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

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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 | . By Triangle Inequality, the right hand side does not exceed | a | + | a n +1 | . By the induction hypothesis, | a | = | n summationdisplay i =1 a i | ≤ n summationdisplay i =1 | a i | .

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