1873_solutions

# 1 0 1 3 2 1 3 1 3 0 2 3 1 by looking at the two

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: bserve that lim Þ0 Sn  n Ý 1 lim nÝ 3 n j1 n nÝ lim j n 3j 3 j1 j3 n3 j 1 3 n3 3j 2  j n4 2  n Ý 3n  22n 1  3 . lim 4 4n In the same way we can show that if s n 3 x at each point of P n and whose constant value in each interval j 1 3 j3 ,3 n3 n is j 1 /n then 1 lim Þ s  3 nÝ 0 n 4 and so the pair of sequences s n and S n squeezes the cube root function on 0, 1 . 4. Prove that the integral Þ 0 xd x 3 1 exists and evaluate it. 304 Solution: This exercise will become obsolete when we reach the theorem on reduction of Riemann-Stieltjes integrals to Riemann integrals. The solution given here is a bare hands approach. We define  x  x 3 for each x. For each positive integer n we define P n  0 , 1 , 2 , , n , nnn n j j we define s n and S n to be the step functions that take the value n at each point n of the partition P n and j1 j j1 j that takes the constant values n and n respectively on each interval n , n of the partition. For each n we see that Þ0 n 1 Sn j1 n Þ 0 S n d  j n j1 and the latter expression approaches 3 4 1 3 j n n j 3 1 n 1 n Ý. Therefore the required integral exists. Now for each n and the latter expression approaches 0 as n we see that 1 j j n s n d  3 j n j 3 1 2  3n  22n 4n n 1 Ý. Therefore as n Þ 0 xd x 3 1  3. 4 5. Suppose that x x  if x 0 2x if x  0 . Prove that the integral Þ 1 1  x d x 1 exists and evaluate it. Solution: For each positive integer n we define P n to be the regular 2n partition of the interval 1, 1 . Thus for each j, the jth point of P n is 1 2j 2n and Pn  1, 1  2 , 1  4 , , 1  4n 2n 2n 2n 2n , 2n  1 , , 1 , 0, 1 , , 2n 1 , 2n . 2n 2n 2n 2n 2n 2n Our reason for taking the 2n-partition instead of the n-partition is that we want to guarantee that the number 0 is a point of the partition. 2j For each n we define s n and S n to be the step functions that take the value 1 1  2n at each point 2j 2j 1 2j 1  2n of the partition P n and that take the constant values 1 1  2n and 1 1  2n respectively in 2j 1 2j each interval 1  2n , 1  2n of the partition. We see that  Þ 1 Sn 1 as n n s n d  j1 2n 2 2n 1 n  jn1 2 2n 2 n 0 Ý. Now for each n we have Þ 1 S n d  1 n j1 2j 2n 2n 1 n 305  jn1 2j 2n 2 n  7n  3 2n and the latter expression approaches 7 2 Ý. as n 6. Suppose that x x  if x 0 x  1 if x  0 . Prove that the integral Þ 1 1  x d x 1 exists and evaluate it. Solution: For each positive integer n we define P n to be the regular 2n partition of the interval 1, 1 . Thus for each j, the jth point of P n is 1 2j 2n and 1, 1  2 , 1  4 , , 1  4n 2n 2n 2n Pn  2n , 2n  1 , , 1 , 0, 1 , , 2n 1 , 2n . 2n 2n 2n 2n 2n 2n Our reason for taking the 2n-partition instead of the n-partition is that we want to guarantee that the number 0 is a point of the partition. 2j For each n we define s n and S n to be the step functions that take the value 1 1  2n at each point 2j 2j 1 2j 1  2n of the partition P n and that take the constant values 1 1  2n and 1 1...
View Full Document

## 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.

Ask a homework question - tutors are online