i | 2 n u003d ue0a2 |0.9 u002b 0.1 i | 2 ue0a3 n u003d ue0a2 0.81 u002b 0.01 ue0a3 n u003d 0.82 n u2192 0 as n u2192 0. Conclude that the sequence converges absolutely to 0. 15 . Boundedness . Let ue0a6 z n ue0a7 be bounded, say, | z n | u003c K for some K and all n . Set z n u003d x n u002b iy n as in the text. Then boundedness of the sequences ue0a6 x n ue0a7 and ue0a6 y n ue0a7 can be seen from | x n | u2264 | z n | u003c K , | y n | u2264 | z n | u003c K . Here it was used that for z u003d x u002b iy you always have x 2 u2264 x 2 u002b y 2 u003d | z | 2 hence | x | u2264 | z | and similarly for the imaginary part y , namely, | y | u2264 | z |. Conversely, let ue0a6 x n ue0a7 and ue0a6 y n ue0a7 be bounded, say, | x n | u003c K , | y n | u003c K . Then x n 2 u003c K 2 , y n 2 u003c K 2 , so that | z n | 2 u003d x n 2 u002b y n 2 u003c 2 K 2 By taking square roots this gives | z n | u003c k ue0a2 k u003d K 2 ue0a3 . Hence ue0a6 z n ue0a7 is bounded.

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144 Complex Analysis Part D 17 . Ratio test . For the ratio test in Theorem 7 you need z n u003d ue0a2 u2212 1 ue0a3 n ue0a2 1 u002b 2 i ue0a3 2 n u002b 1 ue0a2 2 n u002b 1 ue0a3 ! . Calculate the test ratio z n u002b 1 z n u003d ue0a2 u2212 1 ue0a3 n u002b 1 ue0a2 1 u002b 2 i ue0a3 2 ue0a2 n u002b 1 ue0a3 u002b 1 / ue0a2 2 ue0a2 n u002b 1 ue0a3 u002b 1 ue0a3 ! ue0a2 u2212 1 ue0a3 n ue0a2 1 u002b 2 i ue0a3 2 n u002b 1 / ue0a2 2 n u002b 1 ue0a3 !
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