calc 2 webwork 8.7.pdf - Amber Johnson Assignment Section...

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Amber Johnson Ionascu MAT 266 ONLINE A Spring 2018 Assignment Section 8.7 due 02/18/2018 at 11:59pm MST 1. (1 point) The function f ( x ) = sin ( 4 x ) has a Maclaurin series. Find the first 4 nonzero terms in the series, that is write down the Taylor polynomial with 4 nonzero terms. Solution: SOLUTION We use the known Maclaurin series for sin ( x ) , whose first four nonzero terms are sin ( x ) x - x 3 3! + x 5 5! - x 7 7! So the first four nonzero terms of f ( x ) = sin ( 4 x ) are sin ( 4 x ) 4 x - ( 4 x ) 3 3! + ( 4 x ) 5 5! - ( 4 x ) 7 7! Answer(s) submitted: (incorrect) Correct Answers: 4*x - (4**3)/6 * x**3 + (4**5)/120 * x**5 - (4**7)/5040 * x**7 2. (1 point) Find the Maclaurin series of the function f ( x ) = ( 2 x 2 ) e - 6 x ( f ( x ) = n = 0 c n x n ) c 1 = c 2 = c 3 = c 4 = c 5 = Solution: SOLUTION We use the known Maclaurin series for e x , whose first four nonzero terms are e x 1 + x + x 2 2! + x 3 3! So the first four nonzero terms of f ( x ) = e - 6 x are e - 6 x 1 - 6 x + ( - 6 x ) 2 2! + ( - 6 x ) 3 3! And so the first four nonzero terms of f ( x ) = 2 x 2 e - 6 x are 2 x 2 e - 6 x 2 x 2 1 - 6 x + 6 2 x 2 2! - 6 3 x 3 3! = 2 x 2 - 12 x 3 + 2 ( 6 2 ) 2 x 4 - 2 ( 6 3 ) 6 x 5 Equating this to f ( x ) = n = 0 c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + c 4 x 4 + c 5 x 5 + ··· , we get c 0 = 0 c 1 = 0 c 2 = 2 c 3 = - 12 c 4 = 2 ( 6 2 ) 2 = 36 c 5 = - 2 ( 6 3 ) 6 = - 72 for the coefficients up to the x 5 term. Answer(s) submitted: (incorrect) Correct Answers: 0 2 -12 36 -72 3. (1 point) Match the series with the right expression. (Use the Maclaurin series.) 1. n = 0 ( - 1 ) n ( 1 3 ) 2 n + 1 ( 2 n + 1 ) ! 2. n = 0 ( 1 3 ) n n ! 3. n = 0 ( - 1 ) n ( 1 3 ) 2 n + 1 2 n + 1 4. n = 0 ( - 1 ) n ( 1 3 ) 2 n ( 2 n ) ! A. cos ( 1 3 ) B. arctan ( 1 3 ) C. e 1 / 3 D. sin ( 1 3 ) 1
Solution: SOLUTION Analyzing the answers: A. cos ( 1 3 ) , cos ( x ) = n = 0 ( - 1 ) n x 2 n ( 2 n ) ! , so cos 1 3 = n = 0 ( - 1 ) n ( 1 3 ) 2 n ( 2 n ) ! B. arctan ( 1 3 ) , arctan ( x ) = n = 0 ( - 1 ) n x 2 n + 1 2 n + 1 , so arctan 1 3 = n = 0 ( - 1 ) n ( 1 3 ) 2 n + 1 2 n + 1 C. e 1 / 3 , e x = n = 0 x n n ! , so e 1 / 3 = n = 0 ( 1 3 ) n n ! D. sin ( 1 3 ) , sin ( x ) = n = 0 ( - 1 ) n x 2 n + 1 ( 2 n + 1 ) ! , so sin 1 3 = n = 0 ( - 1 ) n ( 1 3 ) 2 n + 1 ( 2 n + 1 ) ! So 1. D , because n = 0 ( - 1 ) n ( 1 3 ) 2 n + 1 ( 2 n + 1 ) ! = sin ( 1 3 ) 2. C , because n = 0 ( 1 3 ) n n ! = e 1 / 3 3. B , because n = 0 ( - 1 ) n ( 1 3 ) 2 n + 1 2 n + 1 = arctan ( 1 3 ) 4. A , because n = 0 ( - 1 ) n ( 1 3 ) 2 n ( 2 n ) ! = cos ( 1 3 ) Answer(s) submitted: (incorrect) Correct Answers: D C B A 4. (1 point) Match each of the Maclaurin series with right function. 1. n = 0 ( - 1 ) n 2 x 2 n + 1 2 n + 1 2. n = 0 ( - 1 ) n 2 x 2 n + 1 ( 2 n + 1 ) ! 3. n = 0 ( - 1 ) n 2 2 n x 2 n ( 2 n ) ! 4. n = 0 2 n x n n ! A. cos ( 2 x ) B. 2arctan ( x ) C. e 2 x D. 2sin ( x ) Solution: SOLUTION Analyzing the answers: A. cos ( 2 x ) , cos ( x ) = n = 0 ( - 1 ) n x 2 n ( 2 n ) ! , so cos ( 2 x ) = n = 0 ( - 1 ) n ( 2 x ) 2 n ( 2 n ) ! = n = 0 ( - 1 ) n 2 2 n x 2 n ( 2 n ) ! B. 2arctan ( x ) , arctan ( x ) = n = 0 ( - 1 ) n x 2 n + 1 2 n + 1 , so 2arctan ( x ) = 2 n = 0 ( - 1 ) n x 2 n + 1 2 n + 1 = n = 0 ( - 1 ) n 2 x 2 n + 1 2 n + 1 C. e 2 x , e x = n = 0 x n n ! , so e 2 x = n = 0 ( 2 x ) n n ! = n = 0 2 n x n n !

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