111_quiz11_spr08_key

# 111_quiz11_spr08_key - lim x → 1-cos x 1 x-e x Solution...

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APMA 111 Quiz 11 April 25, 2008 STUDENT NAME: On my honor as a student, I have neither given nor received aid on this quiz. SIGNATURE: 1. [5 pts] Compute the Maclaurin series for the following function, and determine the interval of convergence: g ( x ) = sinh x . Solution: sinh x = 1 2 ( e x - e - x ) = 1 2 X n =0 x n n ! - X n =0 ( - x ) n n ! ! = X n =0 1 2 ± x n n ! - ( - 1) n x n n ! ² = X n =0 1 2 ( 1 + ( - 1) n +1 ) x n n ! = X n =0 x 2 n +1 (2 n + 1)! 2. [5 pts] Use a series to approximate the deﬁnite inegral to within the indicated accuracy: Z 1 / 2 0 x 2 e - x 2 dx, | error | < 0 . 001 . Solution: e x = X n =0 x n n ! e - x 2 = X n =0 ( - x 2 ) n n ! = X n =0 ( - 1) n x 2 n n ! x 2 e - x 2 = X n =0 ( - 1) n x 2 n +2 n ! Z 1 / 2 0 x 2 e - x 2 dx = Z 1 / 2 0 X n =0 ( - 1) n x 2 n +2 n ! ! dx = X n =0 ( - 1) n x 2 n +3 n !(2 n + 3) ³ ³ ³ ³ ³ 1 / 2 0 = X n =0 ( - 1) n n !2 2 n +3 (2 n + 3)

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Quiz 11, Page 2 of 2 April 25, 2008 Since this is an alternating series, | error | ≤ | a n +1 | = 1 ( n + 1)!2 2 n +5 (2 n + 5) . Setting | a n +1 | < 0 . 001 = 1 / 100 implies that ( n + 1)!2 2 n +5 (2 n + 5) > 1000 . Notice that for n = 1 we get | a n +1 | = | a 2 | = 2!2 7 · 7 = 2 8 · 7 = 256 · 7 > 1000 , which implies that we just have to add the ﬁrst 2 terms (plug in n = 0 and n = 1): 1 2 3 · 3 - 1 2 5 · 5 = 1 24 - 1 160 = 20 - 3 480 = 17 480 . 3. [5 pts] Use series to evaluate the limit:
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Unformatted text preview: lim x → 1-cos x 1 + x-e x . Solution: cos x = ∞ X n =0 (-1) n x 2 n (2 n )! = 1-x 2 2! + x 4 4!- ··· ⇒ 1-cos x = 1-± 1-x 2 2! + x 4 4!- ··· ² = x 2 2!-x 4 4! + x 6 6!- ··· e x = ∞ X n =0 x n n ! = 1 + x + x 2 2! + ··· ⇒ 1 + x-e x = 1 + x-± 1 + x + x 2 2! + x 3 3! + ··· ² =-± x 2 2! + x 3 3! + x 4 4! + ··· ² ⇒ 1-cos x 1 + x-e x = x 2 2!-x 4 4! + x 6 6!- ···-( x 2 2! + x 3 3! + x 4 4! + ··· ) = ± ± x 2 ³ 1 2!-x 2 4! + x 4 6!- ··· ´-± ± x 2 ( 1 2! + x 3! + x 2 4! + ··· ) ⇒ lim x → 1-cos x 1 + x-e x = lim x → 1 2!-x 2 4! + x 4 6!- ···-( 1 2! + x 3! + x 2 4! + ··· ) = 1 2!-0 + 0- ···-( 1 2! + 0 + 0 + ··· ) ⇒ lim x → 1-cos x 1 + x-e x =-1...
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## This note was uploaded on 09/21/2009 for the course MATH 1234 taught by Professor Egcdd during the Spring '09 term at Aarhus Universitet.

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111_quiz11_spr08_key - lim x → 1-cos x 1 x-e x Solution...

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