$F_n$ denotes the $n^{th}$ Fibonacci number. ($F_n = F_{n-2} + F_{n-1}$). Use $F_{13} = 233$ and the golden ratio to approximate the next Fibonacci number $F_{14}$ (round to the nearest whole number). Show all steps here.
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618$. To approximate the next Fibonacci number $F_{14}$, we can multiply $F_{13}$ by the golden ratio. $F_{14} \approx F_{13} \times \phi$ $F_{14} \approx 233 \times 1.618$ Show more…
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The Fibonacci sequence is a well known sequence in mathematics that is obtained following a simple rule: set the first two entries of the sequence to be 0 and 1, and then, inductively, find the next entry by adding two previous entries. This results in the sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, . . . . It is natural to ask whether one can find a formula for the nth Fibonacci number (i.e., the nth entry of this sequence) without having to compute all previous entries. Such rules are called recursions and in this problem we will develop a method to analyze recursion relations. (a) Set F_0 = 0, F_1 = 1, and let F_n denote the nth Fibonacci number (with the convention that 0 is the 0th Fibonacci number). Then the sequence of Fibonacci numbers can be obtained by running the recursion F_n = F_{n-1} + F_{n-2}; F_0 = 0, F_1 = 1. Now, set v_n = [F_n, F_{n-1}]^T. Identify a matrix A such that the recursion above can be written, equivalently, as v_{n+1} = A v_n; v_0 = [F_1, F_0]^T. (b) Using the matrix recursion above A, obtain a (non-recursive) formula for v_{n+1} in terms of powers of A and v_0. (c) Let lambda_1 = (1 + sqrt{5})/2 and lambda_2 = (1 - sqrt{5})/2. Diagonalize A and give a formula for each entry of A^n in terms of lambda_1 and lambda_2. (d) Use your answer from the previous part to obtain a formula for F_n in terms of lambda_1 and lambda_2. (e) Show that lim_{n oinfty} F_{n+1}/F_n = lambda_1. (Note that lambda_1 is the famous "golden ratio").
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2) Let Fn be the nth Fibonacci number. That is, F1 = 1, F2 = 1, F3 = F2 + F1 = 2, F4 = F3 + F2 = 3, F5 = F4 + F3 = 5, ... The general formula for n ≥ 3 is Fn = Fn-1 + Fn-2. Let ϕ = (1+√5)/2. This number is called the golden ratio, and satisfies the equation ϕ^2 = ϕ + 1. (a) Prove that for every n ∈ ℕ we have Fn ≥ ϕ^(n-2). [Hint: use a proof by induction. First check that F1 ≥ ϕ^(1-2) and that F2 ≥ ϕ^(2-2). Then prove that if Fm ≥ ϕ^(m-2) for every 1 ≤ m ≤ k, then Fk+1 ≤ ϕ^(k+1-2).] (b) Prove that the 500th Fibonacci number F500 has at least 105 decimal digits.
Brian B.
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