MATH 115 - HW4 SOLUTIONS
1. Solutions
These solutions are sketched. You should always give a little more detail.
(1) Let xn be the nth element of the sequence in the problem. Then note that xn+1 =
n+1
xn 21/2
for n 1. Therefore xn+1 > xn so (xn ) is an in
MATH 115 - HW3 SOLUTIONS
1. Solutions
These solutions are sketched. You should always give a little more detail.
2
+4a
(1) (Exercise 9.3) In order to show sn := an2 +1n converges we will use the limit theorems
bn
of Secion 9 of our textbook. By Theorem 9.
Math 115 HW 8
Theo Vadpey
December 2, 2015
Ross 19.1
Show if the functions are uniformly continuous on their intervals:
Part e)
f (x) = x13 on (0, 1]
Consider the sequence Sn = n1 : n N on (0,1]. It is well known that
this sequence is Cauchy in this inter
HOMEWORK 5
Math 115
Due Friday 11/06/2014 in class
Chapter 3:
20.13
20.16
Chapter 5:
28.3
28.8
28.16
29.13
18.8
19.2
19.4
19.6 (ignore parts involving derivatives)
19.7 a)
SECTIONS COVERED OR EXPECTED TO BE COVERED
Numbers refer to sections not chapters. That is 9 means the 3rd sections of chapter 2. These numbers
correspond to sections in the second edition. Email me if you are using the first edition and are confused
abou
Math 115, Practice Midterm
Solutions
Problem 1
Suppose that (0, 1] were closed. Then for any converging sequence (xn ) in (0, 1], we should have
limn xn (0, 1]. Take for example xn = 1/n. Then xn 0 and 0 (0, 1], so (0, 1] cant be closed.
/
(0, 1] is also
Solution Keys
Page 1
Note: The homework you turn in must contain each problem statement in its entirety and followed by its solution, as demonstrated in the rst problem below. Others below are sketches, outlines, or hints for solutions. [#1.3] Prove 13 +
Math 115, HW 2
Solutions
Problem 1
1
. We claim an is irrational for all n and further that
+ 2n
lim an = 0. If an were rational, then n2 + n would be an integer, so n2 + 2n = k 2 for some
(a) For n a positive integer, let an =
n2
n
integer k. Then (n +
HW7: Math 115 Functions of a real variable
Due on Dec 1st during class
(1)
(2)
(3)
(4)
(5)
(6)
(7)
32.2
32.6
32.7
33.3 (a)
33.4
33.7
Assume that for all n, the function fn (x) is integrable on [a, b], and there exists function f
on [a, b] satisfying that
HW3: Math 115 Functions of a real variable
Due on Oct 20th during class
(1)
(2)
(3)
(4)
10.6
10.12
11.10
Let a > 0. Define sn such that s1 = a and sn+1 = 12 (sn + san ), prove that sn converges and
find its limit. (Hint: first show that sn a for n 2, then
HOMEWORK 3
Math 115
Due Friday 10/24/2014 in class
Chapter 2:
12.12
13.12 (Prove only for R. Prove straight from the definition of compactness without using the
Heine-Borel Theorem)
Chapter 3:
17.12
17.14
18.5
18.9
HOMEWORK 9
Math 115
Due Friday 12/04/2014 in class
Chapter 32:
32.3, 32.7
Chapter 33:
33.7, 33.9, Prove the dominated convergence theorem.
Chapter 37:
37.1
Math 115, Homework 5 Solutions
Problem 1
Let f (x) =
1
. Then f is continuous on S and
x x0
|f (xn )| =
1
,
|xn x0 |
so f is unbounded on S.
Problem 2
This follows directly from the intermediate value theorem, since f is continuous and 0 lies between
f (
Math 115, Homework 7 Solutions
Problem 1
Suppose that limx0+ f (x) exists and is equal to L. (the cases in which L is innite are dealt
with similarly). Let
> 0. Then there exists > 0 such that whenever 0 < x < , we have
|f (x) L| < . Then there exists M =
Math 115, Homework 6 Solutions
Problem 1
(a) Let
> 0. Then there exists 1 > 0 such that whenever |x a| < 1 , we have |f (x) 3| < /6.
Similarly there is 2 > 0 such that whenever |x a| < 2 , we have |g(x) 2| < 1, hence
|g(x)| < 3. Using the denition again,
Math 115, Homework 9 Solutions
Problem 1
(a) For any partition P , we have L(g, P ) = 0 so L(g) = 0. For U (g, P ) we let h(x) = x2 for
all x and note that on an interval [tj1 , tj ] we have sup g = sup h, so for any partition P ,
U (g, P ) = U (h, P ). H
Math 115, Homework 4
Solutions
Problem 1
(a) The middle inequality is obvious, so well only prove the rst and the third inequality.
