# nphw3 - b holds on the assumption that Φ n is true Taking...

This preview shows page 1. Sign up to view the full content.

Numbers and Polynomials Homework 3 1. Let F 0 = 1, let F 1 = 1, and for each natural number n > 1 let F n = F n - 1 + F n - 2 . Prove that F n 6 2 n for each natural number n . Solution : 1. We may of course use MI2 ; however, we shall simply use MI1 and take into account the two-step nature of the reccurrence relation deﬁning the Fibonacci numbers. Thus, for each natural number n > 0 let Φ n be the statement F n 6 2 n AND F n - 1 6 2 n - 1 . Base step: We are given that F 0 = 1 6 2 0 and F 1 = 1 < 2 1 ; it follows that Φ 1 is true. Inductive step: Assume that Φ n is true. Then F n +1 a = F n + F n - 1 b 6 2 n + 2 n - 1 = (2 + 1)2 n - 1 < 4 · 2 n - 1 = 2 n +1 where equality a is by deﬁnition of the Fibonacci numbers and equality
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: b holds on the assumption that Φ n is true. Taking F n +1 6 2 n +1 and F n 6 2 n together, Φ n +1 is true. The principle of mathematical induction now guarantees that Φ n is true for all natural numbers n > 0 so that F n 6 2 n for all natural numbers n . Remark : The proof actually shows that if n > 0 then F n < 2 n . A further inductive argument shows that if n is a natural number then (perhaps surprisingly) √ 5 F n = ± 1 + √ 5 2 ² n +1-± 1-√ 5 2 ² n +1 . 1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online