HW11 - -1 n ´ 1 z µ n After multiplying both side by 1/z...

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MATH 185 SOLUTIONS OF HW XI (1) (54. 1)(10pts) We already know the Maclaurin series of cosh z . Thus, we have z cosh( z 2 ) = z X n =0 ( z 2 ) 2 n (2 n )! = X n =0 z 4 n +1 (2 n )! (2) (54. 2)(10pts) (a) Since f ( n ) ( x ) = e x for all n , we have f ( n ) (1) = e , so the Taylor series of e x at x = 1 is e z = X n =0 f ( n ) (1)( z - 1) n n ! = e X n =0 ( z - 1) n n ! (b) Since e z = e z - 1 e = e t e where t = z - 1, we have e z = e X n =0 t n n ! = e X n =0 ( z - 1) n n ! (3) (54. 11)(20pts) (a) Just divide both side of the Maclaurin series of e z by z 2 . (b) Similarly as above(54. 1), Just divide both side of the Maclaurin series of sin z 2 by z 4 . (We can obtain the Maclaurin series of sin z 2 by doing similarly as 54. 1.) (4) (54. 13)(10pts) Since 0 < | z | < 4, we have 0 < | z/ 4 | < 1. Hence we have 1 4 z - z 2 = 1 (4 z )(1 - z/ 4) = 1 4 z ± X n =0 ² z 4 ³ n ! = 1 4 z + X n =0 z n 4 n +2 (5) (56. 1)(10pts) Just multiply both side of the Laurent series of sin(1 /z 2 ) by z 2 . We can find the Laurent series of sin(1 /z 2 ) by substituting z by 1 /z 2 to the Maclaurin series of sin x . (6) (56. 3)(10pts) If 1 < | z | < , then | z | < 1, so we have the Laurent series for 1 (1 + 1 /z ) = X n =0
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Unformatted text preview: (-1) n ´ 1 z µ n After multiplying both side by 1 /z , we have the desired expression. 1 2 MATH 185 SOLUTIONS OF HW XI (7) (56. 5)(20pts) (a) If | z | < 1, then 1 1-z = ∞ X n =0 z n Therefore the Maclaurin series of f ( z ) is z + 1 z-1 =-z ∞ X n =0 z n-∞ X n =0 z n =-1-2 ∞ X n =1 z n (b) If | z | > 1, then | 1 /z | < 1 , so we have 1 + 1 /z 1-1 /z = ∞ X n =0 z-n + 1 z ∞ X n =0 z-n = 1 + 2 ∞ X n =1 z-n (8) (56. 6)(10pts) Since 0 < | z-1 | < 2, 0 < | 1 / ( z-1) | < 1, so we have z z-3 =-z/ 2 1-( z-1) / 2 =-z-1 2 ∞ X n =0 ± z-1 2 ² n-1 2 ∞ X n =0 ± z-1 2 ² n Therefore, z ( z-1)( z-3) =-∞ X n =0 ( z-1) n 2 n +1-1 2 ∞ X n =0 ( z-1) n-1 2 n +1 =-3 ∞ X n =0 ( z-1) n 2 n +2-1 2( z-1) ....
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HW11 - -1 n ´ 1 z µ n After multiplying both side by 1/z...

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