1873_solutions

# Analytic on any open interval that contains 0 2 prove

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Unformatted text preview: 390 . 3 1 10. In this exercise we define p   Þ 0 sin p d whenever p  0. a. Prove that  is a decreasing function on the interval 0, Ý . The fact that  is decreasing follows from the fact that whenever 0 0 sin  1 and so sin p  decreases as p increases.   we have b. Prove that  2p  2  1.  2p Given any positive number p we integrate by parts and obtain lim pÝ  2p  2    Þ 0 sin 2p2 d  Þ 0  sin 2p1   2p  1 sin 2p1  sin  d sin 2p1 0 cos   Þ 0 sin 2p  1  Þ0 cos 0 2p  1 sin 2p  cos  sin 2  d  2p  1  2p 2p  1  2p  2 and so lim pÝ  2p  2  2p pÝ lim 2p  1  1. 2p  2 c. Combine the first two parts of this question and deduce that  2p  1  1. lim pÝ  2p For each p  0 we have  2p  2  2p  1  2p and so  2p  2  2p  1 1  2p  2p and the fact that  2p  1 1 lim pÝ  2p follows from the sandwich rule. d. Prove that 4p p lim pÝ 2 p1 2p  1  . This assertion is known as Wallis’s formula From Exercise 7 we know that whenever p  0,  2p  1  Þ0  Þ0  sin 2 p 1  d  2 p 1 1 2 2 p 1 1 2 and  2p  sin 2p d  which gives us 391 2 p 1 2  2p 2 1 cos  d  2p11 2 2p1 1 2 2p1  2 2p 1 2  2p  2 2p   2p  1  2p 2  p 1 2 p1 2 p 1 2 Now from Exercise 8 we have 1 p p 1 2 and so  2p  1  2p    2 p 1 2 2 4p 2  2 p1 2 p p1 p 1 2 2 4p 4 p 1 2 2 2 2  1 2 p p  2p 1 2 p1 2 2p  2 2p 1 p p1 p p  p  1 2p 2p 2 2 2 4p 2 2 p1  2 2p  1 and so lim pÝ 2 4p 4 p  1 p  1  2 2p  1 2 1 from which we deduce that lim pÝ 4p p 2 p1 2p  1 pÝ lim 4p p 2 1 2 p1 2p  1 p p 1 2   e. Assuming that p is restricted to be a positive integer, rewrite Wallis’s formula in terms of factorials. If p is restricted to be a positive integer then Wallis’s formula becomes 4 p p! 2 lim  pÝ p 2p ! 11. The purpose of this exercise is to encourage you to read a proof of an interesting theorem known as Stirling’s formula that states that x1  2 . lim x x xÝxe x Note that if n is a large natural number then Stirling’s formula suggests that an approximate value for n! is 2n n n . en The proof of Stirling’s formula that is provided by the preceding link is based on a proof that is provided on page 195 of Walter Rudin’s classic text Principles of Mathematical Analysis. The proof provided here is actually a little simpler because it makes use of the improper integral form of the bounded convergence theorem. 17 Sets of Measure Zero Some Exercises on the Measure Zero Concept 1. Prove that an elementary set E has measure zero if and only if m E  0. 392 Suppose that E is an elementary set. In the event that m E  0, it is clear that E has (nineteenth century) measure zero. On the other hand, if m E  0 then E includes a closed bounded interval of positive length and it follows from the discussion on closed bounded sets of measure zero that E can’t have measure zero. 2. Prove that every countable set has measure zero. This fact follows at once from the theorem on uni...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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