Analysis2009aug - on [0 , ) but not directly Riemann...

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Analysis Preliminary Exam August 24, 2009 1. If F 1 and F 2 are closed subsets of R 1 and dist( F 1 , F 2 ) = 0 then F 1 F 2 ± = . Prove or give a counterexample. 2. Newton’s method for finding zeroes of a function f : R 1 R 1 is based on the recursion formula x n +1 = x n - f ( x n ) f ± ( x n ) , n 1 . Show that if f C 1 , f ( a ) = 0 and f ± ( a ) ± = 0, then there exists a δ > 0 such that if | x 1 - a | < δ then x n a . (Suggestion: use the Mean Value Theorem.) 3. Let f : [0 , ) [0 , ) and for h > 0 and k 1 set M k ( h ) = sup ( k - 1) h x<kh f ( x ) , m k ( h ) = inf ( k - 1) h x<kh f ( x ) . Let U ( h ) = X k =1 M k ( h ) h, L ( h ) = X k =1 m k ( h ) h. We say f is directly Riemann integrable if U ( h ) < for all h > 0 and lim h 0 ( U ( h ) - L ( h )) = 0 . Recall f is improperly Riemann integrable on [0 , ) if f is Riemann integrable on [0 , a ] for every a > 0, and lim a →∞ Z a 0 f ( t ) dt < . (a) Show that if f is continuous and nonincreasing, then f is directly Riemann integrable whenever f is improperly Riemann integrable on [0 , ). (b) Give an example of a continuous function f which is improperly Riemann integrable
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Unformatted text preview: on [0 , ) but not directly Riemann integrable. 4. Suppose f : [0 , ) [0 , ) is such that for any sequence a n of nonnegative terms we have X n =1 a n < = X n =1 f ( a n ) < Prove that lim sup x + f ( x ) x < 5. Let f be a continuous real valued function dened on the unit square and for each x 1 let f x be the function on the unit interval dened by f x ( y ) = f ( x, y ) . Prove that for any sequence x n in [0,1] there is a subsequence n k such that f x n k converges uniformly on [0,1]. 6. If c is a real parameter prove that x 7 + x + c = 0 has a unique real root and that this root is a dierentiable function of c ....
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This note was uploaded on 06/19/2011 for the course MATH 600 taught by Professor Na during the Spring '11 term at University of North Carolina School of the Arts.

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Analysis2009aug - on [0 , ) but not directly Riemann...

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