This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Final Exam  Math 139 Dec 16th Instructor Mauro Maggioni Web page www.math.duke.edu/˜ mauro/teaching.html You have 3 hours. You may not use books, internet, notes, calculators. The exam should be stapled, written legibly, with your name written at the top of every page, on one side of each page only, and must contain the reaffirmation of the Duke community standard. You can invoke any Theorems that we discussed in class and are in the sections of the book we studied. No partial credit is given for statements for which proof is not provided. Problems are roughly ordered in increasing difficulty, and value (in terms of points). (*) preceeds parts of problems that are for extra credit: most of these are not particularly hard, but do take extra time. 1 . Let f ∈ C 1 ([ a,b ]) and f ′ ( x ) > 0 for all x ∈ [ a,b ]. Prove that f is strictly monotonically increasing, i.e. f ( x ) < f ( y ) for all x < y , x,y ∈ [ a,b ]. [5 pts.] Fix x,y ∈ [ a,b ]. Without loss of generality we may assume that x < y . By the Mean Value Theorem, applied on [ x,y ] to the C 1 function f , we have f ( y ) − f ( x ) = f ′ ( ξ )( y − x ) for some point ξ ∈ ( x,y ). Since f ′ is strictly positive everywhere and y − x > 0, the right hand side is strictly positive, proving that f ( y ) > f ( x ). Since x,y were arbitrary, we are done. 2 . Show that equation arctan( x ) = 1 − x has at least one real solution. [5 pts.] Let f ( x ) = arctan( x ) − (1 − x ). f is continuous on the whole real line, f (0) = − 1 and f ( x ) → π 2 − (1 − π 2 ) =: L > 0 as x → + ∞ since 1 − π 2 < 0. Therefore there exists M > 0 such that f ( x ) > 0 for x ≥ M by definition of limit as x → + ∞ and the fact that the limit is positive (in fact, there exists M such that  f ( x ) − L  < L/ 2 and therefore f ( x ) > L/ 2 > 0 for x ≥ M ). In particular f ( M ) > 0, and by the Intermediate Value Theorem applied in [0 ,M ] to the continuous function f , there exists ξ ∈ (0 ,M ) such that f ( ξ ) = 0, i.e. arctan( ξ ) = 1 − ξ . 3 . Define (a) what a metric space ( X,ρ ) is; (b) what does it mean for a sequence in ( X,ρ ) to converge; (c) what does it mean for a sequence in ( X,ρ ) to be Cauchy; (d) what does it mean for ( X,ρ ) to be complete. State an interesting Theorem that crucially uses completeness. Finally, consider the set A = ∪ + ∞ n =1 { ( x, 1 n ) ∈ R 2 : x ∈ R } in R 2 . Show that A is not complete (with respect to the Euclidean distance in R 2 ). (*) Complete A (with respect to the Euclidean distance in R 2 ). [10 pts.] (a) A metric space ( X,ρ ) is a pair consisting of a set X and a metric ρ , i.e. a function ρ : X × X → [0 , + ∞ ) satisfying the following conditions: • ρ ( x,y ) ≥ 0 for all x,y ∈ X , and ρ ( x,y ) = 0 if and only if x = y ; • ρ ( x,y ) = ρ ( y,x ) for all x,y ∈ X ; • ρ ( x,z ) ≤ ρ ( x,y ) + ρ ( y,z ) for all x,y,z ∈ X ....
View
Full
Document
This note was uploaded on 01/16/2012 for the course MATH 139 taught by Professor Pardon,w during the Fall '08 term at Duke.
 Fall '08
 Pardon,W
 Math

Click to edit the document details