# 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 u u u 1 U n U i /2 n , n u 0,1, ± , are i U i , u 1 U 1 2 i , 1 U 1 4 i , ± . For the absolute value you obtain | z n | 2 u 1 U 1 2 2 n U 1 U 1 u 2, which shows boundedness. Since Im z u 1/2 n ± 0 as n ± ± and Re z u ² 1, you see that ² z n ³ has the two limit points 1 and u 1. Conclude from this that ² z n ³ diverges. 9 . Sequence . Calculate | z n | u |0.9 U 0.1 i | 2 n u u |0.9 U 0.1 i | 2 U n u u 0.81 U 0.01 U n u 0.82 n ± 0 as n ± 0. Conclude that the sequence converges absolutely to 0. 15 . Boundedness . Let ² z n ³ be bounded, say, | z n | ³ K for some K and all n . Set z n u x n U iy n as in the text. Then boundedness of the sequences ² x n ³ and ² y n ³ can be seen from | x n | U | z n | ³ K , | y n | U | z n | ³ K . Here it was used that for z u x U iy you always have x 2 U x 2 U y 2 u | z | 2 hence | x | U | z | and similarly for the imaginary part y , namely, | y | U | z |. Conversely, let ² x n ³ and ² y n ³ be bounded, say, | x n | ³ K , | y n | ³ K . Then x n 2 ³ K 2 , y n 2 ³ K 2 , so that | z n | 2 u x n 2 U y n 2 ³ 2 K 2 By taking square roots this gives | z n | ³ k u k u K 2 U . Hence ² z n ³ 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 u u u 1 U n u 1 U 2 i U 2 n U 1 u 2 n U 1 U ! . Calculate the test ratio
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## This note was uploaded on 04/08/2012 for the course MT 423 taught by Professor M.lee during the Spring '12 term at National Taiwan University.

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

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