ans8 - Math 185. Sample Answers to Problem Set #8 Page 188...

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Unformatted text preview: Math 185. Sample Answers to Problem Set #8 Page 188 5. a . By (1), cos z = ∞ X n =0 i n + (- i ) n 2 · n ! z n = ∞ X n =0 i n (1 + (- 1) n ) 2 · n ! z n . If n is odd then 1 + (- 1) n = 0, so only the even terms remain, and we have cos z = ∞ X n =0 i 2 n (2 n )! z 2 n = ∞ X n =0 (- 1) n (2 n )! z 2 n . b . By successively taking derivatives, we have f ( z ) = cos z , f ( z ) =- sin z , f 00 ( z ) =- cos z , f 000 ( z ) = sin z , f (4) ( z ) = cos z = f ( z ), and the derivatives repeat every four. The odd-numbered derivatives are ± sin z , which vanish at zero, so f (2 n +1) (0) = 0 for all integers n ≥ 0. The even-numbered derivatives are ± cos z , alternating in sign, so f (2 n ) ( z ) = (- 1) n cos z and therefore f (2 n ) (0) = (- 1) n . Thus, by the theorem on Taylor series, cos z = ∞ X n =0 f (2 n ) (0) (2 n )! z 2 n + f (2 n +1) (0) (2 n + 1)! z 2 n +1 = ∞ X n =0 (- 1) n (2 n )! z 2 n . 7. Using the formula in the suggestion and the example on page 180, we have 1 1- z = 1 1- i · 1 1- z- i 1- i = 1 1- i ∞ X n =0 z- i 1- i n = ∞ X n =0 ( z- i ) n (1- i ) n +1 ....
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This note was uploaded on 09/01/2010 for the course MATH 185 taught by Professor Lim during the Fall '07 term at Berkeley.

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ans8 - Math 185. Sample Answers to Problem Set #8 Page 188...

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