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ia2sp10fes - Introductory Analysis 2Spring 2010 Final...

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Introductory Analysis 2–Spring 2010 Final Exam–Solutions May 5, 2010 INSTRUCTIONS: Each exercise is worth 10 points. If an exercise has two or more parts, each part is worth the same fraction of 10 points. The maximum grade you can get for this test is 100 points, so select (at most) 10 of the exercises, and do them. I will only grade 10 of the exercises on your exam. If you tell me which exercises you want considered, I’ll consider only the ones you tell me. If you tell me nothing, I’ll choose the ten that are easiest to grade (and nothing is as easy to grade as a missing exercise). If you tell me more than 10, I will select 10 from those that you mention, beginning with the one I think will be easiest to grade. Some exercises are very easy, for example 1, 2, 5, 7, and 10. Others are of medium difficulty: 4, ,6, 8, 9, 11, 13, 14. Finally, 3 and 12 might be a bit harder (even though 12 was done in class). Please, try to remember that Introductory Analysis is a mathematics course in which proofs play an essential role, and that writing a proof requires writing complete sentences in some language. Because we are in the US, the only acceptable language is English. A complete sentence must have at least one verb. No credit will be given for answers that are left unjustified. 1. Consider the series n =1 e - nx . Determine: (a) The set of all x for which the series converges. (b) All intervals in R in which the series converges uniformly. (c) The sum of the series for all x R for which it converges. Solution. The series is a geometric series of ratio e - x it thus converges for all real values of x such that e - x < 1, hence for all x > 0. A more sophisticated answer, if one allows complex values of x (but this was not required) would be for all complex x such that | e - x | < 1; that is all x C such that Re x > 0. Wherever it converges, the sum is e - x 1 - e - x = 1 e x - 1 . This answers parts (a) and (c). Concerning uniform convergence, the question here is specific for real values of x . It is easy to see that it converges uniformly in all intervals with a positive left endpoint; that is all intervals [ a, b ], ( a, b ), [ a, b ), ( a, b ] where 0 < a b ≤ ∞ . That is by the Weierstrass M-test; if x a then e - nx e - na , and because the series n =1 e - na converges, the original series converges uniformly in the interval of left endpoint a . The only remaining question is what happens in (0 , ). If we consider the tail of the series, X n = N e - nx = e - Nx 1 - e - x → ∞ as x 0 showing that one cannot have a uniform bound on the difference between the sum of the series and any N -th partial sum. The answer to (b) is: The series converges uniformly in all intervals with a positive left endpoint. 2. Assume the power series X n =0 a n z n has radius of convergence equal to 2. Determine the radius of convergence of the series X n =0 a n z 2 n +1 .
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  • Spring '11
  • Speinklo
  • Empty set, Open set, Topological space, Dominated convergence theorem, Lebesgue integration, Non-measurable set

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