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Yellow Final_sln

Yellow Final_sln - Prof Bjorn Poonen MATH 55 FINAL...

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Prof. Bjorn Poonen December 14, 2001 MATH 55 FINAL ANSWERS (yellow) (1) (5 pts. each) (a) FALSE (b) FALSE (c) TRUE (d) FALSE (e) TRUE (2) (10 pts.) { 1 , 3 , 4 , 5 } Also acceptable: 1345. (3) (10 pts.) 9 (4) (15 pts.) 10 (5) (10 pts.) 1 (1 - x )(1 - x 2 )(1 - x 4 ) (6) (20 pts.) 2000 (7) (15 pts.) 80 (8) (10 pts.) 32 (9) (15 pts.) a 1 = 25, a 2 = 626, a n = 25 a n - 1 + a n - 2 for n 3. Alternatively: a 0 = 1, a 1 = 25, a n = 25 a n - 1 + a n - 2 for n 2. (10) (20 pts.) a n = (3 n + 3)3 n + ( - 2) n Alternatively: a n = ( n + 1)3 n +1 + ( - 2) n (11) (15 pts.) 3 / 8 (12) (10 pts. each) (a) 21 / 2 (b) 7 / 2 1
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2 MATH 55 FINAL SOLUTIONS (yellow) Do not write your answers on this sheet. Instead please write your name, your student ID, your TA’s name, your section time, “yellow,” and all your answers in your blue books. IMPORTANT: Write your answers to problems 1–12 on the first page or two of your blue book, without the calculations you did to get the answers. To guarantee full credit and to qualify for partial credit, you should show your work on later pages of the blue book. Total: 200 pts., 170 minutes. (1) (5 pts. each) For each of (a)-(e) below: If the statement is true (always), write TRUE. Otherwise write FALSE. (Please do not use the abbreviations T and F, since in handwriting they are sometimes indistiguishable.) No explanations are required in this problem. (a) Exactly half of the 3-element subsets of { 1 , 2 , 3 , 4 , 5 , 6 , 7 } contain the number 5 . FALSE. There are ( 7 3 ) = 35 three-element subsets of { 1 , 2 , 3 , 4 , 5 , 6 , 7 } . Since 35 is odd, it cannot be that exactly 35 / 2 of them contain 5. In fact, the number of three-element subsets of { 1 , 2 , 3 , 4 , 5 , 6 , 7 } equals the number of two-element subsets of { 1 , 2 , 3 , 4 , 6 , 7 } , which is ( 6 2 ) = 15. (b) If A , B , and C are pairwise disjoint sets (possibly infinite), and | A B | = | A C | , then | B | = | C | . FALSE. Let A = { 2 , 3 , 4 , . . . } , B = { 0 , 1 } , and C = {- 1 } . These are pairwise disjoint. The function f ( n ) = - 1 , if n = 0 n + 1 , if n > 0 defines a bijection between A B and A C , so | A B | = | A C | . But | B | = 2 = 1 = | C | . Thus we have a counterexample. (c) If a function f ( n ) is O (2 n ) as n → ∞ , then f ( n ) is also o ( n !) as n → ∞ . TRUE. First, lim n →∞ 2 n /n ! = 0, by the Ratio Test. Since f ( n ) is O (2 n ), there exists c > 0 such that | f ( n ) | ≤ c 2 n for sufficiently large n . Thus, for sufficiently large n , the ratio f ( n ) /n ! is sandwiched between - c 2 n /n ! and c 2 n /n !. Both expressions on the right tend to 0 as n → ∞ , so lim n →∞ f ( n ) /n ! = 0. This means that f ( n ) = o ( n !), by definition. (d) For any predicate P ( x ) , the propositions ¬∃ xP ( x ) and xP ( x ) are logically equivalent. FALSE. Here is a counterexample. Suppose that P ( x ) is true for all x . Then ¬∃ xP ( x ) is false (provided that the universe of discourse is nonempty), but xP ( x ) is true. So the propositions are not logically equivalent in this case.
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3 (e) The statement p { S } q is true, where p is the assertion “ n = - 1 ”, q is the assertion “ k = - 1 ”, and S is the program segment consisting of the following six lines: k := 1 while n = 0 begin n := n - 1 k := n * k end TRUE. The meaning of “ p { S } q ” is that “if p is true at the beginning of the execution and the program terminates, then q is true at the end of the execution.”
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