make_pdfsoln

# make_pdfsoln - Midterm 1 Math 55 Solutions 1 a False b...

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Midterm 1, Math 55 Solutions July 16, 2002 1. a) False b) False c) False d) True e) False f) True g) True 2. x ( ¬ I ( x ) → ∃ y ( x 6 = y B ( x, y ))) 3. a) a = 1 and b = - 1. b) n X k =1 2 k 2 + 2 k = n X k =1 ± 1 k - 1 k + 2 ² = 1 + 1 2 - 1 n + 1 - 1 n + 2 . 4. Since | sin( x ) | ≤ 1, and for x 1 we have 1 x x 2 , we ±nd from the triangle inequality that | 2 x + sin( x ) | ≤ | 2 x | + | sin( x ) | ≤ 2 | x 2 | + | x 2 | = 3 | x 2 | , so for C = 3 and k = 1 we get that for x > k we have | f ( x ) | ≤ c | g ( x ) | . Hence, by de±nition, f ( x ) is O ( x 2 ). 5. We split the problem up in two cases. Either n is even, or n is odd. If n is even, then n = 2 k for some integer k , so b n/ 2 cd n/ 2 e = k 2 = (2 k ) 2 / 4 = b n 2 / 4 c . If, however, n is odd, then n = 2 k + 1 for some integer k and we get b n/ 2 c = b k + 1 2 c = k , and d n/ 2 e = k + 1, while b n 2 / 4 c = b (4 k 2 + 4 k + 1) / 4 c = k 2 + k . Hence

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## This note was uploaded on 03/08/2010 for the course MATH 55 taught by Professor Strain during the Spring '08 term at Berkeley.

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make_pdfsoln - Midterm 1 Math 55 Solutions 1 a False b...

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