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1 due Monday June 27th. Probelms from the book. Section 1.1: 4. Section 1.2: 7. Section 1.3: 11, 12, 18, 24. Section 1.4: 14. Section

1 due Monday June 27th. Probelms from the book. Section 1.1: 4. Section 1.2: 7. Section 1.3: 11, 12, 18, 24. Section 1.4: 14. Section 2.3: 1, 13, 20. Section 1.5: 8, 14, 16, 36. Section 3.3: 6, 10, 17, 24. Section 3.4: 1abc, 3abc, 9, 10, 11, 12. Extra credit problems from the book. Section 1.3: 34. Section 1.4: 32. Section 1.5: 45. Section 3.3: 39. Section 3.4: 19. Problem 1. The sequence of Fibonacci numbers is defined by recursive relation fn+1 = fn + fn1 , and initial conditions f1 = 1, f2 = 1. The first few terms of this sequence are 1, 1, 2, 3, 5, 8, 13, 21.... a. Compute the first 50 terms of the sequence in Sage and plot them. We want to find the order of growth of that sequence when n . Notice that your sequence looks like exponentially increasing. Thus, we can make a following conjecture: Conjecture. There are constants A and , such that fn An for large n. an = 1.) (Reminder: We say that an bn if lim n bn fn+1 = . n fn b. Prove that if the conjecture is true, then lim Note that the converse of part b. is not true. Therefore, we can't prove the conjecture even if we fn+1 know the existence of the limit lim . Still, the existence of that limit is a good indication that the n fn conjecture is true. c. Plot first 50 values of the sequence fn+1 in Sage and estimate the numerical value of . fn Problem 2. Now, let's try to find theoretically. In order to do that, we want to find explicit formula for the n-th Fibonacci number fn . First, let's forget about initial conditions and guess the form of fn based on just the recursive relation. (If you need help with this problem, see section 1.4 of the book. In particular, theorem 1.7 and exercise 41 should be handy.) a. Consider the sequence gn = n . Find all possible values of so that gn satisfies the recursive relation gn+1 = gn + gn1 . You should get two nonzero values. Call the positive one and the negative one . b. Prove that any linear combination gn = An + B n satisfies the recursive relation gn+1 = gn + gn1 . Now we can account for initial conditions: c. Find the values of A and B so that g1 = 1 and g2 = 1. d. Show that fn = gn . That is, prove that if there are two sequences satisfying relations f1 = 1, f2 = 1, fn+1 = fn + fn1 , they are necessarily equal. So we got an explicit formula for fn . Finally, e. Prove the existence of constants A and , such that fn An for large n. Problems 3. Perform addition with arbitrary large integers in binary representation. That is, given two strings of zeros and ones of arbitrary length, write a function \"Sum\" in Sage that returns the sum of those binary numbers. See example 2.5 in the book for reference. Example: Sum('1010', '11') = '1101' Example: Sum('1010', '111') = '10001

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