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280wk3_x4 - How to Guess What to Prove Sometimes...

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How to Guess What to Prove Sometimes formulating P ( n ) is straightforward; some- times it’s not. This is what to do: Compute the result in some specific cases Conjecture a generalization based on these cases Prove the correctness of your conjecture (by induc- tion) 1 Example Suppose a 1 = 1 and a n = a d n/ 2 e + a b n/ 2 c for n > 1. Find an explicit formula for a n . Try to see the pattern: a 1 = 1 a 2 = a 1 + a 1 = 1 + 1 = 2 a 3 = a 2 + a 1 = 2 + 1 = 3 a 4 = a 2 + a 2 = 2 + 2 = 4 Suppose we modify the example. Now a 1 = 3 and a n = a d n/ 2 e + a b n/ 2 c for n > 1. What’s the pattern? a 1 = 3 a 2 = a 1 + a 1 = 3 + 3 = 6 a 3 = a 2 + a 1 = 6 + 3 = 9 a 4 = a 2 + a 2 = 6 + 6 = 12 a n = 3 n ! 2 Theorem: If a 1 = k and a n = a d n/ 2 e + a b n/ 2 c 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 hypothe- sis. Inductive step: Assume P (1) , . . . , P ( n ); prove P ( n +1). a n +1 = a d ( n +1) / 2 e + a b ( n +1) / 2 c = k d ( n + 1) / 2 e + k b ( n + 1) / 2 c [Induction hypothesis] = k ( d ( n + 1) / 2 e + b ( n + 1) / 2 c ) = k ( n + 1) We used the fact that d n/ 2 e + b n/ 2 c = 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, d n/ 2 e + b n/ 2 c = n/ 2 + n/ 2 = n if n is odd, suppose n = 2 k + 1 ◦ d n/ 2 e + b n/ 2 c = ( k + 1) + k = 2 k + 1 = n This proof has a (small) gap: We should check that d ( n + 1) / 2 e ≤ n 3 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 / ∈ { 0 , 1 } , then r n probably 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 polyno- mial. 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. 4
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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 ...) 5 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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