midterm_2007Ans

# midterm_2007Ans - Fall 2007 Midterm Exam Answer key ARE211...

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Fall 2007 ARE211 Midterm Exam - Answer key Problem 1 : Continuity . (14 points) In the first two parts of this problem, determine whether or not the specified functions are contin- uous. If they are, prove it. If they are not, identify the subset X in the domain of the function such that for every x X , f is continuous at x . a) f ( x ) = braceleftbigg x if x is rational 0 if x is irrational Ans: A function f : X R k is continuous at a point x 0 X if epsilon1> 0 , there exists δ> 0 such that | x - x 0 | implies | f ( x ) - f ( x 0 ) | <epsilon1 . The function f is continuous if it is continuous at x , for every x X . At x 0 = 0 : Fix epsilon1> 0 and set δ = epsilon1 . When | x - 0 | , then | f ( x ) - f (0) | = braceleftbigg | x - 0 | <epsilon1 if x rational | 0 - 0 | <epsilon1 if x irrational Therefore the function is continuous at x = 0 . At x 0 negationslash = 0 , where x 0 is rational: in this case, f ( x 0 ) = x 0 so that | f ( x ) - f ( x 0 ) | = braceleftbigg | x - x 0 | if x rational x 0 if x irrational Set epsilon1 = 0 . 5 x 0 and choose δ> 0 arbitrarily. Regardless of the size of δ , we can choose an irrational x such that | x - x 0 | ; we then have | f ( x ) - f ( x 0 ) | = x 0 >epsilon1 . Hence f is not continuous at x 0 . At x 0 negationslash = 0 , where x 0 is irrational: In this case, f ( x 0 ) = 0 and | f ( x ) - f ( x 0 ) | = braceleftbigg | x | if x rational 0 if x irrational Once again, set epsilon1 = 0 . 5 x 0 , and pick a rational x (0 . 5 x 0 ,x 0 ) , | f ( x ) - f ( x 0 ) | = x> 0 . 5 x 0 = epsilon1 . Hence f is not continuous at x 0 . Conclude that f is continuous only at x 0 = 0 b) f ( x ) = 1 if x = 0 1 /q if x = p/q is rational 0 if x is irrational Ans: At x 0 = 0 : In this case f ( x 0 ) = 1 . Pick epsilon1 = 0 . 5 and choose δ > 0 arbitrarily. There exists an irrational x such that | x | , and for this x | f ( x ) - f ( x 0 ) | = 1 >epsilon1 . Hence f is not continuous at x 0 .

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2 At x 0 = p/q rational: In this case, f ( x 0 ) = 1 /q . Pick epsilon1 = 1 / 2 q , choose δ > 0 arbitrarily and pick an irrational xinB ( x 0 ) Since | f ( x ) - f ( x 0 ) | = 1 /q>epsilon1 , f is not continuous at x 0 . At x 0 irrational: In this case, f ( x 0 ) = 0 . Fix epsilon1> 0 and define q ( epsilon1 ) = min { q N : 1 /q <epsilon1 } . Next, define p ( epsilon1 ) = min { p N : p/q ( epsilon1 ) > x 0 } . Now define R = { x = p/q : p p ( epsilon1 ); q q ( epsilon1 ) } . Since R is a finite set which does not contain x 0 (since, by assumption, x 0 is irrational), we can pick δ > 0 sufficiently small such that B ( x 0 ) R is empty. Finally, note that if x = p/q ( B ( x 0 ) R C ) , then q>q ( epsilon1 ) . (This it true because if q q ( epsilon1 ) and p/q / R then necessarily p>p ( epsilon1 ) , and hence p/q p/q ( epsilon1 ) >p ( epsilon1 ) /q ( epsilon1 ) >x 0 + δ ). It now follows that x B ( x 0 ) implies | f ( x ) - f ( x 0 ) | = | f ( x ) | < epsilon1 : if x is irrational, then f ( x ) = 0 <epsilon1 ; if x is rational, then, as we have shown, x = p/q B ( x 0 ) implies q>q ( epsilon1 ) and hence f ( x ) < 1 /q ( epsilon1 ) <epsilon1 . We have proved, therefore, that x B ( x 0 ) implies | f ( x ) - f ( x 0 | <epsilon1 , verifying that f is continuous at x 0 . Conclude that f is continuous at every irrational number, and discontinuous at every rational number.
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