ineq_ind_review_midterm2 - 1 2-3 since k 1> 1 Case 2 k 1...

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CS173 Induction proof Proving an inequality by induction Consider the recurrence defined as: T ( n ) = ± 1 if n 7 4 T ( b n 2 c ) + 7 if n 8 Prove that n 1, T ( n ) < 7 n 2 - 3 Proof : We prove this by induction on n . Base case ( n = 1): T (1) = 1 < 4 = 7(1 2 ) - 3. Hence the base case is true. Strong Inductive Hypothesis : For some k 1, 1 m k , T ( m ) < 7 m 2 - 3 Inductive step : We need to show that T ( k + 1) < 7( k + 1) 2 - 3. Case 1 ( k + 1 7): Then T ( k + 1) = 1 < 4 = 7(1 2 ) - 3 < 7( k
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Unformatted text preview: + 1) 2-3, since k + 1 > 1. Case 2 ( k + 1 ≥ 8): Now T ( k + 1) = 4 T ( b k +1 2 c ) + 7 by the definition < 4(7( b k +1 2 c ) 2-3) + 7 by IH since b k +1 2 c ≤ k +1 2 ≤ k when k ≥ 1 ≤ 4(7( k +1 2 ) 2-3) + 7 since b k +1 2 c ≤ k +1 2 = 7( k + 1) 2-12 + 7 < 7( k + 1) 2-3 which is what we needed to show 1...
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This note was uploaded on 12/24/2010 for the course CS CS 173 taught by Professor Fleck during the Spring '10 term at University of Illinois, Urbana Champaign.

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