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Unformatted text preview: rove by induction that 3n > n3 for every n > 3. To do so, let p(n) denote the statement (n ≤ 3) or (3n > n3 ). Then clearly p(1), p(2), p(3), p(4) are all true. Suppose now that n > 3 and p(n) is true. Then 3n > n3 . It follows that (note that we are aiming for (n + 1)3 = n3 + 3n2 + 3n + 1 all the way) 3n+1 > 3n3 = n3 + 2n3 > n3 + 6n2 = n3 + 3n2 + 3n2 > n3 + 3n2 + 6n = n3 + 3n2 + 3n + 3n > n3 + 3n2 + 3n + 1 = (n + 1)3 , so that p(n + 1) is true. It now follows from the Principle of induction (Weak form) that 3n > n3 holds for every n > 3. Example 1.2.4. We shall prove by induction the famous De Moivre theorem that (3)
Chapter 1 : The Number System (cos θ + i sin θ)n = cos nθ + i sin nθ
page 3 of 19 First Year Calculus c W W L Chen, 1982, 2005 for every θ ∈ R and every n ∈ N. To do so, let θ ∈ R be ﬁxed, and let p(n) denote the statement (3). Then clearly p(1) is true. Suppose now that p(n) is true, so that (cos θ + i sin θ)n = cos nθ + i sin n...
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 Fall '08
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 Math, Calculus

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