Diagnostic Quiz EECS 401. (a) (1 points) What is the set of all persons who are shorter than 6 feet or wear glasses?

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1 EECS 41 PROBLEM 1 (9 points) Let Januar 9, 27 Total Points: 25 A = set of all persons taller than 6 feet B = set of all persons heavier than 15 lbs. C = set of all persons who wear glasses (a) (1 points) What is the set of all persons who are shorter than 6 feet or wear glasses? A c C (b) (1 points) What is the set of all persons who are taller than 6 feet and are heavier than 15 lbs, but do not wear glasses? A B C c (c) (2 points) Epress these statements in terms of A, B, C : (1) (1 points) There is no one who wears glasses and is shorter than 6 feet. C A c = (2) (1 points) If a person is lighter than 15 lbs., then he or she must either wear glasses or is taller than 6 feet. B c A C (d) (4 points) Can ou relate: (1) (2 points) A B and A B? A B A B. A B A B A B A B If a person is taller than 6 feet and heavier than 15 lbs., then he or she is taller than 6 feet or heavier than 15 lbs. 1 Januar 9, 27

2 s (2) (2 points) A B and A c B c? (A B) c = A c B c. A B A B A B A B The set of people shorter than 6 feet and lighter than 15 lbs. is the same as the set of people who are not taller than 6 feet or heavier than 15 lbs. Note: A B and A c B c are also mutuall eclusive and collectivel ehaustive sets. Draw Venn diagrams and also epress our mathematical statements in terms of plain English. (e) (1 points) If A B, what can ou sa about the relationship between A c and B c? What is the corresponding statement in English? B c A c If everone who is taller than 6 feet is also heavier than 15 lbs., then everone who is lighter than 15 lbs. is also shorter than 6 feet. PROBLEM 2 (16 points) (a) (2 points) Compute p i and ip i. First let s assume p 1. Note that the first summation is a geometic series. If ou do not remember the epression for the sum of a geometric series, it can be derived quickl as follows for p 1: Let S = p + p 2 + p p n Thus, ()S = p p n+1 ps = p 2 + p p n + p n+1 p i = p n The second summation can be summed in different was. It is a arithemetico-geometric series. If ou do not remember the epression for the sum of a arithemetico-geometric series, it can be quickl derived as follows: 2 Januar 9, 27

3 s Thus, S = p + 2p 2 + 3p np n ps = p 2 + 2p (n 1)p n + np n+1 ()S = p + p 2 + p p }{{ n np } n+1 = p pn+1 geometric series ip i = p pn+1 () 2 npn+1 np n+1 Another method is to take the result of the geometric series and differentiate both sides with respect to p = p i = p n ip i 1 = 1 [ pn ()( np n 1 + p ) ( 1)( n ] ) () 2 = n npn + p n () 2 = n () 2 npn ip i = p pn+1 () 2 npn+1 Finall, if p = 1 then the first summation becomes: 1 i = n The second summation becomes: i(1 i ) = n(n + 1) 2 3 Januar 9, 27

4 s (b) (2 points) Let a >. Compute e a d and ea d. (c) (2 points) Compute e d. Use intergration b parts e d = [ e d] e a d = e a a e a d = ea a 1 = 1 a = [ ] d e d d = [ e ] 1 + d e d = 1 2 e (d) (2 points) Is it true that ( ) 1 (1) (1 points) f() + g() d = f()d + g? g()d for all functions f and e d Integration is a linear operator, so the above state is true provided that f and g are integrable in the interval (, 1), that is the integrals on the right hand side eist (2) (1 points)? f()g()dd = This is true. f()g()dd = Notice that g()d = g()d f()d g()d for all functions f and g [ ] [ 1 ] 1 g() f()d d = f()d g()d 4 Januar 9, 27

5 s (e) (4 points) Plot the curves: (1) (2 points) f() = e 2 /2, < <. = e 2 / 2 (2) (2 points) f() = log. = log +1 5 Januar 9, 27

6 s (f) (4 points) Suppose c > and a > b. (1) (2 points) Plot the curve for a 2 + b 2 = c, and label the ke features. For a > b, there is no solution. For a > > b, we have an hperbola c a c a For a > b =, we have 2 = c a, so the plot would be two vertical lines, one at = + c/a and another at = c/a. For a > b >, we have an ellipse c b c a 6 Januar 9, 27

7 s (2) (2 points) Repeat for the curve a 2 + 2a + b 2 2b = c. This equation can be transformed as follows a 2 + 2a + b 2 2b = c a( + 1) 2 + b( 1) 2 = c + a + b Thus we will have the same curves as before, with the origin shifted to ( 1, 1). For > a > b, there is no solution. For = a > c > b, we have ( 1) 2 = 1 + c b. Thus, the plot would be two horizontal lines, one at = c b and the other at = c b. For = a > b > c, then we have ( 1) 2 = 1 + c b, which has no real solution since 1 + c b <. For a > > b, we have an hperbola (-1,1) For a > b =, we have a( + 1) 2 = c + a, which reduces to two vertical lines, one at = c a and the other at = c a. 7 Januar 9, 27

8 s For a > b >, we have an ellipse (-1,1) 8 Januar 9, 27