sole6 - ANALYSIS EXERCISE 6–SOLUTIONS 1 a 1 a 2 = a 1 2 a...

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Unformatted text preview: ANALYSIS EXERCISE 6–SOLUTIONS 1. a + 1 a- 2 = ( a- 1) 2 a ≥ . The equality holds iff a = 1. 2. Multiplying the inequalities a + b ≥ 2 √ ab, a + c ≥ 2 √ ac, c + b ≥ 2 √ cb, we obtain the desired. The equality holds iff a = b = c . 3. For all n ≥ 2 we have 1 n 2 < 1 n ( n- 1) = 1 n- 1- 1 n . Hence 1 2 2 + 1 3 2 + ··· + 1 n 2 < 1- 1 2 + 1 2- 1 3 + ··· + 1 n- 1- 1 n = 1- 1 n < 1 . 4. a) If x ≥ 0 then | - x | =- - x = x = | x | ; while if x < 0, | - x | =- x = | x | . If x ≥ 0 then | x | = x ≥ x ; while if x < 0 then | x | =- x ≥ x . If x ≥ 0 then max { x,- x } = x = | x | while if x < 0, max { x,- x } =- x = | x | . b) If x ≥ 0 and y ≥ 0 then xy ≥ 0 so | xy | = xy = | x || y | . If x ≥ 0 and y < 0 then xy ≤ 0 so | xy | =- xy = x (- y ) = | x || y | . If x < 0 and y ≥ 0 then xy ≤ 0 so | xy | =- xy = (- x ) y = | x || y | . If x < 0 and y < 0 then xy > 0 so | xy | = xy = (- x )(- y ) = | x || y | . c) See lectures. d) By c), | x | = | ( x- y ) + y | ≤ | x- y | + | y | ⇒ | x | - | y | ≤ | x- y | . Similarly, | y | - | x | ≤ | x- y | . By considering the cases | x | ≥ | y | and | x | < | y | we obtain || x |- | y || ≤ | x- y | . 1 2 ANALYSIS EXERCISE 6–SOLUTIONS 5. Base case : n = 2; | a 1 + a 2 | ≤ | a 1 | + | a 2 | , by the triangle inequality. The statement is true. Induction step: Suppose that | a 1 + a 2 + ··· + a n | ≤ | a 1 | + | a 2 | + ··· + | a n | ....
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This note was uploaded on 02/25/2010 for the course MATHEMATIC 11007 taught by Professor Spyros during the Winter '10 term at Bristol Community College.

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sole6 - ANALYSIS EXERCISE 6–SOLUTIONS 1 a 1 a 2 = a 1 2 a...

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