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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 09/29/2011 for the course MATH 3110 at Cornell University (Engineering School).

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online