# Fibonacci series

Jun 27th, 2016
Studypool Tutor
Price: \$30 USD

Tutor description

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 = λ^n. Find all possible values of λ so that gn satisﬁes the recursive relation gn+1 = gn + gn−1. You should get two nonzero values. Call the positive one α and the negative one β. b. Prove that any linear combination gn = Aα^n +Bβ^n satisﬁes the recursive relation gn+1 = gn +gn−1. 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 + fn−1, they are necessarily equal. So we got an explicit formula for fn. Finally, e. Prove the existence of constants A and α, such that fn ∼ Aα^n for large n.

Word Count: 522
Showing Page: 1/4
Now, lets try to nd theoretically. In order to do that, we want to nd explicit formula for the n-thFibonacci number fn. First, lets forget about initial conditions and guess the form of fn based on just therecursive 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 satises the recursive relationgn+1 = gn + gn1.You should get two nonzero values. Call the positive one and the negative one .b. Prove that any linear combination gn = A^n +B^n satises 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 A^n for large n.Solution:a)gn+1 = n+1gn = ngn-1 = n-1plug these into the recursive relation, we haven+1 = n + n-1

## Review from student

Studypool Student
" Was sceptical because of all the poor reviews but I was desparate. Turned out pretty well, so I'm pleasantly surprised. "
Ask your homework questions. Receive quality answers!

Type your question here (or upload an image)

1820 tutors are online

Brown University

1271 Tutors

California Institute of Technology

2131 Tutors

Carnegie Mellon University

982 Tutors

Columbia University

1256 Tutors

Dartmouth University

2113 Tutors

Emory University

2279 Tutors

Harvard University

599 Tutors

Massachusetts Institute of Technology

2319 Tutors

New York University

1645 Tutors

Notre Dam University

1911 Tutors

Oklahoma University

2122 Tutors

Pennsylvania State University

932 Tutors

Princeton University

1211 Tutors

Stanford University

983 Tutors

University of California

1282 Tutors

Oxford University

123 Tutors

Yale University

2325 Tutors