# ans8 - Math 185 Sample Answers to Problem Set#8 Page 188 5...

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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 0 ( 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 . (In the suggestion, the first “=” sign stems from writing the function in terms of z - i , and the second factors out the 1 - i so that the denominator is in the form 1 - something .)

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