Math31B_VanKoten_Fall13_Midterm 2 solutions.pdf

Therefore for all 0 x 2 f 00 x 2 sin x 2 4 x 2 cos x

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Therefore, for all 0 x 2, | f 00 ( x ) | = | 2 sin( x 2 ) + 4 x 2 cos( x 2 ) | 2 | sin( x 2 ) | + 4 x 2 | cos( x 2 ) | (by the triangle inequality: | a + b | ≤ | a | + | b | ) 2 + 4 x 2 (since | sin( u ) | ≤ 1 and | cos( u ) | ≤ 1 for any u ) 2 + 4 · 4 (since x 2 4 for all 0 x 2) = 18 . So we take K 2 = 18. (b) (10 points) Let T N be the trapezoid rule approximation to R 2 0 cos( x 2 ) dx using N trapezoids. Find an N such that Z 2 0 cos( x 2 ) dx - T N 10 - 4 . ( Note: If you are unable to complete part (a), you may explain how you would solve the problem given a correct value of K 2 .) Solution : We know that the error of the trapezoid rule satisfies Z 2 0 cos( x 2 ) dx - T N K 2 (2 - 0) 3 12 N 2 = 18 · 2 3 12 N 2 = 12 N 2 . We then observe that 12 N 2 10 - 4 12 × 10 4 N 2 12 × 10 4 N. (The last equivalence follows since x 7→ x is an increasing function.) Thus, we may take any N 12 × 10 4 .
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Math 31B/4 Exam 2 - Page 5 of 7 November 18, 2013 4. (a) (10 points) Let f ( x ) = ln( x ), and let T n ( x ) be the n th Taylor polynomial of f cen- tered at a = e . Find a general formula expressing T n in terms of n . ( Note: You may write T n either in summation notation or longhand with an ellipsis.) Solution : We recall the formula f ( n ) ( x ) = ( - 1) n +1 ( n - 1)! x n for all n 1 . (On an exam, it would be good to write out a few details; if you had trouble finding this formula, have a look in either the text or your notes from class.) Setting x = e gives f ( n ) ( e ) = ( - 1) n +1 ( n - 1)! e n for all n 1, and f ( e ) = ln( e ) = 1. Therefore, T n ( x ) = 1 + n X j =1 f ( j ) ( e ) j !
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