m25-hw5-solutions

# m25-hw5-solutions - Math 25 Advanced Calculus Fall 2010...

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Unformatted text preview: Math 25: Advanced Calculus Fall 2010 Homework # 5 - Solutions 5.1: Show that max { x,y } = | x- y | 2 + x + y 2 . What expression would give min { x,y } ? Solution 1: We claim that min { x,y } = x + y 2- | x- y | 2 . We know that either x ≥ y or x ≤ y . We will prove the claims by examining these two cases separately. Case 1: x ≥ y . In this case, | x- y | = x- y . We compute that x + y 2 + | x- y | 2 = x + y 2 + x- y 2 = x = max { x,y } , and x + y 2- | x- y | 2 = x + y 2- x- y 2 = y = min { x,y } . Case 2: y ≥ x . In this case, | x- y | = y- x . We compute that x + y 2 + | x- y | 2 = x + y 2 + y- x 2 = y = max { x,y } , and x + y 2- | x- y | 2 = x + y 2- y- x 2 = x = min { x,y } . In either case, we have established that max { x,y } = x + y 2 + | x- y | 2 min { x,y } = x + y 2- | x- y | 2 . Solution 2: First, we observe that | x | = max { x,- x } so that x ≤ | x | and- x ≤ | x | for all x ∈ R . We consider two cases: Either a + b ≥ 0 or a + b < 0. Case 1: a + b ≥ 0. In this case, | a + b | = a + b ≤ | a | + | b | by our observation. Case 2: a + b < 0. In this case, | a + b | =- ( a + b ) =- a +- b ≤ | a | + | b | by our observation. 1 5.2: In class, we saw that for all real numbers a and b , | a + b | ≤ | a | + | b | . Give a careful proof of this fact by examining all possible cases (...
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## This note was uploaded on 03/18/2012 for the course MAT MAT 25 taught by Professor Stevenklee during the Fall '10 term at UC Davis.

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m25-hw5-solutions - Math 25 Advanced Calculus Fall 2010...

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