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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
RiemannStieltjes 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 2npartition instead of the npartition 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 2npartition instead of the npartition 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.
 Fall '08
 STAFF
 Math, Calculus

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