all-quizzes-fa09

# all-quizzes-fa09 - CSCE 750 Fall 2009 Quizzes with Answers...

This preview shows pages 1–3. Sign up to view the full content.

CSCE 750, Fall 2009 — Quizzes with Answers Stephen A. Fenner September 24, 2011 1. Give an exact closed form for X k =1 2 k 3 k +1 . Simplify your answer as much as possible. Answer: We reduce the expression to a form we’ve already seen in class: X k =1 2 k 3 k +1 = 2 3 X k =1 k 3 k = 2 3 X k =1 kr k , where r = 1 / 3. We saw in class 1 that the sum on the right is X k =1 kr k = r (1 - r ) 2 for all r such that | r | < 1. Thus the final answer is 2 3 · 1 / 3 (1 - 1 / 3) 2 = 2 3 · 1 / 3 4 / 9 = 1 2 . 2. Let f a real-valued function defined on R . Recall: f ( n ) is strictly monotone increasing iff x < y = f ( x ) < f ( y ) for all x, y R . f ( n ) is strictly monotone decreasing iff x < y = f ( x ) > f ( y ) for all x, y R . 1 If you forgot the formula, re-derive it as follows: (1) start with the formula for an infinite geometric series, k =0 r k = 1 / (1 - r ); (2) differentiate both sides with respect to r ; (3) multiply both sides by r . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Show that if f ( n ) and g ( n ) are both strictly monotone decreasing , then f ( g ( n )) is strictly monotone increasing . Answer: For all x, y R , we have x < y = g ( x ) > g ( y ) ( g is strictly decreasing) = f ( g ( x )) < f ( g ( y )) . ( f is strictly decreasing) Thus f ( g ( n )) is strictly increasing. 3. Use the substitution method to show that if T ( n ) = 4 T j n 2 k + n 2 , then T = O ( n 2 lg n ) . Only show the inductive step. Don’t worry about any base case(s). Answer: We tacitly assume that T ( n ) is eventually positive. Fix n large enough, and assume (inductive hypothesis) that T ( m ) Cm 2 lg m for all m < n , where C is a constant to be chosen later. Then T ( n ) = 4 T j n 2 k + n 2 4 C j n 2 k 2 lg j n 2 k + n 2 (inductive hypothesis) 4 C n 2 2 lg n 2 + n 2 (monotonicity) = Cn 2 (lg n
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern