1873_solutions

Intervals of e n as c 1 c 2 c 3 c 2 n n we express

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Unformatted text preview:  2n respectively in 2j 1 2j each interval 1  2n , 1  2n of the partition. We see that  Þ 1 Sn 1 as n 2n s n d  0J , 0  j1 2 2n 2 n 1 n 0 Ý. Now for each n we have 2n Þ 1 S n d  1J , 0 1 and the latter expression approaches 3 as n 2j 2n  j1   1  2nn 1 1 n Ý. 7. Prove that the integral Þ 0 xd 1 exists and evaluate it. We define  x  x whenever x Pn  x 0, 1 . For each positive integer n we define 0 2 , 1 2 , 2 2 , , n 2 , n n n n j n we define s n and S n to be the step functions that take the value partition P n and that takes the constant values j1 n 2 , j n 2 2 j1 n and j n 2 2 at each point j n 2 of the respectively on each interval of the partition. For each n we see that Þ0 Sn 1 n s n d  j1 j n 2 and the latter expression approaches 0 as n each n we see that 306 j 1 n 2 j n j 1 n 1 n Ý. Therefore the required integral exists. Now for Þ 0 S n d  n 1 j1 1 3 and the latter expression approaches 2 j n j j n 1 n 2  2n  32n  1 6n Ý. Therefore as n Þ 0 xd 1 x  1. 3 Some Exercises on Riemann-Stieltjes Integrability 1. Suppose that 1 n 1 if x has the form fx  for some positive integer n . 0 otherwise Prove that f is Riemann integrable on the interval 0, 1 and that Þ f  0. 1 0 Solution: For each positive integer n we define P n to be the following partition of the interval 0, 1 : P n  0, 1 , 1 , , 1 , 1 , 1 . nn1 32 For each n we see easily that Þ0 l Pn, f 1 Þ0 u Pn, f 1 0 and 1 n and therefore Þ0 l Pn, f 1 lim nÝ nÝ lim Þ0 u Pn, f 1  0. 2. Suppose that f is defined on the interval in such a way that whenever x 0, 1 and x has the form 1 for some n 1 positive integer n we have f x  0 and whenever x belongs to an interval of the form n1 , 1 for some n positive integer n we have f x  1  1 n. Draw a rough sketch of the graph of this function and explain why it is Riemann integrable on the interval 0, 1 . Solution: 0 11 1 65 4 1 2 1 3 1 For each positive integer n we define P n to be the following partition of the interval 0, 1 : P n  0, 1 , 1 , , 1 , 1 , 1 . nn1 32 For each n we see easily that Þ0 l Pn, f 1 n1  j1 1 j 307 1 j1 1 1 j and n1 Þ0 u Pn, f 1 2 n 1 j1 1 j j1 1 1 j and so Þ0 u Pn, f Þ0 l Pn, f 1 lim nÝ 1 nÝ2 0 lim n and we have shown that f is integrable on 0, 1 . Incidentally, we have also shown that lim Þ0 f  n Ý 1 n1 1 j j1 1 j1 1 1 j n nÝ lim 2 2j 2j  1 j1 n nÝ lim 1 j 2j  1 j1 In the chapter on infinite series you will learn how to show that the latter limit is 2 2 log 2. 3. Given that  is an increasing function f is a bounded function on an interval a, b , prove that the following conditions are equivalent: a. The function f is integrable with respect to  on the interval a, b . b. For every number  0 there exist step functions s and S on the interval a, b such that s Þa S b c. For every number such that if a, b S and f S and s d  .  0 there exist step functions s and S on the interval a, b such that s E x f Sx sx , we have var , E  . d. For eve...
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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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