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solutionshw14-554-2010

solutionshw14-554-2010 - Page 109 Problem 3 Let 0< y< x...

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Solutions homework 14. Page 99 Problem 1. Let p ( x ) be a polynomial of degree 2 k + 1. By multiplying by a constant we can assume that the coefficient of x 2 k +1 is 1. Then p ( x ) = x 2 k +1 + a 2 k x 2 k + ··· + a 1 x + a 0 . Then p ( n ) = n 2 k +1 (1 + a 2 k k + ··· + a 0 n 2 k +1 ) > 0 for n large enough. Similarly p ( - n ) < 0 for n large enough. As p is continuous it follows from the Intermediate Value Theorem that there exists c R such that p ( c ) = 0. Page 99 Problem 2. False. Take I = (0 , 1) and f ( x ) = 1 x . Page 99 Problem 4. Let g ( x ) = x - f ( x ). Then g (0) = - f (0) 0 and g (1) = 1 - f (1) 0. If g (0) = 0, then f (0) = 0 and if g (1) = 0 the f (1) = 1. Otherwise g (0) < 0 and g (1) > 1, so by the Intermediate Value Theorem there exists c (0 , 1) such that g ( c ) = 0, i.e. f ( c ) = c . Page 108 Problem 2. Done in class. (i) is continuity, while (ii) is uniform continuity.
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Unformatted text preview: Page 109 Problem 3. Let 0 < y < x . Then ≤ 1 1 + y 2-1 1 + x 2 = ( x-y ) · x + y (1 + y 2 )(1 + x 2 ) . Hence we need to show x + y ≤ (1 + y 2 )(1 + x 2 ) for proving f ( y )-f ( x ) ≤ ( x-y ). Now ( x-1) 2 ≥ 0 implies that x ≤ 1 2 + 1 2 x 2 . Similarly y ≤ 1 2 + 1 2 y 2 . Adding these two inequalities we get x + y ≤ 1+ 1 2 x 2 + 1 2 y 2 ≤ 1+ x 2 + y 2 + x 2 y 2 = (1+ y 2 )(1+ x 2 ). Hence | f ( x )-f ( y ) | ≤ | x-y | . Now given ± > 0 we can take δ = ± to get from | x-y | < δ that | f ( x )-f ( y ) | < ± , i.e., f is uniformly continuous. 1...
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