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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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) ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online