Fibonacci series

Jun 27th, 2016
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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.

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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

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