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Unformatted text preview: Theorem: If a 1 = k and a n = a ceilingleft n/ 2 ceilingright + a floorleft n/ 2 floorright for n > 1, then a n = kn for n ≥ 1. Proof: By strong induction. Let P ( n ) be the statement that a n = kn . Basis: P (1) says that a 1 = k , which is true by hypothesis. Inductive step: Assume P (1) , . . . , P ( n ); prove P ( n + 1). a n +1 = a ceilingleft ( n +1) / 2 ceilingright + a floorleft ( n +1) / 2 floorright = k ceilingleft ( n + 1) / 2 ceilingright + k floorleft ( n + 1) / 2 floorright [Induction hypothesis] = k ( ceilingleft ( n + 1) / 2 ceilingright + floorleft ( n + 1) / 2 floorright ) = k ( n + 1) We used the fact that ceilingleft n/ 2 ceilingright + floorleft n/ 2 floorright = n for all n (in particular, for n + 1). To see this, consider two cases: n is odd and n is even. • if n is even, ceilingleft n/ 2 ceilingright + floorleft n/ 2 floorright = n/ 2 + n/ 2 = n • if n is odd, suppose n = 2 k + 1 ◦ ceilingleft n/ 2 ceilingright + floorleft n/ 2 floorright = ( k + 1) + k = 2 k + 1 = n This proof has a (small) gap: • We should check that ceilingleft ( n + 1) / 2 ceilingright ≤ n 1 In general, there is no rule for guessing the right inductive hypothesis. However, if you have a sequence of numbers r 1 , r 2 , r 3 , . . . and want to guess a general expression, here are some guidelines for trying to find the type of the expression (exponential, polynomial): • Compute lim n →∞ r n +1 /r n ◦ if it looks like lim n →∞ r n +1 /r n = b / ∈ { , 1 } , then r n proba bly has the form Ab n + ··· . ◦ You can compute A by computing lim n →∞ r n /b n ◦ Try to compute the form of ··· by considering the sequence r n Ab n ; that is, r 1 Ab, r 2 Ab 2 , r 3 Ab 3 , . . . • lim n →∞ r n +1 /r n = 1, then r n is most likely a polynomial. • lim n →∞ r n +1 /r n = 0, then r n may have the form A/b f ( n ) , where f ( n ) /n → ∞ ◦ f ( n ) could be n log n or n 2 , for example Once you have guessed the form of r n , prove that your guess is right by induction. 2 More examples Come up with a simple formula for the sequence 1 , 5 , 13 , 41 , 121 , 365 , 1093 , 3281 , 9841 , 29525 Compute limit of r n +1 /r n : 5 / 1 = 5 , 13 / 5 ≈ 2 . 6 , 41 / 13 ≈ 3 . 2 , 121 / 41 ≈ 2 . 95 , ..., 29525 / 9841 ≈ 3 . 000 Guess: limit is 3 ( ⇒ r n = A 3 n + ··· ) Compute limit of r n / 3 n : 1 / 3 ≈ . 33 , 5 / 9 ≈ . 56 , 13 / 27 ≈ . 5 , 41 / 81 ≈ . 5 , ..., 29525 / 3 10 ≈ . 5000 Guess: limit is 1 / 2 ( ⇒ r n = 1 2 3 n + ··· )+ Compute r n 3 n / 2: (1 3 / 2) , (5 9 / 2) , (13 27 / 2) , (41 81 / 2) ,... = 1 2 , 1 2 , 1 2 , 1 2 ,... Guess: general term is 3 n / 2 + ( 1) n / 2 Verify (by induction ...) 3 One more example Find a formula for 1 1 · 4 + 1 4 · 7 + 1 7 · 10 + ··· + 1 (3 n 2)(3 n + 1) Some values: • r 1 = 1 / 4 • r 2 = 1 / 4 + 1 / 28 = 8 / 28 = 2 / 7 • r 3 = 1 / 4 + 1 / 28 + 1 / 70 = (70 + 10 + 4) / 280 = 84 / 280 = 3 / 10 Conjecture: r n = n/ (3 n + 1). Let this be P ( n...
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 Spring '05
 KLEINBERG
 Mathematical Induction, Inductive Reasoning, induction hypothesis, AXA

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