hw5 - SOLUTIONS FOR HOMEWORK 5 2.18. This is a bonus...

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Unformatted text preview: SOLUTIONS FOR HOMEWORK 5 2.18. This is a bonus problem . Write our polynomial as p ( x ) = ∑ n k =0 a k x k . Denote the set of even and odd integers between 0 and n by E and O , respectively. Then p ( x ) = ∑ k ∈ E a k x k + ∑ k ∈ E a k x k . Let A = ∑ k ∈ E a k and B = ∑ k ∈ O a k . Then p (1) = A + B , while p (- 1) = A- B . Therefore, p (1) p (- 1) = ( A + B )( A- B ) = A 2- B 2 . 3.15. We have to show that ∑ n i =1 (- 1) i i 2 = (- 1) n n ( n + 1) / 2. The basic step consists of verifying the above equality for n = 1. Then we get (- 1) 1 · 1 2 = (- 1) 2 · 1 · (1 +1) / 2, which is true. To perform the inductive step, we show that, for any n ∈ N , ∑ n i =1 (- 1) i i 2 = (- 1) n n ( n +1) / 2 implies ∑ n +1 i =1 (- 1) i i 2 = (- 1) n ( n + 1)( n + 2) / 2 Indeed, by the induction hypothesis, n +1 summationdisplay i =1 (- 1) i i 2 = n summationdisplay i =1 (- 1) i i 2 + (- 1) n +1 ( n + 1) 2 = (- 1) n n ( n + 1) 2 + (- 1) n +1 ( n + 1) 2 = (- 1) n +1 ( n + 1) parenleftBig- n 2 + ( n + 1) parenrightBig = (- 1) n +1 ( n + 1) n + 2 2 , which completes the proof. 3.22. Our goal is to establish that statement P ( n ) = ( ∀ a 1 ,... ,a n ∈ R )( | n summationdisplay i =1 a i | ≥ n summationdisplay i =1 | a i | ) is true for every n . The basic step is easy: P (1) is true. The inductive step consists of showing that if, for any a 1 ,... ,a n ∈ R we have | ∑ n i =1 a i | ≤ ∑ n i =1 | a i | , then also | ∑ n +1 i =1 a i | ≤ ∑ n +1 i =1 | a i | (for any a 1 ,... ,a n +1 ∈ R . Let a = ∑ n i =1 a i . Recall Triangle Inequality (Proposition 1.3): for any x,y ∈ R , | x + y | ≤ | x | + | y | . Then | n +1 summationdisplay i =1 a i | = | n summationdisplay i =1 a i + a n +1 | = | a + a n +1 | ....
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This note was uploaded on 01/27/2010 for the course MATH 347 taught by Professor ? during the Fall '09 term at University of Illinois at Urbana–Champaign.

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hw5 - SOLUTIONS FOR HOMEWORK 5 2.18. This is a bonus...

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