201-1A5-MT - Mathematics Summer 2015 HOMEWORK 2 Deadline : Sunday, August 30, 2015 at 12 :30.

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1 01-1A5-MT - Mathematics Summer 015 HOMEWORK Deadline : Sunday, August 30, 015 at 1 :30. Instructions : The assignment consists of five questions, each worth 0 points. Only hardcopy submissions of your homework will be accepted. It is obligatory to use LaTex and the template file available at http ://ziedzaier.com/01-1a5-mt ; submissions non compliant with this template will not be accepted. Make sure to enter your name and student I.D. number in the appropriate section of the template. Late submissions and submissions which do not comply with these guidelines will not be accepted. Question 1 on induction (0 points) a) Use mathematical induction to prove that for all integers n 0 (n + 1) 1 is divisible by 8. Basis Step : if n = 0, then (n + 1) 1 = 0, and is divisible by 8 since 0 is divisible by 8. Inductive Hypothesis : suppose that (n + 1) 1 is divisible by 8. We need to prove that ((n + 1) + 1) 1 is divisible by 8. We develop first the expression : ((n + 1) + 1) 1 = (n + 3) 1 = 4n + 1n = ((4n + 4n + 1) 1) + 8n + 8 = ((n + 1) 1) + 8(n + 1). The last expression is divisible by 8, since ((n + 1) 1) is divisible by 8, based on the Inductive Hypothesis, and 8(n + 1) is a multiple of 8. b) Use mathematical induction to prove that for all integers n n 3 = n (n+1) 4. Basis Step : pf n = 1, then 1 3 = 1 (1 + 1) /4. Inductive Hypothesis : suppose that n 3 = n (n + 1) /4. We need to prove that n 3 + (n + 1) 3 = (n + 1) (n + ) /4. 1

2 we apply the Inductive Hypothesis to the left side : n 3 + (n + 1) 3 = n (n + 1) /4 + (n + 1) 3 = (n + 1) (n /4 + n + 1) = (n + 1) (n + 4n + 4)/4 = (n + 1) (n + ) /4. c) Use mathematical induction to prove that for all integers n 1 1.1! +.! n.n! = (n + 1)! 1. Basis Step : if n = 1, then 1.1! = (1 + 1)! 1. Inductive Hypothesis : suppose that 1.1! +.! n.n! = (n + 1)! 1. We need to prove that 1.1! +.! n.n! + (n + 1)(n + 1)! = (n + )! 1. we apply the Inductive Hypothesis to the left side : 1.1! +.! n.n! + (n + 1)(n + 1)! = (n + 1)! 1 + (n + 1)(n + 1)! = (n + 1)!(n + ) 1 = (n + )! 1.

3 Question Linear Congruence, Linear and Matrix Algebra (0 points) a) Solve the system : 7x 1 8x + 5x 3 = 5 4x 1 + 5x 3x 3 = 3 x 1 x + x 3 = 0 x1 = 1, x = 1, x3 = b) Solve each of these congruences using the modular inverses : 1. 34x 77(mod89),. 144x 4(mod33), 3. 00x 13(mod1001), 1. a = 34, m = 89 Applying the Euclidean algorithm we get the inverse of 34 modulo 89 is -55. So to solve 34x = 77(mod89), multiply both sides 55). That will give x = 55(77)(mod89).55(77) = 435 = 47(89) + 5.X = 5. a = 144, m = 33 Applying the Euclidean algorithm we get the inverse of 144 modulo 33 is 89. X = a = 00, m = 1001 Applying the Euclidean algorithm we get the inverse of 00 modulo 1001 is 996 (you can use 5 = to simplify the next step). X =

4 Question 3 on Induction and Recursive Definitions (0 points) We are interested in the Fibonacci sequence defined recursively as follows : : f 0 = 1; f 1 = 1; f n = f n 1 + f n, n. a) Give the first 10 terms of the sequence. 1, 1,, 3, 5, 8, 13, 1, 34, 55 b) Show that the number 1+ 5 satisfies the equation x = x + 1. [(1 + 5)/] = ( )/4 = ( + 5)/4 + 4/4 = (1 + 5)/ + 1 c) Use mathematical induction to prove that f n > ( 1+ 5 ) n, n 3. Basis Step : if n = 3, then 3 > ( 1+ 5 ) and is true since 5 < 5. Si n = 4, ) = (1 + 5)/ + 1 > 3 + 1, we have5 > ( 1+ 5 ). since ( 1+ 5 Inductive Hypothesis : suppose that f i > ( 1+ 5 ) i for all integers 3 i n. We need to prove that f n+1 > ( 1+ 5 ) n 1. It will easier if the replace the expression ( 1+ 5 ) with a symbol, a for example. We have then : f n+1 = f n + f n 1 > a n + a n 3 based on Inductive Hypothesis, = a n 3 (a + 1) = a n 3 (a ) based on part b), = a n 1. d) Order, depending on their growth rate, the following functions : f n, n, n and (3/) n. Justify your answer. n, (3/) n, f n, n Based on 3 < 1+ 5 < and that f n < a n, in part c). 4

5 Question 4 on Recursive Definitions (0 points) Consider the{ numerical sequence defined recursively as follows :, if n is odd f n = (f ( n ) ), if n is even a) Give the first 10 terms of the sequence., 4,, 16,, 4,, 64,, 4 b) Use mathematical induction to prove that f n n, n 1. Basis Step : if n = 1, then f n = and n =, Thus f 1 1 is true. Inductive Hypothesis : suppose that f i i for all integers 1 i n. We need to prove that f n+1 n+1. if n + 1 is odd, then f n+1 = and since n 1, we have f n+1 = n+1. if n + 1 is even, then (f ( n+1 )) ( (n+1)/ ) = n+1, based on the Inductive Hypothesis. 5

6 Question 5 on Complexity (0 points) Let the following functions with domain N, the set of natural numbers, and codomain R, the set of real numbers. 1. f 1 (n) = n i=1 (7i + 5),. f (n) = 4log n n, 3. f 3 (n) = n log (4 n ), 4. f 4 (n) = n 1/ 3 n+3, 5. f 5 (n) = 5 + n i=0 3i, 6. f 6 (n) = 5 + 3n+1. a) Find, for each of these functions f i, a function as simple as possible which is the estimate of Big-O as precise as possible for the function f i. b) We say that the function f grows less rapidly than the function g if f is in O(g) and g is not in O(f). It is said that the functions f and g grow at the same speed if f is in O(g) and g is in O(f). Arrange the functions f 1, f, f 3, f 4, f 5, f 6 from top to bottom depending on their growth rate (Big-O). If functions are growing at the same speed, put their number on a same line. Solution a) b) 1. f 1 (n) O(n ),. f (n) O(1), 3. f 3 (n) O(n ), 4. f 4 (n) O(n 1/ 9 n ), 5. f 5 (n) O(3 n ), 6. f 6 (n) O(8 n ). 1. f (n),. f 1 (n) and f 3 (n), 3. f 5 (n), 4. f 6 (n), 5. f 4 (n). 6

Problem Set 5 Solutions

Problem Set 5 Solutions Section 4.. Use mathematical induction to prove each of the following: a) For each natural number n with n, n > + n. Let P n) be the statement n > + n. The base case, P ), is true