a3sol 242

a3sol 242 - SOLUTIONS TO ASSIGNMENT 3 ANALYSIS 1 (MATH...

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Unformatted text preview: SOLUTIONS TO ASSIGNMENT 3 ANALYSIS 1 (MATH 242), FALL 2011 Problem 1. (a). Let > 0. | x 3- c 3 | = | x 3- c 2 x + c 2 x- c 3 | | x 3- c 2 x | + | c 2 x- c 3 | = | x ( x 2- c 2 ) | + | c 2 ( x- c ) | = | x || x- c || x + c | + c 2 | x- c | . If | x- c | < 1, then | x | | c | + 1, and | x + c | 2 | c | + 1. Then, | x || x- c || x + c | + c 2 | x- c | [( | c | + 1)(2 | c | + 1) + c 2 ] | x- 1 | < if = min braceleftbigg 1 , [( | c | + 1)(2 | c | + 1) + c 2 ] bracerightbigg . (b) . Let > 0. vextendsingle vextendsingle vextendsingle vextendsingle x 2- x + 1 x + 1- 1 2 vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 2 x 2- 3 x + 1 2( x + 1) vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ( x- 1 2 )( x- 1) x + 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle If | x- 1 | < 1 , | x + 1 | > 1 and | x- 1 / 2 | | x- 1 | + 1 / 2 < 3 / 2, so vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle ( x- 1 2 )( x- 1) x + 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle x- 1 2 vextendsingle vextendsingle vextendsingle vextendsingle | x- 1 | < 3 2 | x- 1 | , So if = min { 1 , 2 3 } , the last term is smaller than , and the claim is proved....
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This note was uploaded on 03/25/2012 for the course MATH 242 taught by Professor Drury during the Spring '08 term at McGill.

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a3sol 242 - SOLUTIONS TO ASSIGNMENT 3 ANALYSIS 1 (MATH...

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