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Unformatted text preview: θ. Then (cos θ + i sin θ)n+1 = (cos nθ + i sin nθ)(cos θ + i sin θ) = (cos nθ cos θ − sin nθ sin θ) + i(sin nθ cos θ + cos nθ sin θ) = cos(n + 1)θ + i sin(n + 1)θ, so that p(n + 1) is true. It now follows from the Principle of induction (Weak form) that (3) holds for every n ∈ N. Example 1.2.5. Consider the sequence x1 , x2 , x3 , . . . , given by x1 = 5, x2 = 11 and (4) We shall prove by induction that (5) xn = 2n+1 + 3n−1 xn+1 − 5xn + 6xn−1 = 0 if n ≥ 2. for every n ∈ N. To do so, let p(n) denote the statement (5). Then clearly p(1), p(2) are both true. Suppose now that n ≥ 2 and p(m) is true for every m ≤ n, so that xm = 2m+1 + 3m−1 for every m ≤ n. Then xn+1 = 5xn − 6xn−1 = 5(2n+1 + 3n−1 ) − 6(2n−1+1 + 3n−1−1 ) = 2n (10 − 6) + 3n−2 (15 − 6) = 2n+2 + 3n , so that p(n + 1) is true. It now follows from the Principle of induction (Strong form) that (5) holds for every n ∈ N. Example 1.2.6. Suppose that n ∈ N and n > 1. Then n is representable as a p...
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 Fall '08
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 Math, Calculus

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