a2sol - MATH 242 Assignment 2 Solutions 1. Proof by...

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Unformatted text preview: MATH 242 Assignment 2 Solutions 1. Proof by induction on n . In case n = 1 the statement reduces to 1 + a 1 ≤ 1 + a 1 which is clearly true. So the induction starts. Now suppose that the statement is true for n we will establish it for n + 1 . We are assuming that 1 + n X j =1 a j ≤ n Y j =1 (1 + a j ) . Then n +1 Y j =1 (1 + a j ) = (1 + a n +1 ) n Y j =1 (1 + a j ) , ≥ (1 + a n +1 ) 1 + n X j =1 a j , since (1 + a n +1 ) ≥ = 1 + n X j =1 a j + a n +1 + a n +1 n X j =1 a j , ≥ 1 + n +1 X j =1 a j , since a n +1 ≥ and ∑ n j =1 a j ≥ . This completes the induction step. 2. This is a straightforward induction. For n = 1 the assertion reads 1 2 2 ≤ 1 4 , so the induction starts. For the induction step, we assume that the assertion is true for n and establish it for n + 1 . We find that 1 2 · 3 4 · 5 6 · . . . · 2 n- 1 2 n · 2 n + 1 2 n + 2 2 = 1 2 · 3 4 · 5 6 · . . . · 2 n- 1 2 n 2 · 2 n + 1 2 n + 2 2 ≤ 1 3 n + 1 · 2 n + 1 2 n + 2 2 , so it will be enough to establish 1 3 n + 1 · 2 n + 1 2 n + 2 2 ≤ 1 3 n + 4 (1) or equivalently (3 n + 4)(2 n + 1) 2 ≤ (3 n + 1)(2 n + 2) 2 . We find that (3 n + 4)(2 n + 1) 2 = 12 n 3 + 28 n 2 + 19 n + 4 and that (3 n + 1)(2 n + 2) 2 = 12 n 3 + 28 n 2 + 20 n + 4 so (1) follows since ≤ n . This completes the induction step. 3. (i) Clearly for any fixed j ∈ N we have inf ( j,k ) ∈ N × N a j,k ≤ inf k ∈ N a j,k since the inf on the left ranges over a larger set than the one on the right. Since the inf on the left is a lower bound for the gadgets on the right as j varies, we have inf ( j,k ) ∈ N × N a j,k ≤ inf j ∈ N inf k ∈ N a j,k (2) because the right-hand side of (2) is the corresponding greatest lower bound....
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This note was uploaded on 11/05/2009 for the course MATH MATH 242 taught by Professor Drury during the Fall '09 term at McGill.

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a2sol - MATH 242 Assignment 2 Solutions 1. Proof by...

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