1873_solutions

# Prove that if is the cantor function then 1 solution

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Unformatted text preview: s, t, u 3r3s3t3u r1 s1 t1 u1 n n n n n 1 4 3r3s3t3u  r1 s1 t1 u1 n n 1 3r 6 r1 s1 2 n r1 1 3r 4 r1 n 6 r1 13 2 4 6 n 1 2 19 8 19 8  11 81 160 n 1 3 2r n n 1 3s 1 3t t1 n 1 3 2s s1 n u1 1 3u n s1 n 1 3 2r 1 3t n 1 3t n 13 2 1 8 n 1 8 39  16 t1 1 3s 4 n 1 3s 2 1 3r 1 3r 3 r1 s1 n n r1 s1 t1 n n   2 1 3s n 1 2 r1 13 2 13 n 2 19 n 8 13 n 2 1 3 4r 3 1 2 1 8 39 160 n 1 2 13 2 3 n 13 2 n 1 2 1 2 1 81 80 n 1 80 and we conclude that Þ 0 x 4 d x 1 nÝ lim Þ 0 s n d  1 39 . 160 Some Exercises on the Riemann-Stieltjes Integral 302 1. Prove that the integral Þ 1 3x 2 dx 4 exists and has the value 63. Solution: This exercise will become obsolete when we reach the fundamental theorem of calculus later on in the chapter. The solution given here is a bare hands approach and repeats portions of the proof that monotone functions are integrable. We define f x  3x 2 for each number x 1, 4 . Given any positive integer n, if P n  x 0 , x 1 , , x n is the regular n-partition of the interval 1, 4 then we define two step functions s n and S n by making sn x  Sn x  f x whenever x is a point of the partition P n and, in each interval x j 1 , x j of the partition P n we make s n and S n take the constant values f x j 1 and f x j respectively. Since n Þ1 Sn 4 sn  f xj f xj 4 1 j1 n 1 n 3 n 3f 4  as n f xj f xj 1 j1 f1 n  135 n 0 Ý we know that f is integrable and that lim Þ 1 S n  n Ý Þ 1 s n  Þ 1 f. 4 lim nÝ 4 4 Now since 3j xj  1  n for each j  0, , n, we have Þ1 Sn  4  as n n j1 3 n 3 3j 1 n 9 14n 2  15n  3 2n 2 2 63 Ý. 0, 1 . Given a positive integer n, we shall take P n to be the 2. In this exercise we take f x  x for x partition of 0, 1 defined by the equation 2 2 2 2 P n  0 2 , 1 2 , 2 2 , , n 2 . nnn n Prove that if we define a step function S n on 0, 1 by making Sn x  x whenever x is a point of the partition P n and giving S n the constant value j/n in each interval j 1 2 j2 ,2 n2 n of the partition P n , then 1 1 lim Þ 0 x dx  n Ý Þ 0 S n  2 . 3 Solution: For each positive integer n we define the function S n as described in the exercise and we 303 define a step function s n on 0, 1 by making Sn x  x whenever x is a point of the partition P n and giving s n the constant value j j 1 2 j2 ,2 n2 n We see at once that s n f S n for each n and that Þ0 Sn 1 lim nÝ as n n sn  j1 j j n j2 n2 1 n j 1 1 /n in each interval 2 1 n n2 0 Ý. We conclude that the pair of sequences s n and S n squeezes f on the interval 0, 1 and that Þ 0 S n  Þ 0 f. 1 lim nÝ 1 Therefore lim lim Þ0 f  n Ý Þ0 Sn  n Ý 1 1 2  n Ý 4n  32n lim 6n n j1 j2 n2 j n j 1 2 n2 1  2. 3 3. Prove that Þ0 1 x dx  3 . 4 This exercise is very similar to Exercise 2. This time we take 3 3 3 3 P n  0 3 , 1 3 , 3 3 , , n 3 nnn n for each n and, for each n, we define S n to be the step function that takes the value 3 x at each point of P n and whose constant value in each interval j 1 3 j3 ,3 n3 n is j/n. We o...
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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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