Homework 1 solutions
MATH 2283 Spring 2015
1.3
(2pts)
(a) P &Q&R
(b) P Q R
(c) (P &R) Q
(d) P &(Q R)
(e) (P &Q) R
1.4
(2pts)
(a) Q = P
(b) P = Q
(c) P Q
(d) P if not Q, or in other words Q = P
1.6 (2pts) Consider for example P = n is a prime greater than
Homework 4 solutions
MATH 2283 Spring 2015
1.12
(3 pts)
= We show that x = sup S implies statements a) and b). By definition of the
supremum, x must be an upper bound of S, hence a) holds true. Suppose
now that b) is false. Thus
> 0 s.t. s S[x s].
This i
Homework 5 solutions
MATH 2283 Spring 2015
1.7 (2 pts) By the Archimedean property, > 0, n0 N s.t.
But for any n, 2n > n. Thus n > n0 ,
sn L =
We conclude that limn
1
2n
1
n0
< .
1
1
1
< <
< .
n
2
n
n0
= 0.
1.18
(3 pts) Let > 0. Then for all n,
2n +
Homework 11 solutions
MATH 2283 Spring 2015
3.5
(1pt)
3.7
(4pts)
(b) By (), we have
Tn = 1 +
1
1
+ . + ln n 1
2 cfw_z n
ln n
1
1
1
1
+ ln n > 0
Tn = 1 + + . +
2
n1 n
n

cfw_z
ln n
Hence Tn is bounded below by 0 and above by 1.
(c) By the mean value the
Homework 7 solutions
MATH 2283 Spring 2015
4.4
(3 pts)
(b) Note that
sn =
1
1+
n
n
.
We recognize the sequence that converges to ex with x = 1, thus
lim sn = e1 .
n
(c) Note that
1
=
sn
Thus limn
4.5
1
sn
2n + 1
2n
n
=
n
1/2
1+
.
n
= e1/2 , so limn sn = e
Homework 12 solutions
MATH 2283 Spring 2015
5.11
(3pts) Consider the series
X
xn
.
n!
n=0
(1)
We use the ratio test to show that this series converges:
x
xn+1 n!
= lim
= 0.
n
n n + 1
n (n + 1)! x
lim
is less than 1, the series (1) converges. Since
Homework 13 solutions
MATH 2283 Spring 2015
7.8
(5pts)
(g) To find the radius of convergence, we check for absolute convergence, i.e.
we consider the sum
n
X
3n n X 3x
x
=
.
2n
2
n=0
n=0
This is a geometric series, which converges if and only if 3x
1. Match the following using integral theorems we learned this semester. Be sure to check
the hypotheses of the theorem, and if possible, compute the integral.
ZZ
0
Z
Zx3 x
3
2
(a)
x /3, z(zy + x), y z ds
()
1 dA
2
2
2
x +y +z =1
Z1 Z2 Z
(b)
0
0
Z
(c)
1
1. Find the volume of the region between a sphere with radius 1 and a sphere with radius
2, in the region of R3 bounded by x 0, y 0, z 0
Solution: This was meant to be a fairly simple problem, you should get practice
reading a word problem, and thinking i
1. On the moon Pandora, the composition of the atmosphere at the point (x, y, z) is given
by the function
(x2 +y)
z
F (x, y, z) = xez1 + y
x + y 3 3z 2
In which the first coordinate denotes the amount of Ar, the second coordinate denotes
the amount of CO
Homework 6 solutions
MATH 2283 Spring 2015
2.12(b)
(3 pts) We want to find n0 such that the following is true n > n0 :
sn 1 < .
On the left hand side, we have
n
n
2 + n (2n + 1)
2 + n
= n1 .
1 =
sn 1 = n
2n + 1
2 +1
2n + 1
Since 2n > n2 for all
MATH 2283
Sample Midterm Problems
February 18, 2013
INSTRUCTOR: Anar Akhmedov
1. Prove that for any natural number n,
1
1
1
2( n + 1 1) < 1 + + + + < 2 n.
n
2
3
2. Use mathematical induction to prove that 72n 48n1 is divisible by 2304 for every natural
nu
MATH 2283
Solutions to Sample Midterm
February 23, 2016
INSTRUCTOR: Anar Akhmedov
1. Prove that for any natural number n,
1
1
1
2( n + 1 1) < 1 + + + + < 2 n.
n
2
3
1
1
1
Solution: Let P (n) be the following statement: 2( n + 11) < 1+ + + + <
n
2
3
2 n.
O
MATH 2283 SAMPLE MIDTERM II
March 21, 2016
INSTRUCTOR: Anar Akhmedov
The midterm exam II will cover the sections 3.1  3.8.
