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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 | ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 5

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online