HOMEWORK 1
SHUANGLIN SHAO
1. P 22. Ex. 1.3.1
a). E = cfw_3, 1. So
inf E = 3.
sup E = 1,
b). E = cfw_x,
0 < x < 1.5. So
sup E = 1.5,
inf E = 0.
c). E = cfw_r = p/q Q : r2 < 5, r > 0 = cfw_r Q : 0 < r <
sup E = 5, inf E = 0.
5. So
d).
sup E = 1.5,
inf E = 0
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET I
Problem 1.4.8/page 28  3 points
We check this inequality for n = 1, 2, 3, 4. Assuming that it is correct for some n 5,
we check that it also holds for n + 1. We have by the induction hypothesis that
n(n 1)(n
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET II
Problem 1.6.3/page 40  3 points
Let g : A B , which is onto. Assume for a contradiction that A is countable. Then,
there is a function f : N A, which is one to one and onto. Thus, consider the
function h = g
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET III
Problem 3.2.5/page 82  3 points
Let limx f (x) = L. Thus, for > 0, there is M , so that for all x > M , f (x) L <
. Let xn . Find N , so that xn > M , for n > N . Thus, for all n > N ,
f (xn ) L < . Tha
DEPARTMENT OF MATHEMATICS
UNIVERSITY OF KANSAS
MATH 765  Fall 2007  Final Exam
Your Name:
On this exam, you may NOT use books and/or notes.
1
2
3
4
5
6
7
8
9
10
(50)
(50)
(50)
(50)
(50)
(50)
(50)
(50)
(50)
(50)
2
(1) Prove
n
(2k 1)2 =
k=1
n(4n2 1)
3
(2)
DEPARTMENT OF MATHEMATICS
UNIVERSITY OF KANSAS
MIDTERM MATH 765  Fall 2007
Your Name:
1
(75)
2
(75)
3
(75)
4
(75)
BONUS
(50)
Total
(300)
2
(1) Let x0 : 2 and
xn+1 = 1 + xn 1.
Show that the sequence cfw_xn converges and compute its limit.
Hint: First, sh
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET I
Problem 1.3.6/page 22
We use the reection principle and apply the corresponding theorems for supremums.
a) Consider the set A = E = cfw_x : x E . Apply Theorem 1.14 to A. For a given
> 0, there is x A, so tha
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET II
Problem 2.3.3/page 57
By induction, it is easy to show that xn 2. Indeed, x0 2. If xn 2, we have
xn+1 = 2 + xn 2 2.
Case 1: x0 3 In this case, we show that xn 3 and xn is a nonincreasing
sequence (i.e. xn xn
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET III
3.2.6/page 82;
We have that the function f : [0, 1] R is continuous. Let f (q ) = 0 for all q Q.
Then, for each x [0, 1], nd a sequence qn x, where qn Q. We have
f (x) = lim f (qn ) = 0
n
The converse is eas
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET IV
4.1.4/page 104;
Since we have f (x) x , > 1, it follows that f (0) = 0. Thus
f (x)
f (x) f (0)
0
=
x1
x
x
Thus, as x 0, we get limx0 f (x)f (0) = 0, which implies f (0) = 0.
x
If = 1, the funct
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET V
5.2.6/page 151
We use the mean value theorem for integrals (Theorem 5.24/page 147) to conclude
that
b
b
f (x)gn (x)dx = cn
a
gn (x)dx,
a
where cn [inf x[a,b] f, supx[a,b] f ]. Thus,
b

b
b
gn (x)dx M 
f (x)
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET VI
6.2.7/page 198
We have that k ak , k bk converge. In particular, limk ak = 0. Hence, we can nd
N , so that for all k > N , ak < 1. Thus,
0 < ak bk < bk ,
for all k > N . Thus, the series k ak bk is dominated
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET VII
7.2.5/page 236
We have that  sin(y ) y  for all y . Theredore, we have that the functional series
that we consider is bounded by the series
x
k=1
1
,
k (k + 1)
which is convergent for xed x. Moreover,
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET III
Problem 4.3.2/page 116  3 points
By the mean value theorem, there exists I,
f (2) f (0)
f () =
= 1.
2
Thus 1 = f () f (I).
Problem 4.4.4/page 124  3 points
By the formula for the derivatives (sin(x)(n) , w
SOLUTION OF SOME PROBLEMS FROM HOMEWORK SET II
Problem 5.1.5/page 139  3 points
c
If f (x) = 0, then it is clear that a f (x)dx = 0. Conversely, by the fundamental
theorem of calculus, the function
c
f (x)dx = 0
F (c) =
a
has a derivative 0 = F (c) = f (
30 ELEMENTS or THE THEORY CRAP. 1
in the areplane. When we make the change of variables (35), the functional
b
J {3’} = L F(x,y,y') dx
goes into the functional
_ *1 ya + yuv’ ,
J1[v] — Ll F [x(u, U), y(u, v), ———~xu + xuv,] (xu + xav ) du
b
= 1 F101, 1:
HOMEWORK 10
SHUANGLIN SHAO
1. P207. Ex. 6.3.3
1
Proof. a). Since the series is comparable to 2 xp dx, it converges when
p > 1, while it diverges for p 1. Since all ak 0, the series converges
absolutely.
c). Let r = lim supk ak 1/k , where ak = k p /pk .
