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# hw9 - Math 3110 Homework 9 Solutions Exercise 10.1.4 We can...

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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 n =1 a n cos( nx ) k n =1 | a n cos( nx ) | ≤ k n =1 | a n | , so the function is bounded by k 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 = 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 c c = q. This means f ( x ) is bounded by M for all x . 1

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Problem 10-2 As the suggestion recommends, define S by S = { a < 1 | f ( x ) is constant on [0 , a ) } .
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hw9 - Math 3110 Homework 9 Solutions Exercise 10.1.4 We can...

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