Practice Exam 1 with solution

Practice Exam 1 with solution - University of Texas at...

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University of Texas at Dallas Practice Exam #1 Last Name: First Name and Initial: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Course Name: Number: Mathematical Analysis 2 MATH 4302 Instructor: Due Date: Wieslaw Krawcewicz March 1, 2012 E-mail Address: Student’s Signature: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Question Weight Your Score Comments 1. 10 2. 10 3. 10 4. 10 5. 10 6. 10 7. 10 8. 10 9. 10 10. 10 Total: 100 2
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Problem 1. Consider the function f : [ π, π ] R , defined by f ( x ) = x 2 sin( x ). (a): Compute the n -th derivative of the function f . (b): For a given n N , write the Taylor polynomial T n ( x ) of f (centered at x o = 0). (c): For a given ε > 0, find n N such that sup {| f ( x ) T n ( x ) | : x [ a, a ] } < ε. SOLUTION: Notice that we have f ( x ) = 2 x sin( x ) + x 2 cos( x ) , f ′′ ( x ) = 2 sin( x ) + 4 x cos( x ) x 2 sin( x ) , and for n 2 we have (by Leibnitz formula) f ( n ) ( x ) = ( n 0 ) x 2 sin ± x + π 2 n ² + ( n 1 ) 2 x sin ± x + π 2 ( n 1) ² + ( n 2 ) 2 sin ± x + π 2 ( n 2) ² = x 2 sin ± x + π 2 n ² + 2 nx sin ± x + π 2 ( n 1) ² + n ! ( n 2)! sin ± x + π 2 ( n 2) ² Then we have f (0) = f (0) = f ′′ (0) = 0 and for n 3 f ( n ) (0) = n ! ( n 2)! sin ± π 2 ( n 2) ² thus f (2 k +1) (0) = ( 1) k +1 (2 k + 1)! (2 k 1)! . Therefore, we have for n 3 T n ( x ) = n 2 k =1 ( 1) k +1 x 2 k +1 (2 k 1)! . 3
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c ( π, π ) r n ( x ) = f ( n +1) ( c ) ( n + 1)! x n +1 = c 2 sin( x + π 2 ( n + 1)) + 2( n + 1) c sin( c + π 2 n ) + ( n + 1) n sin( c + π 2 ( n 1) ( n + 1)! x n +1 Therefore, by applying the inequality ( n + 1)! ( n +2) n +1 e n +1 , we obtain | r n ( x ) | ≤ c 2 + 2 c ( n + 1) + n ( n + 1) ( n + 1)! π n +1 ( n + 1 + π ) 2 π n +1 ( n + 1)! ( n + 1 + π ) 2 ( πe ) n +1 ( n + 2) n +1 Therefore, if n 8 then πe n +2 < 1, thus (since n +5 n +2 2) | r n ( x ) | ≤ ( n + 5) 2 ( n + 2) 3 ( πe ) 3 4( πe ) 3 n + 2 Therefore, for a given ε > 0 if n max { 8 , 4( πe ) 3 ε 2 } then for x [ π, π ] | f ( x ) T n ( x ) | = | r
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Practice Exam 1 with solution - University of Texas at...

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