HOMEWORK 9
SHUANGLIN SHAO
1. P190. Ex. 6.1.4
Proof. Let bk = ak+1 2ak + ak1 , where k N. Then
n
bk = (a2 2a1 + a0 ) + (a3 2a2 + a1 )+
sn =
k=1
+ (a4 2a3 + a2 ) + + (an+1 2an + an1 )
= a0 a1 + an+1 an a0 a1
as n . So for all n, limn sn exists and equals a0
HOMEWORK 2
SHUANGLIN SHAO
1. P45. 2.1.1
Proof. a). For any > 0, we want to nd N N such that for any n N ,
1
2 2 < .
n
1
In order for n < holds,
1
n> .
So we take N = [ 1 ] + 1, where, for a R, [a] denotes the integer part of a.
d). For any > 0, we need
HOMEWORK 3
SHUANGLIN SHAO
1. P 74. Ex. 3.1.2
Proof. a). The limit does not exist. Since tan x has period , we choose
1
1
xn = n+/6 and yn = n+/4 . Both xn and yn converges to zero as n goes
to innity. However,
tan
1
= tan(n + ) = tan = 3/2,
xn
6
6
and
tan
HOMEWORK 4
SHUANGLIN SHAO
1. P91. Ex. 3.3.3
Proof. If f is continuous on [a, b], so is f . By the extreme value theorem,
f  will attain its maximum at [a, b]. supx[a,b] f (x) = maxx[a,b] f (x).
Thus it is nite.
2. P91. Ex. 3.3.4
Proof. We consider
HOMEWORK 5
SHUANGLIN SHAO
1. P 104. Ex. 4.11
Proof. a). For f (x) = x2 + x,
f (a + h) f (a)
,
h0
h
f (a) = lim
where
f (a + h) f (a)
(a + h)2 + (a + h) a2 a
2ah + h2 + h
=
=
= 2a+h+1.
h
h
h
So
f (a) = lim (2a + h + 1) = 2a + 1.
h0
b). For f (x) =
x and a
HOMEWORK 6
SHUANGLIN SHAO
1. P123. Ex. 4.4.1
Proof. a, b). This follows similarly as in Example 4.26 on page 119.
2. P124. Ex. 4.4.3
Proof. By Example 4.25 on page 119, for any n N and Taylors formula,
n
x
e=
k=0
for some 0 < c < x. Thus
n
k=0
ec
xk
+
k!
HOMEWORK 7
SHUANGLIN SHAO
1. P 139. Ex. 5.1.2
a). For j = 0, 1, , n, we have
n
j
j+1
0
= 0, = 1, and <
.
n
n
n
n
So this is a partition of the interval [0, 1].
b). For a P of [0, 1], P = cfw_xi : 0 i n, xi = 0, xn = 1, x0 < x1 < <
xn . The upper integral
HOMEWORK 8
SHUANGLIN SHAO
1. P 161. Ex. 5.3.3
Proof. a). Let u = x2 + 1. du = 2xdx.
1
2
x3 f (x2 + 1)dx =
0
(u 1)f (u)
1
du
1
=
2
2
2
uf (u)du
1
1
2
2
f (u)du.
1
By hypothesis,
1
3
1
(3 1) = .
2
2
2
=
b). Let u =
1 x2 . Then
du =
So
3/2
0
x
, and x2 =
HOMEWORK 9
SHUANGLIN SHAO
1. P190. Ex. 6.1.4
Proof. Let bk = ak+1 2ak + ak1 , where k N. Then
n
bk = (a2 2a1 + a0 ) + (a3 2a2 + a1 )+
sn =
k=1
+ (a4 2a3 + a2 ) + + (an+1 2an + an1 )
= a0 a1 + an+1 an a0 a1
as n . So for all n, limn sn exists and equals a0
HOMEWORK 10
SHUANGLIN SHAO
1. P207. Ex. 6.3.3
1
Proof. a). Since the series is comparable to 2 xp dx, it converges when
p > 1, while it diverges for p 1. Since all ak 0, the series converges
absolutely.
c). Let r = lim supk ak 1/k , where ak = k p /pk .
HOMEWORK 1
SHUANGLIN SHAO
1. P 22. Ex. 1.3.1
a). E = cfw_3, 1. So
inf E = 3.
sup E = 1,
b). E = cfw_x,
0 < x < 1.5. So
sup E = 1.5,
inf E = 0.
c). E = cfw_r = p/q Q : r2 < 5, r > 0 = cfw_r Q : 0 < r <
sup E = 5, inf E = 0.
5. So
d).
sup E = 1.5,
inf E = 0