Math31B_VanKoten_Fall13_Midterm 2 solutions.pdf

X e j 1 n x j 1 1 j 1 j 1 j e j x e j 1 n x j 1 1 j 1

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( x - e ) j = 1 + n X j =1 ( - 1) j +1 ( j - 1)! j ! e j ( x - e ) j = 1 + n X j =1 ( - 1) j +1 je j ( x - e ) j . (The last equality follows since j ! = j ( j - 1)!.)
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Math 31B/4 Exam 2 - Page 6 of 7 November 18, 2013 (b) (10 points) Let α > 0 be a constant. Find an n so that | ln(2 e ) - T n (2 e ) | ≤ α . ( Note: Of course, your answer will depend on α .) Solution : First, we find a constant K n so that | f ( n ) ( x ) | ≤ K n for all e x 2 e . For n 1 and e x 2 e , we have | f ( n ) ( x ) | = ( - 1) n +1 ( n - 1)! x n = ( n - 1)! x n ( n - 1)! e n . (since e n x n for e x 2 e ) So we take K n = ( n - 1)! e n . Then by the error bound for Taylor polynomials, | ln(2 e ) - T n (2 e ) | ≤ K n +1 | 2 e - e | n +1 ( n + 1)! n ! e n +1 e n +1 ( n + 1)! 1 n + 1 . (Again, the last equality follows from ( n + 1)! = ( n + 1)( n !).) Finally, we observe that 1 n + 1 α 1 α n + 1 1 α - 1 n. Thus, we may choose any n 1 α - 1.
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Math 31B/4 Exam 2 - Page 7 of 7 November 18, 2013 SCRAP
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