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Exam_Final_Solution

Exam_Final_Solution - Ma1a Final Solutions Fall 2005...

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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 , 0 < 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 0 ( 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 (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 ). Solution 3. Note: There are two acceptable solutions to this problem. 1
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2 Solution 1. We have cos 4 x = Re( e 4 xi ) = Re(( e xi ) 4 ) = Re((cos x + i sin x ) 4 ) = Re " 4 X k =0 4 k (cos x ) 4 - k ( i sin x ) k # by the binomial formula. Note that i k is real for k even and imaginary for k odd. Since cos x and sin x
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