Exam_Final_Solution

Exam_Final_Solution - Ma1a Final Solutions Fall 2005...

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Unformatted text preview: Ma1a Final Solutions Fall 2005 Question 1. (Dr. Simon) State and prove a theorem about differentiability implying continuity. Solution 2. This is discussed in the Supplementary Notes (see pp. 14-15). Question 2. (Xiaoyu) Is the series n =1 ( n 4 + 1- n 2 ) absolutely convergent, conditionally but not absolutely conver- gent, or divergent? Explain the tests you use and their requirements. Solution 2. Note that for all positive integers n , < p n 4 + 1- n 2 = ( n 4 + 1- n 2 )( n 4 + 1 + n 2 ) n 4 + 1 + n 2 = n 4 + 1- n 4 n 4 + 1 + n 2 < 1 n 2 . Since X n =1 1 n 2 is convergent, by the comparison test, n =1 ( n 4 + 1- n 2 ) is absolutely convergent. Moreover, the comparison test for n =1 a n being convergent requires that (1) 0 a n b n for all positive integers n ; and (2) n =1 b n is convergent. / Alternate proof for n 4 + 1- n 2 < 1 /n 2 . Let f ( x ) = 1 + x, x (- 1 , ) . Then f ( x ) = 1 2 (1 + x )- 1 / 2 and f 00 ( x ) =- 1 4 (1 + x )- 3 / 2 < 0 for all- 1 < x < . So f is strictly concave on (- 1 , ). Therefore, f ( x ) < f (0) + f (0) x for all x > 0. Take x = 1 /n 4 . Then r 1 + 1 n 4 < 1 + 1 2 1 n 4 . Multiply n 2 on each side, we have p n 4 + 1 < n 2 + 1 2 n 2 , i.e., p n 4 + 1- n 2 < 1 2 n 2 < 1 n 2 . / Question 3. (Chris) Write cos(4 x ) as a polynomial in cos( x )....
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This note was uploaded on 03/30/2010 for the course MA 1a taught by Professor Borodin,a during the Fall '08 term at Caltech.

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Exam_Final_Solution - Ma1a Final Solutions Fall 2005...

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