For n > N , we have
s1 + . . . sn
s1 + . . . sN
N infcfw_sn | n > N
+ infcfw_sn | n > N
n
n
n
Let tN = infcfw_sn | n > N
Math 115, HW 3
Solutions
Problem 1
Let
> 0. Since sn 0, there is some N such that for all n > N , we have |sn | <
|tn | M for all n, it follows that for all n > N , we have |sn tn | < , so sn tn 0.
M
. Since
Problem 2
sn+1
L < 1, there is some a such tha
HOMEWORK 1
Math 115
Due Friday 10/3/2014 in class
Chapter 1:
1.2
1.9
2.2
3.8
4.10
4.16
6.6
Let A=0* union cfw_p in Q : p >= 0 and p^2<2. Prove that A is a Dedekind cut and also that cfw_xy : x is in Q,
y is in Q, x^2<2, x>0, y^2<2 and y>0
= cfw_z : z is i
HW1: Math 115 Functions of a real variable
Due on Oct 6th during class
(1) Mathematical induction: 1.3, 1.9
(2) Rational zeros theorem: 2.5
(3) sup and inf: 4.7, 4.12, 4.14, 4.15
1
Math115 Homework 6 solutions
November 30, 2016
(1) (a) We need to use the fact that if h0 = 0 in some open interval, then h = const in this
interval. Since f 00 = (f 0 )0 , we have f 00 (x) = 0 for all x I implies f 0 is constant
on I. Suppose this consta
HW6: Math 115 Functions of a real variable
Due on Nov 17th during class
(1)
(2)
(3)
(4)
(5)
(6)
29.7
29.10
29.17
29.18
31.2
31.5 (you can use the fact that limx+ ex xn = 0 for any n N)
1
Math 115:
Homework 4
This homework is worth 20 points. Your grade will based on your 5 best problems. Extra points
(maximum of 5) will be saved for the next homework assignment.
q p
p
1. (2 pts) Show that 2, 2 2, 2 2 2, . . . is convergent and find the
MATH 115 - HW6 SOLUTIONS
These solutions are sketched. You should always give a little more detail.
(1) (Exercise 17.4) We need to show that for any sequence xn x in [0, ) that xn
x. As the hint suggests, this is precisely the statement of example 5 of s
Math 115:
Homework 5
Do 20 points.
1. (5 pts) Do problem 14.1 of Ross, items (a), (d), (e).
2. (5 pts) Do problem 14.2 of Ross, items (a), (d), (e).
3. (5 pts) Do problem 14.3 of Ross, items (b), (d), (e).
4. (5 pts) Do problem 14.5 of Ross, items (a), (b
Math115 Homework 3 solutions
October 23, 2016
(1)(a) For any m and n natural numbers m > n, we have
|sm sn | |sm sm1 | + + |sn+2 sn+1 | + |sn+1 sn |
2(m1) + + 2(n+1) + 2n
1
1
1
= n 1 + + + m1n
2
2
2
1
1
1
n 1 + + 2 + .
2
2 2
1
1
= n
2 1 1/2
1
= n1 .
2
S
HW5: Math 115 Functions of a real variable
Due on Nov 10th during class
(1) Check whether the following functions are uniformly continuous and justify your answer.
(i) f (x) = x3 on (0, 1).
(ii) f (x) = x2 on R.
(iii) f (x) = sin x1 on (0, 1).
(2) For the
HW2: Math 115 Functions of a real variable
Due on Oct 13th during class
(1)
(2)
(3)
(4)
(5)
(6)
(7)
8.2 (a) (e)
8.5
8.7 (a)
8.8 (b)
9.11 (c)
9.12
(extra) Assuming limn an = a, define sn =
1
a1 +.+an
,
n
prove limn sn = a.
Math115 Homework 4 solutions
November 1, 2016
(1)(a) For > 0, take 1 > > 0 such that
< .
5
Then for x (2 , 2 + ), we have
|f (x) f (2)| = |x2 22 | = |(x 2)(x + 2)| < 5 < .
where we used the fact that |x + 2| |x| + 2 < 3 + 2 = 5.
Hence, f is continuous at
Math115 Homework 5 solutions
November 16, 2016
1 (a) Given > 0, let < 3 . Then for x and y in (0, 1) such that |x y| < ,
|x3 y 3 | = |x y|x2 + xy + y 2 | < 3|x y| < 3 < .
So, f is uniformly continuous on (0, 1).
(b) Let = 1. For any > 0. Let x > 1 , and l