MIT18_014F10_pset11a - Pset 11 - Part I (Part II will be...

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Unformatted text preview: Pset 11 - Part I (Part II will be available by 11/27.) Due December 3 (4 points each) We first present the following definitions, given for any sequence {an }: lim inf an = lim inf am n n mn and lim sup an = lim n n sup am . mn To give you some intuition, we determine the lim inf, lim sup for a few sequences. (You may want to draw some pictures or write out some terms of the sequence to help yourself.) Let an = (-1)n n-1 . Then lim inf an = -1 and lim sup an = 1. Notice n that limn an does not exist. Let bn = : n even : n odd Then lim inf bn = lim sup bn = lim bn = 0. 1 n 0 Problems (1) Prove a sequence converges if and only if its lim inf equals its lim sup. (2) Use this fact to prove every Cauchy sequence of real numbers converges. (You will find the definition of a Cauchy sequence on the third practice exam.) (3) Carefully prove that an 0 is a necessary condition for an to converge. (Be more clear and thorough than my outline from class.) (4) A function f on R is compactly supported if there exists a constant B > 0 such that f (x) = 0 if |x| B. If f and g are two differentiable, compactly supported functions on R, then we define (f g)(x) = f (x - y)g(y)dy. - Note we define - f (x)dx = limt Prove (f g)(x) = (g f )(x). Prove (f g)(x) = (g f )(x). 0 -t f (x)dx + limt t 0 f (x)dx. 1 MIT OpenCourseWare 18.014 Calculus with Theory Fall 2010 For information about citing these materials or our Terms of Use, visit: ...
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This note was uploaded on 01/18/2012 for the course MATH 18.014 taught by Professor Christinebreiner during the Fall '10 term at MIT.

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MIT18_014F10_pset11a - Pset 11 - Part I (Part II will be...

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