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exam1 - Version 101 Exam01 gilbert(55485 This print-out...

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Version 101 – Exam01 – gilbert – (55485) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points To apply the ratio test to the infinite series summationdisplay n a n , the value λ = lim n → ∞ a n +1 a n has to be determined. Compute λ for the series summationdisplay n =1 4 n 3 n 4 + 8 . 1. λ = 4 correct 2. λ = 0 3. λ = 4 3 4. λ = 1 2 5. λ = 4 11 Explanation: By algebra, a n +1 a n = 4 n +1 4 n bracketleftbigg 3 n 4 + 8 3( n + 1) 4 + 8 bracketrightbigg . But 3 n 4 + 8 3( n + 1) 4 + 8 = 3 + 8 n 4 3 parenleftbigg n + 1 n parenrightbigg 4 + 8 n 4 . Since parenleftbigg n + 1 n parenrightbigg 4 = parenleftbigg 1 + 1 n parenrightbigg 4 1 as n → ∞ , we see that lim n → ∞ a n +1 a n = lim n → ∞ 4 parenleftBig 3 + 8 n 4 parenrightBig 3 parenleftBig n + 1 n parenrightBig 4 + 8 n 4 = 4 . Consequently, λ = 4 . 002 10.0points Compute the degree 2 Taylor polynomial for f centered at x = 4 when f ( x ) = x . 1. T 2 ( x ) = 4 + 1 8 ( x - 4) + 1 32 ( x - 4) 2 2. T 2 ( x ) = 2 + 1 4 ( x - 4) - 1 64 ( x - 4) 2 correct 3. T 2 ( x ) = 2 - 1 8 ( x - 4) + 1 64 ( x - 4) 2 4. T 2 ( x ) = 2 + 1 4 ( x - 4) - 1 32 ( x - 4) 2 5. T 2 ( x ) = 2 - 1 4 ( x - 4) + 1 64 ( x - 4) 2 6. T 2 ( x ) = 4 - 1 8 ( x - 4) - 1 32 ( x - 4) 2 Explanation: The degree 2 Taylor polynomial centered at x = 4 for a general f is given by T 2 ( x ) = f (4)+ f (4) ( x - 4)+ f ′′ (4) 2! ( x - 4) 2 . Now when f ( x ) = x , f ( x ) = 1 2 x , f ′′ ( x ) = - 1 4 x x , in which case, f (4) = 2 , f (4) = 1 4 ,
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Version 101 – Exam01 – gilbert – (55485) 2 while f ′′ (4) 2! = - 1 64 . Consequently, T 2 ( x ) = 2 + 1 4 ( x - 4) - 1 64 ( x - 4) 2 . 003 10.0points Which one of the following properties does the series summationdisplay k =1 ( - 1) k 1 3 1 + k 2 have? 1. divergent 2. conditionally convergent correct 3. absolutely convergent Explanation: The given series has the form of an alter- nating series summationdisplay k =1 ( - 1) k 1 a k , a k = 3 1 + k 2 . To check for absolute convergence we apply the Limit Comparison Test with a k = 3 1 + k 2 , b k = 1 k . For then lim k → ∞ a k b k = lim k → ∞ 3 k 1 + k 2 = 3 > 0 . Thus the series a k converges if and only if the series summationdisplay k =1 1 k converges. But, by the p -series test with p = 1, this last series diverges; in particular, the given series does not converge absolutely. To check if the given series converges con- ditionally, consider first the function f ( x ) = 3 1 + x 2 . Then, by the Chain Rule, f ( x ) = - 3 x (1 + x 2 ) 3 / 2 < 0 for all x > 0, while lim x → ∞ f ( x ) = 0 .
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