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Unformatted text preview: Math 3110 Homework 9 Solutions Exercise 10.1.4 We can use the triangle inequality and that  cos( x )  1 for all x to get k X n =1 a n cos( nx ) k X n =1  a n cos( nx )  k X n =1  a n  , so the function is bounded by k X n =1  a n  . Exercise 10.1.9 (a) True. If f is bounded, then there is some M such that  f ( x )  M for all x . In particular, for every y , if x = g ( y ), we get  f ( g ( y ))  =  f ( x )  M , so f ( g ( x )) is bounded. (b) False. Consider f ( x ) = sin( x ) and g ( x ) = 1 x if x 6 = 0 and g (0) = 0. Pick M > 0. If x = 1 M , then since sin( x ) < x , we get g ( f ( x )) = 1 sin( x ) > 1 x = M , so M is not an upper bound for g ( f ( x )). Since this holds for all M > 0, g ( f ( x )) is not bounded above. Exercise 10.3.2 Let the period of f be c . We are given that there is some q such that f ( x ) is bounded for all x q . Then there is some M such that  f ( x )  M for all x q . So for any x , since f ( x ) has period c ,  f ( x )  = f x + q x c c M, because x + q x c c x + q x...
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This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell University (Engineering School).
 '08
 RAMAKRISHNA
 Math

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