1873_solutions

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

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Unformatted text preview: bserve that lim Þ0 Sn  n Ý 1 lim nÝ 3 n j1 n nÝ lim j n 3j 3 j1 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 j1 n Þ 0 S n d  j n j1 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  j1 2n 2 2n 1 n  jn1 2 2n 2 n 0 Ý. Now for each n we have Þ 1 S n d  1 n j1 2j 2n 2n 1 n 305  jn1 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...
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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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