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Problems
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[APPENDIX
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A.1 A.2 A.3
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Prove Theorem A.1 on page 319. Prove Theorem A.2 on page 320. True or false: g = (f) a. f = o(g) b. f = O(g) g = (f) g = (f) c. f = (g) f = (g) d. f = O(g) e. f = (g) f = (g) f. f = (h) g = (h) g. f = (h) g = (h) h. n 2 O(n lg n) i. n 2 (n lg n) j. n 2 (n lg n) k. lg n (n) l. lg n o(n)
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f + g = (h) f g = (h)
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A.4 A.5 A.6 A.7
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Prove Theorem A.5 on page 323. Prove Theorem A.6 on page 323. Prove Theorem A.7 on page 323. Run a program that tests De Moivre s formula on page 325 by comparing the values obtained from it with those obtained from the recursive definition of the Fibonacci numbers.
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A.1 A.2 A.3 The floor and ceiling functions are idempotent because they return integer values, and according to Theorem A.1 on page 319, the floor or ceiling of an integer is itself. A logarithm is an exponent. It is the exponent on the given base that produces the given value. The First Principle of Mathematical Induction ( weak induction) allows the inductive hypothesis that assumes that the proposition P(n) is true for some single value of n. The Second Principle of Mathematical Induction ( strong induction) allows the inductive hypothesis that assumes that all the propositions P(k) are true for all k less than or equal to some value of n. Use weak induction (the first principle) when the proposition P(n) can be directly related to its predecessor P(n 1). Use strong induction (the second principle) when the proposition P(n) depends upon P(k) for k < n 1. Euler s constant is the limit of the difference (1 + 1/2 + 1/3 + . . . + 1/n) lnn. Its value is approximately 0.5772. Stirling s formula is a useful method for approximating n! for large n (e.g., n > 20).
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A.5 A.6
Solutions to Problems
A.1 Proof of Theorem A.1 on page 319: a. The relationships x = max{m Z | m x}, and restatements of the definitions of x and x .
= min{n
Z | n
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APPENDIX]
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b. Let m = x and n = x . Then by definition, m x < m + 1 and n 1 < x n. Then x 1 < m and n < x + 1. Thus x 1 < m x n < x + 1. x < x + 1 merely summarize those in b above. c. The inequalities x 1 < x x d. Let n Z such that n x < n + 1, and let A = {m Z | m x}. Then n A and x = maxA, x . Now if n < x , then n + 1 x (since both n and x are integers). But n + 1 so n x , by hypothesis. Therefore, n = x . The proof of the second part is analogous. e. Assume that x Z (i.e., x is an integer). Then let n = x in d above: x x < x + 1 so x = x , and x 1 < x x so x = x . f. Assume that x Z (i.e., x is not an integer). Let u = x x and v = x x. Then, by c, u 0 and v 0. Also by c, x 1 < x = x u and v + x = x < x + 1, so u < 1 and v < 1. Thus 0 u < 1 and 0 v < 1. But u and v cannot be integers because if either were, then so would x because x = x + u = x v. Therefore, 0 < u < 1 and 0 < v < 1, so x = x + u > x and x = x v < x . g. Let n = x . Then ( n) = x so by c, ( x) 1 < ( n) ( x), so x n < x + 1, so x n and n 1 < x, so n 1 < x n. Thus by d, n = x . Thus x . = x , so x = x . The second identity follows from the first by replacing x with x. h. Let n = x + 1 . Then by c, (x + 1) 1 < n (x + 1), so x 1 < n 1 x and x = (x + 1) 1 < n. Thus, n 1 x < n; that is, (n 1) x < (n 1) + 1. Thus by d, (n 1) = x , so x + 1 = n = x + 1. The proof of the second identity is similar. A.2 Proof of Theorem A.2 on page 320: a. Let x = b y. Then by definition, log b(b y) = log b(x) = y. b. Let y = log b x. Then by definition, b log b x = b y = x. c. Let y = log b u and z = log b v. Then by definition, u = b y and v = b z, so uv = (b y)(b z) = b y+z, so log b(uv) = y + z = log b u + log b v. d. By Law c above, log b v + log b u/v = log b (v u/v) = log b u, so log b u/v = log b u + log b u. e. Let y = log b u. Then by definition, u = b y, so uv = (b y )v = b vy. Then by definition, log b (uv ) = v y = v log b u. f. Let y = log b x. Then by definition, x = b y, so log c x = log c (b y ) = y log c b, by Law e above. Thus log b x = y = (log c x)/(log c b). a. b. c. d. e. f. g. h. i. j. k. l. True True True False True True False False False True False True + ar , so S rS
Proof of Theorem A.5 on page 323: 2 3 n 1 2 3 4 + ar Let S = a + ar + ar + ar + . Then rS = a r + ar + ar + ar + n, (1 r)S = a(1 r n ), and thus S = a(1 r n )/(1 r). = a ar
Proof of Theorem A.6 on page 323: If 1 < r < 1, then as n increases without bound, rn shrinks down to zero. Let rn =0 in the formula in Theorem A.5. Proof of Theorem A.7 on page 323: Let S = 1 + 2 + 3 + + 3 + 2 + 1 also. Add these two equations, + n . Then S = n + summing the 2n terms on the right pairwise: (1 + n), (2 + (n 1)), etc. There are n pairs, and each pair has the same sum of n + 1. So the total sum on the right is n(n + 1). Then, since the sum on the left is 2S, the correct value of S must be n(n + 1)/2.
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