1. Let sn be the sequence defined recursively by s1 = 2, sn = 6 + sn1 . Determine whether
the sequence sn converges or diverges. If
MATH 2283
Sample Midterm Problems
February 15, 2016
INSTRUCTOR: Anar Akhmedov
1. Prove that for any natural number n,
1
1
1
2( n + 1 1) < 1 + + + + < 2 n.
n
2
3
2. Use mathematical induction to prove that 72n 48n1 is divisible by 2304 for every natural
nu
Homework 9 solutions
MATH 2283 Spring 2015
6.14
(3 pts)
(a) We prove xn > r for all n by induction. The basis case: x1 > r is given in
the problem. Assume xn > r. By the Mean Value Theorem, c [r, xn ]
such that
f (xn )
f (xn ) f (r)
=
,
f 0 (c) =
xn r
xn
Homework 10 solutions
MATH 2283 Spring 2015
2.11, 2.13
2.14
(1 pt)
(5 pts)
(f) Since for all n we have sin n1 sin 1, then
n X
X
1
n
(sin 1) .
sin
n
n=1
n=1
Since the series on the left is geometric with ratio sin 1 < 1 it must converge,
so the series on
Homework 2 solutions
MATH 2283 Spring 2015
2.6
(2pts)
(a) False, since for k = 4, 2k + 1 = 9 is not prime
(b) False, since for k = 3, 2k + 1 = 9 is not prime
(c) False, since for k = 3, 22k + 1 = 65 is not prime
2.12 (2pts) Let P (x, y) be the statement x
Homework 8 solutions
MATH 2283 Spring 2015
6.6, 6.7, 6.8 (1pt)
6.9
(4pts)
(a) Here are the first few terms:
n
xn
1
2.646
2
3.106
3
3.179
4
3.190
5
.
3.192
(b) We show that the sequence is monotonic and bounded then conclude that
it is convergent.
Step 1.
Homework 3 solutions
MATH 2283 Spring 2015
4.6 (2 pts)
Let P (n) be the statement 52n 1 is divisible by 8.
P (1) is clearly true since 52(1) 1 = 24 is divisible by 8.
Assume P (n) is true for some n.
Show P (n + 1) : 52(n+1) 1 is divisible by 8.
52(n+1) 1
1. Lily the ladybug, Bob the beetle, and Sid the Spider all live happily in the xy plane. At
time t Bobbys position is given by the parametric equation
cos(t)
b(t) =
sin(t)
Simultaneously, Lilys position is given by
t1
l(t) =
t
(a) Do the path of Bob and
1. Let S denote the closed cylinder with bottom given by z = 0 and top given by z = 4
and the lateral surface given by x2 + y 2 = 9. Orient S with outward normals. Determine
the Surface Integral
ZZ
y dS
S
(a)
(b)
(c)
(d)
Is this a vector or a scalar surfa
1. Lake Orebegone is a artificial lake outside of Gilbert Minnesota, formed by flooding an
rectangular open pit iron mine. It is 250 meters long, 500 meters wide, and 125 meters
deep.
(a) Use a triple integral to find the volume of water in the lake.
(b
Math 2283
October 10, 2013
Test I
Print Name:
Signature:
ID Number:
SHOW ALL YOUR WORK
1. Let A be a set of real numbers.
(a) (10 points) Express the statement
If there is a largest number in A then there is a rational number in A
in abbreviated form, usi
Math 2283
November 19, 2013
Test II
Print Name:
Signature:
ID Number:
SHOW ALL YOUR WORK
1
Sequences
1. (a) (10 points) Complete the following denition:
Denition. Let L be a real number. Then
lim an = L if . . .
n
Solution: for every > 0
there is some n0
Math 3283
April 3, 2008
Test II
Print Name:
Signature:
Section No:
ID Number:
SHOW ALL YOUR WORK
1. (15 points) Complete the following denitions:
(a) limn an = if .
M n0 s.t. n n0 an M.
(b) The sequence an is increasing if .
n an an+1 .
(c) The sequence a
MATH 2283 Fall 2013
Quiz #3 solutions
1.
(5 pts) Show that for any a, b R with a < b,
sup[a, b) = b.
Solution:
Let S = [a, b).
Clearly x S , x < b, thus b is an upper bound of S .
We need to show that for any upper bound b of S , b b.
Suppose this is not
MATH 2283 Fall 2013
Quiz #4 solutions
1.
(5 pts) Find the supremum of the following set, and use the denition to prove your claim.
(1)n
1 :nN
1+ n
A=
Solution:
Note that n,
(1)n
(1)n n
(1)n
n
= n+1 =
1
1
n+1
n+1
1+ n
n
Thus 1 is an upper bound of the set
MATH 2283 Fall 2013
Quiz #6 solutions
1.
(4 pts) Let sn = 2
1
n
1
and tn = 3 n . Find n0 such that
sn 2
< 103
tn
3
for all n n0
Solution:
Note that
2
sn 2
=
tn
3
3
1
n
1
n
2
2n 1 2
3(2n 1) 2(3n 1)
1
=
=
=
.
3
3n 1 3
3(3n 1)
3(3n 1)
Since 3n 1 2n > 0 for
MATH 2283 Fall 2013
Quiz #5 solutions
1. (2 pts) State precisely what is meant by the statement the sequence sn converges to L.
Solution:
> 0n0 s.t. n n0 sn L <
2.
(4 pts) Using the denition of limit, show that
lim
n
2n + 3
5n 1
=
2
5
Solution:
Let
< 0